本文考虑Orlicz空间中具有散度形式的椭圆方程的很弱解的$ L^{\phi} $型估计. Orlice空间由Orlice^[1]引入.对于Orlice空间的应用参见文献[2]. $ A $ -调和方程具有很强的背景, 且在物理和工程方面有很多应用.近些年来不同形式的$ A $ -调和方程的发展, 参见文献[3-11].

令$ \Omega $为$ {{\Bbb R}} ^{n} $中的有界正则区域, $ n\geq 2 $.在正则区域中, Hodge分解及其估计是满足的.例如, Lipschitz域是正则的.本文考虑如下的$ A $ -调和方程

(1.1) $ \begin{equation} {\rm div} A(x, \nabla u) = {\rm div}(\left | \textbf{f} \right |^{p-2}\textbf{f}), \quad\; x\in \Omega , \end{equation} $

其中$ \textbf{f} = (f^{1}, f^{2}, \cdots , f^{n}) $是给定的向量, 算子$ A = A(x, \xi ):\Omega\times {{\Bbb R}} ^{n}\rightarrow {{\Bbb R}} ^{n} $.对于每一个$ \xi $, $ A $在$ x $上可测, 并且对于几乎所有的$ x $, $ A $在$ \xi $上连续.而且当$ p\in(1, \infty) $时, 算子$ A = A(x, \xi ) $满足如下结构条件

$({\rm H} _{1})\quad\left \langle A(x, \xi )-A(x, \zeta ) , \xi -\zeta \right \rangle\geq C_{1}\left | \xi -\zeta \right |^{p} $,

$ ({\rm H}_{2})\quad | A(x, \xi ) |\leq C_{2}\left | \xi \right |^{p-1} $,

$({\rm H} _{3})\quad A(x, 0) = 0, $

$ ({\rm H}_{4})\quad\left | A(x, \xi )-A(y, \xi ) \right |\leq C_{3}\omega (\left | x-y \right |)\left | \xi \right |^{p-1} $,

其中$ \xi, \zeta\in{{\Bbb R}} ^{n}, x, y\in\Omega, C_{i}> 0, i = 1, 2, 3 $为正常数.这里的连续模$ \omega (x):{{\Bbb R}} ^{+}\rightarrow {{\Bbb R}} ^{+} $非减, 而且满足

(1.2) $ \begin{equation} \omega (r)\rightarrow 0 \qquad \mbox{当} \quad r\rightarrow 0. \end{equation} $

方程$ (1.1) $的原型是$ p $ -调和方程

(1.3) $ \begin{equation} {\rm div} (\left |\nabla u\right |^{p-2}\nabla u) = 0. \end{equation} $

定义1.1^[12]  函数$ u \in W_{loc}^{1, p} (\Omega) $称为方程$ (1.1) $的局部弱解, 若有

(1.4) $ \begin{equation} \int _{\Omega }\left \langle A(x, \nabla u), \nabla \varphi \right \rangle{\rm d}x = \int _{\Omega }\left \langle | \textbf{f} |^{p-2} \textbf{f}, \nabla \varphi \right \rangle{\rm d}x \end{equation} $

对于任意的具有紧支集的$ \varphi \in W_{0}^{1, p} (\Omega ) $都成立.

本文考虑很弱解的概念, 降低了自然可积性假设.

定义1.2  函数$ u \in W_{loc}^{1, r} (\Omega) $, $ \max\{1, p-1\}\leq r < p $, 称为方程$ (1.1) $的局部很弱解, 若对所有具有紧支集的$ \varphi \in W_{0}^{1, \frac{r}{r-p+1}} (\Omega ) $, 方程$ (1.4) $成立.

关于梯度估计, 有下述研究成果. Acerbi-Mingione^[13]证明了$ p(x) $-Laplacian方程组

(1.5) $ \begin{equation} {\rm div}(\left | D u \right |^{p(x)-2} D u) = \rm {div}(\left | F \right |^{p(x)-2}F) \end{equation} $

的梯度估计; DiBenedetto-Manfredi^[14]获得了

(1.6) $ \begin{equation} {\rm div}(\left | D u \right |^{p-2} D u) = \rm {div}(\left | F \right |^{p-2}F) \end{equation} $

的弱解的梯度的更高可积性; Byun-Wang^[15]和Kinnunen-Zhou^[16]获得了

(1.7) $ \begin{equation} {\rm div}((A D u\cdot D u)^{(p-2)/2}A D u) = {\rm div}(\left | F \right |^{p-2}F) \end{equation} $

的弱解的$ L^{q}(q\geq p) $梯度估计; Byun-Wang^[17]获得了一般非线性椭圆方程

(1.8) $ \begin{equation} {\rm div} A(x, \nabla u) = {\rm div}f \end{equation} $

的$ L^{p} $估计; Yao^[12]在2010年获得了方程(1.1)的弱解在Orlicz空间中的梯度估计.

尽管关于弱解的梯度估计有很多出色的成果, 但Orlicz空间中的很弱解的梯度估计尚未研究.本文旨在研究方程(1.1)的很弱解在Orlicz空间中的$ L^{\phi} $估计.下面是本文的主要结论.

定理1.1  假设$ \phi \in\triangle _{2}\bigcap \bigtriangledown_{2}, | {\bf f} |^{r}\in L_{loc}^{\phi}(\Omega) $, $ \max\{1, p-1\}\leq r < p $.如果$ u $是方程$ (1.1) $的局部很弱解, 且算子$ A $满足条件$ (H_{1})-(H_{4}) $, 则有$ \left | \nabla u \right |^{r} \in L_{loc}^{\phi }\left ( \Omega \right ), $且有估计式

(1.9) $ \begin{equation} \int _{B_{R}}\phi \left ( \left | \nabla u \right | ^{r}\right ){\rm d}x\leq C\left \{ \int _{B_{2R}}\phi\left(| {\bf f}| ^{r}\right ){\rm d}x+\phi\left ( \frac{1}{R^{r}}\int_{B_{2R}}\left | u-u_{B_{2R}} \right |^{r}{\rm d}x\right) +1\right \}, \end{equation} $

其中$ B_{2R}\subset \Omega $, 且$ C $是独立于$ u $和$ {\bf f} $的常数.

本文其余部分安排如下.在第二节和第三节, 介绍Orlicz空间中的一些知识和预备引理.在第四节中给出方程(1.1)的很弱解的$ L^{r} $估计.在第五节中给出主要定理(定理1.1)的证明.

用$ \Phi $表示所有函数$ \phi :\left [ 0, + \infty \right )\rightarrow \left [ 0, + \infty \right ) $组成的函数类, 其中$ \phi $是递增的凸函数.

定义2.1^[12]  如果存在一个正常数$ K $, 使得对于任意的$ t> 0 $, 都有

(2.1) $ \begin{equation} \phi (2t)\leq K\phi (t), \end{equation} $

那么, 函数$ \phi \in \Phi $满足全局$ \triangle _{2} $条件, 表示为$ \phi \in \triangle _{2} $.同时, 如果存在一个常数$ a>1 $, 使得对于任意的$ t>0 $, 都有

(2.2) $ \begin{equation} \phi (t)\leq \frac{\phi (at)}{2a}, \end{equation} $

那么, 函数$ \phi \in \Phi $满足全局$ \bigtriangledown _{2} $条件, 表示为$ \phi \in \bigtriangledown _{2} $.

注2.1^[12]   $ (1) $ 注意到全局$ \triangle _{2}\bigcap \bigtriangledown _{2} $条件使函数适当增长.

$ (2) $ 事实上, 如果$ \phi \in \triangle _{2}\bigcap \bigtriangledown _{2} $, 则对于$ 0< \theta _{2}\leq 1\leq \theta _{1}< \infty $, $ \phi $满足

(2.3) $ \begin{equation} \phi (\theta _{1}t)\leq K\theta _{1}^{\alpha _{1}}\phi (t), \qquad \phi (\theta _{2}t)\leq 2a\theta _{2}^{\alpha _{2}}\phi (t), \end{equation} $

其中$ \alpha _{1} = \log _{2}K, \alpha _{2} = \log _{a }2+1 $.

$ (3) $ 在条件(2.3)下, 很容易得到$ \phi \in \Phi $满足$ \phi(0) = 0 $以及

(2.4) $ \begin{equation} \lim\limits_{t\rightarrow 0^{+}}\frac{\phi (t)}{t} = \lim\limits_{t\rightarrow +\infty }\frac{t}{\phi (t)} = 0. \end{equation} $

定义2.2^[12]  令$ \phi \in \Phi $, 则$ \rm Orlicz $类$ K^{\phi }(\Omega ) $是满足

(2.5) $ \begin{equation} \int _{\Omega }\phi (\left | g \right |){\rm d}x< \infty \end{equation} $

的所有可测函数$ g:\Omega \rightarrow {{\Bbb R}} $组成的集合. Orlicz空间$ L^{\phi }(\Omega ) $是$ K^{\phi }(\Omega ) $的线性闭包.

注2.2  注意到, $ \rm Orlicz $空间是$ L^{q } $空间的推广形式.如果$ \phi (t) = t^{q}, t\geq 0 $, 那么$ \phi \in \triangle _{2}\bigcap \bigtriangledown _{2} $, 于是得到一个特例, $ L^{\phi }(\Omega ) = L^{q}(\Omega ) $.

关于Orlicz空间有如下引理.

引理2.1^[12]  假设$ \phi \in\triangle _{2}\bigcap \bigtriangledown _{2} $以及$ g \in L^{\phi }(\Omega ) $, 那么

(1) $ K^{\phi } = L^{\phi }, $且$ C_{0}^{\infty } $在$ L^{\phi } $中稠密;

(2) $ L^{\alpha _{1}}(\Omega )\subset L^{\phi }(\Omega )\subset L^{\alpha _{2}}(\Omega )\subset L^{1}(\Omega ), $其中$ \alpha _{1} = \log _{2}K, \alpha _{2} = \log _{\alpha }2+1; $

(3) $ \int _{\Omega }\phi (\left | g \right |){\rm d}x = \int_{0}^{\infty }\left | \left \{ x \in \Omega :\left | g \right |> \lambda \right \} \right |{\rm d}\left [ \phi (\lambda ) \right ]; $

(4) $ \int_{0}^{\infty }\frac{1}{\mu }\int _{\left \{ x \in \Omega :\left | g \right | > a\mu \right \}}\left | g \right |{\rm d}x{\rm d}\left [ \phi (b\mu ) \right ]\leq C\int _{\Omega }\phi (\left | g \right |){\rm d}x, $其中$ a, b>0, C = C(a, b, \phi ) $.

在本节中, 使用文献[12]中的新标准化方法, 该方法受文献[17]影响较大.

