本文研究了如下带有衰减项的不可压缩的磁流体力学方程组的柯西问题

(1.1) $ \begin{eqnarray} \left\{\begin{array}{ll} {u_t} + u \cdot \nabla u + \nabla p - \mu \Delta u + \eta |u{|^{\beta - 1}}u = b \cdot \nabla b, \\ {b_t} + u \cdot \nabla b - \nu \Delta b = b \cdot \nabla u, \\ \nabla \cdot u = 0, \nabla \cdot b = 0, \\ u{|_{t = 0}} = {u_0}, b{|_{t = 0}} = {b_0}, \\ |u| \to 0, |b| \to 0, \quad {\rm as}\ |x| \to 0. \end{array}\right. \end{eqnarray} $

这里$ x = ({{x_1}, {x_2}, {x_3}}) \in {{\Bbb R} ^3} $, $ t \ge 0 $, $ u = u({x, t}) = ({{u_1}({x, t}), {u_2}({x, t}), {u_3}({x, t})}) $, $ p = p({x, t}) $, $ b = b({x, t}) = ({{b_1}({x, t}), {b_2}({x, t}), {b_3}({x, t})}) $,其中$ u $是速度场, $ b $是磁场, $ p $是压力, $ \mu $是流体粘性系数, $ \eta $是阻尼系数, $ \nu $是磁场的耗散系数并且$ \mu, \eta, \nu \ge 0 $, $ \beta $是常数且满足$ \beta \ge 1 $.已知函数$ {u_0} = {u_0}(x) $和$ {b_0} = {b_0}(x) $分别为初始速度和初始磁场.

当$ \eta = 0 $时, (1.1)式可化为磁流体方程组.磁流体力学是流体力学的一个重要分支,主要研究等离子体这种导电流体在磁场中的运动.磁流体方程模型的具体推导,参见文献[1-3].文献[4-5]证明了空间维数$ n = 2 $时,解的整体适定性以及$ n = 3 $时,解的局部适定性.而关于磁流体力学方程组的全局正则性的最新进展,参见文献[6].众所周知,对于磁流体方程组(1.1),若令$ b = 0 $,则方程组(1.1)可化为Navier-Stokes方程,这正是柯朗研究所提出的七个世纪难题之一,而磁流体方程比Navier-Stokes方程更为复杂,对于它的深入研究无疑会对Navier-Stokes方程起到一定的促进作用.当$ \eta > 0 $时,衰减项$ |u{|^{\beta - 1}}u $是一个外力,在很多物理情景中都有所描述,见文献[7].文献[8]中作者首先研究了带有衰减项的不可压缩的Navier-Stokes方程的适定性问题,采用经典的Galerkin方法构造逼近解,建立相应的先验估计,进一步,由紧性原理可得：当$ \beta \ge 1 $时,存在全局弱解,当$ \beta \ge {7 \over 2} $时,存在全局强解且当$ {7 \over 2} \le \beta \le 5 $时,得到了强解的唯一性.而文献[9]对[8]所得结果进行了进一步完善,将全局强解的存在范围扩充到了$ \beta \ge 3 $,且在$ 3 \le \beta \le 5 $时,得到了强解的唯一性.在文献[8-9]的基础上,文献[10]得到了强解在$ \beta \ge 1 $时的唯一性.受到文献[8-10]的启发,本文将结论推广到了磁流体力学方程组(1.1),得到了在$ \beta \ge 1 $时,全局弱解的存在性以及局部强解的存在性和唯一性.

这篇文章的结构安排如下:第2节证明了方程组(1.1)的全局弱解在$ \beta \ge 1 $时的存在性;第3节证明了方程组(1.1)的局部强解的存在性和唯一性.在这篇文章中,不失一般性,假设$ \mu = \eta = \nu = 1 $,将函数$ f $的$ {L^p} $范数表示为$ {\rm{||}}f|{|_{{L^p}}} $或$ {\rm{||}}f|{|_p} $,另外本文中出现的常数若无特别说明均记为$ C $,不加以区别.

本节将证明问题(1.1)的弱解的全局存在性,下面先给出弱解的定义.

定义2.1  $ (u(x, t), b(x, t), p(x, t)) $被称为问题(1.1)的弱解.如果对于任意的$ T > 0 $,下面的条件都成立:

(1)

$ u \in {L^\infty }(0 , T ;{L^2}({{\Bbb R} ^3} )) \cap {L^2} (0 , T ;H_0^1({{\Bbb R} ^3})) \cap {L^{\beta + 1}}(0 , T ;{L^{\beta + 1}}({{\Bbb R} ^3})), $

$ b \in {L^\infty }(0 , T ;{L^2}({{\Bbb R} ^3} )) \cap {L^2} (0 , T ;H_0^1({{\Bbb R} ^3})). $

(2)对于任意的$ \phi \in C_0^\infty (\left[{0, T} \right] \times {{\Bbb R} ^3}) $且$ \phi (\cdot, T) = 0 $,有

(2.1) $ \begin{eqnarray} && - \int_0^T {(u, {\phi _t})} {\rm d}t + \int_0^T {\int_{{{\Bbb R} ^3}} {(u \cdot \nabla )u\phi } } {\rm d}x{\rm d}t + \int_0^T {\int_{{{\Bbb R} ^3}} {\nabla u:\nabla \phi } } {\rm d}x{\rm d}t \\ &&+ \int_0^T {\int_{{{\Bbb R} ^3}} {|u{|^{\beta - 1}}u\phi } } {\rm d}x{\rm d}t - \int_0^T {\int_{{{\Bbb R} ^3}} b } \nabla b\phi {\rm d}x{\rm d}t = ({u_0}, \phi (0)), \\ & &- \int_0^T {(b, {\phi _t})} {\rm d}t + \int_0^T {\int_{{{\Bbb R} ^3}} {(u \cdot \nabla )b\phi } } {\rm d}x{\rm d}t + \int_0^T {\int_{{{\Bbb R} ^3}} {\nabla b:\nabla \phi } } {\rm d}x{\rm d}t \\ && - \int_0^T {\int_{{{\Bbb R} ^3}} b } \nabla u\phi {\rm d}x{\rm d}t = ({b_0}, \phi (0)). \end{eqnarray} $

(3) $ \nabla \cdot u(x, t) $和$ \nabla \cdot b(x, t) $在$ {{\Bbb R} ^3} \times \left[{0, T} \right] $上几乎处处为$ 0 $.

在(2.1)式中, $ {\nabla u} $表示矩阵$ {({\partial _i}{u_j})_{3 \times 3}} $, $ {\nabla b} $表示矩阵$ {({\partial _i}{b_j})_{3 \times 3}} $.对于两个矩阵$ A = ({a_{ij}}) $和$ B = ({b_{ij}}) $,则$ A:B = \sum\limits_{i, j = 1}^3 {{a_{ij}}{b_{ij}}} $,符号$ (, ) $表示在$ {L^2}({{\Bbb R} ^3}) $上的内积,函数$ \phi (0) = \phi ({ \cdot, 0}) $.

本节的主要结论由下面的定理2.1给出.

