本文中,我们考虑如下带有斜边值条件的Laplace方程

$\begin{equation}\label{1.1}\Delta u=f(x),\ \ \ \ \ \mbox{在$\Omega$内},\end{equation}$

(1.1)

$\begin{equation}\label{1.2}D_{\beta}u=\psi(x),\ \ \ \ \ \mbox{在$\partial\Omega$上},\end{equation}$

(1.2)

其中$\Omega$是有界区域,\ $n \geq2 $,\ $\partial\Omega \in C^{4}$,$\beta$是严格斜的单位向量,即$\beta\cdot\gamma\geq\beta_{0}$,$\gamma$为单位内法向量,$\beta_{0}$为正常数,令$L_{1}=\sup\limits_{\bar{\Omega}} |u|$,$f$,$\psi$均为定义在$\bar{\Omega}\times R$和$\overline{\Omega}$上给定的有界可微函数,且满足如下结构性条件

$\begin{equation}\label{1.3}|f(x)|_{C^{1}(\bar{\Omega})}\leq L_{2},\end{equation}$

(1.3)

$\begin{equation}\label{1.4}|\psi(x)|_{C^{3}(\bar{\Omega})}\leq L_{3},\end{equation}$

(1.4)

其中$L_{2},L_{3}$均为正常数.

Laplace方程是最典型的二阶椭圆方程,它具有广泛的应用背景,譬如静电学中的电势以及牛顿万有引力理论中的引力势均满足这类椭圆方程.Laplace方程由法国数学家皮埃尔-西蒙$\cdot$拉普拉斯首先提出而得名.求解Laplace方程是电磁学、天文学、势力学和流体力学等领域经常遇到的一类重要的问题,因为这种方程以势函数的形式描述了电场、引力场和流场等物理现象的性质.

Laplace方程的Dirichlet边值问题已得到广泛的研究(参见文献[2, 3]).Laplace方程的Neumann边值问题,相应的结论也比较完善(参见文献[1, 3]).1984年Lieberman在文献[4]中通过将整体估计约化到边界梯度估计,再将边界梯度估计约化到切向梯度估计,最后讨论切向梯度估计,得到如下拟线性方程

$\begin{equation}\label{nxx}Qu=a^{ij}(Du)D_{ij}u+a(x,u,Du)=0,\ \ \ \ \ \mbox{在$\Omega$内}\end{equation}$

(1.5)

满足完全非线性边值条件

$\begin{equation}\label{nlx}Nu=F(x,u,Du)=0,\ \ \ \ \ \mbox{在$\partial\Omega$上}\end{equation}$

(1.6)

解的梯度估计.1986年,Lions和Trudinger在文献[5]中利用插值不等式得到了随机理论Bellman方程线性斜边值问题解的梯度估计.1988年,Lieberman在文献[6]中利用极值原理得到拟线性椭圆方程

$\begin{equation}\label{nxx1}Qu=a^{ij}(x,u,Du)D_{ij}u+a(x,u,Du)=0,\ \ \ \ \ \mbox{在$\Omega$内}\end{equation}$

(1.7)

满足完全非线性边值条件

$\begin{equation}\label{nlx1}Nu=F(x,u,Du)=0,\ \ \ \ \ \mbox{在$\partial\Omega$上}\end{equation}$

(1.8)

解的梯度估计.

本文中,我们给出Laplace方程斜边值问题解的梯度估计的两种证明方法.两种证明选取不同的辅助函数来讨论解的梯度内估计,近边梯度估计,以及边界梯度估计,从而得到解的全局梯度估计.第一种证明中,我们采用Lieberman在文献[6]中的办法,利用Cauchy不等式,巧妙地处理二阶导数项,从而得到梯度估计.第二种证明中,我们构造辅助函数,充分利用函数取得极大值点的性质以及斜边值条件得到梯度估计.

下面,我们给出本文的主要结论.

定理1.1 设$u\in C^{2}(\bar{\Omega})\cap C^{3}(\Omega)$为问题(1.1)-(1.2)的解,且$f,\psi$满足结构性条件(1.3),(1.4),则

$\begin{equation}\label{1.5}\sup_{\bar{\Omega}} |Du|\leq C,\end{equation}$

(1.9)

其中$C$为正的常数依赖于$n,\Omega,L_{1},L_{2},L_{3},\beta_0.$

本文由以下几个部分构成:第2节中,我们介绍了后面证明所需的记号并引用Laplace方程的经典估计;第3节中,我们采用两种不同的辅助函数的技巧分边界,近边以及内部3种情况,得到Laplace方程斜边值问题解的整体梯度估计.

本文中,为了后面计算和证明的方便,我们介绍后面章节用到的一些基本概念以及Laplace方程的经典估计.

