0 引言

近几十年来, 数学工作者十分关注偏微分方程尤其是非线性椭圆方程解的存在性、唯一性、正则性等问题.本文在加权Sobolev空间框架下，研究一类带有退化强制项的非线性椭圆方程

$ \left\{ \begin{array}{l} - {\rm{div}}\left( {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla u} \right)}}{{{{\left( {1 + \left| u \right|} \right)}^\gamma }}}} \right) + g\left( {x,u} \right) = f,x \in \Omega ,\\ u = 0,\;\;\;\;\;x \in \partial \Omega , \end{array} \right. $

(1)

其中Ω是R^N(N≥2)中的有界区域，γ > 0, f∈L¹(Ω)，W₀^{1, p}(Ω, ω)为加权的Sobolev空间，ω ={ω_i(x)}_0≤i≤N为权函数，p为满足1 < p < ∞的实数.此外，假设：

H1   a (x, ξ)={a_i(x, ξ)}_1≤i≤N：Ω×R^N→R^N是一个Carathéodory向量值函数，对任意的x∈Ω，每一个ξ∈R^N有以下不等式成立：

$ \left| {{a_i}\left( {x,\xi } \right)} \right| \le \beta \omega _i^{\frac{1}{p}}\left( x \right)\left[ {k\left( x \right) + \sum\limits_{j = 1}^N {\omega _i^{\frac{1}{{p'}}}\left( x \right){{\left| {{\xi _j}} \right|}^{p - 1}}} } \right], $

(2)

$ \mathit{\boldsymbol{a}}\left( {x,\xi } \right) \cdot \xi \ge \alpha \sum\limits_{i = 1}^N {{\omega _i}\left( x \right){{\left| {{\xi _i}} \right|}^p}} , $

(3)

$ \left[ {\mathit{\boldsymbol{a}}\left( {x,\xi } \right) - \mathit{\boldsymbol{a}}\left( {x,\eta } \right)} \right] \cdot \left( {\xi - \eta } \right) > 0,\xi \ne \eta \in {R^N}, $

(4)

其中，k(x)是L^p′(Ω)中一个正函数, $ \frac{1}{p} + \frac{1}{{p'}} = 1 $, α, β为正数.

H2  g(x, s)是一个Carathéodory函数，对几乎所有的x∈Ω，每一个ξ∈R^N，且对任意的k∈R⁺，有

$ g\left( {x,s} \right) \cdot s \ge 0, $

(5)

$ \sup \left| {g\left( {x,s} \right)} \right| = {h_k}\left( x \right) \in {L^1}\left( \Omega \right). $

(6)

H3  ω ={ω_i(x)}_0≤i≤N是一个在Ω上几乎处处严格正的可测权函数向量, 满足

$ {\omega _i} \in L_{{\rm{loc}}}^1\left( \Omega \right),\;\;\;\;\omega _i^{ - \frac{1}{{p - 1}}} \in L_{{\rm{loc}}}^1\left( \Omega \right). $

下文将从两方面来陈述所研究问题的特点.首先，与其他文献最主要的区别是，本研究的椭圆问题(1)在加权的Sobolev空间，权函数的引入，使得嵌入关系发生了很大的变化，这给问题的解决带来了一定的困难.另外，很多文献关注了与式(1)类似的问题^[1-7]，文献[1]考虑了在空间不加权的框架下，当p=2，0≤γ < 1，f∈L^m(Ω)且g(x, u)=0时f可积性的变化对u正则性的影响. BOCCARDO等^[8]同样在空间不加权的情形下，研究了当g(x, u)=u, f∈L^m(Ω)时问题(1)解的存在性与非存在性. CROCE等^[6]证明了在空间不加权的情形下，当g(x, u)=|u|^q-1u时，存在解u∈L^q(Ω)，并讨论了指标q对解u正则性的影响.

其次, 问题(1)的假设条件(2)意味着当u很大时，$ \frac{1}{{{{\left( {1 + |u|} \right)}^\gamma }}} \to 0 $，使得由$ - {\rm{div}}\left( {\frac{{\mathit{\boldsymbol{a}}\left( {x, \nabla u} \right)}}{{{{\left( {1 + |u|} \right)}^\gamma }}}} \right) $所定义的算子项非强制.因逼近问题中定义的g_n(x, u_n)在L¹(Ω)中的强收敛性是非常重要的，但强制性的缺失导致g_n(x, u_n)·u_n在L¹(Ω)中非一致有界，因而无法得到u_n的先验估计.另外，考虑了一个仅满足部分增长性条件(6)的低阶项g(x, u)，同时右端项f仅在L¹(Ω)中.以上困难导致无法直接利用对偶理论和单调算子理论证明解的存在性，必须寻找其他解决方法.因此，本文采用了Marcinkiewicz估计，这不仅有助于克服退化强制所带来的影响，还处理了低阶项.通过选取适当的检验函数得到逼近解序列的截断函数T_k(u_n)强收敛，以此证明了问题(1)重整化解的存在性.