定理2.1  假设$ \beta \ge 1 $且$ {u_0} \in {L^2}({{\Bbb R} ^3}) $, $ {b_0} \in {L^2}({{\Bbb R} ^3}) $.那么对于任意的$ T > 0 $,问题(1.1)存在着一个弱解$ (u(x, t), p(x, t), b(x, t)) $使得

(2.2) $ \begin{equation} \begin{array}{l} u \in {L^\infty }(0 , T ;{L^2}({{\Bbb R} ^3} )) \cap {L^2} (0 , T ;H_0^1({{\Bbb R} ^3})) \cap {L^{\beta + 1}}(0 , T ;{L^{\beta + 1}}({{\Bbb R} ^3})), \\ b \in {L^\infty }(0 , T ;{L^2}({{\Bbb R} ^3} )) \cap {L^2} (0 , T ;H_0^1({{\Bbb R} ^3})), \end{array} \end{equation} $

并且

(2.3) $ \begin{eqnarray} && \mathop {\sup }\limits_{0 \le t \le T} ||u||_{{L^2}}^2 + \mathop {\sup }\limits_{0 \le t \le T} ||b||_{{L^2}}^2 + 2\int_0^T {||\nabla u||_{{L^2}}^2} {\rm d}t + 2\int_0^T {||\nabla b||_{{L^2}}^2} {\rm d}t\\ && + 2\int_0^T {||u||_{{L^{\beta + 1}}}^{\beta + 1}} {\rm d}t \le ||{u_0}||_{{L^2}}^2 + ||{b_0}||_{{L^2}}^2. \end{eqnarray} $

证  因为$ H_0^1 $是可分空间且$ C_0^\infty $在$ H_0^1 $中稠密.因此存在着$ C_0^\infty $中的序列$ {w_1}, {w_2}, \cdots, {w_m} $,对于任意的$ m $,分别定义逼近解$ {u_m} $, $ {b_m} $如下

$ {u_m} = \sum\limits_{j = 1}^m {{g_{jm}}} (t){w_j}(x), \qquad {b_m} = \sum\limits_{j = 1}^m {{h_{jm}}} (t){w_j}(x), $

并且

(2.4) $ \begin{eqnarray} &&({u_m}'(t), {w_j}) + ({u_m}(t) \cdot \nabla {u_m}(t), {w_j}) + (\nabla {u_m}(t), \nabla {w_j}) + (|{u_m}{|^{\beta - 1}}{u_m}(t), {w_j})\\ && = ({b_m}(t) \cdot \nabla {b_m}(t), {w_j}), \end{eqnarray} $

(2.5) $ \begin{eqnarray} && ({b_m}'(t), {w_j}) + ({u_m}(t) \cdot \nabla {b_m}(t), {w_j}) + (\nabla {b_m}(t), \nabla {w_j}) = ({b_m}(t) \cdot \nabla {u_m}(t), {w_j}), \end{eqnarray} $

$ t \in \left[{0, T} \right] $, $ j = 1, 2, 3, \cdots, m $,并且在$ {L^2} $中$ {u_{m0}} \to {u_0} $, $ {b_{m0}} \to {b_0} $当$ m \to \infty $.

这里需要先得出一个对于$ {u_m}, {b_m} $的先验估计.

引理2.1  假设$ {u_m}, {b_m} \in {L^2} $,那么对于任意给定的$ T > 0 $,任意的$ \beta \ge 1 $,有

$ \begin{eqnarray*} &&\mathop {\sup }\limits_{0 \le t \le T} ||{u_m}||_{{L^2}}^2 + \mathop {\sup }\limits_{0 \le t \le T} ||{b_m}||_{{L^2}}^2 + 2\int_0^T {||\nabla u||_2^2} {\rm d}t + 2\int_0^T {||\nabla b||_2^2} {\rm d}t + 2\int_0^T {||u||_{{L^{\beta + 1}}}^{\beta + 1}} {\rm d}t\\ & \le & ||{u_0}||_{{L^2}}^2 + ||{b_0}||_{{L^2}}^2. \end{eqnarray*} $

证 将(2.4)式两端同时乘以$ {g_{jm}}(t) $,然后关于$ j = 1, 2, \cdots, m $求和.将(2.5)式两端同时乘以$ {h_{jm}}(t) $,然后也关于$ j = 1, 2, \cdots, m $求和.最后再将得到的两个结果式相加,可得

$ \begin{eqnarray*} &&{1 \over 2}{{\rm d}\over {{\rm d}t}}||{u_m}||_{{L^2}}^2 + {1 \over 2}{{\rm d}\over {{\rm d}t}}||{b_m}||_2^2 + ||\nabla {u_m}||_{{L^2}}^2 + ||\nabla {b_m}||_{{L^2}}^2 + ||{u_m}||_{{L^{\beta + 1}}}^{\beta + 1}\\ & = & \int_{{{\Bbb R} ^3}} {{b_m}\nabla {b_m}} {u_m}{\rm d}x + \int_{{{\Bbb R} ^3}} {{b_m}\nabla {u_m}} {b_m}{\rm d}x = 0. \end{eqnarray*} $

上式中用到结论:对于$ u \in H_0^1 $和$ v \in {H^1} $,可得$ ((u \cdot \nabla)v, v) = 0 $.

再将上式在$ ({0, T}) $上关于$ t $积分,可得

(2.6) $ \begin{eqnarray} &&\mathop {\sup }\limits_{0 \le t \le T} ||{u_m}||_{{L^2}}^2 + \mathop {\sup }\limits_{0 \le t \le T} ||{b_m}||_{{L^2}}^2 + 2\int_0^T {||\nabla u||_2^2} {\rm d}t + 2\int_0^T {||\nabla b||_2^2} {\rm d}t + 2\int_0^T {||u||_{{L^{\beta + 1}}}^{\beta + 1}} {\rm d}t\\ & \le& ||{u_0}||_{{L^2}}^2 + ||{b_0}||_{{L^2}}^2. \end{eqnarray} $

引理2.1得证.

使用上面的引理2.1,通过一个标准的过程,容易得到逼近解

$ {u_m} \in {L^\infty }(0, T;{L^2}({{\Bbb R} ^3})) \cap {L^2}(0, T;H_0^1({{\Bbb R} ^3})) \cap {L^{\beta + 1}}(0, T;{L^{\beta + 1}}({{\Bbb R} ^3})), $

$ {b_m} \in {L^\infty }(0, T;{L^2}({{\Bbb R} ^3})) \cap {L^2}(0, T;H_0^1({{\Bbb R} ^3})) $

的全局存在性.然后使用文献[11]中定理2.2来证明$ {u_m}, {b_m} $（或者它们的子列）对于任意的$ \Omega \subset {{\mathbb R}^3} $在$ {L^2} \cap {L^\beta }(\left[{0, T} \right] \times \Omega) $上强收敛.将$ {u_m}, {b_m} $从$ \left[{0, T} \right] $上延拓到$ {\Bbb R} $. $ {{\tilde u}_m} $在$ \left[{0, T} \right] $上等于$ {u_m} $,在其补集上等于$ 0 $. $ {{\tilde b}_m} $在$ \left[{0, T} \right] $上等于$ {b_m} $,在其补集上等于$ 0 $.同样地,把$ {g_{jm}}(t) $和$ {h_{jm}}(t) $延拓到$ {\Bbb R} $上.令$ {{\tilde g}_{jm}}(t) = 0 $, $ {{\tilde h}_{jm}}(t) = 0 $,当$ t \in {\Bbb R} \backslash \left[{0, T} \right] $. $ {{\tilde u}_m} $, $ {{\tilde b}_m} $, $ {{\tilde g}_{jm}} $, $ {{\tilde h}_{jm}} $关于时间$ t $的Fourier变换分别为$ {\widehat {\widetilde u}_m} $, $ {\widehat {\widetilde b}_m} $, $ {\widehat {\widetilde g}_{jm}} $, $ {\widehat {\widetilde h}_{im}} $.

