本文考虑二维空间线性化的等熵可压缩Navier-Stokes-Poisson方程, 后面简称为N-S-P方程柯西问题解的逐点估计. 二维N-S-P方程形式为

$ \begin{equation} \begin{cases} \rho_t+divm=0, \\ m_{1t}+\partial_{x_1}\left(\frac{m_1^2}{\rho}\right)+\partial_{x_2}\left(\frac{m_1m_2}{\rho}\right)+\partial_{x_1}P\left(\rho\right)=\rho\partial_{x_1}\phi+\mu_1\Delta\frac{m_1}{\rho}+\mu_2\partial_{x_1}div\frac{m}{\rho}, \\ m_{2t}+\partial_{x_1}\left(\frac{m_1m_2}{\rho}\right)+\partial_{x_2}\left(\frac{m_2^2}{\rho}\right)+\partial_{x_2}P\left(\rho\right)=\rho\partial_{x_2}\phi+\mu_1\Delta\frac{m_2}{\rho}+\mu_2\partial_{x_2}div\frac{m}{\rho}, \\ \Delta\phi=\rho-\overline{\rho}, \phi\to\; 0 \;当 \sqrt{x_1^2+x_2^2}\to+\infty . \end{cases} \end{equation} $

(1.1)

此处$ x=\left(x_1, x_2\right)\in\mathfrak{R}^2 $是空间变量, $ t > 0 $是时间变量, $ div $是散度算子, $ \Delta $是通常的拉普拉斯算子, 即$ \Delta=\partial_ {x_1}^2+\partial_{x_2}^2 $. $ \rho\left(x, t\right) $, $ m\left(x, t\right)=\left(m_1, m_2\right) $, $ \phi\left(x, t\right) $和$ P\left(\rho\right) $分别代表电子流体密度, 动量密度, 静电势和压力. 粘性系数$ \mu_1 $, $ \mu_2 $均大于零, $ \overline{\rho} > 0 $表示正电荷背景离子的规定密度. (1.1)是一个描述等离子体动力学行为的涉及耗散的简单模型, 它是在等熵Navier-Stokes方程基础上加上静电势$ \phi $的作用而得的方程.等熵Navier-Stokes方程是双曲-抛物混合方程. Kawashima[1]研究了一般的双曲-抛物系统, 得到了解的整体存在性和$ L^2 $衰减估计, Liu和Zeng[2]得到了一维一般双曲-抛物系统解的逐点估计, Holf和Zumbrun[3]得到高维等熵Navier-Stokes方程的解的$ L^p $估计, Liu和Wang[4], Wang和Yang[5]通过对格林函数的详细分析, 继续得到了此方程解的逐点估计. 对N-S-P方程, [6,7,8]研究了其局部解和整体弱解的存在性. [9]中用半群的方法得到了三维空间经典解的整体存在性, [10]通过对其线性系统格林函数的详细分析, 得到了解的逐点估计.相较于双曲抛物的等熵Navier-Stokes方程, 加了静电势的N-S-P方程是一个双曲-抛物-椭圆的耦合系统.方程结构越复杂, 格林函数越难计算且形式越复杂, 做逐点估计更难. 另外, 因为有了静电势, 格林函数中含了一个非局部项, 这一项在低频时是奇异的, 因而在估计时遇到与[4,5]不同的新的困难. 从[9,10]中可看出, 方程的衰减与空间维数有关, 维数越低, 衰减越慢, 因而处理更麻烦. 本论文考虑(1.1)在常状态$ \left(\overline{\rho}, \overline{m}_1, \overline{m}_2\right)=\left(1, 0, 0\right) $附近的线性化方程, 研究其格林函数的构造、衰减, 继而得到线性方程的衰减.希望这些工作能给非线性方程的研究提供一个基础和支撑.

本文中, $ \alpha $是多重指标$ \alpha=\left(\alpha_1, \alpha_2\right) $, $ \partial_x^{\alpha}f\left(x\right) $表示$ \partial_{x_1}^{\alpha_1}\partial_{x_2}^{\alpha_2}f $, $ \lvert\alpha\rvert $表示$ \alpha_1+\alpha_2 $. $ F\left(f\right) $或$ \hat{f}\left(\xi\right) $表示函数$ f\left(x\right) $关于变量$ x $的傅立叶变换, 即$ F\left(f\right)\left(\xi, t\right)=\hat{f}\left(\xi, t\right)=\int _{\mathfrak{R}_n}{f}\left(x, t\right)e^{-ix\cdot\xi}dx $, 其中$ i $为虚数单位. $ F^{-1}\left(\hat{f}\right)\left(x, t\right) $或$ f\left(x, t\right) $表示$ \hat{f}\left(\xi, t\right) $关于$ \xi $的逆傅立叶变换, 即$ F^{-1}\left(\hat{f}\right)=f\left(x, t\right)=\left(2\pi\right)^{-2n}\int\hat{f}\left(\xi, t\right)e^{ix\cdot\xi}d\xi $. $ W^{s, p}\left(\mathfrak{R}^n\right) $是通常的索伯列夫空间, 范数为$ \left\|f\right\|_{W^{s, p}\left(\mathfrak{R}^n\right)} $= $ \sum\limits_{\lvert\alpha\rvert=0}^{s}\left\|\partial_x^{\alpha}f\right\|_{L^P\left(\mathfrak{R}^n\right)} $. $ W^{s, 2}\left(\mathfrak{R}^n\right) $记为$ H^s\left(\mathfrak{R}^n\right) $.任意实数$ x $, 定义$ x_+=\begin{cases}x, 若\; x\geq0 , \\0, {若\; x< 0 }.\end{cases} $

首先把方程(1.1)在$ \left(\overline{\rho}, \overline{m}\right)=\left(1, 0, 0\right) $附近做线性化.记$ \tilde{\rho}=\rho-\overline{\rho} $, $ \tilde{m}=m-\overline{m} $, 因为

$ \partial_{x_1}\left(\frac{m_1}{\rho}\right)=\partial_{x_1}\left(\frac{\tilde{m}_1}{\tilde{\rho}+1}\right)=\frac{\partial_{x_1}\tilde{m}_1\cdot\left(\tilde{\rho}+1\right)-\tilde{m}_1\cdot\partial_{x_1}\tilde{\rho}}{\left(\tilde{\rho}+1\right)^2}, $

$ \partial_{x_1}\left(\frac{m_1^2}{\rho}\right)=\partial_{x_1}\left(\frac{\tilde{m}_1^2}{\tilde{\rho}+1}\right)=\frac{2\tilde{m}_1\cdot\partial_{x_1}\tilde{m}_1\cdot\left(\tilde{\rho}+1\right)-\tilde{m}_1^2\cdot\partial_{x_1}\tilde{\rho}}{\left(\tilde{\rho}+1\right)^2}, $

