本文主要研究三维水平周期的平板区域

$ \Omega_h=\Omega \times (0, h)=\bigg\{\tilde{x}=(\tilde{x}', \tilde{x}_3): \tilde{x}'=(\tilde{x}_1, \tilde{x}_2) \in\Big[-\frac{\pi}{\alpha_1}, \frac{\pi}{\alpha_1}\Big)\times \Big[-\frac{\pi}{\alpha_2}, \frac{\pi}{\alpha_2}\Big), 0<\tilde{x}_3<h\bigg\} $

内的可压缩Navier-Stokes方程

$ \left\{% \begin{array}{l} \partial_{\tilde{t}}\tilde{\rho} + {\rm div}(\tilde{\rho} \tilde{v})= 0, \\ \tilde{\rho} \partial_{\tilde{t}}\tilde{v}+ \tilde{\rho} \tilde{v}\cdot\nabla\tilde{v} -\mu \Delta \tilde{v}-(\mu+\overline{\mu})\nabla{\rm div}\tilde{v} +\nabla \tilde{P}=\tilde{\rho}\tilde{{\bf g}}, \end{array}% \right. $

(1.1)

这里$\tilde{\rho}=\tilde{\rho}(\tilde{x}, \tilde{t})$, $\tilde{v}=\big(\tilde{v}_1(\tilde{x}, \tilde{t}), \tilde{v}_2(\tilde{x}, \tilde{t}), \tilde{v}_3(\tilde{x}, \tilde{t})\big)^{\perp}$分别表示密度和速度, $\tilde{P} =\tilde{P}(\tilde{\rho})$表示压力且是$\tilde{\rho}$的光滑函数, 满足对任意给定的$\rho_{\ast}>0$都有

$ \tilde{P}'(\rho_{\ast})>0. $

这里$\cdot^{\perp}$表示转置. $\mu$和$\overline{\mu}$表示粘性系数, 满足

$ \mu>0, \;\;\;\; \frac{2}{3}\mu+\overline{\mu}\geq 0. $

假设外力$\tilde{{\bf g}}=ge_1$, 其中$g$是常数, $e_1=(1, 0, 0)^{\perp}$.

首先, 假设流体是由沿着$\tilde{x}_1$方向运动速度为$V$的上板和外力$\tilde{{\bf g}}$所驱动, 其中$V>0$为常数.假设无滑移的边界条件

$ \tilde{v}=Ve_1, \;\;\;\; \mbox{在$ \Sigma_h$上, } $

(1.2)

$ \tilde{v}=0, \;\;\;\; \mbox{在 $\Sigma_b$上, } $

(1.3)

$\Sigma_h=\{\tilde{x}_3=h\}$表示$\Omega_h$的上边界, $\Sigma_b=\{\tilde{x}_3=0\}$是下边界.

方程(1.1)-(1.3)具有形如$\tilde{u}_s=(\tilde{\rho}_s, \tilde{v}_s)$的稳态解

$ \tilde{\rho}_s=\rho_{\ast}, \tilde{v}_s=\left(\frac{V}{h}\tilde{x}_3 +\frac{\rho_{\ast}g}{2\mu}\tilde{x}_3(h-\tilde{x}_3)\right)e_1, $

这就是所谓的平面Couette-Poiseuille流.

对任意的雷诺数, Romanov^[5]证明了不可压Navier-Stokes方程平面Couette流小扰动问题的稳定性.对于可压缩流, 当马赫数和雷诺数满足

$ \gamma^2 <\frac{1}{280}, \;\;\;\; 2\nu^2+\nu\nu'<30\gamma^2\bigg(\frac{1}{280}-\gamma^2\bigg), $

Kagei^[3]证明了可压Navier-Stokes方程平面Poiseuille流的线性不稳定性.本文的主要目的是找到适当的马赫数和雷诺数使得平面Couette-Poiseuille流是不稳定的.

本节首先给出方程(1.1)的无量纲形式, 然后推导其扰动方程, 最后介绍一些本文用到的函数空间的相关知识.

设$V={\frac{\beta\rho_{\ast}g h^2}{2\mu}}$, 其中$\beta\in\Bbb R$.令

$ V_{\ast}=\left\{% \begin{array}{ll} \frac{(1+\beta)^2\rho_{\ast}g h^2}{\mu}, \quad &\beta\in (-3+2\sqrt{2}, 1), \\ \frac{4|\beta|\rho_{\ast}g h^2}{\mu}, \quad &\beta\in (-\infty, -3+2\sqrt{2}]\cup[1, +\infty). \end{array}% \right. $

首先, 引入下面的无维数变量

$ \tilde{x}=hx, \quad \tilde{t}=\frac{h}{V_{\ast}}t, \quad \tilde{\rho} =\rho_{\ast}\rho, \quad \tilde{v}=V_{\ast}v\quad \tilde{P}=\rho_{\ast}V_{\ast}^2P. $

则当$\beta\in(-\infty, -3+2\sqrt{2}]\cup[1, +\infty)$或$\beta\in(-3+2\sqrt{2}, 1)$时, 问题(1.1})在区域$\Omega=[-\frac{\pi}{\alpha_1h}, \frac{\pi}{\alpha_1h})\times[-\frac{\pi}{\alpha_2h}, \frac{\pi}{\alpha_2h})\times (0, 1)$上分别具有如下形式

