近年来,关于变指数的Lebesgue-Sobolev空间$L^{p(x)}(\Omega)$的研究日益广泛,

$$L^{p(x)}(\Omega) = \left\{ f:\Omega\rightarrow {\mathbf{R}}\big | f \ \mbox{可测,且}\ \int_\Omega |f|^{p(x)}{\rm d}x < \infty\right\},$$ 其上定义范数 $$\|f\|_{L^{p(x)}(\Omega)} = \inf \left\{\lambda>0 : \int_\Omega \left|\frac{f}{\lambda}\right|^{p(x)}{\rm d}x\leq 1\right\},$$ 此时$L^{p(x)}(\Omega)$为Banach空间. $W^{1,p(x)}(\Omega)$空间定义为 $$W^{1,p(x)}(\Omega) =\left\{ u\in L^{p(x)}(\Omega) : \nabla u\in L^{p(x)}(\Omega)\right\},$$ 其上定义范数 $$\|u\|_{W^{1,p(x)}(\Omega)} = \|u\|_{L^{p(x)}(\Omega)} + \|\nabla u\|_{L^{p(x)}(\Omega)},$$ 此时$W^{1,p(x)} (\Omega)$为Banach空间.

变指数空间与非牛顿流的研究相关. 文献^{[1, 2, 3]}在对连续介质中的电流变液模型的研究中产生能量积分 $\int_\Omega |\nabla u|^{p(x)}{\rm d}x$, 并在$W^{1,p(x)}(\Omega)$空间中对这种流体进行分析. 而且,上述能量积分也出现在均匀化问题^[4], 图像恢复^[5]等均匀的物理模型中. 这就引起了对变指数问题的研究,见文献^{[6, 7, 8, 9, 10, 11, 12]}及其参考文献.

设$\Omega$为${\bf R}^n$中的有界开集,$n\geq 2$. 本文考虑下述椭圆方程

$${\rm div} {{\cal A}}(x,\nabla u)={{\cal B}}(x,\nabla u),$$

(1.1)

其中算子${{\cal A}}={{\cal A}}(x,\xi): {\bf R}^n\times{\bf R}^n\rightarrow {\bf R}^n$ 满足: 对每个$\xi$,${{\cal A}}$对$x$可测; 对a.e. $x$,${{\cal A}}$对$\xi$连续. 而且,对给定的$p(x)$,所有$\xi,\eta \in {\bf R}^n$,$x,y \in \Omega$, ${{\cal A}}(x,\xi)$满足如下结构性条件

$$[{{\cal A}}(x,\xi) - {{\cal A}}(x,\eta)]\cdot (\xi-\eta) \geq C_1|\xi-\eta|^{p(x)},$$

(1.2)

$$|{{\cal A}}(x,\xi) - {{\cal A}}(y,\xi)| \leq C_2 w(|x-y|)|\xi|^{p(x)-1},$$

(1.3)

$$|{{\cal A}}(x,\xi) |\leq C_3|\xi|^{p(x)-1},$$

(1.4)

$${{\cal A}}(x,\xi)\cdot\xi \geq C_4|\xi|^{p(x)},$$

(1.5)

其中$C_i > 0$,$i =1,2,3,4$,为正常数. 算子${{\cal B}}={{\cal B}}(x,\xi): {\bf R}^n\times{\bf R}^n\rightarrow {\bf R}$ 满足

$$|{{\cal B}}(x,\xi) |\leq C_5|\xi|^{p(x)-1},$$

(1.6)

这里函数$p(x)$连续,并对任给$\xi,x\in {\bf R}^n$和$1 \leq i,j \leq n$,满足

$$1<\gamma_1=\inf_{\Omega} p(x) \leq \sup_{\Omega} p(x) =\gamma_2 < \infty$$

(1.7)

和

$$p(x)\in C^{0,\alpha_1}_{\rm loc}(\Omega),$$

(1.8)

其中$\alpha_1> 0$为正常数. (1.3)式中的连续模$\omega(x) : {\bf R}^+ \rightarrow {\bf R}^+$非减, 满足 $$|p(x)-p(y)|\leq \omega(|x-y|),$$ 这就要求$p(x)$满足

$$\lim_{R\rightarrow 0}\omega(R)\log(\frac{1}{R})=0.$$

注特别地,当${{\cal A}}(x,\xi)=|\xi|^{p(x)-2}\xi$ 且${{\cal B}}(x,\xi)=0$, 方程 (1.1)即为$p(x)$-Laplace方程

$${\rm div} (|\nabla u|^{p(x)-2}\nabla u)=0 .$$

本文主要考虑方程(1.1)的弱解的梯度的Hölder连续性. 关于$p(x)$-Laplace方程的Hölder估计已有一些结论^{[12, 13, 20, 21, 22, 23]}. Emilio Acerbi和Giuseppe Mingione^[12]考虑了$p(x)$-Laplace方程

$${\rm div} (|\nabla u|^{p(x)-2}\nabla u)={\rm div} ( |{\bf f}|^{p(x)-2}{\bf f}) \ \ \mathrm{in}\ \ \Omega$$

(1.9)

的一类椭圆问题Calderòn-Zygmund估计. Verena Bögelein和Anna Zatorska-Goldstein^[13]证明了方程(1.9)的很弱解的高阶可积性. 姚锋平^[23]获得了$p(x)$-Laplace方程 $${\rm div} \left(\left(A \nabla u \cdot \nabla u\right)^{\frac{p(x)-2}{2}}A \nabla u\right) ={\rm div} ( |{\bf f}|^{p(x)-2}{\bf f})$$ 的弱解的梯度的Hölder正则性,并将其推广到方程 $${\rm div} {{\cal A}}(x,\nabla u)={\rm div} ( |{\bf f}|^{p(x)-2}{\bf f}).$$ 当$p(x)$为常数时,参见文献^{[14, 15, 16, 17, 18, 19]}中的众多结论.

