设$\Omega$为${\Bbb R}^n(n\geq 2)$中的开子集, 对$x_0 \in \Omega$和$t>0$, 记$B_t=B_t(x_0)$是以$x_0$为心, $t$为半径的球.对于可测函数$u(x)$和$k>0$, 设

$ A_k=\{x\in\Omega:|u(x)|>k\}, \ A_{k, t}=A_k\cap B_t, $

$ A^k=\{x\in\Omega: u(x)>k\}, \ A^{k, t}=A^k\cap B_t. $

此外, 如果$m<n$, 则$m^*$表示满足$\frac{1}{m^*}=\frac{1}{m}-\frac{1}{n}$的实数.设$T_k(u)$是在$k>0$水平上$u$的截断函数, 即

$ T_k(u)=\min \left\{1, \frac k {|u|} \right\}u. $

考虑二阶椭圆方程(也称作$\cal A$-调和方程或者Leray-Lions方程)

$ \begin{equation}\label{eq:a1} \mbox {div} {\cal A} \left(x, \nabla u(x) \right)=0, \end{equation} $

(1.1)

其中${\cal A}(x, \xi):\Omega\times {\Bbb R}^n\rightarrow {\Bbb R}^n$是Carathéodory函数, 且满足下面的强制性条件和控制增长条件:对于几乎所有的$x\in \Omega$和所有的$\xi \in {\Bbb R}^n$, 有

$ \begin{equation} \left\langle {\cal A}(x, \xi), \xi \right \rangle \geq \alpha \left( \mu^2+|\xi|^2 \right)^{\frac{p-2}{2}}|\xi|^2, \end{equation} $

(1.2)

$ \begin{equation} |{\cal A} (x, \xi)|\leq\beta \left(\mu^2+|\xi|^2 \right)^{\frac{p-1}{2}}, \end{equation} $

(1.3)

这里$0<\alpha\leq\beta <\infty$, $\mu \ge 0$.

满足(1.2) 式和(1.3) 式的算子${\cal A}(x, \xi)$的原型由下式给出

$ {\cal A}(x, \xi)=\left(\mu^2+|\xi|^2 \right)^{\frac{p-2}{2}}\xi, $

当$\mu=0$时即为$p$-Lapacean情况.

设$\psi$为$\Omega$上取值于${\Bbb R}\cup\{\pm\infty\}$的任意函数, $\theta\in W^{1, p}(\Omega)$.令

$ {\cal K}_{\psi, \theta} (\Omega)= \left\{v\in W^{1, p}(\Omega): v\geq\psi, ~ \mbox {a.e.}~ \Omega, ~ v-\theta\in W_0^{1, p}(\Omega) \right\}. $

函数$\psi$是障碍, $\theta$为边值.

定义1.1  称$u\in {\cal K}_{\psi, \theta} (\Omega)$为方程(1.1) 的${\cal K}_{\psi, \theta}$-障碍问题的解, 若对所有$v\in {\cal K}_{\psi, \theta}(\Omega)$, 有

$ \begin{equation} \int_{\Omega}\langle {\cal A}(x, \nabla u), \nabla(v-u)\rangle \geq 0. \end{equation} $

(1.4)

Meyers和Elcrat于1957年首先考虑了当$\mu=0$时的$\cal A$-调和方程(1.1) 解的高阶可积性, 参见文献[1].对于障碍问题, 1994年李工宝和Martio首先考虑了导数的局部和整体高阶可积性, 参见文献[2].局部正则性和局部有界性在障碍问题的理论研究中很重要.各向异性变分问题的整体有界性参见文献[3].其他相关结果参见文献[4-11].

本文考虑$\cal A$-调和方程(1.1) 障碍问题解的局部正则性和局部有界性.主要结果如下.

定理1.1  设$\cal A$满足(1.2), (1.3) 式, $\psi\in W_{loc}^{1, s}(\Omega)$, $1<p<s<n$.则${\cal K}_{\psi, \theta}$-障碍问题的解$u$属于$L_{loc}^{s^*}(\Omega)$.