$ \left\{% \begin{array}{l} \partial_t\rho + {\rm div} (\rho v)= 0, \\ \rho\partial_t v+\rho v\cdot\nabla v +\nabla P(\rho)-\nu \Delta v-(\nu+\overline{\nu})\nabla {\rm div} v=\frac{\nu\rho e_1}{4|\beta|}, \end{array}% \right. $

(2.1)

或

$ \left\{% \begin{array}{l} \partial_t\rho + {\rm div} (\rho v)= 0, \\ \rho\partial_t v+\rho v\cdot\nabla v +\nabla P(\rho)-\nu \Delta v-(\nu+\overline{\nu})\nabla {\rm div} v=\frac{\nu\rho e_1}{(1+\beta)^2}. \end{array}% \right. $

(2.2)

边界条件

$ v=\frac{V}{V_{\ast}}\;\;\;\; \mbox{在$ \Sigma_h $上, } \quad v=0\;\;\;\; \mbox{在$ \Sigma_b$上, } $

(2.3)

且在新坐标下稳态平面Couette-Poiseuille流变成

$ \rho_s=1, v_s=v_s^1e_1=\frac{(1+\beta)x_3-x_3^2}{2(1+\beta)^2}e_1, \ \beta\in (-3+2\sqrt{2}, 1), $

或

$ \rho_s=1, v_s=v_s^1e_1=\frac{(1+\beta)x_3-x_3^2}{8|\beta|}e_1, \ \beta\in (-\infty, -3+2\sqrt{2}]\cup[1, +\infty). $

$\nu$和$\overline{\nu}$是如下给定的无量纲系数

$ \nu=\frac{\mu}{\rho_{\ast}hV_{\ast}}, \quad \overline{\nu}=\frac{\overline{\mu}}{\rho_{\ast}hV_{\ast}}. $

雷诺数$Re$, 第二雷诺数$\overline{Re}$以及马赫数$Ma$分别为

$ Re=\nu^{-1}, \quad \overline{Re}=\overline{\nu}^{-1}, \quad Ma=\frac{1}{\sqrt{P'(1)}}=\frac{V}{\sqrt{\tilde{P}'(\rho_{\ast})}}. $

本文主要考虑$\beta\in (-3+2\sqrt{2}, 1)$时的方程, 对$\beta\in (-\infty, -3+2\sqrt{2}]\cup[1, +\infty)$证明类似.下面我们先推导扰动方程, 令$u=(\phi, \omega)^{\perp}=(\gamma^2(\rho-\rho_s), v-v_s)^{\perp}$带入(2.2)式, 利用$\gamma=Ma^{-1}$, 可得如下的初边值问题:

$ \partial_t\phi+v_s^1\partial_{x_1}\phi+\gamma^2 {\rm div}\omega=f_0, $

(2.4)

$ \partial_t\omega-\nu\Delta\omega-\nu'\nabla {\rm div}\omega +\nabla\phi-\frac{\nu}{(1+\beta)^2\gamma^2}\phi+v_s^1\partial_{x_1}\omega +\partial_{x_3}v_s^1\omega_3e_1={\bf f}, $

(2.5)

$ \omega|_{\Sigma\cup\Sigma_b}=0, $

(2.6)

$ (\phi, \omega)|_{t=0}=(\phi_0, \omega_0). $

(2.7)

这里$\nu'=\nu+\overline{\nu}$, $f_0$和${\bf f}$是非线性项

$ f_0=-{\rm div}(\phi\omega), $

$ {\bf f}=-\omega\cdot\nabla\omega-\frac{\phi}{\gamma^2+\phi}\{\nu\Delta\omega +\nu'\nabla {\rm div} \omega+(P_2(\gamma, \phi)-1)\nabla\phi\}, $

其中

$ P_2(\gamma, \phi)=\frac{1}{\gamma^2}\int_0^1P''(1+\gamma^2\phi\theta){\rm d}\theta. $

在给出主要结果之前, 先引入本文需要的符号和记法. $L^p(\Omega)$和$W^{k, p}(\Omega)$表示一般的Lebesgue空间.特别地, 范数$\|\cdot\|_{L^p(\Omega)}=\|\cdot\|_{L^p}$可简记为$\|\cdot\|_p$.若$u=(\phi, \omega)^{\perp}$, 其中$\phi\in W^{k, p}(\Omega)$和$\omega=(\omega_1, \omega_2, \omega_3)^{\perp}\in W^{l, q}(\Omega)$, 定义范数

$ \|u\|_{W^{k, p}(\Omega)\times W^{l, q}(\Omega)}=\|\phi\|_{W^{k, p}(\Omega)}+\|\omega\|_{W^{l, q}(\Omega)}, $

若$k=l$, $p=q$, 则简记$\|u\|_{W^{k, p}(\Omega)\times W^{k, p}(\Omega)} =\|u\|_{W^{k, p}(\Omega)}$. $L^2(\Omega)$空间的内积表示成

$ (u_1, u_2)=\frac{1}{\gamma^2}\int_{\Omega}\phi_1(x_3)\overline{\phi_2(x_3)}{\rm d}x_3+\int_{\Omega}\omega_1(x_3)\overline{\omega_2(x_3)}{\rm d}x_3, $

其中$u_j=(\phi_j, \omega_j)\in L^2(\Omega)$, $\overline{f}$表示函数$f$的共轭.