对任意的$ \lambda \ge 1 $, 定义

(3.1) $ \begin{equation} u_\lambda(x) = \frac{u(x)}{\lambda}, \textbf{f}_\lambda(x) = \frac{\textbf{f}(x)}{\lambda}, A_\lambda(x, \xi) = \frac{A(x, \lambda\xi)}{\lambda^{p-1}}. \end{equation} $

引理3.1 (新标准化)  若$ u \in W_{loc}^{1, r}(\Omega) $是方程$ (1.1) $的局部很弱解, $ A $满足条件(H$ _1) $–(H$ _4) $, 那么

(1) $ A_\lambda $满足条件(H$ _1) $–(H$ _4) $, 即有

(3.2) $ \begin{equation} \left \langle A_\lambda(x, \xi )-A_\lambda(x, \zeta ) , \xi -\zeta \right \rangle\geq C_{1}\left | \xi -\zeta \right |^{p}, \end{equation} $

(3.3) $ \begin{equation} |A_\lambda(x, \xi ) |\leq C_{2}\left | \xi \right |^{p-1}, \end{equation} $

(3.4) $ \begin{equation} A_\lambda(x, 0) = 0, \end{equation} $

(3.5) $ \begin{equation} \left | A_\lambda(x, \xi )-A_\lambda(y, \xi ) \right |\leq C_{3}\omega (\left | x-y \right |)\left | \xi \right |^{p-1}, \end{equation} $

其中$ C_{i}(1\le i\le 3) $为常数.

(2) $ u_\lambda $是方程

(3.6) $ \begin{equation} {{\rm div}} A_\lambda(x, \nabla u) = {\rm div}(\left | \textbf{f}_\lambda \right |^{p-2}\textbf{f}_\lambda), \quad x\in \Omega \end{equation} $

的很弱解.

证  使用文献[12, 引理2.1]中的类似方法来证明.

下面给出文献[12]中使用的迭代覆盖方法.令$ u $是方程(1.1)的局部很弱解.先设定理$ 1.1 $中的$ R = 1 $, 再通过缩放论证证明.设

(3.7) $ \begin{equation} \lambda _{0} = \left [-\!\!\!\!\!\!\int_{B_{1}} \left | \nabla u \right |^{r}{\rm d}x+\frac{1}{\varepsilon}-\!\!\!\!\!\!\int_{B_{1}}|\textbf{f}|^{r}{\rm d}x\right ]^{1/r}, \end{equation} $

其中$ \varepsilon> 0 $满足(5.12)式中的条件.对于任意的$ x \in \Omega $和$ \rho > 0 $, 令

(3.8) $ \begin{equation} J_\lambda\left [ B_{\rho } (x)\right ] = -\!\!\!\!\!\!\int _{B_{\rho } (x)}\left | \nabla u_\lambda \right |^{r}{\rm d}y+\frac{1}{\varepsilon}-\!\!\!\!\!\!\int_{B_{\rho } (x)}|\textbf{f}_\lambda|^{r}{\rm d}y, \end{equation} $

(3.9) $ \begin{equation} E_\lambda(1) = \left \{ x \in B_{1}:\left | \nabla u_\lambda \right | ^{r}> 1\right \}. \end{equation} $

由(1.2)式, 选择合适的常数$ R_{0} = R_{0}(\varepsilon ) \in (0, 1) $使得

(3.10) $ \begin{equation} \omega (R_{0})\leq \varepsilon . \end{equation} $

一般情况下, 考虑$ r $趋近于$ p $时, 限制$ 0<p-r<\min\{\frac{1}{2}, \varepsilon \} $.

引理3.2^[12]  令$ \lambda \geq \lambda _{*} = :(10/R_{0})^{n/r}\lambda _{0}+1 $, 则存在一族不相交的球$ \left \{ B_{\rho _{i}}(x_{i}) \right \} $, $ x_{i} \in E_\lambda(1) $, 使得$ 0< \rho _{i} = \rho (x_{i})\leq R_{0}/10 $且有

(3.11) $ \begin{equation} J_\lambda\left [ B_{\rho _{i}} (x_{i})\right ] = 1, \quad J_\lambda\left [ B_{\rho } (x_{i})\right ]< 1 \quad\; \mbox{对任意} \; \rho > \rho _{i} . \end{equation} $

而且有

(3.12) $ \begin{equation} E_\lambda(1)\subset \bigcup\limits_{i \in \mathbb{N}}B_{5\rho _{i}}(x_{i})\cup \mbox{可略集}, \end{equation} $

(3.13) $ \begin{equation} |B_{\rho _{i}} (x_{i})| \leq 3\Big(\int _{\{ x\in B_{\rho _{i}}(x_{i}):| \nabla u_\lambda | ^{r}>1/3 \}}| \nabla u_\lambda |^{r}{\rm d}x +\frac{1}{\varepsilon}\int _{\{ x\in B_{\rho _{i}}(x_{i}):| \textbf{f}_\lambda | ^{r}> \varepsilon/3\}} |\textbf{f}_\lambda |^{r}{\rm d}x \Big). \end{equation} $

以下是对方程$ (1.1) $的很弱解的$ L^{r} $估计.

定理4.1  假设$ \left|{\bf f}\right|^{r}\in L^{\phi}(B_{2R}), B_{2R}\subset \Omega $, 令$ u \in W_{loc}^{1, r}(\Omega) $为方程$ (1.1) $的局部很弱解, 算子$ A $满足条件(H$ _1) $–(H$ _3) $, 则

(4.1) $ \begin{equation} \int _{B_{R}}\left | \nabla u \right |^{r}{\rm d}x\leq C \left\{ \frac{1}{R^{r}} \int _{B_{2R}}\left | u-u_{B_{2R}} \right |^{r}{\rm d}x+\int _{B_{2R}}\left | \bf f\right |^{r}{\rm d}x \right\}, \end{equation} $

其中$ C = C(n, p, C_1, C_2) $.

证  令$ u $是$ A $ -调和方程$ (1.1) $的局部很弱解. $ B_{2R}\subset\subset \Omega $.取截断函数$ \eta \in C_{0}^{\infty }(R_n) $满足

(4.2) $ \begin{equation} 0\leq \eta \leq 1;\qquad \eta \equiv 1, \ x\in B_{R};\qquad \eta \equiv 0, \ x\in {{\Bbb R}}^n \setminus B_{2R}, \qquad \left | \nabla \eta \right | \leq \frac{C}{R}. \end{equation} $

对$ |\nabla [\eta( u-u_{B_{2R}}\;)]|^{r-p}\nabla[\eta( u-u_{B_{2R}}\;)] \in L^{\frac{r}{r-p+1}}\;(\Omega) $进行Hodge分解, 参见文献[18]:

(4.3) $ \begin{equation} |\nabla [\eta( u-u_{B_{2R}})]|^{r-p}\nabla[\eta( u-u_{B_{2R}})] = \nabla\varphi+H, \end{equation} $

其中$ \varphi \in W_{0}^{1, \frac{r}{r-p+1}}\;(\Omega), H \in L^{\frac{r}{r-p+1}}\;(\Omega) $是散度为零向量场且有以下估计式成立

(4.4) $ \begin{equation} ||\nabla \varphi||_{\frac{r}{r-p+1}}\leq C ||\nabla [\eta( u-u_{B_{2R}})]||_{r}^{r-p+1}, \end{equation} $

(4.5) $ \begin{equation} ||H||_{\frac{r}{r-p+1}}\leq C(p-r) ||\nabla [\eta( u-u_{B_{2R}})]||_{r}^{r-p+1}, \end{equation} $

其中$ C = C(n, p) $是只依赖于$ n $和$ p $的常数.令

(4.6) $ \begin{eqnarray} E(\eta, u) = |\nabla [\eta( u-u_{B_{2R}})]|^{r-p}\nabla[\eta( u-u_{B_{2R}})] -|\eta\nabla ( u-u_{B_{2R}})|^{r-p}\eta\nabla( u-u_{B_{2R}}), \end{eqnarray} $

通过文献[19, p271, (4.1)式]

(4.7) $ \begin{equation} \big||X|^{-\varepsilon}X-|Y|^{-\varepsilon}Y\big|\leq 2^{\varepsilon}\frac{1+\varepsilon}{1-\varepsilon}|X-Y|^{1-\varepsilon}, \quad X, Y \in {{\Bbb R}} ^{n}, \quad 0\leq \varepsilon <1, \end{equation} $

可以得到

(4.8) $ \begin{equation} |E(\eta, u)|\leq 2^{p-r}\frac{p-r+1}{r-p+1}| (u-u_{B_{2R}})\nabla \eta|^{r-p+1}. \end{equation} $

由$ (4.3) $和$ (4.6) $式, 有

(4.9) $ \begin{equation} \nabla\varphi = |\eta\nabla(u-u_{B_{2R}})|^{r-p}\eta\nabla(u-u_{B_{2R}})+E-H. \end{equation} $

选取$ \varphi \in W_{0}^{1, \frac{r}{r-p+1}}\;(\Omega) $为定义$ 1.2 $中的检验函数, 则

(4.10) $ \begin{eqnarray} && \int _{B_{2R}}\left \langle A(x, \nabla u), |\eta\nabla(u-u_{B_{2R}})|^{r-p}\eta\nabla(u-u_{B_{2R}})+E-H\right \rangle{\rm d}x \\ & = & \int _{B_{2R}}\left \langle |\textbf{f}|^{p-2}\textbf{f}, |\eta\nabla(u-u_{B_{2R}})|^{r-p}\eta\nabla(u-u_{B_{2R}})+E-H\right \rangle{\rm d}x. \end{eqnarray} $

即

(4.11) $ \begin{eqnarray} && \int _{B_{2R}}\left \langle A(x, \nabla u), |\eta\nabla(u-u_{B_{2R}})|^{r-p}\eta\nabla(u-u_{B_{2R}})\right \rangle{\rm d}x \\ & = & -\int _{B_{2R}}\left \langle A(x, \nabla u), E\right \rangle{\rm d}x+\int _{B_{2R}}\left \langle A(x, \nabla u), H\right \rangle{\rm d}x \\ && +\int _{B_{2R}}\left \langle |\textbf{f}|^{p-2}\textbf{f}, |\eta\nabla(u-u_{B_{2R}})|^{r-p}\eta\nabla(u-u_{B_{2R}})\right \rangle{\rm d}x \\ && +\int _{B_{2R}}\left \langle |\textbf{f}|^{p-2}\textbf{f}, E \right \rangle{\rm d}x - \int _{B_{2R}}\left \langle |\textbf{f}|^{p-2}\textbf{f}, H \right \rangle{\rm d}x . \end{eqnarray} $

上述结果可表示为$ I_1 = I_2+I_3+I_4+I_5+I_6. $估计$ I_1 $.由条件(H$ _1) $和(H$ _3) $, 得

(4.12) $ \begin{eqnarray} I_1 \geq \int _{B_{R}}\left \langle A(x, \nabla u), |\nabla u|^{r-p}\nabla u \right \rangle{\rm d}x \geq C_1\int _{B_{R}}|\nabla u|^{r}{\rm d}x. \end{eqnarray} $

估计$ I_2 $.由条件(H$ _2) $, $ (4.8) $式, Young不等式$ (\tau >0) $以及$ (4.2) $式可得

(4.13) $ \begin{eqnarray} I_2 &\leq& \int _{B_{2R}} |A(x, \nabla u)||E|{\rm d}x \\ &\leq& C_2C(n, p, r)\int _{B_{2R}}|\nabla u|^{p-1}|(u-u_{B_{2R}})\nabla \eta|^{r-p+1}{\rm d}x \\ &\leq& C(n, p, r, C_2)\Big(\tau\int _{B_{2R}}|\nabla u|^{r}{\rm d}x + C{(\tau)}\int _{B_{2R}}|(u-u_{B_{2R}})\nabla \eta|^{r}{\rm d}x\Big) \\ &\leq& C(n, p, r, C_2)\Big(\tau\int _{B_{2R}}|\nabla u|^{r}{\rm d}x + \frac{C(\tau)}{R^{r}}\int _{B_{2R}}|(u-u_{B_{2R}}|^{r}{\rm d}x\Big). \end{eqnarray} $

