磁流体力学(magnetohydrodynamics,MHD)^[1] 是研究等离子体和磁场相互作用的物理学分支,基本思想是在运动的导电流体中,磁场能感应出电流. 它能解释等离子体中的大多数现象,广泛应用于离子体物理学的研究. 磁流体力学的基本方程是流体力学中的 Navier-Stokes 方程和电动力学中的 Maxwell 方程组.磁流体力学由瑞典物理学家 Hannes Alfvén 创立,并且因此获得 1970 年的诺贝尔物理学奖. 本文考虑不可压三维 MHD 方程组

$$\begin{equation} {\bf u}_{t}+({\bf u}\cdot\nabla ){\bf u}-\mu\Delta {\bf u} =-\nabla p+({\bf B}\cdot\nabla) {\bf B},\quad {\bf x}\in {\Bbb R}^{3},t>0,\label{xx2} \end{equation}$$

(1.1)

$$\begin{equation} {\bf B}_{t} + ({\bf u}\cdot\nabla ){\bf B}-\nu \Delta{\bf B}=({\bf B}\cdot \nabla ){\bf u},\quad{\bf x}\in{\Bbb R}^{3},t>0,\label{xx22} \end{equation}$$

(1.2)

$$\begin{equation} \nabla\cdot {\bf u}=0,\quad{\bf x}\in{\Bbb R}^{3},t>0,\label{xx23} \end{equation}$$

(1.3)

$\nabla \cdot {\bf{B}} = 0,\quad {\bf{x}} \in {\Bbb R}{^3},t > 0,$

(1.4)

这里,${\bf B}({\bf x},t)$ 为磁场,${\bf u}({\bf x},t)$ 为速度场, $p({\bf x},t)$ 为压强,$\mu\geq0$ 为流体的粘性系数,$\nu\geq0$ 为电阻率. 为简化计算,设常数 $\mu$ 和 $\nu$ 都为 1.

本文得到如下结论.

定理 1.1 $(u^{\theta}=0,B^{r}=B^{z}=0)$ 设 $0<\delta<1$ 是一个给定可以任意小的数,

$$\begin{equation} B^{\theta}(r,z,0)=\frac{1}{\varepsilon ^{2-\delta}}{\cal B}_{0} (\varepsilon ^{2}r,z),\qquad \omega^{\theta}(r,z,0)=\frac{1}{\varepsilon ^{2-\delta}}W_{0} (\varepsilon ^{2}r,z), \end{equation}$$

(1.5)

其中 ${\cal B}_{0}(r,z)=r{\cal B}_{1}(r,z),W_{0}(r,z)=rW_{1}(r,z)$, ${\cal B}_{0},{\cal B}_{1}\in L^{2p}({\Bbb R}^{2}\times [0,1])$, $W_{0},W_{1}\in L^{2q}({\Bbb R}^{2}\times [0,1])$,$p=2q,q>\frac{1}{\delta}$. 若初始场 ${\bf u}_{0},{\bf B}_{0}\in L^{2}({\Bbb R}^{2}\times [0,1])$, $B^{\theta},\omega^{\theta}$ 和 $\psi^{\theta}$ 的初始条件在 $z$ 轴上是周期为 1 的周期函数且为奇函数,则存在 $\varepsilon_{0}({\cal B}_{0},W_{0},\delta) >0$, 使得对于所有的 $0<\varepsilon\leq\varepsilon_{0}$,三维轴对称 MHD 方程组有一个唯一的整体正则解.

注1.1 这里如果我们选择 $\varepsilon$ 很小,则关于轴对称不可压 MHD 方程组的初值不会小. 因此,经典的关于小初值的正则性分析方法是不适用于各向异性的初始条件的. 此外, 如果我们选择如下的初始条件

$$\begin{equation}B^{\theta}(r,z,0)=\frac{1}{\varepsilon ^{1-\delta}}{\cal B}_{0}(\varepsilon ^{2}r,z),\qquad \omega^{\theta}(r,z,0)=\frac{1}{\varepsilon ^{1-\delta}}W_{0}(\varepsilon ^{2}r,z), \end{equation}$$

(1.6)

则

$$B_{1}(r,z,0)=\varepsilon ^{\delta+1}{\cal B}_{1}(\varepsilon^{2}r,z),\qquad \omega_{1}(r,z,0)=\varepsilon ^{\delta+1}W_{1}(\varepsilon^{2}r,z),$$

$$\| f_{0}\| _{L^{2}(\Omega)}^{\frac{1}{p}} =\varepsilon ^{\delta+1-\frac{2}{p}}\| {\cal B}_{1}\| _{L^{2p}(\Omega)},\quad\| g_{0}\| _{L^{2}(\Omega)}^{\frac{1}{q}} =\varepsilon ^{\delta+1-\frac{2}{q}}\| W_{1}\| _{L^{2q}(\Omega)},$$

这里 ${\cal B}_{1}\in L^{2p}(\Omega),W_{1}\in L^{2q}(\Omega)$ 独立于 $\varepsilon$. 定理 $1.1$ 同样成立.

定理 1.2 $(B^{r}=B^{z}=0)$ 设 $0<\delta<1$ 是一个给定可以任意小的数,

$$\begin{equation} u^{\theta}(r,z,0)=U_{0}(\varepsilon ^{2\delta}r,z),\quad B^{\theta}(r,z,0)={\cal B}_{0}(\varepsilon ^{2\delta}r,z),\quad \omega^{\theta}(r,z,0)=W_{0}(\varepsilon ^{2\delta}r,z), \end{equation}$$

(1.7)

其中 $U_{0}(r,z)=rU_{1}(r,z),{\cal B}_{0}(r,z)=r{\cal B}_{1}(r,z), W_{0}(r,z)=rW_{1}(r,z)$,$U_{0},U_{1},{\cal B}_{0}, {\cal B}_{1}\in L^{2p}({\Bbb R}^{2}\times [0,1])$, $W_{0},W_{1}\in L^{2q}({\Bbb R}^{2}\times [0,1])$,$p=2q,q>\frac{1}{\delta}$. 若初始场 ${\bf u}_{0},{\bf B}_{0}\in L^{2}({\Bbb R}^{2}\times [0,1])$, $u^{\theta},B^{\theta},\omega^{\theta}$ 和 $\psi^{\theta}$ 的初始条件在 $z$ 轴上是周期为 1 的周期函数且为奇函数,则存在 $\varepsilon_{0}(U_{0},{\cal B}_{0}, W_{0},\delta) >0$,使得对于所有的 $0<\varepsilon\leq\varepsilon_{0}$, 三维轴对称 MHD 方程组有唯一整体正则解.