回忆当$\psi \in W_{loc}^{1, s}(\Omega)$, $s>p$时, ${\cal K}_{\psi, \theta}$-障碍问题的解属于$W_{loc}^{1, q}(\Omega)$, $q>p$, 参见文献[2].显然这个可积指数$q\le s$.由Sobolev嵌入定理可知$u\in L_{loc}^{q^*} (\Omega)$, 这里$q^* =\frac {nq}{n-q}$.因为$q^*\le s^*$, 所以定理1.1的结论要好于$u\in L_{loc}^{q^*} (\Omega)$.

本文只考虑$1<p<n$的情况, 因为当$p\geq n$时, 由Sobolev嵌入定理知, 对任意$t>1$, $W_{loc}^{1, p}(\Omega)$中的函数属于$L_{loc}^{t}(\Omega)$.

定理1.2  设$\cal A$满足(1.2), (1.3) 式, $\psi\in W_{loc}^{1, \infty}(\Omega)$.则${\cal K}_{\psi, \theta}$-障碍问题的解$u$在$\Omega$局部上有界.

为证定理1.1和1.2, 我们需要几个预备引理.下面的引理来自文献[4].

引理2.1  设$u\in W_{loc}^{1, p}(\Omega)$, $\phi_0\in L_{loc}^r(\Omega)$, 这里$1<p<n$, $r$满足

$ 1<r<\frac{n}{p}. $

若对于每个$k\in N$和$R_0\leq\rho<R\leq R_1$, 积分估计

$ \int_{{{A}_{k, \rho }}}{|}\nabla u{{|}^{p}}\text{d}x\le {{C}_{0}}[\int_{{{A}_{k, R}}}{{{\phi }_{0}}}\text{d}x+{{(R-\rho )}^{-\sigma }}\int_{{{A}_{k, R}}}{|}u{{|}^{p}}\text{d}x] $

(2.1)

成立, 这里$C_0$为一个只依赖于$N, p, r, R_0, R_1$和$|\Omega|$的正常数, $\sigma$为一个实的正常数, 则$u\in L_{loc}^s(\Omega)$, 这里$s=(pr)^*$.

下面的引理来自文献[5].

引理2.2  设$f(\tau)$是定义在$0\leq R_0\leq t \leq R_1$上的非负有界函数, 若对$R_0\leq\tau<t\leq R_1$有

$ \begin{equation} f(\tau)\leq A(t-\tau)^{-\alpha}+B+\theta f(t), \end{equation} $

(2.2)

这里$A, B, \alpha, \theta$为非负常数且$\theta<1$, 则存在只依赖于$\alpha$和$\theta$的常数$C$, 使得对于每个$\rho, R, R_0\leq \rho < R\leq R_1$, 有

$ \begin{equation} f(\rho)\leq C\left[A(R-\rho)^{-\alpha}+B\right]. \end{equation} $

(2.3)

Fusco和Sbordone在文献[6]中证明了以下引理.

引理2.3  设$p<n$.若$u\in W_{loc}^{1, p}(\Omega)$, 且对于任意球$B_R\subset\subset\Omega$, $0<\rho<R$, $k>0$, 有

$ \begin{equation} \int_{A^{k, \rho}}|\nabla u|^p{\rm d}x\leq C\bigg[\int_{A^{k, R}}\left|\frac{u-k}{R-\rho}\right|^p {\rm d}x+\left| A^{k, R} \right|\bigg], \end{equation} $

(2.4)

则$u(x)$在$\Omega$局部上有界.

下面证明中的$c$表示只依赖于$n, p, \mu, \alpha, \beta$的常数, 在不同的场合取值可能不同.

定理1.1的证明  设$u$是${\cal K}_{\psi, \theta}$-障碍问题的解.由引理2.1, 只需证明$u$满足$\sigma=p$时的积分估计(2.1) 式.

