数学杂志  2021, Vol. 41 Issue (5): 423-434   PDF    
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本文作者相关文章
陈方敏
严畅
柴晓娟
非线性退化椭圆型方程弱解的存在唯一性
陈方敏, 严畅, 柴晓娟    
安徽大学数学科学学院, 安徽 合肥 230601
摘要:本文研究了带有退化强制项的非线性椭圆型方程在分布意义下的弱解.利用适当的紧性定理和光滑逼近,在建立逼近方程解的一致正则性和逼近解序列收敛性的基础上,获得了原方程分布意义下弱解的存在性以及逼近解唯一性的结果.
关键词退化椭圆方程    分布弱解    逼近解    存在性    唯一性    
EXISTENCE AND UNIQUENESS OF WEAK SOLUTIONS FOR NONLINEAR DEGENERATE ELLIPTIC EQUATIONS
CHEN Fang-min, YAN Chang, CHAI Xiao-juan    
School of Mathematical Science, Anhui University, Hefei 230601, China
Abstract: In this paper, we consider weak solutions in the distributional sense for nonlinear elliptic equations with degenerate coercivity. By proper compactness theorem and smooth approximation, on the basis of establishing the uniform regularity of solutions of the approximate equations and the convergence of the sequence of approximate solutions, we derive the existence of the weak solution in the distributional sense and the uniqueness of the approximate solution.
Keywords: Degenerate elliptic equations     distributional weak solution     approximated solution     existence     uniqueness    
1 引言

带低正则值的椭圆方程广泛产生于流体力学、控制问题以及海洋和大气科学的湍流问题中,引起了学者们广泛的关注[1-3]. 在[1] 中, G.Stampacchia最早证明了外力项为Radon测度的线性椭圆方程解的存在唯一性, 即

$ \begin{equation} \left\{ \begin{aligned} -\rm{div}(A(x)|\nabla u|^{p-2}\nabla u) = \mu\; \; \; \; \; \; \; \; &x\in\Omega, \\ u = 0\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; &x\in\partial\Omega, \end{aligned} \right. \end{equation} $ (1.1)

其中$ A(x) $满足一致椭圆假设, 即存在$ \alpha>0 $, 使得对任意$ \xi\in \mathbb{R}^{N} $, 有

$ A(x)\xi\cdot\xi\geq \alpha|\xi|^{2}\; \; \; a.e.\; \; x\in\Omega. $

作者就方程(1.1) 中$ p = 2 $的情况, 应用经典的对偶理论证明解的存在性, 但是此理论仅是由对偶理论来支撑的. 那么对于非线性算子解的存在性问题, Baccardo和Gallou\"{e}t最先在[2, 3] 中提出解决的方法, 他们证明了当$ p>2-\frac{1}{N} $时, 方程(1.1) 解的存在性. 事实上, 若$ p\leq 2-\frac{1}{N} $, 那么$ \frac{N(p-1)}{N-1}\leq1 $. 由于弱解正则性太弱, 无法保证唯一性, Baccardo和Blanchard等提出了椭圆方程熵解和抛物方程重整化解的概念, 研究解的唯一性[4-6].

近年来带有退化强制项的非线性退化椭圆方程引起了学者们的广泛关注, 以如下方程为例

$ \begin{equation} \left\{ \begin{aligned} -\rm{div}(\frac{a(x, \nabla u)}{(1+|u|)^{\theta }})+\lambda u = f \; \; \; &x\in\Omega, \\ u = 0 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; &x\in\partial\Omega, \end{aligned} \right. \end{equation} $ (1.2)