$L^p(0, 1)$空间范数记为$|\cdot|_p$, $W^{k, p}(0, 1)$和$H^k(0, 1)$空间范数分别记为$|\cdot|_{W^{k, p}}$和$|\cdot|_{H^k}$. $L^2(0, 1)$空间的内积表示成

$ \langle u_1, u_2\rangle=\frac{1}{\gamma^2}\int_0^1\phi_1(x_3)\overline{\phi_2(x_3)}{\rm d}x_3+\int_0^1\omega_1(x_3)\overline{\omega_2(x_3)}{\rm d}x_3, $

其中$u_j=(\phi_j, \omega_j)\in L^2(0, 1)$.

更进一步, 对$f\in L^1(0, 1)$, $f$在$(0, 1)$上的平均记为$\langle f\rangle$

$ \langle f\rangle=\int_0^1f(x_3){\rm d}x_3. $

$I_k$表示$k$阶单位矩阵. $Q_0$和$\tilde{Q}$是$4\times 4$对角矩阵

$ Q_0={\rm diag}(1, 0, 0, 0), \quad\tilde{Q}={\rm diag}(0, 1, 1, 1). $

对任意函数$f=f(x')$, $x'\in[-\frac{\pi}{\alpha_1h}, \frac{\pi}{\alpha_1h})\times[-\frac{\pi}{\alpha_2h}, \frac{\pi}{\alpha_2h})$, 它的Fourier级数展开系数为$\hat{f}(\alpha_{n_1, n_2})$

$ \hat{f}(\alpha_{n_1, n_2})=\frac{\alpha_1\alpha_2h^2}{4\pi^2}\int_{-\frac{\pi}{\alpha_1h}}^{\frac{\pi}{\alpha_1h}} \int_{-\frac{\pi}{\alpha_2h}}^{\frac{\pi}{\alpha_2h}}e^{-ix'\cdot\alpha_{n_1, n_2} }f(x'){\rm d}x', $

其中$\alpha_{n_1, n_2}=(\alpha_1n_1h, \alpha_2n_2h)$.

首先, 考虑如下的线性化问题

$ \left\{% \begin{array}{l} \partial_tu+Lu=0, \\ (\phi, \omega)|_{t=0}=(\phi_0, \omega_0), \end{array} \right. $

(2.8)

这里

$ L=\left( \begin{array}{cc} v_s^1\partial_{x_1}&\quad \gamma^2{\rm div}\\ \nabla&\quad-\nu\Delta I_3-\nu'\nabla {\rm div} \end{array} \right) +\left( \begin{array}{cc} 0&\quad 0\\ -\frac{\nu}{\gamma^2(1+\beta)^2}&\quad v_s^1\partial_{x_1}I_3+\partial_{x_3}v_s^1e_1e_3^{\perp} \end{array} \right). $

由文献[1]可知$-L$生成$C_0$半群.定义$\alpha=(\alpha_1h, \alpha_2h)$和$|\alpha|=(\alpha_1^2+\alpha_2^2)^{\frac{1}{2}}h$.

下面给出本文的主要定理.

定理2.1 存在常数$r_0>0$和$\eta_0>0$, 使得如果$|\alpha|\leq r_0$, 则有

$ \sigma(-L)\cap\{\lambda\in {\bf C}: |\lambda|\leq\eta_0\}=\{\lambda_{n_1, n_2}: |n_1|=0, 1, \cdots, k_1, \ |n_2|=0, 1, \cdots, k_2\} $

对某些$n_1, n_2\in {\bf N}$, 其中$\lambda_{n_1, n_2}$是$-L$的特征值且满足

$ \lambda_{n_1, n_2}=-\frac{i}{6}(\alpha_1n_1h)+\kappa(\alpha_1n_1h)^2-\frac{\gamma^2}{12\nu}(\alpha_2n_2h)^2+O(|\alpha_{n_1, n_2}|^3). $

当$|\alpha_{n_1, n_2}|\rightarrow 0$, $\beta\in(-3+2\sqrt{2}, 1)$时

$ \kappa=\frac{1}{12\nu(1+\beta)^4}\left(\frac{1-7\beta^2}{280}-\gamma^2-\frac{\nu^2}{15\gamma^2}-\frac{\nu\nu'}{30\gamma^2} -\frac{\beta^2(\nu+\nu')\nu}{4\gamma^2}\right); $

当$|\alpha_{n_1, n_2}|\rightarrow 0$, $\beta\in (-\infty, -3+2\sqrt{2}]\cup[1, +\infty)$时

$ \kappa=\frac{1}{12\nu|\beta|^2}\left(\frac{1-7\beta^2}{280}-\gamma^2-\frac{\nu^2}{15\gamma^2}-\frac{\nu\nu'}{30\gamma^2} -\frac{\beta^2(\nu+\nu')\nu}{4\gamma^2}\right). $

由此可知, 当$-\frac{\sqrt{7}}{7} <\beta<\frac{\sqrt{7}}{7}$, 且马赫数和雷诺数满足

$ \gamma^2 <\frac{1-7\beta^2}{280}, \frac{15\beta^2(\nu^2+\nu\nu')}{2}+2\nu^2+\nu\nu'<30\gamma^2 \Big(\frac{1-7\beta^2}{280}-\gamma^2\Big), $

则可得$\kappa>0$, 即平面Couette-Poiseuille流是线性不稳定的.

这一节主要是对线性化方程关于$x'\in [-\frac{\pi}{\alpha_1h}, \frac{\pi}{\alpha_1h})\times[-\frac{\pi}{\alpha_2h}, \frac{\pi}{\alpha_2h})$做Fourier级数展开, 且仅考虑其Fourier系数$\hat{u}(\alpha_{n_1, n_2})$的方程.

