一类抛物型界面问题的正则性分析

Regular Analysis of a Class of Parabolic Interface Problems

摘要: 界面问题用于各种工程应用和物理、化学、生物现象的建模，特别是涉及具有不同扩散性、密度、渗透性或导电性的多种不同材料的现象，其在界面上由一定条件耦合。本文考虑具有溶解质输运的线性两相流模型，分别在相界面处耦合非完美界面条件和Henry界面条件。由于解在界面上的跳跃使得解在各自材料区域上具有比在整个区域上更高的正则性，针对这类界面问题的正则性分析，本文给出一个完整的泛函分析过程，采用De Giorgi迭代方法证明该模型弱解的相关性质，进而证明弱解及其梯度的Hölder连续性。此外，对于Henry界面问题，本文给出了梯度的L^q估计(存在q > 2)。

Abstract: Interface problems are used for various engineering applications and modeling of physical, chemi-cal, and biological phenomena, especially those involving a number of different materials with dif-ferent diffusion, density, permeability or conductivity, which are coupled by certain conditions at the interface. In this paper, a linear two-phase flow model with solute transport is considered, cou-pling imperfect interface condition and Henry interface condition at the phase interface, respec-tively. Since the jump of the solution at the interface makes the solution have higher regularity in the respective material region than on the whole region, for the regularity analysis of such interface problems, this paper presents a complete functional analysis process, and uses the De Giorgi itera-tion method to prove the correlation properties of the weak solution of the model, and then prove the Hölder continuity of the weak solution and its gradient. In addition, for the Henry interface problem, this paper gives the L^q estimation of the gradient (q > 2 is present).

1. 引言

当偏微分方程中的系数、或解、或法向流在界面处不连续时，就会产生界面问题。界面问题在自然界以及科学与工程中都有应用。在材料科学中，文献 [1] 提出了一种聚合物基体复合材料的耦合扩散行为模型，在由不同成分构成的复合材料的接触面上产生界面问题。在生物技术中被脂膜所包围的流体所形成的生物膜模型在界面处的相互作用 [2] 。文献 [3] 研究了两相流体动力学界面模型的斯托克斯问题，在界面处具有不连续的密度和粘度系数以及压力溶液。文献 [4] 考虑二维静止热传导椭圆界面问题，其传导系数在光滑的内部界面上是不连续的。文献 [5] 提出一个基于区域分解理论的非完美界面问题的保极值迭代方法，从而使得界面条件自然地嵌入到子域的边界条件中。针对非完美界面问题，文献 [6] 提出了一种保持离散极值原理(DMP)和守恒性的有限体积格式。此外，界面问题也被应用于合金凝固、晶体生长以及在生物系统中 [7] 等。

考虑具有溶解质输运的线性两相流问题(见图1)。设 $\begin{array}{c}\omega \left(\frac{R}{4}\right)=\mu \left(\frac{R}{4}\right)-\tilde{\mu }\left(\frac{R}{4}\right)\le \mu -\min \left\{\underset{Q_{\frac{R}{4},1}}{\text{ess}\inf }{u}_{1,\tilde{k}}^{(-)},\underset{Q_{\frac{R}{4},2}}{\text{ess}\inf }{u}_{2,\tilde{k}}^{(-)}\right\}\\ \le \left(1-\frac{1}{2^s}\right)\omega \left(R\right)+C\left(R^{{\alpha }_1}\left({\left[u_0^1\right]}_{C^{{\alpha }_1}\left({\bar{\Omega }}_1\right)}+{\left[u_0^2\right]}_{C^{{\alpha }_1}\left({\bar{\Omega }}_2\right)}\right)+\left(M+F_0\right)R^{1-\frac{n+2}{p}}\right).\end{array}$ 是一个包含两相不可混溶，不可压缩的流动系统(液–液或液–气)的有界区域，具有光滑边界 $\begin{array}{c}\omega \left(\frac{R}{4}\right)=\mu \left(\frac{R}{4}\right)-\tilde{\mu }\left(\frac{R}{4}\right)\le \max \left\{\underset{Q_{\frac{R}{4},1}}{\text{ess}\sup }{u}_{1,k}^{(+)},\underset{Q_{\frac{R}{4},2}}{\text{ess}\sup }{u}_{2,k}^{(+)}\right\}-\tilde{\mu }\\ \le \left(1-\frac{1}{2^s}\right)\omega \left(R\right)+C\left(R^{{\alpha }_1}\left({\left[u_0^1\right]}_{C^{{\alpha }_1}\left({\bar{\Omega }}_1\right)}+{\left[u_0^2\right]}_{C^{{\alpha }_1}\left({\bar{\Omega }}_2\right)}\right)+\left(M+F_0\right)R^{1-\frac{n+2}{p}}\right).\end{array}$ 。 $0<R\le R_0\le 1$ 是一个开子区域且有边界 $\omega \left(\frac{R}{4}\right)\le \left(1-\frac{1}{2^s}\right)\omega \left(R\right)+C{R}^{\alpha }\left({\left[u_0^1\right]}_{C^{{\alpha }_1}\left({\bar{\Omega }}_1\right)}+{\left[u_0^2\right]}_{C^{{\alpha }_1}\left({\bar{\Omega }}_2\right)}+M+F_0\right),$ ，故有 $0<\alpha \le \min \left\{{\alpha }_1,1-\frac{n+2}{p}\right\}$ 。每个子域中分别包含一个相，这些相通过界面 $\omega \left(R\right)\le C{\left(\frac{R}{R_0}\right)}^{\alpha }\left(\omega \left(R_0\right)+R_0^{\alpha }\left({\left[u_0^1\right]}_{C^{{\alpha }_1}\left({\bar{\Omega }}_1\right)}+{\left[u_0^2\right]}_{C^{{\alpha }_1}\left({\bar{\Omega }}_2\right)}+M+F_0\right)\right),\forall R\in \left(0,R_0\right],$ 分隔开。两相中都含有一种溶解的物质，这种物质由于对流和分子扩散而被输运，并不粘附在界面上。在本文中，假设所考虑的溶解质输运的两相流模型是理想模型，在每个相中都会发生对流传质和扩散传质。假设不会发生相变，反应；界面处没有传质阻力；也不会因为传质而引起界面湍动等。

Figure 1. Two-phase flow model

图1. 两相流模型

在相界面处，考虑定常界面情形。同时对于界面处考虑非完美界面条件 [5] [6] [8] 和Henry界面条件。如果给定的数据，界面 $u\in W\left(0,T;V\right)$ 和外边界 $\forall X_0=\left(x_0,0\right)\in \Gamma$ 光滑，则问题的解在各个区域也非常光滑，但由于界面处的跳跃会使得解的全局正则性降低。在文献 [9] 中，只给出了Henry界面问题弱解的适定性。而在本文中我们着重讨论在上述两种界面条件下的线性两相流模型弱解的性质，我们的主要结论是在给定的Sobolev空间中，利用De Giorgi迭代法来估计线性问题的弱解，得到弱解在界面附近的局部性质，在此基础之上可进一步得到线性模型的弱解及其梯度的Hölder连续性。

设溶解质的浓度为 $Q_R\left(X_0\right)=B_R\left(x_0\right)\times \left(0,R^2\right]$ ，标量 $0<R\le 1$ ，这个问题可以用浓度为 $d=\min \left\{1,dist\left\{x_0,\partial \Omega \right\}\right\}$ 的对流–扩散方程来建模。 $0<R\le d$ 为界面 $0<\alpha \le \min \left\{{\alpha }_1,1-\frac{n+2}{p}\right\}$ 上的单位外法向量，由 $C\ge 1$ 指向 $\max \left\{\underset{Q_{R,1}}{\text{osc}}{u}_1,\underset{Q_{R,2}}{\text{osc}}{u}_2\right\}\le C{R}^{\alpha }\left(M+F_0+{\left[u_0^1\right]}_{C^{{\alpha }_1}\left({\bar{\Omega }}_1\right)}+{\left[u_0^2\right]}_{C^{{\alpha }_1}\left({\bar{\Omega }}_2\right)}\right),$ 。对于 $\alpha ,C$ ， $n,\lambda ,\Lambda ,p$ 为流体速度场， $Q_{R,i}=Q_R\cap Q_i\left(i=1,2\right)$ 为扩散系数矩阵， $F_0={\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert f_i\Vert }_{L^2\left(0,T;L^{\frac{np}{n+p}}\left({\Omega }_i\right)\right)}}<\infty$ 为标量传输系数。

非完美界面条件是指解在界面上的跳跃与界面两侧连续的法向流成正比。可得非完美界面问题的数学模型公式如下：

$Q_T$ (1.1)

其中 $\Gamma \times \left(0,T\right]$ ， $Q_1$ ， $Q_2$ 满足非完美界面条件。

Henry界面条件要求界面在瞬间平衡的情况下，界面两侧的溶质浓度呈恒定比。同时施加另一个界面条件，即要求法向流在界面处是连续的。故有Henry界面问题的数学模型公式如下：

$u$ (1.2)

其中 $u$ 是分片常数，即在 $\Gamma \times \left(0,T\right]$ 中 $\left(\Gamma \times \left[0,T\right]\right)\cap \Omega$ ，一般有 $u\in C^{\alpha }\left({\bar{\Omega }}_1\times \left[0,T\right]\right)\times C^{\alpha }\left({\bar{\Omega }}_2\times \left[0,T\right]\right).$ 。 $0<\alpha \le \min \left\{{\alpha }_1,1-\frac{n+2}{p}\right\}$ 。 ${\partial }_p{Q}_T\in C^{1+\alpha }$ 满足Henry界面条件。

2. 非完美界面模型

在本节中，讨论非完美界面问题(1.1)。

2.1. 函数分析框架

对于 $\Gamma \times \left(0,T\right]\in C^{1+\alpha }$ ，一般Sobolev空间记为 $k_{pg}^i\in C^{\alpha }\left({\bar{\Omega }}_i\times \left[0,T\right]\right)$ 。特别的， $w_i$ 。首先引进一些合适的空间：

$w_i^j\in M^{1,n+1+\alpha }\left({\Omega }_i;d^{*}\right)\left(j=1,\cdots ,n,i=1,2\right)$ ${\Lambda }_1$

令 ${\left[k_{pg}^i\right]}_{\alpha }+{\Vert w_i^j\Vert }_{1,n+1+\alpha }\le {\Lambda }_1.$ ， $u\in W\left(0,T;V\right)$ ，且

$f_i\in M^{1,n+1+\alpha }\left(Q_i;\tilde{\delta }\right)$

由文献 [10] 的定理3.13 (p. 175)可得

$u_0^i\in C^{1+\alpha }\left({\bar{\Omega }}_i\right)$

其中 $u\in C^{1+\alpha }\left({\bar{\Omega }}_1\times \left[0,T\right]\right)\times C^{1+\alpha }\left({\bar{\Omega }}_2\times \left[0,T\right]\right),$ 表示嵌入。

迹算子：

${\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert u_i\Vert }_{1+\alpha }}\le C\left(n,\lambda ,\Lambda ,{\Lambda }_1,\bar{m},\alpha ,Q_T\right)\left({\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert u_0^i\Vert }_{1+\alpha }}+{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert u_i\Vert }_{\alpha }}+{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert f_i\Vert }_{1,n+1+\alpha }}\right).$

是有界的。令 $\tilde{u}=\beta u$ ，并且 ${\left[\tilde{u}\right]}_{\Gamma }=0$ 。

下面给出问题(1.1)中的系数的有关假设。

假设2.1.1区域 $\tilde{u}$ 和 ${\tilde{u}|}_{{\Omega }_i\times \left(0,T\right]}={\tilde{u}}_i$ 是有界区域， $\begin{array}{ll}\frac{1}{\beta }\frac{\partial \tilde{u}}{\partial t}+\frac{1}{\beta }w\cdot \nabla \tilde{u}-div\left(\frac{1}{\beta }K\left(x,t\right)\nabla \tilde{u}\right)=f\left(x,t\right),\hfill & x\in {\Omega }_1\cup {\Omega }_2,t\in \left(0,T\right];\hfill \\ \frac{1}{{\beta }_1}{K}_1\left(x,t\right)\nabla {\tilde{u}}_1\cdot n=\frac{1}{{\beta }_2}{K}_2\left(x,t\right)\nabla {\tilde{u}}_2\cdot n,\hfill & x\in \Gamma ,t\in \left(0,T\right];\hfill \\ {\left[\tilde{u}\right]}_{\Gamma }=0,\hfill & x\in \Gamma ,t\in \left(0,T\right];\hfill \\ \tilde{u}\left(\cdot ,0\right)={\tilde{u}}_0,\hfill & x\in {\Omega }_1\cup {\Omega }_2;\hfill \\ \tilde{u}\left(\cdot ,t\right)=0,\hfill & x\in \partial \Omega ,t\in \left(0,T\right].\hfill \end{array}$ ，界面 ${\tilde{u}}_0$ ， ${\alpha }_1\in \left(0,1\right)$ 为 $\partial \Omega$ 上的 $\Gamma$ 维Hausdroff测度。

假设2.1.2

1) 假设 $C^{1+{\alpha }_1}$ 在 $H=L^2\left(\Omega \right)$ 上是一个充分光滑的速度场，并且满足：

i) ${\left(u,v\right)}_H={\displaystyle {\int }_{\Omega }\frac{1}{\beta }uv\text{d}x},$ ， ${\Vert \cdot \Vert }_H$ ；

ii) 由于流体是不可压缩的，故有在 $V=H_0^1\left(\Omega \right)$ 中 ${\Vert u\Vert }_V^2={\Vert u\Vert }_H^2+{\Vert \nabla u\Vert }_H^2$ ；

iii) 在界面 ${\Vert u\Vert }_V$ 上速度场满足： ${\Vert \nabla u\Vert }_H$ 。

2) 扩散系数矩阵 $V\searrow H\equiv H^{\prime }\searrow V^{\prime }$ 和标量传输系数 $f\in L^2\left(0,T;L^2\left(\Omega \right)\right)$ 如下假设：

i) 扩散系数矩阵 ${\tilde{u}}_0\left(x\right)\in H$ 在 $\mathcal{F}\in L^2\left(0,T;V^{\prime }\right)$ 上是可测的，一致有界的和一致椭圆的，并且 $\tilde{u}\in W\left(0,T;V\right)$ 是对称的，即存在常数 $\{\begin{array}{l}\frac{\text{d}\tilde{u}}{\text{d}t}\left(\tilde{v}\right)+a\left(t;\tilde{u}\left(t\right),\tilde{v}\right)\le (\ge )\mathcal{F}\left(t\right)\left(\tilde{v}\right),\forall \tilde{v}\in V;\hfill \\ \tilde{u}\left(0\right)={\tilde{u}}_0\left(x\right).\hfill \end{array}$ 使得

