椭圆型Monge-Ampère方程已得到广泛研究, 具体可见文献[3-5].在文献[5]中, 作者考虑Monge-Ampère型方程半线性Neumann边值问题, 通过证明二阶导数的先验估计得到该类方程Neumann边值问题经典解的存在性, 唯一性以及正则性.关于完全非线性方程Neumann边值问题也有一些研究成果, 如参考文献[9].抛物型Monge-Ampère型方程是一类典型的完全非线性抛物方程, 它在最优控制理论等方面的研究中具有重要的应用, 许多学者对此类方程进行了深入研究^{[1, 2, 7, 8, 10-12]}.本文考虑抛物型Monge-Ampère型方程的Cauchy-Neumann问题, 形如

$ \frac{\partial u}{\partial t}={\det}^{\frac {1}{n}}[D^{2}u-A(x, u, Du)]-B(x, u, Du), ~~\mbox{在 $\Omega\times[0, T)$ 内} $

(1.1)

$ D_{\nu}u= \varphi(x, u), ~~\mbox{在 $\partial\Omega\times[0, T)$ 上}, $

(1.2)

$ u(x, 0)=u_{0}(x), ~~\mbox{在 $\Omega$ 内}, $

(1.3)

其中$\Omega$是$\mathbb{R}^{n}$中的$A$-凸区域, $n\geq 2$, $\partial\Omega\in C^{4}$, $Du$为梯度向量, $D^{2}u$为$u$的Hessian矩阵, $A\in C^{2}(\bar{\Omega}\times\mathbb{R}\times\mathbb{R}^{n})$是$n\times n$的对称矩阵值函数, $B\in C^{2}(\bar{\Omega}\times\mathbb{R}\times\mathbb{R}^{n})$是正的向量值函数, $\varphi\in C^{2, 1}(\partial\Omega\times\mathbb{R})$是向量值函数, $u_{0}\in C^{2}(\bar{\Omega})$, $\nu$是$\partial\Omega$上的单位内法向量.一般用$x, z, p$分别表示$\Omega, \mathbb{R}, \mathbb{R}^{n}$中的元素.

进一步地, 当$D^{2}u-A(x, u, Du)$正定时, 称$u$是椭圆解.为了得到问题(1.1)-(1.3)椭圆解的先验估计, 要求$A, B, \varphi, u_0$满足适当的光滑性条件和结构性条件:若函数$A$满足

$ A_{ij, kl}(x, z, p)\xi_{i}\xi_{j}\eta_{k}\eta_{l}\geq 0 (> 0), $

(1.4)

则称$A$是正则(严格正则)的, 其中$(x, z, p)\in\Omega\times\mathbb{R}\times \mathbb{R}^{n}, \xi, \eta \in \mathbb{R}^{n}, \xi\perp\eta, A_{ij, kl}=D^{2}_{p_{k}p_{l}}A_{ij}$.若函数$A$满足

$ D_{z}A_{ij}(x, z, p)\xi_{i}\xi_{j}\geq 0 (> 0), \mbox{ $\forall (x, z, p)\in\Omega\times\mathbb{R}\times\mathbb{R}^{n}, \xi\in\mathbb{R}^{n}$}, $

(1.5)

称$A$对$z$是非减(严格递增)的.同样地, 若函数$B$和$\varphi$满足

$ D_{z}B(x, z, p)\geq 0 (> 0), ~~ \mbox{$\forall (x, z, p)\in\Omega\times\mathbb{R}\times\mathbb{R}^{n}$}, $

(1.6)

$ D_{z}\varphi(x, z)\geq 0 (> 0), ~~ \mbox{$\forall (x, z)\in\partial\Omega\times\mathbb{R}$}, $

(1.7)

则称$B$和$\varphi$对$z$是非减(严格递增)的, $u_{0}$满足相容性条件

$ {\det} ^{\frac {1}{n}}[D^{2}u_{0}-A(x, u_{0}, Du_{0})]-B(x, u_{0}, Du_{0})\geq 0. $

(1.8)

另外, 区域$\Omega$满足$A$ -凸条件, 即存在函数$\phi\in C^2(\overline{\Omega})$使得在$\partial\Omega$上$\phi=0$, $D\phi\neq 0$, 在$\Omega$内$\phi< 0$, 且满足不等式

$ D_{ij}\phi-D_{p_{k}}A_{ij}(x, u, Du)D_{k}\phi\geq \delta_{1}I, $

(1.9)

其中$I$为单位矩阵, $\delta_1$为正常数.为了获得问题(1.1)-(1.3)解的$C^2$估计, 还需要假设问题(1.1)-(1.3)的有界上解$\bar{u}$存在且满足

$ \frac{\partial \bar{u}}{\partial t}\geq {\det}^\frac {1}{n}[D^{2}\bar{u}-A(x, \bar{u}, D\bar{u})]-B(x, \bar{u}, D\bar{u}), ~~\mbox{在 $\Omega\times[0, T)$ 内}, $

(1.10)

$ D_{\nu}\bar{u}= \varphi(x, \bar{u}), ~~\mbox{在 $\partial\Omega\times[0, T)$ 上}, $

(1.11)

$ \bar{u}(x, 0)={u}_{0}(x), ~~ \mbox{在 $\Omega$ 上}. $

(1.12)

下面给出本文的主要结论.

定理1.1  设$u\in C^{4, 2}(\bar{\Omega}\times[0, T])$为抛物型Monge-Ampère型方程的Cauchy-Neumann问题(1.1)-(1.3)的椭圆解, $\Omega\in C^{4}$是$\mathbb{R}^{n}$中$A$ -凸区域, $A\in C^{2}(\bar{\Omega}\times\mathbb{R}\times\mathbb{R}^{n})$满足(1.4)-(1.5)式, $0< B\in C^{2}(\bar{\Omega}\times\mathbb{R}\times\mathbb{R}^{n})$满足$(1.6)$式, $\varphi\in C^{2, 1}(\bar{\Omega}\times\mathbb{R})$满足$(1.7)$式, $u_{0}\in C^{4}(\bar{\Omega})$满足$(1.8)$式.假设问题(1.1)-(1.3)的有界上解$\bar{u}\in C^{2, 1}(\bar{\Omega}\times[0, T])$存在且满足(1.10)-(1.12)式, 则有

$ \sup\limits_{\bar{\Omega}\times[0, T]}|D^{2}u|\leq C, $

(1.13)

其中$C$是依赖于$n, A, B, \varphi, u_{0}, \bar{u}, |u|_{1, \Omega}, \delta_1$的常数.

注 定理1.1中假设(1.1)-(1.3)式的上解有界, 因为定理1.1得到(000)式中二阶导数被$C$控制, 而$C$是依赖于$\bar{u}$的常数, 故如果不假设上解有界, 则不能得到解的二阶梯度估计.

为了保证$t=0$处的光滑性, 需要假设$u_{0}$满足相容性条件

$ (\frac{d}{dt})^{m}(\nu^{i}u_{i}-\varphi(x, u))|_{t=0}=0, $

(1.14)

其中$m\geq 0$, $u, u_{i}, \cdots$关于时间的导数可以由$(1.1)$和$(1.3)$式得到.

在梯度估计的证明中, 还需要$A$的结构性条件

$ A(x, z, p)\geq -\mu_0(1+|p|^2)I, $

(1.15)

其中$p\in \mathbb{R}^{n}$, $\mu_0$为正常数.

基于前面的先验估计, 结合连续性方法以及抛物方程的一般理论, 可以得到问题(1.1)-(1.3)光滑解的存在性和正则性如下.

