数学物理学报, 2020, 40(2): 379-394 doi:

论文

自然增长条件下的非齐次A-调和方程弱解的梯度估计

张雅楠, 闫硕, 佟玉霞,

Gradient Estimates for Weak Solutions to Non-Homogeneous A-Harmonic Equations Under Natural Growth

Zhang Yanan, Yan Shuo, Tong Yuxia,

通讯作者: 佟玉霞, E-mail:tongyuxia@126.com

收稿日期: 2018-12-28  

基金资助: 河北省社会科学基金.  HB17YJ094

Received: 2018-12-28  

Fund supported: the Social Science Fund of Hebei Province.  HB17YJ094

摘要

该文主要研究一类自然增长条件下的非齐次A-调和方程弱解的梯度估计,首先获得自然增长条件下的非齐次A-调和方程弱解的Lp估计,然后使用迭代覆盖逼近等方法,将其推广到Orlicz空间.

关键词: A-调和方程 ; 弱解 ; 梯度估计

Abstract

The gradient estimates for weak solutions to non-homogeneous A-harmonic equations under natural growth is obtained. The Lp-type estimates for such equation is derived under natural growth, and then the Lφ-type estimate in Orlicz space is derived by a new iteration-covering approach.

Keywords: A-Harmonic equation ; Weak solution ; Gradient estimate

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本文引用格式

张雅楠, 闫硕, 佟玉霞. 自然增长条件下的非齐次A-调和方程弱解的梯度估计. 数学物理学报[J], 2020, 40(2): 379-394 doi:

Zhang Yanan, Yan Shuo, Tong Yuxia. Gradient Estimates for Weak Solutions to Non-Homogeneous A-Harmonic Equations Under Natural Growth. Acta Mathematica Scientia[J], 2020, 40(2): 379-394 doi:

1 引言

$ 1< p< +\infty $, $ \Omega $$ {{\Bbb R}} ^{n} $中的有界域,本文考虑如下非齐次椭圆方程

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

其中算子$ A = A(x, \xi ):{{\Bbb R}} ^{n}\times {{\Bbb R}} ^{n}\rightarrow {{\Bbb R}} ^{n} $,满足通常的可测性条件(Carathéodory条件),并且对几乎所有的$ x, y \in \Omega $和所有的$ \xi , \eta \in {{\Bbb R}} ^{n} $满足下列条件

$ \begin{equation} A(x, 0) = 0; \end{equation} $

$ \begin{equation} | A(x, \xi ) |\leq C_{1}\left | \xi \right |^{p-1}; \end{equation} $

$ \begin{equation} \left \langle A(x, \xi )-A(x, \eta ) , \xi -\eta \right \rangle\geq C_{2}\left | \xi -\eta \right |^{p}; \end{equation} $

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

算子$ B = B(x, \xi):{{\Bbb R}} ^{n}\times {{\Bbb R}} ^{n}\rightarrow {{\Bbb R}} $满足自然增长条件

$ \begin{equation} \left | B(x, \xi ) \right |\leq C_{4}\left | \xi \right |^{p}; \end{equation} $

其中$ C_{i}> 0, i = 1, 2, 3, 4 $为正常数.这里的连续模$ \omega (x):{{\Bbb R}} ^{+}\rightarrow {{\Bbb R}} ^{+} $非减,而且满足

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

考虑自然增长条件下的非齐次$ A $ -调和方程$ (1.1) $的弱解,给出如下定义.

定义1.1   设算子$ A, B $满足条件(1.2)–(1.6)式,称$ u \in W_{loc}^{1, p} (\Omega)\cap L^{\infty }(\Omega ) $是方程$ (1.1) $的局部弱解,若

$ \begin{equation} \int _{\Omega }\left \langle A(x, \nabla u), \nabla \varphi \right \rangle {\rm d}x = \int _{\Omega }B(x, \nabla u) \varphi {\rm d}x, \end{equation} $

对于任意的$ \varphi \in W_{0}^{1, p} (\Omega ) $都成立.

Dibenedetto等[1]和Iwaniec[2]获得了$ L^{q} $空间($ q\geq p $)$ p $-Laplace型拟线性椭圆方程

弱解的梯度估计; Acerbi和Mingione[3]研究了上述方程在变指数空间中的情况;之后Byun, Kinnunen等[4-5]在不同假设条件下,获得了$ p $-Laplace型拟线性椭圆方程

弱解的梯度估计; Byun和Wang[7]获得了非线性椭圆方程

的弱解的$ W^{1, p} $ ($ 2\leq p< \infty $)正则性; Jia等[8]基于标准的$ W^{2, p} $估计方法及Hardy-Littlewood极大函数,建立了泊松方程

在Orlicz空间中的正则性理论; Yao[7]研究了一类非线性椭圆方程

通过假设算子$ A $满足一些合适条件和向量$ f $满足适当增长条件,得到了其在Orlicz空间弱解的梯度估计;后来又获得了抛物型$ A $ -调和方程

的弱解在Sobolev空间和Orlicz空间的梯度估计[9];最近李慧珍和郑神州[10]研究了具有小的部分BMO系数的非散度型线性椭圆方程

强解的Hesssian矩阵在Orlicz空间的内部估计; Liang和Zheng[11]得到了具有部分BMO系数的非线性椭圆型障碍问题在Orlicz空间中的梯度估计.

Zheng等[12]$ 2008 $年获得了右端为非散度型的自然增长条件下椭圆方程弱解的$ L^{p} $估计,其研究成果中关于自然增长条件下的处理方法,对本文研究的$ A $ -调和方程的$ L^{p} $估计具有很强的参考价值.另外,由于经典的$ L^{\phi } $空间中的极大函数比较繁琐,本文采用了文献[8]中的迭代覆盖逼近方法,将$ L^{p} $估计推广到Orlicz空间,避免了使用极大函数算子.

另外,关于$ A $ -调和方程及其相关问题的结论,可参见文献[13-16].

本文首先得到自然增长条件下非齐次$ A $ -调和方程$ (1.1) $$ L^{p} $估计,再将其扩展到Orlicz空间.