这里$ \theta>0, \; \lambda>0 $. 当$ u $趋于正无穷时, 主部算子的耗散系数趋于零. 因此, 这类方程通常被称为具有退化强制主部算子的椭圆(或抛物) 方程, 其在多孔介质流和等离子体热辐射等领域有着广泛的应用[7, 8]. 由于主部算子没有强制性, 关于椭圆和抛物方程的许多经典理论, 如变分理论和单调算子理论等不再适用. 在[9-12] 中, Boccardo等证明了在$ \lambda = 0, \; a(x, \xi) = A(x)\xi $, 且$ A $是一致有界的椭圆矩阵的假设下, 当$ f\in L^{m}(\Omega) $, $ m\geq1 $, $ 0<\theta\leq1 $时, 解的存在性和正则性结果; 在[13] 中, Boccardo研究了就(1.2) 中$ \lambda = 1 $, $ f\in L^{\frac{\theta+2}{2}}(\Omega), \; \theta>0 $的情形, 得到了解的存在、唯一性. 特别他们证明了当$ f $是凝聚在零容量集上的Radon测度时, 方程无解.

本文主要研究如下带有退化强制项的非线性椭圆型方程

$ \left\{ \begin{array}{l} - {\rm{div}}\left( {\frac{{a(x,\nabla u)}}{{{{(1 + |u|)}^{\theta (p - 1)}}}}} \right) + |u{|^{q - 2}}u = f\;\;\;x \in \Omega \\ \;\;\;\;\;\;u = 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \partial \Omega \end{array} \right. $ (1.3)

分布意义下弱解的存在性、唯一性. 其中$ \Omega $$ \mathbb{R}^{N} $上的有界开集, $ N>2, \; \theta>0 $, $ q>1 $, $ f\in L^{m}(\Omega), \; m\geq1 $. $ a:\Omega \times \mathbb{R}^{N}\rightarrow \mathbb{R}^{N} $是一个Carathéodory函数, 对任意$ \xi\in \mathbb{R}^{N} $, $ a(x, \cdot) $$ \Omega $上是可测的, 对几乎处处的$ x\in\Omega $, $ a(x, \cdot) $$ \mathbb{R}^{N} $上是连续的, 其中$ a(x, \xi) $满足以下条件

$ \begin{align} a(x, \xi)\cdot\xi&\geq \alpha |\xi|^{p}, \; \; \; \; \; \; \; \; \end{align} $ (1.4)
$ \begin{align} |a(x, \xi)|&\leq\beta|\xi|^{p-1}, \; \; \; \; \; \; \; \; \end{align} $ (1.5)
$ \begin{align} [a(x, \xi)-a(x, \eta)]&\cdot(\xi-\eta)>0, \end{align} $ (1.6)

其中$ \xi\neq\eta\in \mathbb{R}^{N} $, $ \alpha, \beta $为正常数. 我们主要关注$ f $的正则性, 非线性项的增长次数$ q $以及描述退化程度的参数$ \theta $这三个因素, 对椭圆方程解的存在性机制和正则性的影响. 当$ \theta>\frac{(pq-p-q)_{+}}{p-1}, \; q>1 $, $ m\geq\frac{\theta(p-1)+q}{q-1} $时, 我们证明了方程(1.3) 在$ {{W_{0}^{1, 1}(\Omega)}\cap {L^{m(q)}}} $中解的存在性. 进一步地, 通过紧性定理和光滑逼近, 证明了方程(1.3) 在$ {{W_{0}^{1, 1}(\Omega)}\cap {L^{m(q)}}} $中逼近解的唯一性. 具体结果如下.

定理 1.1  假设$ \theta>\frac{(pq-p-q)_{+}}{p-1}, \; q>1 $, $ m\geq \frac{\theta(p-1)+q}{p(q-1)} $, $ f\in L^{m}(\Omega) $, 那么(1.3) 存在分布意义下的弱解$ u\in W_0^{1, 1}(\Omega)\cap L^{\frac{\theta(p-1)+q}{p}}(\Omega) $, 使得

$ \begin{equation} \int_{\Omega}\frac{a(x, \nabla u)\cdot\nabla\varphi}{(1+|u|)^{\theta(p-1)}}+\int_{\Omega}|u|^{q-2}u\varphi = \int_{\Omega}f\varphi, \; \; \; \; \forall\varphi\in W_0^{1, \infty}(\Omega). \end{equation} $ (1.7)