注 1.2 类似定理 1.1,如果我们令

$$\begin{equation}u^{\theta}(r,z,0)=\frac{1}{\varepsilon ^{1-\delta}}U_{0}(\varepsilon r,z), \ B^{\theta}(r,z,0)=\frac{1}{\varepsilon ^{1-\delta}}{\cal B}_{0}(\varepsilon r,z), \ \omega^{\theta}(r,z,0)=\frac{1}{\varepsilon ^{1-\delta}}W_{0}(\varepsilon r,z), \end{equation}$$

(1.8)

则

$$u_{1}(r,z,0)=\varepsilon ^{\delta}U_{1}(\varepsilon r,z),\quad B_{1}(r,z,0)=\varepsilon ^{\delta}{\cal B}_{1}(\varepsilon r,z),\quad \omega_{1}(r,z,0)=\varepsilon ^{\delta}W_{1}(\varepsilon r,z),$$

$$\| h_{0}\| _{L^{2}(\Omega)}^{\frac{1}{p}} =\varepsilon ^{\delta-\frac{1}{p}}\| U_{1}\| _{L^{2p}(\Omega)},\ \| f_{0}\| _{L^{2}(\Omega)}^{\frac{1}{p}} =\varepsilon ^{\delta-\frac{1}{p}}\| {\cal B}_{1}\| _{L^{2p}(\Omega)},\ \| g_{0}\| _{L^{2}(\Omega)}^{\frac{1}{q}} =\varepsilon ^{\delta-\frac{1}{q}}\| W_{1}\| _{L^{2q}(\Omega)}.$$

如果

$$(\frac{q}{p}C_{q})^{\frac{1}{2q}}\varepsilon^{2\delta-\frac{1}{q}}\| U_{1}\| _{L^{2p}(\Omega)}^{2}+\varepsilon^{\delta-\frac{1}{q}}\| W_{1}\| _{L^{2q}(\Omega)}^{2} +(\frac{q}{p}C_{q})^{\frac{1}{2q}}\varepsilon^{2\delta-\frac{1}{q}}\| {\cal B}_{1}\| _{L^{2p}(\Omega)}^{2}\leq\frac{7(2p-1)}{8p^{2}},$$

这里 $U_{1},{\cal B}_{1}\in L^{2p}(\Omega),W_{1}\in L^{2q}(\Omega)$ 独立于 $\varepsilon$,则定理 1.2 同样成立.

本文部分借鉴了文献[2]中关于轴对称的 Navier-Stokes 方程组的思想. 下面介绍关于不可压的 MHD 方程组近期的研究成果. 在笛卡儿坐标系下,有许多研究结果,然而在柱面坐标系下的研究结果很少. 在笛卡儿坐标系下,文献[3]推导出弱解的 Hausdorff 维估计及其爆破估计的部分正则性. 文献[4] 证明了如果速度场 $u\in L^{\frac{2}{1-\gamma}}(O,T,\dot{X}_{\gamma}(R^{3})), \gamma\in[0,1]$ 或度场的梯度 $\nabla u\in L^{\frac{2}{2-\gamma}}(O,T,\dot{X}_{\gamma}(R^{3})),\gamma\in[0,1]$, 则解在 $[0,T]$上是光滑的. 文献[5, 6] 给出了两类三维不可压的 MHD 方程组的正则性条件.文献[7]证明了如果初值满足 $\| u_{0}\| _{H^{1}}+\| b_{0}\| _{H^{1}}\leq \varepsilon$,这里 $\varepsilon$ 是任意小的正数,则带部分耗散和磁扩散混合的三维MHD 方程有整体光滑解. 文献[8]证明了如果耗散项为 $-\nu(-\Delta)^{\alpha}u$ 和 $-\kappa(-\Delta)^{\beta}b$, 则在这三种情况下 $\alpha \geq \frac{1}{2},\beta\geq1$; $0\leq\alpha\leq \frac{1}{2},2\alpha+\beta>2$; $\alpha\geq2,\beta =0$, 其光滑解是整体的. 在柱面坐标系下的研究成果如 文献[9] 证明了三维的理想的 MHD 方程组关于一维非平凡磁场 ($u^{\theta}=B^{r}=B^{z}=0$) 的轴对称解的整体正则性. 在定理 1.1 中, 本文考虑了在各向异性初值条件下的这种情况.

以下分为两部分讨论,首先给出轴对称不可压 MHD 方程组 (2.11)-(2.16). 其次,分别给出两类各向异性大初值 (1) $B^{\theta}(r,z,0)= \frac{1}{\varepsilon ^{2-\delta}}{\cal B}_{0}(\varepsilon ^{2}r,z),$ $\omega^{\theta}(r,z,0)=\frac{1}{\varepsilon ^{2-\delta}}W_{0}(\varepsilon ^{2}r,z)$, (2) $u^{\theta}(r,z,0)=U_{0}(\varepsilon ^{2\delta}r,z),$ $B^{\theta}(r,z,0)={\cal B}_{0}(\varepsilon ^{2\delta}r,z),$ $\omega^{\theta}(r,z,0)=W_{0}(\varepsilon ^{2\delta}r,z)$ 情况下的 MHD 方程组的两类特解 (1) $u^{\theta}=0,B^{r}=B^{z}=0$, (2) $B^{r}=B^{z}=0$ 的整体正则性的证明.

设三维轴对称的 MHD 方程组 (1.1)-(1.4) 的一个解 $({\bf u},{\bf B},p)$ 具有形式

$$\begin{equation} %\label{1.1}%-2.1 \left\{\begin{array}{ll} &{\bf u}({\bf x},t)=u^{r}(r,z,t){\bf e}_{r}+u^{\theta}(r,z,t){\bf e}_{\theta}+u^{z}(r,z,t){\bf e}_{z},\\ &{\bf B}({\bf x},t)=B^{r}(r,z,t){\bf e}_{r}+B^{\theta}(r,z,t){\bf e}_{\theta}+B^{z}(r,z,t){\bf e}_{z},\\ &{\bf p}({\bf x},t)=p(r,z,t),\\ \end{array}\right. \end{equation}$$

(2.1)

这里

$${\bf e}_{r}=(\frac{x}{r},\frac{y}{r},0),\qquad {\bf e}_{\theta}= (-\frac{y}{r},\frac{x}{r},0),\qquad {\bf e}_{z}=(0,0,1)$$

为 3 个分别沿着径向方向,角方向和 z 轴方向的相互正交单位向量, $r=\sqrt{x^{2}+y^{2}}$. 则 方程组(1.1)-(1.4) 在柱坐标系下表示为

$$\begin{equation} %\label{1.1}%-2.1 \left\{\begin{array}{ll} (u^{r})_{t}+u^{r}(u^{r})_{r}+u^{z}(u^{r})_{z}-(\nabla ^{2}-\frac{1}{r^{2}})u^{r} -\frac{1}{r}(u^{\theta})^{2}\$ \qquad=-p_{r}+B^{r}(B^{r})_{r}+B^{z}(B^{r})_{z}-\frac{1}{r}(B^{\theta})^{2},\\ (u^{\theta})_{t}+u^{r}(u^{\theta})_{r}+u^{z}(u^{\theta})_{z}-(\nabla ^{2} -\frac{1}{r^{2}})u^{\theta}+\frac{1}{r}u^{\theta}u^{r}\$ \qquad=B^{r}(B^{\theta})_{r}+B^{z}(B^{\theta})_{z}+\frac{1}{r}B^{\theta}B^{r},\\ (u^{z})_{t}+u^{r}(u^{z})_{r}+u^{z}(u^{z})_{z}-\nabla ^{2}u^{z}= -p_{z}+B^{r}(B^{z})_{r}+B^{z}(B^{z})_{z}, \end{array}\right. \end{equation}$$

(2.2)

$\left\{ \begin{array}{l} {({B^r})_t} + {u^r}{({B^r})_r} + {u^z}{({B^r})_z} - ({\nabla ^2} - \frac{1}{{{r^2}}}){B^r} = {B^r}{({u^r})_r} + {B^z}{({u^r})_z},\\ {({B^\theta })_t} + {u^r}{({B^\theta })_r} + {u^z}{({B^\theta })_z} - ({\nabla ^2} - \frac{1}{{{r^2}}}){B^\theta } + \frac{1}{r}{u^\theta }{B^r}\\ \;\;\;\;\;\;\;\;\; = {B^r}{({u^\theta })_r} + {B^z}{({u^\theta })_z} + \frac{1}{r}{u^r}{B^\theta },\\ {({B^z})_t} + {u^r}{({B^z})_r} + {u^z}{({B^z})_z} - {\nabla ^2}{B^z} = {B^r}{({u^z})_r} + {B^z}{({u^z})_z}, \end{array} \right.$