对方程(2.8)做Fourier级数展开, 令$\xi=\alpha_{n_1, n_2}$可得

$ \partial_t\hat{u}+\widehat{L}_{\xi}\hat{u}=0, \quad \hat{u}|_{t=0}=\hat{u}_0, $

其中$\widehat{L}_{\xi}$是$L^2(0, 1)$上的算子, 定义域为

$ D(\widehat{L}_{\xi})=\{\hat{u}=(\hat{\phi}, \hat{\omega})^{\perp}\in L^2(0, 1): \hat{\omega}\in H_0^1(0, 1), \widehat{L}_{\xi}\hat{u}\in L^2(0, 1)\}. $

且有如下形式

$ \widehat{L}_{\xi}=A_{\xi}+B_{\xi}+C_0. $

(3.1)

这里

$ A_{\xi}=\left( \begin{array}{ccc} 0&\quad 0 &\quad 0\\ 0&\quad (\nu(|\xi|^2-\partial_{x_3}^2))I_2+\nu'\xi\xi^{\perp} &\quad-i\nu'\xi\partial_{x_3}\\ 0&\quad-i \nu'\xi^{\perp}\partial_{x_3} &\quad \nu(|\xi|^2-\partial_{x_3}^2)-\nu'\partial_{x_3}^2 \end{array} \right), $

$ B_{\xi}=\left( \begin{array}{ccc} i v_s^1\xi_1&\quad i{\gamma^2}\xi^{\perp} &\quad \gamma^2\partial_{x_3}\\ i\xi&\quad i v_s^1\xi_1I_2 &\quad 0\\ \partial_{x_3} &\quad 0 &\quad i v_s^1 \xi_1 \end{array} \right) $

和

$ C_0=\left( \begin{array}{ccc} 0&\quad 0 &\quad 0\\[2mm] -\frac{\nu}{(1+\beta)^2\gamma^2}e_1'&\quad 0 &\quad \partial_{x_3}v_s^1e_1'\\[2mm] 0 &\quad 0 &\quad 0 \end{array} \right). $

注意到对任意的$\xi\in{\bf R}^2$都有$D(\widehat{L}_{\xi})=D(\widehat{L}_{0})$.

算子$\widehat{L}_{\xi}$关于内积$\langle \cdot, \cdot\rangle$的共轭算子$\widehat{L}^{\ast}_{\xi}$可表示成

$ \widehat{L}^{\ast}_{\xi}=A_{\xi}-B_{\xi}+C_0^{\ast}, $

其中

$ C_0^{\ast}=\left( \begin{array}{ccc} 0&\quad -\frac{\nu}{(1+\beta)^2\gamma^2}e_1'^{\perp} &\quad 0\\[2mm] 0&\quad 0 &\quad 0\\ 0 &\quad \quad \partial_{x_3}v_s^1e_1'^{\perp} &\quad 0 \end{array} \right). $

且算子$\widehat{L}^{\ast}_{\xi}$的定义域为

$ D(\widehat{L}^{\ast}_{\xi})=\{\hat{u}=(\hat{\phi}, \hat{\omega})^{\perp}\in L^2(0, 1): \hat{\omega}\in H_0^1(0, 1), \widehat{L}^{\ast}_{\xi}\hat{u}\in L^2(0, 1)\}. $

为了证明定理2.1, 我们先看当$\xi=0$时算子$-\widehat{L}_{0}$的谱

$ \widehat{L}_0=\left( \begin{array}{ccc} 0&\quad 0 &\quad \gamma^2\partial_{x_3}\\[2mm] -\frac{\nu}{(1+\beta)^2\gamma^2}e_1'&\quad -\nu\partial_{x_3}^2I_2 &\quad \partial_{x_3}v_s^1e_1'\\[2mm] \partial_{x_3} &\quad 0 &\quad -(\nu+\nu')\partial_{x_3}^2 \end{array} \right). $

由于本文是关注于平面Couette-Poiseuille流的不稳定性, 故我们只考虑原点附近的特征值.下面我们给出一个引理, 其证明可参考文献[3].

引理3.1   (1)  存在常数$\eta_1=\eta_1(\nu, \nu', \gamma)>0$, 使得$\{\lambda\in {\bf C}:|\lambda| <\eta_1\}\setminus\{0\}\subset\varrho(-\widehat{L}_{0})$.特别地, 下面的预解式估计对任意的$\lambda\in\{\lambda\in {\bf C}:|\lambda|<\frac{\eta_1}{2}\}\setminus\{0\}$一致成立, 即

$ \mid(\lambda+\widehat{L}_{0})^{-1}f\mid_{L^2}+ \mid \tilde{Q}(\lambda+\widehat{L}_{0})^{-1}f\mid_{L^2} \leq\frac{C}{|\lambda|}\mid f\mid_{L^2}. $

此性质对$\widehat{L}_{0}^{\ast}$同样成立.

(2)   $\lambda=0$是$-\widehat{L}_{0}$的单重特征值, 即${\cal R}(-\widehat{L}_{0})$是闭的且

$ L^2(0, 1)=Ker(-\widehat{L}_{0})\oplus {\cal R}(-\widehat{L}_{0}), \ \dim Ker(-\widehat{L}_{0})=1. $

此性质对$\widehat{L}_{0}^{\ast}$同样成立.