延拓之后的逼近解$ {{\tilde u}_m} $, $ {{\tilde b}_m} $将会满足

(2.7) $ \begin{eqnarray} {{\rm d}\over {{\rm d}t}}( {{{\tilde u}_m}, {w_j}} ) & = & ( {{{\tilde u}_m} \cdot \nabla {{\tilde u}_m}, {w_j}} ) + ( {\nabla {{\tilde u}_m}, \nabla {w_j}} ) + ( {|{{\tilde u}_m}{|^{\beta - 1}}{{\tilde u}_m}, {w_j}} ) - ( {{{\tilde b}_m} \cdot \nabla {{\tilde b}_m}, {w_j}} ) \\ && + ( {{u_{0m}}, {w_j}} ){\delta _0} - ({u_m}(T), {w_j}){\delta _T}, \quad j = 1, 2, \cdots , m, \end{eqnarray} $

(2.8) $ \begin{eqnarray} {{\rm d}\over {{\rm d}t}}( {{{\tilde b}_m}, {w_j}} ) & = & ( {{{\tilde u}_m} \cdot \nabla {{\tilde b}_m}, {w_j}} ) + ( {\nabla {{\tilde b}_m}, \nabla {w_j}} ) - ( {{{\tilde b}_m} \cdot \nabla {{\tilde u}_m}, {w_j}} ) + ( {{b_{0m}}, {w_j}} ){\delta _0} \\ & & - ({b_m}(T), {w_j}){\delta _T}, \quad j = 1, 2, \cdots , m, \end{eqnarray} $

其中$ {\delta _0} $, $ {\delta _T} $分别为$ {0, T} $处的Dirac分布.

令

$ ( {{{\tilde f}_m}, {w_j}} ) = ({{\tilde u}_m} \cdot \nabla {{\tilde u}_m}, {w_j}) + (\nabla {{\tilde u}_m}, \nabla {w_j}), $

$ ( {{{\tilde g}_m}, {w_j}} ) = ({{\tilde u}_m} \cdot \nabla {{\tilde b}_m}, {w_j}) + (\nabla {{\tilde b}_m}, \nabla {w_j}). $

对(2.7)式和(2.8)式两端分别做Fourier变换得

(2.9) $ \begin{eqnarray} 2\pi {\rm i}\tau ({\widehat {\widetilde u}_m}, {w_j})& = & ({\widehat {\widetilde f}_m}, {w_j}) + (\widehat {|{{\tilde u}_m}{|^\beta }{{\tilde u}_m}}, {w_j}) - ( {\widehat {{{\tilde b}_m} \cdot \nabla {{\tilde b}_m}}, {w_j}} )\\ && + ( {{u_{0m}}, {w_j}} )- ({u_m}(T), {w_j})\exp ( - 2\pi {\rm i}T\tau ), \end{eqnarray} $

(2.10) $ \begin{equation} 2\pi {\rm i}\tau ({\widehat {\widetilde b}_m}, {w_j}) = ({\widehat {\widetilde g}_m}, {w_j}) - ( {\widehat {{{\tilde b}_m} \cdot \nabla {{\tilde u}_m}}, {w_j}} ) + ( {{b_{0m}}, {w_j}} )- ({b_m}(T), {w_j})\exp ( - 2\pi {\rm i}T\tau ). \end{equation} $

对(2.9)式和(2.10)式两端分别同时乘以$ {\widehat {\widetilde g}_{im}}(\tau) $, $ {\widehat {\widetilde h}_{im}}(\tau) $并且关于$ j = 1, 2, \cdots, m $分别相加得

(2.11) $ \begin{eqnarray} 2\pi {\rm i}\tau ||{\widehat {\widetilde u}_m}||_{{L^2}}^2 & = & ({\widehat {\widetilde f}_m}, {\widehat {\widetilde u}_m}) + (\widehat {|{{\tilde u}_m}{|^\beta }{{\tilde u}_m}}, {\widehat {\widetilde u}_m}) - ( {\widehat {{{\tilde b}_m} \cdot \nabla {{\tilde b}_m}}, {{\widehat {\widetilde u}}_m}} )\\ && + ( {{u_{0m}}, {{\widehat {\widetilde u}}_m}} ) - ({u_m}(T), {\widehat {\widetilde u}_m})\exp ( - 2\pi {\rm i}T\tau ), \end{eqnarray} $

(2.12) $ \begin{equation} 2\pi {\rm i}\tau ||{\widehat {\widetilde b}_m}||_{{L^2}}^2 = ({\widehat {\widetilde g}_m}, {\widehat {\widetilde b}_m}) - ( {\widehat {{{\tilde b}_m} \cdot \nabla {{\tilde u}_m}}, {{\widehat {\widetilde b}}_m}} ) + ( {{u_{0m}}, {{\widehat {\widetilde b}}_m}} ) - ({u_m}(T), {\widehat {\widetilde b}_m})\exp ( - 2\pi {\rm i}T\tau ). \end{equation} $

对于任意$ v \in {L^2}({0, T; H_0^1}) \cap {L^{\beta + 1}}(0, T; {L^{\beta + 1}}) $,有

$ ( {{f_m}( t ), v} ) = ( {{u_m} \cdot \nabla {u_m}, v} ) + ( {\nabla {u_m}, \nabla v} ) \le C( {||{u_m}||_{{L^2}}^2 + ||\nabla {u_m}||_{{L^2}}^2 + ||\nabla {u_m}|{|_{{L^2}}}} )||v|{|_{{H^1}}}, $

$ ( {{g_m}( t ), v} ) = ( {{u_m} \cdot \nabla {b_m}, v} ) + ( {\nabla {b_m}, \nabla v} ) \le C( {||{u_m}||_{{L^2}}^2 + ||\nabla {b_m}||_{{L^2}}^2 + ||\nabla {b_m}|{|_{{L^2}}}} )||v|{|_{{H^1}}}. $

对于给定的$ T > 0 $,有

$ \int_0^T {||{f_m}(t)|{|_{{H^{ - 1}}}}} {\rm d}t \le \int_0^T {C( {||{u_m}||_{{L^2}}^2 + ||\nabla {u_m}||_{{L^2}}^2 + ||\nabla {u_m}|{|_{{L^2}}}} )} {\rm d}t \le C, $

$ \int_0^T {||{g_m}(t)|{|_{{H^{ - 1}}}}} {\rm d}t \le \int_0^T {C( {||{u_m}||_{{L^2}}^2 + ||\nabla {b_m}||_{{L^2}}^2 + ||\nabla {b_m}|{|_{{L^2}}}} )} {\rm d}t \le C. $

因此

(2.13) $ \begin{eqnarray} \mathop {\sup }\limits_{\tau \in R} ||{\widehat {\widetilde f}_m}( \tau )|{|_{{H^{ - 1}}}} \le \int_0^T {||{f_m}} (t)|{|_{{H^{ - 1}}}}{\rm d}t \le C, \end{eqnarray} $

(2.14) $ \begin{eqnarray} \mathop {\sup }\limits_{\tau \in R} ||{\widehat {\widetilde g}_m}( \tau )|{|_{{H^{ - 1}}}} \le \int_0^T {||{g_m}} (t)|{|_{{H^{ - 1}}}}{\rm d}t \le C. \end{eqnarray} $

对于$ ||{{\tilde b}_m} \cdot \nabla {{\tilde b}_m}|{|_{{L^2}}} $和$ ||{{\tilde b}_m} \cdot \nabla {{\tilde u}_m}|{|_{{L^2}}} $有如下估计

$ \int_0^T {||{{\tilde b}_m} \cdot \nabla {{\tilde b}_m}|{|_{{L^2}}}} {\rm d}t \le C\int_0^T {||{{\tilde b}_m}||_{{L^2}}^2 + ||\nabla {{\tilde b}_m}||_{{L^2}}^2} \le C, $

$ \int_0^T {||{{\tilde b}_m} \cdot \nabla {{\tilde u}_m}|{|_{{L^2}}}} {\rm d}t \le C\int_0^T {||{{\tilde b}_m}||_{{L^2}}^2 + ||\nabla {{\tilde u}_m}||_{{L^2}}^2} \le C, $

易得

(2.15) $ \begin{eqnarray} \mathop {\sup }\limits_{\tau \in R} ||\widehat {{{\tilde b}_m} \cdot \nabla {{\tilde b}_m}}|{|_{{L^2}}} \le C, \end{eqnarray} $

(2.16) $ \begin{eqnarray} \mathop {\sup }\limits_{\tau \in R} ||\widehat {{{\tilde b}_m} \cdot \nabla {{\tilde u}_m}}|{|_{{L^2}}} \le C. \end{eqnarray} $