1 准备知识

首先，介绍常指数情形下加权Sobolev空间的相关知识^[9].

(1) L^p(Ω, γ)空间

$ {L^p}\left( {\Omega ,\mathit{\boldsymbol{\gamma }}} \right) = \left\{ {u = u\left( x \right):u{\mathit{\boldsymbol{\gamma }}^{\frac{1}{p}}} \in {L^p}\left( \Omega \right)} \right\}, $

其中γ为权函数, 赋予以下Luxemburg范数：

$ {\left\| u \right\|_{p,\mathit{\boldsymbol{\gamma }}}} = {\left( {\int_\Omega {{{\left| {u\left( x \right)} \right|}^p}\mathit{\boldsymbol{\gamma }}\left( x \right){\rm{d}}x} } \right)^{\frac{1}{p}}}. $

(2) W₀^{1, p}(Ω, ω)空间

记X=：W₀^{1, p}(Ω, ω)空间为所有实值函数u∈L^p(Ω, ω₀)所构成的空间，对所有的i=1, 2, …, N，其弱导数满足

$ \frac{{\partial u}}{{\partial {x_i}}} \in {L^p}\left( {\Omega ,{\omega _i}} \right). $

赋予范数

$ \begin{array}{l} {\left\| u \right\|_{1,p,\mathit{\boldsymbol{\omega }}}} = \left( {\int_\Omega {{{\left| {u\left( x \right)} \right|}^p}{\omega _0}\left( x \right){\rm{d}}x} + } \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\;{\left. {\sum\limits_{i = 1}^N {\int_\Omega {{{\left| {\frac{{\partial u\left( x \right)}}{{\partial {x_i}}}} \right|}^p}{\omega _i}\left( x \right){\rm{d}}x} } } \right)^{\frac{1}{p}}}. \end{array} $

定义在X上的范数为

$ {\left\| u \right\|_X}: = {\left( {\sum\limits_{i = 1}^N {\int_\Omega {{{\left| {\frac{{\partial u\left( x \right)}}{{\partial {x_i}}}} \right|}^p}{\omega _i}\left( x \right){\rm{d}}x} } } \right)^{\frac{1}{p}}}, $

等价于‖·‖_{1, p, ω}.(X, ‖·‖_{1, p, ω})是一个自反的Banach空间, 其对偶空间为W₀^{-1, p′}(Ω, ω^*), 其中$ {\mathit{\boldsymbol{\omega }}^*} = \{ \omega _i^* = \omega _i^{\frac{{ - p'}}{p}}\} $, 对所有的i=1, 2, …, N, $ p' = \frac{p}{{p - 1}} $成立.

(3) 加权的Hardy型不等式

一般情形下，存在权函数ω (x)和实数q(1 < q < ∞)，满足σ^1-q′∈L¹(Ω), $ q' = \frac{q}{{q - 1}} $, 对每一个u∈X以及与u无关的正常数C, 都有以下形式的Hardy不等式成立：

$ {\left( {\int_\Omega {{{\left| {u\left( x \right)} \right|}^q}\sigma {\rm{d}}x} } \right)^{\frac{1}{q}}} \le C{\left( {\sum\limits_{i = 1}^N {\int_\Omega {{{\left| {\frac{{\partial u\left( x \right)}}{{\partial {x_i}}}} \right|}^p}{\omega _i}\left( x \right){\rm{d}}x} } } \right)^{\frac{1}{p}}}. $

进而，X紧嵌入L^q(Ω, σ)中.有关加权的Hardy不等式例子可参见文献[10-11].

其次, 介绍截断函数的相关知识.一般情形下, 对于在R中的s, k, 其中k≥0, 高度为k的Truncation函数^[12-13]定义为

$ {T_k}\left( s \right) = \max \left( { - k,\min \left( {k,s} \right)} \right) = \left\{ \begin{array}{l} s,\left| s \right| < k,\\ k,s \ge k,\\ - k,s \le - k. \end{array} \right. $

鉴于其重要性，给出其简图，见图 1.

Marcinkiewicz空间的定义由BÉNLIAN等^[14]提出.

定义1  若一个可测函数f：Ω→R相应的分布函数ϕ_f(k)≤meas{x∈Ω：|f(x)| > k}, k > 0, 满足

$ {\phi _f}\left( k \right) \le C{k^{ - q}},C\;为常数, $

则称f属于Marcinkiewicz空间Μ^q(Ω).

下面给出问题(1)重整化解的定义, 此定义在带有扩散项的微分方程理论中是非常经典的.