(3)   算子$-\widehat{L}_{0}$和$\widehat{L}_{0}^{\ast}$的$0$特征值所对应的特征空间分别为$u^{(0)}$和$u^{(0)\ast}$, 这里

$ u^{(0)}=(\phi^{(0)}, \omega^{(0)})^{\perp}, \ \omega^{(0)}=(\omega^{(0), 1}, 0, 0)^{\perp}, $

$ u^{(0)\ast}=(\phi^{(0)\ast}, \omega^{(0)\ast})^{\perp}, \ \omega^{(0)\ast}=(0, 0, 0)^{\perp}, $

其中

$ \phi^{(0)}(x_3)=1, \ \omega^{(0), 1}(x_3)=\frac{1}{2\gamma^2(1+\beta)^2}(x_3-x_3^2), \ \phi^{(0)\ast}(x_3)=\gamma^2. $

(4)   算子$-\widehat{L}_{0}$和$\widehat{L}_{0}^{\ast}$的$0$特征值所对应的特征空间上的投影$\hat{\Pi}^{(0)}$和$\hat{\Pi}^{(0)\ast}$为

$ \hat{\Pi}^{(0)}u=\langle u, u^{(0)\ast}\rangle u^{(0)}=\langle \phi\rangle u^{(0)}, \quad \hat{\Pi}^{(0)\ast}u=\langle u, u^{(0)}\rangle u^{(0)\ast}. $

特别地

$ u=(\phi, \omega)^{\perp}\in {\cal R}(I-\hat{\Pi}^{(0)})\ \mbox{当且仅当}\ \langle \phi\rangle= \langle u, u^{(0)\ast}\rangle=0. $

本节主要计算当$|\xi|\ll 1$时, 算子$-\widehat{L}_{\xi}$在$\{|\lambda|\leq\frac{\eta_1}{2}\}$内的特征值.

先把算子$\widehat{L}_{\xi}$写成如下形式

$ \widehat{L}_{\xi}=\widehat{L}_0+\sum\limits_{j=1}^{2}\xi_j\widehat{L}_j^{(1)}+\sum\limits_{j, k=1}^{2} \xi_j\xi_k\widehat{L}_{jk}^{(2)}, $

其中

$ \widehat{L}_0=\left( \begin{array}{ccc} 0&\quad 0 &\quad \gamma^2\partial_{x_3}\\[2mm] -\frac{\nu}{(1+\beta)^2\gamma^2}e_1'&\quad -\nu\partial_{x_3}^2I_2 &\quad \partial_{x_3}v_s^1e_1'\\[2mm] \partial_{x_3} &\quad 0 &\quad -(\nu+\nu')\partial_{x_3}^2 \end{array} \right), $

$ \widehat{L}_j^{(1)}=\left( \begin{array}{ccc} i v_s^1\delta_{1j}&\quad i{\gamma^2}e'_j{^{\perp}} &\quad 0\\ ie'_j &\quad i v_s^1\delta_{1j}I_2 &\quad -i\nu'e'_j\partial_{x_3}\\ 0 &\quad -i\nu'e'_j{^{\perp}}\partial_{x_3} &\quad i v_s^1\delta_{1j} \end{array} \right), \ j=1, 2, $

$ \widehat{L}_{jk}^{(2)}=\left( \begin{array}{ccc} 0&\quad 0 &\quad 0\\ 0&\quad \nu\delta_{jk}I_2+ \nu'{e'_j}e'_k{^{\perp}}&\quad 0\\ 0 &\quad 0 &\quad \nu\delta_{jk} \end{array} \right), \ j, k=1, 2, $

其中$e_1'=(1, 0)^{\perp}, \ e_2'=(0, 1)^{\perp}$.

基于引理3.1, 下面我们给出当$|\xi|$充分小时, 算子$-\widehat{L}_{\xi}$的预解式估计.

引理4.1   设$\eta_1$是引理3.1中给定的常数, 则存在$r_0=r_0(\eta_0)$, 使得当$|\xi|\leq r_0$时, 下面的预解式估计对$\{|\lambda|=\frac{\eta_1}{2}\}$成立

$ \mid(\lambda+\widehat{L}_{\xi})^{-1}f\mid_{L^2}+ \mid \tilde{Q}(\lambda+\widehat{L}_{\xi})^{-1}f\mid_{L^2} \leq\frac{C}{|\lambda|}\mid f\mid_{L^2}. $

证  引理4.1的证明类似于文献[2]中的定理$3.2$.这里仅给出简要证明思路.

首先, 令

$ \widehat{L}^{(1)}(\xi)=\sum\limits_{j=1}^{2}\xi_j\widehat{L}_j^{(1)}, \quad \widehat{L}^{(2)}(\xi) =\sum\limits_{j, k=1}^{2}\xi_j\xi_k\widehat{L}_{jk}^{(2)}. $

若$\lambda$满足$\{|\lambda|=\frac{\eta_1}{2}\}$, 则由引理3.1可得存在$r_0=r_0(\eta_0)$, 使得当$|\xi|\leq r_0$时

$ \mid \left(\widehat{L}^{(1)}(\xi)+\widehat{L}^{(2)}(\xi)\right)(\lambda+\widehat{L}_{0})^{-1} f\mid_{L^2} \leq \frac{1}{2}\mid f \mid_{L^2}. $

因此, 当$|\xi|\leq r_0$, $\{|\lambda|=\frac{\eta_1}{2}\}$时, 算子

$ I+\left(\widehat{L}^{(1)}(\xi)+\widehat{L}^{(2)}(\xi)\right)(\lambda+\widehat{L}_{0})^{-1} $

是$L^2(0, 1)$上的可逆算子.利用Neumann级数展开可得

$ (\lambda+\widehat{L}_{\xi})^{-1}=(\lambda+\widehat{L}_{0})^{-1}\sum\limits_{N=0}^{\infty}(-1)^N [\left(\widehat{L}^{(1)}(\xi)+\widehat{L}^{(2)}(\xi)\right)(\lambda+\widehat{L}_{0})^{-1}]^N. $

利用引理3.1直接估计即可得证.