$ \partial_{x_2}\left(\frac{m_1 \cdot m_2}{\rho}\right)=\partial_{x_2}\frac{\tilde{m}_1\cdot\tilde{m}_2}{\tilde{\rho}+1}=\frac{\partial_{x_2}\tilde{m}_1\cdot\tilde{m}_2\left(\tilde{\rho}+1\right)+\tilde{m}_1\cdot\partial_{x_2}\tilde{m}_2\cdot\left(\tilde{\rho}+1\right)-\tilde{m}_1\cdot\tilde{m}_2\partial_{x_2}\tilde{\rho}}{\left(\tilde{\rho}+1\right)^2}, $

$ \partial_{x_1}P\left(\rho\right)=P^{\prime}\left(\overline{\rho}\right)\partial_{x_1}\tilde{\rho}+\partial_{x_1}\left(P\left(\rho\right)-P\left(\overline{\rho}\right)-P^{\prime}\left(\overline{\rho}\right)\tilde{\rho}\right), $

所以方程(1.1)在$ \left(1, 0, 0\right) $附近的线性化方程为

$ \begin{equation} \begin{cases} \tilde{\rho}_t+div\tilde{m}=0, \\ \tilde{m}_{1t}+P^{\prime}\left(1\right)\partial_{x_1}\tilde{\rho}=\partial_{x_1}\phi+\mu_1\Delta\tilde{m}_1+\mu_2\partial_{x_1}\left(\partial_{x_1}\tilde{m}_1+\partial_{x_2}\tilde{m}_2\right), \\ \tilde{m}_{2t}+P^{\prime}\left(1\right)\partial_{x_2}\tilde{\rho}=\partial_{x_2}\phi+\mu_1\Delta\tilde{m}_2+\mu_2\partial_{x_2}\left(\partial_{x_1}\tilde{m}_1+\partial_{x_2}\tilde{m}_2\right). \end{cases} \end{equation} $

(2.1)

为减少记号, 不妨还是记$ \rho=\tilde{\rho} $, $ m=\tilde{m} $, $ P^{\prime}\left(1\right)=c^2 > 0 $, 则(2.1)为

$ \begin{equation} \begin{cases} \rho_t+divm=0, \\ m_t+c^2\nabla\rho=\nabla\phi+\mu_1\Delta m+\mu_2\nabla divm, \\ \Delta\phi=\rho, { \phi\to0 , 当 \;\sqrt{x_1^2+x_2^2}\to +\infty }. \end{cases} \end{equation} $

(2.2)

下面考虑(2.2)的格林函数$ G $, 即若(2.2)的初值为$ \left(\rho, m\right)\mid_{t=0}=\left(\rho_0, m_0\right) $, 我们要找出矩阵$ G $满足$ \dbinom{\rho}{m}=G\ast\dbinom{\rho_0}{m_0} $.此处及后面的$ \ast $是表示对空间变量的卷积.

记$ \gamma=\mu_1+\mu_2 $, 由(2.2)得

$ \begin{equation} \begin{aligned} \rho_{tt}&=-\left(divm\right)_t=-divm_t=-div\left(-c^2\nabla\rho+\nabla\phi+\mu_1\Delta m+\mu_2\nabla divm\right)\\ &=c^2\Delta\rho-\Delta\phi-\mu_1\Delta\left(divm\right)-\mu_2\Delta\left(divm\right)\\ &=c^2\Delta\rho-\rho-\mu_1\Delta\left(-\rho_t\right)-\mu_2\Delta\left(-\rho_t\right)=c^2\Delta\rho-\rho+\gamma\Delta\rho_t. \end{aligned} \end{equation} $

(2.3)

对空间变量$ x $作傅立叶变换, 得$ \hat{\rho}_{tt}=c^2\left(-\lvert\xi\rvert^2\right)\hat{\rho}-\hat{\rho}+\gamma\left(-\lvert\xi\rvert^2\right)\hat{\rho}_t $.

解二阶常微分方程初值问题

$ \begin{equation} \begin{cases} \notag \hat{\rho}_{tt}+\left(c^2\lvert\xi\rvert^2+1\right)\hat{\rho}+\gamma\lvert\xi\rvert^2\hat{\rho_t}=0, \\ \hat{\rho}\left(\xi, 0\right)=\hat{\rho}_0\left(\xi\right), \\ \hat{\rho}_t\left(\xi, 0\right)=-i\xi^{\tau}\hat{m}_0\left(\xi\right). \end{cases} \end{equation} $

得到

$ \begin{equation} \hat{\rho}\left(\xi, t\right)=\frac{\lambda_+e^{\lambda_-t}-\lambda_-e^{\lambda_+t}}{\lambda_+-\lambda_-}\hat{\rho}_0\left(\xi\right)+\frac{\left(e^{\lambda_-t}-e^{\lambda_+t}\right)i\xi^\tau}{\lambda_+-\lambda_-}\hat{m}_0\left(\xi\right), \end{equation} $

(2.4)

其中

$ \begin{equation} \begin{aligned} \lambda_+=\frac{-\gamma\lvert\xi\rvert^2+\sqrt{\gamma ^2\left| \xi\right|^4-4\left(c^2\left|\xi\right|^2 +1\right)}}{2}, \lambda_-=\frac{-\gamma\lvert\xi\rvert^2-\sqrt{\gamma ^2\left| \xi\right|^4-4\left(c^2\left|\xi\right|^2 +1\right)}}{2}. \end{aligned} \end{equation} $

(2.5)

由(2.5)易得

$ \begin{equation} \begin{aligned} \lambda_+-\lambda_-=\sqrt{\gamma^2\left|\xi\right|^4-4\left(c^2\left|\xi\right|^2+1\right)}, \\ \lambda_++\gamma\left |\xi\right |^2=-\lambda_-, \lambda_-+\gamma\left |\xi\right |^2=-\lambda_+, \lambda_+\cdot\lambda_-=c^2\left|\xi \right|^2+1. \end{aligned} \end{equation} $

(2.6)

为求解$ \hat{m} $, 将$ \hat{m}\left(\xi, t\right) $分解成与$ \xi $平行和垂直的两个向量，即

$ \begin{equation} \hat{m}\left(\xi, t\right)=a\left(\xi, t\right)\frac{\xi}{\left |\xi\right |}+b\left(\xi, t\right), \end{equation} $

(2.7)

其中$ a\left(\xi, t\right)\in \mathfrak{R}^1 $, $ b\left(\xi, t\right)\in \mathfrak{R}^2 $, 且$ \xi^\tau b=0 $.对(2.2)的第二个方程关于空间向量作傅立叶变换，得

$ \hat{m}_t+c^2i\xi\hat{\rho}=i\xi\hat{\phi}-\mu_1\left|\xi\right|^2\hat{m}-\mu_2\xi\xi^\tau\hat{m}. $