设$B_{R_1}\subset\subset\Omega$, $0\leq R_0\leq\tau<t\leq R_1$任意固定.取截断函数$\phi\in C_0^\infty(B_{R_1})$使得

$ \mbox {supp} \phi\subset B_t, \ 0\leq\phi\leq 1, \ \phi=1 \mbox { in } B_\tau, |\nabla\phi|\leq 2(t-\tau)^{-1}. $

考虑函数

$ v=u-\phi^p\left(u-\max\{T_k(u), \psi\} \right), $

这里$T_k(u)$是$k>0$水平上的截断函数.现证$v\in {\cal K}_{\psi, \theta}(\Omega)$.事实上, 因为$u\in {\cal K}_{\psi, \theta}(\Omega)$, $\phi\in C_0^\infty(\Omega)$, 所以

$ v-\theta=u-\theta-\phi^p \left( u-\max\{T_k(u), \psi\} \right)\in W_0^{1, p}(\Omega). $

并且

$ \begin{eqnarray*} v-\psi &=&\displaystyle u-\phi^p(u-\max\{T_k(u), \psi\})-\psi\\ &\ge &\displaystyle u-\phi^p(u-\psi)-\psi\\ &=&\displaystyle (1-\phi^p)(u-\psi)\\ &\geq &\displaystyle 0 \end{eqnarray*} $

在$\Omega$上几乎处处成立.因为

$ \nabla v=\nabla u-{{\phi }^{p}}(\nabla u-\nabla \max \{{{T}_{k}}(u), \psi \})-p{{\phi }^{p-1}}\nabla \phi \left( u-\max \{{{T}_{k}}(u), \psi \} \right), $

(3.1)

且$u$是${\cal K}_{\psi, \theta}$-障碍问题的解, 由定义1.1得到

$ \int_\Omega \left\langle {\cal A}(x, \nabla u), \nabla v-\nabla u \right\rangle {\rm d}x\geq 0, $

注意到当$x\notin A_k$时$v=u$, 所以由(3.1) 式得到

$ \int_{A_{k, t}} \left\langle {\cal A}(x, \nabla u), \phi^p(\nabla u-\nabla \max\{T_k(u), \psi\})+p\phi^{p-1}\nabla\phi \left( u-\max\{T_k(u), \psi\}\right) \right \rangle {\rm d}x \leq 0. $

由此推出

$ \begin{eqnarray} \int_{A_{k, t}} \left\langle {\cal A}(x, \nabla u), \phi^p\nabla u \right\rangle {\rm d}x &\leq &\displaystyle\int_{A_{k, t}}\left\langle {\cal A}(x, \nabla u), \phi^p\nabla \max\{T_k(u), \psi\}\right\rangle {\rm d}x \\ &&\displaystyle -\int_{A_{k, t}}\left\langle {\cal A}(x, \nabla u), p\phi^{p-1}\nabla\phi\left(u-\max\{T_k(u), \psi\} \right)\right \rangle {\rm d}x\\ &:= &\displaystyle I_1+I_2. \end{eqnarray} $

(3.2)

由(1.2) 式, (3.2) 式左边估计为

$ \begin{eqnarray} \int_{A_{k, t}} \left\langle {\cal A}(x, \nabla u), \phi^p\nabla u \right\rangle {\rm d}x &\geq&\displaystyle \alpha \int_{A_{k, t}} \phi ^p \left( \mu^2+|\nabla u|^2 \right)^{\frac{p-2}{2}}|\nabla u|^2 {\rm d}x\\ &\geq&\displaystyle \alpha \int_{A_{k, \tau}}\left( \mu^2+|\nabla u|^2 \right)^{\frac{p-2}{2}}|\nabla u|^2 {\rm d}x. \end{eqnarray} $

(3.3)

分两种情形.

情形1  $p\ge 2$.

