数学物理学报, 2023, 43(2): 505-514

半无穷柱体上Forchheimer方程组解的Phragmén-Lindelöf型二择一结果

陈雪姣,, 李远飞,*, 侯春娟

广州华商学院应用数学系 广州 511300

Phragmén-Lindelöf Type Results for the Solutions of Forchheimer Equations on a Semi-Infinite Cylinder

Chen Xuejiao,, Li Yuanfei,*, Hou Chunjuan

Department of Apllied Mathematics, Guangzhou Huashang College, Guangdong 511300

通讯作者: *李远飞，E-mail: liqfd@163.com

收稿日期: 2021-09-10   修回日期: 2022-09-20  

基金资助: 广东省普通高校重点项目(自然科学)(2019KZDXM042)
广州华商学院科研团队项目(2021HSKT01)

Received: 2021-09-10   Revised: 2022-09-20  

Fund supported: Key Projects of Universities in Guangdong Province(2019KZDXM042)
Research Team Project of Guangzhou Huashang College(2021HSKT01)

作者简介 About authors

陈雪姣，E-mail:1032170427@qq.com

摘要

考虑了定义在三维半无限柱体上多孔介质中的Forchheimer流体. 建立了一个能量函数, 推导了一个关于该能量函数的微分不等式, 由此不等式得到了解的二择一结果. 在衰减的情况下, 通过设置一个大于零的参数, 得到了解的快速衰减率.

关键词： Forchheimer方程组; 微分不等式; 二择一

Abstract

The Forchheimer fluid defined in porous media on a three-dimensional semi-infinite cylinder is considered. An energy function is established, and a differential inequality about the energy function is derived. From the inequality, an alternative result of the solutions is obtained. In the case of decay, the fast decay rate of the solutions is obtained by setting a positive parameter.

Keywords： Forchheimer matrixs; Differential inequality technique; Alternative

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本文引用格式

陈雪姣, 李远飞, 侯春娟. 半无穷柱体上Forchheimer方程组解的Phragmén-Lindelöf型二择一结果[J]. 数学物理学报, 2023, 43(2): 505-514

Chen Xuejiao, Li Yuanfei, Hou Chunjuan. Phragmén-Lindelöf Type Results for the Solutions of Forchheimer Equations on a Semi-Infinite Cylinder[J]. Acta Mathematica Scientia, 2023, 43(2): 505-514

1 引言

众所周知的Forchheimer方程以及Brinkman, Darcy, Stokes方程组主要用于描述多孔介质中的流体, Nield和Bejan[1]以及Straughan[2]对此做了详细的论述. 在过去的几十年, 这些方程在半无穷柱体上的空间衰减性得到了广泛关注, 出现了大量的成果. Payne和Song[3]研究了定义在半无穷柱体上的多孔Forchheimer流, 得到了Saint-Venant原理型衰减估计. Li等人[4]进一步研究了Brinkman-Forchheimer方程组解的空间衰减估计. 更多的成果见文献[5⇓⇓⇓⇓-10]. 这种类型的研究通常被称为Saint-Venant原理型研究, 在推导解的衰减性时需要解在无穷远处满足一定的先验假设.

经典的Phragmén-Lindelöf型二择一定理不必假设方程组的解在无限端趋近于零, 而是证明方程组的解随距有限端的距离的增大要么呈指数(多项式)增长要么呈指数(多项式)衰减. 由于其在弹性力学、流体力学等领域上的广泛应用, 自上世纪九十年代以来Phragmén-Lindelöf型二择一研究逐步成为了研究的热点(见文献[11⇓⇓⇓⇓⇓⇓⇓⇓-20]). 上述文献大多研究的是线性微分方程, 非线性方程解的二择一性质还没有得到充分关注, 本文研究半无穷柱体上Forchheimer方程控制的流体解的空间性质. 在衰减的情况下, 通过设置一个大于零的任意常数, 得到了解的快速衰减率. 由于Forchheimer方程中具有非线性项, 所以文献中的结果并不能直接推广到本文中来. 因此, 本文的研究对其他类型的偏微分方程解的二择一研究提供借鉴.

2 准备工作

令$R$表示三维区域上母线平行于坐标轴$x_3$的半无限的柱体

$R=\Big\{(x_1,x_2,x_3)|(x_1,x_2)\in D,\ x_3\geq0\Big\},$

其中$D$是坐标平面$x_1Ox_2$上一个有界的区域, 并且光滑的边界$\partial D$. 设$z$是$x_3$轴上的一个动点. 我们令$R_z$表示$R$的一个子区域, $D_z$表示$x_3=z$处的截面, 即

$R_z=\Big\{(x_1,x_2,x_3)|(x_1,x_2)\in D,\ x_3\geq z\geq0\Big\},$
$D_z=\Big\{(x_1,x_2,x_3)|(x_1,x_2)\in D,\ x_3=z\Big\}.$