$\tilde{u}$

并且对于任意 $\tilde{u}$ ，在 $\frac{\text{d}\tilde{u}}{\text{d}t}\left(\tilde{v}\right)={\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{{\Omega }_i}\frac{1}{{\beta }_i}}}\frac{\partial {\tilde{u}}_i}{\partial t}{\tilde{v}}_i\text{d}x,$ 中几乎处处成立

$a\left(t;\tilde{u}\left(t\right),\tilde{v}\right)={\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{{\Omega }_i}\frac{1}{{\beta }_i}{w}_i\cdot \left(\nabla {\tilde{u}}_i\right){\tilde{v}}_i}}+\frac{1}{{\beta }_i}{\left(\nabla {\tilde{v}}_i\right)}^{\top }{K}_i\nabla {\tilde{u}}_i\text{d}x,$

ii) 传输系数 $\mathcal{F}\left(t\right)\left(\tilde{v}\right)={\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{{\Omega }_i}{f}_i}}{\tilde{v}}_i\text{d}x.$ 在 $a\left(t;\tilde{u}\left(t\right),\tilde{v}\right)$ 上是可测的，并且存在常数 $V\times V$ 使得

$Q_T$

传输系数矩阵 $\tilde{u}$ ，则M是半正定矩阵，并且对于任意 $\tilde{u}\in W\left(0,T;V\right)$ ，有

$X_0=\left(x_0,t_0\right)\in \Gamma \times \left(0,T\right]$

在上述假设的基础之上，给出问题 的弱形式。给定 $Q_{\rho ,\tau }\left(X_0\right)=B_{\rho }\left(x_0\right)\times \left(t_0,t_0+\tau \right]\subset Q_T$ 。对于 $k\in ℝ$ ，积分可得：

$\xi \left(x,t\right)\in C^{\infty }\left(\left[t_0,t_0+\tau \right];C_0^{\infty }\left(B_{\rho }\left(x_0\right)\right)\right)$

记

$0\le \xi \le 1$

$\xi \left(\cdot ,t_0\right)=0$

$\begin{array}{l}\underset{t_0<t\le t_0+\tau }{\sup }{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert \xi {\left({\tilde{u}}_i-k\right)}^{\pm }\left(\cdot ,t\right)\Vert }_{L^2\left(B_{\rho ,i}\right)}^2}+{\lambda }_1{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert \nabla \left(\xi {\left({\tilde{u}}_i-k\right)}^{\pm }\right)\Vert }_{L^2\left(t_0,t_0+\tau ;L^2\left(B_{\rho ,i}\right)\right)}^2}\\ \le C^{*}\left[\left({\Vert \frac{\partial \xi }{\partial t}\Vert }_{L^{\infty }\left(Q_{\rho ,\tau }\right)}+{\Vert \nabla \xi \Vert }_{L^{\infty }\left(Q_{\rho ,\tau }\right)}^2\right){\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert {\left({\tilde{u}}_i-k\right)}^{\pm }\Vert }_{L^2\left(t_0,t_0+\tau ;L^2\left(B_{\rho ,i}\right)\right)}^2}+F_{0,\rho ,\tau }^2{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left|Q_{\rho ,\tau ,i}\cap \left[{\left({\tilde{u}}_i-k\right)}^{\pm }>0\right]\right|}^{1-\frac{2}{p}}}\right],\end{array}$

所以(1.1)的变分问题为：设 $B_{\rho ,i}=B_{\rho }\cap {\Omega }_i,Q_{\rho ,\tau ,i}=Q_{\rho ,\tau }\cap Q_i$ 。寻找弱下解(弱上解) $0<\rho ,\tau <1$ 使得

$p>n+2$ (2.1)

如果 ${\lambda }_1>0$ 既是弱下解，又是弱上解，则称 $F_{0,\rho ,\tau }={\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert f_i\Vert }_{L^2\left(t_0,t_0+\tau ;L^{\frac{np}{n+p}}\left(B_{\rho ,i}\right)\right)}}>0$ 是弱解。

由文献 [10] (p. 184定理3.16)可知，若要证明线性问题(2.1)弱解的存在唯一性，只需表明 $C^{*}$ 在 $n,\Lambda ,p,c_0,{\beta }_1,{\beta }_2$ 上是连续的和强制的。事实上，根据假设2.1.2和迹算子的有界性可得 $DG\left(Q_T\right)=DG\left(Q_T;{\lambda }_1,p,n,F_{0,\rho ,\tau },C^{*}\right)$ 是连续的。对于 $\tilde{u}\in W\left(0,T;V\right)$ 的强制性：任意 $\tilde{u}\in D{G}^{+}\left(Q_T\right)$ ，有

$\tilde{u}\in W\left(0,T;V\right)$

其中

$\tilde{u}\in D{G}^{-}(\; Q\; T\; )$

故有

$DG\left(Q_T\right)=D{G}^{+}\left(Q_T\right)\cap D{G}^{-}(\; Q\; T\; )$

由于

$\tilde{u}\in W\left(0,T;V\right)$

$p>n+2$

$f_i\in L^2\left(0,T;L^{\frac{np}{n+p}}\left({\Omega }_i\right)\right)\left(i=1,2\right)$

所以

$\tilde{u}\in D{G}^{+}(\; Q\; T\; )$

其中 $\tilde{u}\in W\left(0,T;V\right)$ 是常数。

2.2. 预备知识

以下定义和定理参考文献 [11] 和 [12] ：对于任意 $\tilde{u}\in D{G}^{-}\left(Q_T\right)$ ，令抛物距离为

$C^{*}$

假设D是 $n,\Lambda ,p,c_0$ 中的有界区域，对于任意 $F_{0,\rho ,\tau }={\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert f_i\Vert }_{L^2\left(t_0,t_0+\tau ;L^{\frac{np}{n+p}}\left(B_{\rho ,i}\right)\right)}}<\infty$ ，令 $\tilde{u}\in DG\left(Q_T\right)$ ，其中 $\begin{array}{l}\underset{t_0<t\le t_0+\tau }{\sup }{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert \xi {\left({\tilde{u}}_i-k\right)}^{\pm }\left(\cdot ,t\right)\Vert }_{L^2\left(B_{\rho ,i}\right)}^2}+{\lambda }_2{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert \xi {\left({\tilde{u}}_i-k\right)}^{\pm }\Vert }_{L^2\left(t_0,t_0+\tau ;L^{2^{*}}\left(B_{\rho ,i}\right)\right)}^2}\\ \le C^{*}\left[\left({\Vert \frac{\partial \xi }{\partial t}\Vert }_{L^{\infty }\left(Q_{\rho ,\tau }\right)}+{\Vert \nabla \xi \Vert }_{L^{\infty }\left(Q_{\rho ,\tau }\right)}^2\right){\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert {\left({\tilde{u}}_i-k\right)}^{\pm }\Vert }_{L^2\left(t_0,t_0+\tau ;L^2\left(B_{\rho ,i}\right)\right)}^2}+F_{0,\rho ,\tau }^2{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left|Q_{\rho ,\tau ,i}\cap \left[{\left({\tilde{u}}_i-k\right)}^{\pm }>0\right]\right|}^{1-\frac{2}{p}}}\right],\end{array}$ ， $0<\alpha \le {\alpha }_1$ 是D关于抛物距离的直径。

定义2.2.1 (Morrey空间)对于 $\tilde{u}\in C^{\alpha }\left(\bar{\Omega }\times \left[0,T\right]\right).$ ，令 $\Gamma \in C^2$ 表示由 $x_0\in \Gamma$ 中满足

$\rho >0$

的所有函数u所组成的赋范线性空间。特别地，若 $C^2$ 是有界区域，对于任意 $\gamma :{ℝ}^{n-1}\to ℝ$ ，记 $B^1={\Omega }_1\cap B_{\rho }\left(x_0\right)=\left\{x\in B_{\rho }\left(x_0\right)|x_n>\gamma \left(x_1,\cdots ,x_{n-1}\right)\right\},$ ， $\Pi =\Gamma \cap B_{\rho }\left(x_0\right)=\left\{x\in B_{\rho }\left(x_0\right)|x_n=\gamma \left(x_1,\cdots ,x_{n-1}\right)\right\},$ ， $B^2={\Omega }_2\cap B_{\rho }\left(x_0\right)=\left\{x\in B_{\rho }\left(x_0\right)|x_n<\gamma \left(x_1,\cdots ,x_{n-1}\right)\right\},$ ，定义 $B_{\rho }\left(x_0\right)$ 是由 $x_0$ 中满足

$\{\begin{array}{l}{y}_j=x_j=:{\Phi }^j\left(x\right),\left(j=1,\cdots ,n-1\right);\hfill \\ y_n=x_n-\gamma \left(x_1,\cdots ,x_{n-1}\right)=:{\Phi }^n\left(x\right).\hfill \end{array}$

的所有函数u所组成的赋范线性空间。

定义2.2.2 (Campanato空间)对于 $y=\Phi \left(x\right).$ ，以 $\{\begin{array}{l}{x}_j=y_j=:{\Psi }^j\left(y\right),\left(j=1,\cdots ,n-1\right);\hfill \\ x_n=y_n+\gamma \left(y_1,\cdots ,y_{n-1}\right)=:{\Psi }^n\left(y\right).\hfill \end{array}$ 表示由 $x=\Psi \left(y\right).$ 中满足

$\Psi ={\Phi }^{-1}$

的所有函数u所组成的赋范线性空间，其上的范数定义为

$\Phi :x\mapsto y$

且 $\Pi$ 表示u在 $J=\nabla \Phi$ 上的积分平均值，即 $J^{-1}=\nabla \Psi$

定义2.2.3 (Hölder空间)对于 $\left|\det \left(J\right)\right|=\left|\det \left(J^{-1}\right)\right|=1$ ，以 $\Phi \left(x_0\right)=y_0\in \Phi \left(\Pi \right)\subset \left\{y_n=0\right\}$ 表示满足

$R\in \left(0,1\right]$

的所有函数u所组成的线性空间，其上的范数定义为

$B_R\left(y_0\right)\subset \Phi \left(B_{\rho }\left(x_0\right)\right)$

显然有

$B_{R,1}=B_R\left(y_0\right)\cap \left\{y_n>0\right\}\subset \Phi \left(B^1\right),$

令 $\Sigma =B_R\left(y_0\right)\cap \left\{y_n=0\right\}\subset \Phi \left(\Pi \right),$ 表示由 $B_{R,2}=B_R\left(y_0\right)\cap \left\{y_n<0\right\}\subset \Phi \left(B^2\right).$ 中满足

$Q_R\left(y_0,t_0\right)=B_R\left(y_0\right)\times \left(t_0-R^2,t_0\right]$

所有函数u所组成的赋范线性空间，其上的范数定义为

$Q_{R,i}=B_{R,i}\times \left(t_0-R^2,t_0\right]$

定义2.2.4 称D是(A)型区域，如果存在常数A，使得对于任意 $0<t_0-R^2<t_0<T$ ，都有

$v\left(y,t\right)=\tilde{u}\left(\Psi \left(y\right),t\right),$

定理2.2.1 设D是(A)型区域， ${v|}_{B_{R,i}}=v_i\left(i=1,2\right)$ ，则当 $\tilde{w}\left(y\right)=w\left(\Psi \left(y\right)\right)$ 时，

$\tilde{K}\left(y,t\right)=K\left(\Psi \left(y\right),t\right)$

其中 $\tilde{f}\left(y,t\right)=f\left(\Psi \left(y\right),t\right)$ ， ${\tilde{v}}_0\left(y\right)={\tilde{u}}_0\left(\Psi \left(y\right)\right)$ 表示 $\tilde{n}$ 与 $B_{R,1}$ 同时成立。

在以后的小节中，记

$B_{R,2}$

$\{\begin{array}{l}\frac{\partial \tilde{u}}{\partial t}=\frac{\partial v}{\partial t};\hfill \\ {\nabla }_x\tilde{u}=J^{\top }{\nabla }_{y}v;\hfill \\ di{v}_x\left(K\left(x,t\right){\nabla }_x\tilde{u}\right)=di{v}_y\left(J\tilde{K}\left(y,t\right)J^{\top }{\nabla }_{y}v\right);\hfill \\ n=\frac{J^{\top }\tilde{n}}{\left|J^{\top }\tilde{n}\right|}.\hfill \end{array}$

$z=J\tilde{w}$

$A=J\tilde{K}\left(y,t\right)J^{\top }$

2.3. 弱解的极值原理

引理2.3.1 (文献( [13] , p. 95)引理5.6)非负序列 $\begin{array}{ll}\frac{1}{\beta }\frac{\partial v}{\partial t}+\frac{1}{\beta }z\cdot {\nabla }_{y}v-di{v}_y\left(\frac{1}{\beta }A{\nabla }_{y}v\right)=\tilde{f}\left(y,t\right),\hfill & y\in B_{R,1}\cup B_{R,2},t\in \left[t_0-R^2,t_0\right];\hfill \\ \frac{1}{{\beta }_1}{A}_1{\nabla }_y{v}_1\cdot \tilde{n}=\frac{1}{{\beta }_2}{A}_2{\nabla }_y{v}_2\cdot \tilde{n},\hfill & y\in \Sigma ,t\in \left[t_0-R^2,t_0\right];\hfill \\ {\left[v\right]}_{\Sigma }=0,\hfill & y\in \Sigma ,t\in \left[t_0-R^2,t_0\right].\hfill \end{array}$ 满足递推关系式 $z,A$ ，其中 $c_0,\lambda ,\Lambda$ ， $r\in \left(0,R\right]$ ，则如果 ${\xi }_{2r}\left(y\right)\in C_0^{\infty }\left(B_{2r}\left(y_0\right)\right)$ ，必有 ${\tau }_{2r}\left(t\right)\in C^{\infty }\left(ℝ\right)$ 。