设$\Omega$是${\Bbb R}^{n}$中有界区域,$n \geq 2$,$\partial\Omega \in C^{4}$,$\gamma$是$\partial\Omega$上的单位内法向量.令

$\begin{equation}\label{1.5a}d(x)={\rm dist}(x,\partial\Omega),\end{equation}$

(2.1)

$\begin{equation}\label{1.5b}\Omega_{\mu}=\{x\in\Omega:d(x)<\mu\},\end{equation}$

(2.2)

则存在常数$\mu_{1} > 0$使得$d(x)\in C^{4}(\overline{\Omega}_{\mu_{1}})$.正如Simon-Spruck^[7]或者Liberman[1,\rm p331]提到,在$\Omega_{\mu_{1}}$内,可取$\gamma=Dd$,并且$\gamma$是一个$C^{2}(\overline{\Omega}_{\mu_{1}})$向量场,且有以下性质:

$\begin{equation}\label{1.5c}|D\gamma|+|D^{2}\gamma| \leq C(n,\Omega),\ \ \ \ \mbox{在$\Omega_{\mu_{1}}$内};\end{equation}$

(2.3)

$\begin{equation}\label{1.5d}|\gamma|=1,\sum\limits_{1 \leq i \leq n}\gamma^{i}D_{j}\gamma^{i}=0,\sum\limits_{1 \leq i \leq n}\gamma^{i}D_{i}\gamma^{j}=0,\ \ \ \ \mbox{在$\Omega$内}.\end{equation}$

(2.4)

引入记号

$\begin{equation}\label{1.5e}c^{ij}=\delta_{ij}-\gamma^{i}\gamma^{j},\ \ \ \ \mbox{在$\Omega$内},\end{equation}$

(2.5)

则$(c^{ij})_{n \times n}$非负定.

对任一${\Bbb R}^{n}$中向量$\xi$,记$\xi'$为$\xi$的切向部分,其第$i$个分量定义为

$\begin{equation}\label{1.5f}\sum\limits_{1 \leq j \leq n}c^{ij}\xi_{j}.\end{equation}$

(2.6)

梯度$Du$的切向量记为$D'u$,则

$\begin{equation}\label{1.5g}|D'u|^{2}=\sum\limits_{1 \leq i,j \leq n}c^{ij}u_{i}u_{j}.\end{equation}$

(2.7)

因为本文中的边界梯度估计依赖于梯度内估计,所以首先叙述Laplace方程标准的梯度内估计,参见文献[2, 3].

引理2.1 设$u \in C^{3}(\Omega)$为方程(1.1)的解,则对任意区域$\Omega'\subset\subset\Omega$,有

$\begin{equation}\label{2.1}\sup_{\Omega'} |Du| \leq M_{1},\end{equation}$

(2.8)

其中$M_{1}$依赖于$n,L_{1},{\rm dist}(\Omega',\partial\Omega),L_{2}.$

为了后面证明中处理边界梯度估计的需要,我们在本节中给出边界上$D_{\beta \gamma}u$与梯度估计的关系.

引理2.2 设$u\in C^{2}(\bar{\Omega})\cap C^{3}(\Omega)$为问题(1.1)-(1.2)的解,且$f,\psi$满足结构性条件(1.3),(1.4),则

$\begin{equation}\label{1.7} \sup_{\partial\Omega}|D_{\beta\gamma}u|\leq \hat{C}_2\sup_{\bar{\Omega}} |Du|+\hat{C}_3,\end{equation}$

(2.9)

其中$\hat{C}_2$和$\hat{C}_3$为正的常数依赖于$n,\Omega,L_{1},L_{2},L_{3}$.

利用Hopf引理的证明思路可得此引理的证明.

本节中,我们分两个小节给出了Laplace方程斜边值问题解的梯度估计的两种不同证明.

定理3.1 设$u\in C^{2}(\overline{\Omega})\bigcap C^{3}(\Omega)$为斜边值问题(1.1)-(1.2)的解.$f,~\varphi$满足结构条件(1.3),(1.4),则存在小的正常数$\mu_{0}$,使得

$\begin{equation}\sup_{\bar{\Omega}_{\mu_{0}}}|Du| \leq C,\end{equation}$

(3.1)

其中$C$依赖于$n,\Omega,\beta_{0},L_{1},L_{2},L_{3}.$

本小节的证明中,我们采用Lieberman在文献[6]中的办法,利用Cauchy不等式,巧妙地处理二阶导数项,从而得到梯度估计.

证 令

$$\Phi(x)=|D'u|^{2}{\rm e}^{\frac{1}{1+L_{1}+u}}{\rm e}^{\alpha_{0}d},$$

其中$\alpha_{0}=\frac{|\psi|+C_{0}(1+\frac{2n}{\beta_{0}^{2}})+1}{\beta_{0}}$,$C_{0}$为正常数依赖于$n,$ $\beta_0,$ $ L_3,\Omega$.又令

$$\varphi(x)=\log \Phi(x)=\log|D'u|^{2}+h(u)+g(d),$$

其中

$\begin{equation}\label{2.3a}h(u)=\frac{1}{1+L_{1}+u},~~ g(d)=\alpha_{0}d.\end{equation}$

(3.2)

假设$\varphi(x)$在$\bar{\Omega}$内的极大值在点$x_{0}$处取得.以下所有的计算都在$x_0$点进行.下面分3种情况证明$|D'u|(x_{0})$有界.

情形1若$x_{0}\in\partial\Omega$,我们将证明$|D'u|(x_{0})$有界.