设$u_i(i=1,2,3), T, p$分别表示流体的速度、温度和压力. 本文研究的多孔介质中的Forchheimer流体可以写为

$-\Delta u_i +(1+\gamma T)u_i=-p_{,i}+g_iT,\ \ \mbox{在}\ R\times\{t>0\},$
$u_{i,i}=0,\ \ \mbox{在}\ R\times\{t>0\},$
$\partial_{t}T+u_iT_{,i}=\Delta T, \ \ \mbox{在}\ R\times\{t>0\},$
$u_i=0, T=0,\ \ \ \mbox{在}\ \partial D\times\{x_3>0\}\times\{t>0\},$
$u_{3}(x_1, x_2, 0, t)=f(x_1, x_2, t), \ \ \ mbox{在}\ D\times\{t>0\},$
$T(x_1, x_2, 0, t)=h(x_1, x_2, t),\ \ \ \mbox{在}\ D\times\{t>0\},$
$T(x_1, x_2, x_3, 0)=0, \ \ \ \mbox{在}\ R, $

其中$\Delta$是拉普拉斯算子, $\gamma$是大于零的常数, $g_i$是重力函数, 不失一般性假设$g_ig_i\leq1$. $f$和$h$是给定的已知函数, 并在柱体的侧面上满足兼容性条件. 在方程(2.1)-(2.7)中我们用$i$表示对$x_i$求导, 利用重复英文字母表示从1到3求和, 重复希腊字母表示从1到2求和. 例如: $u_{i,j}u_{i,j}=\sum\limits_{i, j=1}^3\Big(\frac{\partial u_i}{\partial x_j}\Big)^2, u_{\alpha,\beta}u_{\alpha,\beta}=\sum\limits_{\alpha,\beta=1}^2\Big(\frac{\partial u_\alpha}{\partial x_\beta}\Big)^2$.

为了得到二择一结果, 我们建立以下辅助函数

$\begin{matrix} F(z, t)&=&-\delta_2\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}pu_3{\rm d}A{\rm d}\eta+\delta_2\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_{i,3}u_i{\rm d}A{\rm d}\eta \nonumber\\ &&-\frac{1}{2}\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_3T^2{\rm d}A{\rm d}\eta+\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}T_{,3}T{\rm d}A{\rm d}\eta\nonumber\\ &\doteq& I_1+I_2+I_3+I_4, \end{matrix} $

其中$\delta_1, \delta_2$是大于零的常数. 利用散度定理和方程(2.1)-(2.7), 可得

$\begin{matrix} F(z, t)-F(z_0, t)&=&-\delta_2\int_{0}^{t}\int_{z_0}^z\int_{D_\xi}e^{-\delta_1\eta}p_{,i}u_i{\rm d}A {\rm d}\xi {\rm d}\eta +\delta_2\int_{0}^{t}\int_{z_0}^z\int_{D_\xi}e^{-\delta_1\eta}(u_{i,j}u_i)_{,j}{\rm d}A{\rm d}\xi {\rm d}\eta \nonumber\\ &&-\int_{0}^{t}\int_{z_0}^z\int_{D_\xi}e^{-\delta_1\eta}u_iT_{,i}T{\rm d}A{\rm d}\xi {\rm d}\eta+\int_{0}^{t}\int_{z_0}^z\int_{D_\xi}e^{-\delta_1\eta}(T_{,i}T)_{,i}{\rm d}A{\rm d}\xi {\rm d}\eta\nonumber\\ &=&\int_{0}^{t}\int_{z_0}^z\int_{D_\xi}e^{-\delta_1\eta}\Big[\delta_2 u_{i,j}u_{i,j}+\delta_2(1+\gamma T)|{\textbf{u}}|^2+T_{,i}T_{,i}+\frac{1}{2}\delta_1T^2\Big]{\rm d}A{\rm d}\xi {\rm d}\eta\nonumber\\ &&+\frac{1}{2} e^{-\delta_1 t}\int_{z_0}^z\int_{D_\xi}T^2{\rm d}A{\rm d}\xi-\delta_2\int_0^t\int_{z_0}^z\int_{D_\xi}e^{-\delta_1\eta}g_i u_iT{\rm d}A{\rm d}\xi {\rm d}\eta, \end{matrix}$

其中$z_0$是$x_3$坐标轴上的某个点, 满足$0\leq z_0\leq z$. 对(2.9)式微分, 可得

$\begin{matrix}\frac{\partial}{\partial z}F(z, t) &=&\int_0^t\int_{D_z}e^{-\delta_1\eta}\Big[\delta_2 u_{i,j}u_{i,j}+\delta_2(1+\gamma T)|{\textbf{u}}|^2+T_{,i}T_{,i}+\frac{1}{2}\delta_1T^2\Big]{\rm d}A {\rm d}\eta\nonumber\\ &&+\frac{1}{2} e^{-\delta_1 t}\int_{D_z}T^2{\rm d}A-\delta_2\int_0^t\int_{D_z}e^{-\delta_1\eta}g_i u_iT{\rm d}A{\rm d}\eta. \end{matrix}$

接下来, 我们给出四个本文经常使用的引理.

引理2.1[21] 设$D$是平面上的有界区域, 并具有光滑的边界, $w$是$D$上的充分光滑的函数. 如果$w\big|_{\partial D}=0,$ 则 $\lambda_1\int_Dw^2{\rm d}A\leq\int_Dw_{,\alpha}w_{,\alpha}{\rm d}A, $其中$\lambda_1$是问题 $\varphi_{,\alpha\alpha}+\lambda\varphi=0, $在$ D, \varphi=0,$在$ \partial D$ 的第一特征值.