于是

$ \begin{equation} \begin{aligned} \notag a_t\left(\xi, t\right)\frac{\xi}{\left|\xi\right|}+b_t+ic^2\xi\hat{\rho}&=i\xi\hat{\phi}-\mu_1\left|\xi\right|^2\left(a\left(\xi, t\right) \frac{\xi}{\left|\xi\right|}+b\left(\xi, t\right)\right)-\mu_2\xi\xi^\tau a\left(\xi, t\right)\frac{\xi}{\left|\xi\right|}-\mu_2\xi\xi^\tau b\\ &=i\xi\hat{\phi}-\mu_1\left|\xi\right|a\left(\xi, t\right)\xi-\mu_1\left|\xi\right|^2b\left(\xi, t\right)-\mu_2\left|\xi\right|\xi a\left(\xi, t\right). \end{aligned} \end{equation} $

所以得

$ b_t=-\mu_1\left|\xi\right|^2b\left(\xi, t\right), $

$ a_t\left(\xi, t\right)=-ic^2\left|\xi\right|\hat{\rho}+i\left|\xi\right|\hat{\phi}-\mu_1\left|\xi\right|^2a\left(\xi, t\right)-\mu_2a\left(\xi, t\right)\left|\xi\right|^2=-i\frac{c^2\left|\xi\right|^2+1}{\left|\xi\right|}\hat{\rho}-\gamma a\left(\xi, t\right)\left|\xi\right|^2. $

解常微分方程得

$ \begin{equation} b\left(\xi, t\right)=b\left(\xi, 0\right)e^{-\mu_1\left|\xi\right|^2t}, \end{equation} $

(2.8)

$ \begin{equation} a\left(\xi, t\right)=-i\frac{1+c^2\left|\xi\right|^2}{\left|\xi\right|}e^{-\gamma\left|\xi\right|^2t}\int_{0}^{t}\hat{\rho}\left(\xi, s \right)e^{\gamma\left|\xi\right|^2s}ds+a\left(\xi, 0\right)e^{-\gamma\left|\xi\right|^2t}. \end{equation} $

(2.9)

由(2.5), (2.6)得

$ \begin{equation} \begin{aligned} &\int_{0}^{t}\left [\frac{-\lambda_-}{\lambda_+-\lambda_-}e^{\left(\lambda_++\gamma\left|\xi\right|^2\right)s}+\frac{\lambda_+}{\lambda_+-\lambda_-}e^{\left( \lambda_-+\gamma\left|\xi\right|^2\right)s}\right]ds\\ =&\frac{-\lambda_-}{\lambda_+-\lambda_-}\cdot\frac{1}{\lambda_++\gamma\left|\xi\right|^2} \left(e^{\left(\lambda_++\gamma\left|\xi\right|^2\right)t}-1\right)+\frac{\lambda_+}{\lambda_+-\lambda_-}\cdot\frac{1}{\lambda_-+\gamma\left|\xi\right|^2}\left(e^{\left(\lambda_-+\gamma\left|\xi\right|^2\right)t}-1\right)\\ =&\frac{1}{\lambda_+-\lambda_-}\cdot e^{-\lambda_-t}-\frac{1}{\lambda_+-\lambda_-}\cdot e^{-\lambda_+t}\quad, \end{aligned} \end{equation} $

(2.10)

$ \begin{equation} \begin{aligned} &\int_{0}^{t}\left [\frac{-e^{\left(\lambda_++\gamma\left|\xi\right|^2\right)s}}{\lambda_+-\lambda_-}+\frac{e^{\left(\lambda_-+\gamma\left|\xi\right|^2\right)s}}{\lambda_+-\lambda_-}\right]ds\\ =&\frac{-1}{\lambda_+-\lambda_-}\cdot\frac{1}{\lambda_++\gamma\left|\xi\right|^2}\cdot\left(e^{\left(\lambda_++\gamma\left|\xi\right|^2\right)t}-1\right)+\frac{1}{\lambda_+-\lambda_-}\cdot\frac{1}{\lambda_-+\gamma\left|\xi\right|^2} \cdot\left(e^{\left(\lambda_-+\gamma\left|\xi\right|^2\right)t}-1\right)\\ =&\frac{1}{\left(\lambda_+-\lambda_-\right)\lambda_-}\cdot e^{-\lambda_-t}-\frac{1}{\left(\lambda_+-\lambda_-\right)\lambda_+}\cdot e^{-\lambda_+t}-\frac{1}{\lambda_+\lambda_-} \quad . \end{aligned} \end{equation} $

(2.11)

由(2.4), (2.6), (2.9), (2.10), (2.11)得

$ \begin{equation} \begin{aligned} a\left(\xi, t\right)&=-i\frac{\left(1+c^2\left|\xi\right|^2\right)}{\left|\xi\right|}e^{-\gamma\left|\xi\right|^2t}\cdot\hat\rho_0\cdot\left (\frac{e^{-\lambda _-t}-e^{-\lambda_+t}}{\lambda_+-\lambda_-}\right) -\frac{i\xi^\tau\hat m_0}{\lambda_+\lambda_-}\left (-i\right ) \frac{1+c^2\left|\xi\right|^2}{\left|\xi\right|} e^{-\gamma\left|\xi\right|^2t}\\ &+a\left(\xi, 0\right)e^{-\gamma\left|\xi\right|^2t} +\frac{1+c^2\left|\xi\right|^2}{\left|\xi\right|} e^{-\gamma\left|\xi\right|^2t }\xi^{\tau}\hat m_0\left(\frac{e^{-\lambda _-t}}{\left( \lambda_+-\lambda_-\right)\lambda_-}-\frac{e^{-\lambda_+t}}{\left(\lambda_+-\lambda_- \right )\lambda_+} \right) \\ =&-i\frac{\left(1+c^2\left|\xi\right|^2\right)}{\left|\xi\right|}\hat\rho_0\cdot\frac{e^{\lambda_+t}-e^{\lambda_-t}}{\lambda_+-\lambda_-}+\frac{1}{\left|\xi\right|}\xi^\tau\hat m_0\left(\frac{\lambda_+e^{\lambda _+t}-\lambda_-e^{\lambda_-t}}{\lambda_+-\lambda_-}\right)\\ &-\frac{\xi^\tau}{\left|\xi\right|}\left(a\left(\xi, 0\right)\frac{\xi}{\left|\xi\right|}+b\left(\xi, 0\right)\right) e^{-\gamma\left|\xi\right |^2t}+a\left(\xi, 0\right) e^{-\gamma\left|\xi \right|^2t}\\ =&-i\frac{\left(1+c^2\left|\xi\right|^2\right)}{\left|\xi\right|}\cdot\left (\frac{e^{\lambda_+t}-e^{\lambda_-t}}{\lambda_+-\lambda_-}\right) \cdot\hat\rho_0+\frac{1}{\left|\xi\right|}\left (\frac{\lambda_+e^{\lambda _+t}-\lambda_-e^{\lambda_-t}}{\lambda_+-\lambda_-}\right)\xi^\tau\hat m_0 . \end{aligned} \end{equation} $