首先对$\varphi$沿$\beta$方向求导,由$\gamma=Dd$和边值条件(1.2)可得

$\begin{eqnarray}\label{3.5}D_{\beta}\varphi &=& \frac{1}{|D'u|^{2}}\sum_{1 \leq i \leq n}(|D'u|^{2})_{i}\beta^{i}+h'D_{\beta}u+g'd_{\beta}\\&=&\frac{1}{|D'u|^{2}}\sum_{1\leq i \leq n}(|D'u|^{2})_{i}\beta^{i}+h'\psi+g'\gamma^{i}\beta^{i}.\end{eqnarray}$

(3.3)

由(2.7)式可知

$\begin{eqnarray}\label{3.6}\sum_{1\leq i \leq n}(|D'u|^{2})_{i}\beta^{i}&=&\sum_{1\leq i \leq n}\bigg(\sum_{1 \leq k,l \leq n}c^{kl}u_{k}u_{l}\bigg)_{i}\beta^{i}\\&=&\sum_{1 \leq i,k,l \leq n}(c^{kl})_{i}(u_{k}u_{l})\beta^{i}+2\sum_{1 \leq i,k,l \leq n}c^{kl}u_{ik}u_{l}\beta^{i},\end{eqnarray}$

(3.4)

对边值条件(1.2)关于切向求导可得

$\begin{equation}\label{3.7}\sum_{1\leq k \leq n}c^{kl}(D_{\beta}u)_{k}=\sum_{1\leq k \leq n}c^{kl}\psi_{k},\end{equation}$

(3.5)

由上式可得

$\begin{equation}\label{3.7m}\sum_{1\leq i,k \leq n}c^{kl}u_{ki}\beta^{i}=-\sum_{1\leq i,k \leq n}c^{kl}u_{i}(\beta^{i})_{k}+\sum_{1\leq k \leq n}c^{kl}\psi_{k},\end{equation}$

(3.6)

将(3.6)和(3.4)式代入(3.3)式可得

$\begin{eqnarray}|D'u|^{2}D_{\beta}\varphi&=&(g'\gamma^{i}\beta^{i}+h'\psi)|D'u|^{2}-2\sum_{1\leq i,k,l \leq n}c^{kl}u_{i}u_{l}(\beta^{i})_{k}\\&&+2\sum_{1\leq k,l \leq n}c^{kl}\psi_{k}u_{l}+\sum_{1\leq i,k,l \leq n}(c^{kl})_{i}u_{k}u_{l}\beta^{i}.\end{eqnarray}$

(3.7)

由于边界上任意一点的单位内法向量均可分解为切向和斜向的组合,

$\begin{equation}\label{3.11}\gamma=-\sum_{i=1}^{n}\frac{(\beta,\tau_i)}{(\beta\cdot\gamma)}\tau_i+\frac{1}{(\beta\cdot\gamma)}\beta.\end{equation}$

(3.8)

由(3.8)式可知

$\begin{eqnarray}\label{3.12aa}|Du|^{2}&=&|D'u|^{2}+|u_{\gamma}|^{2}\\&\leq&|D'u|^{2}+|-\sum_{i=1}^{n}\frac{(\beta,\tau_i)}{(\beta\cdot\gamma)}u_i+\frac{1}{(\beta\cdot\gamma)}D_{\beta}u|^{2}\\&\leq &|D'u|^{2}+\frac{n}{(\beta_{0})^{2}}|D_{\beta}u|^{2}-n\sum_{i=1}^{n}\frac{1}{(\beta_{0})^{2}}u_i^{2}\\&\leq &(1+\frac{n}{(\beta_{0})^{2}})|D'u|^{2}+\frac{n}{(\beta_{0})^{2}}|\psi|^{2}.\end{eqnarray}$

(3.9)

不妨设$|D'u|^{2}\geq |\psi|^{2}$,否则$|D'u|^{2} <|\psi|^{2}$,则有

$$|Du|^{2} \leq(1+\frac{2n}{(\beta_{0})^{2}})|\psi|^{2},$$

因此不必证.不妨设

$\begin{equation}|\psi| \leq |Du|^{2} \leq(1+\frac{2n}{(\beta_{0})^{2}})|D'u|^{2},\end{equation}$

(3.10)

因此有

$\begin{eqnarray} |D'u|^{2}D_{\beta}\varphi(x_{0})&\geq&(\alpha_{0}\beta_{0}-|\psi|)|D'u|^{2}-C_{0}|Du|^{2}\\&\geq &\Big(\alpha_{0}\beta_{0}-|\psi|-C_{0}(1+\frac{2n}{(\beta_{0})^{2}})\Big)|D'u|^{2}\\& \geq& |D'u|^{2}\\& >& 0,\end{eqnarray}$

(3.11)

其中$C_0$为正常数依赖于$n,\Omega,\beta_{0},L_3$,由$\alpha_{0}=\frac{|\psi|+C_{0}(1+\frac{2n}{\beta_{0}^{2}})+1}{\beta_{0}}$可使得第三个不等号成立.另一方面,由$\varphi$在$x_{0}$点处取得极大值,故

$\begin{equation}D_{\beta}\varphi(x_{0})\leq 0.\end{equation}$

(3.12)

矛盾.因此

$\begin{equation}|D'u|(x_{0})\leq \frac{|\psi|^2}{(1+\frac{2n}{\beta_{0}^{2}})}.\end{equation}$

(3.13)

情形2若$x_{0} \in \Omega_{\mu_{0}}$,我们证明$|D'u|(x_{0})$有界.

在点$x_{0}$选择坐标系,不妨设$u_{i}(x_{0})=0$,$2 \leq i \leq n$,$u_{1}(x_{0})=|Du|=|D'u|>0$.