定义 1.1 函数$u\in W^{1,p(x)}_{\rm loc}(\Omega)$称为方程(1.1)的局部弱解, 若对任给$\varphi\in W^{1,p(x)}_{0}(\Omega)$,有

$$\int_{\Omega}{{\cal A}}(x,\nabla u)\cdot\nabla\varphi {\rm d}x = \int_{\Omega}{{\cal B}}(x,\nabla u)\varphi {\rm d}x.$$

(1.10)

下面是本文主要结论.

定理 1.2 设$u$为方程(1.1)的局部弱解,满足条件(1.2)-(1.8), 则$\nabla u$是局部Hölder连续的,即 $u\in C^{1,\alpha}_{\rm loc}(\Omega)$,其中

$$\alpha= \frac{(\beta-n\mu-\lambda)\gamma_1}{(1+\mu)\gamma_2},$$ $\beta=1-n\sigma_0$,正常数$\sigma_0< 1/n$, $\mu=\frac{\beta}{n+\beta_1}>0$.

本文用$B_R$表示半径为$R$的立方体. $x^*\in B_R$为定点. 定义

$$p_0=\min_{\overline{B_R}}p(x),\quad \ \ \ p_1=\max_{\overline{B_R}}p(x), \quad \ \ \ p_2=p(x^*).$$ 令$R_1$足够小,使得$0<R\leq R_1<R_0<1$, 且对任给$x,y \in B_{R_1}$,有

$$\begin{equation} |p(x)-p(y)|\leq C_1|x-y|^{\alpha_1} \leq C_1(2R_1)^{\alpha_1}\leq\frac{\sigma_0\gamma_1}{\sigma_0+2},% \end{equation}$$

(2.1)

其中$\sigma_0$为引理2.5中的正常数.

为估计$\nabla u$,首先引入椭圆方程的Dirichlet问题.

令$v$为下述椭圆问题

$$\begin{equation} \left\{\begin{array}{ll} {\rm div} {{\cal A}}(x^*,\nabla v)=0,~~ & {\rm in}\ B_{R},\\ v=u,&{\rm on}\ \partial B_{R} \end{array}\right. % \end{equation}$$

(2.2)

的弱解,其中算子${{\cal A}}$为方程(1.1)中的算子${{\cal A}}$. 此时 对所有$\xi,\eta \in {\bf R}^n$,$x\in B_R$,有

$$\begin{equation} [{{\cal A}}(x,\xi) -{{\cal A}}(x,\eta)]\cdot (\xi-\eta) \geq C_1|\xi-\eta|^{p_2},% \end{equation}$$

(2.3)

$$\begin{equation} |{{\cal A}}(x^*,\xi) - {{\cal A}}(x,\xi)| \leq C_2 w(|x^*-x|)|\xi|^{p_2-1}, % \end{equation}(2.4)$$

(2.4)

$$\begin{equation} |{{\cal A}}(x,\xi) |\leq C_3|\xi|^{p_2-1},% \end{equation}$$

(2.5)

$$\begin{equation} {{\cal A}}(x,\xi)\cdot\xi \geq C_4|\xi|^{p_2}.% \end{equation}$$

(2.6)

定义2.1 设$v\in W^{1,p_2}(B_R)$. 称$v\in W^{1,p_2}(B_R)$为方程(2.2)的在$B_R$中满足 $v-u\in W^{1,p_2}_0(B_R)$的弱解, 若对任给$\varphi\in W^{1,p_2}_{0}(B_R)$,

$$\int_{B_R}{{\cal A}}(x,\nabla v)\cdot\nabla\varphi {\rm d}x =0.$$

我们将使用$v$作为比较函数. 为证明定理1.2需要建立几个关于$\nabla v$的结论.

引理 2.2 (参见文献[24,引理5.1]) 令$g\in C^1$,满足$0\leq \frac{tg'(t)}{g(t)}\leq g_0 (t>0)$, 并对某些正常数$\epsilon$,有$g(t)\geq \epsilon t$. 令${\overline{{A}}} : {\mathbf{R}}^n\rightarrow {\mathbf{R}}^n$, 存在正常数$\Lambda$,使得

$$\overline{a^{ij}}(h)\xi_i\xi_j \geq \frac{g(|h|)}{|h|}|\xi|^2; \hspace{5mm} |\overline{a^{ij}}(h)|\leq \Lambda \frac{g(|h|)}{|h|}|\xi|^2; \hspace{5mm} |\overline{A}(h)|\leq \Lambda g(|h|),$$ 对所有$h,\xi \in {\mathbf{R}}^n$, $\overline{a^{ij}} = \frac{\partial \overline{A^{i}}}{\partial h^{j}}$ 均成立. 若$v\in W^{1,G}(B_R)$为方程 $${\rm div} \overline{A}(Dv)=0$$ 在$B_R$中的有界解,则对某些正数$\sigma(N,g_0,\Lambda)$, $v\in C^{1,\sigma}(B_R)$. 而且,对$0<r<R$,有 $$\sup_{B_{R/2}}G(|Dv|)\leq c(N,g_0,\Lambda) R^{-n} \int_{B_R}G(|Dv|){\rm d}x,$$ $$-\hspace{-4mm}\int_{B_r}G(|Dv - (Dv)_r|){\rm d}x\leq c(N,g_0,\Lambda)(\frac{r}{R})^\sigma -\hspace{-4mm}\int_{B_R}G(|Dv - (Dv)_R|){\rm d}x.$$