下面我们利用解析算子的扰动理论来证明定理2.1.

定理4.2   存在常数$r_1=r_1(\nu, \nu', \gamma)>0$, 使得若$|\xi|\leq r_1$, 则

$ \sigma(-\widehat{L}_{\xi})\cap \Big\{\lambda\in {\cal C}:|\lambda|\leq\frac{\eta_1}{2}\Big\}=\{\lambda_0(\xi)\}, $

这里$\lambda_0(\xi)$是算子$-\widehat{L}_{\xi}$的单重特征值, 且满足

$ \lambda_0(\xi)=-\frac{(2+3\beta)i}{12(1+\beta)^2}\xi_1+\kappa\xi_1^2-\frac{\gamma^2}{12\nu}\xi_2^2+O(|\xi|^3). $

其中

$ \kappa=\frac{1}{12\nu(1+\beta)^4}\left(\frac{1-7\beta^2}{280}-\gamma^2-\frac{\nu^2}{15\gamma^2}-\frac{\nu\nu'}{30\gamma^2} -\frac{\beta^2(\nu+\nu')\nu}{4\gamma^2}\right). $

证  利用引理3.1和引理4.1, 利用解析算子的扰动理论(参考[4])可得, 存在常数$r_1=r_1(\nu, \nu', \gamma)>0$, 使得若$|\xi|\leq r_1$, 则

$ \sigma(-\widehat{L}_{\xi})\cap \Big\{\lambda\in {\cal C}:|\lambda|\leq\frac{\eta_1}{2}\Big\}=\{\lambda_0(\xi)\}, $

这里$\lambda_0(\xi)$是算子$-\widehat{L}_{\xi}$的单重特征值.与此同时

$ \lambda_0(\xi)=\lambda_0+\sum\limits_{j=1}^{2}\xi_j\lambda_j^{(1)} +\sum\limits_{j, k=1}^{2}\xi_j\xi_k\lambda_{jk}^{(2)}+O(|\xi|^3). $

这里

$ \lambda_0=0, \quad \lambda_j^{(1)}=-\langle \widehat{L}_j^{(1)}u^{(0)}, u^{(0)\ast}\rangle, $

$ \lambda_{jk}^{(2)}=-\Big\langle \frac{1}{2}(\widehat{L}_{jk}^{(2)} +\widehat{L}_{kj}^{(2)})u^{(0)}, u^{(0)\ast}\Big\rangle +\Big\langle \frac{1}{2}(\widehat{L}_j^{(1)}\hat{S}\widehat{L}_k^{(1)} +\widehat{L}_k^{(1)}\hat{S}\widehat{L}_j^{(1)})u^{(0)}, u^{(0)\ast}\Big\rangle, $

其中$\hat{S}=[(I-\Pi^{(0)})\widehat{L}_0(I-\Pi^{(0)})]^{-1}$.

只要计算出$\lambda_j^{(1)}$和$\lambda_{jk}^{(2)}$, 定理4.2即可得证.

下面首先计算$\lambda_j^{(1)}$.

命题4.3    $\lambda_j^{(1)}=-\frac{(2+3\beta)i}{12(1+\beta)^2}\delta_{1j}, \quad j=1, 2.$

证  经过计算, 我们有

$ \widehat{L}_1^{(1)}u^{(0)}=i\left( \begin{array}{ccc} v_s^1+\gamma^2\omega^{(0), 1}\\ (1+v_s^1\omega^{(0), 1})e'_1 \\ -\nu'\partial_{x_3}\omega^{(0), 1} \end{array} \right) =i\left( \begin{array}{ccc} \frac{(2+\beta)x_3-2x_3^2}{2(1+\beta)^2}\\[4mm] 1+\frac{(1+\beta)x_3^2-(2+\beta)x_3^3+x_3^4}{4(1+\beta)^4\gamma^2} \\[3mm] 0\\[2mm] -\frac{\nu'(1-2x_3)}{2(1+\beta)^2\gamma^2} \end{array} \right), $

$ \widehat{L}_2^{(1)}u^{(0)}=i\left( \begin{array}{ccc} 0\\ e'_2 \\ 0 \end{array} \right)=i\left( \begin{array}{ccc} 0\\ 0 \\ 1\\ 0 \end{array} \right). $

由此可得

$ \lambda_1^{(1)}=-\langle \widehat{L}_1^{(1)}u^{(0)}, u^{(0)\ast}\rangle= -i\int_0^1\frac{(2+\beta)x_3-2x_3^2}{2(1+\beta)^2}{\rm d}x_3=-\frac{(2+3\beta)i}{12(1+\beta)^2}, $

$ \lambda_2^{(1)}=-\langle \widehat{L}_2^{(1)}u^{(0)}, u^{(0)\ast}\rangle=0. $

命题4.3证毕.

命题4.4    $\lambda_{22}^{(2)}=-\frac{\gamma^2}{12\nu}$.