首先对$\varphi$微分一次,有

$\begin{equation}\label{3.18}\varphi_{i}=\frac{(|D'u|^{2})_{i}}{|D'u|^{2}}+h'u_{i}+g'\gamma^{i},\end{equation}$

(3.14)

由$\varphi_{i}=0$,可得

$\begin{equation}\label{3.18aa}(|D'u|^{2})_{i}=-|D'u|^{2}(h'u_{i}+g'\gamma^{i}),\end{equation}$

(3.15)

$\varphi$对$x_{i}$再微分一次再求和,结合(3.15)式可得

$\begin{eqnarray} \label{3.20a}0 \geq \Delta\varphi&=&\frac{\Delta|D'u|^{2}}{|D'u|^{2}}-\sum_{1 \leq i \leq n}\frac{(|D'u|^{2})_{i}^{2}}{|D'u|^{4}}+h"|Du|^{2}+h'\Delta u+g'\sum_{1 \leq i \leq n}(\gamma^{i})_{i}\\&=&\frac{\Delta|D'u|^{2}}{|D'u|^{2}}+(h"-h'^{2})u_{1}^{2}-2h'g'\gamma^{1}u_{1}+h'f-g'^{2}+g'\sum_{1 \leq i \leq n}(\gamma^{i})_{i}.\end{eqnarray}$

(3.16)

下面计算$\Delta|D'u|^{2}$.记$D_{k}=\frac{\partial}{\partial x_{k}}+\frac{\partial}{\partial z}\frac{\partial z}{\partial x_{k}}$.

由(2.7)式以及坐标系的选取,可得

$\begin{eqnarray}\label{3.21a}\Delta|D'u|^{2}&=&\sum_{1 \leq i \leq n}\bigg[\sum_{1 \leq k,l \leq n}c^{kl}u_{k}u_{l}\bigg]_{ii}\\&=&\sum_{1 \leq i \leq n}\bigg[\sum_{1 \leq k,l \leq n}(c^{kl})_{i}u_{k}u_{l}+2\sum_{1 \leq k,l \leq n}c^{kl}u_{ki}u_{l}\bigg]_{i}\\&=&\sum_{1 \leq k,l \leq n}\Delta(c^{kl})u_{k}u_{l}+4\sum_{1 \leq i,k,l \leq n}(c^{kl})_{i}u_{ki}u_{l}\\&&+2\sum_{1 \leq i,k,l \leq n}c^{kl}u_{ki}u_{li}+2\sum_{1 \leq k,l \leq n}c^{kl}f_ku_{l}\\&=&4u_{1}\sum_{1 \leq i,j \leq n}(c^{j1})_{i}u_{ij}+2\sum_{1 \leq i,k,l \leq n}c^{kl}u_{ki}u_{li}+\Delta(c^{11})u_{1}^{2}\\&&+2u_{1}\sum_{1 \leq k \leq n}c^{k1}f_k\\&\geq&4u_{1}\sum_{1 \leq i,j \leq n}(c^{j1})_{i}u_{ij}+2\sum_{1 \leq i,k,l \leq n}c^{kl}u_{ki}u_{li}-\tilde{C}_{1}u_{1}^{2},\end{eqnarray}$

(3.17)

其中$\tilde{C}_{1}$为正常数依赖于$n,\Omega,L_{2}.$

下面采用Lieberman^[1]的方法处理(3.17)式中的二阶导数项.记

\begin{equation}\label{3.21ab}4u_{1}\sum\limits_{1 \leq i,j \leq n}(c^{j1})_{i}u_{ij}=\sum\limits_{1 \leq i,j \leq n}f^{ij}u_{ij},\end{equation}

(3.18)

其中$f^{ij}=4u_{1}(c^{1j})_{i}$.

引入以下记号:

$ \begin{equation}\begin{array} {l} F=\sum_{1 \leq i,j \leq n}f^{ij}\gamma^{i}\gamma^{j},~~f^{i}=\sum_{1 \leq j \leq n}(f^{ij}-F\delta_{ij})\gamma^{j},\\[4mm]F_{1}=\sum_{1 \leq i,j,k \leq n}f^{ij}c^{jk}u_{ki},~~F_{2}=\sum_{1 \leq i,j,k \leq n}f^{i}c^{ji}u_{kj}\gamma^{k},~~F_{3}=-F\sum_{1 \leq i,k \leq n}c^{ik}u_{ki}.\end{array}\end{equation}$

(3.19)

由方程(1.1),可得

$ \begin{eqnarray}\label{3.23a}\sum_{1 \leq i,j \leq n}f^{ij}u_{ij}=\sum_{1 \leq i,j,k \leq n}f^{ij}c^{kj}u_{ik}+\sum_{1 \leq i,j,k \leq n}f^{i}c^{ji}u_{jk}\gamma^{k}-F\sum_{1 \leq i,k \leq n}c^{ik}u_{ik}+Ff.\end{eqnarray}$

(3.20)

由Cauchy不等式,有

$\begin{eqnarray}\label{3.24a}&&\sum_{1 \leq i,j,k \leq n}f^{ij}c^{kj}u_{ik} \leq \frac{2}{3}\sum_{1 \leq j,k,l \leq n}c^{kj}u_{jl}u_{kl}+\frac{3}{8}\sum_{1 \leq i,j,k \leq n}c^{kj}f^{ij}f^{ik},\\&&\sum_{1 \leq i,j,k \leq n}f^{i}c^{ji}u_{jk}\gamma^{k} \leq \frac{2}{3}\sum_{1 \leq i,j,k \leq n}c^{ij}u_{ik}u_{jk}+\frac{3}{8}\sum_{1 \leq i,j,k \leq n}c^{ij}(f^{i}\gamma^{k})(f^{j}\gamma^{k}),\\&&-F\sum_{1 \leq i,k \leq n}c^{ik}u_{ik} \leq \frac{2}{3}\sum_{1 \leq k,j,l \leq n}c^{jk}u_{jl}u_{kl}+\frac{3}{8}F^{2}, \end{eqnarray}$