由引理2.1中有

$ \int_0^T {|| |{u_m}{|^{\beta - 1}}{u_m}|{|_{{{\beta + 1} \over \beta }}}} {\rm d}t \le \int_0^T {||{u_m}||_{\beta + 1}^\beta } {\rm d}t \le C, $

于是

(2.17) $ \begin{equation} \mathop {\sup }\limits_{\tau \in R} ||\widehat {|{u_m}{|^{\beta - 1}}{u_m}}( \tau )|{|_{{{\beta + 1} \over \beta }}} \le C. \end{equation} $

容易从引理2.1中得到

(2.18) $ \begin{equation} |{u_m}(0)|{|_{{L^2}}} \le C, ||{u_m}(T)|{|_{{L^2}}} \le C, \quad ||{b_m}( 0 )|{|_{{L^2}}} \le C, ||{b_m}( T )|{|_{{L^2}}} \le C. \end{equation} $

由(2.13)–(2.18)式容易得到

$ |\tau |||{\widehat {\widetilde u}_m}( \tau )||_{{L^2}}^2 \le C( {||{{\widehat {\widetilde u}}_m}( \tau )|{|_{{H^1}}} + |{{\widehat {\widetilde u}}_m}( \tau )|{|_{\beta + 1}}} ), $

$ |\tau |||{\widehat {\widetilde b}_m}( \tau )||_{{L^2}}^2 \le C||{\widehat {\widetilde u}_m}( \tau )|{|_{{H^1}}}. $

对于任意给定的$ \gamma $, $ 0 < \gamma < {1 \over 4} $,有

$ |\tau {|^{2\gamma }} \le C{{1 + |\tau |} \over {1 + |\tau {|^{1 - 2\gamma }}}}, \quad \forall \tau \in {\Bbb R}. $

因此

(2.19) $ \begin{eqnarray} \int_{ - \infty }^\infty {|\tau {|^{2\gamma }}||{{\widehat {\widetilde u}}_m}||_{{L^2}}^2} {\rm d}\tau & \le& C\int_{ - \infty }^\infty {{{1 + |\tau |} \over {1 + |\tau {|^{1 - 2\gamma }}}}} ||{\widehat {\widetilde u}_m}||_{{L^2}}^2{\rm d}\tau\\ &\le& C\int_{ - \infty }^\infty {||{{\widehat {\widetilde u}}_m}||_{{L^2}}^2} {\rm d}\tau + C\int_{ - \infty }^\infty {{{||{{\widehat {\widetilde u}}_m}( \tau )|{|_{{H^1}}}} \over {1 + |\tau {|^{1 - 2\gamma }}}}} {\rm d}\tau\\&& + C\int_{ - \infty }^\infty {{{||{{\widehat {\widetilde u}}_m}( \tau )|{|_{\beta + 1}}} \over {1 + |\tau {|^{1 - 2\gamma }}}}} {\rm d}\tau , \end{eqnarray} $

(2.20) $ \begin{eqnarray} \int_{ - \infty }^\infty {|\tau {|^{2\gamma }}||{{\widehat {\widetilde b}}_m}||_{{L^2}}^2} {\rm d}\tau &\le& C\int_{ - \infty }^\infty {{{1 + |\tau |} \over {1 + |\tau {|^{1 - 2\gamma }}}}} ||{\widehat {\widetilde b}_m}||_{{L^2}}^2{\rm d}\tau\\ &\le & C\int_{ - \infty }^\infty {||{{\widehat {\widetilde b}}_m}||_{{L^2}}^2} {\rm d}\tau + C\int_{ - \infty }^\infty {{{||{{\widehat {\widetilde b}}_m}( \tau )|{|_{{H^1}}}} \over {1 + |\tau {|^{1 - 2\gamma }}}}} {\rm d}\tau . \end{eqnarray} $

根据Parseval等式与引理2.1, (2.19)与(2.20)式右边第一个积分项关于$ m $一致有界.由Schwartz不等式, Parseval等式和引理2.1,可得

(2.21) $ \begin{eqnarray} && \int_{ - \infty }^\infty {{{||{{\widehat {\widetilde u}}_m}( \tau )|{|_{{H^1}}}} \over {1 + |\tau {|^{1 - 2\gamma }}}}} {\rm d}\tau \le {\bigg( {\int_{ - \infty }^\infty {{{{\rm d}\tau } \over {{{(1 + |\tau {|^{1 - 2\gamma }})}^2}}}} }\bigg )^{{1 \over 2}}}{\bigg( {\int_0^T {||{u_m}(\tau )||_{{H^1}}^2{\rm d}\tau } }\bigg )^{{1 \over 2}}} \le C, \end{eqnarray} $

(2.22) $ \begin{eqnarray} &&\int_{ - \infty }^\infty {{{||{{\widehat {\widetilde b}}_m}( \tau )|{|_{{H^1}}}} \over {1 + |\tau {|^{1 - 2\gamma }}}}} {\rm d}\tau \le {\bigg( {\int_{ - \infty }^\infty {{{{\rm d}\tau } \over {{{(1 + |\tau {|^{1 - 2\gamma }})}^2}}}} } \bigg)^{{1 \over 2}}}{\bigg( {\int_0^T {||{b_m}(\tau )||_{{H^1}}^2{\rm d}\tau } }\bigg )^{{1 \over 2}}} \le C, \end{eqnarray} $

其中$ 0 < \gamma < {1 \over 4} $.

同样地,当$ 0 < \gamma < {1 \over {2({\beta + 1})}} $,可得

(2.23) $ \begin{eqnarray} \int_{ - \infty }^\infty {{{||{{\widehat {\widetilde u}}_m}( \tau )|{|_{\beta + 1}}} \over {1 + |\tau {|^{1 - 2\gamma }}}}} {\rm d}\tau & \le &{\bigg( {\int_{ - \infty }^\infty {{{{\rm d}\tau } \over {{{(1 + |\tau {|^{1 - 2\gamma }})}^{{{\beta + 1} \over \beta }}}}}} }\bigg )^{{\beta \over {\beta + 1}}}}{\bigg( {\int_{ - \infty }^\infty {||{{\widehat {\widetilde u}}_m}( \tau )||_{\beta + 1}^{\beta + 1}{\rm d}\tau } } \bigg)^{{1 \over {\beta + 1}}}}\\ &\le &C{\bigg( {\int_{ - \infty }^\infty {||{{\tilde u}_m}( \tau )||_{\beta + 1}^{{{\beta + 1} \over \beta }}{\rm d}\tau } }\bigg )^{{\beta \over {\beta + 1}}}}\\ &\le& C{T^{{{\beta - 1} \over {\beta + 1}}}}{\bigg( {\int_0^T {||{u_m}( \tau )||_{\beta + 1}^{\beta + 1}{\rm d}\tau } }\bigg )^{{1 \over {\beta + 1}}}}. \end{eqnarray} $

由(2.19)–(2.23)式可知

(2.24) $ \begin{eqnarray} \int_{ - \infty }^\infty {|\tau {|^{2\gamma }}||{{\widehat {\widetilde u}}_m}||_{{L^2}}^2} {\rm d}\tau \le C , \qquad \int_{ - \infty }^\infty {|\tau {|^{2\gamma }}||{{\widehat {\widetilde b}}_m}||_{{L^2}}^2} {\rm d}\tau \le C . \end{eqnarray} $

根据引理2.1,存在着函数$ u({x, t}) $, $ b({x, t}) $使得

(2.25) $ \begin{equation} \begin{array}{l} u \in {L^\infty }( {0, T;{L^2}( {{{\Bbb R} ^3}} )} ) \cap {L^2}( {0, T;H_0^1( {{{\Bbb R} ^3}} )} ) \cap {L^{\beta + 1}}( {0, T;{L^{\beta + 1}}( {{{\Bbb R} ^3}} )} ), \\ b \in {L^\infty }( {0, T;{L^2}( {{{\Bbb R} ^3}} )} ) \cap {L^2}( {0, T;H_0^1( {{{\Bbb R} ^3}} )} ). \end{array} \end{equation} $