下面是本文主要结论.

定理1.1  假设$ \phi \in\triangle _{2}\cap \triangledown _{2} $,如果$ u \in W_{loc}^{1, p} (\Omega)\cap L^{\infty }(\Omega ) $是方程$ (1.1) $的局部弱解,算子$ A $$ B $满足(1.2)–(1.6)式,那么

$ \begin{equation} \left | \nabla u \right |^{p} \in L_{loc}^{\phi }\left ( \Omega \right ), \end{equation} $

且有估计式

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

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

本文具体内容组织如下:在第二节给出预备知识和迭代覆盖方法;在第三节给出主要结论的证明.

2 预备知识

Orlicz空间是$ L^{ p} $空间的推广,自$ 1932 $年波兰著名数学大师Orlicz引进以来, Orlicz空间理论已应用于诸多领域.下面简单介绍Orlicz空间中的基本定义和一些相关的重要结论.

$ \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[7-8, 17]  如果存在一个正常数$ K $,使得对于任意的$ t> 0 $,都有

$ \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

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

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

注2.1   $ (1) $注意到全局$ \triangle _{2}\cap \triangledown _{2} $条件使函数适当增长.例如, $ \phi (t) = \left | t \right |^{\alpha }(1+\left | \log \left | t \right | \right |)\in \triangle _{2}\cap \triangledown _{2}, \alpha > 1 $.$ t\log (1+t) $这样的例子不满足$ \triangledown _{2} $条件,而像$ \exp (t^{2}) $这样的例子不满足$ \triangle_{2} $条件.

$ (2) $事实上,如果$ \phi \in \triangle _{2}\cap \triangledown _{2} $,那么$ \phi $满足对于$ 0< \theta _{2}\leq 1\leq \theta _{1}< \infty $,都有

$ \begin{equation} \phi (\theta _{1}t)\leq K\theta _{1}^{\alpha _{1}}\phi (t), \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) $在条件(2.3)式下,很容易得到$ \phi \in \Phi $,且满足$ \phi(0) = 0 $以及

$ \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[7-8, 17]    令$ \phi \in \Phi $,则$ \rm Orlicz $$ K^{\phi }(\Omega ) $是满足

$ \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 ) $$ K^{\phi }(\Omega ) $的线性闭包.

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

$ \begin{equation} L^{\phi }(\Omega ) = L^{q}(\Omega ). \end{equation} $

关于Orlicz空间有如下重要引理.

引理2.1[7, 18-19]   假设$ \phi \in\triangle _{2}\cap \triangledown _{2} $以及$ g \in L^{\phi }(\Omega ) $,那么

$ (1) $$ K^{\phi } = L^{\phi } $,且$ C_{0}^{\infty } $$ 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}, \alpha _{2} $取自(2.3)式;

$ (3) $

$ \begin{equation} \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 ]; \end{equation} $

$ (4) $

$ \begin{equation} \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, \end{equation} $

其中$ a, b>0, C = C(a, b, \phi ) $.

$ u $为方程$ (1.1) $的局部弱解.先考虑定理1.1中在$ R = 1 $的情况,然后可通过缩放论证得到(1.10)式.设

$ \begin{equation} \lambda _{0} = : \left [-\!\!\!\!\!\!\int_{B_{1}} \left | \nabla u \right |^{p}{\rm d}x\right ]^{\frac{1}{p}}; \end{equation} $

对于任意的$ x \in \Omega $$ \rho > 0 $,令

$ \begin{equation} J\left [ B_{\rho } (x)\right ] = : -\!\!\!\!\!\!\int _{B_{\rho } (x)}\left | \nabla u \right |^{p}{\rm d}y. \end{equation} $

$ \begin{equation} E(1) = :\left \{ x \in B_{1}:\left | \nabla u \right | ^{p}> \lambda ^{p}\right \}. \end{equation} $

由(7)式,选择合适的常数$ R_{0} = R_{0}(\epsilon ) \in (0, 1) $使得

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

不妨限定$ \epsilon \in (0, 1) $,且由(3.48)式决定.

下面借鉴文献[8],给出迭代覆盖逼近引理.

引理2.2  令$ \lambda \geq \lambda _{*} = :(\frac{10}{R_{0}})^{n/p}\lambda _{0}+1 $,则存在一族不相交的球$ \left \{ B_{\rho _{i}}(x_{i}) \right \} $, $ x_{i} \in E(1) $,使得$ 0< \rho _{i} = \rho (x_{i})\leq R_{0}/10 $,且

$ \begin{equation} J\left [ B_{\rho _{i}} (x_{i})\right ] = \lambda ^{p}, J\left [ B_{\rho } (x_{i})\right ] < \lambda ^{p} \; \mbox{对任意} \; \rho > \rho _{i} \;\mbox{成立}. \end{equation} $

而且,有

$ \begin{equation} E(1)\subset \bigcup\limits_{i \in {\Bbb N}}B_{5\rho _{i}}(x_{i})\cup \mbox{可略集}; \end{equation} $

$ \begin{equation} \left | B_{\rho _{i}} (x_{i})\right |\leq \frac{2}{\lambda ^{p}} \int _{\left \{ x\in B_{\rho _{i}}(x_{i}):\left | \nabla u \right | ^{p}> \frac{\lambda ^{p}}{2}\right \}}\left | \nabla u \right |^{p}{\rm d}x. \end{equation} $

   $ (1) $首先需证

$ \begin{equation} \sup _{\omega \in B_{1}} \sup _{\frac{R_{0}}{10}\leq \rho \leq R_{0}}J\left [ B_{\rho }(\omega ) \right ]\leq \lambda ^{p}. \end{equation} $

为证明上式,给定任意的$ \omega \in B_{1} $$ \frac{R_{0}}{10}\leq \rho \leq R_{0} $,令$ \lambda \geq \lambda _{*} = :(\frac{10}{R_{0}})^{n/p}\lambda _{0}+1 $,则有

$ \begin{eqnarray} J\left [ B_{\rho }(\omega ) \right ] = -\!\!\!\!\!\!\int _{B_{\rho }(\omega)}\left |\nabla u \right |^{p}{\rm d}x &\leq& \frac{\left | B_{1} \right |}{\left | B_{\rho }(\omega ) \right |}-\!\!\!\!\!\!\int _{B_{1}}\left | \nabla u \right |^{p}{\rm d}x \\ &\leq& (\frac{10}{R_{0}})^{n}-\!\!\!\!\!\!\int _{B_{1}}\left | \nabla u \right |^{p}{\rm d}x \\ &\leq& (\frac{10}{R_{0}})^{n}\lambda _{0}^{p}+1 \\ &\leq& \lambda ^{p }. \end{eqnarray} $

即对任意的$ \omega \in B_{1} $$ R_{0}/10\leq \rho \leq R_{0} $, (2.16)式成立.