(2.12)

由(2.7), (2.12)得

$ \begin{equation} \begin{aligned} b\left(\xi, 0\right)=\hat m_0\left(\xi\right)-a\left(\xi, 0\right)\frac{\xi}{\left|\xi\right|} &=\hat m_0\left(\xi\right)-\frac{1}{\left|\xi\right|^2 }\left(\frac{\lambda_+}{\lambda_+-\lambda_-} - \frac{\lambda_-}{\lambda_+-\lambda_-}\right)\xi^\tau\hat m_0\xi\\ &=\hat m_0\left(\xi \right)-\frac{\xi\xi^\tau}{\left|\xi\right|^2}\hat m_0. \end{aligned} \end{equation} $

(2.13)

由(2.8), (2.13)得

$ \begin{equation} b\left(\xi, t\right)=e^{-\mu_1\left|\xi\right|^2t }\left(I-\frac{\xi\xi^\tau}{\left|\xi\right|^2}\right)\hat m_0, \end{equation} $

(2.14)

其中$ I $为单位矩阵.

由(2.7), (2.12), (2.14) 得

$ \begin{equation} \begin{aligned} \hat m\left(\xi, t\right)=&-i\frac{\left(1+c^2\left|\xi\right|^2 \right)\xi}{\left|\xi\right|^2}\hat\rho_0\left(\frac{e^{\lambda _+t}}{\lambda_+-\lambda_-}-\frac{e^{\lambda _-t}}{\lambda_+-\lambda_-}\right)\\ &+\frac{1}{\left|\xi\right|^2}\left(\frac{\lambda_+e^{\lambda _+t}}{\lambda_+-\lambda_-}-\frac{\lambda_-e^{\lambda _-t}}{\lambda_+-\lambda_-}\right)\xi\xi^\tau\hat m_0+e^{-\mu_1\left|\xi\right|^2t}\left(I-\frac{\xi\xi^\tau}{\left|\xi\right|^2}\right) . \end{aligned} \end{equation} $

(2.15)

若记

$ \begin{equation} \begin{pmatrix} \hat\rho\left ( \xi, t \right ) \\ \hat m\left ( \xi, t \right ) \end{pmatrix} =\hat G\left ( \xi, t \right ) \begin{pmatrix} \hat\rho_0\\ \hat m_0 \end{pmatrix}, \end{equation} $

(2.16)

则由(2.4), (2.15) 得

$ \begin{equation} \hat G\left(\xi, t\right)=\begin{pmatrix} \frac{-\lambda_-e^{\lambda_+t}}{\lambda_+-\lambda_-}+\frac{\lambda_+e^{\lambda_-t}}{\lambda_+-\lambda_-} & \frac{e^{\lambda_+t}-e^{\lambda_-t}}{\lambda_+-\lambda_-}\left(-i\xi^\tau\right) \\ -i\frac{\left(1+c^2\left|\xi\right|^2\right)\xi}{\left|\xi\right|^2}\left(\frac{e^{\lambda_+t}-e^{\lambda_-t}}{\lambda_+-\lambda_-}\right) & \frac{\lambda_+e^{\lambda_+t}-\lambda_-e^{\lambda_-t}}{\lambda_+-\lambda_-}\frac{\xi\xi^\tau}{\left|\xi\right|^2} +e^{-\mu_1\left|\xi\right|^2t}\left(I-\frac{\xi\xi^\tau}{\left|\xi\right|^2}\right) \end{pmatrix}. \end{equation} $

(2.17)

我们称$ G\left(x, t\right) $为方程(2.2)的格林函数.

由(2.16)可得

$ \begin{equation} \begin{pmatrix} \rho \\ m \end{pmatrix}=G\ast \begin{pmatrix} \rho_0 \\ m_0 \end{pmatrix}. \end{equation} $

(3.1)

我们希望通过对$ G $的估计及(3.1)式来得到方程(2.2)的解的逐点估计.

为方便，我们记

$ \begin{equation} \hat G\left ( \xi, t \right ) =\begin{pmatrix} \hat G_{11} &\hat G_{12} \\\hat G_{21} &\hat G_{22} \end{pmatrix}. \end{equation} $

(3.2)

因为$ \lambda_+ $, $ \lambda_- $在高、低, 中频时，有不同的表现形式，所以我们定义光滑截断函数

$ \chi _1\left(\xi\right)=\begin{cases}1, \left | \xi \right | \le \varepsilon, \\0, \left | \xi \right |> 2\varepsilon.\end{cases}, \quad \chi _3\left(\xi\right)=\begin{cases}1, \left | \xi \right | \ge R, \\0, \left | \xi \right |< R-1.\end{cases} $

其中$ \varepsilon $充分小, $ R $充分大, $ 2\varepsilon< R $.记$ \chi_2\left(\xi\right)=1-\chi _1\left(\xi\right)-\chi _3\left(\xi\right) $. $ \chi _i\left(D\right)\left(i=1, 2, 3\right) $表示象征为$ \chi _i\left( \xi\right)\left(i=1, 2, 3\right) $的算子.符号$ B_N\left(x, t\right) $表示$ B_N\left(x, t \right)=\left(1+\frac{\left|x\right|^2}{1+t}\right)^{-N} $.

定理3.1   对任意正常数$ N $, 多重指标$ \alpha $, 存在仅依赖于$ N $, $ \alpha $的常数$ C_{N, \alpha} $, 有

$ \begin{equation} \left|\partial _x^\alpha\chi _1\left(D\right)G_{11}\right|\le C_{N, \alpha}\left(1+t\right)^{-\frac{2+\left|\alpha\right|}{2}}B_N\left ( x, t\right), \end{equation} $

(3.3)

$ \begin{equation} \left|\partial _x^\alpha\chi _1\left(D\right)G_{12}\right|\le C_{N, \alpha}\left(1+t\right)^{-\frac{3+\left|\alpha\right|}{2}}B_N\left ( x, t\right), \end{equation} $

(3.4)

$ \begin{equation} \left|\partial _x^\alpha F^{-1}\left(\chi _1\left(\xi\right)\cdot\frac{e^{\lambda_+t}-e^{\lambda_-t}}{\lambda_+-\lambda_-}\right)\right|\le C_{N, \alpha}\left(1+t\right)^{-\frac{2+\left|\alpha\right|}{2}}B_N\left(x, t\right), \end{equation} $

(3.5)

$ \begin{equation} \left|\partial _x^\alpha F^{-1}\left(\chi _1\left(\xi\right)\cdot\frac{\lambda_+e^{\lambda_+t}-\lambda_-e^{\lambda_-t}}{\lambda_+-\lambda_-}\right)\right|\le C_{N, \alpha}\left(1+t\right)^{-\frac{2+\left|\alpha\right|}{2}}B_N\left(x, t\right) . \end{equation} $