引理 2.3 令$u$为方程(1.1)在条件(1.2)-(1.8)下的弱解, $v$为问题(2.2)的弱解. 当$0<\rho<R/2$,$\beta_1\in (0,1)$,有

$$\begin{equation} -\hspace{-4mm}\int_{B_{R}}|\nabla v|^{p_2}{\rm d}x \leq C -\hspace{-4mm}\int_{B_{R}}|\nabla u|^{p_2}{\rm d}x,% \end{equation}$$

(2.7)

$$\begin{equation} \int_{B_\rho} |\nabla v|^{p_2} {\rm d}x\leq c(\frac{\rho}{R})^n \int_{B_R} |\nabla v|^{p_2} {\rm d}x. % \end{equation}$$

(2.8)

$$\begin{equation} -\hspace{-4mm}\int_{B_\rho} |\nabla v - (\nabla v)_{B_\rho}|^{p_2} {\rm d}x\leq c(\frac{\rho}{R})^{\beta_1} -\hspace{-4mm}\int_{B_R}|\nabla v - (\nabla v)_{B_R}|^{p_2} {\rm d}x. % \end{equation}$$

(1)

证 由定义2.1易得(2.7)式. 在定义2.1中取$\varphi=u-v\in W^{1,p_2}_0(B_R)$,可得

$$\begin{equation} \int_{B_R}{{\cal A}}(x^*,\nabla v)\cdot\nabla(u-v) {\rm d}x =0. % \end{equation}$$

(2.10)

此即

$$\begin{equation} \int_{B_R}{{\cal A}}(x^*,\nabla v)\cdot\nabla v {\rm d}x =\int_{B_R}{{\cal A}}(x^*,\nabla v)\cdot\nabla u {\rm d}x . % \end{equation}$$

(2.11)

由(2.6)式,可得

$$\begin{equation} \int_{B_R}{{\cal A}}(x^*,\nabla v)\cdot\nabla v {\rm d}x \geq C_4\int_{B_R}|\nabla v|^{p_2}{\rm d}x. % \end{equation}$$

(2.12)

而且,由(2.5)式和带$\tau$的Young不等式,有

$$\begin{eqnarray} \displaystyle \int_{B_R}{{\cal A}}(x^*,\nabla v)\cdot\nabla u {\rm d}x &\leq& \displaystyle C_3\int_{B_R}|\nabla v|^{p_2-1}|\nabla u|{\rm d}x \nonumber\\ &\leq& \displaystyle C_3\tau\int_{B_R}|\nabla v|^{p_2}{\rm d}x + C_3C(\tau)\int_{B_R}|\nabla u|^{p_2}{\rm d}x . \end{eqnarray}$$

(2.13)

结合估计式(2.11)-(2.13)并选取足够小的常数$\tau > 0$,即可得(2.7)式.

由文献[25,定理 3.17,3.41],可知Dirichlet问题(2.2)的解$v(x)$是局部有界的. 在引理2.2中取$g(t) = pt^{p-1}$. 于是由引理2.2,有

$$\begin{equation} G(t)=\int_0^tg(s){\rm d}s=\int_0^t ps^{p-1}{\rm d}s=t^p. % \end{equation}$$

(2.14)

由文献^[24]中$W^{1,G}$的定义知,$v(x)\in W^{1,G}(B_R)$. 可知Dirichlet问题(2.2)的解$v(x)$满足引理2.2的条件. 于是由引理2.2和$G(t)$的定义,有

$$\begin{equation} \sup_{B_{R/2}}|\nabla v|^p {\rm d}x \leq c R^{-n} \int_{B_R}|\nabla v|^p {\rm d}x. % \end{equation}$$

(2.15)

因$0<\rho<R/2$,于是

$$\begin{equation} \int_{B_\rho}|\nabla v|^p {\rm d}x \leq c|B_\rho|R^{-n} \int_{B_{R/2}}|\nabla v|^p {\rm d}x \leq c(\frac{\rho}{R})^n \int_{B_R} |\nabla v|^p {\rm d}x.% \end{equation}$$

(2.16)

于是(2.8)式成立. 由引理2.2知, $$-\hspace{-4mm}\int_{B_\rho}G(|Dv - (Dv)_\rho|){\rm d}x\leq c(N,g_0,\Lambda)(\frac{\rho}{R})^{\beta_1} -\hspace{-4mm}\int_{B_R}G(|Dv - (Dv)_R|){\rm d}x.$$ 结合$G(t)$的定义知(2.9)式成立.

引理 2.4 (参见文献[12,定理4]) 令立方体$Q_{4_{R_0}}\subset {{\bf R}}^n$,$s,q>1$. 若两函数$f\in L^s(Q_{4_{R_0}}; {{\bf R}}^n)$,$\phi\in L^{sq}(Q_{4_{R_0}}; {{\bf R}}^n)$满足

$$\left(-\hspace{-4mm}\int_{Q_{R/2}} |f|^s{\rm d}x\right)^{1/s} \leq K-\hspace{-4mm}\int_{Q_{R}}|f|{\rm d}x + H\left(-\hspace{-4mm}\int_{Q_{R/2}} |\phi|^s{\rm d}x\right)^{1/s}$$ 对每个立方体$Q_R\subseteq Q_{4_{R_0}}$均成立,$K,H>1$,则有 $$\begin{eqnarray*} \displaystyle \left(-\hspace{-4mm}\int_{Q_{R/2}} |f|^{s(1+\sigma)}{\rm d}x \right)^{1/(1+\sigma)} &\leq&\displaystyle \frac{c(n,s)}{(s-1)^{\frac{1}{1+\sigma}}}-\hspace{-4mm}\int_{Q_{R}} |f|^{s}{\rm d}x \\ &&\displaystyle +\frac{c(n,s)}{(s-1)^{\frac{1}{1+\sigma}}}\left(\frac{H}{K}\right)^s \left(-\hspace{-4mm}\int_{Q_{R}} |\phi|^{s(1+\sigma)}{\rm d}x \right)^{1/(1+\sigma)} \end{eqnarray*}$$ 对每个立方体$Q_R\subseteq Q_{4_{R_0}}$均成立,其中常数$c(n,s)>0$,$\sigma>0$满足 $$\sigma\leq \min\left\{\frac{c(n,s)(s-1)}{K^{sq}},q-1\right\}.$$