(2.3)

$$\begin{equation} %\label{1.1}%-2.1 \left\{\begin{array}{ll} (ru^{r})_{r}+(ru^{z})_{z}=0,\\ (rB^{r})_{r}+(rB^{z})_{z}=0, \end{array}\right. \end{equation}$$

(2.4)

这里 $\nabla ^{2}=\partial ^{2}_{r}+\frac{1}{r}\partial r+\partial ^{2}_{z}$. 注意到一旦初始条件给定, 方程组 (2.2)-(2.4) 就完全决定了轴对称不可压 MHD 方程组. 类似,旋度场 $\omega$ 和流密度 ${\bf J}$ 表示为

$$\begin{equation} \begin{array}{ll} \mbox{ $\omega$} ({\bf x},t)=\nabla \times {\bf u}({\bf x},t)=\omega^{r}(r,z,t){\bf e}_{r} +\omega^{\theta}(r,z,t){\bf e}_{\theta}+\omega^{z}(r,z,t){\bf e}_{z},\\ {\bf J}({\bf x},t)=\nabla \times {\bf B}({\bf x},t)=j^{r}(r,z,t){\bf e}_{r} +j^{\theta}(r,z,t){\bf e}_{\theta}+j^{z}(r,z,t){\bf e}_{z}, \end{array} \end{equation}$$

(2.5)

其中

$$\begin{equation} \begin{array}{l} \omega^{r}=-(u^{\theta})_{z},\quad \omega^{\theta}=(u^{r})_{z}-(u^{z})_{r}, \quad\omega^{z}=\frac{1}{r}u^{\theta}+(u^{\theta})_{r},\$ j^{r}=-(B^{\theta})_{z},\quad j^{\theta}=(B^{r})_{z}-(B^{z})_{r},\quad j^{z}=\frac{1}{r}B^{\theta}+(B^{\theta})_{r}. \end{array} \end{equation}$$

(2.6)

此外,本文引入流函数 $\psi$ 和磁流函数 $\Phi$. $u^{\theta},\omega^{\theta},\psi^{\theta},B^{\theta},j^{\theta}$ 和 $\phi^{\theta}$ 分别被称为速度场 ${\bf u}$,旋度场 $\omega$, 流函数 $\psi$, 磁场 ${\bf B}$,流密度${\bf J}$ 和磁流函数$\Phi$ 的旋涡分 量. 因此,$u^{r},u^{z},B^{r}$ 和 $B^{z}$ 可根据 $\psi^{\theta}$ 和 $\phi^{\theta}$ 被表示为

$$\begin{equation} u^{r}=-(\psi^{\theta})_{z},\quad u^{z}=\frac{1}{r}\psi^{\theta}+(\psi^{\theta})_{r}, \quad B^{r}=-(\phi^{\theta})_{z},\quad B^{z}=\frac{1}{r}\phi^{\theta}+(\phi^{\theta})_{r}. \end{equation}$$

(2.7)

因此,我们得到

$$\begin{equation} %\label{1.1}%-2.1 \left\{\begin{array}{l} (u^{\theta})_{t}+u^{r}(u^{\theta})_{r}+u^{z}(u^{\theta})_{z}-(\nabla ^{2}-\frac{1}{r^{2}})u^{\theta}+\frac{1}{r}u^{\theta}u^{r} \\ \qquad =B^{r}(B^{\theta})_{r}+B^{z}(B^{\theta})_{z}+\frac{1}{r}B^{\theta}B^{r},\\ (\omega^{\theta})_{t}+u^{r}(\omega^{\theta})_{r}+u^{z}(\omega^{\theta})_{z}-(\nabla ^{2}-\frac{1}{r^{2}})\omega^{\theta}-\frac{1}{r}u^{r}\omega^{\theta}-\frac{1}{r}[(u^{\theta})^{2}]_{z}\\ \qquad=B^{r}(j^{\theta})_{r}+B^{z}(j^{\theta})_{z}-\frac{1}{r}B^{r}j^{\theta}-\frac{1}{r}[(B^{\theta})^{2}]_{z},\\ -(\nabla ^{2}-\frac{1}{r^{2}})\psi ^{\theta} =\omega^{\theta},\\ \end{array}\right. \end{equation}$$

(2.8)

$$\begin{equation} %\label{1.1}%-2.1 \left\{\begin{array}{ll} (B^{\theta})_{t}+u^{r}(B^{\theta})_{r}+u^{z}(B^{\theta})_{z}-(\nabla ^{2}-\frac{1}{r^{2}})B^{\theta}+\frac{1}{r}u^{\theta}B^{r}\\ \qquad =B^{r}(u^{\theta})_{r}+B^{z}(u^{\theta})_{z} +\frac{1}{r}u^{r}B^{\theta},\$ (j^{\theta})_{t}+u^{r}(j^{\theta})_{r}+u^{z}(j^{\theta})_{z}-(\nabla ^{2}-\frac{1}{r^{2}})j^{\theta}-B^{r}(\omega^{\theta})_{r}-B^{z}(\omega^{\theta})_{z}\\ \qquad=[(u^{r})_{r}-(u^{z})_{z}][(B^{r})_{z}+(B^{z})_{r}]+[(B^{z})_{z}-(B^{r})_{r}][(u^{r})_{z}+(u^{z})_{r}], \$ -(\nabla ^{2}-\frac{1}{r^{2}})\phi ^{\theta} =j^{\theta}. \end{array}\right. \end{equation}$$

(2.9)

类似于 Navier-Stokes 方程组,对于三维轴对称 MHD 方程组在 $r=0$ 时都满足如下相容性条件

$$u^{\theta}(0,z,t)=\omega^{\theta}(0,z,t)=\psi^{\theta}(0,z,t)=0,$$

$$B^{\theta}(0,z,t)=j^{\theta}(0,z,t)=\phi^{\theta}(0,z,t)=0.$$

因此,重写 $u^{\theta},\omega^{\theta},\psi^{\theta},B^{\theta},j^{\theta}$ 和 $\phi^{\theta}$ 如下

$$\begin{equation} \begin{array}{ll} u^{\theta}(r,z,t)=ru_{1}(r,z,t),\quad \omega^{\theta}(r,z,t)=r\omega_{1}(r,z,t),\quad\psi^{\theta}(r,z,t)=r\psi_{1}(r,z,t),\\ B^{\theta}(r,z,t)=rB_{1}(r,z,t),\quad j^{\theta}(r,z,t)=rj_{1}(r,z,t),\quad\phi^{\theta}(r,z,t)=r\phi_{1}(r,z,t). \end{array} \end{equation}$$

(2.10)

则可推导出关于 $u_{1}(r,z,t),\omega_{1}(r,z,t),\psi_{1}(r,z,t),B_{1}(r,z,t),$ $j_{1}(r,z,t)$ 和 $\phi_{1}(r,z,t)$ 的与方程组 (2.8)-(2.9) 等价的方程组

$$\begin{equation} (u_{1})_{t}+u^{r}(u_{1})_{r}+u^{z}(u_{1})_{z}-2\psi_{1z}u_{1}-(u_{1zz}+u_{1rr}+\frac{3}{r}u_{1r}) =B^{r}(B_{1})_{r}+B^{z}(B_{1})_{z}-2\phi_{1z}B_{1}, \end{equation}$$

(2.11)

$$\begin{equation} (\omega_{1})_{t}+u^{r}(\omega_{1})_{r}+u^{z}(\omega_{1})_{z}-[(u_{1})^{2}]_{z}-(\omega_{1zz}+\omega_{1rr}+\frac{3}{r}\omega_{1r}) =B^{r}(j_{1})_{r}+B^{z}(j_{1})_{z}-[(B_{1})^{2}]_{z}, \end{equation}$$