定理1.2  设$\Omega$是$\mathbb{R}^{n}$中有界光滑的$A$ -凸区域, $n\geq 2$, $A\in C^{\infty}(\bar{\Omega}\times\mathbb{R}\times\mathbb{R}^{n})$且满足(1.4), (1.5)和(As)式, $0< B\in C^{\infty}(\bar{\Omega}\times\mathbb{R}\times\mathbb{R}^{n})$满足$(1.6)$式, $\varphi\in C^{\infty}(\bar{\Omega}\times\mathbb{R})$满足$(1.7)$式, $u_{0}\in C^{\infty}(\bar{\Omega})$满足相容性条件$(1.8)$和(Am)式.假设问题(1.1)-(1.3)的光滑有界上解$\bar{u}\in C^{\infty}(\bar{\Omega}\times[0, T])$存在且满足(1.10)-(1.12)式, 则对任意的$t\geq 0$, 问题(1.1)-(1.3)存在光滑解.另外, 当$t\rightarrow\infty$时, $u$光滑地收敛到光滑函数$u^{\infty}$, 其中$u^{\infty}$满足Neumann边值问题

$\begin{aligned} &{\det}^{\frac {1}{n}}[D^{2}u^{\infty}-A(x, u^{\infty}, Du^{\infty})]=B(x, u^{\infty}, Du^{\infty}), ~~\mbox{在 $\Omega$ 内}, \\\ &D_{\nu}u^{\infty}= \varphi(x, u^{\infty}), ~~\mbox{在 $\partial\Omega$ 上}.\end{aligned} $

这一节, 先介绍与证明相关的一些基本概念和基本引理, 然后给出解的$C^0$和$C^1$估计.定义算子$\mathcal{L}_{t}=-\frac{\partial}{\partial t}+F^{ij}(D_{ij}-D_{p_{l}}A_{ij}D_{l})-D_{p_{l}}BD_{l}, $记$D_{i}=\frac{\partial }{\partial x_{i}}, $ $D_{ij}=\frac{\partial^{2}}{\partial x_{i}\partial x_{j}}, $ $w_{ij}=u_{ij}-A_{ij}(x, u, Du), $ $w_{\xi\xi}=w_{ij}\xi_{i}\xi_{j}, $ $F^{ij}=\frac{\partial{\det}^{\frac {1}{n}}w_{ij}}{\partial w_{ij}}=\frac{1}{n}({\det}^{\frac{1}{n}}w_{ij})w^{ij}, $其中$\{w^{ij}\}$是$\{w_{ij}\}$的逆矩阵.

引理2.1 设$u, v \in C^{2, 1}(\bar{\Omega}\times[0, T])$, 且$u, v$是椭圆解, $A\in C^{2}(\bar{\Omega}\times\mathbb{R}\times\mathbb{R}^{n})$是$n\times n$对称矩阵值函数满足$(1.5)$式, $B\in C^{2}(\bar{\Omega}\times\mathbb{R}\times\mathbb{R}^{n})$是向量值函数满足$(1.6)$式, 假设$u, v$满足如下条件

(1) 在$\Omega\times[0, T)$上,

$ \begin{aligned} &-\frac{\partial u}{\partial t}+{\det}^{\frac {1}{n}}[D^{2}u-A(x, u, Du)]-B(x, u, Du)\\ \geq &-\frac{\partial v}{\partial t}+{\det}^{\frac {1}{n}}[D^{2}v-A(x, v, Dv)]-B(x, v, Dv); \end{aligned} $

(2) 在$\partial\Omega\times[0, T]$上, 如果$u>v$, 那么${u_{\nu}>v_{\nu}}$;

(3) 在$\bar{\Omega}\times\{t=0\}$上, $u\leq v$,

则在$\bar{\Omega}\times[0, T]$内, $u\leq v$.

证  令$w=D^{2}u-A(x, u, Du), \tilde{w}=D^{2}v-A(x, v, Dv)$, 考虑

$ \begin{eqnarray*} && -\frac{\partial u}{\partial t}+{\det}^{\frac {1}{n}}[D^{2}u-A(x, u, Du)]-B(x, u, Du)\\ && -(-\frac{\partial v}{\partial t}+{\det}^{\frac {1}{n}}[D^{2}v-A(x, v, Dv)]-B(x, v, Dv))\\ &=&-(\frac{\partial u}{\partial t}-\frac{\partial v}{\partial t})+\int^{1}_{0}\frac{\partial {\det}^{\frac{1}{n}}[hw+(1-h)\tilde{w}]}{\partial[hw_{ij}+(1-h)\tilde{w}_{ij}]}dh(w-\tilde{w})_{ij}\\ &&-\int^{1}_{0}\frac{\partial B(x, hu+(1-h)v, hDu+(1-h)Dv)}{\partial[hu+(1-h)v]}dh(u-v)\\ &&-\int^{1}_{0}\frac{\partial B(x, hu+(1-h)v, hDu+(1-h)Dv)}{\partial[hu_k+(1-h)v_k]}dh(u-v)_k\\ &=&-\frac{\partial }{\partial t}(u-v)+a_{ij}(u-v)_{ij}-a_{ij}[A_{ij}(x, u, Du)-A_{ij}(x, v, Dv)]\\ &&-b(u-v)-c_k(u-v)_{k}\\ &=&-\frac{\partial }{\partial t}(u-v)+a_{ij}(u-v)_{ij}-(a_{ij}D_{p_{k}}A_{ij}(x, \bar{z}, \bar{p})+c_k)(u-v)_{k}\\ &&-(a_{ij}D_{z}A_{ij}(x, \bar{z}, \bar{p})+b)(u-v), \end{eqnarray*} $

其中$h\in[0, 1]$, $\bar{z}=\vartheta u+(1-\vartheta)v$, $\bar{p}=\vartheta Du+(1-\vartheta)Dv$, $\vartheta \in[0, 1]$,

$ \begin{aligned} a_{ij} &=\int^{1}_{0}\frac{\partial {\det}^{\frac{1}{n}}[hw+(1-h)\tilde{w}]}{\partial[hw_{ij}+(1-h)\tilde{w}_{ij}]}dh,\\ b &=\int^{1}_{0}\frac{\partial B(x, hu+(1-h)v, hDu+(1-h)Dv)}{\partial[hu+(1-h)v]}dh,\\ c_k &=\int^{1}_{0}\frac{\partial B(x, hu+(1-h)v, hDu+(1-h)Dv)}{\partial[hu_k+(1-h)v_k]}dh.\ \end{aligned} $

由$u(x, t), v(x, t)$是椭圆解可知$(a_{ij})$是正定矩阵, 又由$(1.5), (1.6)$式可得$a_{ij}D_{z}A_{ij}(x, \bar{z}, \bar{p})+b \geq0.$由条件(1), 可以得到

$ \begin{aligned} &-\frac{\partial }{\partial t}(u-v)+a_{ij}(u-v)_{ij}-(a_{ij}D_{p_{k}}A_{ij}(x, \bar{z}, \bar{p})+c_k)(u-v)_{k}\\ &-(a_{ij}D_{z}A_{ij}(x, \bar{z}, \bar{p})+b)(u-v)\geq 0, \end{aligned} $

由抛物方程的极值原理得

$ \begin{aligned} \sup\limits_{\bar{\Omega}\times[0, T]}(u-v)\leq \sup\limits_{\partial \Omega\times[0, T]\cup \bar{\Omega}\times\{t=0\}}(u-v)^{+}, \end{aligned} $

这里$(u-v)^{+}=\max\{u-v, 0\}$.

接下来, 假设$u-v$在$\partial\Omega\times[0, T]$上取得正极大值, 则$u-v> 0$, 且$D_{\nu}(u-v)\leq 0$, 这与条件(2)矛盾, 故$u-v$不能在$\partial\Omega\times[0, T]$上取得正极大值.再根据条件(3), 故有$u\leq v$在$\bar{\Omega}\times[0, T]$上恒成立.

为了证明问题(1.1)-(1.3)解的先验估计, 给出以下引理.

引理2.2 设$\Omega$是$\mathbb{R}^{n}$中的$A$ -凸区域, $u\in C^{3, 2}(\bar{\Omega}\times[0, T])$为问题(1.1)-(1.3)的椭圆解, $A\in C^{2}(\bar{\Omega}\times\mathbb{R}\times\mathbb{R}^{n})$满足$(1.5)$式, $B\in C^{2}(\bar{\Omega}\times\mathbb{R}\times\mathbb{R}^{n})$满足$(1.6)$式, $\varphi\in C^{2}(\bar{\Omega}\times\mathbb{R})$满足$(1.7)$式.如果$0\not\equiv\frac{\partial u}{\partial t}\geq 0$对$t=0$成立, 则对任意的$ t > 0$, 有$\frac{\partial u}{\partial t}>0$恒成立.