估计$ I_3 $.由条件(H$ _2) $, Hölder不等式以及$ (4.5) $式有

(4.14) $ \begin{eqnarray} I_3 &\leq & \int _{B_{2R}} |A(x, \nabla u)||H|{\rm d}x \leq C_2\int _{B_{2R}}|\nabla u|^{p-1}|H|{\rm d}x {} \\ &\leq& C_2\left(\int _{B_{2R}}|\nabla u|^{r}{\rm d}x\right)^{\frac{p-1}{r}} \left(\int _{B_{2R}}|H|^{\frac{r}{r-p+1}}\right)^{\frac{r-p+1}{r}} {} \\ &\leq & C_2C(n, p)(p-r)\left(\int _{B_{2R}}|\nabla u|^{r}{\rm d}x\right)^{\frac{p-1}{r}} \left(\int _{B_{2R}}|\nabla[\eta(u-u_{B_{2R}})]|^{r}{\rm d}x\right)^{\frac{r-p+1}{r}}. \end{eqnarray} $

因为$ \nabla[\eta(u-u_{B_{2R}}\;)] = \eta\nabla(u-u_{B_{2R}}\;)+(u-u_{B_{2R}}\;)\nabla\eta $, 由Young不等式$ (\tau >0) $以及$ (4.2) $式, 上面不等式即为

(4.15) $ \begin{eqnarray} I_3 &\leq& C_2C(n, p)(p-r)\left(\int _{B_{2R}}|\nabla u|^{r}{\rm d}x\right)^{\frac{p-1}{r}}\left(\int _{B_{2R}}|\eta\nabla u|^{r}{\rm d}x\right)^{\frac{r-p+1}{r}} \\ && +C_2C(n, p)(p-r)\left(\int _{B_{2R}}|\nabla u|^{r}{\rm d}x\right)^{\frac{p-1}{r}} \left(\int _{B_{2R}}|(u-u_{B_{2R}})\nabla\eta|^{r}{\rm d}x\right)^{\frac{r-p+1}{r}} \\ &\leq& C(n, p, C_2)(p-r)\int _{B_{2R}}|\nabla u|^{r}{\rm d}x +C(n, p, C_2)(p-r)\tau\int _{B_{2R}}|\nabla u|^{r}{\rm d}x \\ && +C(n, p, C_2, \tau)(p-r)\int _{B_{2R}}|(u-u_{B_{2R}})\nabla\eta|^{r}{\rm d}x \\ &\leq& C(n, p, C_2)(p-r)(1+\tau)\int _{B_{2R}}|\nabla u|^{r}{\rm d}x +\frac{C(n, p, C_2, \tau)(p-r)}{R^{r}}\int _{B_{2R}}|(u-u_{B_{2R}})|^{r}{\rm d}x. {}\\ \end{eqnarray} $

估计$ I_4 $.由Young不等式$ (\tau >0) $, 有

(4.16) $ \begin{equation} I_4\leq \int _{B_{2R}} |\textbf{f}|^{p-1}|\nabla u|^{r-p+1}{\rm d}x \leq \tau\int _{B_{2R}}|\nabla u|^{r}{\rm d}x+C(\tau)\int _{B_{2R}}|\textbf{f}|^{r}{\rm d}x. \end{equation} $

估计$ I_5 $.由Young不等式$ (\tau >0) $以及$ (4.8) $式, 有

(4.17) $ \begin{eqnarray} I_5 &\leq& \int _{B_{2R}} |\textbf{f}|^{p-1}|E|{\rm d}x \leq C(n, p, r)\int _{B_{2R}}|\textbf{f}|^{p-1}|(u-u_{B_{2R}})\nabla\eta|^{r-p+1}{\rm d}x \\ &\leq& C(n, p, r)\left(\tau\int _{B_{2R}}|(u-u_{B_{2R}})\nabla\eta|^{r}{\rm d}x+C(\tau)\int _{B_{2R}}|\textbf{f}|^{r}{\rm d}x\right) \\ &\leq& C(n, p, r)\left(\frac{\tau}{R^{r}}\int _{B_{2R}}|(u-u_{B_{2R}})|^{r}{\rm d}x+C(\tau)\int _{B_{2R}}|\textbf{f}|^{r}{\rm d}x\right). \end{eqnarray} $

估计$ I_6 $.由Hölder不等式, $ (4.5) $式, Young不等式$ (\tau >0) $以及$ (4.2) $式, 有

(4.18) $ \begin{eqnarray} I_6 &\leq& \int _{B_{2R}} |\textbf{f}|^{p-1}|H|{\rm d}x \leq C(n, p)\left(\int _{B_{2R}}|\textbf{f}|^{r}{\rm d}x\right)^{\frac{p-1}{r}}\left(\int _{B_{2R}}|H|^{\frac{r}{r-p+1}}{\rm d}x\right)^{\frac{r-p+1}{r}} \\ &\leq& C(n, p)(p-r)\left(\int _{B_{2R}}|\textbf{f}|^{r}{\rm d}x\right)^{\frac{p-1}{r}}\left(\int _{B_{2R}}|\nabla\eta(u-u_{B_{2R}})|^{r}{\rm d}x\right)^{\frac{r-p+1}{r}} \\ &\leq& C(n, p, r)(p-r)\left(\int _{B_{2R}}|\textbf{f}|^{r}{\rm d}x\right)^{\frac{p-1}{r}}\left(\int _{B_{2R}}|\eta\nabla u|^{r}{\rm d}x\right)^{\frac{r-p+1}{r}} \\ && +C(n, p, r)(p-r)\left(\int _{B_{2R}}|\textbf{f}|^{r}{\rm d}x\right)^{\frac{p-1}{r}}\left(\int _{B_{2R}}|(u-u_{B_{2R}})\nabla \eta|^{r}{\rm d}x\right)^{\frac{r-p+1}{r}} \\ &\leq& C(n, p, r)(p-r)\tau \int _{B_{2R}}|\nabla u|^{r}{\rm d}x+C(n, p, r, \tau)(p-r)\int _{B_{2R}}|\textbf{f}|^{r}{\rm d}x \\ && +C(n, p, r)(p-r)\tau \int _{B_{2R}}|(u-u_{B_{2R}})\nabla \eta|^{r}{\rm d}x \\ &\leq& C(n, p, r)(p-r)\tau \int _{B_{2R}}|\nabla u|^{r}{\rm d}x+C(n, p, r, \tau)(p-r)\int _{B_{2R}}|\textbf{f}|^{r}{\rm d}x \\ && +\frac{C(n, p, r)(p-r)\tau}{R^{r}}\int _{B_{2R}}|(u-u_{B_{2R}})|^{r}{\rm d}x. \end{eqnarray} $

结合$ I_i(1\leq i\leq 6) $得到

(4.19) $ \begin{eqnarray} C_1\int _{B_{R}}|\nabla u|^{r}{\rm d}x &\leq& [C\tau +C(p-r)(1+\tau)+\tau+C(p-r)\tau]\int _{B_{2R}}|\nabla u|^{r}{\rm d}x \\ && +\frac{C\tau +C(p-r)+C\tau+C(p-r)\tau}{R^{r}}\int _{B_{2R}}|u-u_{B_{2R}}|^{r}{\rm d}x \\ && +[C\tau +C(p-r)+C(\tau)]\int _{B_{2R}}|\textbf{f}|^{r}{\rm d}x, \end{eqnarray} $

其中$ C = C(n, p, r, C_2, \tau) $.选择足够小的$ \tau $和$ p-r $使得上述不等式右边第一项系数的总和$ C\tau +C(p-r)(1+\tau)+\tau+C(p-r)\tau $远小于1.则上述不等式通过迭代法变为

(4.20) $ \begin{equation} \int _{B_{R}}|\nabla u|^{r}{\rm d}x\leq\frac{C}{R^{r}}\int _{B_{2R}}|u-u_{B_{2R}}|^{r}{\rm d}x+C\int _{B_{2R}}|\textbf{f}|^{r}{\rm d}x, \end{equation} $

其中$ C = C(n, p, C_1, C_2) $.定理$ 4.1 $得证.

令$ v $是下列边值问题的很弱解

(4.21) $ \begin{eqnarray} \left\{\begin{array}{ll} {\rm div} A(x^{*}, \nabla v) = 0, \qquad &x\in B_{\widetilde{R}}, \\ v = u, \qquad \qquad \; \, &x\in \partial B_{\widetilde{R}}, \end{array}\right. \end{eqnarray} $

其中$ x^{*} \in B_{\widetilde{R}} $为定点, 且$ B_{\widetilde{R}} = B_{10\rho_{i}}(x_{i}) $.

下面给出很弱解的定义.

定义4.1  假设$ v \in W^{1, r}(B_{\widetilde{R}}) $, $ v-u \in W_{0}^{1, r}(B_{\widetilde{R}}) $, 称$ v \in W^{1, r}(B_{\widetilde{R}}) $是Dirichlet问题$ (4.21) $在$ B_{\widetilde{R}} $中的很弱解, 如果有

(4.22) $ \begin{equation} \int _{B_{\widetilde{R}}}\left \langle A(x^{*}, \nabla v), \nabla \varphi \right \rangle{\rm d}x = 0 \end{equation} $

对于任意的$ \varphi \in W_{0}^{1, \frac{r}{r-p+1}}\;(B_{\widetilde{R}} ) $成立.

需要下面的定理.

定理4.2  若$ u \in W_{loc}^{1, r}(\Omega) $是方程$ (1.1) $的局部很弱解, $ v \in W^{1, r}(B_{10\rho _{i}}(x_{i})) $是Dirichlet问题$ (4.21) $在$ B_{10\rho _{i}}(x_{i}) $中的很弱解, 其中$ x_{i} \in E_\lambda(1) $, $ \rho _{i} $与在引理$ 3.2 $中的定义相同, 则有

(4.23) $ \begin{equation} -\!\!\!\!\!\!\int _{B_{10\rho _{i}}(x_{i})}\left | \nabla v \right |^{r}{\rm d}x \leq C -\!\!\!\!\!\!\int _{B_{10\rho _{i}}(x_{i})}\left | \nabla u \right |^{r}{\rm d}x, \end{equation} $

其中$ C = C(n, p, C_1, C_2) $.