引理 2.5 设$u$为方程(1.1)满足条件(1.2)-(1.8)的弱解, 则存在正常数$\sigma_0,R_0 < 1$,与$n,\gamma_1,\gamma_2,C_i$ $(i=1,2,\cdots,7)$有关的常数$C$,使得

$$\begin{equation} -\hspace{-4mm}\int_{B_{R/2}} |\nabla u|^{p(x)(1+\sigma)}{\rm d}x \leq C\left( -\hspace{-4mm}\int_{B_{R}} |\nabla u|^{p(x)}{\rm d}x\right)^{1+\sigma} +1 % \end{equation}$$

(2.17)

对$R\leq R_0$和$\sigma\leq\sigma_0<\frac{1}{n}$成立.

证 首先取$R_0$足够小,使得

$$\left\{ \begin{array}{ll} \displaystyle \omega(8nR_0)\leq \sqrt{\frac{n+1}{n}}-1, \\[3mm] \displaystyle 0<\omega(R)\log\left(\frac{1}{R}\right)\leq C_6,\quad \forall R \leq R_0. \end{array}\right.$$ 令$s=\sqrt{\frac{n+1}{n}}>1$. 且$Q_R \subseteq Q_{4R_0}$. 于是有 $$p_2-p_1\leq \omega(2R\sqrt{n})\leq \omega(2nR)=\omega(8nR_0)\leq s-1.$$ 于是由$\frac{p_2-p_1}{p_1} \leq \frac{s-1}{p_1} \leq s-1,$ 可知$\frac{p_2}{p_1} \leq s,$ 且$p_2>p_1>\frac{p_1}{s}$. 为方便记,常数$C$在下文中不一定相等, 仅表示出与$C$有关的参数.

令截断函数$\eta\in C_0^\infty(B_R)$,$0\leq \eta \leq 1$,$|\nabla \eta|\leq C_7/R$, 在$B_{R/2}$上$\eta\equiv1$. 在定义1.1中取$\varphi=\eta^{p_2}(u-u_R)$,则

$$\begin{equation} \nabla\varphi= p_2\eta^{p_2-1}(u-u_R)\nabla\eta + \eta^{p_2}\nabla u,% \end{equation}$$

(2.18)

于是由定义1.1,有

$$\begin{eqnarray} &&\displaystyle \int_{\Omega}p_2\eta^{p_2-1}(u-u_R){{\cal A}}(x,\nabla u)\cdot\nabla\eta {\rm d}x + \int_{\Omega}\eta^{p_2}{{\cal A}}(x,\nabla u)\cdot\nabla u {\rm d}x \nonumber\\ &=&\displaystyle \int_{\Omega}\eta^{p_2}{{\cal B}}(x,\nabla u)(u-u_R) {\rm d}x. \end{eqnarray}$$

(2.19)

从而

$$\begin{eqnarray} &&\displaystyle \int_{B_R}\eta^{p_2}{{\cal A}}(x,\nabla u)\cdot\nabla u {\rm d}x \nonumber\\ &\leq&\displaystyle p_2\int_{B_R}\eta^{p_2-1}|u-u_R||{{\cal A}}(x,\nabla u)||\nabla\eta| {\rm d}x + \int_{B_R}\eta^{p_2}|{{\cal B}}(x,\nabla u)||u-u_R| {\rm d}x. \end{eqnarray}$$

(2.20)

下面分别估计(2.20)式两侧的积分. 由条件(1.5),

$$\begin{eqnarray} \int_{B_R}\eta^{p_2}{{\cal A}}(x,\nabla u)\cdot\nabla u {\rm d}x \geq C_4\int_{B_R}\eta^{p_2}|\nabla u|^{p(x)} {\rm d}x. \end{eqnarray}$$

(2.21)

由条件(1.4)和Young不等式$( \frac{1}{p_2} + \frac{p_2-1}{p_2}=1),$ 并注意到 $|\nabla \eta|\leq C_7/R$,有

$$\begin{eqnarray} &&\displaystyle p_2\int_{B_R}\eta^{p_2-1}|u-u_R||{{\cal A}}(x,\nabla u)||\nabla\eta| {\rm d}x \nonumber\\ &\leq&\displaystyle C_3C_7 p_2\int_{B_R}\eta^{p_2-1}|\nabla u|^{p(x)-1}\frac{|u-u_R|}{R} {\rm d}x \nonumber\\ &\leq&\displaystyle C_3C_7 p_2\tau\int_{B_R}\eta^{p_2}|\nabla u|^{\frac{p_2}{p_2-1}[(p(x)-1)]}{\rm d}x +C_3C_7 C(\tau) p_2\int_{B_R}\eta^{p_2}\left|\frac{u-u_R}{R}\right|^{p_2}{\rm d}x. \end{eqnarray}$$

(2.22)

因为$p(x)\geq \frac{p_2}{p_2-1}[p(x)-1]$,于是由上式可得

$$\begin{eqnarray} &&\displaystyle p_2\int_{B_R}\eta^{p_2-1}|u-u_R||{{\cal A}}(x,\nabla u)||\nabla\eta| {\rm d}x \nonumber\\ &\leq&\displaystyle C_3C_7 p_2\tau\int_{B_R}\eta^{p_2}(|\nabla u|^{p(x)}+1){\rm d}x + C_3C_7C(\tau) p_2\int_{B_R}\left|\frac{u-u_R}{R}\right|^{p_2} {\rm d}x . \end{eqnarray}$$