引理2.2[21] 设$D$是有界的平面区域并具有光滑的边界$\partial D$, $w$是Dirichlet可积函数, 且$w\big|_{\partial D}=0$. 则

$\int_Dw^4{\rm d}A\leq k_1\Big(\int_Dw^2{\rm d}A\Big)\Big(\int_Dw_{,\alpha}w_{,\alpha}{\rm d}A\Big), $

其中$k_1$是大于零的常数.

引理2.3[3] 设$w$是连续可微的函数, 且$w\big|_{\partial D}=0$, 则存在一个向量函数${\textbf{v}}=(v_1, v_2)$满足 $v_{\alpha, \alpha}=w,\ \mbox{在}\ D, \ v_\alpha=0, \mbox{在}\ \partial D,$以及存在一个仅依赖于$D$的大于零的常数$k_2$使得$\int_Dv_{\alpha,\beta}v_{\alpha,\beta}{\rm d}A\leq k_2\int_Dv_{\alpha,\alpha}^2\mbox{d}A.$

利用引理2.1-2.3以及微分不等式技术, 可得以下引理.

引理2.4 如果$w\big|_{\partial D}=0,$ 则对于(2.8)式所定义的$F(z, t)$满足以下微分不等式

$\begin{matrix}|F(z, t)|\leq b_1\Big[\frac{\partial}{\partial z}F(z, t)\Big]+b_2\Big[\frac{\partial}{\partial z}F(z, t)\Big]^\frac{5}{4},\end{matrix}$

其中$b_1=\frac{\sqrt{2}}{2}+\sqrt{\frac{k_2}{\lambda_1}}+\frac{1}{\sqrt{\delta_1}}, b_2=\sqrt[4]{\frac{4k_1k_2^2}{\lambda_1\delta_1}}+\frac{\sqrt{2}}{2\sqrt{\gamma\delta_1\delta_{2}}}.$

证 首先, 利用Hölder不等式和Young不等式, 可得

$\begin{matrix}\Big|-\delta_2\int_0^t\int_{D_z}e^{-\delta_1\eta}g_iu_iT{\rm d}A {\rm d}\eta\Big| &\leq&\frac{1}{2}\delta_2\int_0^t\int_{D_z}e^{-\delta_1\eta}|\textbf{{u}}|^2{\rm d}A {\rm d}\eta +\frac{1}{2}\delta_2\int_0^t\int_{D_z}e^{-\delta_1\eta}T^2{\rm d}A {\rm d}\eta\nonumber\\ &\leq&\frac{1}{2}\delta_2\int_0^t\int_{D_z}e^{-\delta_1\eta}(1+\gamma T)|\textbf{{u}}|^2{\rm d}A {\rm d}\eta\nonumber\\ &&+\frac{1}{2}\delta_2\int_0^t\int_{D_z}e^{-\delta_1\eta}T^2{\rm d}A {\rm d}\eta. \end{matrix}$

取$2\delta_2\leq\delta_1$, 然后把(2.12)式代入到(2.10)式可得

$\begin{matrix}\label{2.13}\frac{\partial}{\partial z}F(z, t)&\geq&\int_0^t\int_{D_z}e^{-\delta_1\eta}\Big[\delta_2 u_{i,j}u_{i,j}+\frac{1}{2}\delta_2(1+\gamma T)|{\textbf{u}}|^2+T_{,i}T_{,i}+\frac{1}{4}\delta_1T^2\Big]{\rm d}A {\rm d}\eta\nonumber\\& &+\frac{1}{2} e^{-\delta_1 t}\int_{D_z}T^2{\rm d}A \end{matrix}$

和

$\begin{matrix}\label{2.14} \frac{\partial}{\partial z}F(z, t) &\leq&\int_0^t\int_{D_z}e^{-\delta_1\eta}\Big[\delta_2 u_{i,j}u_{i,j}+\frac{3}{2}\delta_2(1+\gamma T)|{\textbf{u}}|^2+T_{,i}T_{,i}+\frac{3}{4}\delta_1T^2\Big]{\rm d}A {\rm d}\eta\nonumber\\ &&+\frac{1}{2} e^{-\delta_1 t}\int_{D_z}T^2{\rm d}A. \end{matrix}$

其次, 我们用$\frac{\partial}{\partial z}F(z, t)$来控制$I_i(i=1,2,3,4)$. 我们注意到

$\begin{matrix}\int_{D_z}u_{3}{\rm d}A&=&\int_{D_0}u_{3}{\rm d}A+\int_{0}^{z}\int_{D_\xi}u_{3,3}{\rm d}A{\rm d}\xi\nonumber\\ &=&\int_{D_0}u_{3}{\rm d}A-\int_{0}^{z}\int_{D_\xi}u_{\alpha,\alpha}{\rm d}A{\rm d}\xi=\int_{D_0}f{\rm d}A.\nonumber \end{matrix}$

因为$\int_{D_0}f{\rm d}A=0,$ 所以$\int_{D_z}u_3{\rm d}A=0$. 根据引理2.3, 存在向量函数 $\textbf{{v}}=(v_1, v_2)$满足