证  令$ v \in W^{1, r}(B_{10\rho _{i}}(x_{i})) $是Dirichlet问题$ (4.21) $的很弱解, 令$ \widetilde{\varphi} = v-u $, 则

(4.24) $ \begin{equation} \nabla\widetilde{\varphi} = \nabla v-\nabla u. \end{equation} $

易知$ |\nabla\widetilde{\varphi}|^{r-p}\nabla\widetilde{\varphi}\in L^{\frac{r}{r-p+1}}\;(B_{10\rho _{i}}(x_{i})) $.然后对$ |\nabla\widetilde{\varphi}|^{r-p}\nabla\widetilde{\varphi} $进行Hodge分解(参见文献[18])

(4.25) $ \begin{equation} |\nabla\widetilde{\varphi}|^{r-p}\nabla\widetilde{\varphi} = \nabla\varphi+H, \end{equation} $

其中$ \varphi \in W_{0}^{1, \frac{r}{r-p+1}}\;(B_{10\rho _{i}}(x_{i}) ), H\in L^{\frac{r}{r-p+1}}\;(B_{10\rho _{i}}(x_{i}) ) $是散度为零向量场, 并且有下面的估计成立

(4.26) $ \begin{equation} ||\nabla\varphi||_{\frac{r}{r-p+1}}\leq C||\nabla\widetilde{\varphi}||_{r}^{r-p+1}, \end{equation} $

(4.27) $ \begin{equation} ||H||_{\frac{r}{r-p+1}}\leq C(p-r)||\nabla\widetilde{\varphi}||_{r}^{r-p+1}, \end{equation} $

其中$ C = C(n, p) $是仅依赖于$ n $和$ p $的常数.令

(4.28) $ \begin{equation} E(u, v) = |\nabla(v-u)|^{r-p}\nabla(v-u)-|\nabla v|^{r-p}\nabla v, \end{equation} $

然后由基本不等式$ (4.7) $可得

(4.29) $ \begin{equation} |E(u, v)|\leq2^{p-r}\frac{p-r+1}{r-p+1}|\nabla u|^{r-p+1}, \end{equation} $

因为

(4.30) $ \begin{equation} \nabla\varphi = |\nabla\widetilde{\varphi}|^{r-p}\nabla\widetilde{\varphi}-H = E-H+|\nabla v|^{r-p}\nabla v, \end{equation} $

选择$ \varphi\in W_{0}^{1, \frac{r}{r-p+1}}\;(\Omega) $作为定义$ 4.1 $的检验函数, 则

(4.31) $ \begin{eqnarray} \int _{B_{10\rho _{i}}(x_{i})}\left \langle A(x^{*}, \nabla v), |\nabla v|^{r-p}\nabla v\right \rangle{\rm d}x & = & -\int _{B_{10\rho _{i}}(x_{i})}\left \langle A(x^{*}, \nabla v), E-H\right \rangle{\rm d}x. \end{eqnarray} $

估计上述等式的左右两边.首先由条件(H$ _1) $和(H$ _3) $有

(4.32) $ \begin{equation} \int _{B_{10\rho _{i}}(x_{i})}\left \langle A(x^{*}, \nabla v), |\nabla v|^{r-p}\nabla v\right \rangle{\rm d}x\geq C_1\int _{B_{10\rho _{i}}(x_{i})}|\nabla v|^{r}{\rm d}x, \end{equation} $

再由条件(H$ _2) $和$ (4.29) $式得

(4.33) $ \begin{eqnarray} && -\int _{B_{10\rho _{i}}(x_{i})}\left \langle A(x^{*}, \nabla v), E-H\right \rangle{\rm d}x \\ &\leq& \int _{B_{10\rho _{i}}(x_{i})}|A(x^{*}, \nabla v)||E|{\rm d}x+ \int _{B_{10\rho _{i}}(x_{i})}|A(x^{*}, \nabla v)||H|{\rm d}x \\ &\leq& C_2\int _{B_{10\rho _{i}}(x_{i})}|\nabla v|^{p-1}|E|{\rm d}x+ C_2\int _{B_{10\rho _{i}}(x_{i})}|\nabla v|^{p-1}|H|{\rm d}x \\ &\leq& C_2C(n, p, r)\int _{B_{10\rho _{i}}(x_{i})}|\nabla v|^{p-1}|\nabla u|^{r-p+1}{\rm d}x +C_2\int _{B_{10\rho _{i}}(x_{i})}|\nabla v|^{p-1}|H|{\rm d}x, \end{eqnarray} $

然后利用Hölder不等式$ (\frac{p-1}{r}+\frac{r-p+1}{r} = 1) $, (4.24)式, (4.27)式以及Young不等式$ (\tau >0) $, 上述不等式即为

(4.34) $ \begin{eqnarray} && -\int _{B_{10\rho _{i}}(x_{i})}\left \langle A(x^{*}, \nabla v), E-H\right \rangle{\rm d}x \\ &\leq& C_2C(n, p, r)\left(\int _{B_{10\rho _{i}}(x_{i})}|\nabla v|^{r}{\rm d}x\right)^{\frac{p-1}{r}}\left(\int _{B_{10\rho _{i}}(x_{i})}|\nabla u|^{r}{\rm d}x\right)^{\frac{r-p+1}{r}} \\ && +C_2\left(\int _{B_{10\rho _{i}}(x_{i})}|\nabla v|^{r}{\rm d}x\right)^{\frac{p-1}{r}}\left(\int _{B_{10\rho _{i}}(x_{i})}|H|^{\frac{r}{r-p+1}}{\rm d}x\right)^{\frac{r-p+1}{r}} \\ &\leq& C(n, p, r, C_2)\left\{\tau\int _{B_{10\rho _{i}}(x_{i})}|\nabla v|^{r}{\rm d}x+ C(\tau)\int _{B_{10\rho _{i}}(x_{i})}|\nabla u|^{r}{\rm d}x\right\} \\ && +C(n, p, C_2)(p-r)\tau\int _{B_{10\rho _{i}}(x_{i})}|\nabla v|^{r}{\rm d}x \\ && +C(n, p, C_2, \tau)(p-r)\int _{B_{10\rho _{i}}(x_{i})}(|\nabla v|^{r}+|\nabla u|^{r}){\rm d}x. \end{eqnarray} $

选择足够小的$ \tau $和$ p-r $使得$ C_1 \geq C\tau+C(p-r)\tau+C(p-r) $, 然后结合(4.31)–(4.34)式有

(4.35) $ \begin{equation} \int _{B_{10\rho _{i}}(x_{i})}|\nabla v|^{r}{\rm d}x \leq C\int _{B_{10\rho _{i}}(x_{i})}|\nabla u|^{r}{\rm d}x, \end{equation} $

其中$ C = C(n, p, C_1, C_2) $.定理$ 4.2 $得证.

定理4.3  若$ u \in W_{loc}^{1, r}(\Omega) $是方程$ (1.1) $的局部很弱解, $ v \in W^{1, r}(B_{10\rho _{i}}(x_{i})) $是Dirichlet问题$ (4.21) $在$ B_{10\rho _{i}}(x_{i}) $中的很弱解, 其中$ x_{i} \in E_\lambda(1) $, $ \rho _{i} $与引理$ 3.2 $中的定义相同, 如果

(4.36) $ \begin{equation} -\!\!\!\!\!\!\int _{B_{10\rho _{i}}(x_{i})}| \nabla u|^{r}{\rm d}x \leq 1, \quad -\!\!\!\!\!\!\int _{B_{10\rho _{i}}(x_{i})}|{\bf f} |^{r}{\rm d}x \leq \varepsilon, \end{equation} $

则有

(4.37) $ \begin{equation} -\!\!\!\!\!\!\int _{B_{10\rho _{i}}(x_{i})}| \nabla (u-v)|^{r}{\rm d}x \leq \varepsilon. \end{equation} $

证  选取检验函数$ \widetilde{\varphi} = u-v $.易知$ |\nabla\widetilde{\varphi}|^{r-p}\nabla\widetilde{\varphi}\in L^{\frac{r}{r-p+1}}\;(B_{10\rho _{i}}(x_{i})) $.然后对$ |\nabla\widetilde{\varphi}|^{r-p}\nabla\widetilde{\varphi} $进行Hodge分解

(4.38) $ \begin{equation} |\nabla(u-v)|^{r-p}\nabla(u-v) = \nabla\varphi+H, \end{equation} $

其中$ \varphi \in W_{0}^{1, \frac{r}{r-p+1}}\;(B_{10\rho _{i}}(x_{i}) ), H\in L^{\frac{r}{r-p+1}}\;(B_{10\rho _{i}}(x_{i}) ) $是散度为零向量场, 并且有下面的估计成立

(4.39) $ \begin{equation} ||H||_{\frac{r}{r-p+1}}\leq C(p-r)||\nabla(u-v)||_{r}^{r-p+1}, \end{equation} $

其中$ C = C(n, p) $是仅依赖于$ n $和$ p $的常数.

选择$ \varphi \in W_{0}^{1, \frac{r}{r-p+1}}\;(B_{10\rho _{i}}(x_{i}) ) $作为定义$ 1.1 $和定义$ 4.1 $中的检验函数, 有

(4.40) $ \begin{equation} \int _{B_{10\rho _{i}}(x_{i})}\left \langle A(x, \nabla u) , \nabla \varphi\right \rangle{\rm d}x = \int _{B_{10\rho _{i}}(x_{i})}\left \langle |\textbf{f} |^{p-2}\textbf{f} , \nabla \varphi\right \rangle{\rm d}x. \end{equation} $

(4.41) $ \begin{equation} \int _{B_{10\rho _{i}}(x_{i})}\left \langle A(x^{*}, \nabla v) , \nabla \varphi)\right \rangle{\rm d}x = 0. \end{equation} $

其中$ x^{*} \in B_{10\rho _{i}}(x_{i}) $为定点.然后直接计算得到结果为

(4.42) $ \begin{eqnarray} && \int _{B_{10\rho _{i}}(x_{i})}\left \langle A(x, \nabla u)-A(x, \nabla v), \nabla \varphi\right \rangle{\rm d}x \\ & = & -\int _{B_{10\rho _{i}}(x_{i})}\left \langle A(x, \nabla v)-A(x^{*}, \nabla v), \nabla \varphi\right \rangle{\rm d}x +\int _{B_{10\rho _{i}}(x_{i})}\left \langle |\textbf{f} |^{p-2}\textbf{f} , \nabla \varphi\right \rangle{\rm d}x. \end{eqnarray} $

根据$ (4.38) $式有

(4.43) $ \begin{equation} K_1 = K_2+K_3+K_4+K_5+K_6, \end{equation} $

其中

(4.44) $ \begin{eqnarray} &&K_1 = \int _{B_{10\rho _{i}}(x_{i})}\left \langle A(x, \nabla u)-A(x, \nabla v), |\nabla(u-v)|^{r-p}\nabla(u-v)\right \rangle{\rm d}x, \\ && K_2 = \int _{B_{10\rho _{i}}(x_{i})}\left \langle A(x, \nabla u)-A(x, \nabla v), H\right \rangle{\rm d}x, \\ && K_3 = -\int _{B_{10\rho _{i}}(x_{i})}\left \langle A(x, \nabla v)-A(x^{*}, \nabla v), |\nabla(u-v)|^{r-p}\nabla(u-v)\right \rangle{\rm d}x, \\ && K_4 = \int _{B_{10\rho _{i}}(x_{i})}\left \langle A(x, \nabla v)-A(x^{*}, \nabla v), H\right \rangle{\rm d}x, \\ && K_5 = \int _{B_{10\rho _{i}}(x_{i})}\left \langle |\textbf{f} |^{p-2}\textbf{f}, |\nabla(u-v)|^{r-p}\nabla(u-v)\right \rangle{\rm d}x, \\ && K_6 = -\int _{B_{10\rho _{i}}(x_{i})}\left \langle |\textbf{f} |^{p-2}\textbf{f}, H\right \rangle{\rm d}x. \end{eqnarray} $