$ \int_{A_{k, \tau}}|\nabla u|^p {\rm d}x \le \int_{A_{k, \tau}}(\mu^2+|\nabla u|^2 )^{\frac{p-2}{2}}|\nabla u|^2 {\rm d}x. $

情形2  $1< p< 2$.由Young不等式

$ ab\le \frac 1 p a^p+\frac 1 q b^q, \ \ a, b \ge 0, \ \frac 1 p +\frac 1 q =1 $

得到

$ \begin{eqnarray*} \int_{A_{k, \tau}}|\nabla u|^p {\rm d}x &=&\displaystyle\int_{A_{k, \tau}}\left[\left( \mu^2+|\nabla u|^2 \right )^{\frac{p(2-p)}{4}}\left(\mu^2+|\nabla u|^2 \right)^{\frac{p(p-2)}{4}}|\nabla u|^p \right]{\rm d}x\\ &\leq&\displaystyle \frac {2-p}{2} \int_{A_{k, \tau}} \left( \mu^2+|\nabla u|^2 \right )^{\frac{p}{2}}{\rm d}x +\frac p 2 \int_{A_{k, \tau}}\left( \mu^2+|\nabla u|^2 \right )^{\frac{p-2}{2}}|\nabla u|^2 {\rm d}x\\ &\leq&\displaystyle \frac {2-p}{2} \int_{A_{k, \tau}} \left( \mu^p+|\nabla u|^p \right ){\rm d}x +\frac p 2 \int_{A_{k, \tau}} \left( \mu^2+|\nabla u|^2 \right )^{\frac{p-2}{2}}|\nabla u|^2 {\rm d}x. \end{eqnarray*} $

由此推出

$ \begin{eqnarray} \int_{A_{k, \tau}}|\nabla u|^p {\rm d}x &\le &\displaystyle \frac {2-p}{p} \int_{A_{k, \tau}}\mu^p{\rm d}x +\int_{A_{k, \tau}} \left( \mu^2+|\nabla u|^2 \right )^{\frac{p-2}{2}}|\nabla u|^2 {\rm d}x\\ &<&\displaystyle 2\int_{A_{k, \tau}}\mu^p{\rm d}x +\int_{A_{k, \tau}} \left( \mu^2+|\nabla u|^2 \right )^{\frac{p-2}{2}}|\nabla u|^2 {\rm d}x. \end{eqnarray} $

(3.4)

在这两种情形下, (3.4) 式都成立. (3.4) 式联合(3.3) 式和(3.2) 式得到

$ \begin{equation} \int_{A_{k, \tau}}|\nabla u|^p {\rm d}x \leq 2 \int_{A_{k, t}} \mu^p{\rm d}x +\frac {1}{\alpha} (I_1+I_2). \end{equation} $

(3.5)

由

$ |\nabla \max\{T_k(u), \psi\}|\leq |\nabla \psi|, \ \ \mbox { a.e. in }\ A_{k, t} $

和Young不等式

$ \begin{equation} ab \le \varepsilon a ^p+c(\varepsilon) b^q, \ a, b \ge 0, \ \varepsilon >0, \ \frac 1 p +\frac 1 q =1, \end{equation} $

(3.6)

$|I_1|$可以由(1.3) 式估计为

$ \begin{eqnarray} |I_1| &= &\displaystyle \left|\int_{A_{k, t}}\left \langle {\cal A}(x, \nabla u), \phi^p\nabla \max\{T_k(u), \psi\}\right\rangle {\rm d}x\right|\\ &\leq &\displaystyle \beta\int_{A_{k, t}}\left( \mu^2+|\nabla u|^2 \right) ^{\frac{p-1}{2}}|\nabla\psi| {\rm d}x\\ &\leq&\displaystyle \beta \left[\varepsilon 2 ^{\max \left\{0, \frac p 2-1 \right\}}\int_{A_{k, t}}(\mu^p+|\nabla u|^p) {\rm d}x +c(\varepsilon)\int_{A_{k, t}}|\nabla\psi|^p{\rm d}x\right]. \end{eqnarray} $