定义2  若满足

(1) T_k(u)∈W₀^{1, p}(Ω, ω), k > 0;

(2) g(x, u)∈L¹(Ω)，同时

$ \mathop {\lim }\limits_{j \to \infty } \int_{\Omega \cap \left\{ {j \le \left| u \right| \le j + 1} \right\}} {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla u} \right)}}{{{{\left( {1 + \left| u \right|} \right)}^\gamma }}}\nabla u{\rm{d}}x} = 0; $

(7)

(3) 对每一个具有紧支集的函数S∈W^{1, ∞}(R)，对任何ζ∈W₀^{1, p}(Ω, ω)∩L^∞(Ω)均有

$ \begin{array}{l} \int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla u} \right)}}{{{{\left( {1 + \left| u \right|} \right)}^\gamma }}} \cdot \nabla \left( {S\left( u \right)\zeta } \right){\rm{d}}x} + \\ \;\;\;\;\int_\Omega {g\left( {x,u} \right)S\left( u \right)\zeta {\rm{d}}x} = \int_\Omega {fS\left( u \right)\zeta {\rm{d}}x} , \end{array} $

(8)

则称可测函数u∈W₀^{1, p}(Ω, ω)是问题(1)的重整化解.

2 重整化解的存在性

定理1  假设H1~H3成立，f∈L¹(Ω)，则问题(1)至少存在1个重整化解u.

证明  分5步完成.

第1步  逼近问题及先验估计.

先建立问题(1)的逼近问题.对n∈ N, 设u_n满足

$ \left\{ \begin{array}{l} - {\rm{div}}\left( {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {u_n}} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}} \right) + {g_n}\left( {x,{u_n}} \right) = {f_n},\;\;\;x \in \Omega ,\\ {u_n} = 0,\;\;\;\;\;\;x \in \partial \Omega , \end{array} \right. $

(9)

其中f_n是L^∞(Ω)中的函数序列且在L¹(Ω)中强收敛于f并且‖f_n‖_L¹(Ω)≤‖f‖_L¹(Ω)，g_n(x, s)=T_ng(x, s), 且满足式(5)和(6).由文献[15]的pseudo-monotone算子理论可知, 逼近问题(9)至少存在1个弱解且对任意v∈W₀^{1, p}(Ω, ω)∩L^∞(Ω), 有

$ \int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {u_n}} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}} \cdot \nabla v{\rm{d}}x + \int_\Omega {{g_n}\left( {x,{u_n}} \right)v{\rm{d}}x} = \int_\Omega {{f_n}v{\rm{d}}x} . $

(10)

接下来, 对解序列u_n做一些先验估计, 选取式(9)中T_k(u_n)(k > 0)作为检验函数, 则有

$ \begin{array}{l} \int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {u_n}} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right.}^\gamma }}}} \cdot \nabla {T_k}\left( {{u_n}} \right){\rm{d}}x + \\ \;\;\;\;\;\int_\Omega {{g_n}\left( {x,{u_n}} \right){T_k}\left( {{u_n}} \right){\rm{d}}x} = \int_\Omega {{f_n}{T_k}\left( {{u_n}} \right){\rm{d}}x} . \end{array} $

(11)

由于T_k(s)与s同号，结合式(5)，式(11)左端第2项为非负项，去掉非负项可得

$ \int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {u_n}} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right.}^\gamma }}}} \cdot \nabla {T_k}\left( {{u_n}} \right){\rm{d}}x \le k\int_\Omega {\left| {{f_n}} \right|{\rm{d}}x} . $

考虑式(3)，并选取n > k，有

$ \frac{\alpha }{{{{\left( {1 + k} \right)}^\gamma }}}\sum\limits_{i = 1}^N {\int_\Omega {{\omega _i}\left( x \right){{\left| {\frac{{\partial {T_k}\left( {{u_n}} \right)}}{{\partial {x_i}}}} \right|}^p}{\rm{d}}x} } \le k\int_\Omega {\left| {{f_n}} \right|{\rm{d}}x} . $

(12)

因此，对所有的k > 0，有

$ \sum\limits_{i = 1}^N {\int_\Omega {{\omega _i}\left( x \right){{\left| {\frac{{\partial {T_k}\left( {{u_n}} \right)}}{{\partial {x_i}}}} \right|}^p}{\rm{d}}x} } \le \frac{{k{{\left( {1 + k} \right)}^\gamma }{{\left\| f \right\|}_{{L^1}\left( \Omega \right)}}}}{{{c_0}}}. $

(13)

如果k≥1, 有

$ \sum\limits_{i = 1}^N {\int_\Omega {{\omega _i}\left( x \right){{\left| {\frac{{\partial {T_k}\left( {{u_n}} \right)}}{{\partial {x_i}}}} \right|}^p}{\rm{d}}x} } \le \frac{{{2^\gamma }{k^{\gamma + 1}}{{\left\| f \right\|}_{{L^1}\left( \Omega \right)}}}}{{{c_0}}}. $

(14)