(2.12)

$$\begin{equation} -(\psi_{1zz}+\psi_{1rr}+\frac{3}{r}\psi_{1r}) =\omega_{1}, \end{equation}$$

(2.13)

$$\begin{equation} (B_{1})_{t}+u^{r}(B_{1})_{r}+u^{z}(B_{1})_{z}-(B_{1zz}+B_{1rr}+\frac{3}{r}B_{1r}) =B^{r}(u_{1})_{r}+B^{z}(u_{1})_{z}, \end{equation}$$

(2.14)

$$\begin{eqnarray} &&(j_{1})_{t}+u^{r}(j_{1})_{r}+u^{z}(j_{1})_{z}-\psi_{1z}j_{1}-(j_{1zz}+j_{1rr}+\frac{3}{r}j_{1r})-B^{r}(\omega_{1})_{r} \nonumber\\ &=&B^{z}(\omega_{1})_{z}-\phi_{1z}j_{1}+\frac{1}{r}(u^{r}_{r}-u^{z}_{z})(B^{r}_{z}+B^{z}_{r})+\frac{1}{r}(B^{z}_{z}-B^{r}_{r})(u^{r}_{z}+u^{z}_{r}), \end{eqnarray}$$

(2.15)

$$\begin{equation} -(\phi_{1zz}+\phi_{1rr}+\frac{3}{r}\phi_{1r}) =j_{1}, \end{equation}$$

(2.16)

这里 $u^{r}$ 和 $u^{z}$ 同样被重新表示为

$$\begin{equation} u^{r}=-(r\psi_{1})_{z},\quad u^{z}=\frac{1}{r}(r^{2}\psi_{1})_{r},\quad B^{r}=-(r\phi_{1})_{z},\quad B^{z}=\frac{1}{r}(r^{2}\phi_{1})_{r}. \end{equation}$$

(2.17)

注意到不可压以及磁场散度约束条件 (2.4) 仍成立并可由 (2.17)式 推导出来. 下面,我们回顾 $A_{p}$ 类的定义.

定义 2.1 ^[10] ($A_{p}$ 类) 设局部可积函数 $\omega(x)$ 对于所有在 ${\Bbb R}^{n}$ 内的球 B 满足 $A_{p}$ 不等式

$$\begin{equation}\frac{1}{|B|}\int_{B}\omega({\bf x}){\rm d}{\bf x} \bigg(\frac{1}{|B|}\int_{B} \omega({\bf x})^{-\frac{p'}{p}}{\rm d}{\bf x}\bigg)^{-\frac{p}{p'}}\leq A<\infty, \end{equation}$$

(2.18)

其中,$p'$ 与 $p$ 对偶,即 $\frac{1}{p}+\frac{1}{p'}=1$ $(1<p<\infty)$. 若存在一个小常数 A 对于 (2.18)式成立,则称 $A_{p}$ 是关于 $\omega({\bf x})$ 有界的,并记为$A_{p}(\omega({\bf x}))$.

引理 2.1 ^[2] 设 $\omega({\bf x})=\omega({\bf r})=\frac{1}{r^{2}}$ 在 ${\Bbb R}^{3}$ 内属于 $A_{p}$ 类 ,存在一个正常数 $C>0$ 使得

$$\begin{equation}\int_{{\Bbb R}^{4}\times \Pi_{1}}|\nabla ^{2}u|^{p} \omega({\bf x}){\rm d}x_{1}{\rm d}x_{2}{\rm d}x_{3}{\rm d}x_{4}{\rm d}z\leq C\int_{{\Bbb R}^{4}\times \Pi_{1}} |f|^{p}\omega({\bf x})~{\rm d}x_{1}{\rm d}x_{2}{\rm d}x_{3}{\rm d}x_{4}{\rm d}z, \end{equation}$$

(2.19)

这里 ${\bf x}=(x_{1},x_{2},x_{3},x_{4})\in {\Bbb R}^{4}$, $r=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}}$,$\Pi_{1}$ 是一个周期为 1 的一维环面,则$f=-\Delta u$,$f$ 和 $u$ 关于权函数 $\omega({\bf x})$ 都为 $L^{p}(1<p<\infty)$ 可积.

引理 2.2 设 $\omega_{1},\psi_{1},j_{1}, \phi_{1} \in L^{p}(R^{2}\times\Pi_{1})$,$1<p<\infty$,$\psi_{1}$ 和 $\phi_{1}$ 为(2.12)和(2.16) 的解,且在 $z$ 方向具有周期为 1 的周期性边界条件,则

$$\begin{equation} \| \psi_{1zz}\| _{L^{p}(R^{2}\times\Pi_{1})}\leq C\| \omega _{1}\| _{L^{p}(R^{2} \times\Pi_{1})}, \end{equation}$$

(2.20)

$$\begin{equation} \| \phi_{1zz}\| _{L^{p}(R^{2}\times\Pi_{1})}\leq C\| j_{1}\| _{L^{p}(R^{2}\times\Pi_{1})}. \end{equation}$$

(2.21)

证 证明过程类似于文献 [2,引理 2].

本节考虑两类 MHD 方程组的特解: (1) $u^{\theta}=0,B^{r}=B^{z}=0$,(2) $B^{r}=B^{z}=0$, 给出定理 1.1 和定理 1.2 的证明.

(1) $u^{\theta}=0,B^{r}=B^{z}=0$

本节考虑一族特解

$$\begin{equation} {\bf u}({\bf x},t)=u^{r}(r,z,t){\bf e}_{r}+u^{z}(r,z,t){\bf e}_{z}, \quad{\bf B}({\bf x},t)=B^{\theta}(r,z,t){\bf e}_{\theta}. \end{equation}$$

(3.1)

易知,若 $u^{\theta},B^{r},B^{z}$ 的初值为零,则恒为零. 因此,(2.11)-(2.16)式被简化为

$$\begin{equation} (B_{1})_{t}+u^{r}(B_{1})_{r}+u^{z}(B_{1})_{z}=(B_{1zz}+B_{1rr}+\frac{3}{r}B_{1r}), \end{equation}$$

(3.2)

$$\begin{equation} (\omega_{1})_{t}+u^{r}(\omega_{1})_{r}+u^{z}(\omega_{1})_{z} =-[(B_{1})^{2}]_{z}+(\omega_{1zz}+\omega_{1rr}+\frac{3}{r}\omega_{1r}), \end{equation}$$

(3.3)

$$\begin{equation} -(\psi_{1zz}+\psi_{1rr}+\frac{3}{r}\psi_{1r}) =\omega_{1}. \end{equation}$$

(3.4)

定理 1.1 的证明 记 $\Omega={\Bbb R}^{2}\times[0,1]$,定义 $f=|B_{1}|^{p},g=|W_{1}|^{q}$, $p=2q$. (3.2) 式乘以 $|B_{1}|^{2p-2}B_{1}$ 且在 $\Omega$ 上积分,得

$$\begin{eqnarray} &&\frac{1}{2p}\frac{\rm d}{{\rm d}t}\int_{\Omega}f^{2} r{\rm d}r{\rm d}z+\int_{\Omega}|B_{1}|^{2p-2} B_{1}[u^{r}(B_{1})_{r}+u^{z}(B_{1})_{z}] r{\rm d}r{\rm d}z \nonumber\\ &=&\int_{\Omega}|B_{1}|^{2p-2}B_{1}(B_{1zz}+B_{1rr}+\frac{3}{r}B_{1r}) r{\rm d}r{\rm d}z \nonumber\\ &=&\int_{\Omega}|B_{1}|^{2p-2}B_{1}B_{1zz} r{\rm d}r{\rm d}z+\int_{\Omega}|B_{1}|^{2p-2}B_{1}\frac{(rB_{1r})_{r}}{r} r{\rm d}r{\rm d}z+\int_{\Omega}|B_{1}|^{2p-2}B_{1}\frac{2B_{1r}}{r} r{\rm d}r{\rm d}z \nonumber\\ &:=&I_{1}+I_{2}+I_{3}, \end{eqnarray}$$