(2.23)

由条件(1.6)和Young不等式$( \frac{1}{p_2} + \frac{p_2-1}{p_2}=1),$ 有

$$\begin{eqnarray} &&\displaystyle \int_{B_R}\eta^{p_2}|{{\cal B}}(x,\nabla u)||u-u_R| {\rm d}x \nonumber\\ &\leq&\displaystyle C_5\int_{B_R}\eta^{p_2}|\nabla u|^{p(x)-1}|u-u_R| {\rm d}x \nonumber\\ &\leq&\displaystyle C_5\tau\int_{B_R}\eta^{p_2}|\nabla u|^{\frac{p_2}{p_2-1}[(p(x)-1)]}{\rm d}x + C_5C(\tau)\int_{B_R}\eta^{p_2}\left|u-u_R\right|^{p_2}{\rm d}x \nonumber\\ &\leq&\displaystyle C_5\tau\int_{B_R}\eta^{p_2}(|\nabla u|^{p(x)}+1){\rm d}x + C_5C(\tau)\int_{B_R}\left|u-u_R\right|^{p_2}{\rm d}x . \end{eqnarray}$$

(2.24)

联合(2.20),(2.21),(2.23)和(2.24)式,可得

$$\begin{eqnarray} &&\displaystyle C_4\int_{B_R}\eta^{p_2}|\nabla u|^{p(x)} {\rm d}x \nonumber\\ &\leq&\displaystyle (C_3C_7p_2+C_5)\tau\int_{B_R}\eta^{p_2}|\nabla u|^{p(x)}{\rm d}x + C_5C(\tau)\int_{B_R}\left|u-u_R\right|^{p_2} {\rm d}x \nonumber\\ &&\displaystyle +(C_3C_7p_2+C_5)\tau\int_{B_R}\eta^{p_2}{\rm d}x + C_3C_7p_2C(\tau)\int_{B_R}\left|\frac{u-u_R}{R}\right|^{p_2} {\rm d}x . \end{eqnarray}$$

(2.25)

利用截断函数$\eta$的定义,并取$\tau=\frac{C_4}{2(C_3C_7p_2+C_5)}$. 即取$\tau$足够小,使得$(C_3C_7p_2+C_5)\tau\leq C_1$, 从而有

$$\begin{eqnarray} \int_{B_{R/2}}|\nabla u|^{p(x)} {\rm d}x &\leq& C\int_{B_R}\left|\frac{u-u_R}{R}\right|^{p_2}{\rm d}x +C\int_{B_R}\left|u-u_R\right|^{p_2}{\rm d}x +C\int_{B_R}1 {\rm d}x, \end{eqnarray}$$

(2.26)

其中$C=C(p_2,C_3,C_4,C_5,C_7)$. 下面利用Poincaré不等式,将(2.26)中含$|u-u_R|$的积分用$|\nabla u|$来估计. 注意到$p_2>p_1>\frac{p_1}{s}$, 于是由Poincaré不等式可得

$$\begin{eqnarray} -\hspace{-4mm}\int_{B_{R/2}}|\nabla u|^{p(x)} {\rm d}x &\leq&\displaystyle C\left(-\hspace{-4mm}\int_{B_R}|\nabla u|^{\frac{p_1}{s}}{\rm d}x\right)^{\frac{sp_2}{p_1}} + CR^{p_2}\left(-\hspace{-4mm}\int_{B_R}|\nabla u|^{\frac{p_1}{s}}{\rm d}x\right)^{\frac{sp_2}{p_1}} +C-\hspace{-4mm}\int_{B_R}1 {\rm d}x \nonumber\\ &\leq&\displaystyle C\left(-\hspace{-4mm}\int_{B_R}|\nabla u|^{\frac{p_1}{s}}{\rm d}x\right)^{\frac{sp_2}{p_1}} +C-\hspace{-4mm}\int_{B_R}1 {\rm d}x, \end{eqnarray}$$

(2.27)

其中$C=C (n,\gamma_1,\gamma_2,C_3,C_4,C_5,C_7)$, 这里利用了$R<1$,和$p_2\geq 1$. 又因为$p_2-p_1\leq \omega(2nR)$, 从而$\frac{sp_2}{p_1}\leq \frac{s\omega(2nR)}{p_1} +s$, 并利用$\frac{p_1}{s}\leq \frac{p(x)}{s}$, 由上式可得

$$\begin{eqnarray} && -\hspace{-4mm}\int_{B_{R/2}}|\nabla u|^{p(x)} {\rm d}x \nonumber\\ &\leq&\displaystyle C R^{-2n\omega(2nR)} \left(\int_{B_R}|\nabla u|^{\frac{p(x)}{s}}{\rm d}x\right)^{\frac{s\omega(2nR)}{p_1}} \left(-\hspace{-4mm}\int_{B_R}|\nabla u|^{\frac{p(x)}{s}}{\rm d}x\right)^{s} +C-\hspace{-4mm}\int_{B_R}1 {\rm d}x \nonumber\\ &\leq&\displaystyle CK_0^{\frac{2\omega(8nR_0)}{\gamma_1}} \left(-\hspace{-4mm}\int_{B_R}|\nabla u|^{\frac{p(x)}{s}}{\rm d}x\right)^{s} +C-\hspace{-4mm}\int_{B_R}1 {\rm d}x, \end{eqnarray}$$

(1)

其中$K_0 = \int_{Q_{4R_0}}|\nabla u|^{p(x)}{\rm d}x +1 \geq 1.$ 这里利用了当$0<R\leq R_0$时$R^{-\omega(2nR)}\leq C(n,C_6)$. 于是在引理2.4中 取$f=|\nabla u|^{p(x)/s}$,$\phi = 1$即可得证.