由加权Hardy不等式，同时结合式(14)以及σ^1-q′∈L¹(Ω)，当k≥1，p > γ+1时，有

$ \begin{array}{l} \left| {\left\{ {\left| {{u_n}} \right| > k} \right\}} \right| = \int_{\left\{ {\left| {{u_n}} \right| > k} \right\}} {\frac{{\left| {{T_k}\left( {{u_n}} \right)} \right|{\sigma ^{\frac{1}{q}}}}}{{k{\sigma ^{\frac{1}{q}}}}}{\rm{d}}x} \le \\ \;\;\;\;\frac{1}{k}{\left( {\int_\Omega {{{\left| {{T_k}\left( {{u_n}} \right)} \right|}^q}\sigma {\rm{d}}x} } \right)^{\frac{1}{q}}}{\left( {\int_\Omega {{\sigma ^{ - \frac{{q'}}{q}}}{\rm{d}}x} } \right)^{\frac{1}{q}}} \le \\ \;\;\;\;\frac{1}{k}{\left( {\sum\limits_{i = 1}^N {\int_\Omega {{\omega _i}\left( x \right)\left| {\frac{{\partial {T_k}\left( {{u_n}} \right)}}{{\partial {x_i}}}} \right|{\rm{d}}x} } } \right)^{\frac{1}{p}}}{\left( {\int_\Omega {{\sigma ^{ - \frac{{q'}}{q}}}{\rm{d}}x} } \right)^{\frac{1}{{q'}}}} \le \\ \;\;\;\;C{k^{\frac{{\gamma + 1 - p}}{p}}}{\left\| {{\sigma ^{1 - q'}}} \right\|_{{L^1}\left( \Omega \right)}}, \end{array} $

其中C为常数，当0 < k < 1时，显然

$ {\rm{meas}}\left\{ {\left| {{u_n}} \right| > k} \right\} \le \left| \Omega \right|, $

得到$ {u_n} \in {M^{\frac{{p - \gamma - 1}}{p}}}\left( \Omega \right) $.因此，有

$ \mathop {\lim }\limits_{k \to + \infty } {\rm{meas}}\left\{ {\left| {{u_n}} \right| > k} \right\} = 0,关于\;n\;一致. $

(15)

第2步  u_n在Ω上几乎处处收敛.

先证u_n依测度收敛.注意到对任意的k, ε > 0, 有

$ \begin{array}{l} \left\{ {\left| {{u_n} - {u_m}} \right| > t} \right\} \subset \left\{ {\left| {{u_n}} \right| > k} \right\} \cup \left\{ {\left| {{u_m}} \right| > k} \right\} \cup \\ \;\;\;\;\;\left\{ {{T_k}\left( {{u_n}} \right) - {T_k}\left( {{u_m}} \right)\left| { > t} \right.} \right\}, \end{array} $

即

$ \begin{array}{l} {\rm{meas}}\left\{ {\left| {{u_n} - {u_m}} \right| > t} \right\} \le {\rm{meas}}\left\{ {\left| {{u_n}} \right| > k} \right\} + \\ \;\;\;\;\;\;{\rm{meas}}\left\{ {\left| {{u_m}} \right| > k} \right\} + {\rm{meas}}\left\{ {{T_k}\left( {{u_n}} \right) - {T_k}\left( {{u_m}} \right)\left| { > t} \right.} \right\}. \end{array} $

(16)

由式(13)可知，T_k(u_n)在W₀^{1, p}(Ω, ω)中有界，那么存在ν_k∈W₀^{1, p}(Ω, ω)，使得对任意的k > 0, 有T_k(u_n)→ν_k于W₀^{1, p}(Ω, ω)中弱收敛，T_k(u_n)→ν_k于L^q(Ω, σ)中强收敛，且在Ω上几乎处处收敛.

由此可知，T_k(u_n)是在Ω中依测度收敛的柯西列.那么，对任意的ε > 0，存在k(ε) > 0，当m，n充分大时，并结合式(16)可得，u_n是一个依测度收敛的Cauchy序列，因此，u_n依测度收敛^[16].根据Riesz定理，u_n存在一个子列(仍记其本身)以及一个可测函数u，使得

$ {u_n} \to u\;在\;\Omega \;上几乎处处收敛. $

(17)

联合式(13)和(17)，对任意的k > 0，有

$ {T_k}\left( {{u_n}} \right) \to {T_k}\left( u \right)\; 于\;W_0^{1,p}\left( {\Omega ,\mathit{\boldsymbol{\omega }}} \right)\;中弱收敛, $

(18)

$ {T_k}\left( {{u_n}} \right) \to {T_k}\left( u \right)在\;\Omega \;上几乎处处收敛. $

(19)

因此，对任意的k > 0, 有T_k(u)∈W₀^{1, p}(Ω, ω).

第3步  g_n(x, u_n)在L¹(Ω)中强收敛.