$v_{\alpha,\alpha}=u_3,\ \mbox{在}\ D, v_\alpha=0, \mbox{在}\ \partial D. $

所以利用散度定理和(2.1)式, 有

$\begin{matrix}&&-\delta_2\int_{0}^{t}\int_{z_0}^z\int_{D_\xi}e^{-\delta_1\eta}pu_{3}{\rm d}A{\rm d}\xi {\rm d}\eta\\&=&-\delta_2\int_{0}^{t}\int_{z_0}^z\int_{D_\xi}e^{-\delta_1\eta}pv_{\alpha,\alpha}{\rm d}A{\rm d}\xi {\rm d}\eta=\delta_2\int_{0}^{t}\int_{z_0}^z\int_{D_\xi}e^{-\delta_1\eta} p_{,\alpha}v_\alpha {\rm d}A{\rm d}\xi {\rm d}\eta\nonumber \\&=&\delta_2\int_{0}^{t}\int_{z_0}^z\int_{D_\xi}e^{-\delta_1\eta}\Big[g_\alpha T-\Delta u_\alpha-(1+\gamma T)u_\alpha\Big]v_\alpha {\rm d}A{\rm d}\xi {\rm d}\eta\nonumber\\ &=&\delta_2\int_{0}^{t}\int_{z_0}^z\int_{D_\xi}e^{-\delta_1\eta}g_\alpha v_\alpha T{\rm d}A{\rm d}\xi {\rm d}\eta+\delta_2\int_{0}^{t}\int_{z_0}^z\int_{D_\xi}e^{-\delta_1\eta}u_{\alpha,j} v_{\alpha,j}{\rm d}A{\rm d}\xi {\rm d}\eta \nonumber\\ &&-\delta_2\int_{0}^{t}\int_{z_0}^z\int_{D_\xi}e^{-\delta_1\eta}(1+\gamma T)u_\alpha v_{\alpha} {\rm d}A{\rm d}\xi {\rm d}\eta.\nonumber \end{matrix}$

所以

$\begin{matrix}I_1&=&\delta_2\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}g_\alpha v_\alpha T{\rm d}A{\rm d}\eta+\delta_2\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_{\alpha,j} v_{\alpha,j}{\rm d}A{\rm d}\eta \nonumber\\ &&-\delta_2\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}(1+\gamma T)u_\alpha v_{\alpha} {\rm d}A{\rm d}\eta\nonumber\\ &\doteq &I_{11}+I_{12}+I_{13}. \end{matrix}$

利用Hölder不等式, Young不等式, 引理2.1和引理2.2, 可得

$\begin{matrix}|I_{11}|&\leq&\delta_2\Big[\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}v_{\alpha}v_{\alpha}{\rm d}A{\rm d}\eta \int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}T^2{\rm d}A{\rm d}\eta\Big]^\frac{1}{2}\nonumber\\ &\leq&\frac{\delta_2}{\sqrt{\lambda_1}}\Big[\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}v_{\alpha,\beta}v_{\alpha,\beta}{\rm d}A{\rm d}\eta \int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}T^2{\rm d}A{\rm d}\eta\Big]^\frac{1}{2}\nonumber\\ &\leq&\frac{\delta_2\sqrt{k_2}}{\sqrt{\lambda_1}}\Big[\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}v_{\alpha,\alpha}v_{\alpha,\alpha}{\rm d}A{\rm d}\eta \int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}T^2{\rm d}A{\rm d}\eta\Big]^\frac{1}{2}\nonumber\\ &\leq&\frac{2\sqrt{2\delta_2k_2}}{\sqrt{\delta_1\lambda_1}}\Big[\frac{1}{2}\delta_2\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_{3}^2{\rm d}A{\rm d}\eta \frac{1}{4}\delta_1\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}T^2{\rm d}A{\rm d}\eta\Big]^\frac{1}{2}\nonumber\\ &\leq&\frac{\sqrt{2\delta_2k_2}}{\sqrt{\delta_1\lambda_1}}\Big[\frac{1}{2}\delta_2\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}(1+\gamma T)|\textbf{{u}}|^2{\rm d}A{\rm d}\eta +\frac{1}{4}\delta_1\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}T^2{\rm d}A{\rm d}\eta\Big] \end{matrix}$

和

$\begin{matrix} |I_{12}|&=&\delta_2\Big|\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_{\alpha,\beta} v_{\alpha,\beta}{\rm d}A{\rm d}\eta\Big| +\delta_2\Big|\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_{\alpha,3} v_{\alpha,3}{\rm d}A{\rm d}\eta\Big| \nonumber\\ &\leq&\delta_2\Big[\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_{\alpha,\beta}u_{\alpha,\beta}{\rm d}A{\rm d}\eta\Big]^\frac{1}{2} \Big[\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}v_{\alpha,\beta}v_{\alpha,\beta}{\rm d}A{\rm d}\eta\Big]^\frac{1}{2}\nonumber\\ &&+\delta_2\Big[\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_{\alpha,3}u_{\alpha,3}{\rm d}A{\rm d}\eta\Big]^\frac{1}{2} \Big[\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}v_{\alpha,3}v_{\alpha,3}{\rm d}A{\rm d}\eta\Big]^\frac{1}{2}\nonumber\\ &\leq&\delta_2\sqrt{k_2}\Big[\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_{\alpha,\beta}u_{\alpha,\beta}{\rm d}A{\rm d}\eta\Big]^\frac{1}{2} \Big[\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_{3}^2{\rm d}A{\rm d}\eta\Big]^\frac{1}{2}\nonumber\\ &&+\delta_2\frac{1}{\sqrt{\lambda_1}}\Big[\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_{\alpha,3}u_{\alpha,3}{\rm d}A{\rm d}\eta\Big]^\frac{1}{2} \Big[\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}v_{\alpha,\beta3}v_{\alpha,\beta3}{\rm d}A{\rm d}\eta\Big]^\frac{1}{2} \nonumber \end{matrix}$
$\begin{matrix}&\leq&\delta_2\sqrt{\frac{k_2}{\lambda_1}}\Big[\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_{\alpha,\beta}u_{\alpha,\beta}{\rm d}A{\rm d}\eta\Big]^\frac{1}{2} \Big[\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_{3,\beta}u_{3,\beta}{\rm d}A{\rm d}\eta\Big]^\frac{1}{2}\nonumber\\ &&+\delta_2\sqrt{\frac{k_2}{\lambda_1}}\Big[\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_{\alpha,\beta}u_{\alpha,\beta}{\rm d}A{\rm d}\eta\Big]^\frac{1}{2} \Big[\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u^2_{3,3}{\rm d}A{\rm d}\eta\Big]^\frac{1}{2}\nonumber\\ &\leq&\frac{\delta_2}{2}\sqrt{\frac{k_2}{\lambda_1}}\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_{i,j}u_{i,j}{\rm d}A{\rm d}\eta. \end{matrix} $