注 由引理2.5,并注意到$\nabla u \in L^{p(x)}(\Omega)$,我们有

$$\begin{eqnarray} -\hspace{-4mm}\int_{B_{R/2}} |\nabla u|^{p(x)(1+\sigma_0)}{\rm d}x &\leq&\displaystyle C\left\{\left( -\hspace{-4mm}\int_{B_{R}} |\nabla u|^{p(x)}{\rm d}x\right)^{1+\sigma_0}+1\right\} \nonumber\\ &=& \displaystyle C\left\{\left( -\hspace{-4mm}\int_{B_{R}} |\nabla u|^{p(x)}{\rm d}x\right)^{\sigma_0} \left( -\hspace{-4mm}\int_{B_{R}} |\nabla u|^{p(x)}{\rm d}x\right)+1\right\} \nonumber\\ &=& \displaystyle C\left\{\frac{1}{|B_R|^{\sigma_0}}\left(\int_{B_{R}} |\nabla u|^{p(x)}{\rm d}x\right)^{\sigma_0} \left( -\hspace{-4mm}\int_{B_{R}} |\nabla u|^{p(x)}{\rm d}x\right)+1\right\} \nonumber\\ &\leq& \displaystyle CR^{-n\sigma_0}-\hspace{-4mm}\int_{B_{R}} \left(|\nabla u|^{p(x)} +1\right ){\rm d}x \nonumber\\ &\leq& \displaystyle CR^{-n\sigma_0}-\hspace{-4mm}\int_{B_{R}} \left(|\nabla u|^{p_2} +1\right ){\rm d}x. \end{eqnarray}$$

(2.29)

引理 2.6 令$0<R\leq R_1 \leq R_0 < 1$,$B_{R_0} \subset\Omega$. 若$u$为方程(1.1)满足条件(1.2)-(1.8)的弱解, 则存在仅依赖于$n,\gamma_1,\gamma_2,\alpha_1,\sigma_0$的正常数$\beta$, 使得

$$\begin{equation} -\hspace{-4mm}\int_{B_{R}}|\nabla u - \nabla v|^{p_2}{\rm d}x \leq C R^\beta -\hspace{-4mm}\int_{B_{R}}(|\nabla u|^{p_2}+1){\rm d}x, \end{equation}$$

(2.30)

其中$C$与$n,\gamma_1,\gamma_2,\alpha_1,C_i\ (i=1,2,\cdots ,7)$有关, $v$是问题(2.2)的弱解.

证 由引理2.3,问题(2.2)的弱解$v$满足(2.7)和(2.8)式. 在定义1.1和定义2.1中取$\varphi=u-v$,有

$$\begin{equation} -\hspace{-4mm}\int_{B_R}{{\cal A}}(x,\nabla u)\cdot\nabla(u-v) {\rm d}x = -\hspace{-4mm}\int_{B_R}{{\cal B}}(x,\nabla u)(u-v){\rm d}x \end{equation}$$

(2.31)

和

$$\begin{equation} -\hspace{-4mm}\int_{B_R}{{\cal A}}(x^*,\nabla v)\cdot\nabla(u-v) {\rm d}x =0. \end{equation}$$

(2.32)

令

$$\begin{eqnarray} &&\displaystyle -\hspace{-4mm}\int_{B_R}{{\cal A}}(x,\nabla u)\cdot\nabla(u-v) {\rm d}x - -\hspace{-4mm}\int_{B_R}{{\cal A}}(x^*,\nabla v)\cdot\nabla(u-v) {\rm d}x \nonumber\\ &=& \displaystyle -\hspace{-4mm}\int_{B_R}[{{\cal A}}(x,\nabla u) - {{{\cal A}}}(x^*,\nabla u)]\cdot\nabla(u-v){\rm d}x \nonumber\\ && \displaystyle + -\hspace{-4mm}\int_{B_R}[{{{\cal A}}}(x^*,\nabla u) - {{{\cal A}}}(x^*,\nabla v)]\cdot\nabla(u-v){\rm d}x \nonumber\\ &=& I_1+I_2, \end{eqnarray}$$

(2.33)

$$\begin{equation} -\hspace{-4mm}\int_{B_R}{{\cal B}}(x,\nabla u)(u-v) {\rm d}x = I_3, \end{equation}$$

(2.34)

由(2.32)-(2.35)式,可得

$$\begin{equation} I_1+I_2=I_3. %(2.35) \end{equation}$$

(2.35)

下面分别估计$I_i\,(i=1,2,3)$.

估计I₁ 由(2.4)式,注意到连续模$\omega(x)$当$R\rightarrow 0$时$\omega(x)\rightarrow 0$, 于是存在任意小的常数$\varepsilon$,使$R$充分小时有$\omega(x)<\varepsilon$. 于是结合使用带$\tau$的Young不等式和(2.7)式,我们有

$$\begin{eqnarray} I_1 &=&\displaystyle -\hspace{-4mm}\int_{B_R}[{{{\cal A}}}(x,\nabla u) - {{{\cal A}}}(x^*,\nabla u)]\cdot\nabla(u-v){\rm d}x \nonumber\\ &\leq& \displaystyle C -\hspace{-4mm}\int_{B_R}w(|x-x^*|)|\nabla u|^{p_2-1}|\nabla(u-v)|{\rm d}x \nonumber\\ &\leq& \displaystyle C \varepsilon-\hspace{-4mm}\int_{B_R}|\nabla u|^{p_2-1}|\nabla(u-v)|{\rm d}x \nonumber\\ &\leq& \displaystyle C \varepsilon\left( \tau-\hspace{-4mm}\int_{B_R}|\nabla u|^{p_2}{\rm d}x + c(\tau)\int_{B_R}|\nabla u -\nabla v|^{p_2}{\rm d}x \right) \nonumber\\ &\leq& \displaystyle C\varepsilon -\hspace{-4mm}\int_{B_{R}} \left(|\nabla u|^{p_2}+1\right ){\rm d}x. \end{eqnarray}$$