定义 1.2   假设$ \theta>\frac{(pq-p-q)_{+}}{p-1}, \; q>1 $, $ m\geq \frac{\theta(p-1)+q}{p(q-1)} $, $ f\in L^{m}(\Omega) $, 若$ u\in W_0^{1, 1}(\Omega)\cap L^{\frac{\theta(p-1)+q}{p}}(\Omega) $是方程

$ \left\{ \begin{array}{l} -{\rm{div}}(\frac{a(x, \nabla u)}{(1+|u|)^{\theta (p-1)}})+|u|^{q-2}u = f \; \; \; &x\in\Omega, \\ \;\;\;\;\;\;u = 0 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; &x\in\partial\Omega, \end{array} \right. $ (1.8)

的一个弱解, 且$ u $是如下方程

$ \begin{equation} \left\{ \begin{aligned} -\rm{div}(\frac{a(x, \nabla u_{n})}{(1+|u_{n}|)^{\theta(p-1)}})+|u_{n}|^{q-2}u_{n} = f_{n} \; \; \; &x\in\Omega, \\ u_{n} = 0 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; &x\in\partial\Omega, \end{aligned} \right. \end{equation} $ (1.9)

的解$ u_{n} $的极限, 其中$ f_{n} $$ f $的光滑逼近. 则称$ u $为(1.3) 的逼近解. 相应地称(1.9) 为(1.3) 的逼近方程.

定理 1.3  假设$ \theta>\frac{(pq-p-q)_{+}}{p-1}, \; q>1 $, $ m\geq \frac{\theta(p-1)+q}{p(q-1)} $, $ f\in L^{m}(\Omega) $, 则(1.3) 存在唯一的逼近解$ u\in W_0^{1, 1}(\Omega)\cap L^{\frac{\theta(p-1)+q}{p}}(\Omega) $, 使得

$ \begin{equation} \int_{\Omega}\frac{a(x, \nabla u)\cdot\nabla\varphi}{(1+|u|)^{\theta(p-1)}}+\int_{\Omega}|u|^{q-2}u\varphi = \int_{\Omega}f\varphi, \; \; \; \; \forall\varphi\in W_0^{1, \infty}(\Omega). \end{equation} $ (1.10)
2 定理1.1的证明

为了证明上述定理, 我们先给出几个引理.

引理 2.1  假设$ \Omega\subset \mathbb{R}^{N} $为光滑有界区域, $ \theta>0, \; q>1, \; g\in L^{\infty}(\Omega) $, 那么

$ \left\{\begin{array}{l} -\rm{div}(\frac{a(x, \nabla v)}{(1+|v|)^{\theta (p-1)}})+|v|^{q-2}v = g \; \; \; &x\in\Omega, \\ \; \; \; \; \; \; v = 0 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; &x\in\partial\Omega, \end{array} \right. $ (2.1)

存在唯一的有界弱解$ v\in W_0^{1, p}(\Omega)\cap L^{\infty}(\Omega) $.

  对于$ s\in \mathbb{R} $, $ k\geq0 $, 定义$ T_{k}(s) = {\rm{max}}(-k, {\rm{min}}(s, k)). $$ M^{q-1} = \|g\|_{L^{\infty}(\Omega)}+1 $, 我们考虑如下逼近问题

$ \left\{\begin{array}{l}-\rm{div}(\frac{a(x, \nabla v)}{(1+|T_{M}(v)|)^{\theta (p-1)}})+|v|^{q-2}v = g \; \; \; &x\in\Omega, \\ \; \; \; \; \; \; \; \;v = 0 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; &x\in\partial\Omega, \end{array} \right. $ (2.2)

由经典的变分理论, 对任意$ M>0 $, 方程(2.2) 有一个解$ v\in W_0^{1, p}(\Omega)\cap L^{\infty}(\Omega) $. 选取$ (|v|^{q-1}-\|g\|_{L^{\infty}(\Omega)})_{+}sgn(v) $作为试验函数, 由于第一项是非负的, 可得