(3.21)

将上面(3.21)式代入(3.20)式,再结合(3.17)和(3.18)式,可得

$\begin{equation}\label{3.24ab}\frac{\Delta|D'u|^{2}}{|D'u|^{2}} \geq -\tilde{C}_{2},\end{equation}$

(3.22)

其中$\tilde{C}_{2}$为正常数依赖于$n,\Omega,L_{2}.$

将(3.22)式代入(3.16)式,结合(3.2)式,有

$\begin{eqnarray} 0 &\geq& \Delta\varphi \geq(h"-h')u_{1}^{2}-2h'g'\gamma^1u_{1}+h'f-g'^{2}+g'\sum_{1 \leq i \leq n}(\gamma^{i})_{i}-\tilde{C}_{2}\\&=&\frac{2+(1+L_{1}+u)}{(1+L_{1}+u)^{2}}u_{1}^{2}+\frac{2\alpha_{0}}{(1+L_{1}+u)^{2}}u_{1}\gamma^{1}-\frac{f}{(1+L_{1}+u)^{2}}-\alpha_{0}^{2}+\alpha_{0}\sum_{1 \leq i \leq n}(\gamma^{i})_{i}-\tilde{C}_{2}\\& \geq& \frac{u_{1}^{2}}{(1+2L_{1})^{2}}-\tilde{C}_{3}u_{1},\end{eqnarray}$

(3.23)

其中$\tilde{C}_{3}$为正常数依赖于$L_1,L_2,n,\Omega,\beta_{0}$,我们不妨设

$$u_1\geq \bigg|-\frac{f}{(1+L_{1}+u)^{2}}-\alpha_{0}^{2}+\alpha_{0}\sum\limits_{1 \leq i \leq n}(\gamma^{i})_{i}-\tilde{C}_{2}\bigg|,$$

否则不必证.

进一步得

$\begin{equation}|D'u|(x_{0})\leq \tilde{C}_{4},\end{equation}$

(3.24)

其中$\tilde{C}_{4}$为正常数依赖于$n,\Omega,\beta_{0},L_{1},L_{2}.$

情形3若$x_{0} \in \Omega\setminus\bar{\Omega}_{\mu_{0}}$,则问题归结内部梯度估计.

由引理2.1,我们在其证明中令${\rm dist}(\Omega',\Omega)=\frac{\mu_{0}}{2}$,得到这样区域的内估计.

综合3种情形,可以得到

$\begin{equation}\label{e3.26}\sup_{\overline{\Omega}\mu_{0}}|D'u| \leq \tilde{C},\end{equation}$

(3.25)

其中$\tilde{C}$为正常数依赖于$n,\Omega,L_{1},L_{2},L_{3},\beta_{0}.$

由$\varphi$在$x_0$点取得极大值,可知

$\begin{equation}\sup_{\bar{\Omega}}|D'u| \leq \tilde{C}_{5}.\end{equation}$

(3.26)

令$\tilde{H}(x,\xi)=D_\xi u+\tilde{k}x^2$,其中$\xi$为单位球上的任意方向.则有

$\begin{equation}\Delta \tilde{H}(x,\xi)=\Delta D_\xi u+2n\tilde{k}=f_{\xi}(x)+2n\tilde{k}\geq 0,\end{equation}$

(3.27)

取$\tilde{k} \geq \frac{L_2}{2n}$使得上式中的最后一个不等号成立.

由极值原理,可知

$\begin{equation}\max_{\bar{\Omega}}\tilde{H}(x,\xi)\leq \max_{\partial\Omega}\tilde{H}(x,\xi),\end{equation}$

(3.28)

根据(3.9)和(3.25)式可得

$\begin{equation}\sup_{\overline{\Omega}}|Du| \leq C,\end{equation}$

(3.29)

其中$C$为正常数依赖于$n,\Omega,L_{1},L_{2},L_{3},\beta_{0}.$

本小节中,我们利用文献[8]中的方法,构造辅助函数,充分利用函数取得极大值的性质以及斜边值条件得到梯度估计.

证 令

$\begin{equation}G(x)=|Dw|^{2}{\rm e}^{\frac{1}{1+L_{1}+u}{\rm e}^{\widetilde{\alpha}_{0}d}},\end{equation}$

(3.30)

其中$w(x)=u(x)-\psi(x)d$,$\widetilde{\alpha}_{0}=\frac{L_3+4\bar{C}_{1}+1}{\beta_{0}}$,其中$\bar{C}_{1}$为正常数依赖于$n,\Omega$.

取

$\begin{equation}\phi(x)=\log G(x)=\log|Dw|^{2}+\widetilde{h}(u)+\widetilde{g}(d),\end{equation}$

(3.31)

其中

$\begin{equation}\label{3.28aa}\widetilde{h}(u)=\frac{1}{1+L_{1}+u},~~\widetilde{g}(d)=\widetilde{\alpha}_{0}d.\end{equation}$

(3.32)

设$\phi(x)$在$x_{0}$点达到极大值.以下所有计算均在$x_0$点进行.下面分3种情况证明定理3.1.

情形1若$x_{0}\in\partial\Omega$,我们将证明$|Du|(x_{0})$有界.

