本文研究三维不可压缩广义Navier-Stokes方程组的柯西问题:

$ \left\{\begin{array}{c} \boldsymbol{u}_t+\boldsymbol{u} \cdot \nabla \boldsymbol{u}+\varLambda^\alpha \boldsymbol{u}+\nabla P=0, \\ \nabla \cdot \boldsymbol{u}=0, \\ \boldsymbol{u}(x, 0)=\boldsymbol{u}_0(x) . \end{array}\right. $

(1)

其中: (x, t)∈$\mathbb{R}^3$×$\mathbb{R}$⁺, u=(u₁, u₂, u₃)表示速度场, 标量函数P表示流体的压力, 初始速度u₀=(u₀¹, u₀², u₀³) 满足▽·u₀=0。Λ^α=(-Δ)^{$ \frac{\alpha }{2}$}是关于自变量x的分数阶拉普拉斯算子, 定义为

$ \left(\varLambda^\alpha \boldsymbol{u}\right)(\boldsymbol{x}, t)=\mathcal{F}^{-1}\left(|\boldsymbol{\xi}|^\alpha \hat{u}(\boldsymbol{\xi}, t)\right)(\boldsymbol{x}, t). $

分数阶拉普拉斯算子与现实生活中许多自然现象都有着密切的联系, 出现在等离子体物理、火焰传播、守恒律以及反常扩散问题的一些模型中, 因此在某些领域应用广泛。此外, 分数阶拉普拉斯算子还可以描述很多流体力学中复杂的物理现象, 如水波问题等。

当α=2时, 方程组(1)为经典的Navier-Stokes方程组, 对它的研究, 已经取得了一些重要的成果。首先, Leray^[1]给出了弱解的整体存在性。之后, Fujita和Kato^[2], Kato^[3], Planchon^[4], Cannone^[5], Chemin^[6], Koch和Tataru^[7]等分别在临界的Sobolev空间$ \dot{H}^{\frac{1}{2}}$, Lebesgue空间L^p(其中p>3), Besov空间$\dot{B}_{p, q}^{-1+\frac{3}{p}} $(1≤p < ∞, 1≤q < ∞), 以及BMO^-1空间中对适当小的初始值证明了整体适定性的结论。关于在BMO^-1空间中所证明的整体适定性的结论在某种意义下可能是最佳的, 因为在更大的函数空间中得到了不适定性的结论, 参见文献[8-9]等。注意到如下的关系成立:

$ \dot{H}^{\frac{1}{2}} \subset L^3 \subset \dot{B}_{p, q}^{-1+\frac{3}{p}} \subset \mathrm{BMO}^{-1} \subset \dot{F}_{\infty, r}^{-1} \subset \dot{B}_{\infty, \infty}^{-1}, (r>2) . $

对于一般的α>0, 方程组(1)已经引起了很多学者的关注, 并得到了一些重要的成果。由divu=0, 容易得到下面的能量不等式

$ \begin{aligned} E(\boldsymbol{u})(t): & =\int_{\mathbb{R}^3}|\boldsymbol{u}(\boldsymbol{x}, t)|^2 \mathrm{~d} \boldsymbol{x}+ \\ & 2 \int_0^t \int_{\mathbb{R}^3}\left|\varLambda^{\frac{\alpha}{2}} \boldsymbol{u}(\boldsymbol{x}, s)\right|^2 \mathrm{~d} \boldsymbol{x} \mathrm{d} s \\ & \leqslant \int_{\mathbb{R}^3}\left|\boldsymbol{u}_0(\boldsymbol{x})\right|^2 \mathrm{~d} \boldsymbol{x} . \end{aligned} $

我们观察到如果u(x, t), P(x, t)是方程组(1)的解, 则对任意的λ>0, 按如下方式定义的(u_λ, P_λ):

$ \begin{aligned} \boldsymbol{u}_\lambda(\boldsymbol{x}, t) & =\lambda^{\alpha-1} \boldsymbol{u}\left(\lambda \boldsymbol{x}, \lambda^\alpha t\right) , \\ P_\lambda(\boldsymbol{x}, t) & =\lambda^{2 \alpha-2} P\left(\lambda \boldsymbol{x}, \lambda^\alpha t\right), \end{aligned} $

对相应的初始值u_{λ, 0}(x): =λ^α-1u₀(λx)也满足方程组(1)。从而, 相应的能量E(u_λ)(t)=λ^2α-5E(u)(t)。因此, 对α>5/2, 这是一种次临界的情形。当α=5/2时, 方程组(1)是一种临界的情形。当α < 5/2时, 它是一种超临界的情形, 在这种情况下, 弱解的唯一性仍然是一个公开问题。当α≥5/2时, Lions^[10]在三维情况下证明了经典解的存在性。Katz和Pavlović^[11]已经证明了强解的整体适定性。对α < 5/2, Wu^[12-13]在Besov空间$ \dot{B}_{p, q}^{1-\alpha-\frac{3}{p}}$中研究了适定性问题, 对1≤α≤2, Wang和Wu^[14]在空间χ^1-2α中对适当小的初始值证明了整体弱解的存在性, Yao和Deng^[15]在Triebel-Lizorkin空间$ \dot{F}_{\frac{3}{\frac{\alpha }{2}-1}, r}^{-\frac{\alpha }{2}}$(r>2, 2 < α < 5/2)中证明了方程组(1)是不适定的。

1 结论

本文在临界的Fourier-Triebel-Lizorkin空间$\hat{\dot{F}}_{p, q}^{4-\alpha-\frac{3}{p}} $($\mathbb{R}^3$)中研究广义Navier-Stokes方程组(1)。Fourier-Triebel-Lizorkin空间与其他的一些函数空间是紧密相关的。例如, Cannone和Karch^[16]在空间PM^α中考虑了α=2时解的存在性与唯一性, 其中

$ \begin{aligned} \mathrm{PM}^\alpha= & \left\{\boldsymbol{u} \in D^{\prime}\left(\mathbb{R}^3\right) \mid \hat{\boldsymbol{u}}(\boldsymbol{\xi}) \in L_{\mathrm{loc}}^1\left(\mathbb{R}^3\right), \right. \\ & \left.\underset{\xi \in \mathbb{R}^3}{\operatorname{esssup}}|\xi|^\alpha|\hat{\boldsymbol{u}}(\xi)|<\infty\right\}, \end{aligned} $

它与空间$\hat{\dot{F}}_{\infty, \infty}^\alpha $等价。Lei和Lin^[17]在空间χ^-1中对适当小的初始值证明了α=2时解的整体适定性, 这里$\chi^{-1}=\left\{\left.\boldsymbol{u} \in D^{\prime}\left(\mathbb{R}^3\right)\left|\int_{\mathbb{R}^3}\right| \xi\right|^{-1}|\widehat{\boldsymbol{u}(\xi)}| \mathrm{d} \xi<\infty\right\} $, 它与空间$\hat{F}_{1, 1}^{-1} $等价。具体证明参见注2.2。

下面, 给出本文中的主要结果。首先, 引入2个条件:

$ (p, q) \in\left(\frac{3}{5-\alpha}, \infty\right) \times[1, \infty), $

(2)

$ (p, q) \in\left\{\frac{3}{5-\alpha}\right\} \times\left[\frac{3}{5-\alpha}, \frac{6}{5-\alpha}\right] . $

(3)

定理1.1   设2 < α < 5/2, u₀∈ $ \hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}}$($\mathbb{R}^3$), 满足divu₀=0, (p, q)满足条件(2)或条件(3), 则存在T>0, 使得方程组(1)存在唯一的解u(x, t)∈C([0, T); $ \hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}}$($\mathbb{R}^3$)), 并满足

$ \boldsymbol{u} \in \tilde{L}^{\infty}\left([0, T) ; \hat{\dot{F}}_{p, q}^{4-\alpha-\frac{3}{p}}\left(\mathbb{R}^3\right)\right) \cap \tilde{L}^1\left([0, T) ; \hat{\dot{F}}_{p, q}^{4-\frac{3}{p}}\left(\mathbb{R}^3\right)\right) . $

此外, 存在$ \boldsymbol{\epsilon }$>0, 使得如果${{\left\| {{\boldsymbol{u}}_{0}} \right\|}_{\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}}\left( {{\mathbb{R}}^{3}} \right)}} $ < $ \boldsymbol{\epsilon }$, 则有T=∞成立。其中空间$ \hat{\dot{F}}_{p, q}^s$和$\tilde{L}^r $([0, T); $ \hat{\dot{F}}_{p, q}^s$)的定义将在下一节给出。

注1.1   本文中C和c都表示大于0的常数(不同的地方可能取不同的值)。S($\mathbb{R}^3$)表示$\mathbb{R}^3$上的Schwartz函数空间, S′($\mathbb{R}^3$)表示$\mathbb{R}^3$上的缓增分布空间。D($\mathbb{R}^3$)是$\mathbb{R}^3$上有紧支集的无穷次连续可微函数全体构成的空间, D′($\mathbb{R}^3$)是D($\mathbb{R}^3$)的对偶空间。‖f‖_LT^qLx^p表示f的时空Lebesgue范数, 即

$ \|f\|_{L_T^q L_{\boldsymbol{x}}^p}=\left(\int_0^T\|f(\boldsymbol{x}, t)\|_{L_{\boldsymbol{x}}^p}^q \mathrm{~d} t\right)^{\frac{1}{q}}. $

注1.2   当α=2时, 利用略有不同的方法可以证明类似的结论。

2 预备引理

首先, f的Fourier变换定义为:

$ \mathcal{F} f(\boldsymbol{\xi})=\hat{f}(\boldsymbol{\xi}) \int_{\mathbb{R}^3} \mathrm{e}^{-2 \pi i(\boldsymbol{x} \mid \boldsymbol{\xi})} f(\boldsymbol{x}) \mathrm{d} \boldsymbol{x}. $

这里(x|ξ)表示$\mathbb{R}^3$中的内积。这是从L^p到L^p′上的连续线性映射, 显然有$\|\hat{f}\|_{L^{p^{\prime}}} $≤‖f‖_L^p, p∈[1, 2]。