$ \int_{\Omega}|v|^{q-1}(|v|^{q-1}-\|g\|_{L^{\infty}(\Omega)})_{+}\leq \int_{\Omega}\|g\|_{L^{\infty}(\Omega)}(|v|^{q-1}-\|g\|_{L^{\infty}(\Omega)})_{+}, $

所以

$ \int_{\Omega}(|v|^{q-1}-\|g\|_{L^{\infty}(\Omega)})_{+}(|v|^{q-1}-\|g\|_{L^{\infty}(\Omega)})\leq0. $

$ |v|^{q-1}\leq\|g\|_{L^{\infty}(\Omega)}<M^{q-1} $. 因此$ T_{M}(v) = v $, 所以$ v $是(2.1) 的一个有界弱解.

假设$ {v_{1}} $, $ {v_{2}} $是(2.1) 的解, 则

$ \begin{align*} &-{\rm{div}}[\frac{a(x, \nabla v_{1})}{(1+|v_{1}|)^{\theta(p-1)}}-\frac{a(x, \nabla v_{2})}{(1+|v_{2}|)^{\theta(p-1)}}]+|v_{1}|^{q-2}v_{1}-|v_{2}|^{q-2}v_{2} = 0. \end{align*} $

$ w = {v_{1}}-{v_{2}} $, 选取$ T_{h}(w) $作为试验函数, 有

$ \begin{align*} &\int_{\Omega}[\frac{a(x, \nabla v_{1})}{(1+|v_{1}|)^{\theta(p-1)}}-\frac{a(x, \nabla v_{2})}{(1+|v_{2}|)^{\theta(p-1)}}]\cdot\nabla T_{h}(w)+\int_{\Omega}(|v_{1}|^{q-2}v_{1}-|v_{2}|^{q-2}v_{2})T_{h}(w) = 0. \end{align*} $

$ \begin{align*} &\int_{\Omega}\frac{[a(x, \nabla v_{1})-a(x, \nabla v_{2})]\cdot\nabla T_{h}(w)}{(1+|v_{1}|)^{\theta(p-1)}}+\int_{\Omega}(|v_{1}|^{q-2}v_{1}-|v_{2}|^{q-2}v_{2})T_{h}(w)\\ = &-\int_{\Omega}[\frac{1}{(1+|v_{1}|)^{\theta(p-1)}}-\frac{1}{(1+|v_{2}|)^{\theta(p-1)}}]a(x, \nabla v_{2})\cdot\nabla T_{h}(w). \end{align*} $

由假设(1.6), 等式左边第一项非负, 对右式使用Lagrange中值定理并利用函数$ t\mapsto\frac{1}{(1+|t|)^{\theta(p-1)}} $导数的绝对有界性可得

$ \begin{align*} &\int_{\Omega}(|v_{1}|^{q-2}v_{1}-|v_{2}|^{q-2}v_{2})\frac{T_{h}(w)}{h}\leq\theta(p-1)\int_{\Omega}|a(x, \nabla v_{2})||\nabla T_{h}(w)|. \end{align*} $

由于$ a(x, \xi) $满足假设(1.5). 当$ h\rightarrow0 $时, 有

$ \int_{\Omega}|v_{1}-v_{2}|^{q-1}\leq\int_{\Omega}||v_{1}|^{q-2}v_{1}-|v_{2}|^{q-2}v_{2}|\leq0, $

由此可知$ {v_{1}} = {v_{2}} $, 解的唯一性得证.