(3.7)

由

$ |u-\max\{T_k(u), \psi\}|\leq |u|, \ \ \mbox { a.e. in } A_{k, t} $

并再次使用(1.3) 式和Young不等式(3.6) 得到

$ \begin{eqnarray} \displaystyle |I_2| &= &\displaystyle \left|-\int_{A_{k, t}}\left\langle {\cal A}(x, \nabla u), p\phi^{p-1}\nabla\phi\left( u-\max\{T_k(u), \psi\} \right)\right \rangle {\rm d}x \right|\\ &\leq& \displaystyle 2p\beta\int_{A_{k, t}} \left( \mu^2+|\nabla u|^2 \right)^{\frac {p-1}{2}}\frac{|u|}{t-\tau}{\rm d}x\\ &\leq&\displaystyle 2p\beta \left[\varepsilon\int_{A_{k, t}}(\mu^2+|\nabla u|^2)^{\frac {p}{2}}{\rm d}x +\frac{c(\varepsilon)}{(t-\tau)^p}\int_{A_{k, t}}|u|^p{\rm d}x\right]\\ &\leq&\displaystyle 2p\beta \left[\varepsilon 2 ^{\max \left\{0, \frac p 2-1 \right\}} \int_{A_{k, t}}(\mu^p+|\nabla u|^p){\rm d}x +\frac{c(\varepsilon)}{(t-\tau)^p}\int_{A_{k, t}}|u|^p{\rm d}x\right]. \end{eqnarray} $

(3.8)

结合不等式(3.5), (3.7) 和(3.8), 有

$ \begin{eqnarray*} && \int_{A_{k, \tau}}|\nabla u|^p{\rm d}x \\ &\leq &\displaystyle 2 \int_{A_{k, t}}\mu^p{\rm d}x + \frac {\beta }{\alpha}\left[\varepsilon 2 ^{\max \left\{0, \frac p 2-1 \right\}}\int_{A_{k, t}}(\mu^p+|\nabla u|^p) {\rm d}x +c(\varepsilon)\int_{A_{k, t}}|\nabla\psi|^p{\rm d}x\right] \\ && \displaystyle + \frac {2p\beta }{\alpha} \left[\varepsilon 2 ^{\max \left\{0, \frac p 2-1 \right\}} \int_{A_{k, t}}(\mu^p+|\nabla u|^p){\rm d}x +\frac{c(\varepsilon)}{(t-\tau)^p}\int_{A_{k, t}}|u|^p{\rm d}x\right] \\ &=& \displaystyle \frac {\beta}{\alpha}(1 +2p)2^{\max \left\{0, \frac p 2 -1 \right\}}\varepsilon \int_{A_{k, t}} |\nabla u |^p{\rm d}x +\left(\frac {\beta}{\alpha}(1 +2p)2^{\max \left\{0, \frac p 2 -1 \right\}}\varepsilon +2\right)\int_{A_{k, t}}\mu^p{\rm d}x\\ && \displaystyle +\frac {c(\varepsilon)\beta}{\alpha} \int_{A_{k, t}}|\nabla \psi|^p{\rm d}x +\frac {2p\beta c(\varepsilon)}{\alpha (t-\tau)^p} \int_{A_{k, t}} |u|^p{\rm d}x. \end{eqnarray*} $

取$\varepsilon <1$足够小使得

$ \theta = \frac {\beta}{\alpha}(1 +2p)2^{\max \left\{0, \frac p 2 -1 \right\}}\varepsilon <1. $

由引理2.2, 存在只依赖于$\frac \beta \alpha$和$p$的常数$c$, 使得对任意满足$R_0\le \rho <R\le R_1$的$\rho, R$, 有

$ \int_{A_{k, \rho}}|\nabla u|^p{\rm d}x \le c\left[\int_{A_{k, R}} \left(\mu^p+ |\nabla \psi|^p\right) {\rm d}x + (R-\rho)^{-p}\int_{A_{k, R}} \left|u\right|^p {\rm d}x\right] . $

定理1.1由引理2.1得到.