而且存在着$ \left\{ {{u_m}} \right\}_{m = 1}^\infty $的子列(仍用其自身表示),使得$ {{u_m}} $弱$ * $收敛到$ u \in {L^\infty }({0, T; {L^2}({{{\Bbb R} ^3}})}) $,弱收敛到$ u \in {L^2}({0, T; H_0^1({{{\Bbb R} ^3}})}) $且在$ L^{\beta + 1}({0, T; {L^{\beta + 1}}({{{\Bbb R} ^3}})}) $上有$ u \to u $.存在着$ \left\{ {{b_m}} \right\}_{m = 1}^\infty $的子列(用其自身表示),使$ {{b_m}} $弱$ * $收敛到$ b \in {L^\infty }({0, T; {L^2}({{{\Bbb R} ^3}})}) $,弱收敛到$ b \in {L^2}({0, T; H_0^1({{{\Bbb R} ^3}})}) $.最后,选择出$ {\Omega _1} \subset {\Omega _2} \subset {\Omega _3} \subset, \cdots $,它们都具有光滑的边界且满足$ \cup _{i = 1}^\infty {\Omega _i} = {{\Bbb R} ^3} $.对任意给定的$ i = 1, 2, \cdots $,令$ {X_0} = H_0^1({{\Omega _i}}) $, $ X = {L^2}({{\Omega _i}}) $,通过(2.24)式以及文献[11]中定理2.2和引理2.1容易得到存在着$ \left\{ {{u_m}} \right\}_{m = 1}^\infty $与$ \left\{ {{b_m}} \right\}_{m = 1}^\infty $的子列(仍用它们的本身表示）使得$ {u_m} \to u $(强收敛于$ {L^2}({0, T; {L^2}({{\Omega _i}})}) $), $ {b_m} \to b $(强收敛于$ {L^2}({0, T; {L^2}({{\Omega _i}})}) $).由对角线法则,存在着$ \left\{ {{u_m}} \right\}_{m = 1}^\infty $的子列$ \left\{ {{u_{{m_j}}}} \right\}_{j = 1}^\infty $以及$ \left\{ {{b_m}} \right\}_{m = 1}^\infty $的子列$ \left\{ {{b_{{m_j}}}} \right\}_{j = 1}^\infty $,有$ {{u_{{m_j}}}} $强收敛于$ u \in {L^2}(0, T; {L^2} $$ ({{\Omega _i}})) $以及$ {{b_{{m_j}}}} $强收敛于$ b \in {L^2}({0, T; {L^2}({{\Omega _i}})}) $,其中$ i = 1, 2, \cdots $.因此$ \left\{ {{u_{{m_j}}}} \right\}_{j = 1}^\infty $和$ \left\{ {{b_{{m_j}}}} \right\}_{j = 1}^\infty $强收敛到$ {L^2}(0, T; L_{loc}^2({{{\Bbb R} ^3}})) $.这些收敛保证了$ ({u({x, t}), b({x, t})}) $是问题(1.1)的一个弱解.更进一步, (2.25)式是(2.2)式的直接结果, (2.3)式可由引理2.1直接得到.定理2.1证毕.

本节将证明局部强解的存在性和唯一性.下面先给出强解的定义.

定义3.1  如果$ ({u({x, t}), b({x, t}), p({x, t})}) $是问题(1.1)的一个弱解,并且满足

$ u \in {L^\infty }( {0, T;H_0^1( {{{\Bbb R} ^3}} )} ) \cap {L^2}( {0, T;{H^2}( {{{\Bbb R} ^3}} )} ) \cap {L^\infty }( {0, T;{L^{\beta + 1}}( {{{\Bbb R} ^3}} )} ), $

$ b \in {L^\infty }( {0, T;H_0^1( {{{\Bbb R} ^3}} )} ) \cap {L^2}( {0, T;{H^2}( {{{\Bbb R} ^3}} )} ). $

那么$ ({u({x, t}), b({x, t}), p({x, t})}) $就是问题(1.1)的一个强解.

这里需要说明,类似于Navier-Stokes方程,如果$ ({u({x, t}), b({x, t}), p({x, t})}) $为问题(1.1)的一个强解,那么压力$ {p({x, t})} $能唯一(至多相差一个常数)地被速度场$ {u({x, t})} $和磁场$ {b({x, t})} $确定,详见文献[11].

这一节的主要结论将由下面的定理给出.

定理3.1  假设$ {u_0} \in H_0^1 \cap {L^{\beta + 1}} $, $ {b_0} \in H_0^1 $,则存在着问题(1.1)的强解满足

$ u \in {L^\infty }( {0, T;H_0^1( {{{\Bbb R} ^3}} )} ) \cap {L^2}( {0, T;{H^2}( {{{\Bbb R} ^3}} )} ) \cap {L^\infty }( {0, T;{L^{\beta + 1}}( {{{\Bbb R} ^3}} )} ), $

$ b \in {L^\infty }( {0, T;H_0^1( {{{\Bbb R} ^3}} )} ) \cap {L^2}( {0, T;{H^2}( {{{\Bbb R} ^3}} )} ). $

并且强解具有唯一性.

证 强解的存在性依赖于下面给出的这个先验估计.

引理3.1  若$ ({u({x, t}), b({x, t}), p({x, t})}) $是问题(1.1)的一个光滑解,那么

(3.1) $ \begin{eqnarray} &&\mathop {\sup }\limits_{0 \le t \le T} ( {||\nabla u||_{{L^2}}^2 + {1 \over {\beta + 1}}||u||_{\beta + 1}^{\beta + 1} + ||\nabla b||_{{L^2}}^2} ) + {1 \over 2}\int_0^T {||{u_t}||_{{L^2}}^2} {\rm d}t + {1 \over 2}\int_0^T {||{b_t}||_{{L^2}}^2} {\rm d}t \\ & &+ {3 \over 8}\int_0^T {||\Delta u||_{{L^2}}^2{\rm d}t} + {3 \over 8}\int_0^T {||\Delta b||_{{L^2}}^2{\rm d}t} + \int_0^T {\int_{{{\Bbb R} ^3}} {|u{|^{\beta - 1}}|\nabla u{|^2}{\rm d}x{\rm d}t} } \\ && + {{\beta - 1} \over 4}\int_0^T {\int_{{{\Bbb R} ^3}} {|u{|^{\beta - 3}}|\nabla |u{|^2}{|^2}{\rm d}x{\rm d}t} } \le C, \end{eqnarray} $

其中$ C $与$ ||\nabla {u_0}||_{{L^{^2}}}^2 $, $ ||\nabla {b_0}||_{{L^{^2}}}^2 $, $ ||{u_0}||_{\beta + 1}^{\beta + 1} $, $ T $有关.