注 2.11   令函数列$ f_{n}\in L^{\infty}(\Omega) $, 且$ f_{n} $$ L^{m}(\Omega) $中强收敛到$ f $, 使得在$ \Omega $$ |f_{n}|\leq|f| $几乎处处成立, 那么由引理2.1, 对任意的$ n\in N $, 方程(1.9) 有解$ u_{n}\in W_0^{1, p}(\Omega)\cap L^{\infty}(\Omega). $

引理 2.2   若$ u_{n} $是(1.9) 的解, 那么对于任意$ k\geq0 $,

$ \begin{align} \int_{\{|u_{n}|\geq k\}}{|u_{n}|}^{\frac{\theta(p-1)+q}{p}}&\leq\int_{\{|u_{n}|\geq k\}}|f|^{{\frac{\theta(p-1)+q}{p(q-1)}}};\; \; \; \; \; \; \; \; \; \; \; \; \; \; \end{align} $ (2.3)
$ \begin{align} \int_{\{|u_{n}|\geq k\}}\frac{|\nabla u_{n}|^{p}}{(1+|u_{n}|)^{\frac{\theta(p-1)^{2}+q(p-1)}{p}}}&\leq C[\int_{\{|u_{n}|\geq k\}}|f|^{\frac{\theta(p-1)+q}{p(q-1)}}]^{\frac{p(q-1)}{\theta(p-1)+q}};\; \; \; \; \end{align} $ (2.4)
$ \begin{align} \int_{\{|u_{n}|\geq k\}}|\nabla u_{n}|&\leq C[\int_{\{|u_{n}|\geq k\}}|f|^{\frac{\theta(p-1)+q}{p(q-1)}}]^{\frac{q-1}{\theta(p-1)+q}};\; \; \; \; \; \; \; \; \; \; \; \end{align} $ (2.5)
$ \begin{align} \alpha\int_{\Omega}|\nabla T_{k}(u_{n})|^{p}&\leq k(k+1)^{\theta(p-1)}\int_{\Omega}|f|.\; \; \; \; \; \; \; \; \; \; \; \; \; \end{align} $ (2.6)

其中$ C $是一个正常数且$ C = C(\alpha, \; \theta, \; {\rm{meas}}(\Omega), \; \|f\|_{L^{\frac{\theta(p-1)+q}{q-1}}(\Omega)}) $.

  对任意给定的$ k\geq0, \; h>0, $$ \psi_{h, k}(s) $是如下定义的函数,

$ \begin{equation*} \psi_{h, k}(s) = \left\{\begin{array}{l} \; \; \; \; \; \; \; \; \; \; \; \; \; 0\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; 0\leq s\leq k, \\ \; \; \; \; \; \; \; h(s-k)\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; k\leq s\leq k+\frac{1}{h}, \\ \; \; \; \; \; \; \; \; \; \; \; \; \; 1\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; s>k+\frac{1}{h}, \\ \psi_{h, k}(s) = -\psi_{h, k}(-s)\; \; \; s<0. \end{array} \right. \end{equation*} $

$ \begin{equation*} \lim\limits_{h\rightarrow +\infty}\psi_{h, k}(s) = \left\{\begin{array}{l} 1\; \; \; \; \; \; \; \; \; \; \; \; \; s>k, \\ 0\; \; \; \; \; \; \; \; \; \; \; \; |s|\leq k, \\ -1\; \; \; \; \; \; \; \; \; \; s<-k. \end{array} \right. \end{equation*} $

$ \epsilon>0 $, 选取$ (\epsilon+|u_{n}|)^{\frac{\theta(p-1)+p+q-pq}{p}}\psi_{h, k}(u_{n}) $作为(1.9) 的试验函数, 则有

$ \begin{align} &\frac{\theta(p-1)+p+q-pq}{p}\int_{\Omega}a(x, \nabla u_{n})\nabla u_{n}\cdot\frac{(\epsilon+|u_{n}|)^{\frac{\theta(p-1)+q-pq}{p}}}{(1+|u_{n}|)^{\theta (p-1)}}\cdot|\psi_{h, k}(u_{n})|\\ &+\int_{\Omega}{a(x, \nabla u_{n})\nabla u_{n}}{(1+|u_{n}|)^{\theta (p-1)}}\psi'_{h, k}(u_{n})(\epsilon+|u_{n}|)^{\frac{\theta(p-1)+p+q-pq}{p}}\\ &+\int_{\Omega}|u_{n}|^{q-2}u_{n}(\epsilon+|u_{n}|)^{\frac{\theta(p-1)+p+q-pq}{p}}\psi_{h, k}(u_{n})\\ = &\int_{\Omega}f_{n}(\epsilon+|u_{n}|)^{\frac{\theta(p-1)+p+q-pq}{p}}. \end{align} $ (2.7)