$ (2) $对几乎处处$ \omega \in E(1) $,由Lebesgue微分定理得

$ \begin{equation} \lim _{\rho \rightarrow 0}J\left [ B_{\rho }(\omega ) \right ]> \lambda ^{p}, \end{equation} $

则存在$ \rho > 0 $,满足

$ \begin{equation} J\left [ B_{\rho }(\omega ) \right ]> \lambda ^{p}; \end{equation} $

由(2.16)式,选择半径$ \rho _{\omega } \in (0, \frac{R_{0}}{10}] $使得

$ \begin{equation} \rho _{\omega } = :\max \left \{ \rho |J\left [ B_{\rho } (\omega )\right ] = \lambda ^{p} , 0< \rho \leq \frac{R_{0}}{10}\right \}. \end{equation} $

由于$ J\left [ B_{\rho } (\omega )\right ] $是关于$ \rho $的连续函数,则

$ \begin{equation} J\left [ B_{\rho_{\omega } } (\omega )\right ] = \lambda ^{p}, \end{equation} $

于是,对$ \rho _{\omega }< \rho \leq R_{0} $

$ \begin{equation} J\left [ B_{\rho } (\omega )\right ]<\lambda ^{p}. \end{equation} $

对于几乎处处$ \omega \in E(1) $,存在一个球$ B_{\rho _{\omega }}(\omega ) $满足上述论证.因此,应用Vitali覆盖引理,有一族互不相交的球$ \left \{ B_{\rho _{i }}(x_{i}) \right \}_{i \in {\Bbb N}} $,其中$ x_{i} \in E(1), \rho _{i} = \rho (x_{i})\in (0, R_{0}/10] $,使得(2.13)式和(2.14)式成立.

$ (3) $由(2.13)式有

$ \begin{equation} J\left [ B_{\rho _{i}}(x_{i}) \right ] = -\!\!\!\!\!\!\int _{B_{\rho _{i}}(x_{i})}\left | \nabla u \right |^{p}{\rm d}x = \lambda ^{p}, \end{equation} $

$ \begin{equation} \lambda ^{p}\left | B_{\rho _{i}}(x_{i})\right | = \int _{B_{\rho _{i}}(x_{i})}\left | \nabla u \right |^{p}{\rm d}x. \end{equation} $

于是将(2.24)式进行积分区域分割,得

$ \begin{equation} \lambda ^{p}\left | B_{\rho _{i}}(x_{i})\right |\leq \int _{\left \{ x \in B_{\rho _{i}}(x_{i}):\left | \nabla u \right |^{p} > \frac{\lambda ^{p}}{2}\right \}}\left | \nabla u \right |^{p}{\rm d}x+\frac{\lambda ^{p}}{2}\left | B_{\rho _{i}}(x_{i})\right |, \end{equation} $

移项整理得

$ \begin{equation} \left | B_{\rho _{i}}(x_{i})\right |\leq \frac{2}{\lambda ^{p}}\int _{\left \{ x \in B_{\rho _{i}}(x_{i}):\left | \nabla u \right |^{p} > \frac{\lambda ^{p}}{2}\right \}}\left | \nabla u \right |^{p}{\rm d}x. \end{equation} $

引理2.2证毕.

下面的引理来自文献[20-21].

引理2.3  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} $

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

这里$ A, B, \alpha, \theta $为非负常数且$ \theta < 1 $,则存在只依赖于$ \alpha $$ \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} $

3 主要定理的证明

本节内容组织如下:在3.1小节,通过添加假设条件$ \left | \nabla u \right |^{p} \in L_{loc}^{\infty }(\Omega ) \subset L_{loc}^{\phi}(\Omega) $给出定理$ 1.1 $的证明;在3.2小节,利用逼近方法将上述添加的假设条件移除.

3.1 假设条件下定理1.1的证明

本小节在假设$ \left | \nabla u \right |^{p} \in L_{loc}^{\infty }(\Omega )\subset L_{loc}^{\phi}(\Omega) $的条件下,考虑定理$ 1.1 $的证明.首先建立自然增长条件下方程$ (1.1) $的局部$ L^{p} $估计.

引理3.1   假设$ B_{2R}\subset \Omega $,令$ u \in W_{loc}^{1, p}(\Omega) \cap L^{\infty }(\Omega ) $为方程$ (1.1) $的局部弱解,算子$ A $$ B $满足(1.2)–(1.6)式,则

$ \begin{equation} \int _{B_{R}}\left | \nabla u \right |^{p}{\rm d}x\leq \frac{C}{R^{p}} \int _{B_{2R}}\left | u-u_{2R} \right |^{p}{\rm d}x. \end{equation} $

  选取适当的检验函数$ \varphi = (u-u_{2R})e^{\beta \left |u-u_{2R} \right |}\eta ^{p} $,其中$ \beta $为待定常数, $ u_{2R} = \frac{1}{|B_{2R}|}\int _{B_{2R}}u{\rm d}x $, $ \eta \in C_{0}^{\infty }(\Omega) $为截断函数,满足

$ \begin{equation} 0\leq \eta \leq 1, \eta \equiv 1 \; in \;B_{R}, \eta \equiv 0 \; in \;\Omega \setminus B_{2R}, \left | \nabla \eta \right | \leq \frac{C}{R}. \end{equation} $