证 将方程组(1.1)中第一个方程乘以$ {u_t} $,第二个方程乘以$ {b_t} $,然后分别在$ {{\Bbb R} ^3} $关于$ x $积分,再相加可得

(3.2) $ \begin{eqnarray} & &||{u_t}||_{{L^2}}^2 + ||{b_t}||_{{L^2}}^2 + {1 \over 2}{{\rm d}\over {{\rm d}t}}||\nabla u||_{{L^2}}^2 + {1 \over 2}{{\rm d}\over {{\rm d}t}}||\nabla b||_{{L^2}}^2 + {1 \over {\beta + 1}}{{\rm d}\over {{\rm d}t}}||u||_{\beta + 1}^{\beta + 1}\\ & = & \int_{{{\Bbb R} ^3}} {b \cdot \nabla b{u_t}} {\rm d}x - \int_{{{\Bbb R} ^3}} {u \cdot \nabla u{u_t}} {\rm d}x - \int_{{{\Bbb R} ^3}} {u \cdot \nabla b{b_t}} {\rm d}x + \int_{{{\Bbb R} ^3}} {b \cdot \nabla u{b_t}} {\rm d}x\\ &\le & \int_{{{\Bbb R} ^3}} {|b \cdot \nabla b||{u_t}|} {\rm d}x + \int_{{{\Bbb R} ^3}} {|u \cdot \nabla u||{u_t}|} {\rm d}x + \int_{{{\Bbb R} ^3}} {|u \cdot \nabla b||{b_t}} |{\rm d}x + \int_{{{\Bbb R} ^3}} {|b \cdot \nabla u||{b_t}} |{\rm d}x . \end{eqnarray} $

对于(3.2)式不等号右边使用Hölder不等式和Young不等式可得

(3.3) $ \begin{eqnarray} &&\int_{{{\Bbb R} ^3}} {|b \cdot \nabla b||{u_t}|} {\rm d}x + \int_{{{\Bbb R} ^3}} {|u \cdot \nabla u||{u_t}|} {\rm d}x + \int_{{{\Bbb R} ^3}} {|b \cdot \nabla u||{b_t}} |{\rm d}x + \int_{{{\Bbb R} ^3}} {|u \cdot \nabla b||{b_t}} |{\rm d}x\\ &\le &||u \cdot \nabla u|{|_{{L^2}}}||{u_t}|{|_{{L^2}}} + ||b \cdot \nabla b|{|_{{L^2}}}||{u_t}|{|_{{L^2}}} + ||b \cdot \nabla u|{|_{{L^2}}}||{b_t}|| + ||u \cdot \nabla b|{|_{{L^2}}}||{b_t}|{|_{{L^2}}}\\ &\le& {1 \over 4}||{u_t}||_{{L^2}}^2 + C||u \cdot \nabla u||_{{L^2}}^2 + {1 \over 4}||{u_t}||_{{L^2}}^2 + C||b \cdot \nabla b||_{{L^2}}^2 + {1 \over 4}||{b_t}||_{{L^2}}^2 + C||b \cdot \nabla u||_{{L^2}}^2 \\ &&+ {1 \over 4}||{b_t}||_{{L^2}}^2 + C||u \cdot \nabla b||_{{L^2}}^2\\ &\le & {1 \over 2}||{u_t}||_{{L^2}}^2 + {1 \over 2}||{b_t}||_{{L^2}}^2 + C( {||u \cdot \nabla u||_{{L^2}}^2 + ||b \cdot \nabla b||_{{L^2}}^2 + ||u \cdot \nabla b||_{{L^2}}^2 + ||b \cdot \nabla u||_{{L^2}}^2} )\\ &\le& {1 \over 2}||{u_t}||_{{L^2}}^2 + {1 \over 2}||{b_t}||_{{L^2}}^2 + C( {{I_1} + {I_2} + {I_3} + {I_4}} ). \end{eqnarray} $

对于(3.3)式中的$ {{I_1}} $, $ {{I_2}} $, $ {{I_3}} $, $ {{I_4}} $,使用Hölder不等式, Gagliardo-Nirenberg不等式和Young不等式有

(3.4) $ \begin{eqnarray} {I_1}& \le& {( {||u|{|_{{L^4}}}||\nabla u|{|_{{L^4}}}} )^2}\\ &\le& C{( {||u||_{{L^2}}^{{1 \over 4}}||\nabla u||_{{L^2}}^{{3 \over 4}}||\nabla u||_{{L^2}}^{{1 \over 4}}||\Delta u||_{{L^2}}^{{3 \over 4}}} )^2}\\ &\le& C||u||_{{L^2}}^{{1 \over 2}}||\nabla u||_{{L^2}}^2||\Delta u||_{{L^2}}^{{3 \over 2}}\\ &\le & C||u||_{{L^2}}^2||\nabla u||_{{L^2}}^8 + {1 \over 8}||\Delta u||_{{L^2}}^2\\ &\le& C||\nabla u||_{{L^2}}^8 + {1 \over 8}||\Delta u||_{{L^2}}^2. \end{eqnarray} $

(3.5) $ \begin{eqnarray} {I_3} &\le& {( {||u|{|_{{L^4}}}||\nabla b|{|_{{L^4}}}} )^2}\\ &\le& C{( {||u||_{{L^2}}^{{1 \over 4}}||\nabla u||_{{L^2}}^{{3 \over 4}}||\nabla b||_{{L^2}}^{{1 \over 4}}||\Delta b||_{{L^2}}^{{3 \over 4}}} )^2}\\ &\le & C||u||_{{L^2}}^{{1 \over 2}}||\nabla u||_{{L^2}}^{{3 \over 2}}||\nabla b||_{{L^2}}^{{1 \over 2}}||\Delta b||_{{L^2}}^{{3 \over 2}}\\ &\le& C||u||_{{L^2}}^2||\nabla u||_{{L^2}}^6||\nabla b||_{{L^2}}^2 + {1 \over 8}||\Delta b||_{{L^2}}^2\\ &\le& C||\nabla u||_{{L^2}}^8 + C||\nabla b||_{{L^2}}^8 + {1 \over 8}||\Delta b||_{{L^2}}^2. \end{eqnarray} $

类似于(3.4)和(3.5)式,可得

(3.6) $ \begin{equation} {I_2} \le C||\nabla b||_{{L^2}}^8 + {1 \over 8}||\Delta b||_{{L^2}}^2, \end{equation} $

(3.7) $ \begin{equation} {I_4} \le C||\nabla b||_{{L^2}}^8 + C||\nabla u||_{{L^2}}^8 + {1 \over 8}||\Delta u||_{{L^2}}^2. \end{equation} $

由(3.2)–(3.7)式可得

(3.8) $ \begin{eqnarray} & &{1 \over 2} ||{u_t}||_{{L^2}}^2 + {1 \over 2} ||{b_t}||_{{L^2}}^2 + {1 \over 2}{{\rm d}\over {{\rm d}t}}||\nabla u||_{{L^2}}^2 + {1 \over 2}{{\rm d}\over {{\rm d}t}}||\nabla b||_{{L^2}}^2 + {1 \over {\beta + 1}}{{\rm d}\over {{\rm d}t}}||u||_{\beta + 1}^{\beta + 1}\\ &\le& C||\nabla u||_{{L^2}}^8 + C||\nabla b||_{{L^2}}^8 + {1 \over 4}||\Delta u||_{{L^2}}^2 + {1 \over 4}||\Delta b||_{{L^2}}^2. \end{eqnarray} $

再将方程组(1.1)中第一个方程乘以$ - \Delta u $,第二个方程乘以$ - \Delta b $,然后在$ {{\mathbb R}^3} $上关于$ x $积分再相加可得

(3.9) $ \begin{eqnarray} & &{1 \over 2}{{\rm d}\over {{\rm d}t}}||\nabla u||_{{L^2}}^2 + {1 \over 2}{{\rm d}\over {{\rm d}t}}||\nabla b||_{{L^2}}^2 + ||\Delta u||_{{L^2}}^2 + ||\Delta b||_{{L^2}}^2 + \int_{{{\Bbb R} ^3}} {|u{|^{\beta - 1}}|\nabla u{|^2}} {\rm d}x \\ &&+ {{\beta - 1} \over 4}\int_{{{\Bbb R} ^3}} {|u{|^{\beta - 3}}|\nabla |u{|^2}} {|^2}{\rm d}x\\ & = & \int_{{{\Bbb R} ^3}} {u \cdot \nabla u\Delta u} {\rm d}x - \int_{{{\Bbb R} ^3}} {b \cdot \nabla b\Delta u} {\rm d}x + \int_{{{\Bbb R} ^3}} {u \cdot \nabla b\Delta b} {\rm d}x - \int_{{{\Bbb R} ^3}} {b \cdot \nabla b\Delta b{\rm d}x}\\ &\le& \int_{{{\Bbb R} ^3}} {|u \cdot } \nabla u||\Delta u|{\rm d}x + \int_{{{\Bbb R} ^3}} {|b \cdot \nabla b||\Delta u|} {\rm d}x + \int_{{{\Bbb R} ^3}} {|u \cdot \nabla b||\Delta b|} {\rm d}x + \int_{{{\Bbb R} ^3}} {|b \cdot \nabla u||\Delta b|} {\rm d}x. \end{eqnarray} $