(3.6)

证    当$ \left|\xi\right| $充分小时，由$ Taylor $展开有

$ \begin{equation} \begin{split} \sqrt{\gamma^2\left|\xi\right|^4-4\left(c^2\left|\xi\right|^2+1\right)}=2i+ic^2\left|\xi\right|^2+o\left(\left|\xi\right|^2\right), \\ \frac{1}{\sqrt{\gamma^2\left|\xi\right|^4-4\left(c^2\left|\xi\right|^2+1\right)}}=\frac{1}{2i}+\frac{i}{4}c^2\left|\xi\right|^2+o\left( \left|\xi\right|^2\right) . \end{split} \end{equation} $

(3.7)

所以

$ \begin{equation} \lambda _+=i+\frac{i}{2} c^2\left | \xi \right | ^2-\frac{\gamma }{2} \left | \xi \right | ^2+o\left ( \left | \xi \right | ^2 \right ), \end{equation} $

(3.8)

$ \begin{equation} \lambda _-=-i-\frac{i}{2} c^2\left | \xi \right | ^2-\frac{\gamma }{2} \left | \xi \right | ^2+o\left ( \left | \xi \right | ^2 \right ), \end{equation} $

(3.9)

$ \begin{equation} \frac{\lambda _+}{\lambda _+-\lambda _-} =\frac{1}{2} +\frac{i}{4} \gamma \left | \xi \right |^2+o\left ( \left | \xi \right | ^2 \right ), \end{equation} $

(3.10)

$ \begin{equation} \frac{\lambda _-}{\lambda _+-\lambda _-} =-\frac{1}{2} +\frac{i}{4} \gamma \left | \xi \right |^2+o\left ( \left | \xi \right | ^2 \right ) . \end{equation} $

(3.11)

于是对任意多重指标$ \alpha $, $ \beta $, 存在正常数$ b $, $ C $, 有

$ \left | \partial _\xi^\beta\left ( \xi^\alpha e^{\lambda_+t}\right ) \right | \le C\left ( \left | \xi \right |^{\left(\left | \alpha \right |-\left | \beta \right | \right) _+} +\left | \xi \right | ^{\left | \alpha \right | } t^{\frac{\left | \beta \right | }{2} }\right ) \left ( 1+\left | \xi \right | ^2t \right )^{\left | \beta \right | +1} e^{-b\left | \xi \right |^2t }, $

$ \left | \partial _\xi^\beta\left ( \xi^\alpha e^{\lambda_-t}\right ) \right | \le C\left ( \left | \xi \right |^{\left(\left | \alpha \right |-\left | \beta \right | \right)_+} +\left | \xi \right | ^{\left | \alpha \right | } t^{\frac{\left | \beta \right | }{2} }\right ) \left ( 1+\left | \xi \right | ^2t \right )^{\left | \beta \right | +1} e^{-b\left | \xi \right |^2t }, $

所以

$ \left | \partial_\xi^\beta\left ( \xi^\alpha\left ( \frac{\lambda_+e^{\lambda_-t}-\lambda_-e^{\lambda_+t}}{\lambda_+-\lambda_-} \right ) \right ) \right | \le C\left ( \left | \xi \right | ^{\left ( \left | \alpha \right |+\left | \beta \right | \right )_+ }+\left | \xi \right | ^{\left | \alpha \right |} t^{\frac{\left | \beta \right | }{2}}\right )\left ( 1+\left | \xi \right | ^2t \right )^{\left | \beta\right |+1 } e^{-b\left | \xi \right | ^2t} , $

$ \left | \partial_\xi^\beta\left ( \xi^\alpha\left ( \frac{e^{\lambda_+t}-e^{\lambda_-t}}{\lambda_+-\lambda_-} \right ) \right ) \right | \le C\left ( \left | \xi \right |^{\left(\left | \alpha \right | -\left | \beta \right |\right) _+} +\left | \xi \right |^{\left | \alpha \right |}t^{\frac{\left | \beta \right | }{2} } \right )\left ( 1+\left | \xi \right |^2t \right ) ^{\left | \beta \right | +1}e^{-b\left | \xi \right |^2t }, $

$ \left | \partial_\xi^\beta\left ( \xi^\alpha\left ( \frac{\lambda_+e^{\lambda_+t}-\lambda_-e^{\lambda_-t}}{\lambda_+-\lambda_-} \right ) \right ) \right | \le C\left ( \left | \xi \right |^{\left(\left | \alpha \right | -\left | \beta \right |\right) _+} +\left | \xi \right |^{\left | \alpha \right |}t^{\frac{\left | \beta \right | }{2} } \right )\left ( 1+\left | \xi \right |^2t \right ) ^{\left | \beta \right | +1}e^{-b\left | \xi \right |^2t }. $

由[11]的引理3.1, (2.17), 本定理得证.

因为$ \hat{G}_{21} $, $ \hat{G}_{22} $中有非局部算子$ \frac{1}{\left|\xi\right|^2} $, 所以我们需要用新的技巧来估计它们.

定理3.2   存在正常数$ C $, 有

$ \begin{equation} \left | \partial _x^\alpha\chi _1\left ( D \right ) G_{21}\right|\le C\left ( 1+t \right ) ^{-\frac{1+\left | \alpha \right | }{2} }B_\frac{1}{2}\left ( x, t \right ), \end{equation} $

(3.12)

$ \begin{equation} \left | \partial _x^\alpha\chi _1\left ( D \right ) G_{22}\right|\le C\left ( 1+t \right ) ^{-\frac{2+\left | \alpha \right | }{2} }B_\frac{1}{2}\left ( x, t \right ) . \end{equation} $

(3.13)

证   把$ \left|\partial_x^\alpha\chi_1\left(D\right)G_{21}\right| $估计分成两部分,

$ \left|\partial _x^\alpha\chi_1\left(D\right)G_{21}\right|\le\left| \partial_x^\alpha\chi_1\left(D\right)G_{21}\right|_{\left\{x\mid\left|x\right|^2\le 1+t\right\}}+\left|\partial _x^\alpha\chi_1\left(D\right) G_{21}\right|_{\left\{x\mid\left |x\right|^2>1+t\right\}}:=I_1+I_2. $

由(2.17), (2.17), (3.2) (3.7), (3.8), (3.9)得

$ \begin{equation} \begin{aligned} \notag I_1\le \int \left | \xi^\alpha\chi _1\left ( \xi \right ) \hat G_{21} \right | d\xi&\le C\int \left | \xi \right |^{\left | \alpha \right | -1} e^{-\frac{\gamma \left | \xi \right |^2t }{4}}d\xi\le C_\alpha\left ( 1+t \right ) ^{-\frac{\left | \alpha \right |+1 }{2} }\\&\le C_{N, \alpha}\left ( 1+t \right ) ^{-\frac{\left | \alpha \right | +1}{2} }B_N\left ( x, t \right ). \end{aligned} \end{equation} $