利用引理2.1-2.3和算术几何平均不等式, 可得

$\begin{matrix} |I_{13}|&\leq &\delta_2\Big(\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_\alpha u_{\alpha} {\rm d}A{\rm d}\eta\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}v_\alpha v_{\alpha} {\rm d}A{\rm d}\eta\Big)^\frac{1}{2}\nonumber\\& &+\delta_{2}\gamma \Big(\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}Tu_\alpha u_{\alpha} {\rm d}A{\rm d}\eta\Big)^\frac{1}{2}\Big(\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}T^2{\rm d}A{\rm d}\eta\Big)^\frac{1}{4}\nonumber\\ &&\cdot\Big(\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}(v_\alpha v_{\alpha})^2{\rm d}A{\rm d}\eta\Big)^\frac{1}{4}\nonumber\\ &\leq&\delta_2\sqrt{\frac{k_2}{\lambda_1}}\Big(\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_\alpha u_{\alpha} {\rm d}A{\rm d}\eta\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_3^2{\rm d}A{\rm d}\eta\Big)^\frac{1}{2}\nonumber\\ &&+\delta_{2}\sqrt[4]{\frac{k_1k_2^2}{\lambda_1}} \Big(\gamma\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}Tu_\alpha u_{\alpha} {\rm d}A{\rm d}\eta\Big)^\frac{1}{2}\nonumber\\ &&\cdot\Big(\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}T^2{\rm d}A{\rm d}\eta\Big)^\frac{1}{4} \Big(\gamma\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_3^2{\rm d}A{\rm d}\eta\Big)^\frac{1}{2}\nonumber\\ &\leq&\sqrt{\frac{k_2}{\lambda_1}}\Big(\frac{\delta_2}{2}\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}(1+\gamma T)|\textbf{{u}}|^2{\rm d}A{\rm d}\eta\Big)\nonumber\\ &&+\sqrt[4]{\frac{4k_1k_2^2}{\lambda_1\delta_1}} \Big(\frac{\delta_2}{2}\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}(1+\gamma T)|\textbf{{u}}|^2 {\rm d}A{\rm d}\eta\Big)\nonumber\\ &&\cdot\Big(\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}\frac{1}{4}\delta_1T^2{\rm d}A{\rm d}\eta\Big)^\frac{1}{4}. \end{matrix}$

把(2.16)-(2.18)式代入到(2.15)式, 可得

$\begin{matrix}|I_1|&\leq&\sqrt{\frac{k_2}{\lambda_1}}\Big[\frac{\partial}{\partial z}F(z, t)\Big]+\sqrt[4]{\frac{4k_1k_2^2}{\lambda_1\delta_1}}\Big[\frac{\partial}{\partial z}F(z, t)\Big]^\frac{5}{4}. \end{matrix}$

利用Schwarz不等式, 可得

$\begin{matrix} |I_2|&\leq&\delta_2\Big[\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_{i,3}u_{i,3}{\rm d}A{\rm d}\eta\Big]^\frac{1}{2} \Big[\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}|{\textbf{u}}|^2{\rm d}A{\rm d}\eta\Big]^\frac{1}{2} \nonumber\\ &\leq&\sqrt{2}\Big[\delta_2\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}u_{i,3}u_{i,3}{\rm d}A{\rm d}\eta\Big]^\frac{1}{2} \Big[\frac{1}{2}\delta_2\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}(1+\gamma T)|{\textbf{u}}|^2{\rm d}A{\rm d}\eta\Big]^\frac{1}{2}\nonumber\\ &\leq&\frac{\sqrt{2}}{2}\Big[\frac{\partial}{\partial z}F(z, t)\Big]. \end{matrix}$