证明

$ {g_n}\left( {x,{u_n}} \right) \to g\left( {x,u} \right)\;于\;{L^1}\left( \Omega \right)\;中强收敛. $

(20)

直观起见，给出θ_l¹(s)的简图如图 2所示.在式(9)中选取θ_l¹(u_n)=T_l+1(u_n)-T_l(u_n) (l > 0)作为检验函数，则有

$ \begin{array}{l} \int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {u_n}} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}} \cdot \nabla \left( {{T_{l + 1}}\left( {{u_n}} \right) - {T_{l + 1}}\left( {{u_n}} \right)} \right){\rm{d}}x + \\ \;\;\;\;\;\;\int_\Omega {{g_n}\left( {x,{u_n}} \right)\left( {{T_{l + 1}}\left( {{u_n}} \right) - {T_{l + 1}}\left( {{u_n}} \right)} \right){\rm{d}}x} = \\ \;\;\;\;\;\;\int_\Omega {{f_n}\left( {{T_{l + 1}}\left( {{u_n}} \right) - {T_l}\left( {{u_n}} \right)} \right){\rm{d}}x} . \end{array} $

整理后可得

$ \begin{array}{l} \int_{\left\{ {l \le \left| {{u_n}} \right| \le l + 1} \right\}} {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {u_n}} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}\nabla {u_n}{\rm{d}}x} + \\ \;\;\;\;\;\int_{\left\{ {\left| {{u_n}} \right| \ge l} \right\}} {\left| {{g_n}\left( {x,{u_n}} \right)} \right|{\rm{d}}x} \le \int_{\left\{ {\left| {{u_n}} \right| \ge l} \right\}} {\left| {{f_n}} \right|{\rm{d}}x} . \end{array} $

由式(3)可知，上式左端第1项为非负项，去掉非负项，有

$ \int_{\left\{ {\left| {{u_n}} \right| \ge l} \right\}} {\left| {{g_n}\left( {x,{u_n}} \right)} \right|{\rm{d}}x} \le \int_{\left\{ {\left| {{u_n}} \right| \ge l} \right\}} {\left| {{f_n}} \right|{\rm{d}}x} . $

注意到式(15)以及f_n在L¹(Ω)中强紧，有

$ \mathop {\lim }\limits_{k \to + \infty } \mathop {\sup }\limits_{n \in N} \int_{\left\{ {\left| {{u_n}} \right| \ge k} \right\}} {\left| {{f_n}} \right|{\rm{d}}x = 0} . $

设ε > 0，存在l(ε)≥1，使得

$ \int_{\left\{ {\left| {{u_n}} \right| \ge l\left( \varepsilon \right)} \right\}} {\left| {{g_n}\left( {x,{u_n}} \right)} \right|{\rm{d}}x} \le \frac{\varepsilon }{2}. $

(21)

对于Ω中的任何可测子集E，有

$ \begin{array}{l} \int_E {\left| {{g_n}\left( {x,{u_n}} \right)} \right|{\rm{d}}x} \le \int_{E \cap \left\{ {\left| {{u_n}} \right| \ge l\left( \varepsilon \right)} \right\}} {\left| {{g_n}\left( {x,{u_n}} \right)} \right|{\rm{d}}x} + \\ \;\;\;\;\int_{E \cap \left\{ {\left| {{u_n}} \right| > l\left( \varepsilon \right)} \right\}} {\left| {{g_n}\left( {x,{u_n}} \right)} \right|{\rm{d}}x} \le \int_E {\left| {{h_{l\left( \varepsilon \right)}}\left( x \right)} \right|{\rm{d}}x} + \\ \;\;\;\;\int_{E \cap \left\{ {\left| {{u_n}} \right| > l\left( \varepsilon \right)} \right\}} {\left| {{g_n}\left( {x,{u_n}} \right)} \right|{\rm{d}}x} . \end{array} $

注意到式(6)，存在η(ε) > 0，满足meas(E) < η(ε)，使得

$ \int_E {\left| {{h_{l\left( \varepsilon \right)}}\left( x \right)} \right|{\rm{d}}x} \le \frac{\varepsilon }{2}. $

(22)

综上，结合式(21)和(22)，易得对所有的E，若满足meas(E) < η(ε)，则有∫_E|g_n(x, u_n)|dx≤ε.同时由式(17)可知，g_n(x, u_n)→g(x, u)在Ω上几乎处处收敛.根据Vitali定理，有

g_n(x, u_n)→g(x, u)于L¹(Ω)中强收敛.

第4步  T_k(u_n)在W₀^{1, p}(Ω, ω)中强收敛于T_k(u).