当$ N > 1 $时,

$ \begin{equation} \begin{aligned} \int\left(1+\frac{\left|x-y\right|^2}{1+t}\right)^{-N}dy &=\int_{0}^{2\pi }d\theta \int_{0}^{+\infty}\left(1+\frac{r^2}{1+t}\right)^{-N}rdr\\ &=\int_{0}^{2\pi}d\theta\int_{0}^{+\infty}\left(1+r^2\right)^{-N}\sqrt{1+tr}\cdot\sqrt{1+t}dr\le C\left(1+t\right). \end{aligned} \end{equation} $

(3.14)

当$ N > \frac{1}{2} $时,

$ \begin{equation} \begin{aligned} \int \left | y \right | ^{-1}\left ( 1+\frac{\left | y \right |^2 }{1+t} \right )^{-N}dy&=\int_{0}^{2\pi}d\theta \int_{0}^{+\infty} \left ( 1+\frac{r^2}{1+t}\right ) ^{-N} \frac{1}{r}\cdot rdr\\ &=2\pi\int_{0}^{+\infty}\left ( 1+r^2 \right ) ^{-N}\sqrt{1+t}dr\le C\left ( 1+t \right ) ^\frac{1}{2} . \end{aligned} \end{equation} $

(3.15)

当$ \left | y \right | \le \frac{\left | x \right | }{2} $时，有

$ \begin{equation} \left | y-x \right | \ge \left | x \right | -\left | y \right | \ge \frac{\left | x \right | }{2} \ge \left | y \right | . \end{equation} $

(3.16)

又

$ \begin{equation} F^{-1}\left ( \frac{i\xi_j}{\left | \xi \right | ^2} \right ) =-\partial_{x_j}\Delta^{-1}\delta \left ( x \right )=C\partial_{x_j}ln\left | x \right |=C\frac{x_j}{\left | x \right | ^2}, \end{equation} $

(3.17)

取(3.5)中$ N > 2 $, 则由(3.5), (3.14), (3.15), (3.16), (3.17) 得

$ \begin{equation} \begin{aligned} I_2\le&\int_{\mathfrak{R} ^2}C\frac{\left|y_j\right|}{\left|y\right|^2}\cdot\left(1+t\right)^{-\frac{2+\left|\alpha\right|}{2}}B_N\left(x-y, t\right)dy\\ \le& C\int _{\mathfrak{R} ^2\cap\left\{y\mid\left|y\right|\ge\frac{\left|x\right|}{2}\right\}}\frac{1}{\left|y\right|}\left(1+t\right)^{-\frac{2+\left| \alpha\right|}{2}}B_N\left(x-y, t\right)dy\\ &+C\int_{\mathfrak{R} ^2\cap\left\{y\mid\left|y\right|\le\frac{\left|x\right|}{2}\right\}}\frac{1}{\left|y\right|}\left(1+t\right)^{-\frac{2+\left|\alpha \right|}{2}}B_N\left(x-y, t\right )dy\\ \le& C\left ( \left | x \right | ^2 \right ) ^{-\frac{1}{2} }\left ( 1+t \right ) ^{-\frac{2+\left | \alpha \right | }{2} }\cdot\left ( 1+t \right ) +C\left ( 1+t \right ) ^{-\frac{2+\left | \alpha \right | }{2} }\left ( 1+t \right )^{\frac{1}{2}} B_{\frac{1}{2} }\left ( x, t \right ) \\ \le& C\left ( 1+t \right ) ^{-\frac{\left | \alpha \right | }{2} } \left ( 1+t \right ) ^{-\frac{1}{2} } B_\frac{1}{2} \left ( x, t \right ) +C\left ( 1+t \right ) ^{-\frac{1+\left | \alpha \right | }{2} }B_\frac{1}{2} \left ( x, t \right )\\ =&C\left ( 1+t \right ) ^{-\frac{1+\left | \alpha \right | }{2} }B_\frac{1}{2} \left ( x, t \right ). \end{aligned} \end{equation} $

(3.18)

由$ I_1 $, $ I_2 $的估计, 得到(3.12).用同样的方法可得到(3.13).

对中频的部分，我们有下面定理.

定理3.3   对任意正整数$ N $, 存在正常数$ C $, $ b $, 有$ \left | \partial _x^\alpha \chi _2\left ( D \right ) G\left ( x, t \right ) \right | \le Ce^{-bt}B_N\left ( x, t \right ) $.

证   因为

$ \frac{e^{\lambda_+t}-e^{\lambda_-t}}{\lambda _+-\lambda_-} =e^{\lambda_-t}\cdot\frac{e^{\left ( \lambda _+-\lambda_- \right )t }-1}{\lambda _+-\lambda_-}, $

$ \begin{equation} \begin{aligned} \notag \frac{\lambda_+e^{\lambda_-t}-\lambda _-e^{\lambda_+t}}{\lambda _+-\lambda_-} =&\frac{\lambda _+e^{\lambda_+t}\left ( e^{\left ( \lambda _–\lambda_+\right )t }-1 \right )-\lambda _-e^{\lambda_-t}\left ( e^{\left ( \lambda _+-\lambda_-\right )t }-1 \right ) }{\lambda _+-\lambda_-} \\ &-\frac{\lambda _+e^{\lambda_-t}\left ( e^{\left(\lambda_+-\lambda_-\right)t} -1\right )}{\lambda _+-\lambda_-} +e^{\lambda_-t}, \end{aligned} \end{equation} $

所以当$ \varepsilon \le\left|\xi \right | \le R $时, $ \hat G $无奇点, 无需考虑对$ \xi $的可积性问题, 又$ \left|Re\lambda_{\pm}\right|\le -b $, 所以$ \left | \partial_x^\alpha\chi _2\left (D\right ) G\left ( x, t \right ) \right | \le \int \xi^\alpha\chi _2\left ( \xi \right )\hat G\left ( \xi, t \right ) d\xi\le Ce^{-bt}B_N\left ( x, t \right ) $.

定理3.3得证.