将检验函数代入定义1.1,得

$ \begin{equation} \int _{\Omega }\left \langle A(x, \nabla u), \nabla \varphi \right \rangle {\rm d}x = \int _{\Omega }B(x, \nabla u)\varphi {\rm d}x, \end{equation} $

并移项整理得

$ \begin{eqnarray} &&\int _{\Omega }\left \langle A(x, \nabla u), e^{\beta \left | u-u_{2R} \right |}\eta ^{p} \nabla u +\beta \eta ^{p}\left | u-u_{2R} \right | e^{\beta \left | u-u_{2R} \right |} \nabla u \right \rangle {\rm d}x \\ & = & -p\int _{\Omega }\left \langle A(x, \nabla u), \eta ^{p-1} ( u-u_{2R} ) e^{\beta \left | u-u_{2R} \right |} \nabla \eta \right \rangle {\rm d}x +\int _{\Omega }B(x, \nabla u)\varphi {\rm d}x. \end{eqnarray} $

上述等式左边利用(1.2)及(1.4)式得

$ \begin{eqnarray} &&\int _{\Omega }\left \langle A(x, \nabla u), e^{\beta \left | u-u_{2R} \right |}\eta ^{p} \nabla u +\beta \eta ^{p}\left | u-u_{2R} \right | e^{\beta \left | u-u_{2R} \right |} \nabla u \right \rangle {\rm d}x \\ &\geq & C_{2}\int _{\Omega }e^{\beta \left | u-u_{2R} \right |}\left | \eta \nabla u \right |^{p}{\rm d}x +C_{2}\beta \int _{\Omega }\left | \eta \nabla u \right |^{p} \left | u-u_{2R} \right |e^{\beta \left | u-u_{2R} \right |}{\rm d}x. \end{eqnarray} $

将(3.4)式右边利用(1.3)式和(1.6)式得

$ \begin{eqnarray} &&-p\int _{\Omega }\left \langle A(x, \nabla u), \eta ^{p-1} ( u-u_{2R} ) e^{\beta \left | u-u_{2R} \right |} \nabla \eta \right \rangle {\rm d}x +\int _{\Omega }B(x, \nabla u)\varphi {\rm d}x \\ &\leq & pC_{1}\int _{\Omega }\left | \eta \nabla u \right | ^{p-1}\left | \nabla \eta \right |\left | u-u_{2R} \right |e^{\beta \left | u-u_{2R} \right |} {\rm d}x+ C_{4}\int _{\Omega }\left | \eta \nabla u \right |^{p}\left | u-u_{2R} \right |e^{\beta \left | u-u_{2R} \right |} {\rm d}x \\ & = & I_{1}+I_{2}. \end{eqnarray} $

估计$ I_{1} $.由Young不等式得

$ \begin{eqnarray} I_{1} \leq pC_{1}\tau \int _{\Omega }e^{\beta \left | u-u_{2R} \right |}\left | \eta \nabla u \right |^{p}{\rm d}x +pC_{1}C(\tau )\int _{\Omega }e^{\beta \left | u-u_{2R} \right |} \left | (u-u_{2R}) \nabla \eta \right |^{p}{\rm d}x. \end{eqnarray} $

综合(3.4), (3.5), (3.6)及(3.7)式,得

$ \begin{eqnarray} && C_{2}\int _{\Omega }e^{\beta \left | u-u_{2R} \right |}\left | \eta \nabla u \right |^{p}{\rm d}x +C_{2}\beta \int _{\Omega }\left | \eta \nabla u \right |^{p} \left | u-u_{2R} \right |e^{\beta \left | u-u_{2R} \right |}{\rm d}x \\ &\leq & pC_{1}\tau \int _{\Omega }e^{\beta \left | u-u_{2R} \right |}\left | \eta \nabla u \right |^{p}{\rm d}x +pC_{1}C(\tau)\int _{\Omega }e^{\beta \left | u-u_{2R} \right |} \left | (u-u_{2R}) \nabla \eta \right |^{p}{\rm d}x \\ & & +C_{4}\int _{\Omega }\left | \eta \nabla u \right |^{p}\left | u-u_{2R} \right |e^{\beta \left | u-u_{2R} \right |} {\rm d}x. \end{eqnarray} $

$ \beta $满足$ C_{2} \beta > C_{4} $,将上式移项整理得

$ \begin{eqnarray} && C_{2}\int _{\Omega }e^{\beta \left | u-u_{2R} \right |}\left | \eta \nabla u \right |^{p}{\rm d}x \\ &\leq & pC_{1}\tau \int _{\Omega }e^{\beta \left | u-u_{2R} \right |}\left | \eta \nabla u \right |^{p}{\rm d}x +pC_{1}C(\tau )\int _{\Omega }e^{\beta \left | u-u_{2R} \right |} \left | (u-u_{2R}) \nabla \eta \right |^{p}{\rm d}x. \end{eqnarray} $

$ \tau >0 $且满足$ C_{2} - p C_{1}\tau > 0 $.上式利用(3.2)式得

$ \begin{eqnarray} \int _{B_{R}}e^{\beta \left | u-u_{2R} \right |}\left | \nabla u \right |^{p}{\rm d}x \leq \frac{C}{R^{p}}\int _{B_{2R}}e^{\beta \left | u-u_{2R} \right |} \left | u-u_{2R} \right |^{p}{\rm d}x. \end{eqnarray} $

由于$ u \in L^{\infty }(\Omega ), $故存在充分大的正常数$ M $,使得$ \left | u \right |\leq M $.代入上式,得

$ \begin{eqnarray} \int _{B_{R}}\left | \nabla u \right |^{p}{\rm d}x \leq \frac{C}{R^{p}}\int _{B_{2R}}\left | u-u_{2R} \right |^{p}{\rm d}x, \end{eqnarray} $

其中$ C = C(C_{1}, C_{2}, C_{4}, p, M) $,引理3.1证毕.

$ v $是下列边值问题的弱解:

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

下面给出全局弱解的定义.