证  对$(1.1)$式关于$t$求导可得

$ \begin{aligned} \frac{\partial}{\partial t}(\frac{\partial u}{\partial t})=& \frac{\partial}{\partial t}({\det}^\frac{1}{n}w_{ij})-\frac{\partial}{\partial t}B=F^{ij}\frac{\partial w_{ij}}{\partial t}-(D_{z}B)\frac{\partial u}{\partial t}-(D_{p_{k}}B)\frac{\partial u_{k}}{\partial t}\\ =&F^{ij}(\frac{\partial u}{\partial t})_{ij}-F^{ij}\frac{\partial A_{ij}}{\partial t}-(D_{z}B)\frac{\partial u}{\partial t}-(D_{p_{k}}B)\partial_{k}(\frac{\partial u}{\partial t})\\ =&F^{ij}(\frac{\partial u}{\partial t})_{ij}-F^{ij}(D_{z}A_{ij})(\frac{\partial u}{\partial t})-F^{ij}(D_{p_{k}}A_{ij})\partial_{k}(\frac{\partial u}{\partial t})-(D_{z}B)\frac{\partial u}{\partial t}-\\(D_{p_{k}}B)\partial_{k}(\frac{\partial u}{\partial t})\\ =&F^{ij}(\frac{\partial u}{\partial t})_{ij}-(F^{ij}D_{p_{k}}A_{ij}+D_{p_{k}}B)\partial_{k}(\frac{\partial u}{\partial t})-(F^{ij}D_{z}A_{ij}+D_{z}B)(\frac{\partial u}{\partial t}), \end{aligned} $

由椭圆解知$F^{ij}$为正定矩阵, 由$(1.5), (1.6)$式得$D_{z}A_{ij}\geq 0, D_{z}B\geq 0$, 从而可以得到$F^{ij}D_{z}A_{ij}+D_{z}B\geq 0$, 则由抛物方程的极值原理得

$ \begin{aligned} \inf\limits_{\bar{\Omega}\times[0, T]}(\frac{\partial u}{\partial t})\geq \inf\limits_{\partial\Omega\times[0, T]\cup\bar{\Omega}\times\{t=0\}}(\frac{\partial u}{\partial t})^{-}, \end{aligned} $

其中$(\frac{\partial u}{\partial t})^{-}= \min \{\frac{\partial u}{\partial t}, 0\}$.

假设$\frac{\partial u}{\partial t}$在$\partial\Omega\times[0, T]$上取得负极小值, 则$\frac{\partial u}{\partial t} <0$, 又有

$ \begin{aligned} (\frac{\partial u}{\partial t})_{\nu}=\frac{\partial }{\partial t}(\frac{\partial u}{\partial \nu})=(D_{z}\varphi)\frac{\partial u}{\partial t}\leq 0, \end{aligned} $

这与$(\frac{\partial u}{\partial t})_{\nu}> 0$矛盾.

根据相容性条件(cc)知在$\bar{\Omega}\times\{t=0\}$上, $\frac{\partial u}{\partial t}\geq 0$.由上面的计算得

$ \begin{aligned} \inf\limits_{\bar{\Omega}\times[0, T]}(\frac{\partial u}{\partial t})\geq \inf\limits_{\partial\Omega\times[0, T)\cup\bar{\Omega}\times\{t=0\}}(\frac{\partial u}{\partial t})^{-}=0. \end{aligned} $

若存在点$(x_{0}, t_{0})\in\Omega\times(0, T)$使得$\frac{\partial u}{\partial t}=0$, 则由抛物方程的强极值原理得$\frac{\partial u}{\partial t}\equiv 0$, 这与$t=0$时$\frac{\partial u}{\partial t}\not\equiv 0$矛盾; 若存在点$(x_{0}, t_{0})\in\partial\Omega\times(0, T)$使得$\frac{\partial u}{\partial t}=0$, 则$(\frac{\partial u}{\partial t})_{\nu}=D_{z}\varphi\frac{\partial u}{\partial t}=0$, 与Hopf引理矛盾.

综上可得, 对任意的$ t > 0$, $\frac{\partial u}{\partial t}>0$恒成立.

接下来给出问题(1.1)-(1.3)解的$C^0$和$C^1$估计.

定理2.1  设$\Omega$是$\mathbb{R}^{n}$中有界的$A$-凸区域, $u\in C^{2, 1}(\bar{\Omega}\times[0, T])$为问题(1.1)-(1.3)的椭圆解, $A\in C^{2}(\bar{\Omega}\times\mathbb{R}\times\mathbb{R}^{n})$满足$(1.5)$式, $B\in C^{2}(\bar{\Omega}\times\mathbb{R}\times\mathbb{R}^{n})$满足$(1.6)$式, $\varphi\in C^{2, 1}(\bar{\Omega}\times\mathbb{R})$满足$(1.7)$式, 假设问题(1.1)-(1.3)的有界上解$\bar{u}\in C^{2, 1}(\bar{\Omega}\times[0, T])$存在且满足(1.10)-(1.12)式, 则有$\sup\limits_{\bar{\Omega}\times[0, T]}|u| \leq C, $其中$C$是依赖于$n, A, B, \varphi, u_{0}, \bar{u}$的常数.

证  由相容性条件$(1.8)$和引理2.2可得对任意的$t>0$有$\frac{\partial u}{\partial t} > 0 $, 则有$u(x, t)\geq u(x, 0)=u_{0}(x)$, 已知问题(1.1)-(1.3)存在上解$\bar{u}$, 根据引理2.1有$u\leq \bar{u}$, 从而可以得到$\sup\limits_{\bar{\Omega}\times[0, T]}|u| \leq C, $其中$C$是依赖于$n, A, B, \varphi, u_{0}, \bar{u}$的常数.

定理2.2  设$\Omega$是$\mathbb{R}^{n}$中有界的$A$ -凸区域, $u\in C^{3, 2}(\bar{\Omega}\times[0, T])$为问题(1.1)-(1.3)的$A$ -凸解, $A\in C^{2}(\bar{\Omega}\times\mathbb{R}\times\mathbb{R}^{n})$满足$(1.5)$式, $B\in C^{2}(\bar{\Omega}\times\mathbb{R}\times\mathbb{R}^{n})$满足$(1.6)$式, $\varphi\in C^{2, 1}(\bar{\Omega}\times\mathbb{R})$满足$(1.7)$式.假设问题(1.1)-(1.3)的有界上解$\bar{u}\in C^{2, 1}(\bar{\Omega}\times[0, T])$存在且满足(1.10)-(1.12)式.如果对$t=0$有$0\not\equiv\frac{\partial u}{\partial t}\geq 0$成立, 则有$\sup\limits_{\bar{\Omega}\times[0, T]}|\frac{\partial u}{\partial t}| \leq C, $其中$C$是依赖于$n, A, B, \varphi, u_{0}, \bar{u}, \Omega$的常数.

证  首先对$(\frac{\partial u}{\partial t})^{2}$关于$t$求导得

$ \begin{aligned} \frac{\partial}{\partial t}[(\frac{\partial u}{\partial t})^{2}] =&2\frac{\partial u}{\partial t}\frac{\partial ^{2}u}{\partial t^{2}}=2\frac{\partial u}{\partial t}\frac{\partial[{\det}^{\frac{1}{n}}w_{ij}-B]}{\partial t}\\ =&2\frac{\partial u}{\partial t}[F^{ij}\frac{\partial w_{ij}}{\partial t}-(D_{z}B)\frac{\partial u}{\partial t}-(D_{p_{k}}B)\frac{\partial u_{k}}{\partial t}]\\ =&2F^{ij}\frac{\partial u}{\partial t}\partial_{ij}(\frac{\partial u}{\partial t})-2F^{ij}(D_{z}A_{ij})(\frac{\partial u}{\partial t})^2-2F^{ij}(D_{p_{k}}A_{ij})\frac{\partial u}{\partial t}\partial_{k}(\frac{\partial u}{\partial t})\\ &-2(D_{z}B)(\frac{\partial u}{\partial t})^{2}-2(D_{p_{k}}B)\frac{\partial u}{\partial t}\partial_{k}(\frac{\partial u}{\partial t})\\ =&F^{ij}[(\frac{\partial u}{\partial t})^{2}]_{ij}-2F^{ij}(\frac{\partial u}{\partial t})_{i}(\frac{\partial u}{\partial t})_{j}-2F^{ij}(D_{z}A_{ij})(\frac{\partial u}{\partial t})^{2}\\ &-F^{ij}(D_{p_{k}}A_{ij})\partial_{k}(\frac{\partial u}{\partial t})^{2}-2(D_{z}B)(\frac{\partial u}{\partial t})^{2}-(D_{p_{k}}B)\partial_{k}(\frac{\partial u}{\partial t})^{2}, \end{aligned} $