当$ \left|\xi\right| $充分大时, 由泰勒展开有

$ \lambda_+-\lambda_-=\gamma \left | \xi \right | ^2\cdot\sqrt{1-4\frac{c^2}{\gamma ^2} \frac{1}{\left | \xi \right | ^2} -\frac{4}{\gamma ^2\left | \xi \right |^4 } } = \gamma \left | \xi \right | ^2-\frac{2c^2}{\gamma } -\frac{2}{\left | \xi \right |^2 } -\frac{2c^4}{\gamma ^3} \frac{1}{\left | \xi \right | ^2} +o\left ( \frac{1}{\left | \xi \right |^2 } \right ), $

$ \frac{1}{\lambda _+-\lambda_-} =\frac{1}{\gamma \left | \xi \right |^2 } +\frac{2}{\gamma ^3\left | \xi \right |^4 } +o\left ( \frac{1}{\left | \xi \right |^4 } \right ), $

所以

$ \begin{equation} \begin{aligned} \frac{\lambda_+e^{\lambda_-t}-\lambda _-e^{\lambda_+t}}{\lambda _+-\lambda_-} &=\left ( -\frac{1}{\gamma ^2\left | \xi \right | ^2} +o\left ( \frac{1}{\left | \xi \right |^2 } \right ) \right )e^{\left ( {-\gamma \left | \xi \right | ^2+\frac{c^2}{\gamma }+o\left ( \frac{1}{\left | \xi \right | } \right ) } \right )t }\\ &+\left ( 1+\frac{1}{\gamma ^2\left | \xi \right | ^2} +o\left ( \frac{1}{\left | \xi \right |^2 } \right ) \right )e^{-\frac{c^2}{\gamma } t+o\left ( \frac{1}{\left | \xi \right | } \right )t }, \end{aligned} \end{equation} $

(3.19)

$ \begin{equation} \frac{e^{\lambda_-t}-e^{\lambda_+t}}{\lambda _+-\lambda_-} =\left ( \frac{1}{\gamma \left | \xi \right | ^2} +o\left ( \frac{1}{\left | \xi \right |^3 } \right ) \right ) \cdot \left ( e^{\left ( -\gamma \left | \xi \right |^2+\frac{c^2}{\gamma}+o\left ( \frac{1}{\left | \xi\right | } \right ) \right )t } -e^{\left ( -\frac{c^2}{\gamma} +o\left ( \frac{1}{\left | \xi \right | } \right )t \right ) }\right ), \end{equation} $

(3.20)

$ \begin{equation} \begin{aligned} \frac{\lambda_+e^{\lambda_+t}-\lambda_-e^{\lambda_-t}}{\lambda_+-\lambda_-}&=\left(-\frac{1}{\gamma^2\left|\xi\right|^2}+o\left(\frac{1}{\left|\xi\right|^2}\right)\right)e^{\left({-\frac{c^2}{\gamma}+o\left(\frac{1}{\left|\xi\right|}\right)}\right)t}\\ &+\left(1+\frac{1}{\gamma^2}\frac{1}{\left|\xi\right|^2}+o\left(\frac{1}{\left|\xi\right|^2}\right)\right)e^{\left(-\gamma\left|\xi\right|^2+\frac{c^2}{\gamma}+o\left(\frac{1}{\left|\xi\right|}\right)\right)t}. \end{aligned} \end{equation} $

(3.21)

所以高频时, 若$ \left |\alpha \right | $大, $ \xi ^\alpha\hat G $关于$ \xi $是不可积分的, 只有(2.2)的初值有光滑性时, 才能得到(2.2)解的光滑性.这就是下面的定理.

定理3.4   当$ \rho_0\in H^{l+s} $, $ m_0\in H^{l+s} $, $ s> 1 $, $ max\left \{ \left \| \rho_0 \right \|_{H^{l+s}}, \left \| m_0 \right \|_{H^{l+s}} \right \}=E $, 当$ \left | \alpha \right | \le l $时, 对任意正整数$ N $, 存在正常数$ C_N $, $ b $, 有$ \left | \partial _x^\alpha \chi _3\left ( D \right ) \rho \right |+\left | \partial _x^\alpha \chi _3\left ( D \right ) m \right | \le CEe^{-bt}B_N\left ( x, t \right ). $

证  由(3.1), (3.19), (3.20), 当$ \left | \alpha \right | \le l $时, 有

$ \begin{equation} \begin{aligned} \notag &\left | x^\beta\partial_x^\alpha\chi _3\left ( D \right )\rho \right | \le\left | \int_{\left | \xi \right |\ge R } \partial_\xi^\beta\left ( \left | \xi \right |^{ \alpha } \hat\rho\right)d\xi\right|\le\int_{\left | \xi \right |\ge R }\left|\xi\right|^{\left | \alpha \right |-\left | \beta \right|}\left(\hat G_{11}\hat \rho_0 +\hat G_{12}\hat m_0\right) d\xi\\ \le& Ce^{-bt}\int_{\left | \xi \right |\ge R } \left|\xi\right|^{\left | \alpha \right |-\left | \beta \right|}\cdot\hat \rho_0+\left | \xi \right |^{\left | \alpha \right |-1-\left | \beta \right | }\hat m_0d\xi \le Ce^{-bt}\int_{\left | \xi \right |\ge R } \left | \xi \right | ^{\left | \alpha \right | }\cdot\hat \rho_0+\left | \xi \right | ^{\left | \alpha \right | -1}\hat m_0d\xi\\ \le& Ce^{-bt}\left ( \left \| \left | \xi \right | ^{-s} \right \| _{L^2\left ( \left | \xi \right |\ge R \right ) } \cdot\left \| \left | \xi \right |^{\left | \alpha \right |+s } \cdot\hat \rho_0 \right \|_{L^2}+\left \| \left | \xi \right | ^{-s} \right \|_{L^2\left ( \left | \xi \right |\ge R \right )} \cdot\left \| \left | \xi \right |^{\left | \alpha \right |+s } \cdot\hat m_0\right \|_{L^2} \right ) \\ \le& Ce^{-bt}\left(\left\|\rho_0\right\|_{H^{l+s}}+\left\| m_0\right\|_{H^{l+s}}\right) . \end{aligned} \end{equation} $

由(3.1), (3.19), (3.20), 当$ \left|\alpha\right|<l $, 有

$ \begin{equation} \begin{aligned} \notag &\left | x^\beta\partial _x^\alpha \chi _3\left ( D \right ) m \right | \le\int_{\left | \xi \right |\ge R } \partial _\xi ^\beta \left ( \xi^\alpha\hat m \right ) d\xi \le Ce ^{-bt}\int _{\left | \xi \right |\ge R }\left | \xi \right | ^{\left | \alpha \right |-\left | \beta \right | -3 }\cdot\hat \rho_0+\left | \xi \right | ^{\left | \alpha \right |-\left | \beta \right | }\hat m_0d\xi \\ \le& Ce^{-bt}\left ( \left \| \left | \xi \right |^{-2} \right \| _{L^2\left({\left | \xi \right |\ge R }\right)} \left \| \left | \xi \right | \cdot\hat \rho_0 \right \| _{L^2}+\left \| \left | \xi \right |^{-s} \right \|_{L^2\left({\left | \xi \right |\ge R }\right)} \left \| \left | \xi \right |^{\left | \alpha \right |+s }\cdot\hat m_0 \right \|_{L^2}\right )\\ \le& Ce^{-bt}E. \end{aligned} \end{equation} $

所以$ \left|\partial_x^\alpha\rho\right|\le CEe^{-bt}B_N\left(x, t\right ) $, $ \left|\partial_x^\alpha m\right|\le CEe^{-bt}B_N\left(x, t\right) $.