定义3.1  假设$ v \in W^{1, p}(B_{\widetilde{R}}), $于是$ v-u \in W_{0}^{1, p}(B_{\widetilde{R}}) $,则称$ v \in W^{1, p}(B_{\widetilde{R}}) $是边值问题(3.12)在$ B_{\widetilde{R}} $中的弱解,如果有

$ \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, p}(B_{\widetilde{R}} ) $成立.

下面借鉴文献[12],给出$ v $的有界性引理.

引理3.2  令$ v \in W^{1, p}(B_{\widetilde{R}}) $是边值问题(3.12)的弱解,则有

$ \begin{equation} \sup _{ B_{\widetilde{R}}}\left | v \right |\leq \sup _{\partial B_{\widetilde{R}}}\left | u \right |\leq M. \end{equation} $

  设$ { } M = \sup _{\partial B_{\widetilde{R}}}u $, $ { } m = \inf _{\partial B_{\widetilde{R}}}u $,则$ s = \min \left \{ v, M \right \} $是边值问题(3.12)的弱上解.取非负函数$ \varphi _{1} = v- s \in W_{0}^{1, p}(B_{\widetilde{R}}) $为检验函数,代入边值问题(3.12),结合$ (1.4) $式得

于是有

由于$ \varphi _{1} = v- \min \left \{ v, M \right \} \in W_{0}^{1, p}(B_{\widetilde{R}}) $,则$ v = \min \left \{ v, M \right \} $,于是

同理可证得$ v\geq m = \inf _{\partial B_{\widetilde{R}}}u $.引理3.2得证.

由定义3.1很容易获得如下引理.

引理3.3  若$ v \in W^{1, p}(B_{10\rho _{i}}(x_{i})) $是边值问题(3.12)在$ B_{10\rho _{i}}(x_{i}) $中的弱解,其中$ x_{i} \in E(1) $, $ \rho _{i} $与在引理$ 2.2 $中的定义相同,则有

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

  取检验函数$ \phi = u-v \in W_{0}^{1, p}(B_{10\rho _{i}}(x_{i})) $,由定义3.1,得

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

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

由(1.2)及(1.4)式得

$ \begin{equation} \int _{B_{10\rho _{i}}(x_{i})}\left \langle A(x^{*}, \nabla v) , \nabla v\right \rangle {\rm d}x\geq C_{2}\int _{B_{10\rho _{i}}(x_{i})}\left | \nabla v \right |^{p} {\rm d}x. \end{equation} $

由(1.3)式及Young不等式,得

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

综合(3.17), (3.18)和(3.19)式,得

$ \begin{eqnarray} C_{2}\int _{B_{10\rho _{i}}(x_{i})}\left | \nabla v \right |^{p} {\rm d}x \leq C_{1}\left \{ \tau_{1} \int _{B_{10\rho _{i}}(x_{i})} \left | \nabla v \right |^{p}{\rm d}x +C(\tau _{1})\int _{B_{10\rho _{i}}(x_{i})} \left | \nabla u \right |^{p}{\rm d}x \right \}. \end{eqnarray} $

令常数$ \tau_{1} >0 $足够小,满足$ C_2>C_1\tau_1 $,则有

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

其中$ C = C(C_{1}, C_{2}) $,于是引理3.3证毕.

在证明主要结论之前,给出如下引理.

引理3.4  若$ v \in W^{1, p}(B_{10\rho _{i}}(x_{i})) $是边值问题(3.12)在$ B_{10\rho _{i}}(x_{i}) $中的弱解, $ u \in W_{loc}^{1, p} (\Omega)\cap L^{\infty }(\Omega ) $是方程$ \rm (1.1) $的局部弱解,算子$ A $$ B $满足(1.2)–(1.6)式.若

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

则存在$ N_{0}> 1 $,使得

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

$ \begin{equation} \sup _{B_{5\rho _{i}}(x_{i})}\left | \nabla v \right |\leq N_{0}. \end{equation} $

  若结论(3.23)式成立,则结论(3.24)式可证(参见文献[22,引理5.1]).

下面只需证明$ (3.24) $式.选取检验函数$ \varphi_{2} = u-v \in W_{0}^{1, p}(B_{10\rho _{i}}(x_{i})) $,代入定义1.1及定义3.1,得

$ \begin{equation} \int _{B_{10\rho _{i}}(x_{i})} \left \langle A(x, \nabla u), \nabla (u-v) \right \rangle {\rm d}x = \int _{B_{10\rho _{i}}(x_{i})} B(x, \nabla u)(u-v) {\rm d}x, \end{equation} $

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

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

$ \begin{eqnarray} K_{1} = K_{2}+K_{3}, \end{eqnarray} $

其中

$ \begin{eqnarray} && K_{1} = \int _{B_{10\rho _{i}}(x_{i})} \left \langle A(x, \nabla u)-A(x, \nabla v), \nabla (u-v) \right \rangle {\rm d}x, \\ && K_{2} = - \int _{B_{10\rho _{i}}(x_{i})} \left \langle A(x, \nabla v)-A(x^{*}, \nabla v), \nabla (u-v) \right \rangle {\rm d}x, \\ && K_{3} = \int _{B_{10\rho _{i}}(x_{i})} B(x, \nabla u )(u-v) {\rm d}x. \end{eqnarray} $

先估计$ K_{1}. $由(1.4)式得

$ \begin{eqnarray} K_{1}\geq C_{2}\int _{B_{10\rho _{i}}(x_{i})} \left | \nabla (u-v) \right |^{p} {\rm d}x. \end{eqnarray} $

再估计$ K_{2}. $$ \rho _{i} \in (0, R_{0}/10] $, (1.5)式, Young不等式和(2.12)式,得

$ \begin{eqnarray} K_{2} &\leq& C_{3}\omega (10\rho _{i}) \int _{B_{10\rho _{i}}(x_{i}) } \left | \nabla v \right |^{p-1} (\left | \nabla u \right |+\left | \nabla v \right | ) {\rm d}x \\ &\leq& C_{3}\omega (R_{0}) \left \{ \int _{B_{10\rho _{i}}(x_{i}) } \left | \nabla v \right | ^{p}{\rm d}x +\int _{B_{10\rho _{i}}(x_{i}) } (\left | \nabla u \right |^{p} + \left | \nabla v \right |^{p} ) {\rm d}x \right \} \\ &\leq& C_{3}\epsilon \left \{ \int _{B_{10\rho _{i}}(x_{i}) } \left | \nabla v \right |^{p} {\rm d}x +\int _{B_{10\rho _{i}}(x_{i}) } \left | \nabla u \right |^{p} {\rm d}x \right \}, \end{eqnarray} $