由假设(1.4) 和$ \psi'_{h, k}(s)\geq0 $, 上面等式左边的前两项均为非负的, 又因为$ |f_{n}|\leq|f| $, 故

$ \int_{\Omega}|u_{n}|^{q-2}u_{n}(\epsilon+|u_{n}|)^{\frac{\theta(p-1)+p+q-pq}{p}}\psi_{h, k}(u_{n})\leq \int_{\Omega}|f|(\epsilon+|u_{n}|)^{\frac{\theta(p-1)+p+q-pq}{p}}|\psi_{h, k}(u_{n})|. $

$ \epsilon\rightarrow0, \; h\rightarrow +\infty $, 不等式的左边利用Fatou引理, 右边利用Lebesgue控制收敛定理, 可得

$ \int_{\{|u_{n}|\geq k\}}|u_{n}|^{\frac{\theta(p-1)+q}{p}}\leq\int_{\{|u_{n}|\geq k\}}|f||u_{n}|^{\frac{\theta(p-1)+p+q-pq}{p}}, $

由Hölder不等式, 可得

$ \int_{\{|u_{n}|\geq k\}}|u_{n}|^{\frac{\theta(p-1)+q}{p}}\leq[\int_{\{|u_{n}|\geq k\}}|f|^{{\frac{\theta(p-1)+q}{p(q-1)}}}]^{\frac{p(q-1)}{\theta(p-1)+q}}[\int_{\{|u_{n}|\geq k\}}|u_{n}|^{\frac{\theta(p-1)+q}{p}}]^{\frac{\theta(p-1)+p+q-pq}{\theta(p-1)+q}}, $

进而有

$ \int_{\{|u_{n}|\geq k\}}{|u_{n}|}^{\frac{\theta(p-1)+q}{p}}\leq\int_{\{|u_{n}|\geq k\}}|f|^{{\frac{\theta(p-1)+q}{p(q-1)}}}. $

(2.3) 得证. 特别地, 当$ k = 0 $时, 有

$ \begin{align} \int_{\Omega}{|u_{n}|}^{\frac{\theta(p-1)+q}{p}}\leq\int_{\Omega}|f|^{{\frac{\theta(p-1)+q}{p(q-1)}}} = \|f\|_{L^{\frac{\theta(p-1)+q}{p(q-1)}}(\Omega)} ^{\frac{\theta(p-1)+q}{p(q-1)}}. \end{align} $ (2.8)

在(2.7) 中取$ \epsilon = 1 $, 可得

$ \begin{align*} &\frac{\theta(p-1)+p+q-pq}{p}\int_{\Omega}\frac{a(x, \nabla u_{n})\cdot\nabla u_{n}}{(1+|u|)^{\frac{\theta(p-1)^{2}+q(p-1)}{p}}}|\psi_{h, k}(u_{n})|\leq\int_{\Omega}|f|(1+|u_{n}|)^{\frac{\theta(p-1)+p+q-pq}{p}}|\psi_{h, k}(u_{n})|. \end{align*} $

由假设(1.4), 令$ h\rightarrow +\infty $, 再次由Fatou引理和Lebesgue控制收敛定理, 可得

$ \begin{align*} &\alpha\frac{\theta(p-1)+p+q-pq}{p}\int_{|u_{n}|\geq k}\frac{|\nabla u_{n}|^{p}}{(1+|u_{n}|)^{\frac{\theta(p-1)^{2}+q(p-1)}{p}}}\\ \leq&[\int_{\{|u_{n}|\geq k\}}|f|^{{\frac{\theta(p-1)+q}{p(q-1)}}}]^{\frac{p(q-1)}{\theta(p-1)+q}}[\int_{\{|u_{n}|\geq k\}}(1+|u_{n}|)^{\frac{\theta(p-1)+q}{p}}]^{\frac{\theta(p-1)+p+q-pq}{\theta(p-1)+q}}\\ \leq&[\int_{\{|u_{n}|\geq k\}}|f|^{{\frac{\theta(p-1)+q}{p(q-1)}}}]^{\frac{p(q-1)}{\theta(p-1)+q}}[\int_{\Omega}(1+|u_{n}|)^{\frac{\theta(p-1)+q}{p}}]^{\frac{\theta(p-1)+p+q-pq}{\theta(p-1)+q}}. \end{align*} $