(3.5)

由分部积分及 (2.4)式,得

$$\int_{\Omega}|B_{1}|^{2p-2}B_{1}[u^{r}(B_{1})_{r}+u^{z}(B_{1})_{z}] r{\rm d}r{\rm d}z=0.$$

$$\begin{eqnarray*} I_{1}&=&\int_{\Omega}|B_{1}|^{2p-1} r{\rm d}r{\rm d}B_{1z} =-(2p-1)\int_{\Omega}|B_{1}|^{2p-2}(|B_{1}|_{z})^{2} r{\rm d}r{\rm d}z\\ &=&-\frac{2p-1}{p^{2}}\int_{\Omega}(f_{z})^{2} r{\rm d}r{\rm d}z. \end{eqnarray*}$$

$$\begin{eqnarray*} I_{2}&=&\int_{\Omega}|B_{1}|^{2p-1}{\rm d}rB_{1r}{\rm d}z =-(2p-1)\int_{\Omega}|B_{1}|^{2p-2}(|B_{1}|_{r})^{2} r{\rm d}r{\rm d}z \\ &=&-\frac{2p-1}{p^{2}}\int_{\Omega}(f_{r})^{2} r{\rm d}r{\rm d}z. \end{eqnarray*}$$

$$\begin{eqnarray*} I_{3}&=&\frac{1}{p}\int_{\Omega}(|B_{1}|^{2p})_{r}{\rm d}r{\rm d}z =\frac{1}{p}\int_{0}^{1}\int_{0}^{\infty}d|B_{1}|^{2p}{\rm d}z\\ &=&-\frac{1}{p}\int_{0}^{1}|B_{1}|^{2p}(0,z,t){\rm d}z=-\frac{1}{p}\int_{0}^{1}f^{2}(0,z,t){\rm d}z. \end{eqnarray*}$$

因此,得到

$$\begin{equation} \frac{1}{2p}\frac{\rm d}{{\rm d}t}\int_{\Omega}f^{2}r{\rm d}r{\rm d}z=-\frac{(2p-1)}{p^{2}}\| \nabla f\| _{L^{2}(\Omega)}^{2}-\frac{1}{p}\int_{0}^{1}f^{2}(0,z,t){\rm d}z. \end{equation}$$

(3.6)

类似地,(3.3)式乘以 $|\omega_{1}|^{2q-2}\omega_{1}$ 且在 $\Omega$ 上积分,得

$$\begin{eqnarray} &&\frac{1}{2q} \frac{\rm d}{{\rm d}t}\int_{\Omega}g^{2} r{\rm d}r{\rm d}z+\int_{\Omega}| \omega_{1}|^{2q-2}\omega_{1}[u^{r}(\omega_{1})_{r}+u^{z}(\omega_{1})_{z}] r{\rm d}r{\rm d}z \nonumber\\ &=&-\int_{\Omega}|\omega_{1}|^{2q-2}[(B_{1})^{2}]_{z} r{\rm d}r{\rm d}z+\int_{\Omega} |\omega_{1}|^{2q-2}\omega_{1}\bigg(\omega_{1zz}+\frac{(r\omega_{1r})_{r}}{r} +\frac{2}{r}\omega_{1r}\bigg) r{\rm d}r{\rm d}z \nonumber\\ &:=&J_{1}+J_{2}+J_{3}+J_{4},\end{eqnarray}$$

(3.7)

这里同样由分部积分和(2.4)式,得

$$\int_{\Omega}|\omega_{1}|^{2q-2}\omega_{1} [u^{r}(\omega_{1})_{r}+u^{z}(\omega_{1})_{z}] r{\rm d}r{\rm d}z=0.$$

关于 $J_{1}$,我们利用 Hölder 不等式,Young's 不等式及 $p=2q$有

$$\begin{eqnarray*} J_{1}&\leq &(2q-1)\int_{\Omega}f^{\frac{2}{p}}|\omega_{1}|^{2q-2}|\omega _{1}|_{z} r{\rm d}r{\rm d}z\\ &=&\bigg(2-\frac{1}{q}\bigg)\int_{\Omega}f^{\frac{2}{p}}g^{1-\frac{1}{q}}|g_{z}| r{\rm d}r{\rm d}z\\ &\leq&\bigg(2-\frac{1}{q}\bigg) \bigg(\int _{\Omega}f^{\frac{4}{p}}g^{2(1-\frac{1}{q})} r{\rm d}r{\rm d}z\bigg)^{\frac{1}{2}}\| g_{z}\| _{L^{2}(\Omega)} \\ &\leq&\bigg(2-\frac{1}{q}\bigg) \Big(\| f\| _{L^{2}(\Omega)}^{\frac{2}{p}}\| g\| _{L^{2}(\Omega)}^{1-\frac{2}{p}}\Big)\| g_{z}\| _{L^{2}(\Omega)}\\ &\leq &C_{q}\| f\| _{L^{2}(\Omega)}^{\frac{4}{p}}\| g\| _{L^{2}(\Omega)}^{2(1-\frac{2}{p})}+\frac{2q-1}{4q^{2}}\| \nabla g\| _{L^{2}(\Omega)}^{2}\\ &\leq &C_{q}\frac{2p-1}{2p^{2}}\| f\| _{L^{2}(\Omega)}^{2}+\frac{2q-1}{4q^{2}}\| g\| _{L^{2}(\Omega)}^{2}+\frac{2q-1}{4q^{2}}\| \nabla g\| _{L^{2}(\Omega)}^{2}. \end{eqnarray*}$$

与 $I_{1}-I_{3}$ 类似,得

$$J_{1}=\int_{\Omega}|\omega_{1}|^{2q-1} r{\rm d}r{\rm d}\omega_{1z}=-(2q-1)\int_{\Omega}|\omega_{1}|^{2q-2}(|\omega_{1}|_{z})^{2} r{\rm d}r{\rm d}z=-\frac{2q-1}{q^{2}}\int_{\Omega}(g_{z})^{2} r{\rm d}r{\rm d}z.$$

$$J_{3}=\int_{\Omega}|\omega_{1}|^{2q-1}{\rm d}r\omega_{1r}{\rm d}z=-(2q-1)\int_{\Omega}|\omega_{1}|^{2q-2}(|B_{1}|_{r})^{2} r{\rm d}r{\rm d}z=-\frac{2q-1}{q^{2}}\int_{\Omega}(g_{r})^{2} r{\rm d}r{\rm d}z.$$

$$J_{4}=\frac{1}{q}\int_{\Omega}(|\omega_{1}|^{2q})_{r} {\rm d}r{\rm d}z=-\frac{1}{q}\int_{0}^{1}|\omega_{1}|^{2q}(0,z,t){\rm d}z=-\frac{1}{q}\int_{0}^{1}g^{2}(0,z,t){\rm d}z.$$