对于(3.9)式使用Hölder不等式和Young不等式,类似于(3.3)式,可得

(3.10) $ \begin{eqnarray} &&\int_{{{\Bbb R} ^3}} {|u \cdot } \nabla u||\Delta u|{\rm d}x + \int_{{{\Bbb R} ^3}} {|b \cdot \nabla b||\Delta u|} {\rm d}x + \int_{{{\Bbb R} ^3}} {|u \cdot \nabla b||\Delta b|} {\rm d}x + \int_{{{\Bbb R} ^3}} {|b \cdot \nabla u||\Delta b|} {\rm d}x\\ &\le& {2 \over 8}||\Delta u||_{{L^2}}^2 + {2 \over 8}||\Delta b||_{{L^2}}^2 + C( {||u \cdot \nabla u||_{{L^2}}^2 + ||u \cdot \nabla b||_{{L^2}}^2 + ||b \cdot \nabla b||_{{L^2}}^2 + ||b \cdot \nabla u||_{{L^2}}^2} ). \qquad \end{eqnarray} $

对于(3.10)式使用Hölder不等式, Gagliardo-Nirenberg不等式和Young不等式.类似于(3.4)–(3.7)式,易得

(3.11) $ \begin{eqnarray} &&\int_{{{\Bbb R} ^3}} {|u \cdot } \nabla u||\Delta u|{\rm d}x + \int_{{{\Bbb R} ^3}} {|b \cdot \nabla b||\Delta u|} {\rm d}x + \int_{{{\Bbb R} ^3}} {|u \cdot \nabla b||\Delta b|} {\rm d}x + \int_{{{\Bbb R} ^3}} {|b \cdot \nabla u||\Delta b|} {\rm d}x\\ &\le& C( {||\nabla u||_{{L^2}}^8 + ||\nabla b||_{{L^2}}^8} ) + {3 \over 8}||\Delta u||_{{L^2}}^2 + {3 \over 8}||\Delta b||_{{L^2}}^2. \end{eqnarray} $

最后用(3.8)式加上(3.11)式得

(3.12) $ \begin{eqnarray} &&{1 \over 2} ||{u_t}||_{{L^2}}^2 + {1 \over 2} ||{b_t}||_{{L^2}}^2 + {{\rm d}\over {{\rm d}t}}||\nabla u||_{{L^2}}^2 + {{\rm d}\over {{\rm d}t}}||\nabla b||_{{L^2}}^2 + {1 \over {\beta + 1}}{{\rm d}\over {{\rm d}t}}||u||_{\beta + 1}^{\beta + 1}\\ & &{\rm{ + }}{3 \over 8}||\Delta u||_{{L^2}}^2{\rm{ + }}{3 \over 8}||\Delta {\rm{b}}||_{{L^2}}^2+ \int_{{{\Bbb R} ^3}} {|u{|^{\beta - 1}}|\nabla u{|^2}} {\rm d}x + {{\beta - 1} \over 4}\int_{{{\Bbb R} ^3}} {|u{|^{\beta - 3}}|\nabla |u{|^2}} {|^2}{\rm d}x \\ &\le& C( {||\nabla u||_{{L^2}}^8 + ||\nabla b||_{{L^2}}^8} )\le C{( {||\nabla u||_{{L^2}}^2 + ||\nabla b||_{{L^2}}^2 + ||u||_{\beta + 1}^{\beta + 1}} )^4}. \end{eqnarray} $

令

$ I( t ) = ||\nabla u||_{{L^2}}^2 + ||\nabla b||_{{L^2}}^2 + ||u||_{\beta + 1}^{\beta + 1}, $

由(3.12)式可得

(3.13) $ \begin{eqnarray} {{\rm d}\over {{\rm d}t}}I( t ) \le C{( {I( t )} )^4}, \qquad I( t ) \le {1 \over {{{( {I{{( 0 )}^{ - 3}} - Ct} )}^{{1 \over 3}}}}}, \end{eqnarray} $

将(3.13)式带入(3.12)式,并关于$ t $在$ \left[{0, T} \right] $上积分可得

(3.14) $ \begin{eqnarray} &&{1 \over 2} \int_0^T {||{u_t}||_{{L^2}}^2{\rm d}t} + {1 \over 2} \int_0^T {||{b_t}||_{{L^2}}^2} {\rm d}t + \mathop {\sup }\limits_{0 \le t \le T} ( {||\nabla u||_{{L^2}}^2 + ||\nabla b||_{{L^2}}^2 + {1 \over {\beta + 1}}||u||_{\beta + 1}^{\beta + 1}} )\\ &&{\rm{ + }}{3 \over 8}\int_0^T {||\Delta u||_{{L^2}}^2} {\rm{{\rm d}t}}{\rm{ + }}{3 \over 8}\int_0^T {||\Delta b||_{{L^2}}^2} {\rm{{\rm d}t}} + \int_0^T {\int_{{{\Bbb R} ^3}} {|u{|^{\beta - 1}}|\nabla u{|^2}} {\rm d}x} {\rm d}t\\ && + {{\beta - 1} \over 4}\int_0^T {\int_{{{\Bbb R} ^3}} {|u{|^{\beta - 3}}|\nabla |u{|^2}} {|^2}{\rm d}x} {\rm d}t \le C( T ). \end{eqnarray} $

引理3.1得证.

现在证明定理3.1中强解的唯一性,假设在相同的初始条件下,同时存在着两个问题(1.1)的强解满足

(3.15) $ \begin{equation} ( {{u_t}, \phi } ) + \int_{{{\Bbb R} ^3}} {( {u \cdot \nabla } )} u\phi {\rm d}x + \int_{{{\Bbb R} ^3}} {\nabla u:\nabla \phi } {\rm d}x + \int_{{{\Bbb R} ^3}} {|u{|^{\beta - 1}}u\phi } {\rm d}x = \int_{{{\Bbb R} ^3}} {( {b \cdot \nabla } )b} \phi {\rm d}x, \end{equation} $

(3.16) $ \begin{equation} ( {{{\rm{b}}_t}, \phi } ) + \int_{{{\Bbb R} ^3}} {( {u \cdot \nabla } )} {\rm{b}}\phi {\rm d}x + \int_{{{\Bbb R} ^3}} {\nabla {\rm{b}}:\nabla \phi } {\rm d}x = \int_{{{\Bbb R} ^3}} {( {b \cdot \nabla } ){\rm{u}}} \phi {\rm d}x, \end{equation} $

(3.17) $ \begin{equation} ( {{{\overline u }_t}, \phi } ) + \int_{{{\Bbb R} ^3}} {( {\overline u \cdot \nabla } )} \overline u \phi {\rm d}x + \int_{{{\Bbb R} ^3}} {\nabla \overline u :\nabla \phi } {\rm d}x + \int_{{{\mathbb R}^3}} {|\overline u {|^{\beta - 1}}\overline u \phi } {\rm d}x = \int_{{{\Bbb R} ^3}} {( {\overline {\rm{b}} \cdot \nabla } )\overline b } \phi {\rm d}x, \end{equation} $

(3.18) $ \begin{equation} ( {{{\overline {\rm{b}} }_t}, \phi } ) + \int_{{{\Bbb R} ^3}} {( {\overline u \cdot \nabla } )} \overline {\rm{b}} \phi {\rm d}x + \int_{{{\Bbb R} ^3}} {\nabla \overline {\rm{b}} :\nabla \phi } {\rm d}x = \int_{{{\Bbb R} ^3}} {( {\overline {\rm{b}} \cdot \nabla } )\overline u } \phi {\rm d}x, \end{equation} $

其中$ \phi \in C_0^\infty ({\left[{0, T} \right] \times {{\mathbb R}^3}}) $,由于$ C_0^\infty $在$ {L^2} $和$ H_0^1 $中稠密,所以当$ \phi \in {L^2}({0, T; H_0^1({{{\mathbb R}^3}})}) $, (3.15)式到(3.18)式仍然成立.