下面简单地介绍一下Littlewood-Paley理论, 参见文献[18]。

设$\mathscr{C}$是一个环, $\mathscr{C}$=$\left\{\xi \in \mathbb{R}^3\left|\frac{3}{4} \leqslant\right| \xi \mid \leqslant \frac{8}{3}\right\} $。存在径向函数χ和φ, 0≤χ(ξ)≤1, 0≤φ(ξ)≤1, χ∈D(B(0, 43)), φ∈D($\mathscr{C}$), 使得

$ \begin{gathered} \forall \xi \in \mathbb{R}^3, \chi(\xi)+\sum\limits_{j \geqslant 0} \varphi\left(2^{-j} \xi\right)=1, \\ \forall \xi \in \mathbb{R}^3 \backslash \{0\}, \sum\limits_{j \in \mathbb{Z}} \varphi\left(2^{-j} \xi\right)=1 . \end{gathered} $

对∀j∈$\mathbb{Z}$, 定义齐次二进块$ \dot{\Delta}_j$以及齐次的低频截断算子$ \dot{S}_j$如下:

$ \begin{aligned} & \dot{\Delta}_j \boldsymbol{u}=\varphi\left(2^{-j} D\right) \boldsymbol{u}=2^{3 j} \int_{\mathbb{R}^3} h\left(2^j \boldsymbol{y}\right) \boldsymbol{u}(\boldsymbol{x}-\boldsymbol{y}) \mathrm{d} \boldsymbol{y}, \\ & \dot{S}_j \boldsymbol{u}=\chi\left(2^{-j} D\right) \boldsymbol{u}=2^{3 j} \int_{\mathbb{R}^3} \tilde{h}\left(2^j \boldsymbol{y}\right) \boldsymbol{u}(\boldsymbol{x}-\boldsymbol{y}) \mathrm{d} \boldsymbol{y} . \end{aligned} $

其中h=$\mathcal{F}$^-1φ, $ \tilde{h}$=$\mathcal{F}$^-1χ.

显然, 由上面的定义可知下列事实成立:

$ \begin{gathered} \dot{\Delta}_k \dot{\Delta}_j \boldsymbol{u}=0, |j-k| \geqslant 2, \\ \dot{\Delta}_k\left(\dot{S}_{j-1} \boldsymbol{u} \dot{\Delta}_j \boldsymbol{u}\right)=0, |j-k| \geqslant 5 . \end{gathered} $

记

$ \begin{aligned} & S_h^{\prime}\left(\mathbb{R}^3\right)= \\ & \left\{\boldsymbol{u} \in S^{\prime}\left(\mathbb{R}^3\right) \lim _{\lambda \rightarrow \infty}\left\|\mathcal{F}^{-1}(\theta(\lambda \cdot) \hat{\boldsymbol{u}})\right\|_{L^{\infty}}=0, \forall \theta \in \mathcal{D}\left(\mathbb{R}^3\right)\right\}, \end{aligned} $

则

$ \boldsymbol{u}=\sum\limits_{j \in \mathbb{Z}} \dot{\Delta}_j \boldsymbol{u}, \dot{S}_j \boldsymbol{u}=\sum\limits_{i \leqslant j-1} \dot{\Delta}_i \boldsymbol{u}, \forall \boldsymbol{u} \in S_h^{\prime} \text {. } $

设u, v∈S′_h, 则乘积uv有如下的Bony分解:

$ \boldsymbol{u} \boldsymbol{v}=\dot{T}_{\boldsymbol{u}} \boldsymbol{v}+\dot{T}_{\boldsymbol{v}} \boldsymbol{u}+\dot{R}(\boldsymbol{u}, \boldsymbol{v}), $

其中,

$ \begin{aligned} & \dot{T}_{\boldsymbol{u}} \boldsymbol{v}:=\sum\limits_{j \in \mathbb{Z}} \dot{S}_{j-1} \boldsymbol{u} \dot{\Delta}_j \boldsymbol{v}, \dot{T}_{\boldsymbol{v}} \boldsymbol{u}:=\sum\limits_{j \in \mathbb{Z}} \dot{S}_{j-1} \boldsymbol{v} \dot{\Delta}_j \boldsymbol{u}, \\ & \dot{R}(\boldsymbol{u}, \boldsymbol{v}):=\sum\limits_{j \in \mathbb{Z}} \dot{\Delta}_j \boldsymbol{u} \widetilde{\dot{\Delta}}_j \boldsymbol{v}, \tilde{\Delta}_j {\boldsymbol{v}}:=\sum\limits_{|j-k| \leqslant 1} \dot{\Delta}_k \boldsymbol{v}. \end{aligned} $

设$\mathbb{P}$是作用于$\mathbb{R}^3$中向量场上的Leray投影算子, 则

$ \mathcal{F}(\mathbb{P} \boldsymbol{f})^j(\boldsymbol{\xi})=\sum\limits_{k=1}^3\left(\delta_{j, k}-\frac{\xi_j \xi_k}{|\boldsymbol{\xi}|^2}\right) \hat{f}^k(\boldsymbol{\xi}) $

成立

定义2.1(Fourier-Triebel-Lizorkin空间的定义)  设s∈$\mathbb{R}$, (p, q)∈[1, ∞]², 齐次Fourier-Triebel-Lizorkin空间$ \hat{\dot{F}}_{p, q}^s$定义如下:

$ \hat{\dot{F}}_{p, q}^s\left(\mathbb{R}^3\right):=\left\{\boldsymbol{u} \in S_h^{\prime}\left(\mathbb{R}^3\right) ;\|\boldsymbol{u}\|_{\hat{\dot{F}}_{p, q}^s\left(\mathbb{R}^3\right)}<\infty\right\}. $

其中

$ {{\left\| \boldsymbol{u} \right\|}_{\hat{\dot{F}}_{p,q}^{s}\left( {{\mathbb{R}}^{3}} \right)}}={{\left\| {{\left( \sum\limits_{j\in \mathbb{Z}}{{{2}^{jsq}}}{{\left| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right|}^{q}} \right)}^{\frac{1}{q}}} \right\|}_{p}},q<\infty ,$

以及

$ {{\left\| \boldsymbol{u} \right\|}_{\hat{\dot{F}}_{p, q}^{s}\left( {{\mathbb{R}}^{3}} \right)}}={{\left\| {\sup \limits_{j\in \mathbb{Z}}}\, {{2}^{js}}\left| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right| \right\|}_{p}}, q=\infty . $

注2.1(Fourier-Herz空间的定义^[19])   设s∈$\mathbb{R}$, (p, q)∈[1, ∞]², 齐次Fourier-Herz空间$\hat{\dot{B}}_{p, q}^s $定义如下:

$ \hat{\dot{B}}_{p, q}^s\left(\mathbb{R}^3\right):=\left\{\boldsymbol{u} \in S_h^{\prime}\left(\mathbb{R}^3\right) ;\|\boldsymbol{u}\|_{\hat{\dot{B}}_{p, q}^s\left(\mathbb{R}^3\right)}<\infty\right\}. $

其中

$ {{\left\| \boldsymbol{u} \right\|}_{\hat{\dot{B}}_{p, q}^{s}\left( {{\mathbb{R}}^{3}} \right)}}={{\left( \sum\limits_{j\in \mathbb{Z}}{{{2}^{jsq}}}\left\| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right\|_{p}^{q} \right)}^{\frac{1}{q}}}, 1\le q<\infty , $

以及

$ {{\left\| \boldsymbol{u} \right\|}_{\hat{\dot{B}}_{p, q}^{s}\left( {{\mathbb{R}}^{3}} \right)}}=\sup \limits_{j \in \mathbb{Z}}\, {{2}^{js}}{{\left\| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right\|}_{p}}, q=\infty . $

注2.2   由Fourier-Triebel-Lizorkin空间的定义以及函数φ的选取可得

$ \begin{aligned} \|\boldsymbol{u}\|_{\hat{\dot{F}}_{\infty, \infty}^\alpha} & =\left\|\sup \limits_{j \in \mathbb{Z}} 2^{j \alpha}\left|\varphi_j \hat{\boldsymbol{u}}(\boldsymbol{\xi})\right|\right\|_{L^{\infty}} \\ & \approx\left\||\boldsymbol{\xi}|^\alpha|\hat{\boldsymbol{u}}(\boldsymbol{\xi})|\right\|_{L^{\infty}} . \end{aligned} $

根据之前关于空间PM^α的定义可知, 空间$ \hat{\dot{F}}_{\infty , \infty }^{\alpha }$与PM^α等价。同理, 由

$ \begin{aligned} \|\boldsymbol{u}\|_{\hat{\dot{F}}_{1, 1}^{-1}} & =\left\|\sum\limits_{j \in \mathbb{Z}} 2^{-j}\left|\varphi_j \hat{\boldsymbol{u}}\right|\right\|_{L^1} \\ & \approx \int_{\mathbb{R}^3} \sum\limits_{j \in \mathbb{Z}} \varphi_j(\boldsymbol{\xi})|\boldsymbol{\xi}|^{-1}|\hat{\boldsymbol{u}}(\boldsymbol{\xi})| \mathrm{d} \boldsymbol{\xi} \approx\|\boldsymbol{u}\|_{\chi^{-1}} \end{aligned} $

可知空间$ \hat{\dot{F}}_{1, 1}^{-1}$与Lei-Lin空间χ^-1等价。

引理2.1   由Fourier-Triebel-Lizorkin空间的定义可知, 下列不等式成立:

1) ${{\left\| \boldsymbol{u} \right\|}_{\hat{\dot{F}}_{p, {{q}_{_{1}}}}^{s}\left( {{\mathbb{R}}^{3}} \right)}} $≤ ${{\left\| \boldsymbol{u} \right\|}_{\hat{\dot{F}}_{p, {{q}_{_{2}}}}^{s}\left( {{\mathbb{R}}^{3}} \right)}} $, (1≤p < ∞, 1≤q₂≤q₁≤∞.)