定理1.2的证明  设$u$为${\cal K}_{\psi, \theta}$-障碍问题的解.由引理2.3, 只需要证明$u$满足积分估计(2.4) 式.设$\phi$为定理1.1证明中的截断函数.考虑函数

$ v=u-\phi^p\left( u-\max\{\psi, \min\{k, u\}\} \right). $

如定理1.1的证明, 可证$v\in {\cal K}_{\psi, \theta}(\Omega)$.因为

$ \nabla v=\nabla u-\phi^p\left( \nabla u-\nabla \max\{\psi, \min\{k, u\}\} \right)-p\phi ^{p-1} \nabla \phi \left( u-\max\{\psi, \min\{k, u\}\} \right), $

所以由定义1.1联合当$x\notin A^{k}$时$v=u$的事实, 可得

$ \begin{eqnarray*} &&\int_{A^{k, t}} \langle {\cal A} (x, \nabla u), \phi^p\left( \nabla u-\nabla \max\{\psi, \min\{k, u\}\} \right)\\ &&+p\phi ^{p-1} \nabla \phi \left( u-\max\{\psi, \min\{k, u\}\} \right)\rangle {\rm d}x \leq 0. \end{eqnarray*} $

此即

$ \begin{eqnarray} && \displaystyle\int_{A^{k, t}}\langle {\cal A}(x, \nabla u), \phi^p\nabla u\rangle {\rm d}x\\ &\leq &\displaystyle \int_{A^{k, t}} \left\langle {\cal A}(x, \nabla u), \phi^p \nabla \max\{\psi, \min\{k, u\}\} \right \rangle {\rm d}x\\ &&\displaystyle -\int_{A^{k, t}} \left \langle {\cal A}(x, \nabla u), p\phi ^{p-1} \nabla \phi \left( u-\max\{\psi, \min\{k, u\}\} \right) \right \rangle {\rm d}x\\ &:=&\displaystyle I_3+I_4. \end{eqnarray} $

(3.9)

如定理1.1的证明, 可以得到

$ \begin{equation} \int_{A^{k, \tau}} |\nabla u|^p {\rm d}x \le 2 \int _{A^{k, t}} \mu^p{\rm d}x +\frac 1 \alpha (I_3+I_4). \end{equation} $

(3.10)

利用

$ |\nabla \max\{\psi, \min\{k, u\}\} |\leq |\nabla \psi|, ~|u-\max\{\psi, \min\{k, u\}\} |\leq |u-k|, \ \mbox { a.e. in }\ A^{k, t}, $

Young不等式(3.6) 和(1.3) 式得到

$ \begin{eqnarray} \displaystyle |I_3| &= &\displaystyle \left|\int_{A^{k, t}} \left\langle {\cal A}(x, \nabla u), \phi^p \nabla \max\{\psi, \min\{k, u\}\} \right \rangle {\rm d}x\right|\\ &\leq &\displaystyle \beta\int_{A^{k, t}}\left( \mu^2+|\nabla u|^2 \right) ^{\frac{p-1}{2}}|\nabla\psi| {\rm d}x\\ &\leq&\displaystyle \beta \left[\varepsilon 2 ^{\max \left\{0, \frac p 2-1 \right\}}\int_{A^{k, t}}(\mu^p+|\nabla u|^p) {\rm d}x +c(\varepsilon)\int_{A^{k, t}}|\nabla\psi|^p{\rm d}x\right]. \end{eqnarray} $

(3.11)