(2.36)

估计$I_2$由(2.3)式可知

$$\begin{eqnarray} I_2 &=& -\hspace{-4mm}\int_{B_R}[{{{\cal A}}}(x^*,\nabla u) - {{{\cal A}}}(x^*,\nabla v)]\cdot\nabla(u-v){\rm d}x \nonumber\\ &\geq& C -\hspace{-4mm}\int_{B_R}|\nabla(u-v)|^{p_2}{\rm d}x. \end{eqnarray}$$

(2.37)

估计$I_3$ 因$p_1 \geq\gamma_1>1$,由(2.1)式有

$$\begin{equation} p_2 = p_1+ (p_2-p_1) \leq p_1 + \frac{\sigma_0\gamma_1}{\sigma_0+2},% (2.38) \end{equation}$$

(2.38)

于是有

$$\begin{eqnarray} p_2 &\leq&\displaystyle p_2(1+\sigma_0/2) \nonumber\\ &\leq& \displaystyle \left(p_1 + \frac{\sigma_0\gamma_1}{\sigma_0+2}\right)(1+\sigma_0/2) \nonumber\\ &\leq& \displaystyle p_1(1+\sigma_0) \nonumber\\ &\leq& \displaystyle p(x)(1+\sigma_0). \end{eqnarray}$$

(2.39)

由(2.7),(2.39)式,可得

$$\begin{eqnarray} \displaystyle -\hspace{-4mm}\int_{B_R}(|\nabla u|^{p_2}+|\nabla v|^{p_2}){\rm d}x &\leq&\displaystyle C -\hspace{-4mm}\int_{B_R}|\nabla u|^{p_2}{\rm d}x \nonumber\\ &\leq& \displaystyle C -\hspace{-4mm}\int_{B_R}(1+|\nabla u|)^{p_2}{\rm d}x \nonumber\\ &\leq& \displaystyle C -\hspace{-4mm}\int_{B_R}(1+|\nabla u|)^{p(x)(1+\sigma_0)}{\rm d}x. \end{eqnarray}$$

(2.40)

由(1.6)式,带$\tau$的Young不等式$(\frac{p_2-1}{p_2}+\frac{1}{p_2}=1),$ Poincaré不等式,(2.40)和(2.29)式,有

$$\begin{eqnarray} I_3 &=&\displaystyle -\hspace{-4mm}\int_{B_R}{{\cal B}}(x,\nabla u)(u-v){\rm d}x \nonumber\\ &\leq& \displaystyle C-\hspace{-4mm}\int_{B_R}|\nabla u|^{p(x)-1}|u-v|{\rm d}x \nonumber\\ &\leq& \displaystyle \tau C-\hspace{-4mm}\int_{B_R}|\nabla u|^{\frac{[p(x)-1]p_2}{p_2-1}}{\rm d}x +C(\tau)-\hspace{-4mm}\int_{B_R}|u-v|^{p_2}{\rm d}x \nonumber\\ &\leq& \displaystyle \tau C -\hspace{-4mm}\int_{B_R} |\nabla u|^{p(x)} {\rm d}x +C(\tau)R-\hspace{-4mm}\int_{B_R}|\nabla u-\nabla v|^{p_2}{\rm d}x \nonumber\\ &\leq& \displaystyle \tau C -\hspace{-4mm}\int_{B_R} [C(\tau)|\nabla u|^{p(x)(1+\sigma_0)}+\tau] {\rm d}x +C(\tau)R-\hspace{-4mm}\int_{B_R}(1+|\nabla u|)^{p(x)(1+\sigma_0)}{\rm d}x. \nonumber\\ &\leq& \displaystyle \tau CR^{-n\sigma_0} -\hspace{-4mm}\int_{B_R}\left(|\nabla u|^{p_2} +1\right ){\rm d}x +CR^{1-n\sigma_0}-\hspace{-4mm}\int_{B_{2R}} \left(|\nabla u|^{p_2} +1\right ){\rm d}x+C\tau^2. \end{eqnarray}$$

(2.41)

于是,联立(2.36),(2.37)和 (2.41)式,可得

$$\begin{eqnarray} -\hspace{-4mm}\int_{B_R}|\nabla(u-v)|^{p_2}{\rm d}x \leq C \left(\varepsilon + \tau R^{-n\sigma_0}+R^{1-n\sigma_0}\right) -\hspace{-4mm}\int_{B_{2R}}(|\nabla u|^{p_2}+1){\rm d}x+C\tau^2. \end{eqnarray}$$

(2.42)

令$\beta=1-n\sigma_0$,并取$\varepsilon,\tau$足够小使$\varepsilon,\tau\leq R^{\beta}$, 则(2.42)式即为

$$\begin{eqnarray} \displaystyle -\hspace{-4mm}\int_{B_R}|\nabla(u-v)|^{p_2}{\rm d}x \leq CR^{\beta}-\hspace{-4mm}\int_{B_{2R}}(|\nabla u|^{p_2}+1){\rm d}x. \end{eqnarray}$$

(2.43)

证毕.

引理 2.7 若$u$为方程(1.1)满足条件(1.2)-(1.8)的弱解, 则对任给$\lambda\in (0,n)$, 存在正常数$R_2$和仅与$n,\gamma_1,\gamma_2,\alpha_1,\sigma_0, {\rm dist}\{\Omega_0,\Omega\}$有关的正常数$C$, 使得

$$\begin{equation} -\hspace{-4mm}\int_{B_{\rho}}|\nabla u |^{p_2}{\rm d}x \leq CR^{-\lambda}, \end{equation}$$

(2.44)

其中$0<\rho\leq R\leq R_2\leq \frac{R_1}{16}$,$\Omega_0\subset\Omega$ 和$B_{4R_1}\subset\Omega_0$.