先估计$ K_1 $.利用条件(H$ _1) $得

(4.45) $ \begin{eqnarray} K_1 &\geq& C_1\int _{B_{10\rho _{i}}(x_{i})}|\nabla(u-v)|^{r}{\rm d}x. \end{eqnarray} $

估计$ K_2 $.由条件(H$ _2) $, Hölder不等式, $ (4.39) $式以及Young不等式$ (\tau >0) $得

(4.46) $ \begin{eqnarray} K_2 &\leq& \int _{B_{10\rho _{i}}(x_{i})}(|A(x, \nabla u)|+|A(x, \nabla v)|)|H|{\rm d}x \\ &\leq& C_2\int _{B_{10\rho _{i}}(x_{i})}\Big(|\nabla u|^{p-1}+|\nabla v|^{p-1}\Big)|H|{\rm d}x \\ &\leq& C_2\left(\int _{B_{10\rho _{i}}(x_{i})}(|\nabla u|^{r}+|\nabla v|^{r}){\rm d}x\right)^{\frac{p-1}{r}}\left(\int _{B_{10\rho _{i}}(x_{i})}|H|^{\frac{r}{r-p+1}}{\rm d}x\right)^{\frac{r-p+1}{r}} \\ &\leq& C(n, p, C_2)(p-r)\left(\int _{B_{10\rho _{i}}(x_{i})}(|\nabla u|^{r}+|\nabla v|^{r}){\rm d}x\right)^{\frac{p-1}{r}} \left(\int _{B_{10\rho _{i}}(x_{i})}|\nabla(u-v)|^{r}{\rm d}x\right)^{\frac{r-p+1}{r}} \\ &\leq& C(n, p, C_2)(p-r)\left(\int _{B_{10\rho _{i}}(x_{i})}(|\nabla (u-v)|^{r}+|\nabla v|^{r}){\rm d}x\right)^{\frac{p-1}{r}} \\ && \cdot\left(\int _{B_{10\rho _{i}}(x_{i})}|\nabla(u-v)|^{r}{\rm d}x\right)^{\frac{r-p+1}{r}} \\ &\leq& C(n, p, \tau , C_2)(p-r)\left(\int _{B_{10\rho _{i}}(x_{i})}|\nabla (u-v)|^{r}{\rm d}x+\int _{B_{10\rho _{i}}(x_{i})}|\nabla v|^{r}{\rm d}x\right) \\ && +C(n, p)(p-r)\tau\int _{B_{10\rho _{i}}(x_{i})}|\nabla(u-v)|^{r}{\rm d}x. \end{eqnarray} $

估计$ K_{3}. $由条件(H$ _4) $, (3.10)式, Young不等式$ (\tau >0) $以及$ \rho _{i} \in (0, R_{0}/10] $得

(4.47) $ \begin{eqnarray} K_3 &\leq& C_3\omega(|x-x^{*}|)\int _{B_{10\rho _{i}}(x_{i})}|\nabla v|^{p-1}|\nabla(u-v)|^{r-p+1}{\rm d}x \\ &\leq& C_3\omega(R_0)\int _{B_{10\rho _{i}}(x_{i})}|\nabla v|^{p-1}|\nabla(u-v)|^{r-p+1}{\rm d}x \\ &\leq& C_3\varepsilon\left\{C(\tau)\int _{B_{10\rho _{i}}(x_{i})}|\nabla v|^{r}{\rm d}x+\tau\int _{B_{10\rho _{i}}(x_{i})}|\nabla(u-v)|^{r}{\rm d}x\right\}. \end{eqnarray} $

估计$ K_{4}. $由条件(H$ _4) $, Hölder不等式, (4.39)式, (3.10)式, Young不等式$ (\tau >0) $以及$ \rho _{i} \in (0, R_{0}/10] $可得

(4.48) $ \begin{eqnarray} K_4 &\leq& C_3\omega(|x-x^{*}|)\int _{B_{10\rho _{i}}(x_{i})}|\nabla v|^{p-1}|H|{\rm d}x \\ &\leq& C_3\omega(R_0)\left(\int _{B_{10\rho _{i}}(x_{i})}|\nabla v|^{r}{\rm d}x\right)^{\frac{p-1}{r}}\left(\int _{B_{10\rho _{i}}(x_{i})}|H|^{\frac{r}{r-p+1}}{\rm d}x\right)^{\frac{r-p+1}{r}} \\ &\leq& C_3C(n, p)(p-r)\varepsilon\left(\int _{B_{10\rho _{i}}(x_{i})}|\nabla v|^{r}{\rm d}x\right)^{\frac{p-1}{r}} \left(\int _{B_{10\rho _{i}}(x_{i})}|\nabla(u-v)|^{r}{\rm d}x\right)^{\frac{r-p+1}{r}} \\ &\leq& C_3C(n, p)(p-r)\varepsilon \Bigg\{C(\tau)\int _{B_{10\rho _{i}}(x_{i})}|\nabla v|^{r}{\rm d}x +\tau\int _{B_{10\rho _{i}}(x_{i})}|\nabla(u-v)|^{r}{\rm d}x \Bigg\}. \end{eqnarray} $

估计$ K_{5}. $由Young不等式$ (\tau >0) $得

(4.49) $ \begin{eqnarray} K_5 &\leq& \int _{B_{10\rho _{i}}(x_{i})} |\textbf{f} |^{p-1}|\nabla(u-v)|^{r-p+1}{\rm d}x \\ &\leq& C(\tau)\int _{B_{10\rho _{i}}(x_{i})} |\textbf{f} |^{r}{\rm d}x+\tau\int _{B_{10\rho _{i}}(x_{i})}|\nabla(u-v)|^{r}{\rm d}x . \end{eqnarray} $

估计$ K_{6}. $由Hölder不等式, $ (4.39) $式以及Young不等式$ (\tau >0) $得

(4.50) $ \begin{eqnarray} K_6 &\leq& \int _{B_{10\rho _{i}}(x_{i})} |\textbf{f} |^{p-1}|H|{\rm d}x \\ &\leq& \left(\int _{B_{10\rho _{i}}(x_{i})} |\textbf{f} |^{r}{\rm d}x\right)^{\frac{p-1}{r}} \left(\int _{B_{10\rho _{i}}(x_{i})} |H|^{\frac{r}{r-p+1}}{\rm d}x\right)^{\frac{r-p+1}{r}} \\ &\leq& C(n, p)(p-r)\left(\int _{B_{10\rho _{i}}(x_{i})} |\textbf{f} |^{r}{\rm d}x\right)^{\frac{p-1}{r}} \left(\int _{B_{10\rho _{i}}(x_{i})} |\nabla(u-v)|^{r}{\rm d}x\right)^{\frac{r-p+1}{r}} \\ &\leq& C(n, p)(p-r)\Bigg\{C(\tau)\int _{B_{10\rho _{i}}(x_{i})} |\textbf{f} |^{r}{\rm d}x +\tau\int _{B_{10\rho _{i}}(x_{i})} |\nabla(u-v)|^{r}{\rm d}x\Bigg\}. \end{eqnarray} $

综合估计式$ K_i(1 \leq i\leq 6) $, 且选取足够小的$ \tau $和$ (p-r) $, 使得

$ C_1>C(n, p)(p-r)\tau+C(n, p, \tau)(p-r)+C_3\varepsilon\tau+C_3C(n, p)(p-r)\varepsilon\tau+\tau+C(n, p)(p-r)\tau, $

则有

(4.51) $ \begin{eqnarray} \int _{B_{10\rho _{i}}(x_{i})} |\nabla(u-v)|^{r}{\rm d}x &\leq& C\Bigg\{\varepsilon\int _{B_{10\rho _{i}}(x_{i})} |\nabla v|^{r}{\rm d}x+ \int _{B_{10\rho _{i}}(x_{i})}|\textbf{f} |^{r}{\rm d}x\Bigg\}, \end{eqnarray} $

即

(4.52) $ \begin{equation} -\!\!\!\!\!\!\int _{B_{10\rho _{i}}(x_{i})} |\nabla(u-v)|^{r}{\rm d}x\leq C\Bigg\{\varepsilon-\!\!\!\!\!\!\int _{B_{10\rho _{i}}(x_{i})} |\nabla v|^{r}{\rm d}x+ -\!\!\!\!\!\!\int _{B_{10\rho _{i}}(x_{i})}|\textbf{f} |^{r}{\rm d}x\Bigg\}. \end{equation} $

然后根据定理$ 4.2 $得

(4.53) $ \begin{equation} -\!\!\!\!\!\!\int _{B_{10\rho _{i}}(x_{i})} |\nabla(u-v)|^{r}{\rm d}x\leq C\Bigg\{\varepsilon-\!\!\!\!\!\!\int _{B_{10\rho _{i}}(x_{i})} |\nabla u|^{r}{\rm d}x+ -\!\!\!\!\!\!\int _{B_{10\rho _{i}}(x_{i})}|\textbf{f} |^{r}{\rm d}x\Bigg\}, \end{equation} $

由$ (4.36) $式得

(4.54) $ \begin{equation} -\!\!\!\!\!\!\int _{B_{10\rho _{i}}(x_{i})} |\nabla(u-v)|^{r}{\rm d}x\leq C\varepsilon, \end{equation} $

其中$ C = C(n, p, C_1, C_2, C_3) $.定理$ 4.3 $证毕.

接下来, 给出如下Dirichlet问题$ (4.21) $很弱解$ v $的有界性引理.

引理4.1^[20]  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

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

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 = c(\alpha, \theta) $, 使得对于每个$ \rho , R, R_{0}\leq \rho < R\leq R_{1} $, 有

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

定义4.2^[21]  函数$ u \in W_{loc}^{1, m}(\Omega) $属于$ B(\Omega, \gamma, m, k_0) $类, 若对于每一个$ k>k_0, k_0>0 $和$ B_{\rho} = B_{\rho}(x_0), B_{\rho-\rho\sigma} = B_{\rho-\rho\sigma}(x_0), B_R = B_R(x_0) $, 有

$ \int_{A_{k, \rho-\rho\sigma}^{+}}|\nabla u|^{m}{\rm d}x\leq\gamma\left\{\sigma^{-m}\rho^{-m} \int_{A_{k, \rho}^{+}}(u-k)^{m}{\rm d}x+|A_{k, \rho}^{+}|\right\}, $

其中$ R/2\leq\rho-\rho\sigma<\rho<R, m<n, |A_{k, \rho}^{+}| $是集合$ A_{k, \rho}^{+} $的$ n $维勒贝格测度.

引理4.2^[21]  假设$ u(x) $是属于$ B(\Omega, \gamma, m, k_0) $类的任意函数, 且$ B_R\subset\subset\Omega $.则有

$ \begin{eqnarray*} \max\limits_{ B_{R/2}}u(x)\leq c, \end{eqnarray*} $

其中常数$ c $仅由$ \gamma, m, k_0, R, ||\nabla u||_{m} $确定.