因此,得到

$$\begin{eqnarray} \frac{1}{2q}\frac{\rm d}{{\rm d}t}\int_{\Omega}g^{2}r{\rm d}r{\rm d}z&\leq& C_{q}\frac{2p-1}{2p^{2}}\| f\| _{L^{2}(\Omega)}^{2}+\frac{2q-1}{4q^{2}}\| g\| _{L^{2}(\Omega)}^{2}+\frac{2q-1}{4q^{2}}\| \nabla g\| _{L^{2}(\Omega)}^{2} \nonumber\\ &&-\frac{(2q-1)}{q^{2}}\| \nabla g\| _{L^{2}(\Omega)}^{2}-\frac{1}{q}\int_{0}^{1}g^{2}(0,z,t){\rm d}z \nonumber\\ &\leq & C_{q}\frac{2p-1}{2p^{2}}\| \nabla f\| _{L^{2}(\Omega)}^{2}-\frac{(2q-1)}{2q^{2}}\| \nabla g\| _{L^{2}(\Omega)}^{2}, \end{eqnarray}$$

(3.8)

这里我们使用沿着 $z$ 方向的 Poincare 不等式

$$\| f\| _{L^{2}}\leq \| f_{z}\| _{L^{2}},\| g\| _{L^{2}}\leq \| g_{z}\| _{L^{2}}.$$

(3.6)式乘以 $C_{q}$ 后与 (3.8) 式相加,得到

$$\begin{eqnarray} &&\frac{\rm d}{{\rm d}t}\bigg(\frac{C_{q}}{2p}\int_{\Omega}f^{2} r{\rm d}r{\rm d}z+\frac{1}{2q} \int_{\Omega}g^{2} r{\rm d}r{\rm d}z\bigg) \nonumber\\ &\leq&-C_{q}\frac{(2p-1)}{p^{2}}\| \nabla f\| _{L^{2}(\Omega)}^{2}+ C_{q}\frac{2p-1}{2p^{2}}\| \nabla f\| _{L^{2}(\Omega)}^{2}-\frac{(2q-1)}{2q^{2}}\| \nabla g\| _{L^{2}(\Omega)}^{2} \nonumber\\ &\leq&-C_{q}\frac{(2p-1)}{2p^{2}}\| \nabla f\| _{L^{2}(\Omega)}^{2}-\frac{(2q-1)}{2q^{2}}\| \nabla g\| _{L^{2}(\Omega)}^{2}\leq 0. \end{eqnarray}$$

(3.9)

利用 Gronwall 不等式,得

$$\begin{equation} \frac{C_{q}}{2p}\int_{\Omega}f^{2} r{\rm d}r{\rm d}z+\frac{1}{2q}\int_{\Omega}g^{2} r{\rm d}r{\rm d}z \leq\frac{C_{q}}{2p}\int_{\Omega}f_{0}^{2} r{\rm d}r{\rm d}z+\frac{1}{2q}\int_{\Omega}g_{0}^{2} r{\rm d}r{\rm d}z. \end{equation}$$

(3.10)

根据初值条件(1.5),有

$$\begin{equation} B_{1}(r,z,0)=\varepsilon ^{\delta}{\cal B}_{1}(\varepsilon^{2}r,z),\quad \omega_{1}(r,z,0)=\varepsilon ^{\delta}W_{1}(\varepsilon^{2}r,z), \end{equation}$$

(3.11)

其中 ${\cal B}_{1}\in L^{2p}(\Omega),W_{1}\in L^{2q}(\Omega)$ 独立于 $\varepsilon$. 此外,我们有

$$\begin{equation} \| f_{0}\| _{L^{2}(\Omega)}^{\frac{1}{p}} =\bigg(\int_{\Omega}\frac{1}{\varepsilon ^{4}}|\varepsilon ^{\delta}{\cal B}_{1} (\varepsilon^{2} r,z)|^{2p}\varepsilon^{2} r{\rm d}\varepsilon^{2} r{\rm d}z \bigg)^{\frac{1}{2p}}=\varepsilon ^{\delta-\frac{2}{p}}\| {\cal B}_{1}\| _{L^{2p}(\Omega)}. \end{equation}$$

(3.12)

类似地,得到

$$\begin{equation} \| g_{0}\| _{L^{2}(\Omega)}^{\frac{1}{q}} =\varepsilon ^{\delta-\frac{2}{q}}\| W_{1}\| _{L^{2q}(\Omega)}. \end{equation}$$

(3.13)

将(3.10),(3.12) 和 (3.13) 式相加,得到 $\| B_{1}\| _{L^{2p}(\Omega)}$, $\| \omega_{1}\| _{L^{2q}(\Omega)}$ 整体有界以及 $\| B^{\theta}\| _{L^{2p}(\Omega)},$ $\| \omega^{\theta}\| _{L^{2q}(\Omega)}$ 整体有界. 定理 1.1 证毕.

(2) $B^{r}=B^{z}=0$

本节考虑一族特解

$$\begin{equation} {\bf u}({\bf x},t)=u^{r}(r,z,t){\bf e}_{r}+ u^{\theta}(r,z,t){\bf e}_{\theta}+u^{z}(r,z,t){\bf e}_{z},\quad{\bf B}({\bf x},t) =B^{\theta}(r,z,t){\bf e}_{\theta}. \end{equation}$$

(3.14)

易知,若 $B^{r},B^{z}$ 的初值为零,则恒为零. 因此,(2.11)-(2.16)式 被简化为

$$\begin{equation} (u_{1})_{t}+u^{r}(u_{1})_{r}+u^{z}(u_{1})_{z}=2(\psi _{1})_{z} u_{1}+(u_{1zz}+u_{1rr}+\frac{3}{r}u_{1r}), \end{equation}$$

(3.15)

$$\begin{equation} (\omega_{1})_{t}+u^{r}(\omega_{1})_{r}+u^{z}(\omega_{1})_{z}=[(u_{1})^{2}]_{z}-[(B_{1})^{2}]_{z}+(\omega_{1zz}+\omega_{1rr}+\frac{3}{r}\omega_{1r}), \end{equation}$$

(3.16)

$$\begin{equation} (B_{1})_{t}+u^{r}(B_{1})_{r}+u^{z}(B_{1})_{z}=(B_{1zz}+B_{1rr}+\frac{3}{r}B_{1r}), \end{equation}$$

(3.17)

$$\begin{equation} -(\psi_{1zz}+\psi_{1rr}+\frac{3}{r}\psi_{1r}) =\omega_{1}. \end{equation}$$

(3.18)

定理 1.2 的证明 记 $\Omega={\Bbb R}^{2}\times[0,1]$,且定义 $h=|u_{1}|^{p}, f=|B_{1}|^{p},g=|W_{1}|^{q}$,$p=2q$. (3.15) 式乘以 $|u_{1}|^{2p-2}u_{1}$ 且在 $\Omega$ 上积分,得

$$\begin{eqnarray} &&\frac{1}{2p} \frac{\rm d}{{\rm d}t}\int_{\Omega}h^{2} r{\rm d}r{\rm d}z+\int_{\Omega}|u_{1}|^{2p-2}u_{1} [u^{r}(u_{1})_{r}+u^{z}(u_{1})_{z}] r{\rm d}r{\rm d}z \nonumber\\ &=&\int_{\Omega}|u_{1}|^{2p}(\psi_{1})_{z} r{\rm d}r{\rm d}z+\int_{\Omega}|u_{1}|^{2p-2}u_{1} \bigg[u_{1zz} +\frac{(ru_{1r})_{r}}{r}+\frac{2}{r}u_{1r}\bigg] r{\rm d}r{\rm d}z \nonumber\\ &:=&K_{1}+K_{2}, \end{eqnarray}$$

(3.19)

这里由分部积分及(2.4)式,得

$$\int_{\Omega}|u_{1}|^{2p-2}u_{1}[u^{r}(u_{1})_{r}+u^{z}(u_{1})_{z}] r{\rm d}r{\rm d}z=0.$$