由(3.15)式减去(3.17)式,再令$ \phi = u - \overline u $可得

(3.19) $ \begin{eqnarray} & &{1 \over 2}{{\rm d}\over {{\rm d}t}}||u - \overline u ||_{{L^2}}^2 + ||\nabla ( {u - \overline u } )||_{{L^2}}^2 + |||u{|^{{{\beta - 1} \over 2}}}|u - \overline u |||_{{L^2}}^2 + \int_{{{\mathbb R}^3}} {( {|u{|^{\beta - 1}}u - |\overline u {|^{\beta - 1}}\overline u } )} ( {u - \overline u } ){\rm d}x\\ &\le& \int_{{{\Bbb R} ^3}} {|u - \overline u {|^2}|\nabla } \overline u |{\rm d}x + \int_{{{\Bbb R} ^3}} {|b||\nabla ( {b - \overline {\rm{b}} } )||u - } \overline u |{\rm d}x + \int_{{{\Bbb R} ^3}} {|b - \overline {\rm{b}} ||\nabla \overline {\rm{b}} ||u - } \overline u |{\rm d}x. \end{eqnarray} $

由(3.16)式减去(3.18)式,再令$ \phi = b - \overline b $可得

(3.20) $ \begin{eqnarray} & &{1 \over 2}{{\rm d}\over {{\rm d}t}}||b - \overline b ||_{{L^2}}^2 + ||\nabla ( {b - \overline b } )||_{{L^2}}^2\\ &\le & \int_{{{\Bbb R} ^3}} {|u - \overline u ||\nabla \overline {\rm{b}} ||b - } \overline b |{\rm d}x + \int_{{{\Bbb R} ^3}} {|b||\nabla ( {u - \overline u } )||b - } \overline b |{\rm d}x + \int_{{{\Bbb R} ^3}} {|b - \overline b {|^2}|\nabla } \overline u |{\rm d}x. \end{eqnarray} $

将(3.19)式加上(3.20)式可得

(3.21) $ \begin{eqnarray} & &{1 \over 2}{{\rm d}\over {{\rm d}t}}||u - \overline u ||_{{L^2}}^2 + {1 \over 2}{{\rm d}\over {{\rm d}t}}||b - \overline b ||_{{L^2}}^2 + ||\nabla ( {u - \overline u } )||_{{L^2}}^2 + ||\nabla ( {b - \overline b } )||_{{L^2}}^2 + |||u{|^{{{\beta - 1} \over 2}}}|u - \overline u |||_{{L^2}}^2\\ & &+ \int_{{{\Bbb R} ^3}} {( {|u{|^{\beta - 1}}u - |\overline u {|^{\beta - 1}}\overline u } )} ( {u - \overline u } ){\rm d}x\\ &\le & \int_{{{\Bbb R} ^3}} {|u - \overline u {|^2}|\nabla } \overline u |{\rm d}x + \int_{{{\Bbb R} ^3}} {|b - \overline b {|^2}|\nabla } \overline u |{\rm d}x + 2\int_{{{\Bbb R} ^3}} {|b - \overline {\rm{b}} ||\nabla \overline {\rm{b}} ||u - } \overline u |{\rm d}x\\ &\le& {J_1} + {J_2} + 2{J_3} + {J_4}, \end{eqnarray} $

其中使用了结论:对于任意$ u \in H_0^1 $,任意$ v \in {H^1} $,有$ ({({u \cdot \nabla })v, v}) = 0 $.

由文献[8]可得

$ \int_{{{\mathbb R}^3}} {( {|u{|^{\beta - 1}}u - |\overline u {|^{\beta - 1}}\overline u } )} ( {u - \overline u } ){\rm d}x \ge 0. $

对于$ {J_1} $, $ {J_2} $,使用Hölder不等式, Gagliardo-Nirenberg不等式和Young不等式可得

(3.22) $ \begin{eqnarray} {J_1}& \le &||u - \overline u ||_{{L^4}}^2||\nabla \overline u |{|_{{L^2}}}\\ &\le& C( {||\nabla ( {u - \overline u } )||_{{L^2}}^{{3 \over 2}}||u - \overline u ||_{{L^2}}^{{1 \over 2}}} )||\nabla \overline u |{|_{{L^2}}}\\ &\le &\varepsilon ||\nabla ( {u - \overline u } )||_{{L^2}}^2 + C||u - \overline u ||_{{L^2}}^2||\nabla \overline u ||_{{L^2}}^4, \end{eqnarray} $

(3.23) $ \begin{eqnarray} {J_2}& \le &||b - \overline b ||_{{L^4}}^2||\nabla \overline u |{|_{{L^2}}}\\ &\le& C( {||\nabla ( {b - \overline b } )||_{{L^2}}^{{3 \over 2}}||b - \overline b ||_{{L^2}}^{{1 \over 2}}} )||\nabla \overline u |{|_{{L^2}}}\\ &\le& \varepsilon ||\nabla ( {b - \overline b } )||_{{L^2}}^2 + C||b - \overline b ||_{{L^2}}^2||\nabla \overline u ||_{{L^2}}^4, \end{eqnarray} $

对于$ {J_3} $,使用Hölder不等式, Gagliardo-Nirenberg不等式和Young不等式可得

(3.24) $ \begin{eqnarray} {J_3} &\le& ||b - \overline b |{|_{{L^4}}}||\nabla \overline b |{|_{{L^2}}}||u - \overline u |{|_{{L^4}}}\\ &\le & C||\nabla ( {b - \overline b } )||_{{L^2}}^{{3 \over 4}}||\nabla ( {u - \overline u } )||_{{L^2}}^{{3 \over 4}}||b - \overline b ||_{{L^2}}^{{1 \over 4}}||u - \overline u ||_{{L^2}}^{{1 \over 4}}\\ &\le& \varepsilon ||\nabla ( {b - \overline b } )|{|_{{L^2}}}||\nabla ( {u - \overline u } )|{|_{{L^2}}} + C||b - \overline b |{|_{{L^2}}}||u - \overline u |{|_{{L^2}}}\\ &\le& {\varepsilon \over 2}||\nabla ( {b - \overline b } )||_{{L^2}}^2 + {\varepsilon \over 2}||\nabla ( {u - \overline u } )||_{{L^2}}^2 + C||b - \overline b ||_{{L^2}}^2 + C||u - \overline u ||_{{L^2}}^2. \end{eqnarray} $

取$ \varepsilon = {1 \over 5} $,将(3.22)–(3.24)式代入到(3.21)式中可以得到

(3.25) $ \begin{eqnarray} &&{1 \over 2}{{\rm d}\over {{\rm d}t}}||u - \overline u ||_{{L^2}}^2 + {1 \over 2}{{\rm d}\over {{\rm d}t}}||b - \overline b ||_{{L^2}}^2 + {7 \over {10}}||\nabla ( {u - \overline u } )||_{{L^2}}^2 + {7 \over {10}}||\nabla ( {b - \overline b } )|| + |||u{|^{{{\beta - 1} \over 2}}}|u - \overline u |||_{{L^2}}^2\\ & &+ \int_{{{\mathbb R}^3}} {( {|u{|^{\beta - 1}}u - |\overline u {|^{\beta - 1}}\overline u } )} ( {u - \overline u } ){\rm d}x \le C||u - \overline u ||_{{L^2}}^2 + C||b - \overline b || _{{L^2}}^2. \end{eqnarray} $

对(3.25)式使用Gronwall不等式,易得$ u = \overline u $, $ b = \overline b $在$ ({x, t}) \in {{\mathbb R}^3} \times \left[{0, T} \right] $上几乎处处成立.定理3.1证毕.