推论2.3.1令 $0\le {\xi }_{2r}\le 1$ 是定义在 $B_r\left(y_0\right)$ 上的非增非负函数，并且存在 ${\xi }_{2r}=1$ 使得对于任意 ${\left|\nabla {\xi }_{2r}\right|}^2\le \frac{C}{r^2}$ ，有 $\left(t_0-r^2,+\infty \right)$ ，则当 ${\tau }_{2r}=1$ ，有 $\left(-\infty ,t_0-4r^2\right]$ 。

定理2.3.1 (极值原理)令问题 的系数满足假设2.1.2。如果 ${\tau }_{2r}=0$ 是问题的弱下解，且对于某个常数 $0\le {\tau }_{2r}\le 1$ ， $\left|\frac{\text{d}{\tau }_{2r}}{\text{d}t}\right|\le \frac{C}{r^2}$ ，则

$v\left(y,t\right)\in L^1(\; Q\; R\; )$

其中C仅依赖于 ${\tilde{v}}_{2r}\left(t\right)={\left({\displaystyle {\int }_{B_{2r}}{\xi }_{2r}^2}\left(y\right)\text{d}y\right)}^{-1}{\displaystyle {\int }_{B_{2r}}v}\left(y,t\right){\xi }_{2r}^2\left(y\right)\text{d}y,t\in \left(t_0-4r^2,t_0\right],$ ， ${\displaystyle {\int }_{B_{2r}}\left(v\left(y,t\right)-{\tilde{v}}_{2r}\left(t\right)\right){\xi }_{2r}^2\left(y\right)\text{d}y}=0.$ ，且 $v\left(y,t\right)=\tilde{u}\left(x,t\right)\in W\left(0,T;V\right)$ 。

证明 令 ${\tilde{v}}_{2r}\left(t\right)$ ，for $\frac{\text{d}{\tilde{v}}_{2r}\left(t\right)}{\text{d}t}\in L^2\left(\left(t_0-4r^2,t_0\right]\right)$ ，取测试函数 $p>n+2$ ，则 $q_1>\frac{np}{n+p}$ ，且 $\tilde{f}\in L^{q_1}\left(Q_R\right)$ 。由于 $v\left(y,t\right)$ 是弱下解，有

$\begin{array}{l}\underset{t_0-r^2<t\le t_0}{\sup }{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{r,i}}{\left|v_i-{\tilde{v}}_r\left(t\right)\right|}^2\text{d}y}}+{\lambda }_0{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{r,i}}{\left|\nabla v_i\right|}^2\text{d}y\text{d}t}}\\ \le C\left\{\frac{1}{r^2}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{2r,i}}{\left|v_i-{\tilde{v}}_{2r}\left(t\right)\right|}^2\text{d}y\text{d}t}}+r^{\left(n+2\right)\left(1-\frac{2}{p}\right)-\frac{4}{n}}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left({\displaystyle {\int }_{Q_{2r,i}}{\left|{\tilde{f}}_i\right|}^{\frac{np}{n+p}}\text{d}y\text{d}t}\right)}^{\frac{2\left(n+p\right)}{np}}}\right\},\end{array}$

即，

$n,{\beta }_1,{\beta }_2,c_0,\Lambda ,p$

由假设2.1.2可得

${\lambda }_0$

类似强制性的证明可得

$\lambda$

其中 $Q_{2r}\subset Q_R$ 。

记 $\phi \left(y,t\right)=\left(v-{\tilde{v}}_{2r}\left(t\right)\right){\xi }_{2r}^2\left(y\right){\tau }_{2r}^2\left(t\right)$ ，故有

$\xi ,\tau$

取 $\begin{array}{l}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{2r,i}}\frac{1}{{\beta }_i}}\frac{\partial \left(\left(v_i-{\tilde{v}}_{2r}\left(t\right)\right)\xi \tau \right)}{\partial t}\left(v_i-{\tilde{v}}_{2r}\left(t\right)\right)\xi \tau +\frac{1}{{\beta }_i}{A}_i\left(\xi \tau \nabla v_i\right)\cdot \left(\xi \tau \nabla v_i\right)\text{d}y}\\ ={\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{2r,i}}{\tilde{f}}_i\left(v_i-{\tilde{v}}_{2r}\left(t\right)\right){\xi }^2{\tau }^2\text{d}y}}+{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{2r,i}}\frac{1}{{\beta }_i}}}{\left(v_i-{\tilde{v}}_{2r}\left(t\right)\right)}^2{\xi }^2\tau \frac{\text{d}\tau }{\text{d}t}\text{d}y-{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{2r,i}}\frac{1}{{\beta }_i}}}\frac{\partial {\tilde{v}}_{2r}\left(t\right)}{\partial t}\left(v_i-{\tilde{v}}_{2r}\left(t\right)\right){\xi }^2{\tau }^2\text{d}y\\ +{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{2r,i}}\frac{1}{{\beta }_i}{z}_i}}\cdot \left(v_i-{\tilde{v}}_{2r}\left(t\right)\right)\tau \nabla \xi \left(\left(v_i-{\tilde{v}}_{2r}\left(t\right)\right)\xi \tau \right)\text{d}y-{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{2r,i}}\frac{1}{{\beta }_i}{A}_i\nabla v_i}}\cdot {\tau }^{2}2\xi \left(\nabla \xi \right)\left(v_i-{\tilde{v}}_{2r}\left(t\right)\right)\text{d}y.\end{array}$ 充分小，可得

$\begin{array}{l}\left|{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{2r,i}}\frac{1}{{\beta }_i}}}\frac{\partial {\tilde{v}}_{2r}\left(t\right)}{\partial t}\left(v_i-{\tilde{v}}_{2r}\left(t\right)\right){\xi }^2{\tau }^2\text{d}y\right|\\ \le {\tau }^2\left|\frac{\partial {\tilde{v}}_{2r}\left(t\right)}{\partial t}\right|\max \left\{\frac{1}{{\beta }_1},\frac{1}{{\beta }_2}\right\}\left|{\displaystyle {\int }_{B_{2r}}\left(v-{\tilde{v}}_{2r}\left(t\right)\right){\xi }^2\text{d}y}\right|=0,\end{array}$

上式两边对t从0到T积分

$\begin{array}{l}\left|{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{2r,i}}\frac{1}{{\beta }_i}{z}_i\cdot \left(v_i-{\tilde{v}}_{2r}\left(t\right)\right)\tau \nabla \xi \left(\left(v_i-{\tilde{v}}_{2r}\left(t\right)\right)\xi \tau \right)\text{d}y}}\right|\\ \le 2\epsilon {\displaystyle \underset{i=1}{\overset{2}{\sum }}\frac{1}{{\beta }_i}}{\displaystyle {\int }_{B_{2r,i}}{\left|\xi \tau \nabla v_i\right|}^2\text{d}y}+C{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{2r,i}}\frac{1}{{\beta }_i}}}{\left|\nabla \xi \right|}^2{\left|v_i-{\tilde{v}}_{2r}\left(t\right)\right|}^2\text{d}y,\end{array}$

因此

$\begin{array}{l}\left|{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{2r,i}}\frac{1}{{\beta }_i}{A}_i\nabla v_i}}\cdot {\tau }^2\left(2\xi \right)\left(\nabla \xi \right)\left(v_i-{\tilde{v}}_{2r}\left(t\right)\right)\text{d}y\right|\\ \le \epsilon {\displaystyle \underset{i=1}{\overset{2}{\sum }}\frac{1}{{\beta }_i}}{\displaystyle {\int }_{B_{2r,i}}{\left|\xi \tau \nabla v_i\right|}^2\text{d}y}+C{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{2r,i}}\frac{1}{{\beta }_i}}}{\left|\nabla \xi \right|}^2{\left|v_i-{\tilde{v}}_{2r}\left(t\right)\right|}^2\text{d}y,\end{array}$

对于 $\begin{array}{l}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{2r,i}}{\tilde{f}}_i\left(v_i-{\tilde{v}}_{2r}\left(t\right)\right){\xi }^2{\tau }^2\text{d}y}}\\ \le 2\epsilon {\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{2r,i}}{\left|\xi \tau \nabla v_i\right|}^2\text{d}y}}+2\epsilon {\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{2r,i}}{\left|\nabla \xi \right|}^2}}{\left|v_i-{\tilde{v}}_{2r}\left(t\right)\right|}^2\text{d}y+C{r}^{n\left(1-\frac{2}{p}\right)}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left({\displaystyle {\int }_{B_{2r,i}}{\left|{\tilde{f}}_i\right|}^{\frac{np}{n+p}}\text{d}y}\right)}^{\frac{2\left(n+p\right)}{np}}},\end{array}$ ，

$\epsilon$

则

$\begin{array}{l}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{2r,i}}\frac{\partial \left(\left(v_i-{\tilde{v}}_{2r}\left(t\right)\right)\xi \tau \right)}{\partial t}\left(v_i-{\tilde{v}}_{2r}\left(t\right)\right)\xi \tau \text{d}y}}+\left(\lambda -5\epsilon \right){\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{2r,i}}{\left|\xi \tau \nabla v_i\right|}^2\text{d}y}}\\ \le C\left\{{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{2r,i}}\left(\left|\frac{\text{d}\tau }{\text{d}t}\right|+{\left|\nabla \xi \right|}^2\right){\left|v_i-{\tilde{v}}_{2r}\left(t\right)\right|}^2\text{d}y}}+r^{n\left(1-\frac{2}{p}\right)}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left({\displaystyle {\int }_{B_{2r,i}}{\left|{\tilde{f}}_i\right|}^{\frac{np}{n+p}}\text{d}y}\right)}^{\frac{2\left(n+p\right)}{np}}}\right\}.\end{array}$

令 $t\in \left(t_0-4r^2,t_0\right]$ ，故有

$\begin{array}{l}\underset{t_0-4r^2<t\le t_0}{\sup }{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{2r,i}}{\left|\left(v_i-{\tilde{v}}_{2r}\left(t\right)\right)\xi \tau \right|}^2\text{d}y}}+{\lambda }_0{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{2r,i}}{\left|\xi \tau \nabla v_i\right|}^2\text{d}y\text{d}t}}\\ \le C\left\{\frac{1}{r^2}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{2r,i}}{\left|v_i-{\tilde{v}}_{2r}\left(t\right)\right|}^2\text{d}y\text{d}t}}+r^{\left(n+2\right)\left(1-\frac{2}{p}\right)-\frac{4}{n}}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left({\displaystyle {\int }_{Q_{2r,i}}{\left|{\tilde{f}}_i\right|}^{\frac{np}{n+p}}\text{d}y\text{d}t}\right)}^{\frac{2\left(n+p\right)}{np}}}\right\}.\end{array}$

由推论2.3.1可得，对于 ${\xi }_r$ ，有 $B_r$ ，即，

${\displaystyle {\int }_{B_r}{\xi }_r^2}\text{d}y\ge \left|B_{\frac{r}{2}}\right|=\frac{\left|B_r\right|}{2^n}$

所以在 $g\left(l\right)={\displaystyle {\int }_{B_r}{\left(v\left(y,t\right)-l\right)}^2{\xi }_r^2\text{d}y}$ 中， $g\left(l\right)$ ，因此

$l={\left({\displaystyle {\int }_{B_r}{\xi }_r^2}\left(y\right)\text{d}y\right)}^{-1}{\displaystyle {\int }_{B_r}v}\left(y,t\right){\xi }_r^2\left(y\right)\text{d}y={\tilde{v}}_r(\; t\; )$

□

推论2.3.2令问题(1.1)的系数满足假设2.1.2。如果 ${\displaystyle {\int }_{B_r}{\left|{\tilde{v}}_{2r}\left(t\right)-{\tilde{v}}_r\left(t\right)\right|}^2}{\xi }_r^2\text{d}y\le 4{{\displaystyle {\int }_{B_r}\left|v-{\tilde{v}}_{2r}\left(t\right)\right|}}^2{\xi }_r^2\text{d}y,$ 是问题的弱解，且对于某个常数 ${\displaystyle {\int }_{B_r}{\left|v-{\tilde{v}}_r(t)\right|}^{2}dy}\le 2{\displaystyle {\int }_{B_r}{\left|v-{\tilde{v}}_{2r}\left(t\right)\right|}^2\text{d}y}+2{\displaystyle {\int }_{B_r}{\left|{\tilde{v}}_{2r}\left(t\right)-{\tilde{v}}_r\left(t\right)\right|}^2\text{d}y}\le 2\left(1+2^{n+2}\right){\displaystyle {\int }_{B_{2r}}{\left|v-{\tilde{v}}_{2r}\left(t\right)\right|}^2{\xi }_{2r}^2\text{d}y,}$ ， $w\left(y\right)\in W^{1,q}\left(B_r\right)\left(1\le q<\infty \right)$ 则

${\Vert w\Vert }_{W^{1,q}\left(B_r\right)}\le C\left\{{\Vert \nabla w\Vert }_{L^q\left(B_r\right)}+{\left({\displaystyle {\int }_{B_r}{\xi }_r^2}\left(y\right)\text{d}y\right)}^{-1}{\displaystyle {\int }_{B_r}w}{\xi }_r^2\text{d}y\right\},$

其中C仅依赖于 $1\le q<n$ ，且 ${\Vert w\Vert }_{L^s\left(B_r\right)}\le C{r}^{1+\frac{n}{s}-\frac{n}{q}}{\Vert w\Vert }_{W^{1,q}\left(B_r\right)},\left(1\le s\le \frac{nq}{n-q}\right),$ 。

2.4. 弱解的局部性质

定义2.4.1 称定义于 $q=n$ 上的函数 ${\Vert w\Vert }_{L^s\left(B_r\right)}\le C{r}^{1+\frac{n}{s}-\frac{n}{q}}{\Vert w\Vert }_{W^{1,q}\left(B_r\right)},\left(1\le s<\infty \right).$ 属于De Giorgi类，如果 ${\Vert w-{\tilde{w}}_r\Vert }_{L^s\left(B_r\right)}\le C{r}^{1+\frac{n}{s}-\frac{n}{q}}{\Vert \nabla w\Vert }_{L^q\left(B_r\right)},$ ， $1\le q<n$ ，且对于 $1\le s\le \frac{nq}{n-q}$ ， $q=n$ ， $1\le s<\infty$ 满足 $p>n+2$ ，并且 $q_1>\frac{np}{n+p}$ ，有下式成立：

$\tilde{f}\in L^{q_1}\left(Q_R\right)$ (2.2)