由上面的计算可得

$ \begin{aligned} 0\leq &2F^{ij}(\frac{\partial u}{\partial t})_{i}(\frac{\partial u}{\partial t})_{j}=-\frac{\partial}{\partial t}[(\frac{\partial u}{\partial t})^{2}]+F^{ij}[(\frac{\partial u}{\partial t})^{2}]_{ij}\\ &-[F^{ij}D_{p_{k}}A_{ij}+D_{p_{k}}B]\partial_{k}(\frac{\partial u}{\partial t})^{2}-[2F^{ij}D_{z}A_{ij}+2D_{z}B](\frac{\partial u}{\partial t})^{2}, \end{aligned} $

由椭圆解知$F^{ij}$为正定矩阵, 由$(1.5), (1.6)$式得$D_{z}A_{ij}\geq 0, D_{z}B \geq 0$, 从而可以得到$2F^{ij}D_{z}A_{ij}+2D_{z}B\geq 0$, 则由抛物方程的极值原理得

$ \begin{aligned} \sup\limits_{\bar{\Omega}\times[0, T]}(\frac{\partial u}{\partial t})^{2}\leq \sup\limits_{\partial\Omega\times[0, T]\cup\bar{\Omega}\times\{t=0\}}[(\frac{\partial u}{\partial t})^{2}]^{+}, \end{aligned} $

其中$[(\frac{\partial u}{\partial t})^{2}]^{+}=\max\{(\frac{\partial u}{\partial t})^{2}, 0\}$.假设$(\frac{\partial u}{\partial t})^2$在$\partial\Omega\times[0, T]$上取得正极大值, 计算可得

$ \begin{aligned} D_{\nu}(\frac{\partial u}{\partial t})^2=2\frac{\partial u}{\partial t}D_{\nu}(\frac{\partial u}{\partial t})=2(\frac{\partial u}{\partial t})^2D_{z}\varphi\geq 0, \end{aligned} $

这与已知$D_{\nu}(\frac{\partial u}{\partial t})^2< 0$相矛盾, 故

$ \begin{aligned} \sup\limits_{\bar{\Omega}\times[0, T]}(\frac{\partial u}{\partial t})^{2} \leq \sup\limits_{\bar{\Omega}\times\{t=0\}}[(\frac{\partial u}{\partial t})^{2}]^{+}. \end{aligned} $

综合上面的计算可得$\sup\limits_{\bar{\Omega}\times[0, T]}|\frac{\partial u}{\partial t}|\leq M, $其中$M$是依赖于$n, A, B, \varphi, u_{0}, \bar{u}, \Omega$的常数.

由文献[11]中的定理$9$可得对任意的$t_{0}\in[0, T]$, 有

$ \sup\limits_{\bar{\Omega}\times[0, T]}|Du(x, t_{0})| \leq M_{1}, $

其中$M_{1}$是依赖于$n, A, B, \varphi, \Omega$的常数.由$t_{0}$的任意性可得

$ \sup\limits_{\bar{\Omega}\times[0, T]}|Du| \leq C, $

(2.1)

其中$C$是依赖于$n, A, B, \varphi, \Omega, M_{1}$的常数.

本节给出定理1.1的证明, 在证明二阶梯度估计之前先介绍一个引理, 此引理对证明二阶梯度估计非常重要, 其证明过程和文献[6]中的引理2.1类似, 省略证明细节.

引理3.1 设$u\in C^2(\bar{\Omega}\times[0, T])$是问题(1.1)-(1.3)的椭圆解, $\tilde{u}\in C^2(\bar{\Omega}\times[0, T])$是问题(1.1)-(1.3)的椭圆有界上解, 其中$A$满足(1.4)-(1.5)式, 则有$\mathcal{L}_{t}e^{N(\tilde{u}-u)}\geq \varepsilon\sum_{i}F^{ii}-C, $其中$N$是正常数, $\varepsilon, C$是依赖于$A, B, \Omega, |u|_{1, \Omega}, \tilde{u}, N$的正常数.

首先定义辅助函数$v, G$如下

$ \begin{equation}\label{f} \begin{split} v=v(x, t, \xi)=e^{\alpha|Du|^{2}+\beta \Phi}(w_{\xi\xi}-v'),\\ G=\log v=\log(w_{\xi\xi}-v')+\alpha |Du|^{2}+\beta \Phi, \end{split} \end{equation} $

(3.1)

其中$(x, t)\in\bar{\Omega}\times[0, T]$, $|\xi|=1$, 且$\alpha, \beta$是待定的正常数.另外, $(3.1)$式中的$\Phi$和$v'$分别给出如下: $\Phi=e^{N(\tilde{u}-u)}, $其中$\Phi$是引理$3.1$中的辅助函数, $\tilde{u}=\bar{u}-a\phi$, $\bar{u}$满足(1.10)-(1.12)式, $a$是足够小的常数, $\phi$是$A$ -凸区域的定义函数, 满足$(1.9)$式且在$\partial\Omega$上$D_{\nu}\phi=-1$.接着定义$v'$如下

$ v'(x, t, \xi)=2(\xi, \nu)\xi_{i}'(D_{i}\varphi(x, u)-D_{k}uD_{i}\nu_{k}-A_{ij}\nu_{j}), $

其中$\xi'=\xi-(\xi, \nu)\nu$, $\nu\in C^{3}(\bar{\Omega})$是从$\partial\Omega\times[0, T]$延拓到$\bar{\Omega}\times[0, T]$的单位内法向量.

定理1.1的证明 本定理的证明通过对$v$在$\bar{\Omega}\times[0, T]$上的最大值的估计, 得到对应的$D^{2}u$在$\bar{\Omega}\times[0, T]$上的估计, 从而得到本定理的结论.

证  下面分两种情形证明定理1.1.

假设$v(x, t)$在点$(x_{0}, t_{0})$处取得最大值, 且以下所有的计算都在点$(x_{0}, t_{0})$处进行.

情形一 若$(x_{0}, t_{0})\in \Omega \times (0, T)$, 根据$(3.1)$式中$v$和$G$的定义知, $G$也在点$(x_{0}, t_{0})$处取得最大值.对$G$关于$x_{i}$求偏导得

$ \begin{equation}\label{3.2} 0=D_{i}G=\frac{D_{i}(w_{\xi\xi}-v')}{w_{\xi\xi}-v'}+2\alpha D_{k}uD_{ki}u+\beta D_{i}\Phi, \end{equation} $

(3.2)

再对$D_{i}G$关于$x_{j}$求偏导得

$ \begin{equation}\label{g} \begin{split} 0\geq D_{ij}G =&\frac{D_{ij}(w_{\xi\xi}-v')}{w_{\xi\xi}-v'}-\frac{D_{i}(w_{\xi\xi}-v')D_{j}(w_{\xi\xi}-v')}{(w_{\xi\xi}-v')^{2}}\\&+2\alpha (D_{ki}uD_{kj}u+D_{k}uD_{kij}u)+\beta D_{ij}\Phi. \end{split} \end{equation} $

(3.3)

将算子$\mathcal{L}_{t}$作用到$G$上得

$ \begin{equation}\label{3.4} \begin{split} 0\geq&\mathcal{L}_{t}G =-\frac{\partial}{\partial t}G+F^{ij}(D_{ij}G-D_{p_{l}}A_{ij}D_{l}G)-D_{p_{l}}BD_{l}G\\ =& \mathcal{L}_{t}\log(w_{\xi\xi}-v')+\alpha\mathcal{L}_{t}|Du|^{2}+\beta\mathcal{L}_{t} \Phi, \end{split} \end{equation} $

(3.4)

其中第一个不等式运用了$(3.2)$, $(3.3)$式和$\frac{\partial G}{\partial t}=0$.