首先对$\phi$沿$\beta$方向求导数,有

$\begin{equation}\label{3.28}D_{\beta}\phi=\frac{1}{|Dw|^{2}}\sum_{1 \leq i \leq n }(|Dw|^{2})_{i}\beta^{i}+\widetilde{h}'D_{\beta}u+\widetilde{\beta}_{0}\gamma^{i}\beta^{i}.\end{equation}$

(3.33)

因为

$\begin{equation}\label{a}\begin{array}{l}w_{i}=u_{i}-\psi_{i}d-\psi\gamma^{i}=u_{i}-\psi\gamma^{i},\\|Dw|^{2}=|D'w|^{2}+w_{\gamma}^{2},\\w_{\gamma}=u_{\gamma}-d\psi_{\gamma}-\psi=u_{\gamma}-\psi(x),\end{array}\end{equation}$

(3.34)

所以,在$\partial\Omega$上有

$\begin{equation}(|Dw|^{2})_{i}=(|D'w|^{2})_{i}+(w_{\gamma}^{2})_{i}.\end{equation}$

(3.35)

因此

$\begin{eqnarray} \label{3.31}\sum_{1 \leq i \leq n}(|Dw|^{2})_{i}\beta^{i}&=&\sum_{1 \leq i \leq n}(|D'w|^{2})_{i}\beta^{i}+\sum_{1 \leq i \leq n}(w_{\gamma}^{2})_{i}\beta^{i}\\&=&2\sum_{1 \leq i,k,l \leq n}c^{kl}w_{ki}w_{l}\beta^{i}+\sum_{1 \leq i,k,l \leq n}(c^{kl})_{i}w_{k}w_{l}\beta^{i}+\sum_{1 \leq i \leq n}(w_{\gamma}^{2})_{i}\beta^{i}\\&=&2\sum_{1 \leq i,k,l \leq n}c^{kl}(u_{ki}-\psi_{k}\gamma^i-\psi(\gamma^i)_k)(u_l-\psi\gamma^l)\beta^{i}\\&&+\sum_{1 \leq i,k,l \leq n}(c^{kl})_{i}(u_k-\psi\gamma^k)(u_l-\psi\gamma^l)\beta^{i}\\&&+\sum_{1 \leq i \leq n}(u_{\gamma}^{2}-2u_\gamma\psi+\psi^2)_{i}\beta^{i}\\&=&2\sum_{1 \leq i,k,l \leq n}c^{kl}u_{ki}u_{l}\beta^{i}+\sum_{1 \leq i,k \leq n}2(u_{\gamma}-\psi)u_{ki}\gamma^{k}\beta^{i}\\&&-2\sum_{1 \leq i,k,l \leq n}c^{kl}\psi_{k}\gamma^iu_l\beta^i-2\sum_{1 \leq i,k \leq n}(\gamma^i)_ku_k\psi\beta^i\\&&+\sum_{1 \leq i,k,l \leq n}(c^{kl})_{i}u_{k}u_{l}\beta^{i}+2\sum_{1 \leq i,k \leq n}(\gamma^k)_iu_k\psi\beta^i\\&&+\sum_{1 \leq i,k \leq n}2(u_{\gamma}-\psi)u_{k}(\gamma^{k})_{i}\beta^{i}-2u_{\gamma}\sum_{1 \leq i \leq n}\psi_{i}\beta^{i}+2\sum_{1 \leq i \leq n}\psi\psi_{i}\beta^{i}.~~\end{eqnarray}$

(3.36)

对边界条件关于切向求导有

$\begin{equation}\label{3.32}(D_{\beta}u)_{k}=\psi_{k},\end{equation}$

(3.37)

因此

$\begin{equation}\label{3.33}u_{ik}\beta^{i}+u_{i}(\beta^{i})_{k}=\psi_{k},\end{equation}$

(3.38)

计算可得

$\begin{equation}\label{3.34}u_{ik}\beta^{i}=-u_{i}(\beta^{i})_{k}+\psi_{k},\end{equation}$

(3.39)

所以

$\begin{equation}\label{3.34ab}\sum_{1 \leq i,k \leq n}c^{kl}u_{ki}\beta^{i}=-\sum_{1 \leq i,k \leq n}c^{kl}u_{i}(\beta^{i})_{k}+\sum_{1 \leq k \leq n}c^{kl}\psi_{k}.\end{equation}$

(3.40)