其中 $v\left(y,t\right)$ ， $\begin{array}{l}\underset{t_0-r^2<t\le t_0}{\sup }{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{r,i}}{\left|v_i-{\tilde{v}}_r\left(t\right)\right|}^2\text{d}y}}+{\lambda }_0{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{r,i}}{\left|\nabla v_i\right|}^2\text{d}y\text{d}t}}\\ \le C\left\{{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{2r,i}}{\left|\nabla v_i\right|}^2\text{d}y\text{d}t}}+r^{\left(n+2\right)\left(1-\frac{2}{p}\right)-\frac{4}{n}}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left({\displaystyle {\int }_{Q_{2r,i}}{\left|{\tilde{f}}_i\right|}^{\frac{np}{n+p}}\text{d}y\text{d}t}\right)}^{\frac{2\left(n+p\right)}{np}}}\right\},\end{array}$ ，常数 $n,{\beta }_1,{\beta }_2,c_0,\Lambda ,p$ ， ${\lambda }_0$ 只与 $\lambda$ 有关， ${\bar{Q}}_{2r}\subset Q_R$ ， $Q\subset {ℝ}^{n+1}$ ， $F\in L_{loc}^m\left(Q\right)$ 依赖于 $G\in L_{loc}^{m_1}\left(Q\right)\left(1<m<m_1\right)$ ，记De Giorgi类为 $F,G\ge 0$ 。如果 $Q_r\subset {\bar{Q}}_{2r}\subset Q$ ，且满足(2.2)⁺，则记 $\frac{1}{\left|Q_r\right|}{\displaystyle {\int }_{Q_r}{F}^m}\text{d}y\text{d}t\le a\left\{\frac{1}{\left|Q_{2r}\right|}{\displaystyle {\int }_{Q_{2r}}{G}^m}\text{d}y\text{d}t+{\left(\frac{1}{\left|Q_{2r}\right|}{\displaystyle {\int }_{Q_{2r}}F}\text{d}y\text{d}t\right)}^m\right\}+\theta \frac{1}{\left|Q_{2r}\right|}{\displaystyle {\int }_{Q_{2r}}{F}^m}\text{d}y\text{d}t,$ ；如果 $a\ge 1$ ，且满足(2.2)^-，则记 $\theta \in \left[0,1\right)$ 。显然 $\epsilon =\epsilon \left(a,\theta ,m,n\right)>0$ 。

定理2.4.1 设问题(1.1)的系数满足假设2.1.2。如果 $F\in L_{loc}^{m_0}\left(Q\right)\left(\forall m<m_0<\min \left\{m+\epsilon ,m_1\right\}\right),$ 是问题的弱下解，且对于某个常数 $\frac{1}{\left|Q_r\right|}{\displaystyle {\int }_{Q_r}{F}^{m_0}}\text{d}y\text{d}t\le c\left\{\frac{1}{\left|Q_{2r}\right|}{\displaystyle {\int }_{Q_{2r}}{G}^{m_0}}\text{d}y\text{d}t+{\left(\frac{1}{\left|Q_{2r}\right|}{\displaystyle {\int }_{Q_{2r}}{F}^m}\text{d}y\text{d}t\right)}^{\frac{m_0}{m}}\right\},$ ， $n,m,a,\theta$ ，则 $a\to \infty$ ；如果 $\epsilon \to 0$ 是问题的弱上解，则 $Q_{4r}$ 。其中 $Q_{2r}$ 依赖于 $L^q$ ，并且 $p>n+2$ 。

证明 如果 $q_1>\frac{np}{n+p}$ 是问题的弱下解，令 $\tilde{f}\in L^{q_1}\left(Q_R\right)$ 是 $q>2$ 上的截断函数， $\nabla v_i\in L_{loc}^q\left(Q_{R,i}\right),\left(i=1,2\right)$ ，并且 $Q_{4r}\subset {\bar{Q}}_{4r}\subset Q_R$ 。记 ${\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{r,i}}{\left|\nabla v_i\right|}^q\text{d}y\text{d}t}}\le C\left\{{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{4r,i}}{\left|{\tilde{f}}_i\right|}^{q_1}\text{d}y\text{d}t}}+r^{\left(n+2\right)\left(1-\frac{q}{2}\right)}{\left({\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{4r,i}}{\left|\nabla v_i\right|}^2\text{d}y\text{d}t}}\right)}^{\frac{q}{2}}\right\},$ 。取测试函数为 $\begin{array}{l}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{2r,i}}{\left|v_i-{\tilde{v}}_{2r}\left(t\right)\right|}^2\text{d}y\text{d}t}}={\displaystyle {\int }_{Q_{2r}}{\left|v-{\tilde{v}}_{2r}\left(t\right)\right|}^2\text{d}y\text{d}t}\\ \le {\left(\underset{t_0-4r^2<t\le t_0}{\sup }{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{2r,i}}{\left|v_i-{\tilde{v}}_{2r}\left(t\right)\right|}^2\text{d}y}}\right)}^{\frac{1}{2}}{\displaystyle {\int }_{t_0-4r^2}^{t_0}{\left({\displaystyle {\int }_{B_{2r}}{\left|v-{\tilde{v}}_{2r}\left(t\right)\right|}^2\text{d}y}\right)}^{\frac{1}{2}}\text{d}t}\\ \le C{\left\{{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{4r,i}}{\left|\nabla v_i\right|}^2\text{d}y\text{d}t}}+r^{\left(n+2\right)\left(1-\frac{2}{p}\right)-\frac{4}{n}}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left({\displaystyle {\int }_{Q_{4r,i}}{\left|f_i\right|}^{\frac{np}{n+p}}\text{d}y\text{d}t}\right)}^{\frac{2\left(n+p\right)}{np}}}\right\}}^{\frac{1}{2}}\\ \times {\displaystyle {\int }_{t_0-4r^2}^{t_0}{\left({\displaystyle {\int }_{B_{2r}}{\left|v-{\tilde{v}}_{2r}\left(t\right)\right|}^{\frac{2n}{n-2}}\text{d}y}\right)}^{\frac{n-2}{4n}}}{\left({\displaystyle {\int }_{B_{2r}}{\left|v-{\tilde{v}}_{2r}\left(t\right)\right|}^{\frac{2n}{n+2}}\text{d}y}\right)}^{\frac{n+2}{4n}}\text{d}t\end{array}$ ，则

$s=\frac{2n}{n-2},q=2$

其中

${\left({\displaystyle {\int }_{B_{2r}}{\left|v-{\tilde{v}}_{2r}\left(t\right)\right|}^{\frac{2n}{n-2}}\text{d}y}\right)}^{\frac{n-2}{4n}}\le C{\left({\displaystyle {\int }_{B_{2r}}{\left|\nabla v\right|}^2\text{d}y}\right)}^{\frac{1}{4}}.$

$s=q=\frac{2n}{n+2}$

则

${\left({\displaystyle {\int }_{B_{2r}}{\left|v-{\tilde{v}}_{2r}\left(t\right)\right|}^{\frac{2n}{n+2}}\text{d}y}\right)}^{\frac{n+2}{4n}}\le C{r}^{\frac{1}{2}}{\left({\displaystyle {\int }_{B_{2r}}{\left|\nabla v\right|}^{\frac{2n}{n+2}}\text{d}y}\right)}^{\frac{n+2}{4n}}.$

由于

$\begin{array}{l}{r}^{\frac{1}{2}}{\displaystyle {\int }_{t_0-4r^2}^{t_0}{\left({\displaystyle {\int }_{B_{2r}}{\left|\nabla v\right|}^2\text{d}y}\right)}^{\frac{1}{4}}{\left({\displaystyle {\int }_{B_{2r}}{\left|\nabla v\right|}^{\frac{2n}{n+2}}\text{d}y}\right)}^{\frac{n+2}{4n}}}\text{d}t\\ \le C{r}^{\frac{1}{2}}{\left\{{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{4r,i}}{\left|\nabla v_i\right|}^2}}\text{d}y\text{d}t+r^{\left(n+2\right)\left(1-\frac{2}{p}\right)-\frac{4}{n}}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left({\displaystyle {\int }_{Q_{4r,i}}{\left|{\tilde{f}}_i\right|}^{\frac{np}{n+p}}\text{d}y\text{d}t}\right)}^{\frac{2\left(n+p\right)}{np}}}\right\}}^{\frac{1}{4}}{r}^{1-\frac{1}{n}}{\left({\displaystyle {\int }_{Q_{2r}}{\left|\nabla v\right|}^{\frac{2n}{n+2}}\text{d}y\text{d}t}\right)}^{\frac{n+2}{4n}},\end{array}$

$\begin{array}{l}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{2r,i}}{\left|v_i-{\tilde{v}}_{2r}\left(t\right)\right|}^2\text{d}y\text{d}t}}\\ \le C{r}^{\frac{3}{2}-\frac{1}{n}}{\left\{{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{4r,i}}{\left|\nabla v_i\right|}^2\text{d}y\text{d}t}}+r^{\left(n+2\right)\left(1-\frac{2}{p}\right)-\frac{4}{n}}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left({\displaystyle {\int }_{Q_{4r,i}}{\left|{\tilde{f}}_i\right|}^{\frac{np}{n+p}}\text{d}y\text{d}t}\right)}^{\frac{2\left(n+p\right)}{np}}}\right\}}^{\frac{3}{4}}{\left({\displaystyle {\int }_{Q_{2r}}{\left|\nabla v\right|}^{\frac{2n}{n+2}}\text{d}y\text{d}t}\right)}^{\frac{n+2}{4n}}.\end{array}$ (2.3)

则

$\begin{array}{l}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{r,i}}{\left|\nabla v_i\right|}^2\text{d}y\text{d}t}}\\ \le C{r}^{-\frac{1}{2}-\frac{1}{n}}{\left\{{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{4r,i}}{\left|\nabla v_i\right|}^2\text{d}y\text{d}t}}+r^{\left(n+2\right)\left(1-\frac{2}{p}\right)-\frac{4}{n}}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left({\displaystyle {\int }_{Q_{4r,i}}{\left|{\tilde{f}}_i\right|}^{\frac{np}{n+p}}\text{d}y\text{d}t}\right)}^{\frac{2\left(n+p\right)}{np}}}\right\}}^{\frac{3}{4}}{\left({\displaystyle {\int }_{Q_{2r}}{\left|\nabla v\right|}^{\frac{2n}{n+2}}\text{d}y\text{d}t}\right)}^{\frac{n+2}{4n}}\\ +C{r}^{\left(n+2\right)\left(1-\frac{2}{p}\right)-\frac{4}{n}}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left({\displaystyle {\int }_{Q_{4r,i}}{\left|{\tilde{f}}_i\right|}^{\frac{np}{n+p}}\text{d}y\text{d}t}\right)}^{\frac{2\left(n+p\right)}{np}}}\\ \le \epsilon {\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{4r,i}}{\left|\nabla v_i\right|}^2\text{d}y\text{d}t}}+C_1{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{4r,i}}{\left|{\tilde{f}}_i\right|}^{\frac{np}{n+p}}\text{d}y\text{d}t}}+C\left(\epsilon \right)r^{-\frac{2\left(n+2\right)}{n}}{\left({\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{4r,i}}{\left|\nabla v_i\right|}^{\frac{2n}{n+2}}\text{d}y\text{d}t}}\right)}^{\frac{n+2}{n}},\end{array}$

取 $C_1$ 充分小，对于任意 ${\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert {\tilde{f}}_i\Vert }_{L^{\frac{np}{n+p}}\left(Q_{R,i}\right)}}$ ，积分可得

$\left|Q_r\right|$ (2.4)

对于 ，在 $\left|Q_{4r}\right|=4^{n+2}\left|Q_r\right|$ 上关于t取上确界，有 $\begin{array}{c}\frac{1}{\left|Q_r\right|}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{r,i}}{\left|\nabla v_i\right|}^2\text{d}y\text{d}t}}\le \epsilon 4^{n+2}\frac{1}{\left|Q_{4r}\right|}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{4r,i}}{\left|\nabla v_i\right|}^2\text{d}y\text{d}t}}+C_1\frac{1}{\left|Q_{4r}\right|}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{4r,i}}{\left|{\tilde{f}}_i\right|}^{\frac{np}{n+p}}\text{d}y\text{d}t}}\\ +C\left(\epsilon \right){\left(\frac{1}{\left|Q_{4r}\right|}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{Q_{4r,i}}{\left|\nabla v_i\right|}^{\frac{2n}{n+2}}}\text{d}y\text{d}t}\right)}^{\frac{n+2}{n}}.\end{array}$ 成立。

如果 $\epsilon \in \left(0,1\right]$ 是问题的弱上解，同理可证 $\epsilon 4^{n+2}\le \frac{1}{2}$ 。

□

注2.4.1 如果 $r\in \left(0,R\right]$ ，则由 可得

$\frac{1}{\left|Q_r\right|}{\displaystyle {\int }_{Q_r}{\left|\nabla v\right|}^2\text{d}y\text{d}t}\le \frac{1}{2}\frac{1}{\left|Q_{4r}\right|}{\displaystyle {\int }_{Q_{4r}}{\left|\nabla v\right|}^2\text{d}y\text{d}t}+C_1\frac{1}{\left|Q_{4r}\right|}{\displaystyle {\int }_{Q_{4r}}{\left|\tilde{f}\right|}^{\frac{np}{n+p}}\text{d}y\text{d}t}+C{\left(\frac{1}{\left|Q_{4r}\right|}{\displaystyle {\int }_{Q_{4r}}{\left|\nabla v\right|}^{\frac{2n}{n+2}}\text{d}y\text{d}t}\right)}^{\frac{n+2}{n}}.$ (2.5)

定理2.4.2 (局部极值原理)令 $m=\frac{n+2}{n},m_1=\frac{q_1\left(n+p\right)\left(n+2\right)}{n^{2}p},$ ， $F={\left|\nabla v\right|}^{\frac{2n}{n+2}},G={\left({\left|\tilde{f}\right|}^{\frac{np}{n+p}}\right)}^{\frac{n}{n+2}}.$ ，常数 $F\in L_{loc}^m\left(Q_R\right),G\in L_{loc}^{m_1}\left(Q_R\right)\left(1<m<m_1\right).$ ，则对于任意 $\frac{1}{\left|Q_r\right|}{\displaystyle {\int }_{Q_r}{F}^m\text{d}y\text{d}t}\le \frac{1}{2}\frac{1}{\left|Q_{4r}\right|}{\displaystyle {\int }_{Q_{4r}}{F}^m\text{d}y\text{d}t}+C\left\{\frac{1}{\left|Q_{4r}\right|}{\displaystyle {\int }_{Q_{4r}}{G}^m\text{d}y\text{d}t}+{\left(\frac{1}{\left|Q_{4r}\right|}{\displaystyle {\int }_{Q_{4r}}F\text{d}y\text{d}t}\right)}^m\right\},$ ，有