由引理3.3得

$ \begin{eqnarray} K_{2} \leq C_{3}\epsilon \int _{B_{10\rho _{i}}(x_{i}) } \left | \nabla u \right |^{p} {\rm d}x, \end{eqnarray} $

再利用(3.22)式得

$ \begin{eqnarray} K_{2} \leq C_{3}\left | B_{10\rho _{i}}(x_{i}) \right | \epsilon . \end{eqnarray} $

最后估计$ K_{3} $.利用(1.6)式得

$ \begin{eqnarray} K_{3} \leq C_{4} \int _{B_{10\rho _{i}}(x_{i}) }\left | \nabla u \right |^{p}\left | u-v \right |{\rm d}x , \end{eqnarray} $

$ \left | u \right |\leq M $和引理3.2得

$ \begin{eqnarray} K_{3}\leq C \int _{B_{10\rho _{i}}(x_{i}) }\left | \nabla u \right |^{p} {\rm d}x, \end{eqnarray} $

其中$ C = C(C_{4}, M) $.综合所有$ K_{i}(i = 1, 2, 3) $,得

$ \begin{eqnarray} \int _{B_{10\rho _{i}}(x_{i}) }\left | \nabla (u-v) \right |^{p}{\rm d}x \leq C\left | B_{10\rho _{i}}(x_{i}) \right | \epsilon +C \int _{B_{10\rho _{i}}(x_{i}) }\left | \nabla u \right |^{p} {\rm d}x, \end{eqnarray} $

再由(3.22)式得

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

其中$ C = C(C_{2}, C_{3}, C_{4}, M). $引理3.4证毕.

现在开始证明定理1.1.

  对于任意的$ \lambda \geq \lambda _{*} = (10/R_{0}) ^{n/p}\lambda _{0}+1 $,取$ N_{0}>1 $如引理3.4所述.利用引理3.4有

$ \begin{eqnarray} && \left | \left \{ x \in B_{5\rho _{i}}(x_{i}) :\left | \nabla u \right | > 2N_{0}\lambda \right \} \right | \\ &\leq& \left | \left \{ x \in B_{5\rho _{i}}(x_{i}) :\left | \nabla (u-v) \right | > N_{0}\lambda \right \} \right |+ \left | \left \{ x \in B_{5\rho _{i}}(x_{i}) :\left | \nabla v\right | > N_{0}\lambda \right \} \right | \\ & = & \left | \left \{ x \in B_{5\rho _{i}}(x_{i}) :\left | \nabla (u-v) \right | > N_{0}\lambda \right \} \right | \\ &\leq& \int _{B_{5\rho _{i}}(x_{i})\cap \left \{ x: \frac{\nabla (u-v)}{N_{0}}> \lambda\right \} } \lambda ^{p} {\rm d}x \\ &\leq& \frac{1}{N_{0}^{p}}\int _{B_{5\rho _{i}}(x_{i})}\left | \nabla (u-v) \right |^{p} {\rm d}x \\ &\leq& C\epsilon \left | B_{\rho _{i}}(x_{i}) \right |, \end{eqnarray} $

于是,由(2.15)式得

$ \begin{eqnarray} \left | \left \{ x \in B_{5\rho _{i}}(x_{i}) :\left | \nabla u \right | > 2N_{0}\lambda \right \} \right |\leq \frac{C\epsilon }{\lambda ^{p}} \int _{ \left \{ x \in B_{\rho _{i}}(x_{i}) :\left | \nabla u \right |^{p}> \lambda ^{p}/2 \right \} } \left | \nabla u \right |^{p} {\rm d}x, \end{eqnarray} $

其中$ C = C(n, p, N_{0}). $考虑一族互不相交的球$ \left \{ B_{\rho _{i}}(x_{i}) \right \} $,由引理2.2有

$ \begin{eqnarray} \bigcup _{i \in {\Bbb N}}B_{5\rho _{i}}(x_{i}) \supset E(1) = \left \{ x \in B_{1}:\left | \nabla u \right |^{p} > \lambda ^{p}\right \}, \end{eqnarray} $

对任意的$ \lambda \geq \lambda _{*} = (10/R_{0})^{n/p}\lambda _{0}+1 $成立.综合(3.39)和(3.38)式,并注意到$ \rho_{i} \leq \frac{R_{0}}{10} $$ R_{0} \in (0, 1) $,得

$ \begin{eqnarray} \left | \left \{ x \in B_{1}:\left | \nabla u \right |^{p}> (2N_{0})^{p}\lambda ^{p} \right \} \right | & = & \left | \left \{ x \in B_{1}:\left | \nabla u \right |> 2N_{0} \lambda \right \} \right | \\ &\leq & \sum _{i}\left | \left \{ x \in B_{5\rho _{i}}(x_{i}): \left | \nabla u \right |> 2N_{0} \lambda \right \} \right | \\ &\leq & \frac{C\epsilon }{\lambda ^{p}} \sum _{i}\int _{\left \{ x \in B_{\rho _{i}}(x_{i}):\left | \nabla u \right |^{p} > \lambda ^{p} /2 \right \}}\left | \nabla u \right |^{p} {\rm d}x \\ &\leq & \frac{C\epsilon }{\lambda ^{p}} \int _{\left \{ x \in B_{2}:\left | \nabla u \right |^{p} > \lambda ^{p} /2 \right \} }\left | \nabla u \right |^{p} {\rm d}x. \end{eqnarray} $