2) ${{\left\| \boldsymbol{u} \right\|}_{\hat{\dot{F}}_{p, {{q}_{_{1}}}}^{s_1}\left( {{\mathbb{R}}^{3}} \right)}} $≤ ${{\left\| \boldsymbol{u} \right\|}_{\hat{\dot{F}}_{{{p}_{_{2}}}, q}^{{{s}_{1}}+3\left( \frac{1}{{{p}_{1}}}-\frac{1}{{{p}_{2}}} \right)}\left( {{\mathbb{R}}^{3}} \right)}} $, (1≤p₁≤p₂ < ∞, 1≤q≤∞.)

类似于Besov空间和Fourier-Herz空间, 对于负指标的Fourier-Triebel-Lizorkin空间, 有同样的等价刻画。具体结果如下:

引理2.2   设s < 0, (p, q)∈[1, ∞]×[1, ∞], 则u∈ $ \hat{\dot{F}}_{p, q}^s$当且仅当$ {{\left\| {{\left( \sum\nolimits_{j\in \mathbb{Z}}{{{2}^{jsq}}}{{\left| \widehat{{{{\dot{S}}}_{j}}\boldsymbol{u}} \right|}^{q}} \right)}^{\frac{1}{q}}} \right\|}_{{{L}^{p}}}}<\infty $, 并且

$ \begin{align} & {{2}^{-|s|-1}}{{\left\| \boldsymbol{u} \right\|}_{\hat{\dot{F}}_{p, q}^{s}}}\le {{\left\| {{\left( \sum\limits_{j\in \mathbb{Z}}{{{2}^{jsq}}}{{\left| \widehat{{{{\dot{S}}}_{j}}\boldsymbol{u}} \right|}^{q}} \right)}^{\frac{1}{q}}} \right\|}_{{{L}^{p}}}} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \le \frac{{{2}^{s}}}{1-{{2}^{s}}}{{\left\| \boldsymbol{u} \right\|}_{\hat{\dot{F}}_{p, q}^{s}}} \\ \end{align} $

成立。证明过程类似于文献[20]中对负指标的Fourier-Herz空间刻画的证明。

定义2.2   设u(x, t)∈S′_h($\mathbb{R}^3$×$\mathbb{R}$⁺), s∈$\mathbb{R}$, p∈[1, ∞), (q, r)∈[1, ∞]×[1, ∞], 当q < ∞时, 记

$ {{\left\| \boldsymbol{u} \right\|}_{{{{\tilde{L}}}^{r}}\left( {{\mathbb{R}}^{+}};\hat{\dot{F}}_{p, q}^{s} \right)}}=\left\|\left(\sum\limits_{j \in \mathbb{Z}} 2^{j s q} \widehat{\dot{\Delta}_j \boldsymbol{u}(t)_{L_t^r}^q}\right)^{\frac{1}{q}}\right\|_{L^p}, $

以及当q=∞时

$ {{\left\| \boldsymbol{u} \right\|}_{{{{\tilde{L}}}^{r}}\left( {{\mathbb{R}}^{+}};\hat{\dot{F}}_{p, \infty }^{s} \right)}}={{\left\| \sup \limits_{j \in \mathbb{Z}}\, {{2}^{js}}{{\left\| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}(t)} \right\|}_{L_{t}^{r}}} \right\|}_{{{L}^{p}}}}. $

空间$ {{{\tilde{L}}}^{r}}\left( {{\mathbb{R}}^{+}};\hat{\dot{F}}_{p, q }^{s} \right)$定义为:

$ \begin{align} & {{{\tilde{L}}}^{r}}\left( {{\mathbb{R}}^{+}};\hat{\dot{F}}_{p, q}^{s} \right)= \\ & \left\{ \boldsymbol{u}\in S_{h}^{\prime }\left( {{\mathbb{R}}^{3}}\times {{\mathbb{R}}^{+}} \right)\mid {{\left\| \boldsymbol{u} \right\|}_{{{{\tilde{L}}}^{r}}\left( {{\mathbb{R}}^{+}};\hat{\dot{F}}_{p, q}^{s} \right)}}<\infty \right\}. \\ \end{align} $

引理2.3(Banach不动点定理^[21])   设E是一个Banach空间, $\mathcal{B}$是E×E|→E的连续双线性映射, α是一个正实数, 使得

$ \alpha<\frac{1}{4\|\mathcal{B}\|} $

成立, 其中

$ \|\mathcal{B}\|:=\sup \limits_{\|\boldsymbol{u}\|, \|\boldsymbol{v}\| \leqslant 1}\|\mathcal{B}(\boldsymbol{u}, \boldsymbol{v})\|. $

则对于∀a∈B(0, α)⊆E(其中B(0, α)表示以0为心, 以α为半径的开球), 方程

$ \boldsymbol{x}=\boldsymbol{a}+\mathcal{B}(\boldsymbol{x}, \boldsymbol{x}) $

存在唯一的解x∈B(0, 2α).

引理2.4   设s∈$\mathbb{R}$, p∈[1, ∞), (q, r)∈[1, ∞]², u是下面方程组的解:

$ \left\{\begin{array}{c} \boldsymbol{u}_t+\varLambda^\alpha \boldsymbol{u}=\boldsymbol{f}, \\ \boldsymbol{u}(\boldsymbol{x}, 0)=\boldsymbol{u}_0(\boldsymbol{x}) . \end{array}\right. $

(4)

则

$ {{\left\| \boldsymbol{u} \right\|}_{{{{\tilde{L}}}^{r}}\left( {{\mathbb{R}}^{+}};\hat{\dot{F}}_{^{p, q}}^{s+\frac{\alpha }{r}} \right)}}C\left( {{\left\| {{\boldsymbol{u}}_{0}} \right\|}_{\hat{\dot{F}}_{p, q}^{s}}}+{{\left\| \boldsymbol{f} \right\|}_{{{{\tilde{L}}}^{{{r}_{1}}}}\left( {{\mathbb{R}}^{+}};\hat{\dot{F}}_{^{p, q}}^{s-\alpha +\frac{\alpha }{{{r}_{1}}}} \right.}} \right). $

成立。其中, 1≤r₁≤r≤∞, C=C(r, r₁)。

证明   根据Duhamel公式, 首先将方程组(4)写成积分方程的形式

$ \boldsymbol{u}(\boldsymbol{x}, t)=\mathrm{e}^{-t \varLambda^\alpha} \boldsymbol{u}_0+\int_0^t \mathrm{e}^{-(t-s) \varLambda^\alpha} \boldsymbol{f}(\boldsymbol{x}, s) \mathrm{d} s, $

再由Young不等式, 可得

$ \begin{align} & {{\left\| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right\|}_{L_{t}^{r}}}\le {{\left\| \mathcal{F}\left[ {{{\dot{\Delta }}}_{j}}{{\text{e}}^{-t{{\varLambda }^{\alpha }}}}{{\boldsymbol{u}}_{0}} \right] \right\|}_{L_{t}^{r}}}+ \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ {{\left\| \mathcal{F}\left[ {{{\dot{\Delta }}}_{j}}\left( {{\text{e}}^{-(\cdot ){{\varLambda }^{\alpha }}}}{{*}_{t}}\boldsymbol{f}(\boldsymbol{x}, \cdot ) \right)(t) \right] \right\|}_{L_{t}^{r}}} \\ & \ \ \ \ \ \ \ \ \ \ \ \le \left\| {{\text{e}}^{-ct{{2}^{{{j}{\alpha }}}}}}\widehat{{{{\dot{\Delta }}}_{j}}{{\boldsymbol{u}}_{0}}{{\|}_{L_{t}^{r}}}} \right.+{{\left\| {{\text{e}}^{-ct{{2}^{j\alpha }}}} \right\|}_{L_{t}^{m}}}\left\| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{f}\|{_{^{_{{L_t}}^{{r_1}}}}}} \right. \\ & \ \ \ \ \ \ \ \ \ \ \ \le C{{2}^{-j\frac{\alpha }{r}}}\mid \widehat{{{{\dot{\Delta }}}_{j}}{{\boldsymbol{u}}_{0}}\mid }+C{{2}^{j\alpha \left( -1-\frac{1}{r}+\frac{1}{{{r}_{1}}} \right)}}\|\widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{f}}{{\|}{_{^{_{{L_t}}^{{r_1}}}}}}. \\ \end{align} $

其中$1+\frac{1}{r}=\frac{1}{m}+\frac{1}{r_1} $.

从而, 在两边先乘以$ 2^{j\left(s+\frac{\alpha}{r}\right)}$, 再取$ {{\ell }^{q}}$范数, 可得

$ \begin{align} & {{\left( \sum\limits_{j\in \mathbb{Z}}{{{2}^{j\left( s+\frac{\alpha }{r} \right)q}}}\left\| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right\|_{L_{t}^{r}}^{q} \right)}^{\frac{1}{q}}}\le C{{\left( \sum\limits_{j\in \mathbb{Z}}{{{2}^{jsq}}}{{\left| \widehat{{{{\dot{\Delta }}}_{j}}{{\boldsymbol{u}}_{0}}} \right|}^{q}} \right)}^{\frac{1}{q}}}+ \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ C{{\left( \sum\limits_{j\in \mathbb{Z}}{{{2}^{j\left( s-\alpha +\frac{\alpha }{{{r}_{1}}} \right)q}}}\left\| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{f}} \right\|_{_{{{L}_{t}}}^{{{r}_{1}}}}^{q} \right)}^{\frac{1}{q}}}. \\ \end{align} $

两边取L^p范数, 以及由Minkowski不等式, 可得

$ {{\left\| \boldsymbol{u} \right\|}_{\tilde{L}_{t}^{r}\left( {{\mathbb{R}}^{+}};\hat{\dot{F}}_{^{p, q}}^{s+\frac{\alpha }{r}} \right)}}\le C\left( {{\left\| {{\boldsymbol{u}}_{0}} \right\|}_{\hat{\dot{F}}_{p, q}^{s}}}+{{\left\| \boldsymbol{f} \right\|}_{{{\widetilde{L}}^{{{r}_{1}}}}\left( {{\mathbb{R}}^{+}};\hat{\dot{F}}_{p, q}^{s-\alpha +\frac{\alpha }{{{r}_{1}}}} \right)}} \right). $

引理2.4得证。

引理2.5   设s∈$\mathbb{R}$, p∈[1, ∞), q∈[1, ∞]。则有下列嵌入关系成立:

$ \hat{\dot{B}}_{p, \min (p, q)}^s \hookrightarrow \hat{\dot{F}}_{p, q}^s \hookrightarrow \hat{\dot{B}}_{p, \max (p, q)}^s. $

证明   首先假设p≥q, 则有${{\ell }^{q}} $$\hookrightarrow$$ {{\ell }^{p}}$, 从而

$ \begin{aligned} \left\| \boldsymbol{u} \right\|_{\hat{\dot{B}}_{p, p}^{s}}^{p} &=\sum\limits_{j\in \mathbb{Z}}{{{2}^{jsp}}}\left\| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right\|_{{{L}^{p}}}^{p}\\ & =\sum\limits_{j \in \mathbb{Z}} \int_{\mathbb{R}^n} 2^{j s p}\left|\widehat{\dot{\Delta}_j \boldsymbol{u}}\right|^p \mathrm{~d} x \\ & =\int_{\mathbb{R}^n} \sum\limits_{j \in \mathbb{Z}} 2^{j s p}\left|\widehat{\dot{\Delta}_j \boldsymbol{u}}\right|^p \mathrm{~d} x \\ & =\int_{\mathbb{R}^n}\left\|\left\{2^{j s}\left|\widehat{\dot{\Delta}_j \boldsymbol{u}}\right|\right\}\right\|_{\ell^p}^p \mathrm{~d} x \\ & \leqslant \int_{\mathbb{R}^n}\left\|\left\{2^{j s}\left|\widehat{\dot{\Delta}_j \boldsymbol{u}}\right|\right\}\right\|_{\ell^q}^p \mathrm{~d} x \\ & =\|\boldsymbol{u}\|_{\hat{\dot{F}}_{p, q}^s}^p, \end{aligned} $

且

$ \begin{align} & \left\| \boldsymbol{u} \right\|_{\hat{\dot{F}}_{p, q}^{s}}^{q}=\left\| {{\left( \sum\limits_{j\in \mathbb{Z}}{{{2}^{jsq}}}{{\left| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right|}^{q}} \right)}^{\frac{1}{q}}} \right\|_{{{L}^{p}}}^{q} \\ & \ \ \ \ \ \ \ \ \ \ \ =\left\| \sum\limits_{j\in \mathbb{Z}}{{{2}^{jsq}}}\left| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right| \right\|_{{{L}^{_{\frac{p}{q}}}}}^{q} \\ & \ \ \ \ \ \ \ \ \ \ \ \le \sum\limits_{j\in \mathbb{Z}}{{{\left\| {{2}^{jsq}}\left| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right| \right\|}_{{{L}^{_{\frac{p}{q}}}}}^{q}}} \\ & \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{j \in \mathbb{Z}}\left\|2^{j \boldsymbol{s}}\left|\widehat{\dot{\Delta}_j \boldsymbol{u}}\right|\right\|_{L^p}^q \\ & \ \ \ \ \ \ \ \ \ \ \ =\left\| \boldsymbol{u} \right\|_{\hat{\dot{B}}_{p,q}^{s}}^{q}. \end{align} $

接下来证明p < q的情形, 由Minkowski不等式可得

$ \begin{align} & \|\boldsymbol{u}\|_{\hat{\dot{B}}_{p, q}^{s}}^{q}=\sum\limits_{j\in \mathbb{Z}}{{{2}^{jsq}}}\left\| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right\|_{{{L}^{p}}}^{q} \\ & \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{j\in \mathbb{Z}}{\left\| {{2}^{js}}\left| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right| \right\|_{{{L}^{p}}}^{q}} \\ & \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{j\in \mathbb{Z}}{\left\| {{2}^{jsq}}\left| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right| \right\|_{{{L}^{\frac{p}{q}}}}^{q}} \\ & \ \ \ \ \ \ \ \ \ \ \ \le \left\| \sum\limits_{j\in \mathbb{Z}}{{{2}^{jsq}}}\left| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right| \right\|_{{{L}^{\frac{p}{q}}}}^{q} \\ & \ \ \ \ \ \ \ \ \ \ \ =\left\| \left( \sum\limits_{j\in \mathbb{Z}}{{{2}^{jsq}}}{{\left| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right|}^{q}} \right) \right\|_{{{L}^{p}}}^{\frac{1}{q}q} \\ & \ \ \ \ \ \ \ \ \ \ \ =\|\boldsymbol{u}\|_{\hat{\dot{F}}_{p, q}^{s}}^{q}, \\ \end{align} $

由p < q可知, ${{\ell }^{p}}\hookrightarrow {{\ell }^{q}}$。所以

$ \begin{align} & \|\boldsymbol{u}\|_{\hat{\dot{F}}_{p, q}^{s}}^{p}=\left\| {{\left\| \left\{ {{2}^{js}}\left| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right| \right\} \right\|}_{{{\ell }^{q}}}} \right\|_{{{L}^{p}}}^{p} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \le \left\| {{\left\| \left\{ {{2}^{js}}\left| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right| \right\} \right\|}_{{{\ell }^{p}}}} \right\|_{{{L}^{p}}}^{p} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\left\| \sum\limits_{j\in \mathbb{Z}}{{{2}^{jsp}}}{{\left| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right|}^{p}} \right\|}_{{{L}^{1}}}} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{j\in \mathbb{Z}}{{{2}^{jsp}}}\left\| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right\|_{{{L}^{p}}}^{p} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ =\|\boldsymbol{u}\|_{\hat{\dot{B}}_{p, p}^{s}}^{p}. \\ \end{align} $

3 主要定理的证明

本节证明定理1.1。首先根据Duhamel公式, 将方程组(1)写成积分方程的形式

$ \boldsymbol{u}(\boldsymbol{x}, t)=\mathrm{e}^{-t \varLambda^\alpha} \boldsymbol{u}_0-\int_0^t \mathrm{e}^{-(t-s) \Lambda^\alpha} \mathbb{P}(\boldsymbol{u} \cdot \nabla \boldsymbol{u}) \mathrm{d} s, $

(5)

记

$ B(\boldsymbol{u}, \boldsymbol{v}):=-\int_0^t \mathrm{e}^{-(t-s) \Lambda^\alpha} \mathbb{P}(\boldsymbol{u} \cdot \nabla \boldsymbol{v}) \mathrm{d} s, $

(6)

则B(·, ·)为双线性算子。在本节中, 始终假设2 < α < 5/2。

接下来, 根据积分方程(5), 将应用Banach不动点定理证明定理1.1。首先, 给出工作空间:

$ X=\left\{ \boldsymbol{u}(\boldsymbol{x}, t)\in S_{h}^{\prime }\mid \|\boldsymbol{u}(\boldsymbol{x}, t){{\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}<\infty \right\}. $

此处,

$ \|\boldsymbol{u}{{\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}\triangleq {{\left\| {{\left( \sum\limits_{j\in \mathbb{Z}}{{{2}^{j\left( 4-\frac{\alpha }{2}-\frac{3}{p} \right)q}}}\left\| \widehat{{{{\dot{\Delta }}}_{j}}\boldsymbol{u}} \right\|_{L_{T}^{2}}^{q} \right)}^{\frac{1}{q}}} \right\|}_{{{L}^{p}}}}. $

其次, 将会在工作空间X中估计方程(5)的线性项e^-tΛαu₀和非线性项B(u, u)。

考虑热方程组

$ \left\{\begin{array}{c} \boldsymbol{u}_t+\varLambda^\alpha \boldsymbol{u}=0 , \\ \boldsymbol{u}(\boldsymbol{x}, 0)=\boldsymbol{u}_0(\boldsymbol{x}). \end{array}\right. $

(7)

则u_L: =e^-tΛαu₀为方程组(7)的解。

命题3.1   假设u₀∈ $ \hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}}$($\mathbb{R}^3$), 则

$ \lim\limits_{T\to {{0}^{+}}} {{\left\| {{\boldsymbol{u}}_{L}}(\cdot , \cdot ) \right\|}_{\tilde{L}_{T}^{r}\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}+\frac{\alpha }{r}}}}=0. $

其中T∈$\mathbb{R}$⁺, (p, q, r)∈[1, ∞)³。

证明   根据引理2.4, 可得

$ {{\left\| {{\boldsymbol{u}}_{L}}(\cdot , \cdot ) \right\|}_{\tilde{L}_{T}^{r}\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}+\frac{\alpha }{r}}}}\le C{{\left\| {{\boldsymbol{u}}_{0}} \right\|}_{\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}}}}. $

因为u₀∈ $ \hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}}$($\mathbb{R}^3$), 则对∀$ \boldsymbol{\epsilon }$>0, ∃N, 使得

$ {{\left\| {{\left( \sum\limits_{|j|\ge N}{{{2}^{j\left( 4-\alpha -\frac{3}{p}+\frac{\alpha }{r} \right)q}}}\left\| \mathcal{F}\left[ {{{\dot{\Delta }}}_{\text{j}}}{{e}^{-\text{t}{{\Lambda }^{\alpha }}}}{{\boldsymbol{u}}_{0}} \right] \right\|_{L_{T}^{r}}^{q} \right)}^{\frac{1}{q}}} \right\|}_{L_{\xi }^{p}}}<\frac{\boldsymbol{\epsilon }}{2}. $

下面估计余项:

$ \begin{aligned} & {{\left\| {{\left( \sum\limits_{|j| <N}{{{2}^{j\left( 4-\alpha -\frac{3}{p}+\frac{\alpha }{r} \right)q}}}\left\| \mathcal{F}\left[ {{{\dot{\varDelta }}}_{\text{j}}}{{e}^{-\text{t }{{\Lambda }^{\alpha }}}}{{\boldsymbol{u}}_{0}} \right] \right\|_{L_{T}^{r}}^{q} \right)}^{\frac{1}{q}}} \right\|}_{L_{\xi }^{p}}} \\ & \le C{{\left\| {{\left( \sum\limits_{|j|<N}{{{2}^{j\left( 4-\alpha -\frac{3}{p}+\frac{\alpha }{r} \right)q}}}\left\| {{\text{e}}^{-ct{{2}^{j\alpha }}}}\widehat{{{{\dot{\Delta }}}_{j}}{{\boldsymbol{u}}_{0}}} \right\|_{L_{T}^{r}}^{q} \right)}^{\frac{1}{q}}} \right\|}_{L_{\xi }^{p}}} \\ & =C{{\left\| {{\left( \sum\limits_{|j|<N}{{{2}^{j\left( 4-\alpha -\frac{3}{p}+\frac{\alpha }{r} \right)q}}}{{\left| \widehat{{{{\dot{\Delta }}}_{j}}{{\boldsymbol{u}}_{0}}} \right|}^{q}}\left\| {{\text{e}}^{-ct{{2}^{j\alpha }}}} \right\|_{L_{T}^{r}}^{q} \right)}^{\frac{1}{q}}} \right\|}_{L_{\xi }^{p}}} \\ & \leqslant C\left(1-\text{e}^{-c rT 2^{N \alpha}}\right)^{\frac{1}{r}}\left\|\boldsymbol{u}_0\right\|_{\hat{\dot{F}}_{p, q}^{4-\alpha-\frac{3}{p}}} \\ & <\frac{\boldsymbol{\epsilon}}{2}, \left(T \rightarrow 0^{+}\right) . \end{aligned} $

接下来, 在工作空间X中估计双线性项B(u, v)。假设(p, q)满足条件(2)或条件(3)。

命题3.2   设$ (p, q)\in \left( \frac{3}{5-\alpha }, \infty \right)\times [1, \infty )$。则存在常数C=C(α, p), 使得

$ \begin{align} & \ \ \ \ \ \ \ \ \ \ \ \ \|B(\boldsymbol{u}, \boldsymbol{v}){{\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}} \\ & \le C(\alpha , p)\|\boldsymbol{u}{{\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}\|\boldsymbol{v}{{\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}. \\ \end{align} $

(8)

证明   由Bony仿积分解, 可得

$ \boldsymbol{u}\cdot \nabla \boldsymbol{v}=\sum\limits_{k=1}^{3}{\left[ {{{\dot{T}}}_{{{u}^{k}}}}{{\partial }_{k}}\boldsymbol{v}+{{{\dot{T}}}_{{{\partial }_{k}}\boldsymbol{v}}}{{u}^{k}}+\dot{R}\left( {{u}^{k}}, {{\partial }_{k}}\boldsymbol{v} \right) \right]}. $

因为divu=0, 从而

$ \boldsymbol{u}\cdot \nabla \boldsymbol{v}=\sum\limits_{k=1}^{3}{\left[ {{{\dot{T}}}_{{{u}^{k}}}}{{\partial }_{k}}\boldsymbol{v}+{{{\dot{T}}}_{{{\partial }_{k}}\boldsymbol{v}}}{{u}^{k}}+{{\partial }_{k}}\dot{R}\left( {{u}^{k}}, \boldsymbol{v} \right) \right]}. $

(9)

将上式代入式(6)中, 可得

$ \begin{aligned} B(\boldsymbol{u}, \boldsymbol{v})= & -\sum\limits_{k=1}^3 \int_0^t \mathrm{e}^{-(t-s) \Lambda^\alpha} \mathbb{P}\left(\dot{T}_{u^k} \partial_k \boldsymbol{v}\right) \mathrm{d} s- \\ & \sum\limits_{k=1}^3 \int_0^t \mathrm{e}^{-(t-s) \Lambda^\alpha} \mathbb{P}\left(\dot{T}_{\partial_k \boldsymbol{v}} u^k\right) \mathrm{d} s- \\ & \sum\limits_{k=1}^3 \int_0^t \mathrm{e}^{-(t-s) \Lambda^\alpha} \mathbb{P}\left(\partial_k \dot{R}\left(u^k, \boldsymbol{v}\right)\right) \mathrm{d} s \\ : & =I_1+I_2+I_3 . \end{aligned} $

(10)

所以,

$ \begin{align} & \|B(\boldsymbol{u}, \boldsymbol{v}){{\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}\le {{\left\| {{I}_{1}} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}+ \\ & \ \ \ \ \ \ {{\left\| {{I}_{2}} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}+{{\left\| {{I}_{3}} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}. \\ \end{align} $

第1步, 估计$ {{\left\| {{I}_{1}} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}$。由Leray投影算子的性质, Young不等式以及Hölder不等式可得

$ \begin{aligned} & \left\|\widehat{\dot{\Delta}_j I_1}\right\|_{L_T^2}=\left\|\varphi_j \hat{I}_1\right\|_{L_T^2} \\ \leqslant & \sum\limits_{k=1}^3\left\|\mathrm{e}^{-c(\cdot) 2^{j \alpha}} *{ }_t \varphi_j \mathcal{F}\left(\mathbb{P} \dot{T}_{u^k} \partial_k \boldsymbol{v}\right)\right\|_{L_T^2} \\ \leqslant & C 2^{-j \frac{\alpha}{2}} \sum\limits_{k=1}^3\left\|\varphi_j \mathcal{F}\left(\mathbb{P} \dot{T}_{u^k} \partial_k \boldsymbol{v}\right)\right\|_{L_T^1} \\ \leqslant & C 2^{-j \frac{\alpha}{2}} \sum\limits_{k=1}^3\left\|\sum\limits_{|i-j| \leqslant 5} \mathcal{F}\left[\dot{S}_{i-1} u^k \dot{\Delta}_i \partial_k \boldsymbol{v}\right]\right\|_{L_T^1} \\ \leqslant & C 2^{j\left(1-\frac{\alpha}{2}\right)} \sum\limits_{k=1}^3 \sum\limits_{|i-j| \leqslant 5}\left\|\widehat{\dot{S}_{i-1} u^k} * \widehat{\dot{\Delta}_i \boldsymbol{v}}\right\|_{L_T^1} \\ \leqslant & C 2^{j\left(1-\frac{\alpha}{2}\right)} \sum\limits_{k=1}^3 \sum\limits_{|i-j| \leqslant 5}\left\|\widehat{S_{i-1} u^k}\right\|_{L_T^2} *\left\|\widehat{\dot{\Delta}_i \boldsymbol{v}}\right\|_{L_T^2}. \end{aligned} $

两边乘以$2^{j\left(4-\frac{\alpha}{2}-\frac{3}{p}\right)} $, 再取$ {{\ell }^{q}}$范数, 由Hölder不等式可得

$ \begin{align} & \ \ \ \ \ {{\left\| \left\{ {{2}^{j\left( 4-\frac{\alpha }{2}-\frac{3}{p} \right)}}\left\| \widehat{{{{\dot{\Delta }}}_{j}}{{I}_{1}}} \right\|{{\|}_{L_{T}^{2}}} \right\} \right\|}_{{{\ell }^{q}}}} \\ & \le C\int_{{{\mathbb{R}}^{3}}}{\left( \sum\limits_{j\in \mathbb{Z}}{\sum\limits_{k=1}^{3}{\sum\limits_{\left| i-j \right|\le 5}{{{2}^{j\left( 5-\alpha -\frac{3}{p} \right)q}}}}}\times \right.} \\ & \ \ \ \ {{\left. \left\| \widehat{{{{\dot{S}}}_{i-1}}{{u}^{k}}}(\boldsymbol{x}-\boldsymbol{y}, t) \right\|_{L_{T}^{2}}^{q}\left\| \widehat{{{{\dot{\Delta }}}_{i}}\boldsymbol{v}}(\boldsymbol{y}, t) \right\|_{L_{T}^{2}}^{q} \right)}^{\frac{1}{q}}}~\text{d}y \\ & \le C\int_{{{\mathbb{R}}^{3}}}{\sum\limits_{k=1}^{3}{{{\left\| \left\{ {{2}^{i\left( 1-\frac{\alpha }{2} \right)}}{{\left\| \widehat{{{{\dot{S}}}_{i-1}}{{u}^{k}}}(\boldsymbol{x}-\boldsymbol{y}, t) \right\|}_{L_{T}^{2}}} \right\} \right\|}_{{{\ell }^{\infty }}}}}}\times \\ & \ \ \ \ {{\left\| \left\{ {{2}^{i\left( 4-\frac{\alpha }{2}-\frac{3}{p} \right)}}{{\left\| \widehat{{{{\dot{\Delta }}}_{i}}\boldsymbol{v}}(\boldsymbol{y}, t) \right\|}_{L_{T}^{2}}} \right\} \right\|}_{{{\ell }^{q}}}}~\text{d}y. \\ \end{align} $

上式两边取L^p范数并由Young不等式可得

$ \begin{align} & \ \ \ \ \ {{\left\| {{I}_{1}} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}} \\ & =C(\alpha , p)\sum\limits_{k=1}^{3}{\|\|}\left\{ {{2}^{i\left( 1-\frac{\alpha }{2} \right)}}{{\left\| \widehat{{{{\dot{S}}}_{i-1}}{{u}^{k}}} \right\|}_{{{L}_{T}}}} \right\}{{\|}_{{{\ell }^{\infty }}}}^{*} \\ & \ \ \ \ \ {{\left\| \left\{ {{2}^{i\left( 4-\frac{\alpha }{2}-\frac{3}{p} \right)}}{{\left\| \widehat{{{{\dot{\Delta }}}_{i}}\boldsymbol{v}} \right\|}_{L_{T}^{2}}} \right\} \right\|}_{{{\ell }^{q}}}}{{\|}_{L_{\xi }^{p}}} \\ & \le C(\alpha , p){{\sum\limits_{k=1}^{3}{\left\| {{\left\| \left\{ {{2}^{i\left( 1-\frac{\alpha }{2} \right)}}{{\left\| \widehat{{{{\dot{S}}}_{i-1}}{{u}^{k}}} \right\|}_{L_{T}^{2}}} \right\} \right\|}_{{{\ell }^{\infty }}}} \right\|}}_{L_{\xi }^{1}}}\times \\ & \ \ \ \ \ {{\left\| {{\left\| \left\{ {{2}^{i\left( 4-\frac{\alpha }{2}-\frac{3}{p} \right)}}{{\left\| \widehat{{{{\dot{\Delta }}}_{i}}\boldsymbol{v}} \right\|}_{L_{T}^{2}}} \right\} \right\|}_{{{\ell }^{q}}}} \right\|}_{L_{\xi }^{p}}} \\ & \le C(\alpha , p)\|\boldsymbol{u}{{\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{^{1, \infty }}^{1-\frac{\alpha }{2}}}}\|\boldsymbol{v}{{\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}. \\ \end{align} $