下面证明对每一个k > 0, T_k(u_n)在W₀^{1, p}(Ω, ω)中强收敛于T_k(u).设h > k，选取T_2k(u_n-T_h(u_n)+ T_k(u_n)-T_k(u))作为式(9)的1个检验函数，有

$ \begin{array}{l} \int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {u_n}} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}} \cdot \nabla {T_{2k}}\left( {{u_n} - {T_h}\left( {{u_n}} \right) + {T_k}\left( {{u_n}} \right) - } \right.\\ \;\;\;\;\;\;\left. {{T_k}\left( u \right)} \right){\rm{d}}x + \int_\Omega {{g_n}\left( {x,{u_n}} \right){T_{2k}}\left( {{u_n} - {T_h}\left( {{u_n}} \right) + } \right.} \\ \;\;\;\;\;\;\left. {{T_k}\left( {{u_n}} \right) - {T_k}\left( u \right)} \right){\rm{d}}x = \int_\Omega {{f_n}{T_{2k}}\left( {{u_n} - {T_h}\left( {{u_n}} \right) + } \right.} \\ \;\;\;\;\;\;\left. {{T_k}\left( {{u_n}} \right) - {T_k}\left( u \right)} \right){\rm{d}}x. \end{array} $

(23)

方便起见，将式(23)记为

$ {I_1} + {I_2} = {I_3}, $

同时定义ε_{n, h}为

$ \mathop {\lim }\limits_{h \to + \infty } \mathop {\lim }\limits_{n \to + \infty } {\varepsilon _{n,h}} = 0, $

类似地，ε_n表示$ \mathop {\lim }\limits_{n \to \infty } $ ε_n=0.

关于I₂，I₃，结合式(17)和(20)，以及f_n在L¹(Ω)中强紧，有

$ \begin{array}{l} \int_\Omega {{g_n}\left( {x,{u_n}} \right){T_{2k}}\left( {{u_n} - {T_h}\left( {{u_n}} \right) + {T_k}\left( {{u_n}} \right) - } \right.} \\ \;\;\;\;\;\;\;\left. {{T_k}\left( u \right)} \right){\rm{d}}x = {\varepsilon _{n,h}}, \end{array} $

(24)

$ \int_\Omega {{f_n}{T_{2k}}\left( {{u_n} - {T_h}\left( {{u_n}} \right) + {T_k}\left( {{u_n}} \right) - {T_k}\left( u \right)} \right){\rm{d}}x} = {\varepsilon _{n,h}}. $

(25)

关于I₁，令M=4k+h，易知当|u_n| > M时，

$ \nabla {T_{2k}}\left( {{u_n} - {T_h}\left( {{u_n}} \right) + {T_k}\left( {{u_n}} \right) - {T_k}\left( u \right)} \right) = 0, $

I₁可化为

$ \begin{array}{l} {I_1} = \int_{\left\{ {\left| {{u_n}} \right| < k} \right\}} {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {T_M}\left( {{u_n}} \right)} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}} \cdot \nabla {T_{2k}}\left( {{u_n} - {T_h}\left( {{u_n}} \right) + } \right.} \\ \;\;\;\;\;\;\left. {{T_k}\left( {{u_n}} \right) - {T_k}\left( u \right)} \right){\rm{d}}x + \int_{\left\{ {\left| {{u_n}} \right| < k} \right\}} {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {T_M}\left( {{u_n}} \right)} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}} \cdot } \\ \;\;\;\;\;\;\nabla {T_{2k}}\left( {{u_n} - {T_h}\left( {{u_n}} \right) + {T_k}\left( {{u_n}} \right) - {T_k}\left( u \right)} \right){\rm{d}}x. \end{array} $

考虑到在集合{|u_n| < k}上，u_n-T_h(u_n)=0，集合{|u_n|≥k}上，∇T_k(u_n)=0，并注意到a(x, ξ)·ξ≥0，有

$ \begin{array}{l} {I_1} \ge \int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {T_k}\left( {{u_n}} \right)} \right)\nabla \left( {{T_k}\left( {{u_n}} \right) - {T_k}\left( u \right)} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}{\rm{d}}x} + \\ \;\;\;\;\;\;\int_{\left\{ {\left| {{u_n}} \right| \ge k} \right\}} {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {T_M}\left( {{u_n}} \right)} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}\nabla \left( {{u_n} - {T_h}\left( {{u_n}} \right)} \right){\rm{d}}x} - \\ \;\;\;\;\;\;\int_{\left\{ {\left| {{u_n}} \right| \ge k} \right\}} {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {T_M}\left( {{u_n}} \right)} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}\nabla {T_h}\left( u \right){\rm{d}}x} . \end{array} $

去掉上式右端第2个非负项，有

$ \begin{array}{l} {I_1} \ge \int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {T_k}\left( {{u_n}} \right)} \right)\nabla \left( {{T_k}\left( {{u_n}} \right) - {T_k}\left( u \right)} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}{\rm{d}}x} - \\ \;\;\;\;\;\;\int_{\left\{ {\left| {{u_n}} \right| \ge k} \right\}} {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {T_M}\left( {{u_n}} \right)} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}\nabla {T_k}\left( u \right){\rm{d}}x} . \end{array} $