首先将算子$\mathcal{L}_{t}$作用到$\log(w_{\xi\xi}-v')$上得

$ \begin{eqnarray}\label{3.8} \mathcal{L}_{t}\log(w_{\xi\xi}-v') &=&-\frac{\partial}{\partial t}\log(w_{\xi\xi}-v')+F^{ij}(D_{ij}-D_{p_{l}}A_{ij}D_{l})[\log(w_{\xi\xi}-v')]\nonumber\\ &&-D_{p_{l}}BD_{l}[\log(w_{\xi\xi}-v')]\nonumber\\ &=&-\frac{1}{w_{\xi\xi}-v'}\frac{\partial}{\partial t}(w_{\xi\xi}-v')+F^{ij}\frac{1}{w_{\xi\xi}-v'}D_{ij}(w_{\xi\xi}-v')\nonumber\\ &&-F^{ij}\frac{D_{i}(w_{\xi\xi}-v')D_{j}(w_{\xi\xi}-v')}{(w_{\xi\xi}-v')^{2}}-F^{ij}D_{p_{l}}A_{ij}\frac{D_{l}(w_{\xi\xi}-v')}{w_{\xi\xi}-v'}\nonumber\\ &&-D_{p_{l}}B\frac{D_{l}(w_{\xi\xi}-v')}{(w_{\xi\xi}-v')}\nonumber\\ &=&\frac{1}{w_{\xi\xi}-v'}\mathcal{L}_{t}(w_{\xi\xi}-v')-F^{ij}\frac{D_{i}(w_{\xi\xi}-v')D_{j}(w_{\xi\xi}-v')}{(w_{\xi\xi}-v')^{2}}, \end{eqnarray} $

(3.5)

其中第三个等式运用了算子$\mathcal{L}_{t}(w_{\xi\xi}-v')$的定义.

接着将算子$\mathcal{L}_{t}$作用到$|Du|^{2}$上得

$ \begin{equation}\label{3.9} \begin{split} \mathcal{L}_{t}|Du|^{2} =&-\frac{\partial}{\partial t}(|Du|^{2})+F^{ij}(D_{ij}-D_{p_{l}}A_{ij}D_{l})(|Du|^{2})-D_{p_{l}}BD_{l}(|Du|^{2})\\ =&-2D_{k}u\frac{\partial D_{k}u}{\partial t}+2F^{ij}D_{ki}uD_{kj}u+2F^{ij}D_{k}uD_{kij}u\\&-2F^{ij}(D_{p_{l}}A_{ij})D_{k}uD_{kl}u-2 (D_{p_{l}}B)D_{k}uD_{kl}u\\ =&2D_{k}u\mathcal{L}_{t}u_{k}+2F^{ij}D_{ki}uD_{kj}u. \end{split} \end{equation} $

(3.6)

同理, 第三个等式运用了算子$\mathcal{L}_{t}u_{k}$的定义.将$(3.5)$, $(3.6)$式代入$(3.4)$式得

$ \begin{equation}\label{m} \begin{split} 0\geq&\mathcal{L}_{t}G =\frac{1}{w_{\xi\xi}-v'}\mathcal{L}_{t}(w_{\xi\xi}-v')-F^{ij}\frac{D_{i}(w_{\xi\xi}-v^{'})D_{j}(w_{\xi\xi}-v')}{(w_{\xi\xi}-v')^{2}}\\ &+2\alpha D_{k}u\mathcal{L}_{t}u_{k}+2\alpha F^{ij}D_{ki}uD_{kj}u+\beta\mathcal{L}_{t} \Phi. \end{split} \end{equation} $

(3.7)

首先估计$(3.7)$式中的$\mathcal{L}_{t}(w_{\xi\xi}-v')$, 将$(1.1)$式沿$\xi$方向求导得

$ \begin{equation}\label{3.11} \begin{split} \frac{\partial u_{\xi}}{\partial t}=& F^{ij}[D_{ij}u_{\xi}-D_{\xi}A_{ij}-D_{z}A_{ij}u_{\xi}-D_{p_{l}}A_{ij}D_{l}u_{\xi}]-D_{\xi}B\\ &-D_{z}Bu_{\xi}-D_{p_{l}}BD_{l}u_{\xi}, \end{split} \end{equation} $

(3.8)

再次对$(3.8)$式沿$\xi$方向求导得

$ \begin{equation}\label{0} \begin{split} \frac{\partial u_{\xi\xi}}{\partial t} =&-F^{ik}F^{jl}D_{\xi}w_{ij}D_{\xi}w_{kl}+F^{ij}(D_{ij}u_{\xi\xi}-D_{\xi\xi}A_{ij}-D_{\xi z}A_{ij}u_{\xi}\\ &-D_{\xi p_{l}}A_{ij}D_{l}u_{\xi}-D_{z\xi}A_{ij}u_{\xi}-D_{zz}A_{ij}u_{\xi}^{2}-D_{zp_{l}}A_{ij}u_{\xi}D_{l}u_{\xi}-D_{z}A_{ij}u_{\xi\xi}\\ &-D_{p_{l}\xi}A_{ij}D_{l}u_{\xi}-D_{p_{l}z}A_{ij}u_{\xi}D_{l}u_{\xi}-D_{p_{l}p_{k}}A_{ij}D_{l}u_{\xi}D_{k}u_{\xi}\\ &-D_{p_{l}}A_{ij}D_{l}u_{\xi\xi})-D_{\xi\xi}B-D_{\xi z}Bu_{\xi}-D_{\xi p_{l}}BD_{l}u_{\xi}-D_{z\xi}Bu_{\xi}-D_{zz}Bu_{\xi}^{2}\\ &-D_{zp_{l}}Bu_{\xi}D_{l}u_{\xi}-D_{z}Bu_{\xi\xi}-D_{p_{l}\xi}BD_{l}u_{\xi}-D_{p_{l}z}D_{l}u_{\xi}u_{\xi}\\ &-D_{p_{l}p_{k}}BD_{l}u_{\xi}D_{k}u_{\xi}-D_{p_{l}}BD_{l}u_{\xi\xi}\\ =&-F^{ik}F^{jl}D_{\xi}w_{ij}D_{\xi}w_{kl}+F^{ij}(D_{ij}u_{\xi\xi}-D_{\xi\xi}A_{ij}-D_{zz}A_{ij}u_{\xi}^{2}\\ &-D_{p_{k}p_{l}}A_{ij}D_{k}u_{\xi}D_{l}u_{\xi}-D_{z}A_{ij}u_{\xi\xi}-D_{p_{l}}A_{ij}D_{l}u_{\xi\xi}-2D_{\xi z}A_{ij}u_{\xi}\\ &-2D_{\xi p_{l}}A_{ij}D_{l}u_{\xi}-2D_{zp_{l}}A_{ij}u_{\xi}D_{l}u_{\xi})-D_{\xi\xi}B-D_{zz}Bu_{\xi}^{2}\\ &-D_{p_{l}p_{l}}BD_{l}u_{\xi}D_{l}u_{\xi}-2D_{\xi z}Bu_{\xi}-2D_{\xi p_{l}}BD_{l}u_{\xi}\\ &-2D_{z p_{l}}Bu_{\xi}D_{l}u_{\xi}-D_{z}Bu_{\xi\xi}-D_{p_{l}}BD_{l}u_{\xi\xi}. \end{split} \end{equation} $

(3.9)