${\beta }_1,{\beta }_2,n,\lambda ,\Lambda ,c_0,p,{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert {\tilde{f}}_i\Vert }_{L^{\frac{np}{n+p}}\left(Q_{R,i}\right)}}$

如果 $m_0\in \left(m,m_1\right]$ ，则有

$F\in L_{loc}^{m_0}(\; Q\; R\; )$

其中C仅依赖于 $\frac{1}{\left|Q_r\right|}{\displaystyle {\int }_{Q_r}{F}^{m_0}\text{d}y\text{d}t}\le C\left\{\frac{1}{\left|Q_{4r}\right|}{\displaystyle {\int }_{Q_{4r}}{G}^{m_1}\text{d}y\text{d}t}+{\left(\frac{1}{\left|Q_{4r}\right|}{\displaystyle {\int }_{Q_{4r}}{F}^m\text{d}y\text{d}t}\right)}^{\frac{m_0}{m}}\right\}.$ 的参数， $q=\frac{2n{m}_0}{n+2}$ ，并且 $q>2$ 。

证明 下面只证明第一种情况。令 ${\displaystyle {\int }_{Q_r}{\left|\nabla v\right|}^q\text{d}y\text{d}t}\le C\left\{{\displaystyle {\int }_{Q_{4r}}{\left|\tilde{f}\right|}^{q_1}\text{d}y\text{d}t}+r^{\left(n+2\right)\left(1-\frac{q}{2}\right)}{\left({\displaystyle {\int }_{Q_{4r}}{\left|\nabla v\right|}^2\text{d}y\text{d}t}\right)}^{\frac{q}{2}}\right\}.$ ； $L^q$ ，其中 $Q_i={\Omega }_i\times \left(0,T\right]\left(i=1,2\right)$ 待定。令 $L^q$ ，且取 $\left(i=1,2\right)$ 是 $C^{\mu ,k}\left({\bar{\Omega }}_i\times \left[0,T\right]\right)=\{g\left(x,t\right):g\left(x,t\right)$ 上的截断函数，并且满足

$C^k$

应用公式(2.5)⁺，有

$\mu$

令 $\mu \in \left[0,1\right],k\in {\mathbb{N}}_{+}\}$ 。首先取 ${\Vert g\left(x,t\right)\Vert }_{C^{\mu ,k}\left({\bar{\Omega }}_i\times \left[0,T\right]\right)}={\displaystyle \underset{s=0}{\overset{k}{\sum }}\underset{{\bar{\Omega }}_i\times \left[0,T\right]}{\sup }}\left|D_t^{s}g\left(x,t\right)\right|+\underset{x,y\in {\Omega }_i,t\in \left[0,T\right]}{\sup }\frac{\left|g\left(x,t\right)-g\left(y,t\right)\right|}{{\left|x-y\right|}^{\mu }}.$ ，则有

$K\left(x,t\right)$

令 $K_i\left(x,t\right)\in C^{\mu ,\infty }\left({\bar{\Omega }}_i\times \left[0,T\right]\right)$ ，则有

$\mu \in \left(0,1\right)$

又因为 $C^{\prime }$ ，则

$\left|K_i\left(x,t\right)-K_i\left(y,t\right)\right|\le C^{\prime }{\left|x-y\right|}^{\mu },\forall \left(x,t\right),\left(y,t\right)\in {\Omega }_i\times \left(0,T\right).$

故有

$l\ge 1$

所以

$\begin{array}{c}{\phi }_{m+1}\le {\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert {\xi }_m{\left(u_i-k_{m+1}\right)}^{+}\Vert }_{L^2\left(t_0-{\rho }_m^2,t_0;L^2\left(B_{{\rho }_m,i}\right)\right)}^2}\\ \le C{\left|A_m\left(k_{m+1}\right)\right|}^{\frac{2}{n}}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert {\xi }_m{\left(u_i-k_{m+1}\right)}^{+}\Vert }_{L^2\left(t_0-{\rho }_m^2,t_0;L^{2^{*}}\left(B_{{\rho }_m,i}\right)\right)}^2}\\ \le C{\left|A_m\left(k_{m+1}\right)\right|}^{\frac{2}{n}}\left(\frac{C{2}^{2m}}{R^2}{\phi }_m+\frac{C{k}^2}{R^{2\left(1-\frac{n}{p}\right)}}{\left|A_m\left(k_{m+1}\right)\right|}^{1-\frac{2}{p}}\right)\\ \le C{2}^{2m\left(1+\frac{2}{n}\right)}\left(\frac{{\phi }_m^{1+\frac{2}{n}}}{R^2{k}^{\frac{4}{n}}}+\frac{{\phi }_m^{1-\frac{2}{p}+\frac{2}{n}}}{R^{2\left(1-\frac{n}{p}\right)}{k}^{\frac{4}{n}-\frac{4}{p}}}\right).\end{array}$

令 ${\displaystyle \underset{s=0}{\overset{l}{\sum }}\left|D_t^s{K}_i\left(x,t\right)\right|}\le {\Lambda }_{2l},{\Omega }_i\times \left(0,T\right);$ ，则

${\displaystyle \underset{s=0}{\overset{l}{\sum }}\left|D_t^s{K}_i\left(x,t\right)-D_t^s{K}_i\left(y,t\right)\right|}\le {\Lambda }_{2l}{\left|x-y\right|}^{\mu },{\Omega }_i\times \left(0,T\right).$

因此

$ϵ>0$

令 ${\Omega }_{ϵ}=\left\{x\in \Omega |dist\left(x,\partial \Omega \right)>ϵ\right\}.$ ，则

$\tilde{f}\in L^{\infty }(\; Q\; T\; )$

由引理2.3.1可得，如果

$\tilde{u}\in W\left(0,T;V\right)$ (2.6)

其中 $0<ϵ<\frac{1}{2}$ ，即

${\alpha }^{*}<\min \left\{\mu ,\frac{\alpha }{2\left(1+\alpha \right)}\right\}$ (2.7)

则 ${\Vert \tilde{u}\Vert }_{L^{\infty }\left({\Omega }_{ϵ}\times \left(ϵT,T\right)\right)}+{\Vert {\nabla }_x\tilde{u}\Vert }_{C^{{\alpha }^{*},0}\left(\left({\Omega }_{ϵ}\cap {\bar{\Omega }}_i\right)\times \left(ϵT,T\right)\right)}\le C\left({\Vert \tilde{u}\Vert }_{L^2\left(Q_T\right)}+{\Vert \tilde{f}\Vert }_{L^{\infty }\left(Q_T\right)}\right),$ ，所以在 $n,{\beta }_1,{\beta }_2,c_0,\lambda ,\Lambda ,C^{\prime },\mu ,\alpha ,ϵ,T,{\Vert K_i\Vert }_{C^{{\alpha }^{*},1}\left({\bar{\Omega }}_i\times \left[0,T\right]\right)}$ 中，我们有 ${\Omega }_i$ 。现在取k满足 $C^{1+\alpha }$ ，则条件(2.7)成立，即(2.6)满足。总结以上关于k的选取，最终取k为

$\left(i=1,2\right)$

所以结论得证。

□

有了局部极值原理之后，我们总假设弱解在界面附近是有界的。在此前提之下，将会得到弱解的一些局部性质。为了适应方程的需要，下面对De Giorgi类的定义稍加修改。

定义2.4.2我们称定义于 ${\Vert {\nabla }_x\tilde{u}\Vert }_{L^{\infty }\left({\Omega }_{ϵ}\times \left(ϵT,T\right)\right)}\le C\left({\Vert \tilde{u}\Vert }_{L^2\left(Q_T\right)}+{\Vert \tilde{f}\Vert }_{L^{\infty }\left(Q_T\right)}\right).$ 上的函数 $L^q$ 属于De Giorgi类，如果 $q>2$ ， $L^q$ ， $L^q$ ，并且存在 $\Omega \subset {ℝ}^n\left(n>2\right)$ ，使得对于任意k满足：

$\partial \Omega$ (2.8)

有 ${\Omega }_1\subset \Omega$ 满足 ，记 $\Gamma \in C^2$ ，同样可定义 ${\Omega }_2=\Omega \setminus {\Omega }_1$ 。

引理2.4.1 (De Giorgi lemma) (文献 [13] )设 $\Gamma =\partial {\Omega }_1$ ，记 $\Gamma$ ，则对于 $\partial \Omega$ ，有

$u=\left(u_1,u_2\right)$

其中 $u_1:{\Omega }_1\times \left(0,T\right]\to ℝ,u_2:{\Omega }_2\times \left(0,T\right]\to ℝ$ 只依赖于n。

定理2.4.3令 $u\left(x,t\right)$ ， $n$ ， $\Gamma$ ， ${\Omega }_1$ ，则存在 ${\Omega }_2$ ，使得对于 $i=1,2$ ，如果

$w_i=w_i\left(x\right)$ (2.9)

$K_i=K_i\left(x,t\right)$ (2.10)

则

$m_{\Gamma }=m_{\Gamma }\left(x,t\right)$

其中 $\begin{array}{ll}\frac{\partial u_i}{\partial t}+w_i\cdot \nabla u_i-div\left(K_i\left(x,t\right)\nabla u_i\right)=f_i\left(x,t\right),\hfill & x\in {\Omega }_i,i=1,2,t\in \left(0,T\right];\hfill \\ m_{\Gamma }\left(x,t\right){\left[u\right]}_{\Gamma }=K_1\left(x,t\right)\nabla u_1\cdot n,\hfill & x\in \Gamma ,t\in \left(0,T\right];\hfill \\ K_1\left(x,t\right)\nabla u_1\cdot n=K_2\left(x,t\right)\nabla u_2\cdot n,\hfill & x\in \Gamma ,t\in \left(0,T\right];\hfill \\ u_i\left(\cdot ,0\right)=u_0^i,\hfill & x\in {\Omega }_i,i=1,2;\hfill \\ u_2\left(\cdot ,t\right)=0,\hfill & x\in \partial \Omega ,t\in \left(0,T\right].\hfill \end{array}$ 仅依赖于 ${\left[u\right]}_{\Gamma }={u_2|}_{\Gamma }-{u_1|}_{\Gamma }$ 的参数，并且 $u_0\left(x\right)=\left(u_0^1,u_0^2\right)$ 。

证明令 $u_0\left(x\right)$ ； $\begin{array}{ll}\frac{\partial u_i}{\partial t}+w_i\cdot \nabla u_i-div\left(K_i\left(x,t\right)\nabla u_i\right)=f_i\left(x,t\right),\hfill & x\in {\Omega }_i,i=1,2,t\in \left(0,T\right];\hfill \\ K_1\left(x,t\right)\nabla u_1\cdot n=K_2\left(x,t\right)\nabla u_2\cdot n,\hfill & x\in \Gamma ,t\in \left(0,T\right];\hfill \\ {\left[\beta u\right]}_{\Gamma }=0,\hfill & x\in \Gamma ,t\in \left(0,T\right];\hfill \\ u_i\left(\cdot ,0\right)=u_0^i,\hfill & x\in {\Omega }_i,i=1,2;\hfill \\ u_2\left(\cdot ,t\right)=0,\hfill & x\in \partial \Omega ,t\in \left(0,T\right].\hfill \end{array}$ 。记 $\beta$ ，取与定理2.4.2中相同的截断函数 ${\Omega }_i$ 。在 ⁺中分别取 $\beta ={\beta }_i>0$ 代替 ${\beta }_1\ne {\beta }_2$ ，则

${\left[\beta u\right]}_{\Gamma }={{\beta }_2{u}_2|}_{\Gamma }-{{\beta }_1{u}_1|}_{\Gamma }$

令 $u_0\left(x\right)$ ，则

$p\in \left[1,\infty \right]$

又因为

$W^{1,p}(\; \Omega \; )$

所以

$H^1\left(\Omega \right):=W^{1,2}(\; \Omega \; )$

由条件 可得， $\{\begin{array}{l}{V}_1=H^1\left({\Omega }_1\right);\hfill \\ V_2=\left\{v_2\in H^1\left({\Omega }_2\right)|{v_2|}_{\partial \Omega }=0\right\};\hfill \\ V=V_1\times V_2,{\Vert v\Vert }_V^2={\Vert v_1\Vert }_{H^1\left({\Omega }_1\right)}^2+{\Vert v_2\Vert }_{H^1\left({\Omega }_2\right)}^2\hfill \end{array}$ ，且 $$ ，则有

$H=L^2\left({\Omega }_1\right)\times L^2(\; \Omega \; 2\; )$

令 $W\left(0,T;V\right):=\left\{v\in L^2\left(0,T;V\right)|v_t\in L^2\left(0,T;V^{\prime }\right)\right\}$ ，且 ${\Vert v\Vert }_H^2={\Vert v_1\Vert }_{L^2\left({\Omega }_1\right)}^2+{\Vert v_2\Vert }_{L^2\left({\Omega }_2\right)}^2,{\Vert v\Vert }_W^2={\Vert v\Vert }_{L^2\left(0,T;V\right)}^2+{\Vert v_t\Vert }_{L^2\left(0,T;V^{\prime }\right)}^2.$ ，则

$W\left(0,T;V\right)\searrow C\left(\left[0,T\right];H\right),$

注意 $\searrow$ ，由引理2.3.1可得，存在 ${\text{tr}}_{\Gamma }^i:V_i\to L^2\left(\Gamma \right),\left(i=1,2\right),$ 使得当 ${\text{tr}}_{\Gamma }u={\left({\text{tr}}_{\Gamma }^1{u}_1,{\text{tr}}_{\Gamma }^2{u}_2\right)}^{\top }$ ，有 ${\Vert {\text{tr}}_{\Gamma }u\Vert }_{L^2{\left(\Gamma \right)}^2}={\Vert {\text{tr}}_{\Gamma }^1{u}_1\Vert }_{L^2\left(\Gamma \right)}+{\Vert {\text{tr}}_{\Gamma }^2{u}_2\Vert }_{L^2\left(\Gamma \right)}$ 。所以在 ${\Omega }_1$ 中， ${\Omega }_2$ 。因此结论得证。