对于$I_3$, 利用Hölder不等式、引理2.2和算术几何平均不等式, 可得

$\begin{matrix}|I_3|&\leq&\frac{1}{2}\Big(\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}T u_3^2{\rm d}A{\rm d}\eta\Big)^\frac{1}{2}\Big(\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}T^2{\rm d}A{\rm d}\eta\Big)^\frac{1}{4} \Big(\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}T^4{\rm d}A{\rm d}\eta\Big)^\frac{1}{4} \nonumber\\ &\leq&\frac{\sqrt{2}}{\sqrt{\gamma\delta_1\delta_{2}}}\Big(\frac{\delta_2}{2}\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}(1+\gamma T) u_3^2{\rm d}A{\rm d}\eta\Big)^\frac{1}{2}\Big(\frac{1}{4}\delta_1\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}T^2{\rm d}A{\rm d}\eta\Big)^\frac{1}{2}\nonumber\\ &&\cdot\Big(\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}T_{,\alpha}T_{,\alpha}{\rm d}A{\rm d}\eta\Big)^\frac{1}{4} \nonumber\\ &\leq&\frac{\sqrt{2}}{2\sqrt{\gamma\delta_1\delta_{2}}}\Big[\frac{\partial}{\partial z}F(z, t)\Big]^\frac{5}{4}. \end{matrix} $

类似地, 对于$I_4$, 有

$\begin{matrix}|I_4|&\leq&\Big(\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}T_{,3}^2{\rm d}A{\rm d}\eta\Big)^\frac{1}{2} \Big(\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}T^2{\rm d}A{\rm d}\eta\Big)^\frac{1}{2}\nonumber\\ &\leq&\frac{2}{\sqrt{\delta_1}}\Big(\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}T_{,i}T_{,i}{\rm d}A{\rm d}\eta\Big)^\frac{1}{2} \Big(\frac{1}{4}\delta_1\int_{0}^{t}\int_{D_z}e^{-\delta_1\eta}T^2{\rm d}A{\rm d}\eta\Big)^\frac{1}{2}\nonumber\\ &\leq&\frac{1}{\sqrt{\delta_1}}\Big[\frac{\partial}{\partial z}F(z, t)\Big]. \end{matrix} $

最后把(2.19)-(2.22)式代入到(2.8)式即可完成引理2.4的证明.

3 二择一结果

对引理2.4进行分析, 可得以下定理.

定理3.1 假设$(u_i, T)$是问题(2.1)-(2.7)的解, 并且$\int_Df{\rm d}A=0$. 则要么

$\begin{matrix} &&\lim_{z\rightarrow\infty}\Big\{z^{-5}\Big[\int_0^t\int_{z_0}^z\int_{D_\xi}e^{-\delta_1\eta}\Big(\delta_2 u_{i,j}u_{i,j}+\frac{3}{2}\delta_2(1+\gamma T)|{\textbf{u}}|^2+T_{,i}T_{,i}+\frac{3}{4}\delta_1T^2\Big){\rm d}A{\rm d}\xi {\rm d}\eta\nonumber\\& &+\frac{1}{2}e^{-\delta_1 t}\int_{z_0}^z\int_{D_\xi}T^2{\rm d}A{\rm d}\xi\Big]\Big\}\geq c_1 \end{matrix}$

成立, 要么

$\begin{matrix} &&\int_0^t\int_z^\infty\int_{D_\xi}e^{-\delta_1\eta}\Big[\delta_2 u_{i,j}u_{i,j}+\frac{1}{2}\delta_2(1+\gamma T)|{\textbf{u}}|^2+T_{,i}T_{,i}+\frac{1}{4}\delta_1T^2\Big]{\rm d}A{\rm d}\xi {\rm d}\eta\nonumber\\ &&+\frac{1}{2}e^{-\delta_1 t}\int_z^\infty\int_{D_\xi}T^2{\rm d}A{\rm d}\xi\nonumber\\ &\leq &m_4Q^2(0, t)e^{-\frac{2z}{m_3\delta_3}}+m_3\delta_3Q(0, t)e^{-\frac{z}{m_3\delta_3}} \end{matrix}$

成立, 其中$c_1, m_3, m_4$是大于零的常数, $\delta_3$是大于零的任意常数以及$Q(0, t)$将由(3.17)式定义.

证 我们对(2.11)式分两种情形进行分析.

I 如果存在一个$z_0\geq0$, 使得$F(z_0,t)>0$. 由(2.13)式可知$\frac{\partial}{\partial z}F(z,t)\geq0$, 所以$F(z,t)>0,$$ z\geq z_0$. 则(2.11) 式可以写为

$\begin{matrix}F(z, t)\leq b_1\Big[\frac{\partial}{\partial z}F(z, t)\Big]+b_2\Big[\frac{\partial}{\partial z}F(z, t)\Big]^\frac{5}{4}.\end{matrix}$

在(3.3)式右边的第一项应用Young不等式, 可得

$\begin{matrix}\Big[\frac{\partial F}{\partial z}(z, t)\Big]=\Big[\frac{\partial F}{\partial z}(z, t)\Big]^{\frac{5}{8}\cdot\frac{2}{5}}\Big[\frac{\partial F}{\partial z}(z, t)\Big]^{\frac{5}{4}\cdot\frac{3}{5}}\leq\frac{2}{5}\Big[\frac{\partial F}{\partial z}(z, t)\Big]^{\frac{5}{8}}+\frac{3}{5}\Big[\frac{\partial F}{\partial z}(z, t)\Big]^{\frac{5}{4}}.\end{matrix}$

把(3.4)式代入到(3.3)式, 可得

$\begin{matrix}F(z, t)\leq m_1\Big[\frac{\partial}{\partial z}F(z, t)\Big]^\frac{5}{8}+m_2\Big[\frac{\partial}{\partial z}F(z, t)\Big]^\frac{5}{4}, \end{matrix}$