接着将算子$\mathcal{L}_{t}$作用到$u_{\xi\xi}$上得

$ \begin{equation}\label{3.13} \begin{split} \mathcal{L}_{t}u_{\xi\xi} =&-\frac{\partial}{\partial t}u_{\xi\xi}+F^{ij}(D_{ij}u_{\xi\xi}-D_{p_{l}}A_{ij}D_{l}u_{\xi\xi})-D_{p_{l}}BD_{l}u_{\xi\xi}\\ =&F^{ik}F^{jl}D_{\xi}w_{ij}D_{\xi}w_{kl}+F^{ij}D_{\xi\xi}A_{ij}+F^{ij}(D_{zz}A_{ij})u_{\xi}^{2}\\ &+F^{ij}(D_{p_{k}p_{l}}A_{ij})D_{k}u_{\xi}D_{l}u_{\xi}+F^{ij}(D_{z}A_{ij})u_{\xi\xi}+2F^{ij}(D_{\xi z}A_{ij})u_{\xi}\\ &+2F^{ij}(D_{\xi p_{l}}A_{ij})D_{l}u_{\xi}+2F^{ij}(D_{zp_{l}}A_{ij})(D_{l}u_{\xi})u_{\xi}+D_{\xi\xi}B+(D_{zz}B)u_{\xi}^{2}\\ &+(D_{p_{k}p_{l}}B)D_{k}u_{\xi}D_{l}u_{\xi}+2(D_{\xi z}B)u_{\xi}+2(D_{\xi p_{l}}B)D_{l}u_{\xi}\\ &+2(D_{z p_{l}}B)(D_{l}u_{\xi})u_{\xi}+(D_{z}B)u_{\xi\xi}, \end{split} \end{equation} $

(3.10)

其中第二个等式是将$(3.9)$式代入所得.由$(3.10)$和$(1.4)$式可以得到

$ \begin{equation}\label{3.14} \mathcal{L}_{t}u_{\xi\xi}\geq F^{ik}F^{jl}D_{\xi}w_{ij}D_{\xi}w_{kl}-C[(1+w_{ii})\mathcal{J}+(w_{ii})^{2}], \end{equation} $

(3.11)

其中$\mathcal{J}={\rm tr} F^{ii}$, $C$是依赖于$n, A, B, |u|_{1, \Omega}$的常数.同理可得

$ \begin{equation}\label{3.15} |\mathcal{L}_{t}A_{\xi\xi}|\leq C\{(1+w_{ii})\mathcal{J}+w_{ii}\}, \end{equation} $

(3.12)

$ \begin{equation}\label{3.16} |\mathcal{L}_{t}v'|\leq C\{(1+w_{ii})\mathcal{J}+w_{ii}\}. \end{equation} $

(3.13)

结合$(3.11)$, $(3.12)$, $(3.13)$式可得

$ \begin{equation}\label{3.17} \mathcal{L}_{t}(w_{\xi\xi}-v')\geq F^{ik}F^{jl}D_{\xi}w_{ij}D_{\xi}w_{kl}-C[(1+w_{ii})\mathcal{J}+(w_{ii})^{2}], \end{equation} $

(3.14)

其中$C$是依赖于$n, A, B, |u|_{1, \Omega}$的常数.

接着估计$F^{ij}D_{i}(w_{\xi\xi}-v')D_{j}(w_{\xi\xi}-v')$, 由柯西不等式得

$ \begin{equation}\label{3.18} \begin{split} F^{ij}D_{i}(w_{\xi\xi}-v')D_{j}(w_{\xi\xi}-v')\leq (1+\theta)F^{ij}D_{i}w_{\xi\xi}D_{j}w_{\xi\xi}+C(\theta)F^{ij}D_{i}v'D_{j}v', \end{split} \end{equation} $

(3.15)

其中$\theta > 0$, $C(\theta)$是依赖于$\theta$的正常数.进一步, 估计$2\alpha D_{k}u\mathcal{L}_{t}u_{k}$, 将算子$\mathcal{L}_{t}$作用在$u_{k}$上可得

$ \begin{equation}\label{3.19} \begin{split} \mathcal{L}_{t}u_{k} =&-\frac{\partial}{\partial t}u_{k}+F^{ij}(D_{ij}u_{k}-D_{pl}A_{ij}D_{l}u_{k})-D_{p_{l}}BD_{l}u_{k}\\ =&-F^{ij}[D_{ij}u_{k}-D_{k}A_{ij}-D_{z}A_{ij}u_{k}-D_{p_{l}}A_{ij}D_{l}u_{k}]+D_{k}B+D_{z}Bu_{k}\\ &+D_{p_{l}}BD_{l}u_{k}+F^{ij}(D_{ij}u_{k}-D_{pl}A_{ij}D_{l}u_{k})-D_{p_{l}}BD_{l}u_{k}\\ =&F^{ij}D_{k}A_{ij}+F^{ij}D_{z}A_{ij}u_{k}+D_{k}B+D_{z}Bu_{k}, \end{split} \end{equation} $

(3.16)

其中第二个等式是将$(1.1)$式关于$x_{k}$求导代入得到的.则由$(3.16)$和$(2.1)$式得

$ \begin{equation}\label{3.20} 2\alpha D_{k}u\mathcal{L}_{t}u_{k}\leq C\alpha (\mathcal{J}+1), \end{equation} $

(3.17)

其中$C$是依赖于$n, A, B, |u|_{1, \Omega}$的常数.由引理$3.1$得

$ \begin{equation}\label{3.21} \begin{split} \mathcal{L}_{t}\Phi &=-\beta\frac{\partial}{\partial t}\Phi+\beta F^{ij}(D_{ij}\Phi-\beta D_{p_{l}}A_{ij}D_{l}\Phi)-\beta D_{p_{l}}BD_{l}\Phi \geq \varepsilon\mathcal{J}-C. \end{split} \end{equation} $

(3.18)

合并$(3.14)$, $(3.15)$, $(3.17)$, $(3.18)$式得

$ \begin{equation}\label{3.22} \begin{split} 0\geq \mathcal{L}_{t}G \geq&\frac{1}{w_{\xi\xi}-v'}F^{ik}F^{jl}D_{\xi}w_{ij}D_{\xi}w_{kl}-\frac{C}{w_{\xi\xi}-v'}[(1+w_{ii})\mathcal{J}+w_{ii}^{2}]\\ &-\frac{1+\theta}{(w_{\xi\xi}-v')^{2}}F^{ij}D_{i}w_{\xi\xi}D_{j}w_{\xi\xi}-\frac{C(\theta)}{(w_{\xi\xi}-v')^{2}}F^{ij}D_{i}v'D_{j}v'\\ &+\alpha w_{ii}-C\alpha(\mathcal{J}+1) +\varepsilon\beta \mathcal{J}-\beta C. \end{split} \end{equation} $

(3.19)

假设$\{w_{ij}\}$在$(x_{0}, t_{0})$处为对角矩阵函数, 有最大特征值$w_{11}$, 且$w_{11}>1$, 否则结论已证.首先估计$(3.19)$式的三阶导数项.由文献[11]中的$(3.48)$式可以得到

$ \begin{equation}\label{3.23} F^{ik}F^{jl}D_{\xi}w_{ij}D_{\xi}w_{kl}-\frac{1}{w_{11}}F^{ij}D_{i}w_{\xi\xi}D_{j}w_{\xi\xi}\geq 0. \end{equation} $

(3.20)

由于$v'$是有界的, $w_{11}$, $w_{\xi\xi}$是可比较的, 则对任意的$\theta>0$, 存在更大的常数$C(\theta)$, 若$w_{11}>C(\theta)$, 则有

$ |w_{11}-w_{\xi\xi}+v'|<\theta w_{11}. $

(3.21)

由$(3.20)$, $(3.21)$式可得

$ \begin{equation}\label{3.25} \begin{split} F^{ik}F^{jl}D_{\xi}w_{ij}D_{\xi}w_{kl}\geq \frac{1-\theta}{w_{\xi\xi}-v'}F^{ij}D_{i}w_{\xi\xi}D_{j}w_{\xi\xi}. \end{split} \end{equation} $

(3.22)

由$(3.2)$式中的$D_{i}G=0$可得

$ \begin{equation}\label{01} D_{i}w_{\xi\xi}=(w_{\xi\xi}-v')(2\alpha D_{k}uD_{ki}u+\beta D_{i}\Phi)+D_{i}v'. \end{equation} $

(3.23)

由$(3.23)$式和柯西不等式可得

$ \begin{equation} \begin{split}\label{3.26} F^{ij}D_{i}w_{\xi\xi}D_{j}w_{\xi\xi}&\leq 2F^{ii}[|D_{i}v'|^{2}+(w_{\xi\xi}-v')^{2}(2\alpha D_{k}uD_{ki}u+\beta D_{i}\Phi)^{2}]\\ &\leq 2F^{ii}|D_{i}v'|^{2}+C(w_{\xi\xi}-v')^{2}(\alpha^{2}w_{ii}+\beta^{2}\mathcal{J}+\alpha\beta), \end{split} \end{equation} $

(3.24)

其中$C$是依赖于$n, a, N, \bar{u}, \phi, |u|_{1, \Omega}$的常数.结合$(3.19)$, $(3.21)$, $(3.22)$, $(3.24)$式得对$w_{11}>C(\theta)$, 有

$ \begin{aligned} (\alpha-C(1+\alpha^2)\theta)w_{ii}\leq C[\alpha+\beta+\alpha\beta\theta +(\theta+\alpha-\beta+\theta \beta^{2})\mathcal{J}], \end{aligned} $

先选择$\alpha, \beta$足够大, 然后选择$\theta$是充分小的正常数, 从而可以得到估计$ w_{ii}(x_{0}, t_{0})\leq C, $其中$C$是依赖于$A, B, \Omega, |u|_{1, \Omega}$的常数.从而可以得到$|D^{2}u(x, t)|$对应的估计.