由(3.32),(3.33),(3.36)及(3.40)式得

$\begin{eqnarray}|Dw|^{2}D_\beta\phi(x_{0})&=&(\widetilde{h'}\psi+\widetilde{\alpha}_{0}\gamma^{i}\beta^{i})|Dw|^{2}+2\sum_{1 \leq i,k,l \leq n}c^{kl}u_{ki}u_{l}\beta^{i}\\&&+\sum_{1 \leq i,k \leq n}2(u_{\gamma}-\psi)u_{ki}\gamma^{k}\beta^{i}-2\sum_{1 \leq i,k,l \leq n}c^{kl}\psi_{k}\gamma^iu_l\beta^i\\&&-2\sum_{1 \leq i,k \leq n}(\gamma^i)_ku_k\psi\beta^i+\sum_{1 \leq i,k,l \leq n}(c^{kl})_{i}u_{k}u_{l}\beta^{i}\\&&+2\sum_{1 \leq i,k \leq n}(\gamma^k)_iu_k\psi\beta^i+\sum_{1 \leq i,k \leq n}2(u_{\gamma}-\psi)u_{k}(\gamma^{k})_{i}\beta^{i}\\&&-2u_{\gamma}\sum_{1 \leq i \leq n}\psi_{i}\beta^{i}+2\sum_{1 \leq i \leq n}\psi\psi_{i}\beta^{i}\\&=&(\widetilde{h'}\psi+\widetilde{\alpha}_{0}\gamma^{i}\beta^{i})|Dw|^{2}-2\sum_{1 \leq i,k \leq n}c^{kl}u_{i}u_l(\beta^{i})_{k}\\&&+2\sum_{1 \leq k \leq n}c^{kl}\psi_{k}u_l+\sum_{1 \leq i,k \leq n}2(u_{\gamma}-\psi)u_{ki}\gamma^{k}\beta^{i}\\&&-2\sum_{1 \leq i,k,l \leq n}c^{kl}\psi_{k}\gamma^iu_l\beta^i-2\sum_{1 \leq i,k \leq n}(\gamma^i)_ku_k\psi\beta^i\\&&+\sum_{1 \leq i,k,l \leq n}(c^{kl})_{i}u_{k}u_{l}\beta^{i}+2\sum_{1 \leq i,k \leq n}(\gamma^k)_iu_k\psi\beta^i\\&&+\sum_{1 \leq i,k \leq n}2(u_{\gamma}-\psi)u_{k}(\gamma^{k})_{i}\beta^{i}-2u_{\gamma}\sum_{1 \leq i \leq n}\psi_{i}\beta^{i}+2\sum_{1 \leq i \leq n}\psi\psi_{i}\beta^{i}\\&\geq&(\widetilde{h'}\psi+\widetilde{\alpha}_{0}\gamma^{i}\beta^{i})|Dw|^{2}-2\sum_{1 \leq i,k \leq n}c^{kl}u_{i}u_l(\beta^{i})_{k}\\&&+2\sum_{1 \leq k \leq n}c^{kl}\psi_{k}u_l-\sum_{1 \leq i,k \leq n}2|(u_{\gamma}-\psi)|(\hat{C}_2\sup_{\bar{\Omega}} |Du|+\hat{C}_3)\\&&-2\sum_{1 \leq i,k,l \leq n}c^{kl}\psi_{k}\gamma^iu_l\beta^i-2\sum_{1 \leq i,k \leq n}(\gamma^i)_ku_k\psi\beta^i\\&&+\sum_{1 \leq i,k,l \leq n}(c^{kl})_{i}u_{k}u_{l}\beta^{i}+2\sum_{1 \leq i,k \leq n}(\gamma^k)_iu_k\psi\beta^i\\&&+\sum_{1 \leq i,k \leq n}2(u_{\gamma}-\psi)u_{k}(\gamma^{k})_{i}\beta^{i}-2u_{\gamma}\sum_{1 \leq i \leq n}\psi_{i}\beta^{i}+2\sum_{1 \leq i \leq n}\psi\psi_{i}\beta^{i}\\& \geq &\Big(\widetilde{\alpha}_{0}\beta_{0}-\frac{|\psi|}{(1+L_{1}+u)^{2}}\Big)|Dw|^{2}-\bar{C}_{1}|Du|^{2},\end{eqnarray}$

(3.41)

第一个等式是利用(3.36)式,第二个等式是利用(3.40)式,第一个不等式是利用引理2.2以及$|Du|$的连续性得到的.最后一个不等式我们假设$|Du|$充分大,否则可得到估计,其中$\bar{C}_{1}$依赖于$n$,$\Omega$,$\beta_0$,$L_3.$

由(3.34)式,可得在$x_{0}$点,有

$\begin{equation}\label{3.37}|Dw|^{2}=\sum_{1 \leq k \leq n}|u_{k}-\psi\gamma^{k}|^{2},\end{equation}$

(3.42)

$\begin{equation}\label{3.37a}|Du|^{2}=\sum_{1 \leq k \leq n}|w_{k}+\psi\gamma^{k}|^{2},\end{equation}$

(3.43)

由Cauchy不等式,可得

$\begin{equation}\label{3.38}|Du|^{2} \leq 2|Dw|^{2}+2|\psi|^{2}.\end{equation}$

(3.44)

不妨设$|Du|^{2} \geq 2|\psi|^{2}$,$|Dw|^{2} \geq |\psi|^{2},$否则$|Du|^{2} \leq 4|\psi|^{2}$,得到估计.因此不妨设

$$2|\psi|^{2}_{c^{0}(\partial\Omega)} \leq |Du|^{2} \leq 4|Dw|^{2}.$$

因此

$\begin{equation}\label{3.39}|Dw|^{2}D_\beta\phi(x_{0})\geq(\widetilde{\alpha}_{0}{\beta_{0}}-|\psi|-4\bar{C}_{1})|Dw|^{2}\geq |Dw|^{2} > 0.\end{equation}$

(3.45)

另一方面,由$\phi$在$x_{0}$点取得极大值,则

$\begin{equation}D_\beta \phi(x_{0})\leq 0,\end{equation}$

(3.46)

与(3.45)式矛盾.因此,$|Du|(x_{0})$有界.

情形2若$x_{0}\in\Omega_{\mu_{0}}$,我们证明$|Du|(x_{0})$有界.

在$x_{0}$点选择标准坐标系,不妨设$u_{i}(x_{0})=0$,$2 \leq i \leq n$,$u_{1}(x_{0})=|Du| > 0$.