定理4.4  假设$ v \in W^{1, r}(B_{\widetilde{R}}) $是方程$ (4.21) $的很弱解, 则$ v $局部有界.

证  令$ v $是Dirichlet问题$ (4.21) $的很弱解.由于$ v-u \in W_{0}^{1, r}(B_{\widetilde{R}}) $, 则在$ {\Omega} \backslash B_{\widetilde{R}} $上$ v = u $.令$ B_{2\widetilde R}\subset\subset\Omega, 0<\widetilde R\leq \widetilde\tau <t\leq2\widetilde R $.考虑函数$ v(x) $和$ k>0 $有

$ \begin{eqnarray*} A_k = \{x\in\Omega:|v(x)|>k\}, {\quad} A_k^{+} = \{x\in\Omega:v(x)>k\}, \end{eqnarray*} $

$ \begin{eqnarray*} A_{k, t} = A_k\cap B_t, {\quad} A_{k, t}^{+} = A_k^{+}\cap B_t. \end{eqnarray*} $

选取截断函数$ \eta \in C_0^{\infty}(B_{2\widetilde R}) $使得

(4.55) $ \begin{equation} {\rm supp}\eta \subset B_t, 0\leq \eta \leq 1, \eta \equiv 1 \; {\rm in} \;B_{\tilde{\tau}}, \left | \nabla \eta \right | \leq 2(t-\widetilde\tau)^{-1}. \end{equation} $

考虑函数

(4.56) $ \begin{equation} \phi = \eta^{r}(v-t_k(v)), \end{equation} $

其中

(4.57) $ \begin{equation} t_k(v) = \min\left\{v, k\right\}, \quad k \geq 0. \end{equation} $

易知$ |\nabla\phi|^{r-p}\nabla\phi \in L^{\frac{r}{r-p+1}}\;(B_{2\widetilde R}) $.对$ |\nabla\phi|^{r-p}\nabla\phi $进行Hodge分解

(4.58) $ \begin{equation} |\nabla[\eta^{r}(v-t_k(v))]|^{r-p}\nabla[\eta^{r}(v-t_k(v))] = \nabla\varphi+H, \end{equation} $

其中$ \varphi\in W_0^{1, \frac{r}{r-p+1}}\;(B_{2\widetilde R}), H\in L^{\frac{r}{r-p+1}}\;(B_{2\widetilde R}) $是散度为零向量场, 且有以下估计成立

(4.59) $ \begin{equation} ||H||_{\frac{r}{r-p+1}}\leq C(p-r)||\nabla[\eta^{r}(v-t_k(v))]||_r^{r-p+1}, \end{equation} $

其中$ C = C(n, p) $是只依赖于$ n $和$ p $的常数.

选取$ \varphi \in W_0^{1, \frac{r}{r-p+1}}\;(B_{2\widetilde R}) $作为定义$ 4.1 $中的检验函数, 则有

(4.60) $ \begin{equation} \int_{B_{2\widetilde R}}\langle A(x^*, \nabla v), |\nabla[\eta^{r}(v-t_k(v))]|^{r-p}\nabla[\eta^{r}(v-t_k(v))]-H\rangle{\rm d}x\geq 0, \end{equation} $

则有

(4.61) $ \begin{eqnarray} && \int_{A_{k, t}^{+}}\langle A(x^*, \nabla v), H\rangle{\rm d}x \\ &\leq & \int_{B_{2\widetilde R}}\langle A(x^*, \nabla v), H \rangle{\rm d}x \\ &\leq & \int_{B_{2\widetilde R}}\langle A(x^*, \nabla v), |\nabla[\eta^{r}(v-t_k(v))]|^{r-p}\nabla[\eta^{r}(v-t_k(v))] \rangle{\rm d}x \\ & = & \Big(\int_{B_{2\widetilde R} \cap \{v\leq k\}}+\int_{B_{2\widetilde R} \cap \{v> k\}} \Big)\langle A(x^*, \nabla v), |\nabla[\eta^{r}(v-t_k(v))]|^{r-p}\nabla[\eta^{r}(v-t_k(v))] \rangle{\rm d}x \\ & = & \int_{B_{2\widetilde R} \cap \{v> k\}} \langle A(x^*, \nabla v), |\nabla[\eta^{r}(v-t_k(v))]|^{r-p}\nabla[\eta^{r}(v-t_k(v))] \rangle{\rm d}x \\ & = & \int_{A_{k, t}^{+}} \langle A(x^*, \nabla v), |\nabla[\eta^{r}(v-t_k(v))]|^{r-p}\nabla[\eta^{r}(v-t_k(v))] \rangle{\rm d}x. \end{eqnarray} $

令

(4.62) $ \begin{equation} E(\eta, v) = |\nabla[\eta^{r}(v-t_k(v))]|^{r-p}\nabla[\eta^{r}(v-t_k(v))]-|\eta^{r}\nabla v|^{r-p}\eta^{r}\nabla v. \end{equation} $

由基本不等式$ (4.7) $和

(4.63) $ \begin{equation} \nabla[\eta^{r}(v-t_k(v))] = \eta^{r}(\nabla v-\nabla t_k(v))+r\eta^{r-1}(v-t_k(v))\nabla \eta, \end{equation} $

可得

(4.64) $ \begin{equation} |E(\eta, v)|\leq2^{p-r}\frac{p-r+1}{r-p+1}|\eta^{r}\nabla t_k(v)-r\eta^{r-1}(v-t_k(v))\nabla \eta|^{r-p+1}. \end{equation} $

由$ E(\eta, v) $的定义和$ (4.61) $式得

(4.65) $ \begin{eqnarray} &&\int_{A_{k, t}^{+}} \langle A(x^*, \nabla v), |\eta^{r}\nabla v|^{r-p} \eta^{r}\nabla v\rangle{\rm d}x \\ & = & \int_{A_{k, t}^{+}} \langle A(x^*, \nabla v), E(\eta, v) \rangle{\rm d}x -\int_{A_{k, t}^{+}} \langle A(x^*, \nabla v), |\nabla[\eta^{r}(v-t_k(v))]|^{r-p}\nabla[\eta^{r}(v-t_k(v))]\rangle{\rm d}x \\ &\leq& \int_{A_{k, t}^{+}} \langle A(x^*, \nabla v), E(\eta, v) \rangle{\rm d}x - \int_{A_{k, t}^{+}} \langle A(x^*, \nabla v), H \rangle{\rm d}x \\ & = & I_1+I_2. \end{eqnarray} $

下面估计$ (4.65) $式的左右两边.先由条件$ (H_1) $和$ (H_3) $得出

(4.66) $ \begin{eqnarray} \int_{A_{k, t}^{+}} \langle A(x^*, \nabla v), |\eta^{r}\nabla v|^{r-p} \eta^{r}\nabla v\rangle{\rm d}x &\geq& \int_{A_{k, \widetilde \tau}^{+}} \langle A(x^*, \nabla v), |\nabla v|^{r-p} \nabla v\rangle{\rm d}x \\ &\geq& C_1\int_{A_{k, \widetilde \tau}^{+}}|\nabla v|^{r}{\rm d}x, \end{eqnarray} $

再由条件$ (H_2) $和$ (4.64) $式得出

(4.67) $ \begin{eqnarray} |I_1|& = & \left|\int_{A_{k, t}} \langle A(x, \nabla v), E(\eta, v) \rangle{\rm d}x\right| \leq C_{2}\int_{A_{k, t}^{+}} |\nabla v|^{p-1}|E(v, u)|{\rm d}x \\ &\leq& C_{2} 2^{p-r}\frac{p-r+1}{r-p+1}\int_{A_{k, t}^{+}}|\nabla v|^{p-1}|\eta^{r}\nabla t_k(v)-r\eta^{r-1}\nabla \eta(v-t_k(v))|^{r-p+1} {\rm d}x \\ &\leq& C\int_{A_{k, t}^{+}} |\nabla v|^{p-1} |\eta^{r}\nabla t_k(v)|^{r-p+1}{\rm d}x{}\\ && + C\int_{A_{k, t}^{+}} |\nabla v|^{p-1}|r\eta^{r-1}(v-t_k(v))\nabla \eta|^{r-p+1}{\rm d}x, \end{eqnarray} $

其中$ C = C_{2} 2^{p-r}\frac{p-r+1}{r-p+1} $.因为在$ {A_{k, t}^{+}} $上$ t_k(v) = k $, 利用Young不等式$ (\tau>0) $得到

(4.68) $ \begin{eqnarray} |I_1|&\leq& C\int_{A_{k, t}^{+}} |\nabla v|^{p-1}|r\eta^{r-1}\nabla \eta(v-t_k(v))|^{r-p+1}{\rm d}x \\ &\leq& C\left[\tau\int_{A_{k, t}^{+}} |\nabla v|^{r}{\rm d}x+C(\tau)\int_{A_{k, t}^{+}}|r\eta^{r-1}(v-t_k(v))\nabla \eta|^{r}{\rm d}x\right] , \end{eqnarray} $

根据$ |\nabla\eta|\leq2(t-\widetilde \tau)^{-1} $以及在$ {A_{k, t}^{+}} $上$ |v-t_k(v)| = |v-k| $, 可得

(4.69) $ \begin{equation} |I_1|\leq C\tau \int_{A_{k, t}^{+}} |\nabla v|^{r}{\rm d}x+\frac{C(p, r, \tau)}{(t-\widetilde \tau)^{r}}\int_{A_{k, t}^{+}}|v-k|^{r}{\rm d}x. \end{equation} $

现估计$ I_2 $, 由条件$ (H_2) $, Hölder不等式, $ (4.59) $式以及Young不等式可得

(4.70) $ \begin{eqnarray} |I_2|& = &\left|\int_{A_{k, t}^{+}} \langle A(x^*, \nabla v), H \rangle{\rm d}x \right| \leq C_2\int_{A_{k, t}^{+}}|\nabla v|^{p-1}|H|{\rm d}x \\ &\leq& C_2\bigg(\int_{A_{k, t}^{+}}|\nabla v|^{r}{\rm d}x\bigg)^{\frac{p-1}{r}} \bigg(\int_{A_{k, t}^{+}}|H|^{\frac{r}{r-p+1}}{\rm d}x\bigg)^{\frac{r-p+1}{r}} \\ &\leq& C(n, p, C_2)(p-r)\bigg(\int_{A_{k, t}^{+}}|\nabla v|^{r}{\rm d}x\bigg)^{\frac{p-1}{r}} \bigg(\int_{A_{k, t}^{+}}|\nabla[\eta^ {r}(v-t_k(v))]|^{r}{\rm d}x\bigg)^{\frac{r-p+1}{r}} \\ &\leq& C(n, p, C_2)(p-r)\tau\int_{A_{k, t}^{+}}|\nabla v|^{r}{\rm d}x + C(n, p, \tau, C_2)(p-r)\int_{A_{k, t}^{+}}|\nabla[\eta^{r}(v-t_k(v))]|^{r}{\rm d}x. {}\\ \end{eqnarray} $

由$ (4.63) $式有

(4.71) $ \begin{equation} \int_{A_{k, t}^{+}}|\nabla[\eta^{r}(v-t_k(v))]|^{r}{\rm d}x\leq \int_{A_{k, t}^{+}}|\nabla v|^{r}{\rm d}x+\frac{2^{r}r}{(t-\widetilde \tau)^{r}}\int_{A_{k, t}^{+}}|v-k|^{r}{\rm d}x, \end{equation} $