由(2.8), 可得

$ \alpha\frac{\theta(p-1)+p+q-pq}{p}\int_{\{|u_{n}|\geq k\}}\frac{|\nabla u_{n}|^{p}}{(1+|u_{n}|)^{\frac{\theta(p-1)+q}{p}}}\leq C[\int_{\{|u_{n}|\geq k\}}|f|^{\frac{\theta(p-1)+q}{p(q-1)}}]^{\frac{p(q-1)}{\theta(p-1)+q}}. $

(2.4) 得证. 再次由$ H\ddot{o}lder $不等式, (2.4) 和(2.8) 可得

$ \begin{align*} \int_{\{|u_{n}|\geq k\}}|\nabla u_{n}|& = \int_{\{|u_{n}|\geq k\}}\frac{|\nabla u_{n}|}{(1+|u_{n}|)^{\frac{\theta(p-1)^{2}+q(p-1)}{p^{2}}}}{(1+|u_{n}|)^{\frac{\theta(p-1)^{2}+q(p-1)}{p^{2}}}}\\ &\leq[\int_{\{|u_{n}|\geq k\}}\frac{|\nabla u_{n}|^{p}}{(1+|u_{n}|)^{\frac{\theta(p-1)^{2}+q(p-1)}{p}}}]^{\frac{1}{p}}[\int_{\{|u_{n}|\geq k\}}(1+|u_{n}|)^{\frac{\theta(p-1)+q}{p}}]^{\frac{p-1}{p}}\\ &\leq[\int_{\{|u_{n}|\geq k\}}\frac{|\nabla u_{n}|^{p}}{(1+|u_{n}|)^{\frac{\theta(p-1)^{2}+q(p-1)}{p}}}]^{\frac{1}{p}}[\int_{\Omega}(1+|u_{n}|)^{\frac{\theta(p-1)+q}{p}}]^{\frac{p-1}{p}}\\ &\leq C[\int_{\{|u_{n}|\geq k\}}|f|^{\frac{\theta(p-1)+q}{p(q-1)}}]^{\frac{q-1}{\theta(p-1)+q}}. \end{align*} $

那么(2.5) 得证.

最后选取$ T_{k}(u_{n}) $作为(1.9) 的试验函数, 由假设(1.4) 知

$ \alpha\int_\Omega{{{| {\nabla {T_k}( {{u_n}} )} |}^p}}\leq k(k+1)^{\theta(p-1)}\int_{\Omega}|f|, $

故(2.6) 得证.

引理 2.3   假设$ {u_{n}} $是(1.9)的解, 则存在一个子列(依旧记作$ {u_{n}} $)以及函数$ u\in L^{\frac{\theta(p-1)+q}{p}}(\Omega) $, 使得对任意$ k>0 $, $ T_{k}(u_{n}) $$ W_0^{1, p}(\Omega) $中弱收敛到$ T_{k}(u) $, $ {u_{n}} $$ \Omega $中几乎处处收敛到$ u $.