最后一个不等式可由引理2.2得到。再由引理2.1以及$ {{\ell }^{q}} \hookrightarrow {{\ell }^{\infty }}$, q∈[1, ∞), 可得

$ {{\left\| {{I}_{1}} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}\le C(\alpha , p){{\left\| \boldsymbol{u} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}{{\left\| \boldsymbol{v} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}. $

(11)

对I₂, 同理可得

$ {{\left\| {{I}_{2}} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}\le C(\alpha , p){{\left\| \boldsymbol{u} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}{{\left\| \boldsymbol{v} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}. $

(12)

接下来, 估计$ {{\left\| {{I}_{3}} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}$。由Leray投影算子的性质, Minkowski不等式以及Young不等式可得

$ \begin{aligned} \left\|\widehat{\dot{\Delta}_j I_3}\right\|_{L_T^2} & =\left\|\varphi_j \hat{I}_3\right\|_{L_T^2} \\ & \leqslant \sum\limits_{k=1}^3\left\|\int_0^t \mathrm{e}^{-(t-s){{\left| \xi \right|}^{\alpha }}} \varphi_j \mathcal{F}\left[\mathbb{P} \partial_k \dot{R}\left(u^k, \boldsymbol{v}\right)\right] \mathrm{d} s\right\|_{L_T^2} \\ & \leqslant C\sum\limits_{k=1}^{3}{{{\left\| {{\text{e}}^{-c(\cdot ){{2}^{j\alpha }}}}{{*}_{t}}{{\varphi }_{j}}\mathcal{F}\left[ {{\partial }_{k}}\dot{R}\left( {{u}^{k}}, \boldsymbol{v} \right) \right](\cdot , \boldsymbol{\xi}) \right\|}_{L_{T}^{2}}}} \\ & \leqslant C \sum\limits_{k=1}^3 2^j\left\|\mathrm{e}^{-ct 2^{j \alpha}}\right\|_{L_T^2}\left\|\varphi_j \dot{R} \widehat{\left(u^k, \boldsymbol{v}\right)}\right\|_{L_T^1} \\ & \leqslant C 2^{j\left(1-\frac{\alpha}{2}\right)} \sum\limits_{k=1}^3\left\|\varphi_j \dot{R} \widehat{\left(u^k, \boldsymbol{v}\right)}\right\|_{L_T^1}. \end{aligned} $

上式两边乘以$ 2^{j\left(4-\frac{\alpha}{2}-\frac{3}{p}\right)}$, 依次取$ {{\ell }^{q}}$和L^p范数, 由Hölder不等式, Young不等式以及$ p>\frac{3}{5-\alpha}$可得

$ \begin{align} & \ \ \ \ \ \ {{\left\| {{I}_{2}} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}} \\ & \le C\sum\limits_{k=1}^{3}{{{\left\| {{\left\| \left\{ {{2}^{j\left( 5-\alpha -\frac{3}{p} \right)}}{{\left\| {{\varphi }_{j}}\dot{R}\left( \widehat{{{u}^{k}}, \boldsymbol{v}} \right) \right\|}_{L_{T}^{1}}} \right\} \right\|}_{{{\ell }^{q}}}} \right\|}_{{{L}^{p}}}}} \\ & \le C{{\sum\limits_{k=1}^{3}{\left\| {{\left\| \left\{ \sum\limits_{i\ge j-3}{{{2}^{i\left( 5-\alpha -\frac{3}{p} \right)}}}{{\left\| \widehat{{{{\dot{\Delta }}}_{i}}{{u}^{k}}} \right\|}_{L_{T}^{2}}}*\widehat{\|{{{\widetilde{{\dot{\Delta }}}}}_{i}}\boldsymbol{v}}{{\|}_{L_{T}^{2}}} \right\} \right\|}_{{{\ell }^{q}}}} \right\|}}_{{{L}^{p}}}} \\ & \le C\sum\limits_{k=1}^{3}{\sum\limits_{j-i\le 3}{{{2}^{(j-i)\left( 5-\alpha -\frac{3}{p} \right)}}}}\left\| {{\left\| \left\{ {{2}^{i\left( 1-\frac{\alpha }{2} \right)}}{{\left\| \widehat{{{{\dot{\Delta }}}_{i}}{{u}^{k}}} \right\|}_{L_{T}^{2}}} \right\} \right\|}_{{{\ell }^{\infty }}}} \right.* \\ & \ \ \ \ \ {{\left. {{\left\| \left\{ {{2}^{i\left( 4-\frac{\alpha }{2}-\frac{3}{p} \right)}}{{\left\| \widehat{{{\widetilde{{\dot{\Delta }}}}_{i}}\boldsymbol{v}} \right\|}_{L_{T}^{2}}} \right\} \right\|}_{{{\ell }^{q}}}} \right\|}_{{{L}^{p}}}} \\ & \le C\sum\limits_{k=1}^{3}{{{\left\| {{u}^{k}} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{1, \infty }^{1-\frac{\alpha }{2}}}}}\|\boldsymbol{v}{{\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}} \\ & \le C\|\boldsymbol{u}{{\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}\|\boldsymbol{v}{{\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}. \\ \end{align} $

(13)

其中, 最后一个不等式可以由引理2.1得到。因此, 由不等式(11)~(13), 命题3.2得证。

接下来, 当(p, q)满足条件(3)时, 估计双线性项B(u, v)。

命题3.3   设$p=\frac{3}{5-\alpha}, q \in\left[\frac{3}{5-\alpha}, \frac{6}{5-\alpha}\right] $, 则存在常数C=C(α), 使得

$ \begin{align} & \|B(u, v){{\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{\frac{3}{5-\alpha }, q}^{\frac{\alpha }{2}-1}}}\le \\ & C(\alpha )\|\boldsymbol{u}{{\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{\frac{3}{5-\alpha }, q}^{\frac{\alpha }{2}-1}}}\|\boldsymbol{v}{{\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{\frac{3}{5-\alpha }, q}^{\frac{\alpha }{2}-1}}}. \\ \end{align} $

(14)

证明   由等式(10)可得

$ \begin{align} & \|B(\boldsymbol{u}, \boldsymbol{v}){{\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{\frac{3}{5-\alpha }, q}^{\frac{\alpha }{2}-1}}}\le {{\left\| {{I}_{1}} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{\frac{3}{5-\alpha }, q}^{\frac{\alpha }{2}-1}}}+ \\ & \ \ \ \ \ {{\left\| {{I}_{2}} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{\frac{3}{5-\alpha }, q}^{\frac{\alpha }{2}-1}}}+{{\left\| {{I}_{3}} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{\frac{3}{5-\alpha }, q}^{\frac{\alpha }{2}-1}}}. \\ \end{align} $

由命题3.2的证明可知, 当$ p=\frac{3}{5-\alpha}, q \in \left[\frac{3}{5-\alpha}, \frac{6}{5-\alpha}\right]$时, 有

$ {{\left\| {{I}_{1}} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{\frac{3}{5-\alpha }, q}^{\frac{\alpha }{2}-1}}}\le C\left( \alpha \right){{\left\| \boldsymbol{u} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{\frac{3}{5-\alpha }, q}^{\frac{\alpha }{2}-1}}}.{{\left\| \boldsymbol{v} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{\frac{3}{5-\alpha }, q}^{\frac{\alpha }{2}-1}}}, $

(15)

$ {{\left\| {{I}_{2}} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{\frac{3}{5-\alpha }, q}^{\frac{\alpha }{2}-1}}}\le C\left( \alpha \right){{\left\| \boldsymbol{u} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{\frac{3}{5-\alpha }, q}^{\frac{\alpha }{2}-1}}}.{{\left\| \boldsymbol{v} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{\frac{3}{5-\alpha }, q}^{\frac{\alpha }{2}-1}}}. $

(16)

下证:

$ {{\left\| {{I}_{3}} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{\frac{3}{5-\alpha }, q}^{\frac{\alpha }{2}-1}}}\le C\left( \alpha \right){{\left\| \boldsymbol{u} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{\frac{3}{5-\alpha }, q}^{\frac{\alpha }{2}-1}}}.{{\left\| \boldsymbol{v} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{\frac{3}{5-\alpha }, q}^{\frac{\alpha }{2}-1}}}. $

首先由Leray投影算子的性质以及Young不等式可得

$ \begin{aligned} \widehat{{{\left\| {{{\dot{\Delta }}}_{j}}{{I}_{3}} \right\|}_{L_{T}^{2}}}}\le & C\sum\limits_{k=1}^{3}{{{\left\| {{\text{e}}^{-c{{(\cdot )}^{2^{j\alpha }}}}}{{*}_{t}}{{\varphi }_{j}}\mathcal{F}\left[ {{\partial }_{k}}\dot{R}\left( {{u}^{k}}, \boldsymbol{v} \right) \right](\cdot , \boldsymbol{\xi}) \right\|}_{L_{T}^{2}}}} \\ & \leqslant C 2^j \sum\limits_{k=1}^3\left\|\mathrm{e}^{-c t 2^{j \alpha}}\right\|_{L_T^2}\left\|\varphi_j \dot{R}\left(\widehat{u^k, \boldsymbol{v}}\right)\right\|_{L_T^1} \\ & \leqslant C \sum\limits_{k=1}^3 2^{j\left(1-\frac{\alpha}{2}\right)}\left\|\varphi_j \dot{R} \widehat{\left(u^k, \boldsymbol{v}\right)}\right\|_{L_T^1}. \end{aligned} $

由$ \ell^{\frac{3}{5-\alpha}} \hookrightarrow \ell^q, q \in\left[\frac{3}{5-\alpha}, \frac{6}{5-\alpha}\right]$, 应用Minkowski不等式, Hölder不等式以及Young不等式可得