情形二 若$(x_{0}, t_{0})\in\partial\Omega\times[0, T]$, 考虑当$\xi$属于三种不同的方向, 分别来估计$v(x_{0}, t_{0}, \xi)$.

首先将切算子$\delta_{i}$作用到$(1.2)$式上可以得到

$ (D_{k}u)\delta_{i}\nu_{k}+\nu_{k}\delta_{i}(D_{k}u)=\delta_{i}\varphi, \mbox{在 $\partial\Omega$ 上}, $

其中$\delta_{i}=(\delta_{ij}-\tau_{i}\tau_{j})D_{j}$, 可以得到

$ D_{\tau\nu}u\leq C, \mbox{在$\partial\Omega$上}, $

(3.25)

其中$\tau$为任意的切向量.

(ⅰ) $\xi$是点$(x_{0}, t_{0})$处的单位法向量.首先定义辅助函数$g$如下: $\begin{aligned} g=\nu_{k}D_{k}u-\varphi(x, u). \end{aligned}$将算子$\mathcal{L}_{t}$作用到$g$上得

$ \begin{equation}\label{3.30} \begin{split} \mathcal{L}_{t}g =&-\frac{\partial}{\partial t}g+F^{ij}(D_{ij}g-D_{pl}A_{ij}D_{l}g)-D_{p_{l}}BD_{l}g\\ =&-\nu_{k}D_{kt}u+D_{z}\varphi D_{t}u+F^{ij}[\nu_{k}D_{ijk}u+2D_{i}\nu_{k}D_{jk}u+D_{ij}\nu_{k}D_{k}u\\ &-D_{ij}\varphi-2D_{iz}\varphi D_{j}u-D_{zz}uD_{i}uD_{j}u-D_{z}\varphi D_{ij}u-D_{p_{l}}A_{ij}D_{l}\nu_{k}D_{k}u\\ &-D_{p_{l}}A_{ij}\nu_{k}D_{kl}u+D_{p_{l}}A_{ij}D_{l}\varphi+D_{p_{l}}A_{ij}D_{z}\varphi D_{l}u]-D_{p_{l}}BD_{l}\nu_{k}D_{k}u\\ &-D_{p_{l}}B\nu_{k}D_{kl}u+D_{p_{l}}BD_{l}\varphi+D_{p_{l}}BD_{z}\varphi D_{l}u. \end{split} \end{equation} $

(3.26)

接下来, 将算子$\mathcal{L}_{t}$作用到$u$上可得

$ \begin{equation}\label{3.31} \begin{split} \mathcal{L}_{t}u &=-\frac{\partial}{\partial t}u+F^{ij}(D_{ij}u-D_{pl}A_{ij}D_{l}u)-D_{p_{l}}BD_{l}u\\ &=-\frac{\partial}{\partial t}u+F^{ij}(w_{ij}+A_{ij}-D_{pl}A_{ij}D_{l}u)-D_{p_{l}}BD_{l}u\\ &=-\frac{\partial}{\partial t}u+{\det}^{\frac{1}{n}}w_{ij}+F^{ij}A_{ij}-F^{ij}D_{pl}A_{ij}D_{l}u-D_{p_{l}}BD_{l}u\\ &=-\frac{\partial}{\partial t}u+\frac{\partial}{\partial t}u+B+F^{ij}A_{ij}-F^{ij}D_{pl}A_{ij}D_{l}u-D_{p_{l}}BD_{l}u\\ &=B+F^{ij}A_{ij}-F^{ij}D_{pl}A_{ij}D_{l}u-D_{p_{l}}BD_{l}u. \end{split} \end{equation} $

(3.27)

将$(3.16)$, $(3.27))$式代入$(3.26)$式可得

$ \begin{equation}\label{3.32} \begin{split} \mathcal{L}_{t}g =&\nu_{k}\mathcal{L}_{t}u_{k}-D_{z}\varphi\mathcal{L}_{t}u+F^{ij}(2D_{i}\nu_{k}D_{jk}u+D_{ij}\nu_{k}D_{k}u-D_{ij}\varphi-2\varphi_{zi}D_{j}u\\ &-D_{zz}\varphi D_{i}uD_{j}u-D_{p_{l}}A_{ij}D_{l}\nu_{k}D_{k}u+D_{p_{l}}A_{ij}D_{l}\varphi)-D_{p_{l}}BD_{l}\nu_{k}D_{k}u\\ &+D_{p_{l}}BD_{l}\varphi. \end{split} \end{equation} $

(3.28)

由$(3.16)$, $(3.27))$和$(3.28)$式可得$ |\mathcal{L}_{t}g|\leq C(1+\mathcal{J}+|D^{2}u|), $其中$C$是依赖于$\Omega, A, B, \varphi, |u|_{1, \Omega}$的常数.又因为$1\leq Cw^{ii}, (w_{ii})^{\frac{1}{n-1}}\leq Cw^{ii}$, 所以

$ \begin{equation}\label{3.34} |\mathcal{L}_{t}g|\leq C(1+|D^{2}u|^{\frac{n-2}{n-1}})\mathcal{J}. \end{equation} $

(3.29)

又由于$\phi$是$A$ -凸区域的定义函数, 由$(1.9)$式可得

$ \mathcal{L}_{t}\phi\geq \delta_{1}\mathcal{J}. $

(3.30)

结合$(3.29)$, $(3.30)$式, 并选取$-\phi$为闸函数, 由Hopf引理的证明可以得到

$ D_{\nu}g\leq C(1+M_{2}^{\frac{n-2}{n-1}}), \mbox{在 $\partial\Omega$ 上}, $

(3.31)

其中$M_{2}=\sup\limits_{\bar{\Omega}}|D^{2}u|$.计算$D_{\nu\nu}u$可得

$ \begin{equation}\label{12} D_{\nu\nu}u=D_{\nu}g+\nu_{i}D_{z}\varphi D_{i}u+\nu_{i}D_{i}\varphi-\nu_{i}D_{i}\nu_{k}D_{k}u. \end{equation} $

(3.32)

结合$(3.31)$, $(3.32)$式可以得到

$ D_{\nu\nu}u\leq C(1+M_{2})^{\frac{n-2}{n-1}}, \mbox{在 $\partial\Omega$ 上}. $

(3.33)

则由$(3.33)$式可得

$ v(x_{0}, t_{0}, \xi)= v(x_{0}, t_{0}, \nu)\leq C(1+M_{2})^{\frac{n-2}{n-1}}, \mbox{在 $\partial\Omega$ 上.} $

(3.34)

(ⅱ) $\xi$是点$(x_{0}, t_{0})$处的非切非法向量.单位向量$\xi$可以被写成$\xi=(\xi\cdot\tau)\tau+(\xi\cdot\nu)\nu$, 且$\tau\cdot\nu=0, (\xi\cdot\tau)^{2}+(\xi\cdot\nu)^{2}=1$.由$v'$的定义可得