证 若$v$为问题(2.2)的弱解,则由(2.30),(2.7)和(2.8)式,有

$$\begin{eqnarray} \int_{B_{\rho}}|\nabla u |^{p_2}{\rm d}x &\leq&\displaystyle C\int_{B_{\rho}}|\nabla u -\nabla v|^{p_2}{\rm d}x + C\int_{B_{\rho}}|\nabla v |^{p_2}{\rm d}x \nonumber\\ &\leq& \displaystyle C\rho^{\beta}\int_{B_{2\rho}}(|\nabla u|^{p_2}+1){\rm d}x + C\left(\frac{\rho}{R}\right)^{n}\int_{B_{R}}|\nabla v |^{p_2}{\rm d}x \nonumber\\ &\leq& \displaystyle CR^{\beta}\int_{B_{2\rho}}(|\nabla u|^{p_2}+1){\rm d}x + C\left(\frac{\rho}{R}\right)^{n}\int_{B_{R}}|\nabla u |^{p_2}{\rm d}x \nonumber\\ &\leq& \displaystyle C \left(R^{\beta}+ \left(\frac{\rho}{R}\right)^{n}\right)\int_{B_R}(|\nabla u|^{p_2}{\rm d}x +CR^{n}. \end{eqnarray}$$

(2.45)

于是由覆盖迭代讨论(见文献[26,引理3.2]),对每个$\lambda\in(0,n)$, 存在正常数$R_2,C>0$ 使得对所有$0<\rho\leq R\leq R_2\leq \frac{R_1}{16}$,有

$$-\hspace{-4mm}\int_{B_{\rho}}|\nabla u |^{p_2}{\rm d}x \leq C\rho^{-\lambda}. (2.46)$$

(2.46)

证毕.

证 注意到$R<1$. 根据引理2.3,引理2.6 和引理2.7,可得

$$\begin{eqnarray} \int_{B_\rho}|\nabla u - (\nabla u)_{B_\rho}|^{p_2}{\rm d}x &\leq& \displaystyle C \int_{B_\rho}|\nabla u - (\nabla v)_{B_\rho}|^{p_2}{\rm d}x \nonumber\\ &\leq& \displaystyle \int_{B_\rho}|\nabla u - \nabla v|^{p_2}{\rm d}x +\int_{B_\rho}|\nabla v - (\nabla v)_{B_\rho}|^{p_2}{\rm d}x \nonumber\\ &\leq& \displaystyle CR^{\beta}\int_{B_{2R}}(|\nabla u|^{p_2}+1){\rm d}x + C(\frac{\rho}{R})^{\beta_1}\rho^n -\hspace{-4mm}\int_{B_R}|\nabla u|^{p_2}{\rm d}x \nonumber\\ &\leq& \displaystyle CR^{n+\beta}-\hspace{-4mm}\int_{B_{2R}}|\nabla u|^{p_2}{\rm d}x + CR^{n+\beta} + C(\frac{\rho}{R})^{\beta_1}\rho^n R^{-\lambda}. \end{eqnarray}$$

(3.1)

令$\rho=\frac{R^{1+\mu}}{2}$,其中$\mu=\frac{\beta}{n+\beta_1}>0$. 则上述不等式右侧的$R$有相同的指数$n+\beta-\lambda$. 有

$$\begin{eqnarray} \int_{B_\rho}|\nabla u - (\nabla u)_{B_\rho}|^{p_2}{\rm d}x &\leq& \displaystyle CR^{n+\beta-\lambda} + CR^{n+\beta} + CR^{n+\beta-\lambda} \nonumber\\ &\leq& \displaystyle CR^{n+\beta-\lambda} \nonumber\\ &\leq& \displaystyle C\rho^{\frac{n+\beta-\lambda}{1+\mu}} \nonumber\\ &=& \displaystyle C\rho^{n+\frac{\beta-n\mu-\lambda}{1+\mu}}. \end{eqnarray}$$

(3.2)

令$\lambda=\frac{\beta\beta_1}{2(n+\beta_1)}>0$. 则有 $\beta-n\mu-\lambda>0$. (3.2)式即为

$$\begin{equation} -\hspace{-4mm}\int_{B_\rho}|\nabla u - (\nabla u)_{B_\rho}|^{p_2}{\rm d}x \leq C\rho^{\frac{\beta-n\mu-\lambda}{1+\mu}}. \end{equation}$$

(3.3)

最后,由Hölder不等式可得

$$\begin{eqnarray} \left( -\hspace{-4mm}\int_{B_\rho}|\nabla u - (\nabla u)_{B_\rho}|^{\gamma_1}{\rm d}x\right)^{\frac{1}{\gamma_1}} &\leq& \displaystyle \left( -\hspace{-4mm}\int_{B_\rho}|\nabla u - (\nabla u)_{B_\rho}|^{p_2}{\rm d}x\right)^{\frac{1}{p_2}} \left( -\hspace{-4mm}\int_{B_\rho}1{\rm d}x\right)^{\frac{p_2-1}{p_2}} \nonumber\\ &\leq& \displaystyle C\rho^{\frac{\beta-n\mu-\lambda}{(1+\mu)p_2}}. \end{eqnarray}$$

(3.4)

即

$$\begin{equation} -\hspace{-4mm}\int_{B_\rho}|\nabla u - (\nabla u)_{B_\rho}|^{\gamma_1}{\rm d}x \leq C\rho^{\frac{(\beta-n\mu-\lambda)\gamma_1}{(1+\mu)\gamma_2}}=C\rho^{\alpha} . \end{equation}$$

(3.5)

证毕.