□

定理2.4.4令 $\partial \Omega \in \text{Lip}$ ， $\Gamma =\partial {\Omega }_1\in C^2$ ，如果 ${\mathcal{H}}_{n-1}$ ， $\Gamma$ ， $n-1$ ，对于 $w_i=w_i\left(x\right)\left(i=1,2\right)$ ， ${\Omega }_i$ ， $w_i(\cdot )\in {\left[H^1\left({\Omega }_i\right)\right]}^n$ 满足

${\Vert w_i(\cdot )\Vert }_{L^{\infty }\left({\Omega }_i\right)}\le c_0<\infty$ (2.11)

其中 ${\Omega }_i$ ， $div{w}_i=0$ ，则存在 $\Gamma$ ，使得或者

$w_1\cdot n=w_2\cdot n=0$ (2.12)

或者

$K_i=K_i\left(x,t\right)\left(i=1,2\right)$ (2.13)

成立，其中 $m_{\Gamma }=m_{\Gamma }\left(x,t\right)$ ，s仅依赖于 $K_i\left(\cdot ,t\right)={\left(k_{pq}^i\left(\cdot ,t\right)\right)}_{pq}$ ，并且 ${\Omega }_i$ 。

在证明定理2.4.4之前，先给出以下两个辅助引理。

引理2.4.2令 $K_i$ ， $\lambda ,\Lambda >0$ ，如果 ${\displaystyle \underset{p,q=1}{\overset{n}{\sum }}{\Vert k_{pq}^i\Vert }_{L^{\infty }\left({\Omega }_i\right)}}\le \Lambda ,a.e.t\in \left[0,T\right];$ ， $\xi \in {ℝ}^n$ ， ${\Omega }_i$ ， $\lambda {\left|\xi \right|}^2\le {\xi }^{\top }{K}_i\left(\cdot ,t\right)\xi \le \Lambda {\left|\xi \right|}^2,a.e.t\in \left[0,T\right].$ ，对于 $m_{\Gamma }=m_{\Gamma }\left(\cdot ,t\right)$ ， $\Gamma$ ， $\bar{m}\ge \underset{\_}{m}>0$ 满足

$\underset{\_}{m}\le m_{\Gamma }\left(x,t\right)\le \bar{m},a.e.\left(x,t\right)\in \Gamma \times \left[0,T\right].$ (2.14)

其中 $M=\left(\begin{array}{cc}{m}_{\Gamma }& -m_{\Gamma }\\ -m_{\Gamma }& m_{\Gamma }\end{array}\right)$ ， $r=\left(r_1,r_2\right)\in {ℝ}^2$ ，则对于任意正整数 $r^{\top }Mr\stackrel{a.e.}{=}0\iff r_1=r_2.$ ，或者

$f_i\in L^2\left(0,T;L^2\left({\Omega }_i\right)\right)$ (2.15)

或者

$\forall v=\left(v_1,v_2\right)\in V$ (2.16)

成立，其中 ${\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{{\Omega }_i}\frac{\partial u_i}{\partial x}{v}_i}}+w_i\cdot \left(\nabla u_i\right)v_i+{\left(\nabla v_i\right)}^{\top }{K}_i\nabla u_i\text{d}x+{\displaystyle {\int }_{\Gamma }{\left({\text{tr}}_{\Gamma }u\right)}^{\top }}M\left({\text{tr}}_{\Gamma }v\right)\text{d}{\mathcal{H}}_{n-1}={\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{{\Omega }_i}{f}_i}}{v}_i\text{d}x.$ ， $\frac{\text{d}u}{\text{d}t}\left(v\right)={\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{{\Omega }_i}\frac{\partial u_i}{\partial t}{v}_i\text{d}x}},$ ，C依赖于 $a\left(t;u\left(t\right),v\right)={\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{{\Omega }_i}{w}_i}}\cdot \left(\nabla u_i\right)v_i+{\left(\nabla v_i\right)}^{\top }{K}_i\nabla u_i\text{d}x+{\displaystyle {\int }_{\Gamma }{\left({\text{tr}}_{\Gamma }u\right)}^{\top }}M\left({\text{tr}}_{\Gamma }v\right)\text{d}{\mathcal{H}}_{n-1},$ 。

证明对于 $\mathcal{F}\left(t\right)\left(v\right)={\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{{\Omega }_i}{f}_i}}{v}_i\text{d}x,$ ，记 $u_0\left(x\right)\in H,\mathcal{F}\in L^2\left(0,T;V^{\prime }\right)$ ， $u\in W\left(0,T;V\right)$ ，所以 $\{\begin{array}{l}\frac{\text{d}u}{\text{d}t}\left(v\right)+a\left(t;u\left(t\right),v\right)\le (\ge )\mathcal{F}\left(t\right)\left(v\right),\forall v\in V;\hfill \\ u\left(0\right)=u_0\left(x\right).\hfill \end{array}$ 。取 $u$ ，则 $u$ 。对于 $a\left(t;u\left(t\right),v\right)$ ，由引理2.4.1可以得到

$V\times V$ (2.17)

应用条件(2.14)可知， $a\left(t;u\left(t\right),v\right)$ ，则 $a\left(t;u\left(t\right),v\right)$ ，因此对 应用Hölder不等式，可以得到

$v\in V$ (2.18)

对 两边关于 $a\left(t;v,v\right)={\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{{\Omega }_i}{w}_i\cdot \left(\nabla v_i\right)v_i+{\left(\nabla v_i\right)}^{\top }}}{K}_i\nabla v_i\text{d}x+{\displaystyle {\int }_{\Gamma }{\left({\text{tr}}_{\Gamma }v\right)}^{\top }}M\left({\text{tr}}_{\Gamma }v\right)\text{d}{\mathcal{H}}_{n-1},$ 积分，进而可得

${\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{{\Omega }_i}{w}_i\cdot \left(\nabla v_i\right)v_i\text{d}x=0.}}$ (2.19)

令 $a\left(t;v,v\right)\ge \min \left\{\lambda ,\underset{\_}{m}\right\}\left[{\Vert \nabla v_1\Vert }_{L^2\left({\Omega }_1\right)}^2+{\Vert \nabla v_2\Vert }_{L^2\left({\Omega }_2\right)}^2+{\displaystyle {\int }_{\Gamma }{\left(v_1-v_2\right)}^2}\text{d}{\mathcal{H}}_{n-1}\right].$ ， ${\Vert v_2\Vert }_{H^1\left({\Omega }_2\right)}^2\le c_2\left({\Vert \nabla v_2\Vert }_{L^2\left({\Omega }_2\right)}^2+{\displaystyle {\int }_{\partial \Omega }{v}_2^2}\text{d}s\right)=c_2{\Vert \nabla v_2\Vert }_{L^2\left({\Omega }_2\right)}^2,$ ，并且在 两边对i求和，有

${\Vert v_1\Vert }_{H^1\left({\Omega }_1\right)}^2\le c_1\left({\Vert \nabla v_1\Vert }_{L^2\left({\Omega }_1\right)}^2+{\displaystyle {\int }_{\Gamma }{v}_1^2}\text{d}{\mathcal{H}}_{n-1}\right),$

取 ${\displaystyle {\int }_{\Gamma }{v}_1^2}\text{d}{\mathcal{H}}_{n-1}\le 2{\displaystyle {\int }_{\Gamma }{\left(v_1-v_2\right)}^2}\text{d}{\mathcal{H}}_{n-1}+2{\displaystyle {\int }_{\Gamma }{v}_2^2}\text{d}{\mathcal{H}}_{n-1}\le 2{\displaystyle {\int }_{\Gamma }{\left(v_1-v_2\right)}^2}\text{d}{\mathcal{H}}_{n-1}+2c_3^2{\Vert v_2\Vert }_{H^1\left({\Omega }_2\right)}^2,$ 是 $a\left(t;v,v\right)\ge \frac{\min \left\{\lambda ,\underset{\_}{m}\right\}}{\max \left\{c_1,2c_1,2c_3^2{c}_1\left(c_2+1\right)\right\}}{\Vert v\Vert }_V^2={\alpha }_0{\Vert v\Vert }_V^2,$ 上的截断函数，使得在 $c_1,c_2,c_3$ 上有 $X=\left(x,t_X\right),Y=\left(y,t_Y\right)\in Q_T$ 。由于 $\tilde{\delta }\left(X,Y\right)=\max \left\{\left|x-y\right|,{\left|t_x-t_y\right|}^{\frac{1}{2}}\right\}.$ ，类似定理2.4.1的证明，取 ${ℝ}^{n+1}$ ， $X\in D$ ，则有

$D\left(X,r\right)=D\cap Q_r(\; X\; )$

其中 $Q_r\left(X\right)=B_r\left(x\right)\times \left(t_X-r^2,t_X+r^2\right)$ 。如果 不成立，则 $d=diam\left(D\right)$ ，进而可得 $p\ge 1,\theta \ge 0$ ，所以

$M^{p,\theta }\left(D;\tilde{\delta }\right)$ (2.20)

对(2.20)关于l从 $L^p\left(D\right)$ 到 ${\Vert u\Vert }_{M^{p,\theta }\left(D;\tilde{\delta }\right)}:={\left(\underset{X\in \bar{D},d\ge \rho >0}{\sup }{\rho }^{-\theta }{\displaystyle {\int }_{D\left(X,\rho \right)}{\left|u\left(Y\right)\right|}^p}\text{d}Y\right)}^{\frac{1}{p}}<\infty ,$ 求和，可以得到

$\Omega \subset {ℝ}^n$

因此

$x\in \Omega$

□

引理2.4.3 令 $\Omega \left(x,r\right)=\Omega \cap B_r\left(x\right)$ ， $d_1=diam\left(\Omega \right)$ ，如果 $d^{*}\left(x,y\right)=\left|x-y\right|$ ， $M^{p,\theta }\left(\Omega ;d^{*}\right)$ ， $L^p\left(\Omega \right)$ ，对于 ${\Vert u\Vert }_{M^{p,\theta }\left(\Omega ;d^{*}\right)}:={\left(\underset{x\in \bar{\Omega },d_1\ge \rho >0}{\sup }{\rho }^{-\theta }{\displaystyle {\int }_{\Omega \left(x,\rho \right)}{\left|u\left(y\right)\right|}^p}\text{d}y\right)}^{\frac{1}{p}}<\infty ,$ ， $p\ge 1,\theta \ge 0$ ， ${\mathcal{L}}^{p,\theta }\left(D;\tilde{\delta }\right)$ 满足

$L^p\left(D\right)$ (2.21)

其中 ${\left[u\right]}_{{\mathcal{L}}^{p,\theta }\left(D;\tilde{\delta }\right)}:={\left(\underset{X\in D,d\ge \rho >0}{\sup }{\rho }^{-\theta }{\displaystyle {\int }_{D\left(X,\rho \right)}{\left|u\left(Y\right)-u_{X,\rho }\right|}^p}\text{d}Y\right)}^{\frac{1}{p}}<\infty ,$ ， ${\Vert u\Vert }_{{\mathcal{L}}^{p,\theta }\left(D;\tilde{\delta }\right)}={\left({\Vert u\Vert }_{L^p\left(D\right)}^p+{\left[u\right]}_{{\mathcal{L}}^{p,\theta }\left(D;\tilde{\delta }\right)}^p\right)}^{\frac{1}{p}},$ ，则存在 $u_{X,\rho }$ 使得或者

$D\left(X,\rho \right)$ (2.22)

或者

$u_{X,\rho }=\frac{1}{\left|D\left(X,\rho \right)\right|}{\displaystyle {\int }_{D\left(X,\rho \right)}u\left(Y\right)\text{d}Y}$ (2.23)

成立，其中 $0<\alpha \le 1$ 仅依赖于 $C^{\alpha }\left(\bar{D};\tilde{\delta }\right)$ ，并且 ${\left[u\right]}_{\alpha ;D}:=\underset{X,Y\in D,X\ne Y}{\sup }\frac{\left|u\left(X\right)-u\left(Y\right)\right|}{\tilde{\delta }{\left(X,Y\right)}^{\alpha }}<\infty ,$ 。

证明 令 ${\Vert u\Vert }_{\alpha ;D}=\underset{D}{\sup }\left|u\right|+{\left[u\right]}_{\alpha ;D}.$ 是 $C^{\alpha }\left(\bar{D};\tilde{\delta }\right)\searrow C\left(\bar{D}\right).$ 上的截断函数，使得 $C^{1+\alpha }\left(\bar{D};\tilde{\delta }\right)$ 上有 $C^1\left(\bar{D};\tilde{\delta }\right)$ ，其中 ${\left[u\right]}_{1+\alpha ;D}:={\displaystyle \underset{i=1}{\overset{n}{\sum }}\underset{X,Y\in D,X\ne Y}{\sup }}\frac{\left|D_{i}u\left(X\right)-D_{i}u\left(Y\right)\right|}{\tilde{\delta }{\left(X,Y\right)}^{\alpha }}+\underset{\left(x,t_1\right),\left(x,t_2\right)\in D,t_1\ne t_2}{\sup }\frac{\left|u\left(x,t_1\right)-u\left(x,t_2\right)\right|}{{\left|t_1-t_2\right|}^{\frac{1+\alpha }{2}}}<\infty ,$ 未知，对于 ${\Vert u\Vert }_{1+\alpha ;D}=\underset{D}{\sup }\left|u\right|+{\displaystyle \underset{i=1}{\overset{n}{\sum }}\underset{D}{\sup }}\left|D_{i}u\right|+{\left[u\right]}_{1+\alpha ;D}.$ ，在 $X\in D,0<\rho \le diam\left(D\right)$ 上类似定理2.4.1的证明，记 $\left|D\left(X,\rho \right)\right|\ge A\left|Q_{\rho }\left(X\right)\right|.$ ，故有

$\begin{array}{l}\underset{t_0<t\le t_0+a{R}^2}{\sup }{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert \xi {\left(u_i-k\right)}^{+}\left(\cdot ,t\right)\Vert }_{L^2\left(B_{R,i}\right)}^2}\\ \le \left(1+\epsilon \right){\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert \xi {\left(u_i-k\right)}^{+}\left(\cdot ,t_0\right)\Vert }_{L^2\left(B_{R,i}\right)}^2}+\gamma \left(\epsilon \right)\left[\frac{C{H}^2}{{\left(1-\beta \right)}^2{R}^2}\left|A_R^a\left(k\right)\right|+{\left(M+F_{0,R}^a\right)}^2{\left|A_R^a\left(k\right)\right|}^{1-\frac{2}{p}}\right],\end{array}$