$ \begin{equation}\label{a} \begin{split} w_{\xi\xi} &=(\xi\cdot\tau)^{2}w_{\tau\tau}+(\xi\cdot\nu)^{2}w_{\nu\nu}+2(\xi\cdot\tau)(\xi\cdot\nu)w_{\tau\nu}\\ &=(\xi\cdot\tau)^{2}w_{\tau\tau}+(\xi\cdot\nu)^{2}w_{\nu\nu}+v'(x, \xi), \end{split} \end{equation} $

(3.35)

由$v$的定义得

$ \begin{equation}\label{c} \begin{split} v(x_{0}, t_{0}, \xi) &=e^{\alpha|Du|^{2}+\beta \Phi}(w_{\xi\xi}-v'(x, \xi))\\ &=e^{\alpha|Du|^{2}+\beta \Phi}((\xi\cdot\tau)^{2}w_{\tau\tau}+(\xi\cdot\nu)^{2}w_{\nu\nu}+v'(x, \xi)-v'(x, \xi))\\ &=(\xi\cdot\tau)^{2}v(x_{0}, t_{0}, \tau)+(\xi\cdot \nu)^{2}v(x_{0}, t_{0}, \nu)\\ &\leq (\xi\cdot\tau)^{2}v(x_{0}, t_{0}, \xi)+(\xi\cdot v)^{2}v(x_{0}, t_{0}, \nu), \end{split} \end{equation} $

(3.36)

其中第二个等式是将$(3.35)$式代入所得, 第四个不等式是由$v$在点$(x_{0}, t_{0})$和向量$\xi$处取得最大值所得.再结合$(3.34)$, $(3.36)$式得

$ v(x_{0}, t_{0}, \xi)\leq v(x_{0}, t_{0}, \nu)\leq C(1+M_{2})^{\frac{n-2}{n-1}}, \mbox{在 $\partial\Omega$ 上.} $

(ⅲ) $\xi$是点$(x_{0}, t_{0})$处的切向量, 则$(\xi\cdot\nu)=0$, 由$v'$的定义知$v'(x_{0}, t_{0}, \xi)=0$.假设在点$(x_{0}, t_{0})$的法向量为$\nu=(0, \cdots, 0, 1)$, ${w_{ij}(x_{0}, t_{0})}$是对角阵, 且有最大特征值$w_{11}(x_{0}, t_{0})>1$, 否则结论已证.计算$D_{\nu}\Phi$可得

$ \begin{equation}\label{1} \begin{split} D_{\nu}\Phi &=D_{\nu}(e^{N(\tilde{u}-u)}) =Ne^{N(\bar{u}-u-a\phi)}D_{\nu}(\bar{u}-u-a\phi)\\ &=Ne^{N(\bar{u}-u-a\phi)}[\varphi(x, \bar{u})-\varphi(x, u)+a] \geq aN, \end{split} \end{equation} $

(3.37)

其中第三个等式运用了$D_{\nu}\phi=-1$, 第四个不等式运用$(1.7)$式.在计算$D_{\nu}v$之前不妨设$D_{k}u> 0$, 否则取$|D_{k}u+C|^2$, 其中$C=\max\limits_{\bar{\Omega}}|Du|$.接着计算$D_{\nu}v$可得

$ \begin{equation}\label{aa} \begin{split} 0\geq& D_{\nu}v =D_{\nu}[e^{\alpha |Du|^{2}+\beta \Phi}(w_{\xi\xi}-v')]\\ \geq &e^{\alpha |Du|^{2}+\beta \Phi}\{[\beta D_{\nu}\Phi+2\alpha D_{k}u(D_{k}\varphi+D_{z}\varphi D_{k}u-D_{i}uD_{k}\nu_{i})]w_{\xi\xi}\\ &+D_{\nu}u_{\xi\xi}-D_{\nu}(A_{\xi\xi}+v')\}\\ \geq &e^{\alpha |Du|^{2}+\beta \Phi}\{(\beta c_{0}-2\alpha M)w_{\xi\xi}+D_{\nu}u_{\xi\xi}-D_{\nu}(A_{\xi\xi}+v')\}, \end{split} \end{equation} $

(3.38)

其中第二个不等式利用了$v'(x_{0}, t_{0}, \xi)=0$, 由(3.37)式可得$c_{0}=aN$,

$ M=\max\limits_{x\in\partial\Omega}|D_{k}u(D_{k}\varphi+D_{z}\varphi D_{k}u-D_{i}uD_{k}\nu_{i})|. $

由(3.38)式可得

$\begin{equation}\label{d} D_{\nu}u_{\xi\xi}\leq -(\beta c_{0}-2\alpha M)w_{\xi\xi}+D_{\nu}(A_{\xi\xi}+v'). \end{equation} $

(3.39)

另外, 对$(1.2)$式沿切向求二阶导得

$ (D_{k}u)\delta_{i}\delta_{j}\nu_{k}+(\delta_{i}D_{k}u)\delta_{j}\nu_{k}+(\delta_{j}D_{k}u)\delta_{i}\nu_{k}+\nu_{k}\delta_{i}\delta_{j}D_{k}u=\delta_{i}\delta_{j}\varphi, \mbox{在 $\partial\Omega$ 上}. $

在点$(x_{0}, t_{0})$处对切向量$\xi$, 有

$ \begin{equation}\label{e} \begin{split} D_{\nu}u_{\xi\xi} &\geq D_{z}\varphi D_{ij}u\xi_{i}\xi_{j}-2(\delta_{i}\nu_{k})D_{jk}u\xi_{i}\xi_{j}+(\delta_{i}\nu_{j})\xi_{i}\xi_{j}D_{\nu\nu}u-C\\ &\geq D_{z}\varphi D_{ij}u\xi_{i}\xi_{j}-2(\delta_{i}\nu_{k})D_{jk}u\xi_{i}\xi_{j}-C(1+M_{2})^{\frac{n-2}{n-1}}\\ &\geq D_{z}\varphi w_{\xi\xi}-2(\delta_{i}\nu_{k})D_{jk}u\xi_{i}\xi_{j}-C(1+M_{2})^{\frac{n-2}{n-1}}, \end{split} \end{equation} $

(3.40)

第二个不等式由$(3.33)$式得到.结合(3.39), (3.40)式可以得到

$ (\beta c_{0}-2\alpha M+D_{z}\varphi)w_{\xi\xi}\leq 2(\delta_{i}\nu_{k})D_{jk}u\xi_{i}\xi_{j}+D_{\nu}(A_{\xi\xi}+v')+C(1+M_{2})^{\frac{n-2}{n-1}}. $

(3.41)

观察(3.41)式右边的第一项的二阶导数项, 由$(3.25)$式可以得到在点$(x_{0}, t_{0})$的估计

$ \begin{aligned} (\beta c_{0}-2\alpha M+D_{z}\varphi )w_{\xi\xi} &\leq C(w_{\xi\xi}+|DD_{\nu}u|)+C(1+M_{2})^{\frac{n-2}{n-1}}\\ &\leq Cw_{\xi\xi}+C(1+M_{2})^{\frac{n-2}{n-1}}, \end{aligned} $

取$\beta$满足$\begin{aligned} \beta\geq \frac{2}{c_{0}}[2\alpha M-\inf D_{z}\varphi+C], \end{aligned}$从而可以得到$w_{\xi\xi}\leq C(1+M_{2})^{\frac{n-2}{n-1}}, $故可以得到$v(x_{0}, t_{0}, \xi)\leq C(1+M_{2})^{\frac{n-2}{n-1}}, $其中$C$是依赖于$n, A, a, N, \varphi, |u|_{1, \Omega}$的常数.

从上面(ⅰ), (ⅱ), (ⅲ)三种情况得, 若$v$在边界点$(x_{0}, t_{0})$处取得$\bar{\Omega}\times[0, T]$上的最大值, 则$v(x_{0}, t_{0}, \xi)$在$\bar{\Omega}\times[0, T]$上是有界的, 从而$D_{\xi\xi}u(x_{0}, \xi)$在$\bar{\Omega}\times[0, T]$上是有界的.

综合以上两种情形, 运用柯西不等式可以得到$\sup\limits_{\bar{\Omega}\times[0, T]}|D^{2}u|\leq C, $其中$C$是依赖于$n, A, B, \varphi, u_{0}, \bar{u}, |u|_{1, \Omega}$的常数.