并对右端第1项进行整理，可得

$ \begin{array}{l} {I_1} \ge \\ \int_\Omega {\frac{{\left[ {\mathit{\boldsymbol{a}}\left( {x,\nabla {T_k}\left( {{u_n}} \right)} \right) - \mathit{\boldsymbol{a}}\left( {x,\nabla {T_k}\left( u \right)} \right)} \right]\nabla \left( {{T_k}\left( {{u_n}} \right) - {T_k}\left( u \right)} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}{\rm{d}}x} + \\ \int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {T_k}\left( u \right)} \right)\nabla \left( {{T_k}\left( {{u_n}} \right) - {T_k}\left( u \right)} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}{\rm{d}}x} - \\ \int_{\left\{ {\left| {{u_n}} \right| \ge k} \right\}} {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {T_M}\left( {{u_n}} \right)} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}\nabla {T_k}\left( u \right){\rm{d}}x} = \\ {I_{11}} + {I_{12}} + {I_{13}}. \end{array} $

关于I₁₂，由式(2)可知，$\frac{{\mathit{\boldsymbol{a}}(x, \nabla {T_k}(u))}}{{{{(1 + |{T_n}({u_n})|)}^\gamma }}} $在(L^p(Ω, ω))^N中强紧，结合式(18)，有

$ {I_{12}} = {\varepsilon _n}. $

(26)

关于I₁₃，可整理为

$ {I_{13}} = - \int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {T_M}\left( {{u_n}} \right)} \right)\nabla {T_k}\left( u \right){{\rm{ \mathsf{ χ} }}_{\left\{ {\left| {{u_n}} \right| \ge k} \right\}}}}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}{\rm{d}}x} . $

令n > M，当n→∞时，

$\frac{{\nabla {T_k}(u)\chi \{ |{u_n}| \ge k\} }}{{{{(1 + |{T_n}({u_n})|)}^\gamma }}} \to 0 $于(L^p(Ω, ω))^N中强收敛.并且由式(2)可知，a (x, ∇T_M(u_n))在(L^p′(Ω, ω^*))^N中有界，则有

$ {I_{13}} = {\varepsilon _n}. $

因而，结合I₁₂，I₁₃的估计式，I₁可整理为

$ \begin{array}{l} {I_1} \ge \\ \int_\Omega {\frac{{\left[ {\mathit{\boldsymbol{a}}\left( {x,\nabla {T_k}\left( {{u_n}} \right)} \right) - \mathit{\boldsymbol{a}}\left( {x,\nabla {T_k}\left( u \right)} \right)} \right]\nabla \left( {{T_k}\left( {{u_n}} \right) - {T_k}\left( u \right)} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}{\rm{d}}x} . \end{array} $

综上，注意到式(24)和(25)，对n > M > h > k，有

$ \begin{array}{l} {\varepsilon _{n,h}} = {I_1} \ge \\ \int_\Omega {\frac{{\left[ {\mathit{\boldsymbol{a}}\left( {x,\nabla {T_k}\left( {{u_n}} \right)} \right) - \mathit{\boldsymbol{a}}\left( {x,\nabla {T_k}\left( u \right)} \right)} \right]\nabla \left( {{T_k}\left( {{u_n}} \right) - {T_k}\left( u \right)} \right)}}{{{{\left( {1 + k} \right)}^\gamma }}}{\rm{d}}x} . \end{array} $

考虑到式(4)，则有

$ \begin{array}{l} \mathop {\lim }\limits_{n \to \infty } \int_\Omega {\left[ {\mathit{\boldsymbol{a}}\left( {x,\nabla {T_k}\left( {{u_n}} \right)} \right) - \mathit{\boldsymbol{a}}\left( {x,\nabla {T_k}\left( u \right)} \right)} \right]\nabla \left( {{T_k}\left( {{u_n}} \right) - } \right.} \\ \;\;\;\;\;\;\left. {{T_k}\left( u \right)} \right){\rm{d}}x = 0. \end{array} $

于是可得，当n→∞时，有

$ {T_k}\left( {{u_n}} \right) \to {T_k}\left( u \right)\;于\;{\left( {{L^p}\left( {\Omega ,\mathit{\boldsymbol{\omega }}} \right)} \right)^N}\;中强收敛. $

(27)

第5步  u为重整化解.

选取u_nζ作为式(9)的一个检验函数，其中ζ∈D(Ω)，且S∈W^{1, ∞}(Ω)具有紧支集，有

$ \begin{array}{l} \int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {u_n}} \right)}}{{\left( {1 + {T_n}\left( {{u_n}} \right)} \right)}^{\gamma }}\zeta } \cdot \nabla S\left( {{u_n}} \right){\rm{d}}x + \\ \;\;\;\;\;\int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {u_n}} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}S\left( {{u_n}} \right)} \cdot \nabla \zeta {\rm{d}}x + \\ \;\;\;\;\;\int_\Omega {{g_n}\left( {x,{u_n}} \right)S\left( {{u_n}} \right)\zeta {\rm{d}}x} = \int_\Omega {{f_n}S\left( {{u_n}} \right)\zeta {\rm{d}}x} . \end{array} $

(28)