即有

(4.72) $ \begin{equation} |I_2|\leq C(p-r) \int_{A_{k, t}^{+}} |\nabla v|^{r}{\rm d}x+\frac{C(p, r, \tau)}{(t-\widetilde \tau)^{r}}\int_{A_{k, t}^{+}}|v-k|^{r}{\rm d}x. \end{equation} $

故由不等式$ (4.65), (4.66), (4.69) $以及$ (4.72) $, 得到

(4.73) $ \begin{equation} \int_{A_{k, \widetilde \tau}^{+}} |\nabla v|^{r}{\rm d}x \leq \frac{C\tau+C(p-r)}{C_1}\int_{A_{k, t}^{+}} |\nabla v|^{r}{\rm d}x+\frac{C(p, r, \tau)}{C_1(t-\widetilde \tau)^{r}} \int_{A_{k, t}^{+}}|v-k|^{r}{\rm d}x. \end{equation} $

选取足够小的$ \tau $和$ p-r $, 使得$ (4.73) $式右边第一项的系数总和$ \theta $小于$ 1 $.因此, 对每一个$ t $和$ \widetilde \tau $, 满足$ \widetilde R\leq \widetilde \tau<t<2\widetilde R $, 有

(4.74) $ \begin{eqnarray} \int_{A_{k, \widetilde \tau}^{+}} |\nabla v|^{r}{\rm d}x &\leq& \theta\int_{A_{k, t}^{+}} |\nabla v|^{r}{\rm d}x+\frac{C(p, r)}{C_1(t-\widetilde \tau)^{r} }\int_{A_{k, R}^{+}}|v-k|^{r}{\rm d}x. \end{eqnarray} $

给定任意的$ \rho, \sigma $, 满足$ \widetilde R\leq \rho <\sigma\leq 2\widetilde R. $利用引理$ 4.1 $可得出

(4.75) $ \begin{eqnarray} \int_{A_{k, \rho}^{+}} |\nabla v|^{r}{\rm d}x &\leq& \frac{cC(p, r)}{C_1(\sigma-\rho)^{r} }\int_{A_{k, \sigma}^{+}}|v-k|^{r}{\rm d}x, \end{eqnarray} $

其中$ c $是引理$ 4.1 $中给定的常数.因此, $ v $属于$ B $类, 且有$ \gamma = cC(p, r)/C_1, m = r $.于是由引理$ 4.2 $可得

$ \max\limits_{ B_{\widetilde R}}v(x)\leq c. $

定理4.4证毕.

引理4.3^[22]  令$ g\in C^{1} $且满足$ 0\leq \frac{tg'(t)}{g(t)}\leq g_0(t>0) $, 假设对某正常数$ \varepsilon $有$ g(t)\geq\varepsilon t $.令$ \overline{A}:{{\Bbb R}} ^{n}\rightarrow {{\Bbb R}} ^{n} $, 假设存在一个正常数$ \Lambda $使得

$ \overline{a^{ij}}(h)\xi_{i}\xi_{j}\geq \frac{g(|h|)}{|h|}|\xi|^{2};\quad |\overline{a^{ij}}(h)|\leq \Lambda \frac{g(|h|)}{|h|};\quad |\overline{A}(h)|\leq \Lambda g(|h|), $

其中$ h, \xi \in {{\Bbb R}}^{n}, \overline{a^{ij}} = \frac{\partial{\overline {A^{i}}}}{\partial{h^{j}}} $.若$ v $是方程$ {\rm div}\overline{A}(Dv) = 0 $在$ W^{1, G}(B_R) $中的有界解, 则$ v \in C^{1, \sigma}(B_R), \sigma(N, g_0, \Lambda) $是正数.又有

$ \sup\limits_{B_{R/2}}G(|Dv|) \leq c(N, g_0, \Lambda)R^{-n}\int_{B_{R}} G(|Dv|){\rm d}x, $

$ -\!\!\!\!\!\!\int _{B{r}}G(|Dv-(Dv)_{r}|){\rm d}x\leq c(N, g_0, \Lambda)(\frac{r}{R})^{\sigma}-\!\!\!\!\!\!\int _{B_{R}}G(|Dv-(Dv)_{R}|){\rm d}x, $

其中$ 0<r<R $.

定理4.5  假设$ v\in W^{1, r}(B_{10\rho _{i}}(x_{i})) $是Dirichlet问题$ (4.21) $在$ B_{10\rho _{i}}(x_{i}) $上的很弱解, 则存在$ N_0>1 $使得

(4.76) $ \begin{equation} \sup\limits_{B_{5\rho _{i}}(x_{i})}|\nabla v| \leq N_0. \end{equation} $

证  令$ v $是方程$ (4.21) $的很弱解.由定理$ 4.4 $可知$ v $是局部有界的.令$ g(t) = rt^{r-1} $.然后根据引理$ 4.3 $, 有

(4.77) $ \begin{equation} G(t) = \int_{0}^{ t }g(s){\rm d}s = \int_{0}^{ t }rs^{r-1}{\rm d}s = t^{r}. \end{equation} $

根据文献[22]中$ W^{1, G} $的定义, $ v(x)\in W^{1, G}(B_{R}) $, 则Dirichlet问题$ (4.21) $的很弱解$ v(x) $满足引理$ 4.3 $的条件.然后根据引理$ 4.3 $和$ G(t) $的定义, 有

(4.78) $ \begin{equation} \sup\limits_{B_{\widetilde R/2}}|\nabla v|^r \leq c\widetilde R^{-n}\int_{B_{\widetilde R}} |\nabla v|^r{\rm d}x \leq c -\!\!\!\!\!\!\int _{B{r}}(1+|\nabla v|^r){\rm d}x \leq N_0. \end{equation} $

根据引理$ 3.2 $, 给定$ \lambda \geq \lambda _{*} = :(10/R_{0}) ^{n/r}\lambda _{0}+1 $, 构造一族互不相交的球$ \left \{ B_{\rho _{i}}(x_{i}) \right \}_{i\in N} $, 其中$ x_i\in E_\lambda(1), i\in N $.则对任意的$ \rho>\rho_i $, 有

(4.79) $ \begin{equation} -\!\!\!\!\!\!\int _{B_{\rho}(x_i)}|\nabla u_{\lambda}|^r{\rm d}x\leq 1, \quad -\!\!\!\!\!\!\int _{B_{\rho}(x_i)}|\textbf{f}_{\lambda}|^r{\rm d}x\leq \varepsilon. \end{equation} $

此外, 根据引理$ 3.1 $的新标准化, 可得出以下定理$ 4.3 $和定理$ 4.5 $的推论.

推论4.1  假设$ v_{\lambda}\in W^{1, r}(B_{10\rho _{i}}(x_{i})) $是方程$ (4.21) $在$ B_{10\rho _{i}}(x_{i}) $上的很弱解, $ A_{\lambda} $满足(H$ _1) $–(H$ _4) $.则存在$ N_0>1 $使得

(4.80) $ \begin{equation} \sup\limits_{B_{5\rho _{i}}(x_{i})}|\nabla v_{\lambda}| \leq N_0. \end{equation} $

推论4.2  假设$ v_{\lambda}\in W^{1, r}(B_{10\rho _{i}}(x_{i})) $是方程$ (4.21) $的很弱解, 其中$ x^{*}\in B_{10\rho _{i}}(x_{i}) $, 且$ A_{\lambda} $满足条件$ (H_1) $–$ (H_4) $.则有

(4.81) $ \begin{equation} -\!\!\!\!\!\!\int _{B_{10\rho _{i}}(x_{i})}|\nabla (u_{\lambda}-v_{\lambda})|^r{\rm d}x \leq \varepsilon. \end{equation} $

本节证明定理$ 1.1 $.首先假设$ \left | \nabla u \right |^{r} \in L_{loc}^{\infty }(\Omega )\subset L_{loc}^{\phi}(\Omega). $该假设可以如文献[23-24]通过逼近的标准方法移除.

假设$ \left | \nabla u \right |^{r} \in L_{loc}^{\infty }(\Omega )\subset L_{loc}^{\phi}(\Omega) $.根据引理$ 2.1(3) $, 有

(5.1) $ \begin{eqnarray} \int_{B_1}\phi (| \nabla u|^{r}){\rm d}x & = &\int_{0}^{\infty}\Big|\{x \in B_{1}:| \nabla u|^{r}>\mu\}\Big|{\rm d}[\phi(\mu)] \\ & = & \int_{0}^{(2N_0)^{r}\lambda_{*}^{r}}\Big|\{x \in B_{1}:| \nabla u|^{r}>\mu\}\Big|{\rm d}[\phi(\mu)] \\ && +\int_{(2N_0)^{r}\lambda_{*}^{r}}^{\infty}\Big|\{x \in B_{1}:| \nabla u|^{r}>\mu\}\Big|{\rm d}[\phi(\mu)] \\ & = & \int_{0}^{(2N_0)^{r}\lambda_{*}^{r}}\Big|\{x \in B_{1}:| \nabla u|^{r}>\mu\}\Big|{\rm d}[\phi(\mu)] \\ && +\int_{\lambda_{*}}^{\infty}\Big|\{x \in B_{1}:| \nabla u|^{r}>(2N_0)^{r}\lambda^{r}\}\Big|{\rm d}[\phi((2N_0)^{r}\lambda^{r})] \\ & = & J_1+J_2. \end{eqnarray} $

估计$ J_1 $.