证  利用$\widehat{L}_{jk}^{(2)}u^{(0)}=(0, \ast, \ast, \ast)^{\perp}$, 可得对$j, k=1, 2$有

$ \langle \widehat{L}_{jk}^{(2)}u^{(0)}, u^{(0)\ast}\rangle=0. $

(4.1)

下面计算$\widehat{L}_j^{(1)}\hat{S}\widehat{L}_2^{(1)}u^{(0)}$.易知$\widehat{L}_2^{(1)}u^{(0)}\in {\cal R}(I-\hat{\Pi}^{(0)})$, 所以$\hat{S}\widehat{L}_2^{(1)}u^{(0)}$是下面关于$u=(\phi, \omega)^{\perp}$的方程唯一解

$ \widehat{L}_0u=\widehat{L}_2^{(1)}u^{(0)}, \quad \langle \phi\rangle=0. $

解上述方程可得$\phi=0, \omega_1=0, \ \omega_3=0$, $\omega_2$满足方程$ -\nu\partial_{x_3}^2\omega_2=i, $ $ \omega_2\mid_{x_3=0, 1}=0. $由此可得

$ \hat{S}\widehat{L}_2^{(1)}u^{(0)}=\Big(0, 0, \frac{i}{2\nu}(x_3-x_3^2), 0\Big)^{\perp}. $

计算可得

$ \widehat{L}_1^{(1)}\hat{S}\widehat{L}_2^{(1)}u^{(0)}=(0, \ast, \ast, \ast)^{\perp}, $

(4.2)

$ \widehat{L}_2^{(1)}\hat{S}\widehat{L}_2^{(1)}u^{(0)}= \Big(\frac{i}{2\nu}(x_3-x_3^2), \ast, \ast, \ast\Big)^{\perp}. $

(4.3)

利用(4.1)和(4.3)式可得

$ \lambda_{22}^{(2)}=-\frac{\gamma^2}{2\nu}\int_0^1(x_3-x_3^2){\rm d}x_3=-\frac{\gamma^2}{12\nu}. $

命题4.4证毕.

为了得到$\lambda_{1j}^{(2)}$, 下面先计算$\hat{S}\widehat{L}_1^{(1)}u^{(0)}$.

命题4.5

$ \hat{S}\widehat{L}_1^{(1)}u^{(0)}=(\phi^{(1)}, \omega_1^{(1)}, \omega_2^{(1)}, \omega_3^{(1)}), $

其中

$ \phi^{(1)}=\frac{(\nu+\nu')i}{\gamma^2}\left(-\frac{x_3^2}{(1+\beta)^2}+\frac{(2+\beta)x_3}{2(1+\beta)^2} -\frac{2+3\beta}{12(1+\beta)^2}\right)-\frac{\nu'i}{2(1+\beta)^2\gamma^2} \bigg(x_3-x_3^2-\frac{1}{6}\bigg), $

$ \omega_1^{(1)}=-\frac{i}{2\nu}(x_3^2-x_3) +\frac{i}{24\gamma^2(1+\beta)^4\nu} \bigg(-\frac{1}{15}x_3^6+\frac{1+\beta}{5}x_3^5\\ \;\;\;\;\;\;\;\;\;-\frac{3\beta^2+6\beta+2}{12}x_3^4 +\frac{\beta(2+3\beta)}{6}x_3^3 -\frac{15\beta^2+2\beta-2}{60}x_3\bigg)\\ \;\;\;\;\;\;\;\;\;+\frac{i}{12\gamma^4(1+\beta)^4}\bigg(\nu\Big(-(2+\beta)x_3^3+x_3^4+ \frac{2+3\beta}{2}x_3^2-\frac{\beta x_3}{2}\Big) \\ \;\;\;\;\;\;\;\;\;+\nu'\Big(\frac{x_3^4}{2}-(1+\beta)x_3^3+\frac{1+3\beta}{2}x_3^2-\frac{\beta x_3}{2} \Big)\bigg), $

$ \omega_2^{(1)}=0, $

$ \omega_3^{(1)}=\frac{i}{\gamma^2}\left(-\frac{x_3^3}{3(1+\beta)^2}+ \frac{(2+\beta)x_3^2}{4(1+\beta)^2} -\frac{(2+3\beta)x_3}{12(1+\beta)^2}\right). $

证 $\hat{S}\widehat{L}_1^{(1)}u^{(0)}$是下面关于$u$的方程唯一解

$ \widehat{L}_0u=(I-\hat{\Pi}^{(0)})\widehat{L}_1^{(1)}u^{(0)}, \ \langle\phi\rangle=0. $

(4.4)

通过计算得

$ (I-\hat{\Pi}^{(0)})\widehat{L}_1^{(1)}u^{(0)}=i\left( \begin{array}{ccc} \frac{(2+\beta)x_3-2x_3^2}{2(1+\beta)^2}-\frac{2+3\beta}{12(1+\beta)^2}\\ 1+\frac{(8+9\beta)x_3^2-6(2+\beta)x_3^3+6x_3^4-(2+3\beta)x_3}{24(1+\beta)^4\gamma^2} \\[3mm] 0\\[2mm] -\frac{\nu'(1-2x_3)}{2(1+\beta)^2\gamma^2} \end{array} \right), $

则方程(4.4)可写成如下方程组的形式

$ \gamma^2\partial_{x_3}\omega_3=\frac{(2+\beta)x_3-2x_3^2}{2(1+\beta)^2}i-\frac{2+3\beta}{12(1+\beta)^2}i, $