其中 $n+2<\theta \le n+2+p$ 。对于任意整数 ${\mathcal{L}}^{p,\theta }\left(D;\tilde{\delta }\right)\cong C^{\alpha }\left(\bar{D};\tilde{\delta }\right),$ ，我们有

$\alpha =\frac{\theta -\left(n+2\right)}{p}$

如果(2.22)不成立，并且应用条件(2.21)，故可以得到

$A\cong B$

其中 $A\searrow B$ 。显然对于 $B\searrow A$ ，可得

$Q_T=\Omega \times \left(0,T\right],Q_i={\Omega }_i\times \left(0,T\right]\left(i=1,2\right);$

首先取 ${\partial }_p{Q}_T=\left\{\left(x,t\right)|x\in \bar{\Omega },t=0\right\}\cup \left(\partial \Omega \times \left(0,T\right]\right);$ 使得 $\underset{{\partial }_p{Q}_T}{\sup }u=\max \left\{\underset{{\partial }_p{Q}_T}{\sup }{u}_1,\underset{{\partial }_p{Q}_T}{\sup }{u}_2\right\};$ ，则

$v^{+}:=\max \left\{v,0\right\},v^{-}:={\left(-v\right)}^{+}=\max \left\{-v,0\right\}.$

由于 $y_h\left(h=0,1,\cdots \right)$ 是关于 $y_{h+1}\le C{b}^h{y}_h^{1+\epsilon }$ 的严格递增函数，令 $b>1$ 是其逆函数。当 $\epsilon >0$ ，有 $y_0\le \theta :=C^{-\frac{1}{\epsilon }}{b}^{-\frac{1}{{\epsilon }^2}}$ ，则 $\underset{h\to \infty }{\lim }{y}_h=0$ 。取 $\phi \left(t\right)$ ，故有

$\left[k_0,\infty \right)$ (2.24)

又因为 $C>0,\alpha >0,\beta >1$ ，取 $h>k\ge k_0$ 使得 $\phi \left(h\right)\le \frac{C}{{\left(h-k\right)}^{\alpha }}{\left[\phi \left(k\right)\right]}^{\beta }$ 。对于如此确定的常数a，不妨设 $d\ge C^{\frac{1}{\alpha }}{\left[\phi \left(k_0\right)\right]}^{\frac{\beta -1}{\alpha }}{2}^{\frac{\beta }{\beta -1}}$ 是整数N，记 $\phi \left(k_0+d\right)=0$ 。

我们将会通过数学归纳法证明以下结论：存在 $u\in W\left(0,T;V\right)$ 使得

$p>n+2$ (2.25)

其中 $f_i\in L^2\left(0,T;L^{\frac{np}{n+p}}\left({\Omega }_i\right)\right)\left(i=1,2\right)$ 仅依赖于 $\max \left\{\underset{Q_1}{\text{ess}\sup }{u}_1,\underset{Q_2}{\text{ess}\sup }{u}_2\right\}\le \underset{{\partial }_p{Q}_T}{\sup }{u}^{+}+C{F}_0{\left|\Omega \right|}^{\frac{1}{n}-\frac{1}{p}},$ 。

Step (1) 当 $n,\lambda ,\underset{\_}{m},p,T$ ，已知

$\underset{{\partial }_p{Q}_T}{\sup }{u}^{+}=\max \left\{\underset{{\partial }_p{Q}_T}{\sup }{u}_1^{+},\underset{{\partial }_p{Q}_T}{\sup }{u}_2^{+}\right\}$

取 $F_0={\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert f_i\Vert }_{L^2\left(0,T;L^{\frac{np}{n+p}}\left({\Omega }_i\right)\right)}}<\infty$ 充分大，可以得到

$k_0=\underset{{\partial }_p{Q}_T}{\sup }{u}^{+}$

应用引理2.4.2，对于任意 $k\ge k_0$ ，有

$v_1={\left(u_1-k\right)}^{+},v_2={\left(u_2-k\right)}^{+}$ (2.26)

其中 ${v_2|}_{\partial \Omega \times \left(0,T\right]}=0$ ，C只依赖于 $v={\left(v_1,v_2\right)|}_{t=0}=0$ 。在 中取 $u$ 分别代替 ${\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{{\Omega }_i}\frac{\partial u_i}{\partial t}{v}_i}}+w_i\cdot \left(\nabla u_i\right)v_i+{\left(\nabla v_i\right)}^{\top }{K}_i\nabla u_i\text{d}x+{\displaystyle {\int }_{\Gamma }{m}_{\Gamma }}\left(u_1-u_2\right)\left(v_1-v_2\right)\text{d}{\mathcal{H}}_{n-1}\le {\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{{\Omega }_i}{f}_i}}{v}_i\text{d}x,$ ，并且将不等式 代入 的右侧，则有

${\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{{\Omega }_i}\frac{\partial v_i}{\partial t}{v}_i}}+w_i\cdot \left(\nabla v_i\right)v_i+{\left(\nabla v_i\right)}^{\top }{K}_i\nabla v_i\text{d}x+{\displaystyle {\int }_{\Gamma }{m}_{\Gamma }}{\left(v_1-v_2\right)}^2\text{d}{\mathcal{H}}_{n-1}\le {\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{{\Omega }_i}{f}_i}}{v}_i\text{d}x.$ (2.27)

首先取 ${\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{{\Omega }_i}\frac{\partial v_i}{\partial t}{v}_i}}+\lambda {\left|\nabla v_i\right|}^2\text{d}x+\underset{\_}{m}{\displaystyle {\int }_{\Gamma }{\left(v_1-v_2\right)}^2\text{d}{\mathcal{H}}_{n-1}}\le {\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{{\Omega }_i}{f}_i}}{v}_i\text{d}x.$ 足够大使得 的右端方括号中的第二项不大于 $\lambda {\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert \nabla v_i\Vert }_{L^2\left({\Omega }_i\right)}^2}+\underset{\_}{m}{\displaystyle {\int }_{\Gamma }{\left(v_1-v_2\right)}^2\text{d}{\mathcal{H}}_{n-1}}\ge C\left[{\Vert v_1\Vert }_{L^{2^{*}}\left({\Omega }_1\right)}^2+{\Vert v_2\Vert }_{L^{2^{*}}\left({\Omega }_2\right)}^2\right],$ ，然后取 $2^{*}=\frac{2n}{n-2}$ 使得方括号中的第一项不大于 ${\Omega }_i\cap \left[u_i\left(\cdot ,t\right)>k\right]=\left\{x\in {\Omega }_i|u_i\left(x,t\right)>k\right\}\left(i=1,2\right)$ ，对于选定的 ${\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{{\Omega }_i}\frac{\partial v_i}{\partial t}{v}_i\text{d}x}}+C{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert v_i\Vert }_{L^{2^{*}}\left({\Omega }_i\right)}^2}\le {\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{{\Omega }_i}{f}_i}}{v}_i\text{d}x\le {\displaystyle \underset{i=1}{\overset{2}{\sum }}\left(\epsilon {\Vert v_i\Vert }_{L^{2^{*}}\left({\Omega }_i\right)}^2+C\left(\epsilon \right){\Vert f_i\Vert }_{L^{\frac{np}{n+p}}\left({\Omega }_i\right)}^2{\left|{\Omega }_i\cap \left[u_i\left(\cdot ,t\right)>k\right]\right|}^{1-\frac{2}{p}}\right)},$ ，我们有

$\epsilon$

Step (2) 在 ${\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{{\Omega }_i}\frac{\partial v_i}{\partial t}{v}_i\text{d}x}}+C_0{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert v_i\Vert }_{L^{2^{*}}\left({\Omega }_i\right)}^2}\le C\left(\epsilon \right){\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert f_i\Vert }_{L^{\frac{np}{n+p}}\left({\Omega }_i\right)}^2}{\left|{\Omega }_i\cap \left[u_i\left(\cdot ,t\right)>k\right]\right|}^{1-\frac{2}{p}}$ 上应用 ⁺，然后重复以上步骤，可证明(2.25)对 ${\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert v_i\left(\cdot ,T\right)\Vert }_{L^2\left({\Omega }_i\right)}^2}+C_0{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert v_i\Vert }_{L^2\left(0,T;L^{2^{*}}\left({\Omega }_i\right)\right)}^2}\le C{F}_0^2{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left|Q_i\cap \left[u_i>k\right]\right|}^{1-\frac{2}{p}}}$ 成立。

因此有结论成立。

□

定理2.4.4的证明：

根据定理2.4.4的条件，引理2.4.3成立，令 ${\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert v_i\Vert }_{L^2\left(0,T;L^{2^{*}}\left({\Omega }_i\right)\right)}^2}\le C{F}_0^2{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left|Q_i\cap \left[u_i>k\right]\right|}^{1-\frac{2}{p}}}.$ 是由引理2.4.3确定的常数。对于待定的 $h>k$ ，令(2.12)不成立，则(2.15)，(2.22)也不成立，而(2.10)成立。所以由引理2.4.3可得，

${\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert v_i\Vert }_{L^2\left(0,T;L^{2^{*}}\left({\Omega }_i\right)\right)}^2}={\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_0^T{\left({\displaystyle {\int }_{{\Omega }_i}{\left|v_i\right|}^{2^{*}}}\text{d}x\right)}^{\frac{2}{2^{*}}}}}\text{d}t\ge C{\left(h-k\right)}^2{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left|Q_i\cap \left[u_i>h\right]\right|}^{\frac{n-2}{n}}},$

因此(2.14)成立。所以可由引理2.4.2得到以下不等式，

${\left(h-k\right)}^2{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left|Q_i\cap \left[u_i>h\right]\right|}^{\frac{n-2}{n}}}\le C{F}_0^2{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left|Q_i\cap \left[u_i>k\right]\right|}^{1-\frac{2}{p}}}.$

取s充分大使得 $\psi \left(k\right)={\displaystyle \underset{i=1}{\overset{2}{\sum }}\left|Q_i\cap \left[u_i>k\right]\right|}$ ，故(2.9)成立。则由定理2.4.3可得以下结论，

$\psi \left(h\right)\le {\left(\frac{C{F}_0}{h-k}\right)}^{2^{*}}{\left[\psi \left(k\right)\right]}^{\frac{n\left(p-2\right)}{p\left(n-2\right)}}.$

□

对于 $d={\left(C{F}_0\right)}^{\frac{n-2}{2n}}{\left|\psi \left(k_0\right)\right|}^{\frac{1}{n}-\frac{1}{p}}{2}^{\frac{n\left(p-2\right)}{2\left(p-n\right)}}\le C{F}_0{\left|Q_T\right|}^{\frac{1}{n}-\frac{1}{p}}$ ，我们有相似的性质。

定理2.4.5令 $\psi \left(k_0+d\right)=0$ ， ${\displaystyle \underset{i=1}{\overset{2}{\sum }}\left|Q_i\cap [u_i>\underset{{\partial }_p{Q}_T}{\sup }{u}^{+}+d\right|}=0.$ ，如果 $Q_i$ ， $u_i\le \underset{{\partial }_p{Q}_T}{\sup }{u}^{+}+C{F}_0{\left|Q_T\right|}^{\frac{1}{n}-\frac{1}{p}}$ ， $\max \left\{\underset{Q_1}{\text{ess}\sup }{u}_1,\underset{Q_2}{\text{ess}\sup }{u}_2\right\}\le \underset{{\partial }_p{Q}_T}{\sup }{u}^{+}+C{F}_0{\left|\Omega \right|}^{\frac{1}{n}-\frac{1}{p}}.$ ，对于 $u\in W\left(0,T;V\right)$ ， $p>n+2$ ， $f_i\in L^2\left(0,T;L^{\frac{np}{n+p}}\left({\Omega }_i\right)\right)\left(i=1,2\right)$ 满足

$\max \left\{{\Vert u_1\Vert }_{L^{\infty }\left(Q_1\right)},{\Vert u_2\Vert }_{L^{\infty }\left(Q_2\right)}\right\}\le \max \left\{\underset{{\partial }_p{Q}_T}{\sup }\left|u_1\right|,\underset{{\partial }_p{Q}_T}{\sup }\left|u_2\right|\right\}+C{F}_0{\left|\Omega \right|}^{\frac{1}{n}-\frac{1}{p}},$ (2.28)

其中 $n,\lambda ,\underset{\_}{m},p,T$ ， $F_0={\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert f_i\Vert }_{L^2\left(0,T;L^{\frac{np}{n+p}}\left({\Omega }_i\right)\right)}}<\infty$ ，则存在 $Q_T$ 使得或者

$u$ (2.29)

或者

$u\in W\left(0,T;V\right)$ (2.30)

成立。

2.5. 弱解的局部Hölder连续性

下面考虑 $X_0=\left(x_0,t_0\right)\in \Gamma \times \left(0,T\right]$ 在界面上 $Q_{\rho ,\tau }\left(X_0\right)=B_{\rho }\left(x_0\right)\times \left(t_0,t_0+\tau \right]\subset Q_T$ 和界面与初值层 $k\in ℝ,\epsilon \in \left(0,1\right]$ 的相交处的Hölder连续性。

引理2.5.1 (文献 [14] , p. 140引理4.1)令 $\xi \left(x,t\right)\in C^{\infty }\left(\left[t_0,t_0+\tau \right];C_0^{\infty }\left(B_{\rho }\left(x_0\right)\right)\right)$ 是定义于 $0\le \xi \le 1$ 上的非减非负函数，如果它满足