(5.2) $ \begin{equation} J_1 = \int_{0}^{(2N_0)^{r}\lambda_{*}^{r}}\Big|\{x \in B_{1}:| \nabla u|^{r}>\mu\}\Big|{\rm d}[\phi(\mu)]\leq |B_1|\phi\Big[(2N_0)^{r}\lambda_{*}^{r}\Big]. \end{equation} $

下面考虑$ \lambda_{*}^{r} $.利用$ (3.7) $式中$ \lambda_{0} $的定义和定理$ 4.1 $得到

(5.3) $ \begin{eqnarray} \lambda_{*}^{r}&\leq&C[\lambda_{0}^{r}+1] \\ &\leq& C\left\{-\!\!\!\!\!\!\int _{B_{1}}| \nabla u|^{r}{\rm d}x+\frac{1}{\varepsilon}-\!\!\!\!\!\!\int _{B_{1}}|\textbf{f}|^{r}{\rm d}x+1\right \} \\ &\leq& C\left\{-\!\!\!\!\!\!\int _{B_{2}}| u-u_{B_{2R}}|^{r}{\rm d}x+-\!\!\!\!\!\!\int _{B_{2}}|\textbf{f}|^{r}{\rm d}x+\frac{1}{\varepsilon}-\!\!\!\!\!\!\int _{B_{1}}|\textbf{f}|^{r}{\rm d}x+1\right \} \\ &\leq& C\left\{-\!\!\!\!\!\!\int _{B_{2}}| u-u_{B_{2R}}|^{r}{\rm d}x+-\!\!\!\!\!\!\int _{B_{2}}|\textbf{f}|^{r}{\rm d}x+1\right \}, \end{eqnarray} $

其中$ C = C(n, r, \varepsilon). $最后利用(5.2)和(2.3)式, 上述不等式及Jensen不等式可得

(5.4) $ \begin{eqnarray} J_1&\leq& C\phi\Big(\lambda_{*}^{r} \Big)|B_1| \\ &\leq& C\left\{\phi\Big(-\!\!\!\!\!\!\int _{B_{2}}| u-u_{B_{2R}}|^{r}{\rm d}x\Big)+\phi\Big(-\!\!\!\!\!\!\int _{B_{2}}|\textbf{f}|^{r}{\rm d}x\Big)+1\right \} \\ &\leq& C\left\{\phi\Big(\int _{B_{2}}| u-u_{B_{2R}}|^{r}{\rm d}x\Big)+\Big(\int _{B_{2}}\phi(|\textbf{f}|^{r}){\rm d}x\Big)+1\right \}, \end{eqnarray} $

其中$ C = C(n, r, \phi, \varepsilon, C_1, C_2, N_0). $

估计$ J_2 $.由$ (5.1) $式可得

(5.5) $ \begin{eqnarray} J_2& = &\int_{\lambda_{*}}^{\infty}\Big|\{x \in B_{1}:| \nabla u|^{r}>(2N_0)^{r}\lambda^{r}\}\Big|{\rm d}[\phi((2N_0)^{r}\lambda^{r})] \\ &\leq& \int_{0}^{\infty}\Big|\{x \in B_{1}:| \nabla u|>(2N_0)\lambda\}\Big|{\rm d}[\phi((2N_0)^{r}\lambda^{r})]. \end{eqnarray} $

现在考虑上面不等式右边最后一个积分的被积函数.

对任意的$ \lambda \geq \lambda _{*} = (10/R_0)^{n/r}\lambda _{0}+1 $, 利用推论4.1, 4.2以及引理$ 3.2 $中的$ (3.13) $式, 可得

(5.6) $ \begin{eqnarray} &&\Big|\{x \in B_{5\rho_{i}}(x_{i}):| \nabla u|>2N_0\lambda\}\Big| \\ & = & \Big|\{x \in B_{5\rho_{i}}(x_{i}):| \nabla u_{\lambda}|>2N_0\}\Big| \\ &\leq& \Big|\{x \in B_{5\rho_{i}}(x_{i}):| \nabla (u_{\lambda}-v_{\lambda})|>N_0\}\Big|+\Big|\{x \in B_{5\rho_{i}}(x_{i}):| \nabla v_{\lambda}|>N_0\}\Big| \\ & = & \Big|\{x \in B_{5\rho_{i}}(x_{i}):| \nabla (u_{\lambda}-v_{\lambda})|>N_0\}\Big| \\ &\leq& \frac{1}{N_0^{r}}\int_{B_{5\rho_{i}}(x_{i})}| \nabla (u_{\lambda}-v_{\lambda})|^{r}{\rm d}x \\ &\leq& C\varepsilon \Big|B_{\rho_{i}}(x_{i})\Big| \\ &\leq& C\varepsilon\Big(\int_{\{x \in B_{\rho_{i}}(x_{i}):| \nabla u_{\lambda}|^{r}>1/3\}}|\nabla u_{\lambda}|^{r}{\rm d}x +\frac{1}{\varepsilon}\int_{\{x \in B_{\rho _{i}}(x_{i}):|\textbf{f}_{\lambda}|^{r}>\varepsilon /3\}}| \textbf{f}_{\lambda}|^{r}{\rm d}x\Big), \end{eqnarray} $

其中$ C = C(n, r, N_0). $即可知球$ B_{\rho_i}{(x_i)} $不相交及对于任意的$ \lambda \geq \lambda _{*} = (10/R_0)^{n/r}\lambda _{0}+1 $, 有

(5.7) $ \begin{eqnarray} \mathop{\bigcup}_{i\in N}B_{5\rho _{i}}(x_{i})\supset E_{\lambda}(1) = \{x\in B_1:|\nabla u_{\lambda}|>1\}. \end{eqnarray} $

然后对上述不等式在$ i\in N $上求和, 可得到

(5.8) $ \begin{eqnarray} &&\Big|\{x \in B_{1}:| \nabla u|>2N_0\lambda\}\Big| \\ &\leq& \sum\limits_{i}\Big|\{x \in B_{5\rho_{i}}(x_{i}):| \nabla u|>2N_0\lambda\}\Big| \\ &\leq& C\varepsilon\Big(\int_{\{x \in B_2:| \nabla u_{\lambda}|^{r}>1/3\}}|\nabla u_{\lambda}|^{r}{\rm d}x +\frac{1}{\varepsilon}\int_{\{x \in B_2:| \textbf{f}_{\lambda}|^{r}>\varepsilon /3\}}| \textbf{f}_{\lambda}|^{r}{\rm d}x\Big). \end{eqnarray} $

然后将$ (5.8) $式代入$ (5.5) $式, 并令$ \mu = \lambda ^{r} $可得

(5.9) $ \begin{eqnarray} J_2&\leq& C\varepsilon\int_{0}^{\infty}\int_{\{x \in B_2:| \nabla u_{\lambda}|^{r}>1/3\}}|\nabla u_{\lambda}|^{r}{\rm d}x{\rm d}[\phi((2N_0)^{r}\lambda^{r})] \\ && +C\int_{0}^{\infty}\int_{\{x \in B_2:|\textbf{f}_{\lambda}|^{r}>\varepsilon /3\}}| \textbf{f}_{\lambda}|^{r}{\rm d}x{\rm d}[\phi((2N_0)^{r}\lambda^{r})] \\ &\leq& C\varepsilon\int_{0}^{\infty}\frac{1}{\mu}\int_{\{x \in B_2:| \nabla u|^{r}>\mu/3\}}|\nabla u|^{r}{\rm d}x{\rm d}[\phi((2N_0)^{r}\mu)] \\ && +C\int_{0}^{\infty}\frac{1}{\mu}\int_{\{x \in B_2:|\textbf{f}|^{r}>\varepsilon\mu /3\}}| \textbf{f}|^{r}{\rm d}x{\rm d}[\phi((2N_0)^{r}\mu)]. \end{eqnarray} $

回顾引理$ 2.1(4) $有

(5.10) $ \begin{eqnarray} J_2\leq C_4\varepsilon\int_{B_2}\phi (| \nabla u|^{r}){\rm d}x+C_5\int_{B_2}\phi (| \textbf{f}|^{r}){\rm d}x, \end{eqnarray} $

其中$ C_4 = C(n, r, \phi, N_0), C_5 = C(n, r, \phi, \varepsilon , N_0, C_1, C_2, C_3). $

综合估计$ J_1 $和$ J_2 $可得出

(5.11) $ \begin{equation} \int_{B_1}\phi (| \nabla u|^{r}){\rm d}x\leq C_4\varepsilon\int_{B_2}\phi (| \nabla u|^{r}){\rm d}x+C_6\int_{B_2}\phi (| \textbf{f}|^{r}){\rm d}x +C_7\phi \Big( \int_{B_{2}}\left | u-u_{B_{2R}} \right |^{r}{\rm d}x \Big)+1, \end{equation} $

其中$ C_6 = C(n, r, \phi, \varepsilon , C_1, C_2, C_3, N_0), C_7 = C(n, r, \phi, \varepsilon , C_1, C_2, N_0). $选取合适的$ \varepsilon $使得

(5.12) $ \begin{eqnarray} C_4\varepsilon = \frac{1}{2}. \end{eqnarray} $

通过迭代覆盖(参见文献[20, 第三章, 引理2.1])将上面不等式右边第一个积分吸收, 可得到

(5.13) $ \begin{eqnarray} \int_{B_1}\phi (| \nabla u|^{r}){\rm d}x&\leq& C\left\{\int_{B_2}\phi (| \textbf{f}|^{r}){\rm d}x+ \phi \Big( \int_{B_{2}}\left | u-u_{B_{2R}} \right |^{r}{\rm d}x \Big)+1\right\}. \end{eqnarray} $

故通过基本缩放论证, 定理$ 1.1 $得证.

条件$ \left | \nabla u \right |^{r} \in L_{loc}^{\infty }(\Omega )\subset L_{loc}^{\phi}(\Omega) $可以采用逼近的标准方法移除.

根据文献[23], 因为$ \textbf{f}^{r}\in L_{loc}^{\phi}(\Omega), $即有

(5.14) $ \begin{eqnarray} \lim\limits_{k\rightarrow \infty} \int _{B_{R}(x_0)} \phi ( |\textbf{f}_k|^{r} ){\rm d}x = \int _{B_{R}(x_0)} \phi ( |\textbf{f}|^{r} ){\rm d}x. \end{eqnarray} $

考虑Dirichlet问题

(5.15) $ \begin{eqnarray} \left\{\begin{array}{ll} {\rm div}A(x, \nabla u_{k}) = {\rm div}(|\textbf{f}|^{p-2}\textbf{f}), \quad x\in B_{2R}, \\ u_{k}-u \in W_{0}^{1, r}(B_{2R}). \end{array}\right. \end{eqnarray} $

对所有$ k = 1, 2, 3, \cdot\cdot\cdot $, 存在唯一解$ u_{k} \in W^{1, r}(B_{2R}) $.由$ u_{k} $的梯度的正则性理论知, $ \nabla u_{k} \in L^{\infty}(B_{2R}). $经过初步推断, 可得到当$ k\rightarrow \infty $时

(5.16) $ \begin{equation} \| u_{k}-u \| _{W^{1, r} (B_{2R})}\rightarrow 0 . \end{equation} $

因此, 存在$ \{u_{k}\}_{k = 1}^{\infty} $的子序列(由$ \{u_{k}\} $表示), 使得

(5.17) $ \begin{equation} \nabla u_{k}\rightarrow \nabla u \quad a.e. \quad\ {\rm in} \quad B_{2R} . \end{equation} $

因此, 根据Fatou引理, (1.9), (5.14), (5.16)和$ (5.17) $式可得

(5.18) $ \begin{eqnarray} \int _{B_{R}} \phi ( |\nabla u|^{r} ){\rm d}x &\leq& \liminf\limits_{k\rightarrow \infty} \int _{B_{R}} \phi ( |\nabla u_{k}|^{r} ){\rm d}x \\ &\leq& C\liminf\limits_{k\rightarrow \infty} \left \{ \phi\left ( \frac{1}{R^{r}}\int _{B_{2R}}|u_{k}-(u_{k})_{2R}|^{r}{\rm d}x \right )+\int _{B_{2R}}\phi(|\textbf{f}_k|^{r}){\rm d}x +1 \right \} \\ &\leq& C\left \{ \phi\left ( \frac{1}{R^{r}} \int _{B_{2R}}|u-u_{2R}|^{r}{\rm d}x \right )+\int _{B_{2R}}\phi(|\textbf{f}|^{r}){\rm d}x +1 \right \}. \end{eqnarray} $

所以, 在附加假设$ \nabla u $是局部有界时可证明定理$ 1.1. $只要在一般情况下证明$ (1.9) $式, 即可通过标准覆盖论证得到$ |\nabla u|^{r}\in L_{loc}^{\phi}(\Omega) $.