其中$m_1=\frac{2}{5}b_1,m_2=\frac{3}{5}b_1+b_2.$ 由(3.5)可得

$\begin{matrix}\label{3.6} F(z, t)+\frac{m_1^2}{4m_2}\leq m_2\Big[\Big(\frac{\partial F}{\partial z}(z, t)\Big)^\frac{5}{8}+\frac{m_1}{2m_2}\Big]^2,\ z\geq z_0. \end{matrix}$

在(3.6)式中解出$\frac{\partial F}{\partial z}(z, t)$, 可得

$ \frac{\partial F}{\partial z}(z, t)\geq \Big[\sqrt{\frac{1}{m_2}F(z, t)+\frac{m_1^2}{4m_2^2}}-\frac{m_1}{2m_2}\Big]^\frac{8}{5},\ z\geq z_0.$

于是有

$\begin{matrix} &&\Big\{2m_2\frac{1}{\Big[\sqrt{\frac{1}{m_2}F(z, t)+\frac{m_1^2}{4m_2^2}}-\frac{m_1}{2m_2}\Big]^\frac{3}{5}}+m_1\frac{1}{\Big[\sqrt{\frac{1}{m_2}F(z, t)+\frac{m_1^2}{4m_2^2}}-\frac{m_1}{2m_2}\Big]^\frac{8}{5}}\Big\}\nonumber\\ &&\cdot d\Big[\sqrt{\frac{1}{m_2}F(z, t)+\frac{m_1^2}{4m_2^2}}-\frac{m_1}{2m_2}\Big]\geq1,\ z\geq z_0. \end{matrix}$

对(3.7)式从$z_0$到$z$积分, 可得

$\begin{matrix}&&5m_2\Big\{\Big[\sqrt{\frac{1}{m_2}F(z, t)+\frac{m_1^2}{4m_2^2}}-\frac{m_1}{2m_2}\Big]^\frac{2}{5}-\Big[\sqrt{\frac{1}{m_1}F(z_0, t)+\frac{m_1^2}{4m_2^2}}-\frac{m_1}{2m_2}\Big]^\frac{2}{5}\Big\}\nonumber\\ &&-\frac{5}{3}m_1\Big\{\Big[\sqrt{\frac{1}{m_2}F(z, t)+\frac{m_1^2}{4m_2^2}}-\frac{m_1}{2m_2}\Big]^{-\frac{3}{5}} -\Big[\sqrt{\frac{1}{m_2}F(z_0, t)+\frac{m_1^2}{4m_2^2}}-\frac{m_1}{2m_2}\Big]^{-\frac{3}{5}}\Big\}\nonumber\\ &\geq &z-z_0,\ z\geq z_0. \end{matrix}$

在(3.8)式的左边舍弃第二项和第三项, 在第一项中利用不等式

$\sqrt{a+b}\leq\sqrt{a}+\sqrt{b},\ a,b\geq0,$

可得

$\begin{matrix}\label{3.9} F(z, t)\geq m_2\Big\{\frac{1}{5m_2}(z-z_0) -\frac{m_1}{3m_2}\Big[\sqrt{\frac{1}{m_2}F(z_0, t)+\frac{m_1^2}{4m_2^2}}-\frac{m_1}{2m_2}\Big]^{-\frac{3}{5}}\Big\}^5. \end{matrix}$

另一方面, 对(2.14)式从$z_0$到$z$积分, 可得

$\begin{matrix}\label{3.10} F(z, t)-F(z_0, t)&\leq&\int_0^t\int_{z_0}^z\int_{D_\xi}e^{-\delta_1\eta}\Big[\delta_2 u_{i,j}u_{i,j}+\frac{3}{2}\delta_2(1+\gamma T)|{\textbf{u}}|^2+T_{,i}T_{,i}+\frac{3}{4}\delta_1T^2\Big]{\rm d}A{\rm d}\xi {\rm d}\eta\nonumber\\& &+\frac{1}{2}e^{-\delta_1 t}\int_{z_0}^z\int_{D_\xi}T^2{\rm d}A{\rm d}\xi. \end{matrix}$

结合(3.9)和(3.10)式可以完成(3.1)式的证明.

II 如果$\forall z>0,$$F(z,t)<0$, 则(2.11)式可以写为

$\begin{matrix} -F(z, t)\leq b_1\Big[\frac{\partial}{\partial z}F(z, t)\Big]+b_2\Big[\frac{\partial}{\partial z}F(z, t)\Big]^\frac{5}{4}, \end{matrix} $

对(3.11)式右边的第二项应用Young不等式, 可得

$\begin{matrix} \Big[\frac{\partial F}{\partial z}(z, t)\Big]^\frac{5}{4}=\Big[\frac{\partial F}{\partial z}(z, t)\Big]^{1\cdot\frac{3}{4}} \Big[\frac{\partial F}{\partial z}(z, t)\Big]^{2\cdot\frac{1}{4}}\leq\frac{3}{4}\delta_3\Big[\frac{\partial F}{\partial z}(z, t)\Big]+ \frac{1}{4}\delta_3^{-3}\Big[\frac{\partial F}{\partial z}(z, t)\Big]^2, \end{matrix}$