由于f_n在L¹(Ω)中是强紧的，且S(u_n)ζ在L^∞(Ω)中有界，结合式(17)和(20)，当n→∞时，可得

$ \begin{array}{*{20}{c}} {\int_\Omega {{g_n}\left( {x,{u_n}} \right)S\left( {{u_n}} \right)\zeta {\rm{d}}x} \to \int_\Omega {g\left( {x,u} \right)S\left( u \right)\zeta {\rm{d}}x} ,}\\ {\int_\Omega {{f_n}S\left( {{u_n}} \right)\zeta {\rm{d}}x} \to \int_\Omega {fS\left( u \right)\zeta {\rm{d}}x} } \end{array} $

因为函数S，S′具有紧支集，那么存在L > 0，使得suppS⊂[-L, L]，suppS′⊂[-L, L].当n充分大时，结合式(9)和(17)，有

$ \begin{array}{l} \int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {u_n}} \right)}}{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^{\gamma }}\zeta } \cdot \nabla S\left( {{u_n}} \right){\rm{d}}x = \\ \;\;\;\;\;\;\;\int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {T_L}\left( {{u_n}} \right)} \right)}}{{{{\left( {1 + \left| {{T_L}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}\zeta } \cdot \nabla S\left( {{u_n}} \right){\rm{d}}x \to \\ \;\;\;\;\;\;\;\int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {T_L}\left( u \right)} \right)}}{{{{\left( {1 + \left| {{T_L}\left( u \right)} \right|} \right)}^\gamma }}}\zeta } \cdot \nabla S\left( u \right){\rm{d}}x = \\ \;\;\;\;\;\;\;\int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla u} \right)}}{{{{\left( {1 + u} \right)}^\gamma }}}\zeta } \cdot \nabla S\left( u \right){\rm{d}}x. \end{array} $

注意到式(2)和(27)，当n充分大时，有

$ \begin{array}{l} \int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {u_n}} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}} S\left( {{u_n}} \right) \cdot \nabla \zeta {\rm{d}}x = \\ \;\;\;\;\;\;\;\int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {T_L}\left( {{u_n}} \right)} \right)}}{{{{\left( {1 + \left| {{T_L}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}S\left( {{u_n}} \right)} \cdot \nabla \zeta {\rm{d}}x \to \\ \;\;\;\;\;\;\;\int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {T_L}\left( u \right)} \right)}}{{{{\left( {1 + \left| {{T_L}\left( u \right)} \right|} \right)}^\gamma }}}S\left( u \right)} \cdot \nabla \zeta {\rm{d}}x = \\ \;\;\;\;\;\;\;\int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla u} \right)}}{{{{\left( {1 + u} \right)}^\gamma }}}S\left( u \right)} \cdot \nabla \zeta {\rm{d}}x. \end{array} $

综上，对式(28)在n→∞时取极限，可得

$ \begin{array}{l} \int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla u} \right)}}{{{{\left( {1 + \left| u \right|} \right)}^\gamma }}}\zeta } \cdot \nabla S\left( u \right){\rm{d}}x + \\ \;\;\;\;\;\int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla u} \right)}}{{{{\left( {1 + \left| u \right|} \right)}^\gamma }}}S\left( u \right)} \cdot \nabla \zeta {\rm{d}}x + \\ \;\;\;\;\;\int_\Omega {g\left( {x,u} \right)S\left( u \right)\zeta {\rm{d}}x} = \int_\Omega {fS\left( u \right)\zeta {\rm{d}}x} , \end{array} $

此式等价于式(8).

现证明重整化解存在的条件，即当j→∞时，$ {\smallint _{\{ j < \left| u \right| < j + 1\} }}\frac{{\mathit{\boldsymbol{a}}(x, \nabla u)}}{{{{\left( {1 + |u|} \right)}^\gamma }}} \cdot \nabla u{\rm{d}}x \to 0 $.对n > j+1，运用Lebesgue控制收敛定理以及式(27)和(17)可知，当n→∞时，有

$ \begin{array}{l} \int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla {u_n}} \right)}}{{{{\left( {1 + \left| {{T_n}\left( {{u_n}} \right)} \right|} \right)}^\gamma }}}\zeta } \cdot \left| {\nabla \left( {{T_{j + 1}}\left( {{u_n}} \right) - } \right.} \right.\\ \;\;\;\;\;\;\left. {\left. {{T_j}\left( {{u_n}} \right)} \right)} \right|{\rm{d}}x \to \int_\Omega {\frac{{\mathit{\boldsymbol{a}}\left( {x,\nabla u} \right)}}{{{{\left( {1 + \left| u \right|} \right)}^\gamma }}}\zeta } \cdot \\ \;\;\;\;\;\;\left| {\nabla \left( {{T_{j + 1}}\left( {{u_n}} \right) - {T_j}\left( u \right)} \right)} \right|{\rm{d}}x, \end{array} $

令j→∞，即可得到

$ \int_{\left\{ {j < \left| u \right| < j + 1} \right\}} {\mathit{\boldsymbol{a}}\left( {x,\nabla u} \right) \cdot \nabla u{\rm{d}}x} \to 0. $