为得到解的逐点估计, 还需要下述两个引理.

引理3.1   当$ N $, $ s>1 $, 且$ N>s $, 有$ \int B_N\left(x-y, t\right)\left(1+\left|y\right|^2\right)^{-s}dy\le CB_s\left(x, t\right). $

证  当$ \left | x \right | ^2\le 1+t $, 有

$ \int B_N\left ( x-y, t \right ) \left ( 1+\left | y \right | ^2 \right ) ^{-s}dy\le \int \left ( 1+\left | y \right | ^2 \right )^{-s}dy\le C\le CB_s\left ( x, t \right ). $

当$ \left | x \right | ^2> 1+t $, $ \left | y \right | \le \frac{\left | x \right | }{2} $, 有$ \left | x-y \right | \ge \left | x \right | -\left | y \right | \ge \frac{\left | x \right | }{2}, $所以

$ \int B_N\left ( x-y, t \right ) \left ( 1+\left | y \right | ^2 \right ) ^{-s}dy\le B_N\left ( x, t \right ) \int \left ( 1+\left | y \right | ^2 \right )^{-s}dy\le CB_s\left ( x, t \right ). $

当$ \left | x \right | ^2> 1+t $, $ \left | y \right | > \frac{\left | x \right | }{2} $, 由(3.14), 有

$ \begin{equation} \begin{aligned} \notag \int B_N\left ( x-y, t \right ) \left ( 1+\left | y \right |^2 \right ) ^{-s}dy&\le C\left ( 1+t \right ) \left ( 1+\left | x \right | ^2 \right )^{-s}\\ &\le C \frac{1}{\left ( \frac{1+\left | x \right |^2 }{1+t} \right )^s } \le C\frac{1}{\left ( \frac{\left | x \right |^2 }{1+t} \right )^s }\le CB_s\left ( x, t \right ) . \end{aligned} \end{equation} $

引理3.1得证.

引理3.2  当$ s> 1 $, 有$ \int B_\frac{1}{2}\left ( x-y, t \right ) \left ( 1+\left | y \right | ^2 \right ) ^{-s}dy\le CB_\frac{1}{2}\left ( x, t \right ) $.

证   当$ \left | x \right | ^2\le 1+t $, 有$ \int B_\frac{1}{2}\left ( x-y, t \right ) \left ( 1+\left | y \right | ^2 \right ) ^{-s}dy\le C\le CB_\frac{1}{2}\left ( x, t \right ). $当$ \left | x \right | ^2> 1+t $, $ \left | y \right | \le \frac{\left | x \right | }{2} $, 有$ \left | x-y \right | \ge \left | x \right | -\left | y \right | \ge \frac{\left | x \right | }{2} . $于是

$ \int B _\frac{1}{2} \left ( x-y, t \right )\left ( 1+\left | y \right | ^2 \right ) ^{-s}dy\le CB_\frac{1}{2}\left ( x, t \right ) \int \left ( 1+\left | y \right | ^2 \right ) ^{-s}dy \le CB_\frac{1}{2} \left ( x, t \right ) . $

当$ \left | x \right | ^2> 1+t, \left | y \right | > \frac{\left | x \right | }{2} $, 有$ \left | x-y \right | \le \left | x \right | +\left | y \right | \le 3\left | y \right |, $于是

$ \begin{equation} \begin{aligned} \notag \int B_\frac{1}{2} \left ( x-y, t \right ) \left ( 1+\left | y \right | ^2 \right ) ^{-s}dy& \le \left ( 1+\left | x \right |^2 \right )^{-\frac{1}{2} } \int B_\frac{1}{2} \left ( x-y, t \right ) \left ( 1+\frac{\left | x-y \right |^2 }{3} \right )^{\frac{1}{2} -s} dy\\ &\le C\left ( \left | x \right | ^2 \right ) ^{-\frac{1}{2} }\int_{0}^{+\infty} \left ( 1+r^2 \right ) ^{-\frac{1}{2}}\left ( 1+t\right )^{\left ( \frac{1}{2} -s \right ) }r^{1-2s}\cdot r \left ( 1+t\right )^\frac{1}{2} dr \\ &\le C\left ( \left | x \right |^2 \right ) ^{-\frac{1}{2} }\left ( 1+t \right ) ^{\left ( 1-s \right )}\int_{0}^{+\infty} \left ( 1+r^2 \right ) ^{- \frac{1}{2} }r^{2-2s}dr\\ &\le C\left ( \left | x \right |^2 \right ) ^{-\frac{1}{2} }\le CB_\frac{1}{2} \left ( x, t \right ) . \end{aligned} \end{equation} $

引理3.2得证.

由定理3.1、定理3.2、定理3.3、引理3.1、引理3.2、式(3.1)得如下定理.

定理3.5   当初值$ \left | \rho _0 \right | +\left | m_0\right | \le C\left ( 1+\left | x \right | ^2 \right ) ^{-s} $, $ s> 1 $, 有

$ \left | \partial_x^\alpha\left ( 1-\chi _3\left ( D \right ) \right ) \rho \right | \le C_{N, \alpha }\left ( 1+t \right )^{-\frac{2+\left | \alpha \right | }{2} }B_s\left ( x, t \right), $

$ \left | \partial_x^\alpha\left ( 1-\chi _3\left ( D \right ) \right ) m \right | \le C_{N, \alpha }\left ( 1+t \right )^{-\frac{1+\left | \alpha \right | }{2} }B_\frac{1}{2} \left ( x, t \right). $

综合定理3.4、定理3.5、式(3.1), 得到本文主要结论.

定理3.6   当初值$ \left | \rho _0 \right | +\left | m_0\right | \le C\left ( 1+\left | x \right | ^2 \right ) ^{-s_1} $, $ s_1>1 $, 且$ \rho _0\in H^{l+s_2} $, $ m _0\in H^{l+s_2} $, max$ \left \{ \left \| \rho_0 \right \|_{H^{l+s_2} }, \quad\left \| m_0 \right \|_{H^{l+s_2} }\right \}\le E $, $ s_2>1 $, 当$ \left | \alpha \right | \le l $, 有

$ \left | \partial_x^\alpha\rho \right | \le C\left ( 1+t \right ) ^{-\frac{2+\left | \alpha \right | }{2} }B_{s_1}\left ( x, t \right )\quad, \quad \left | \partial_x^\alpha m \right | \le C\left ( 1+t \right ) ^{-\frac{1+\left | \alpha \right | }{2} }B_\frac{1}{2} \left ( x, t \right ). $

注：由(3.14)及定理3.6, 知$ \left \| m \right \| _{L^2} $关于时间$ t $不衰减, 这是非线性方程组(1.1)存在性做不下去的重要原因之一.