$ \begin{align} & {{\left\| {{I}_{3}} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{F}}_{\frac{3}{5-\alpha },q}^{\frac{\alpha }{2}-1}}}={{\left\| {{\left\| \left\{ {{2}^{j\left( \frac{\alpha }{2}-1 \right)}}{{\left\| \widehat{{{{\dot{\Delta }}}_{j}}{{I}_{3}}} \right\|}_{L_{T}^{2}}} \right\} \right\|}_{{{\ell }^{q}}}} \right\|}_{L_{\xi }^{\frac{3}{5-\alpha }}}} \\ & \le C\sum\limits_{k=1}^{3}{{{\left\| {{\left\| \left\{ {{\varphi }_{j}}{{\left\| \dot{R}\widehat{\left( {{u}^{k}},\boldsymbol{v} \right.}) \right\|}_{L_{T}^{1}}} \right\} \right\|}_{{{\ell }^{q}}}} \right\|}_{L_{\xi }^{5-\alpha }}}} \\ & \le C\sum\limits_{k=1}^{3}{{{\left\| {{\left\| \left\{ {{\varphi }_{j}}{{\left\| \dot{R}\widehat{\left( {{u}^{k}},\boldsymbol{v} \right.}) \right\|}_{L_{T}^{1}}} \right\} \right\|}_{L_{\xi }^{\frac{3}{5-\alpha }}}} \right\|}_{{{\ell }^{\frac{3}{5-\alpha }}}}}} \\ & \le C\sum\limits_{k=1}^{3}{\left\| {{\left\| {{\varphi }_{j}} \right\|}_{L_{\xi }^{\frac{3}{\alpha -2}}}} \right\|}{{\left\| \sum\limits_{i\ge j-3}{\widehat{{{{\dot{\Delta }}}_{i}}{{u}^{k}}}}*\widehat{{{{\tilde{\dot{\Delta }}}}_{i}}\boldsymbol{v}}{{\left\| _{L_{T}^{1}} \right\|}_{\frac{3}{L_{\xi }^{7-2\alpha }}}} \right\|}_{{{\ell }^{\frac{3}{5-\alpha }}}}} \\ & \le C\sum\limits_{k=1}^{3}{{{\left\| \left\{ {{2}^{i(\alpha -2)}}\sum\limits_{i\ge j-3}{{{\left\| \widehat{{{{\dot{\Delta }}}_{i}}{{u}^{k}}} \right\|}_{L_{T}^{2}L_{\xi }^{\frac{3}{5-\alpha }}}}\times }{{\left\| {{{\tilde{\dot{\Delta }}}}_{i}}\boldsymbol{v} \right\|}_{L_{T}^{2}L_{\xi }^{\frac{3}{5-\alpha }}}} \right\} \right\|}_{{{\ell }^{\frac{3}{5-\alpha }}}}}} \\ & \le C\sum\limits_{k=1}^{3}{{{\left\| {{u}^{k}} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{B}}_{\frac{3}{5-\alpha },q}^{\frac{\alpha }{2}-1}}}{{\left\| \boldsymbol{v} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{B}}_{\frac{3}{5-\alpha },r}^{\frac{\alpha }{2}-1}}}} \\ & \le C{{\left\| \boldsymbol{u} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{B}}_{\frac{3}{5-\alpha },q}^{\frac{\alpha }{2}-1}}}{{\left\| \boldsymbol{v} \right\|}_{\tilde{L}_{T}^{2}\hat{\dot{B}}_{\frac{3}{5-\alpha },r}^{\frac{\alpha }{2}-1}}}. \\ \end{align} $

其中, $ \frac{1}{r}+\frac{1}{q}=\frac{5-\alpha}{3}$。由$q \in\left[\frac{3}{5-\alpha}, \frac{6}{5-\alpha}\right] $可知, $ {{\ell }^{q}}$$\hookrightarrow$$ {{\ell }^{r}}$。

由引理2.5可得

$ \begin{array}{l} {\left\| {{I_3}} \right\|_{\widetilde {L}_T^2\hat {\dot F}_{\frac{3}{{5 - \alpha }}, q}^{\frac{\alpha }{2} - 1}}} \le C{\left\| \mathit{\boldsymbol{u}} \right\|_{\widetilde {L}_T^2\hat {\dot B}_{\frac{3}{{5 - \alpha }}, q}^{\frac{\alpha }{2} - 1}}}{\left\| \mathit{\boldsymbol{v}} \right\|_{\widetilde {L}_T^2\hat {\dot B}_{\frac{3}{{5 - \alpha }}, q}^{\frac{\alpha }{2} - 1}}}\\ \;\;\;\;\;\;\;\;\;\;\;\;\; \le C{\left\| \mathit{\boldsymbol{u}} \right\|_{\widetilde {L}_T^2\hat {\dot F}_{\frac{3}{{5 - \alpha }}, q}^{\frac{\alpha }{2} - 1}}}{\left\| \mathit{\boldsymbol{v}} \right\|_{\widetilde {L}_T^2\hat {\dot F}_{\frac{3}{{5 - \alpha }}, q}^{\frac{\alpha }{2} - 1}}}. \end{array} $

结合式(15)和式(16), 命题得证。

注3.1   由上面的证明易知不等式(8)可以推广为

$ \begin{array}{l} {\left\| {B(\mathit{\boldsymbol{u}}, \mathit{\boldsymbol{v}})} \right\|_{\widetilde {L}_T^r\hat {\dot F}_{p, q}^{4 - \alpha - \frac{3}{p} + \frac{\alpha }{r}}}} \le C{\left\| \mathit{\boldsymbol{u}} \right\|_{\widetilde {L}_T^2\hat {\dot F}_{p, q}^{4 - \frac{\alpha }{2} - \frac{3}{p}}}} \times \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\left\| \mathit{\boldsymbol{v}} \right\|_{\widetilde {L}_T^2\hat {\dot F}_{p, q}^{4 - \frac{\alpha }{2} - \frac{3}{p}}}}. \end{array} $

其中C=C(α, p), r≥1, $\frac{1}{r}+\frac{1}{r^{\prime}}=1$。

注3.2   显然, 由命题3.3的证明可知, 关于$q \in\left[\frac{3}{5-\alpha}, \frac{6}{5-\alpha}\right] $的假设主要是由余项决定的。

接下来证明定理1.1。首先, 对(p, q)∈ $\left(\frac{3}{5-\alpha}, \infty\right) \times[1, \infty) $, 由命题3.2, 可得

$ \begin{array}{l} \;\;\;{\left\| {B(\mathit{\boldsymbol{u}}, \mathit{\boldsymbol{v}})} \right\|_{\widetilde {L}_T^2\hat {\dot F}_{p, q}^{4 - \frac{\alpha }{2} - \frac{3}{p}}}}\\ \le C(\alpha , p){\left\| \mathit{\boldsymbol{u}} \right\|_{\widetilde {L}_T^2\hat {\dot F}_{p, q}^{4 - \frac{\alpha }{2} - \frac{3}{p}}}}{\left\| \mathit{\boldsymbol{v}} \right\|_{\widetilde {L}_T^2\hat {\dot F}_{p, q}^{4 - \frac{\alpha }{2} - \frac{3}{p}}}}. \end{array} $

以及由命题3.1可知, 存在T>0, 使得

$ {\left\| {{{\rm{e}}^{ - t{\varLambda ^\alpha }}}{\mathit{\boldsymbol{u}}_0}} \right\|_{\widetilde {L}_T^2\hat {\dot F}_{p, q}^{4 - \frac{\alpha }{2} - \frac{3}{p}}}} < \frac{1}{{4C(\alpha , p)}}. $

因此, 在空间$ X=\tilde{L}_T^2 \hat{\dot{F}}_{p, q}^{4-\frac{\alpha}{2}-\frac{3}{p}}$($\mathbb{R}^3$)中应用Banach不动点定理可知, 方程组(1)存在唯一的局部解u。记T^*为使得方程组(1)的解存在的最大时间点, 则对任意的T < T^*, 有

$ \boldsymbol{u} \in \tilde{L}_T^2 \hat{\dot{F}}_{p, q}^{4-\frac{\alpha}{2}-\frac{3}{p}}\left(\mathbb{R}^3\right). $

同理, 对$ (p, q) \in\left\{\frac{3}{5-\alpha}\right\} \times\left[\frac{3}{5-\alpha}, \frac{6}{5-\alpha}\right]$, 由Banach不动点定理及命题3.1和命题3.3, 方程组(1)存在唯一的局部解

$ \boldsymbol{u}\in \tilde{L}_{T}^{2}\hat{\dot{F}}_{\frac{3}{5-\alpha }, q}^{\frac{\alpha }{2}-1}\left( {{\mathbb{R}}^{3}} \right). $

此外, 对适当小的初始值, 可以得到整体解。这是因为, 对任意的t>0,

$ {{\left\| {{\text{e}}^{-t{{\varLambda }^{\alpha }}}}{{\boldsymbol{u}}_{0}} \right\|}_{\widetilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}\le {{\left\| {{\boldsymbol{u}}_{0}} \right\|}_{\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}}}}. $

所以, 存在充分小的$ \boldsymbol{\epsilon }$>0, 使得如果

$ {{\left\| {{\boldsymbol{u}}_{0}} \right\|}_{\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}}}}\le \epsilon \le \min \left\{ \frac{1}{4C(\alpha , p)}, \frac{1}{4C(\alpha )} \right\}. $

那么再次由Banach不动点定理可知, 对适当小的初始值, 方程组(1)存在唯一整体解

$ \boldsymbol{u} \in \tilde{L}^2\left(\mathbb{R}^{+} ; \hat{\dot{F}}_{p, q}^{4-\frac{\alpha}{2}-\frac{3}{p}}\left(\mathbb{R}^3\right)\right). $

其中, p, q满足条件(2)或条件(3)。

接下来证明: 如果$ \boldsymbol{u} \in \tilde{L}^2\left([0, T] ; \hat{\dot{F}}_{p, q}^{4-\frac{\alpha}{2}-\frac{3}{p}}\right)$, 则对任意的r∈[1, ∞], 以及0 < T < T^*, u∈ $ \tilde{L}^r\left([0, T] ; \hat{\dot{F}}_{p, q}^{4-\alpha-\frac{3}{p}+\frac{\alpha}{r}}\right)$成立。