(4.5)

$ -\frac{\nu}{\gamma^2(1+\beta)^2}\phi-\nu\partial_{x_3}^2\omega_1+(\partial_{x_3}v_s^1)\omega_3 =i+i\frac{(8+9\beta)x_3^2-6(2+\beta)x_3^3+6x_3^4-(2+3\beta)x_3}{24(1+\beta)^4\gamma^2}, $

(4.6)

$ -\nu\partial_{x_3}^2\omega_2=0, $

(4.7)

$ \partial_{x_3}\phi-(\nu+\nu')\partial_{x_3}^2\omega_3=-\frac{\nu'(1-2x_3)i}{2(1+\beta)^2\gamma^2}, $

(4.8)

$ \omega\mid_{x_3=0, 1}=0, $

(4.9)

$ \langle\phi\rangle=0. $

(4.10)

由(4.7)和(4.9)式可得$\omega_2=0$.

对(4.5)式关于$x_3$从$0$到$x_3$直接积分可得

$ \omega_3=\frac{i}{\gamma^2}\left(-\frac{x_3^3}{3(1+\beta)^2}+\frac{(2+\beta)x_3^2}{4(1+\beta)^2} -\frac{(2+3\beta)x_3}{12(1+\beta)^2}\right). $

(4.11)

利用(4.8)-(4.11)式可得

$ \phi=\frac{(\nu+\nu')i}{12(1+\beta)^2\gamma^2}\big(6(2+\beta)x_3-12x_3^2 -2-3\beta\big)-\frac{\nu'i}{2(1+\beta)^2\gamma^2}\bigg(x_3-x_3^2-\frac{1}{6}\bigg). $

(4.12)

由方程(4.6)可得

$ \partial_{x_3}^2\omega_1 = -\frac{i}{\nu}-i\frac{(8+9\beta)x_3^2-6(2+\beta)x_3^3+6x_3^4-(2+3\beta)x_3}{24(1+\beta)^4\gamma^2\nu} \\ \;\;\;\;\;\;\;\;\;\;\;\;-\frac{1}{\gamma^2(1+\beta)^2}\phi +\frac{(\partial_{x_3}v_s^1)}{\nu}\omega_3. $

利用(4.11)-(4.12)式, 以及边界条件(4.9), 经过复杂计算可得

$ \omega_1^{(1)} = -\frac{i}{2\nu}(x_3^2-x_3) -\frac{i}{24\gamma^2(1+\beta)^4\nu} \bigg(-\frac{1}{15}x_3^6+\frac{1+\beta}{5}x_3^5 \\ \;\;\;\;\;\;\;\;\;\;-\frac{3\beta^2+6\beta+2}{12}x_3^4+\frac{\beta(2+3\beta)}{6}x_3^3 -\frac{15\beta^2+2\beta-2}{60}x_3\bigg)\\ \;\;\;\;\;\;\;\;\;\;+\frac{i}{12\gamma^4(1+\beta)^4}\bigg(\nu \Big(-(2+\beta)x_3^3+x_3^4+\frac{2+3\beta}{2}x_3^2-\frac{\beta x_3}{2}\Big) \\ \;\;\;\;\;\;\;\;\;\;+\nu'\Big(\frac{x_3^4}{2}-(1+\beta)x_3^3+\frac{1+3\beta}{2}x_3^2-\frac{\beta x_3}{2}\Big)\bigg). $

命题4.5证毕.

命题4.6    $\lambda_{1j}^{(2)}=\lambda_{j1}^{(2)}=\kappa\delta_{1j}, \ j=1, 2.$这里

$ \kappa=\frac{1}{12\nu(1+\beta)^4}\left(\frac{1-7\beta^2}{280}-\gamma^2-\frac{\nu^2}{15\gamma^2}-\frac{\nu\nu'}{30\gamma^2} -\frac{\beta^2(\nu+\nu')\nu}{4\gamma^2}\right). $

证  由命题4.5可知$\widehat{L}_2^{(1)}\hat{S}\widehat{L}_1^{(1)}u^{(0)}=0$, 所以下面式子成立

$ \langle\widehat{L}_2^{(1)}\hat{S}\widehat{L}_1^{(1)}u^{(0)}, u^{(0)\ast}\rangle=0. $

联立上式和(4.1)-(4.2)式可得

$ \lambda_{12}^{(2)}=\lambda_{21}^{(2)}=0. $

下面计算$\lambda_{11}^{(2)}$, 注意到(4.1)式, 则有

$ \lambda_{11}^{(2)}=\langle\widehat{L}_1^{(1)}\hat{S}\widehat{L}_1^{(1)}u^{(0)}, u^{(0)\ast}\rangle= i\langle v_s^1\phi^{(1)}+\gamma^2\omega_1^{(1)}\rangle. $

利用命题4.5中的$\phi^{(1)}$和$\omega_1^{(1)}$的表达式, 经过复杂计算可得

$ \lambda_{11}^{(2)}=\frac{1}{12\nu(1+\beta)^4}\left(\frac{1-7\beta^2}{280}-\gamma^2-\frac{\nu^2}{15\gamma^2}-\frac{\nu\nu'}{30\gamma^2} -\frac{\beta^2(\nu+\nu')\nu}{4\gamma^2}\right). $

命题4.6证毕.

定理2.1的证明  结合命题4.3-4.6, 即可证明定理4.2, 再利用线性算子Fourier变换的基本性质和定理4.2即可证明定理2.1.