$ \begin{eqnarray} \displaystyle |I_4| &= &\displaystyle \left|-\int_{A^{k, t}} \left \langle {\cal A}(x, \nabla u), p\phi ^{p-1} \nabla \phi \left( u-\max\{\psi, \min\{k, u\}\} \right) \right \rangle {\rm d}x\right|\\ &\leq& \displaystyle 2p\beta \int_{A_{k, t}} \left( \mu^2+|\nabla u|^2 \right)^{\frac {p-1}{2}}\frac{|u-k|}{t-\tau}{\rm d}x\\ &\leq&\displaystyle 2p\beta \left[\varepsilon\int_{A_{k, t}}(\mu^2+|\nabla u|^2)^{\frac {p}{2}}{\rm d}x +c(\varepsilon)\int_{A_{k, t}}\left|\frac {u-k}{t-\tau}\right|^p{\rm d}x\right]\\ &\leq&\displaystyle 2p\beta \left[\varepsilon 2 ^{\max \left\{0, \frac p 2-1 \right\}} \int_{A_{k, t}}(\mu^p+|\nabla u|^p){\rm d}x +c(\varepsilon)\int_{A_{k, t}}\left|\frac {u-k}{t-\tau}\right|^p{\rm d}x\right]. \end{eqnarray} $

(3.12)

联合不等式(3.10), (3.11) 和(3.12) 有

$ \begin{eqnarray} &&\displaystyle \int_{A^{k, \tau}}|\nabla u|^p{\rm d}x\\ &\leq &\displaystyle 2 \int _{A^{k, t}} \mu^p{\rm d}x +\frac \beta \alpha \left[\varepsilon 2 ^{\max \left\{0, \frac p 2-1 \right\}}\int_{A^{k, t}}(\mu^p+|\nabla u|^p) {\rm d}x +c(\varepsilon)\int_{A^{k, t}}|\nabla\psi|^p{\rm d}x\right]\\ &&\displaystyle + \frac {2p\beta}\alpha \left[\varepsilon 2 ^{\max \left\{0, \frac p 2-1 \right\}} \int_{A_{k, t}}(\mu^p+|\nabla u|^p){\rm d}x +c(\varepsilon)\int_{A_{k, t}}\left|\frac {u-k}{t-\tau}\right|^p{\rm d}x\right]\\ &=& \displaystyle \frac {\beta}{\alpha}(1 +2p)2^{\max \left\{0, \frac p 2 -1 \right\}}\varepsilon \int_{A^{k, t}} |\nabla u |^p{\rm d}x % \\ +\left(\frac {\beta}{\alpha}(1 +2p)2^{\max \left\{0, \frac p 2 -1 \right\}}\varepsilon +2\right)\int_{A^{k, t}}\mu^p{\rm d}x\\ && \displaystyle +\frac {c(\varepsilon)\beta}{\alpha} \int_{A^{k, t}}|\nabla \psi|^p{\rm d}x +\frac {2p\beta c(\varepsilon)}{\alpha } \int_{A^{k, t}} \left|\frac {u-k}{t-\tau}\right|^p{\rm d}x . \end{eqnarray} $

(3.13)

取$\varepsilon <1$足够小使得

$ \theta = \frac {\beta}{\alpha}(1 +2p)2^{\max \left\{0, \frac p 2 -1 \right\}}\varepsilon <1. $

由引理2.2知, 存在只依赖于$\frac \beta \alpha$和$p$的常数$c$, 使得对任意满足$R_0\le \rho <R\le R_1$的$\rho, R$, 有

$ \begin{eqnarray*} \displaystyle \int_{A^{k, \rho}}|\nabla u|^p{\rm d}x & \le &\displaystyle c \left[\int_{A^{k, R}} \left(\mu^p+ |\nabla \psi|^p\right) {\rm d}x + \int_{A^{k, R}} \left|\frac {u-k}{R-\rho}\right|^p{\rm d}x\right] \\ &\le&\displaystyle c \left[\|\mu^p+ |\nabla \psi|^p\|_{L^\infty (\Omega)} \left|A^{k, R} \right| + \int_{A^{k, R}} \left|\frac {u-k}{R-\rho}\right|^p{\rm d}x\right]. \end{eqnarray*} $

定理1.2由引理2.3得到.