命题3.4   假设u是方程组(1)的解, p, q满足条件(2)或条件(3)。如果

$ \boldsymbol{u}(\boldsymbol{x}, t) \in \tilde{L}^2\left([0, T] ; \hat{\dot{F}}_{p, q}^{4-\frac{\alpha}{2}-\frac{3}{p}}\left(\mathbb{R}^3\right)\right), $

则对任意的r∈[1, ∞],

$ \boldsymbol{u}(\boldsymbol{x}, t) \in \tilde{L}^r\left([0, T] ; \hat{\dot{F}}_{p, q}^{4-\alpha-\frac{3}{p}+\frac{\alpha}{r}}\left(\mathbb{R}^3\right)\right) $

成立。

证明   由Duhamel公式

$ \boldsymbol{u}(\boldsymbol{x}, t)=\mathrm{e}^{-t \varLambda^\alpha} \boldsymbol{u}_0+B(\boldsymbol{u}, \boldsymbol{u}). $

由注3.1

$ \begin{align} & \ \ \ \ \ \ \ \ {{\left\| B(\boldsymbol{u}, \boldsymbol{v}) \right\|}_{\widetilde{L}_{T}^{r}\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}+\frac{\alpha }{r}}}} \\ & \le C{{\left\| \boldsymbol{u} \right\|}_{\widetilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}{{\left\| \boldsymbol{v} \right\|}_{\widetilde{L}_{T}^{2}\hat{\dot{F}}_{p, q}^{4-\frac{\alpha }{2}-\frac{3}{p}}}}. \\ \end{align} $

故$ B(\boldsymbol{u}, \boldsymbol{v})\in \widetilde{L}_{T}^{r}\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}+\frac{\alpha }{r}}$。再根据引理2.4, 对线性项e^-tΛαu₀, 有

$ {{\left\| {{\text{e}}^{-t{{\varLambda }^{\alpha }}}}{{\boldsymbol{u}}_{0}} \right\|}_{\widetilde{L}_{T}^{r}\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}+\frac{\alpha }{r}}}}\le C{{\left\| {{\boldsymbol{u}}_{0}} \right\|}_{\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}}}}<\infty . $

综上, 有u(x, t)∈ $ \widetilde{L}_{T}^{r}\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}+\frac{\alpha }{r}}$, 其中r∈[1, ∞]。

最后证明$\|\boldsymbol{u}(t){{\|}_{\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}}}} $关于时间t的连续性。

命题3.5   设u是方程组(1)的解。如果

$ \begin{gathered} \boldsymbol{u} \in \tilde{L}^1\left([0, T] ; \hat{\dot{F}}_{p, q}^{4-\frac{3}{p}}\left(\mathbb{R}^3\right)\right) \cap \tilde{L}^{\infty}([0, T] ; \\ \left.\hat{\dot{F}}_{p, q}^{4-\alpha-\frac{3}{p}}\left(\mathbb{R}^3\right)\right), q \in[1, \infty), \end{gathered} $

则

$ \boldsymbol{u} \in C\left([0, T] ; \hat{\dot{F}}_{p, q}^{4-\alpha-\frac{3}{p}}\left(\mathbb{R}^3\right)\right). $

证明   假设u是方程组(1)的解, 则

$ \partial_t \boldsymbol{u}=-\mathbb{P}(\boldsymbol{u} \cdot \nabla) \boldsymbol{u}-\varLambda^\alpha \boldsymbol{u} . $

从而, 有

$ \partial_t \dot{\Delta}_k \boldsymbol{u}=-\dot{\Delta}_k \mathbb{P}(\boldsymbol{u} \cdot \nabla) \boldsymbol{u}-\varLambda^\alpha \dot{\varDelta}_k \boldsymbol{u}. $

所以

$ \begin{align} & {{\partial }_{t}}{{{\dot{\Delta }}}_{k}}{{\boldsymbol{u}}_{\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}}}}\le {{\left\| {{{\dot{\Delta }}}_{k}}\mathbb{P}(\boldsymbol{u}\cdot \boldsymbol{\nabla })\boldsymbol{u} \right\|}_{\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}}}}+ \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{\left\| {{\varLambda }^{\alpha }}{{{\dot{\Delta }}}_{k}}\boldsymbol{u} \right\|}_{\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}}}}. \\ \end{align} $

由u的正则性可知

$ {{\left\| {{\partial }_{t}}{{{\dot{\Delta }}}_{k}}\boldsymbol{u} \right\|}_{\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}}}}\in L_{\text{loc}}^{1}(0, T). $

(17)

因此, ${{\left\| {{{\dot{\Delta }}}_{k}}\boldsymbol{u} \right\|}_{\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}}}} $关于时间变量t是连续的。对任意的t₀, t₁∈[0, T],

$ \begin{align} & \ \ \ \ \ \ \ \ \ {{\left\| \boldsymbol{u}\left( {{t}_{1}} \right)-\boldsymbol{u}\left( {{t}_{0}} \right) \right\|}_{\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}}}} \\ & ={{\left\| {{\left( \sum\limits_{i\in \mathbb{Z}}{{{2}^{i\left( 4-\alpha -\frac{3}{p} \right)q}}}{{\left| \widehat{{{{\dot{\Delta }}}_{i}}\boldsymbol{u}\left( {{t}_{1}} \right)}-\widehat{{{{\dot{\Delta }}}_{i}}\boldsymbol{u}\left( {{t}_{0}} \right)} \right|}^{q}} \right)}^{\frac{1}{q}}} \right\|}_{{{L}^{p}}}} \\ & =\|\left( \sum\limits_{|i|<N}{{{2}^{i\left( 4-\alpha -\frac{3}{p} \right)q}}}{{\left| \widehat{{{{\dot{\Delta }}}_{i}}\boldsymbol{u}\left( {{t}_{1}} \right)}-\widehat{{{{\dot{\Delta }}}_{i}}\boldsymbol{u}\left( {{t}_{0}} \right)} \right|}^{q}}+ \right. \\ & {{\left. \sum\limits_{|i|\ge N}{{{2}^{i\left( 4-\alpha -\frac{3}{p} \right)q}}}{{\left| \widehat{{{{\dot{\Delta }}}_{i}}\boldsymbol{u}\left( {{t}_{1}} \right)}-\widehat{{{{\dot{\Delta }}}_{i}}\boldsymbol{u}\left( {{t}_{0}} \right)} \right|}^{q}} \right)}^{\frac{1}{q}}}{{\|}_{{{L}^{p}}}}\text{. } \\ \end{align} $

由q∈[1, ∞], 以及Minkowski不等式可得

$ \begin{align} & \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{\left\| \boldsymbol{u}\left( {{t}_{1}} \right)-\boldsymbol{u}\left( {{t}_{0}} \right) \right\|}_{\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}}}} \\ & \le {{\left\| {{\left( \sum\limits_{i<N}{{{2}^{i\left( 4-\alpha -\frac{3}{p} \right)q}}}{{\left| \widehat{{{{\dot{\Delta }}}_{i}}\boldsymbol{u}\left( {{t}_{1}} \right)}-\widehat{{{{\dot{\Delta }}}_{i}}\boldsymbol{u}\left( {{t}_{0}} \right)} \right|}^{q}} \right)}^{\frac{1}{q}}} \right\|}_{{{L}^{p}}}}+ \\ & \ \ \ {{\left\| {{\left( \sum\limits_{i\ge N}{{{2}^{i\left( 4-\alpha -\frac{3}{p} \right)q}}}{{\left| \widehat{{{{\dot{\Delta }}}_{i}}\boldsymbol{u}\left( {{t}_{1}} \right)}-\widehat{{{{\dot{\Delta }}}_{i}}\boldsymbol{u}\left( {{t}_{0}} \right)} \right|}^{q}} \right)}^{\frac{1}{q}}} \right\|}_{{{L}^{p}}}} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ :=M_{N}^{1}+M_{N}^{2}. \\ \end{align} $

由于u∈ $ \tilde{L}^{\infty}$([0, T]; $ \hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}}$), 则对任意的$ \boldsymbol{\epsilon }$>0, 存在N₀, 使得

$ M_{N_0}^2<\frac{\boldsymbol{\epsilon}}{2} . $

下面估计$M_{N_0}^1 $, 由(17)

$ \begin{align} & M_{{{N}_{0}}}^{1}={{\left\| {{\left( \sum\limits_{i<{{N}_{0}}}{{{2}^{i\left( 4-\alpha -\frac{3}{p} \right)q}}}{{\left| \widehat{{{{\dot{\Delta }}}_{i}}\boldsymbol{u}\left( {{t}_{1}} \right)}-\widehat{{{{\dot{\Delta }}}_{i}}\boldsymbol{u}\left( {{t}_{0}} \right)} \right|}^{q}} \right)}^{\frac{1}{q}}} \right\|}_{{{L}^{p}}}} \\ & \ \ \ \ \ \ \ \ \le {{\left\| {{\left( \sum\limits_{|i|<{{N}_{0}}}{{{\left( {{2}^{i\left( 4-\alpha -\frac{3}{p} \right)}}\int_{{{t}_{0}}}^{{{t}_{1}}}{\left| {{\partial }_{t}}\widehat{{{{\dot{\Delta }}}_{i}}\boldsymbol{u}\left( t \right)} \right|}\text{d}t \right)}^{q}}} \right)}^{\frac{1}{q}}} \right\|}_{{{L}^{p}}}} \\ & \ \ \ \ \ \ \ \ \le {{\int_{{{t}_{0}}}^{{{t}_{1}}}{\sum\limits_{|i|<{{N}_{0}}}{\left\| {{\partial }_{t}}{{{\dot{\Delta }}}_{i}}\boldsymbol{u}(t) \right\|}}}_{\hat{\dot{F}}_{p, q}^{4-\alpha -\frac{3}{p}}}}~\text{d}t. \end{align} $

显然当|t₁-t₀|→0时, 由上面的不等式可知M_N0¹→0。故

$ \boldsymbol{u} \in C\left([0, T] ; \hat{\dot{F}}_{p, q}^{4-\alpha-\frac{3}{p}}\left(\mathbb{R}^3\right)\right). $

命题得证。

因此, 定理1.1得证。