其中$\delta_3$是一个大于零的任意常数. 把上式代入到(3.11)式, 可得

$\begin{matrix} -F(z,t)\leq m_3\Big[\frac{\partial}{\partial z}F(z, t)\Big]+m_4\Big[\frac{\partial}{\partial z}F(z, t)\Big]^2, \end{matrix}$

其中$m_3=b_1+\frac{3}{4}b_2\delta_{3},\ m_2=\frac{1}{4}b_2\delta_3^{-3}.$ 由(3.13)式可得

$\frac{\partial F}{\partial z}(z, t)\geq \sqrt{-\frac{F(z,t)}{m_4}+\frac{m_3^2}{4m_4^2}}-\frac{m_3}{2m_4}.\nonumber $

于是, 有

$\begin{matrix}\Big\{-2m_4-m_3\frac{1}{\sqrt{-\frac{F(z,t)}{m_4}+\frac{m_3^2}{4m_4^2}}-\frac{m_3}{2m_4}}\Big\}d\Big\{\sqrt{-\frac{F(z,t)}{m_4}+\frac{m_3^2}{4m^2_4}}-\frac{m_3}{2m_4}\Big\}\geq 1,\ z\geq 0. \end{matrix}$

对(3.14)式从$0$到$z$积分, 可得

$\begin{matrix} &&2m_4\Big[\sqrt{-\frac{F(z,t)}{m_4}+\frac{m_3^2}{4m_4^2}}-\sqrt{-\frac{F(0,t)}{m_4}+\frac{m_3^2}{4m_4^2}}\Big]\nonumber\\ &&+m_3\ln\Big[\sqrt{-\frac{F(z,t)}{m_4}+\frac{m_3^2}{4m_4^2}}-\frac{m_3}{2m_4}\Big]-m_3 \ln\Big[\sqrt{-\frac{F(0,t)}{m_4}+\frac{m_3^2}{4m_4^2}} -\frac{m_3}{2m_4}\Big]\nonumber\\ &\leq&-z. \end{matrix}$

舍弃(3.15)式左边的第一项, 可得

$\begin{matrix} &&m_3\ln\Big[\sqrt{-\frac{F(z,t)}{m_4}+\frac{m_3^2}{4m_4^2}}-\frac{m_3}{2m_4}\Big]\\&\leq&-z +2m_4\sqrt{-\frac{F(0,t)}{m_4}+\frac{m_3^2}{4m_4^2}}+m_3\ln\Big[\sqrt{-\frac{F(0,t)}{m_4}+\frac{m_3^2}{4m_4^2}}-\frac{m_3}{2m_4}\Big].\nonumber \end{matrix}$

所以

$\begin{matrix} \sqrt{-\frac{F(z,t)}{m_4}+\frac{m_3^2}{4m_4^2}}\leq Q(0, t)e^{-\frac{z}{m_3}}+\frac{m_3}{2m_4}, \end{matrix}$

其中

$\begin{matrix} Q(0, t)=\Big[\sqrt{-\frac{F(0,t)}{m_4}+\frac{m_3^2}{4m_4^2}}-\frac{m_3}{2m_4}\Big]e^{\frac{2m_4}{m_3} \sqrt{-\frac{F(0,t)}{m_4}+\frac{m_3^2}{4m_4^2}}}.\end{matrix}$

在(3.16)式两边平方, 可得

$\begin{matrix}-F(z,t)\leq m_4Q^2(0, t)e^{-\frac{2z}{m_3}}+m_3Q(0, t)e^{-\frac{z}{m_3}}, \end{matrix}$

所以$ \lim_{z\rightarrow\infty}\Big[-F(z, t)\Big]=0$. 现对(2.13)式从$z$到$\infty$积分, 可得

$\begin{matrix} -F(z, t)&\geq&\int_0^t\int_z^\infty\int_{D_\xi}e^{-\delta_1\eta}\Big[\delta_2 u_{i,j}u_{i,j}+\frac{1}{2}\delta_2(1+\gamma T)|{\textbf{u}}|^2+T_{,i}T_{,i}+\frac{1}{4}\delta_1T^2\Big]{\rm d}A{\rm d}\xi {\rm d}\eta\nonumber\\& &+\frac{1}{2}e^{-\delta_1 t}\int_z^\infty\int_{D_\xi}T^2{\rm d}A{\rm d}\xi. \end{matrix}$

联合(3.18)和(3.19)式可以完成对(3.2)式的证明. 证毕.

注3.1 定理3.1表明当$z\rightarrow\infty$时方程(2.1)-(2.7)的解要么多项式增长要么指数式衰减, 其中增长速度至少和$z^{5}$一样快.

注3.2 由于$\delta_3$是大于零的任意常数, 只要选择$\delta_3$足够大使得$m_{3}\geq b_{1}$, 则衰减速度至多和$e^{-\frac{z}{b_{1}}}$一样快.

注3.3 在衰减的情形, 为了使得衰减结果有意义, 还必须推导$-F(0, t)$的显式上界. 我们可以使用文献[3,4]中的方法, 易得该上界, 故本文略去.

4 总结

本文考虑了半无穷柱体上多孔介质中的Forchheimer方程组, 成功的控制了非线性项, 获得了解的Phragmén-Lindelöf型二择一结果. 本文的研究可以为其它类型的偏微分方程(例如Brinkman方程组,Stokes方程组等)提供借鉴.

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