关于$K_{1}$,利用沿 $z$ 方向的 Poincare 不等式

$$\| \psi_{1z}\| _{L^{2q}(\Omega)}\leq\| \psi_{1zz}\| _{L^{2q}(\Omega)},$$

引理 2.2 和 Hölder 不等式,有

$$K_{1}\leq\| \psi_{1z}\| _{L^{2q}(\Omega)}\| h^{2}\| _{L^{\frac{2q}{2q-1}}(\Omega)}=\| g\| _{L^{2}(\Omega)}^{\frac{1}{q}}\| h\| _{L^{\frac{4q}{2q-1}}(\Omega)}^{2}\leq C_{p}\| g\| _{L^{2}(\Omega)}^{\frac{1}{q}}\| h\| _{L^{2}(\Omega)}^{2-\frac{3}{2q}}\| \nabla h\| _{L^{2}(\Omega)}^{\frac{3}{2q}}.$$

类似于 $I_{1}-I_{3}$,估计 $K_{2}$ 为

$$K_{2}\leq-\frac{(2p-1)}{p^{2}}\| \nabla h\| _{L^{2}(\Omega)}^{2}-\frac{1}{p}\int_{0}^{1}h^{2}(0,z,t){\rm d}z\leq-\frac{(2p-1)}{p^{2}}\| \nabla h\| _{L^{2}(\Omega)}^{2}.$$

因此,得到

$$\begin{eqnarray} \frac{1}{2p}\frac{\rm d}{{\rm d}t}\int_{\Omega}h^{2} r{\rm d}r{\rm d}z&\leq& C_{p}\| g\| _{L^{2}(\Omega)}^{\frac{1}{q}}\| h\| _{L^{2}(\Omega)}^{2-\frac{3}{2q}}\| \nabla h\| _{L^{2}(\Omega)}^{\frac{3}{2q}}-\frac{(2p-1)}{p^{2}}\| \nabla h\| _{L^{2}(\Omega)}^{2} \nonumber\\ &\leq& C_{p}\| g\| _{L^{2}(\Omega)}^{\frac{1}{q}}\| \nabla h\| _{L^{2}}^{2}-\frac{(2p-1)}{p^{2}}\| \nabla h\| _{L^{2}(\Omega)}^{2}, \end{eqnarray}$$

(3.20)

这里利用了沿$z$ 方向的 Poincare 不等式

$$\| h\| _{L^{2}}\leq \| h_{z}\| _{L^{2}}.$$

类似的,(3.3)式乘以 $|\omega_{1}|^{2q-2}\omega_{1}$ 且在 $\Omega$ 上积分,得

$$\begin{eqnarray} &&\frac{1}{2q} \frac{\rm d}{{\rm d}t}\int_{\Omega}g^{2} r{\rm d}r{\rm d}z+\int_{\Omega}|\omega_{1}|^{2q-2}\omega_{1}[u^{r}(\omega_{1})_{r}+u^{z}(\omega_{1})_{z}] r{\rm d}r{\rm d}z \nonumber\\ &=&\int_{\Omega}|\omega_{1}|^{2q-2}\omega_{1}[(u_{1})^{2}]_{z} r{\rm d}r{\rm d}z-\int_{\Omega}|\omega_{1}|^{2q-2}\omega_{1}[(B_{1})^{2}]_{z} r{\rm d}r{\rm d}z \nonumber\\ &&+\int_{\Omega}|\omega_{1}|^{2q-2}\omega_{1} \bigg(\omega_{1zz}+\frac{(r\omega_{1r})_{r}}{r}+\frac{2}{r}\omega_{1r}\bigg) r{\rm d}r{\rm d}z \nonumber\\ &:=&L_{1}+L_{2}+L_{3}, \end{eqnarray}$$

(3.21)

这里由分部积分及 (2.4)式,得

$$\int_{\Omega}|\omega_{1}|^{2q-2}\omega_{1}[u^{r}(\omega_{1})_{r}+u^{z}(\omega_{1})_{z}] r{\rm d}r{\rm d}z=0.$$

$$\begin{eqnarray*} L_{1}&\leq &(2q-1)\int_{\Omega}h^{\frac{2}{p}}|\omega_{1}|^{2q-2}|\omega _{1}|_{z} r{\rm d}r{\rm d}z\\ &=&\bigg(2-\frac{1}{q}\bigg)\int_{\Omega}h^{\frac{2}{p}}|\omega_{1}|^{q-1}(g|\omega_{1}|^{q-1}|\omega _{1z}|) r{\rm d}r{\rm d}z\\ &\leq&\bigg(2-\frac{1}{q}\bigg)\bigg (\int _{\Omega}h^{\frac{4}{p}}g^{2(1-\frac{1}{q})} r{\rm d}r{\rm d}z\bigg)^{\frac{1}{2}}\| g_{z}\| _{L^{2}(\Omega)} \\ &\leq&\bigg(2-\frac{1}{q}\bigg) \Big(\| h\| _{L^{2}(\Omega)}^{\frac{2}{p}}\| g\| _{L^{2}(\Omega)}^{1-\frac{2}{p}}\Big)\| g_{z}\| _{L^{2}(\Omega)}\\ &\leq &C_{q}\| h\| _{L^{2}(\Omega)}^{\frac{4}{p}}\| g\| _{L^{2}(\Omega)}^{2(1-\frac{2}{p})}+\frac{2q-1}{8q^{2}}\| \nabla g\| _{L^{2}(\Omega)}^{2}\\ &\leq &C_{q}\frac{2p-1}{8p^{2}}\| h\| _{L^{2}(\Omega)}^{2}+\frac{2q-1}{8q^{2}}\| g\| _{L^{2}(\Omega)}^{2}+\frac{2q-1}{8q^{2}}\| \nabla g\| _{L^{2}(\Omega)}^{2}, \end{eqnarray*}$$

这里我们利用了分部积分,Hölder 不等式,Young's 不等式和 $p=2q$. 类似于 $J_{1}-J_{4}$,得

$$L_{2}\leq C_{q}\frac{2p-1}{8p^{2}}\| f\| _{L^{2}(\Omega)}^{2}+\frac{2q-1}{8q^{2}}\| g\| _{L^{2}(\Omega)}^{2}+\frac{2q-1}{8q^{2}}\| \nabla g\| _{L^{2}(\Omega)}^{2}.$$

$$L_{3}\leq-\frac{(2q-1)}{q^{2}}\| \nabla g\| _{L^{2}(\Omega)}^{2}-\frac{1}{q}\int_{0}^{1}g^{2}(0,z,t){\rm d}z\leq-\frac{(2q-1)}{q^{2}}\| \nabla g\| _{L^{2}(\Omega)}^{2}.$$

因此,我们有

$$\begin{eqnarray} \frac{1}{2q} \frac{\rm d}{{\rm d}t}\int_{\Omega}g^{2}r{\rm d}r{\rm d}z &\leq & C_{q}\frac{2p-1}{8p^{2}}\| h\| _{L^{2}(\Omega)}^{2}+\frac{2q-1}{8q^{2}}\| g\| _{L^{2}(\Omega)}^{2}+\frac{2q-1}{8q^{2}}\| \nabla g\| _{L^{2}(\Omega)}^{2} \nonumber\\ &&+C_{q}\frac{2p-1}{8p^{2}}\| f\| _{L^{2}(\Omega)}^{2} +\frac{2q-1}{8q^{2}}\| g\| _{L^{2}(\Omega)}^{2}\nonumber\\ &&+\frac{2q-1}{8q^{2}}\| \nabla g\| _{L^{2}(\Omega)}^{2} -\frac{(2q-1)}{q^{2}}\| \nabla g\| _{L^{2}(\Omega)}^{2} \nonumber\\ &\leq & C_{q}\frac{2p-1}{8p^{2}}[~\| \nabla h\| _{L^{2}(\Omega)}^{2}+\| \nabla f\| _{L^{2}(\Omega)}^{2}~]-\frac{(2q-1)}{2q^{2}}\| \nabla g\| _{L^{2}(\Omega)}^{2}. \end{eqnarray}$$