利用(2.7)式得

$ \begin{eqnarray} \int _{B_{1}}\phi \left ( \left | \nabla u \right |^{p} \right ) {\rm d}x & = & \int_{0}^{\infty }\left | \left \{ x \in B_{1}: \left | \nabla u \right |^{p}> \mu \right \} \right | {\rm d}\left [ \phi (\mu ) \right ] \\ & = & \int_{0}^{(2N_{0})^{p}\lambda _{*}^{p}} \left | \left \{ x \in B_{1} : \left | \nabla u \right |^{p} > \mu \right \} \right | {\rm d}\left [ \phi (\mu ) \right ] \\ &&+\int_{(2N_{0})^{p}\lambda _{*}^{p}}^{\infty } \left | \left \{ x \in B_{1} : \left | \nabla u \right |^{p} > \mu \right \} \right | {\rm d}\left [ \phi (\mu ) \right ] \\ & = & \int_{0}^{(2N_{0})^{p}\lambda _{*}^{p}} \left | \left \{ x \in B_{1} : \left | \nabla u \right |^{p} > \mu \right \} \right | {\rm d}\left [ \phi (\mu ) \right ] \\ &&+\int_{\lambda _{*}}^{ \infty } \left | \left \{ x \in B_{1} : \left | \nabla u \right |^{p} > (2N_{0})^{p}\lambda ^{p} \right \} \right | {\rm d}\left [ \phi \left ( (2N_{0})^{p}\lambda^{p} \right ) \right ] \\ & = & J_{1}+J_{2}. \end{eqnarray} $

估计$ J_{1} $.$ \lambda _{*} $的定义及(2.9)式,得

$ \begin{eqnarray} \lambda _{*}^{p}\leq C\left [ \lambda _{0}^{p} +1 \right ]\leq C\left [ -\!\!\!\!\!\!\int _{B_{1}}\left | \nabla u \right |^{p}{\rm d}x+1 \right ], \end{eqnarray} $

其中$ C = C(n, p) $.由引理3.1得

$ \begin{eqnarray} \lambda _{*}^{p}\leq C\left \{ -\!\!\!\! \!\! \int _{B_{2}}\left | u-u_{2R} \right |^{p} {\rm d}x +1 \right \}, \end{eqnarray} $

因此,由(2.3)和(3.43)式得

$ \begin{eqnarray} J_{1} \leq \phi \left [ (2N_{0})^{p}\lambda _{*}^{p} \right ]\left | B_{1} \right | \leq C\left \{ \phi \left ( \int _{B_{2}} \left | u-u_{2R} \right |^{p}{\rm d}x \right )+1 \right \}, \end{eqnarray} $

其中$ C = C(n, p, \phi, N_{0} ). $

再估计$ J_{2}. $由(3.40)式得

$ \begin{eqnarray} J_{2} &\leq & \int_{0}^{ \infty } \left | \left \{ x \in B_{1} : \left | \nabla u \right |^{p} > (2N_{0})^{p}\lambda ^{p} \right \} \right | {\rm d}\left [ \phi \left ((2N_{0})^{p}\lambda^{p} \right ) \right ] \\ &\leq & C\epsilon \int_{0}^{\infty } \frac{1}{\lambda ^{p}} \int _{ \left \{ x \in B_{2} : \left | \nabla u \right |^{p} > \lambda ^{p}/2 \right \} } \left | \nabla u \right |^{p} {\rm d}x {\rm d}\left [ \phi \left ( (2N_{0})^{p}\lambda^{p} \right ) \right ], \end{eqnarray} $

$ \mu = \lambda ^{p} $代入上述不等式,再利用(2.8)式得

$ \begin{eqnarray} J_{2} &\leq & C\epsilon \int_{0}^{\infty }\frac{1}{\mu } \int _{\left \{ x \in B_{2} :\left | \nabla u \right |^{p} > \mu /2\right \}} \left | \nabla u \right |^{p}{\rm d}x {\rm d}\left [ \phi \left ((2N_{0})^{p} \mu \right ) \right ] \\ &\leq & C\epsilon \int _{B_{2}}\phi ( \left | \nabla u \right |^{p}){\rm d}x, \end{eqnarray} $

其中$ C = C(n, p, \phi , N_{0}). $

于是综合估计$ J_{1} $$ J_{2} $,得

$ \begin{eqnarray} \int _{B_{1}} \phi (\left | \nabla u \right |^{p}){\rm d}x \leq C\left \{ \phi \left ( \int _{B_{2}} \left | u-u_{2R} \right |^{p}{\rm d}x \right )+1 \right \} + C\epsilon \int _{B_{2}}\phi ( \left | \nabla u \right |^{p}){\rm d}x, \end{eqnarray} $

选择适当的$ \epsilon $,使得

$ \begin{eqnarray} C\epsilon = \frac{1}{2}, \end{eqnarray} $

再通过引理2.3,将上述不等式最后一项积分重新吸收,得

$ \begin{eqnarray} \int _{B_{1}}\phi \left ( \left | \nabla u \right |^{p} \right ){\rm d}x \leq C\left \{ \phi \left ( \int _{B_{2}}\left | u-u_{2R} \right |^{p} {\rm d}x \right )+1 \right \}. \end{eqnarray} $

最后通过缩放讨论,证毕.

3.2 逼近

现参考[18, 23]采用逼近方法将3.1小节添加的假设条件$ \left | \nabla u \right |^{p} \in L_{loc}^{\infty }(\Omega )\subset L_{loc}^{\phi}(\Omega) $移除.

考虑Dirichlet问题

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

对所有$ k = 1, 2, 3, \cdot\cdot\cdot $,存在唯一弱解$ u_{k} \in W^{1, p}(B_{2R}) $,且$ u_{k} \in L^{\infty}(B_{2R}) $.由梯度的正则性理论知, $ \nabla u_{k} \in L^{\infty}(B_{2R}) $ (参见文献[23-24]).于是上一节在Orlicz空间的估计成立,即

$ \begin{equation} \int _{B_{R}}\phi (|\nabla u_{k}|^{p}){\rm d}x\leq C \left \{ \phi\left ( \int _{B_{2R}}|u_{k}-(u_{k})_{2R}|^{p}{\rm d}x \right ) +1 \right \}. \end{equation} $

由于$ u_{k} \in L^{\infty}(B_{2R}) $$ \nabla u_{k} \in L^{\infty}(B_{2R}) $,因此存在一个子列$ \left \{ u_{k} \right \} $ (仍表示为$ \left \{ u_{k} \right \} $)和函数$ v \in W^{1, p}( B_{2R} ) $,使得

$ \begin{eqnarray} \left\{\begin{array}{ll} u_{k}\mathop{\longrightarrow}\limits^{\mbox强} v \; \; \;\; \, & x\in L^{p}(B_{2R} ), \\ \nabla u_{k} \mathop{\longrightarrow}\limits^{\mbox弱} \nabla v & x\in L^{p}(B_{2R} ). \end{array}\right. \end{eqnarray} $

则有

$ \begin{equation} \nabla u_{k}\mathop{\longrightarrow}\limits^{\mbox强} \nabla v x\in L^{p}(B_{2R}) . \end{equation} $

稍后给出证明过程.