由$w(x)=u(x)-\psi(x)d$可知

$\begin{eqnarray} \label{3.42}|Dw|^{2}&=&\sum_{1 \leq k \leq n}(u_{k}-(d\psi)_{k})^{2}\\&\leq& 2|Du|^{2}+2\sum_{1 \leq k \leq n}(d\psi)_{k}^{2}\\&\leq &2|Du|^{2}+4\sum_{1 \leq k \leq n}(\gamma^{k})^{2}|\psi|^{2}+4\sum_{1 \leq k \leq n}d^{2}\psi_{k}^{2}\\&\leq &\bar{C}_{2}|Du|^{2},\end{eqnarray}$

(3.47)

其中$\bar{C}_{2}$为正常数依赖于$n,\Omega,L_3$,假设(3.47)式最后一个不等式成立,否则可得到估计.

同理可得

$\begin{equation}\label{3.43}|Du|^{2} \leq \bar{C}_{3}|Dw|^{2}.\end{equation}$

(3.48)

因此

$\begin{equation}\label{3.44}\frac{1}{\bar{C}_{4}}|Dw|^{2} \leq |Du|^{2} \leq \bar{C}_{4}|Dw|^{2},\end{equation}$

(3.49)

其中$\bar{C}_{3}$和$\bar{C}_{4}$为正常数依赖于$n,\Omega,L_3$.

对$\phi$求导可得

$\begin{equation}\label{3.45}\phi_{i}=\frac{(|Dw|^{2})_{i}}{|Dw|^{2}}+h'u_{i}+g'\gamma^{i},\end{equation}$

(3.50)

由$\phi_{i}=0$,可得

$\begin{equation}\label{3.46}(|Dw|^{2})_{i}=-|Dw|^{2}(h'u_{i}+g'\gamma^{i}),\end{equation}$

(3.51)

对$\phi$求两次导可得

$\begin{eqnarray}\label{3.47}0&\geq& \Delta\phi\\&=&\frac{\Delta|Dw|^{2}}{|Dw|^{2}}+h'\Delta u+(h"-h'^{2})|Du|^{2}-2h'g'\sum_{1 \leq i \leq n}\gamma^{i}u_{i}-g'^{2}+g'\sum_{1 \leq i \leq n}(\gamma^{i})_{i}\\&=&\frac{\Delta|Dw|^{2}}{|Dw|^{2}}+(h"-h'^{2})u_{1}^{2}-2h'g'\gamma^{1}u_{1}+h'f-g'^{2}+g'\sum_{1 \leq i \leq n}(\gamma^{i})_{i}.\end{eqnarray}$

(3.52)

接下来计算$\Delta|Dw|^{2}$.由$w=u-d\psi$,可得

$\begin{eqnarray}\label{3.48}\Delta|Dw|^{2}&=&\sum_{1 \leq i \leq n}\bigg[\sum_{1 \leq k \leq n}w_{k}^{2}\bigg]_{ii}\\&=&\sum_{1 \leq i \leq n}\bigg[\sum_{1 \leq k \leq n}2w_{k}w_{ki}\bigg]_{i}\\&=&2\sum_{1 \leq k,i \leq n}w_{ki}^{2}+2\sum_{1 \leq k \leq n}w_{k}(\Delta w)_{k}\\&=&2\sum_{1 \leq k,i \leq n}w_{ki}^{2}+2\sum_{1 \leq k \leq n}w_{k}(\Delta u)_{k}-2\sum_{1 \leq k \leq n}w_{k}(\Delta(d\psi))_{k}\\&=&2\sum_{1 \leq k,i \leq n}w_{ki}^{2}+2\sum_{1 \leq k \leq n}w_{k}f_k-2\sum_{1 \leq k \leq n}w_{k}(\Delta(d\psi))_{k}\\&=&2\sum_{1 \leq k,i \leq n}w_{ki}^{2}+2\sum_{1 \leq k \leq n}w_{k}f_k-2\sum_{1 \leq k \leq n}w_{k}(\Delta(d\psi))_{k}\\& \geq& -\bar{C}_{5}|Dw|^{2},\end{eqnarray}$

(3.53)

其中$\bar{C}_{5}$为正常数依赖于$n,\Omega,L_{2},L_{3}.$将(3.53)式代入(3.52)式,结合(3.32)式有

$\begin{eqnarray}\label{3.49}0 &\geq& \Delta\phi\\&\geq &(h"-h'^{2})u_{1}^{2}-2h'g'\gamma^{1}u_{1}+h'f-g'^{2}+g'\sum_{1 \leq i \leq n}(\gamma^{i})_{i}-\bar{C}_{5}\\ &\geq&(h"-h'^{2})u_{1}^{2}-\bar{C}_{6}u_{1}\\&\geq& \frac{u_{1}^{2}}{(1+2L_1)^{2}}-\bar{C}_{6}u_{1},\end{eqnarray}$

(3.54)

其中$\bar{C}_{6}$为正常数依赖于$n,\Omega,L_{2},L_{3}$,第三个不等式取$u_1$充分大,否则可得到估计.

因此

$$|Du|(x_{0})\leq \bar{C}_{7},$$

其中$\bar{C}_{7}$为正常数依赖于$n,\Omega,L_{1},L_{2},L_{3}.$

情形3若$x_{0} \in \Omega\setminus\bar{\Omega}_{\mu_{0}}$,则问题归结内部梯度估计.

由引理2.1,我们在其证明中令

$${\rm dist}(\Omega',\Omega)=\frac{\mu_{0}}{2},$$

得到这样区域的内估计.

综合3种情形,可以得到

$\begin{equation}\label{3.53}\sup_{{\overline{\Omega}_{\mu_{0}}}}|Du| \leq C,\end{equation}$

(3.55)

其中$C$为正常数依赖于$n,\Omega,L_{1},L_{2},L_{3},\beta_{0}.$