$\xi \left(\cdot ,t_0\right)=0$

其中 $\begin{array}{l}\underset{t_0<t\le t_0+\tau }{\sup }{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert \xi {\left(u_i-k\right)}^{\pm }\left(\cdot ,t\right)\Vert }_{L^2\left(B_{\rho ,i}\right)}^2}+{\lambda }_1{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert \nabla \left(\xi {\left(u_i-k\right)}^{\pm }\right)\Vert }_{L^2\left(t_0,t_0+\tau ;L^2\left(B_{\rho ,i}\right)\right)}^2}\\ +m_1{\displaystyle {\int }_{t_0}^{t_0+\tau }{\displaystyle {\int }_{\Gamma \cap B_{\rho }}{\left(\xi {\left(u_1-k\right)}^{\pm }-\xi {\left(u_2-k\right)}^{\pm }\right)}^2\text{d}{\mathcal{H}}_{n-1}\text{d}t}}\\ \le C^{*}\left[\left({\Vert \frac{\partial \xi }{\partial t}\Vert }_{L^{\infty }\left(Q_{\rho ,\tau }\right)}+{\Vert \nabla \xi \Vert }_{L^{\infty }\left(Q_{\rho ,\tau }\right)}^2\right){\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert {\left(u_i-k\right)}^{\pm }\Vert }_{L^2\left(t_0,t_0+\tau ;L^2\left(B_{\rho ,i}\right)\right)}^2}+F_{0,\rho ,\tau }^2{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left|Q_{\rho ,\tau ,i}\cap \left[{\left(u_i-k\right)}^{\pm }>0\right]\right|}^{1-\frac{2}{p}}}\right],\end{array}$ ， $B_{\rho ,i}=B_{\rho }\cap {\Omega }_i,Q_{\rho ,\tau ,i}=Q_{\rho ,\tau }\cap Q_i$ ， $0<\rho ,\tau <1$ 是常数，则存在 $p>n+2$ ， ${\lambda }_1>0$ 使得

$\lambda$

其中 $m_1=2\underset{\_}{m}$ 仅依赖于 $F_{0,\rho ,\tau }={\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert f_i\Vert }_{L^2\left(t_0,t_0+\tau ;L^{\frac{np}{n+p}}\left(B_{\rho ,i}\right)\right)}}>0$ 。

定理2.5.1令 $C^{*}$ ， $n,\Lambda ,p,c_0$ ， $DG\left(Q_T\right)=DG\left(Q_T;{\lambda }_1,m_1,p,n,F_{0,\rho ,\tau },C^{*}\right)$ ， $u\in W\left(0,T;V\right)$ ，则对于 $u\in D{G}^{+}\left(Q_T\right)$ ，有

$u\in W\left(0,T;V\right)$

其中 $u\in D{G}^{-}\left(Q_T\right)$ ， $DG\left(Q_T\right)=D{G}^{+}\left(Q_T\right)\cap D{G}^{-}\left(Q_T\right)$ 仅依赖于 $u\in W\left(0,T;V\right)$ 的参数， $p>n+2$ ，并且 $f_i\in L^2\left(0,T;L^{\frac{np}{n+p}}\left({\Omega }_i\right)\right)\left(i=1,2\right)$ ， $u\in D{G}^{+}\left(Q_T\right)$ ， $u\in W\left(0,T;V\right)$ 。

证明 记

$u\in D{G}^{-}(\; Q\; T\; )$

$C^{*}$

对于 $n,\Lambda ,p,c_0$ ， $F_{0,\rho ,\tau }={\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert f_i\Vert }_{L^2\left(t_0,t_0+\tau ;L^{\frac{np}{n+p}}\left(B_{\rho ,i}\right)\right)}}<\infty$ ，可以看出以下两种情况中必有一种成立 $u\in W\left(0,T;V\right)$

$\xi \left(x,t\right)\in C^{\infty }\left(\left[t_0,t_0+\tau \right];C_0^{\infty }\left(B_{\rho }\left(x_0\right)\right)\right)$ (2.31)

$Q_{\rho ,\tau }\left(X_0\right)$ (2.32)

Step (1) 如果(2.31)成立，则由定理2.4.4可得，存在 $0\le \xi \left(x,t\right)\le 1$ 使得当 $\xi \left(\cdot ,t_0\right)=0$ 时，

$B_{\rho ,i}=B_{\rho }\cap {\Omega }_i$ (2.33)

或者

$v_i={\xi }^2{\left(u_i-k\right)}^{+}\ge 0\left(i=1,2\right)$ (2.34)

成立，其中 $\begin{array}{l}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{\rho ,i}}\frac{\partial {\left(u_i-k\right)}^{+}}{\partial t}{\xi }^2{\left(u_i-k\right)}^{+}}}+w_i\cdot \left(\nabla {\left(u_i-k\right)}^{+}\right){\xi }^2{\left(u_i-k\right)}^{+}+{\left(\nabla \left({\xi }^2{\left(u_i-k\right)}^{+}\right)\right)}^{\top }{K}_i\nabla {\left(u_i-k\right)}^{+}\text{d}x\\ +\underset{\_}{m}{\displaystyle {\int }_{\Gamma \cap B_{\rho }}{\left(\xi {\left(u_1-k\right)}^{+}-\xi {\left(u_2-k\right)}^{+}\right)}^2\text{d}{\mathcal{H}}_{n-1}}\le {\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{\rho ,i}}{f}_i}}{\xi }^2{\left(u_i-k\right)}^{+}\text{d}x.\end{array}$ ，并且由 可得当 ${\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{\rho ,i}}{w}_i}}\cdot \left(\nabla {\left(u_i-k\right)}^{+}\right){\xi }^2{\left(u_i-k\right)}^{+}\text{d}x=-{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{\rho ,i}}{w}_i}}\cdot {\left(u_i-k\right)}^{+}\left(\nabla \xi \right)\left(\xi {\left(u_i-k\right)}^{+}\right)\text{d}x,$ 时，有 ${\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{\rho ,i}}{\left(\nabla \left({\xi }^2{\left(u_i-k\right)}^{+}\right)\right)}^{\top }{K}_i\nabla {\left(u_i-k\right)}^{+}\text{d}x}}\ge \lambda {\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert \nabla \left(\xi {\left(u_i-k\right)}^{+}\right)\Vert }_{L^2\left(B_{\rho ,i}\right)}^2}-{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{\rho ,i}}{\left({\left(u_i-k\right)}^{+}\right)}^2}}{\left(\nabla \xi \right)}^{\top }{K}_i\nabla \xi \text{d}x.$ 。

Step (2) 如果(2.32)成立，则由定理2.4.5可得，存在 $\begin{array}{l}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{\rho ,i}}\frac{\partial \left(\xi {\left(u_i-k\right)}^{+}\right)}{\partial t}\xi {\left(u_i-k\right)}^{+}\text{d}x}}+\lambda {\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert \nabla \left(\xi {\left(u_i-k\right)}^{+}\right)\Vert }_{L^2\left(B_{\rho ,i}\right)}^2}+\underset{\_}{m}{\displaystyle {\int }_{\Gamma \cap B_{\rho }}{\left(\xi {\left(u_1-k\right)}^{+}-\xi {\left(u_2-k\right)}^{+}\right)}^2\text{d}{\mathcal{H}}_{n-1}}\\ \le {\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{\rho ,i}}\frac{\partial \xi }{\partial t}\xi {\left({\left(u_i-k\right)}^{+}\right)}^2\text{d}x}}+{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{\rho ,i}}{w}_i}}\cdot {\left(u_i-k\right)}^{+}\left(\nabla \xi \right)\left(\xi {\left(u_i-k\right)}^{+}\right)\text{d}x\\ +{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{\rho ,i}}{\left({\left(u_i-k\right)}^{+}\right)}^2{\left(\nabla \xi \right)}^{\top }{K}_i\nabla \xi \text{d}x}}+{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{\rho ,i}}{f}_i}}{\xi }^2{\left(u_i-k\right)}^{+}\text{d}x.\end{array}$ 使得当 ${\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{\rho ,i}}{w}_i}}\cdot {\left(u_i-k\right)}^{+}\left(\nabla \xi \right)\left(\xi {\left(u_i-k\right)}^{+}\right)\text{d}x\le \epsilon {\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert \nabla \left(\xi {\left(u_i-k\right)}^{+}\right)\Vert }_{L^2\left(B_{\rho ,i}\right)}^2}+\gamma \left(\epsilon \right){\Vert \nabla \xi \Vert }_{L^{\infty }\left(Q_{\rho ,\tau }\right)}^2{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert {\left(u_i-k\right)}^{+}\Vert }_{L^2\left(B_{\rho ,i}\right)}^2},$ 时，

${\Vert \xi {\left(u_i-k\right)}^{+}\Vert }_{L^{2^{*}}\left(B_{\rho ,i}\right)}^2\le C\left({\Vert \nabla \left(\xi {\left(u_i-k\right)}^{+}\right)\Vert }_{L^2\left(B_{\rho ,i}\right)}^2+{\displaystyle {\int }_{\partial B_{\rho ,i}\setminus \left(\Gamma \cap B_{\rho }\right)}{\left|\xi {\left(u_i-k\right)}^{+}\right|}^2\text{d}{\mathcal{H}}_{n-1}}\right)=C{\Vert \nabla \left(\xi {\left(u_i-k\right)}^{+}\right)\Vert }_{L^2\left(B_{\rho ,i}\right)}^2,$ (2.35)

或者

$\begin{array}{c}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\displaystyle {\int }_{B_{\rho ,i}}{f}_i}}{\xi }^2{\left(u_i-k\right)}^{+}\text{d}x\le {\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert f_i\Vert }_{L^{\frac{np}{n+p}}\left(B_{\rho ,i}\right)}}{\Vert \xi {\left(u_i-k\right)}^{+}\Vert }_{L^{2^{*}}\left(B_{\rho ,i}\right)}{\left|B_{\rho ,i}\cap \left[u_i\left(\cdot ,t\right)>k\right]\right|}^{\frac{1}{2}-\frac{1}{p}}\\ \le \epsilon {\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert \nabla \left(\xi {\left(u_i-k\right)}^{+}\right)\Vert }_{L^2\left(B_{\rho ,i}\right)}^2}+\gamma \left(\epsilon \right){\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert f_i\Vert }^2{}_{L^{\frac{np}{n+p}}\left(B_{\rho ,i}\right)}}{\left|B_{\rho ,i}\cap \left[u_i\left(\cdot ,t\right)>k\right]\right|}^{1-\frac{2}{p}}.\end{array}$ (2.36)

成立，并且由(2.36)可得当 $\epsilon$ 时，有 $t\in \left(t_0,t_0+\tau \right]$ 。

已知 $\begin{array}{l}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert \xi {\left(u_i-k\right)}^{\pm }\left(\cdot ,t\right)\Vert }_{L^2\left(B_{\rho ,i}\right)}^2}+{\lambda }_1{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert \nabla \left(\xi {\left(u_i-k\right)}^{\pm }\right)\Vert }_{L^2\left(t_0,t;L^2\left(B_{\rho ,i}\right)\right)}^2}+m_1{\displaystyle {\int }_{t_0}^t{\displaystyle {\int }_{\Gamma \cap B_{\rho }}{\left(\xi {\left(u_1-k\right)}^{\pm }-\xi {\left(u_2-k\right)}^{\pm }\right)}^2\text{d}{\mathcal{H}}_{n-1}\text{d}t}}\\ \le C^{*}\left[\left({\Vert \frac{\partial \xi }{\partial t}\Vert }_{L^{\infty }\left(Q_{\rho ,\tau }\right)}+{\Vert \nabla \xi \Vert }_{L^{\infty }\left(Q_{\rho ,\tau }\right)}^2\right){\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert {\left(u_i-k\right)}^{\pm }\Vert }_{L^2\left(t_0,t_0+\tau ;L^2\left(B_{\rho ,i}\right)\right)}^2}+F_{0,\rho ,\tau }^2{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left|Q_{\rho ,\tau ,i}\cap \left[{\left(u_i-k\right)}^{\pm }>0\right]\right|}^{1-\frac{2}{p}}}\right].\end{array}$ ，则当 $\left(t_0,t_0+\tau \right]$ 时， ${}^{+}$ ，即(2.33)， 成立，故 $u\in W\left(0,T;V\right)$ ，所以 $u\in D{G}^{-}\left(Q_T\right)$ 。因此，当 $u\in DG\left(Q_T\right)$ ，我们有

$\begin{array}{l}\underset{t_0<t\le t_0+\tau }{\sup }{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert \xi {\left(u_i-k\right)}^{\pm }\left(\cdot ,t\right)\Vert }_{L^2\left(B_{\rho ,i}\right)}^2}+\bar{\lambda }{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert \xi {\left(u_i-k\right)}^{\pm }\Vert }_{L^2\left(t_0,t_0+\tau ;L^{2^{*}}\left(B_{\rho ,i}\right)\right)}^2}\\ +m_1{\displaystyle {\int }_{t_0}^{t_0+\tau }{\displaystyle {\int }_{\Gamma \cap B_{\rho }}{\left(\xi {\left(u_1-k\right)}^{\pm }-\xi {\left(u_2-k\right)}^{\pm }\right)}^2\text{d}{\mathcal{H}}_{n-1}\text{d}t}}\\ \le C^{*}\left[\left({\Vert \frac{\partial \xi }{\partial t}\Vert }_{L^{\infty }\left(Q_{\rho ,\tau }\right)}+{\Vert \nabla \xi \Vert }_{L^{\infty }\left(Q_{\rho ,\tau }\right)}^2\right){\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert {\left(u_i-k\right)}^{\pm }\Vert }_{L^2\left(t_0,t_0+\tau ;L^2\left(B_{\rho ,i}\right)\right)}^2}+F_{0,\rho ,\tau }^2{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\left|Q_{\rho ,\tau ,i}\cap \left[{\left(u_i-k\right)}^{\pm }>0\right]\right|}^{1-\frac{2}{p}}}\right].\end{array}$

其中 $u\in D{G}^{+}\left(Q_T\right)$ 仅依赖于 $X_0=\left(x_0,t_0\right)\in \Gamma \times \left(0,T\right]$ 。由引理2.5.1可得，

$p>n+2$

其中 $Q_R\left(X_0\right)=B_R\left(x_0\right)\times \left(t_0-R^2,t_0\right]\subset Q_T,0<R\le 1$ 。所以结论成立。

□

定理2.5.2 令 $\max \left\{\underset{Q_{\frac{R}{2},1}}{\text{ess}\sup }{u}_1,\underset{Q_{{}_{\frac{R}{2},2}}}{\text{ess}\sup }{u}_2\right\}\le C\left(\frac{1}{\sqrt{R^n}}{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\Vert u_i\Vert }_{L^2\left(t_0-R^2,t_0;L^2\left(B_{R,i}\right)\right)}}+F_{0,R}{R}^{1-\frac{n}{p}}\right),$ ，Q与界面 $u\in D{G}^{-}($