由(3.53)式可知, $ v \in W^{1, p}(B_{2R} ) $是Dirichlet问题(3.50)的弱解.已知$ u \in W^{1, p}(B_{2R} ) $也是Dirichlet问题(3.50)的弱解.从而由弱解的唯一性得$ v = u $,则(3.53)式即为

于是

$ \begin{equation} \| u_{k}-u \| _{W^{1, p} (B_{2R})}\rightarrow 0 \mbox{当} \;\; k\rightarrow \infty. \end{equation} $

因此存在$ \{u_{k}\}_{k = 1}^{\infty} $的子列(用$ \{u_{k}\} $表示),使得

$ \begin{eqnarray} && u_{k}\rightarrow u \; \; \; a.e. \; in \; B_{2R} , \\ && \nabla u_{k}\rightarrow \nabla u a.e. \; in \; B_{2R} . \end{eqnarray} $

于是由Fatou引理, $ |u_{k}|^{p} \in L^{\infty} (B_{2R}) \subset L^{\phi} (B_{2R}) $, (3.51)式以及(3.55)式,得

这就是说,为证明定理1.1,只需增加一个假设($ \nabla u_{k} $是局部有界的)证明(1.10)式即可.一旦(1.10)式在一般情况下成立,则可经由标准覆盖讨论得到$ |\nabla u|^{p} \in L_{loc}^{\phi} (\Omega) $.

现证明(3.53)式.下面主要考虑$ p\geq 2 $的情形, $ 1<p<2 $时,可参考文献[25]中相同的方法处理.取检验函数$ \varphi _{3} = (u_{k}-v)e^{\theta |u_{k}-v|} $,其中$ \theta $为待定正常数,代入Dirichlet问题(3.50)弱解的定义,得

两式相减整理得

$ \begin{eqnarray} J_{3}+J_{4} = J_{5}, \end{eqnarray} $

其中

估计$ J_{3} $.由(1.4)式得

$ \begin{eqnarray} J_{3}\geq C_{2}\int _{B_{2R}} |\nabla (u_{k}-v)|^{p} e^{\theta |u_{k}-v|}{\rm d}x. \end{eqnarray} $

估计$ J_{4} $.由(1.4)式得

$ \begin{eqnarray} J_{4}\geq C_{2} \theta \int _{B_{2R}} |\nabla (u_{k}-v)|^{p} |u_{k}-v| e^{\theta |u_{k}-v|}{\rm d}x. \end{eqnarray} $

最后估计$ J_{5} $.由(1.6)式得

$ \begin{eqnarray} J_{5}&\leq& C_{4} \int _{B_{2R}} (|\nabla u_{k}|^{p} + |\nabla v|^{p} ) |u_{k}-v| e^{\theta |u_{k}-v|}{\rm d}x \\ &\leq& C_{4} \int _{B_{2R}} |\nabla (u_{k}-v)|^{p} |u_{k}-v| e^{\theta |u_{k}-v|}{\rm d}x + C_{4} \int _{B_{2R}} |\nabla u_{k}|^{p} |u_{k}-v| e^{\theta |u_{k}-v|}{\rm d}x. \end{eqnarray} $

由于$ u_{k}, v \in L^{\infty }( B_{2R} ) $,故存在充分大的正常数$ M $,使得$ |u_{k}|\leq M, |v| \leq M $.综合(3.56), (3.57), (3.58)和(3.59)式,并且令$ \theta $满足$ C_{2} \theta > C_{4} $,移项整理得

$ \begin{eqnarray} \int _{B_{2R}} |\nabla (u_{k}-v)|^{p} {\rm d}x \leq C \int _{B_{2R}} |\nabla u_{k}|^{p} |u_{k}-v| {\rm d}x. \end{eqnarray} $

利用Hölder不等式得

$ \begin{eqnarray} \int _{B_{2R}} |\nabla (u_{k}-v)|^{p} {\rm d}x \leq C \left( \int _{B_{2R}} |\nabla u_{k}|^{p\alpha_{2}} {\rm d}x \right )^{\frac{1}{\alpha_{2}}} \left( \int _{B_{2R}} | u_{k} -v|^{\frac{\alpha_{2}}{\alpha_{2}-1}} {\rm d}x \right )^{\frac{\alpha_{2}-1}{\alpha_{2}}}, \end{eqnarray} $

其中$ \alpha_{2} $如(2.3)式所述.由引理2.1及(2.3)式,对任意$ g \in L^{\phi } (B_{2R}) $,有

其中常数$ a $取自定义2.1, $ a>1 $.由于$ |\nabla u_{k}|^{p} \in L^{\phi}(B_{2R}) $,于是

代入(3.61)式,再利用(3.52)式得

$ \begin{eqnarray} \int _{B_{2R}} |\nabla (u_{k}-v)|^{p} {\rm d}x \leq C \left( \int _{B_{2R}} | u_{k} -v|^{\frac{\alpha_{2}}{\alpha_{2}-1}} {\rm d}x \right )^{\frac{\alpha_{2}-1}{\alpha_{2}}} \rightarrow 0, \mbox{当} \;\; k \rightarrow \infty, \end{eqnarray} $

其中$ C = C( C_{2}, C_{4}, M, n, p, a) $.于是(3.53)式得证.

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