(3.22)

用 (3.2) 式乘以 $|B_{1}|^{2p-2}B_{1}$ 且在 $\Omega$ 上积分,得

$$\begin{eqnarray} \frac{1}{2p} \frac{\rm d}{{\rm d}t}\int_{\Omega}f^{2}r{\rm d}r{\rm d}z &=& \int_{\Omega}|B_{1}|^{2p-2}B_{1}(B_{1zz}+B_{1rr}+\frac{3}{r}B_{1r}) r{\rm d}r{\rm d}z \nonumber\\ &=&-\frac{(2p-1)}{p^{2}}\| \nabla f\| _{L^{2}(\Omega)}^{2}-\frac{1}{p}\int_{0}^{1}f^{2} (0,z,t){\rm d}z\nonumber\\ &\leq&-\frac{(2p-1)}{p^{2}}\| \nabla f\| _{L^{2}(\Omega)}^{2}. \end{eqnarray}$$

(3.23)

将 (3.20),(3.21) 和 (3.22)式相加,有

$$\begin{eqnarray} &&\frac{\rm d}{{\rm d}t} \bigg(\frac{C_{q}}{2p}\int_{\Omega}h^{2} r{\rm d}r{\rm d}z+\frac{1}{2q}\int_{\Omega}g^{2} r{\rm d}r{\rm d}z+\frac{C_{q}}{2p}\int_{\Omega}f^{2} r{\rm d}r{\rm d}z\bigg) \nonumber\\ &\leq & C_{q}C_{p}\| g\| _{L^{2}(\Omega)}^{\frac{1}{q}} \| \nabla h\| _{L^{2}}^{2}-C_{q}\frac{(2p-1)}{p^{2}}\| \nabla h\| _{L^{2}(\Omega)}^{2} \nonumber\\ &&+C_{q}\frac{2p-1}{8p^{2}}[\| \nabla h\| _{L^{2}(\Omega)}^{2}+\| \nabla f\| _{L^{2}(\Omega)}^{2}] -\frac{(2q-1)}{2q^{2}}\| \nabla g\| _{L^{2}(\Omega)}^{2}-C_{q}\frac{(2p-1)}{p^{2}}\| \nabla f\| _{L^{2}(\Omega)}^{2} \nonumber\\ &\leq & C_{q}\bigg[C_{p}\| g\| _{L^{2}(\Omega)}^{\frac{1}{q}}-\frac{7(2p-1)}{8p^{2}}\bigg] \| \nabla h\| _{L^{2}}^{2}. \end{eqnarray}$$

(3.24)

若要 (3.24)式 的右边为负,只需

$$\begin{equation} C_{p}\| g\| _{L^{2}(\Omega)}^{\frac{1}{q}}-\frac{7(2p-1)}{8p^{2}}\leq 0. \end{equation}$$

(3.25)

如果 (3.25) 式成立,有

$$\frac{\rm d}{{\rm d}t} \bigg(\frac{C_{q}}{2p}\int_{\Omega}h^{2} r{\rm d}r{\rm d}z+\frac{1}{2q} \int_{\Omega}g^{2} r{\rm d}r{\rm d}z+\frac{C_{q}}{2p}\int_{\Omega}f^{2} r{\rm d}r{\rm d}z\bigg)\leq0.$$

由 Gronwall 不等式,得

$$\begin{eqnarray} &&\frac{C_{q}}{2p}\int_{\Omega}h^{2} r{\rm d}r{\rm d}z+\frac{1}{2q}\int_{\Omega}g^{2} r{\rm d}r{\rm d}z+\frac{C_{q}}{2p}\int_{\Omega}f^{2} r{\rm d}r{\rm d}z \nonumber\\ &\leq& \frac{C_{q}}{2p}\int_{\Omega}h_{0}^{2} r{\rm d}r{\rm d}z+\frac{1}{2q}\int_{\Omega}g_{0}^{2} r{\rm d}r{\rm d}z+\frac{C_{q}}{2p}\int_{\Omega}f_{0}^{2} r{\rm d}r{\rm d}z. \end{eqnarray}$$

(3.26)

根据初值 条件(1.6),有

$$u_{1}(r,z,0)=\varepsilon ^{2\delta}U_{1}(\varepsilon^{2\delta} r,z),\quad \omega_{1}(r,z,0)=\varepsilon ^{2\delta}W_{1}(\varepsilon^{2\delta} r,z),\quad B_{1}(r,z,0)=\varepsilon ^{2\delta}{\cal B}_{1}(\varepsilon^{2\delta} r,z),$$

且

$$\| h_{0}\| _{L^{2}(\Omega)}^{\frac{1}{p}} =\varepsilon ^{2\delta(1-\frac{1}{p})}\| U_{1}\| _{L^{2p}(\Omega)}, \quad \| g_{0}\| _{L^{2}(\Omega)}^{\frac{1}{q}} =\varepsilon ^{2\delta(1-\frac{1}{q})}\| W_{1}\| _{L^{2q}(\Omega)},$$

$$\| f_{0}\| _{L^{2}(\Omega)}^{\frac{1}{p}} =\varepsilon ^{2\delta(1-\frac{1}{p})}\| {\cal B}_{1}\| _{L^{2p}(\Omega)},$$

这里 $U_{1},{\cal B}_{1}\in L^{2p}(\Omega),W_{1}\in L^{2q}(\Omega)$ 独立于 $\varepsilon$. 根据 (3.26)式,有

$$\begin{eqnarray*} \| g\| _{L^{2}(\Omega)}^{\frac{1}{q}} & \leq& (\frac{q}{p}C_{q})^{\frac{1}{2q}} \varepsilon^{4\delta (1-\frac{1}{q})}\| U_{1}\| _{L^{2p}(\Omega)}^{2}+ \varepsilon^{2\delta (1-\frac{1}{q})}\| W_{1}\| _{L^{2q}(\Omega)}^{2} \\ &&+(\frac{q}{p}C_{q})^{\frac{1}{2q}}\varepsilon^{4\delta (1-\frac{1}{q})}\| {\cal B}_{1}\| _{L^{2p}(\Omega)}^{2}. \end{eqnarray*}$$

因此,如果 $\varepsilon$ 充分小,使

$$\begin{eqnarray*} &&(\frac{q}{p}C_{q})^{\frac{1}{2q}}\varepsilon^{4\delta (1-\frac{1}{q})}\| U_{1}\| _{L^{2p}(\Omega)}^{2}+\varepsilon^{2\delta (1-\frac{1}{q})}\| W_{1}\| _{L^{2q}(\Omega)}^{2} \\ &&+(\frac{q}{p}C_{q})^{\frac{1}{2q}}\varepsilon^{4\delta (1-\frac{1}{q})}\| {\cal B}_{1}\| _{L^{2p}(\Omega)}^{2}\leq\frac{7(2p-1)}{8p^{2}}, \end{eqnarray*}$$

其中 $\delta>\frac{1}{q}$. 则 $\| u_{1}\| _{L^{2p}(\Omega)}$, $\| B_{1}\| _{L^{2p}(\Omega)}$,$\| \omega_{1}\| _{L^{2q}(\Omega)}$ 整体有界和 $\| U^{\theta}\| _{L^{2p}(\Omega)},$ $\| B^{\theta}\| _{L^{2p}(\Omega)},$ $\| \omega^{\theta}\| _{L^{2q}(\Omega)}$ 整体有界. 定理 1.2 证毕.