  由(2.4), 令$ k = 0 $, 可得

$ \begin{align} \int_{\Omega}\frac{|\nabla u_{n}|^{p}}{(1+|u_{n}|)^{\frac{\theta(p-1)^{2}+q(p-1)}{p}}}\leq C\|f\|_{L^{\frac{\theta(p-1)+q}{p(q-1)}}(\Omega)}. \end{align} $ (2.9)

$ \theta\neq\frac{p^{2}-q(p-1)}{(p-1)^{2}} $, 则

$ \frac{|\nabla u_{n}|^{p}}{(1+|u_{n}|)^{\frac{\theta(p-1)^{2}+q(p-1)}{p}}} = [\frac{p^{2}}{p^{2}-\theta(p-1)^{2}-q(p-1)}]^{p}|\nabla[(1+|u_{n}|)^\frac{p^{2}-\theta(p-1)^{2}-q(p-1)}{p^{2}}-1]|^{p}. $

由(2.9), 序列$ v_{n} = \frac{p^{2}}{p^{2}-\theta(p-1)^{2}-q(p-1)}[(1+|u_{n}|)^\frac{p^{2}-\theta(p-1)^{2}-q(p-1)}{p^{2}}-1]sgn(u_{n}) $$ W_0^{1, p}(\Omega) $内有界.

$ \theta = \frac{p^{2}-q(p-1)}{(p-1)^{2}} $, 那么

$ \frac{|\nabla u_{n}|^{p}}{(1+|u_{n}|)^{\frac{\theta(p-1)^{2}+q(p-1)}{p}}} = \frac{|\nabla u_{n}|^{p}}{(1+|u_{n}|)^{p}} = |\nabla\log(1+|u_{n}|)|^{p}, $

$ v_{n} = [\nabla\log(1+|u_{n}|)]sgn(u_{n}) $$ W_0^{1, p}(\Omega) $内有界.

综合以上两种情况, $ v_{n} $$ W_0^{1, p}(\Omega) $中弱收敛到某个$ v $, $ v_{n} $$ L^{p}(\Omega) $中强收敛到某个$ v $, 并且$ v_{n} $$ \Omega $中几乎处处收敛到某个$ v $.

$ \theta<\frac{p^{2}-q(p-1)}{(p-1)^{2}} $, 根据$ v(x) $的取值来定义$ u(x) $:当$ |v(x)|<\frac{p^{2}}{\theta(p-1)^{2}+q(p-1)-p^{2}} $时, 定义$ u(x) = [(\frac{p^{2}-\theta(p-1)^{2}-q(p-1)}{p^{2}}|v(x)|+1)^{\frac{p^{2}}{p^{2}-\theta(p-1)^{2}-q(p-1)}}-1]sgn(v(x)) $; 当$ v(x) = \frac{p^{2}}{\theta(p-1)^{2}+q(p-1)-p^{2}} $时, 定义$ u(x) = +\infty $; 当$ v(x) = -\frac{p^{2}}{\theta(p-1)^{2}+q(p-1)-p^{2}} $时, 定义$ u(x) = -\infty $.

$ \theta = \frac{p^{2}-q(p-1)}{(p-1)^{2}} $, 定义

$ u(x) = [\exp^{|v(x)|-1}]sgn(v(x)). $

由以上定义知, $ u_{n} $几乎处处收敛到$ u $.

由(2.3), 当$ k = 0 $

$ \int_{\Omega}{|u_{n}|}^{\frac{\theta(p-1)+q}{p}}\leq\int_{\Omega}|f|^{{\frac{\theta(p-1)+q}{p(q-1)}}}\leq C. $

因为$ u_{n} $几乎处处收敛到$ u $, 再由Fatou引理$ \int_{\Omega}{|u|}^{\frac{\theta(p-1)+q}{p}}\leq C, $因此$ u\in L^{\frac{\theta(p-1)+q}{p}}(\Omega) $, 可得$ u $几乎处处有限.

$ k>0 $, 由(2.6), 序列$ {T_{k}(u_{n})} $$ W_0^{1, p}(\Omega) $中有界, 存在子序列$ {T_{k}(u_{nj})} $$ W_0^{1, p}(\Omega) $中弱收敛到某个$ v_{k} $. 由于$ u_{n} $几乎处处收敛到$ u $, 有$ v_{k} = T_{k}(u) $. 因为此极限不依赖于子序列的选取, 故对任意$ k>0 $, 序列$ {T_{k}(u_{n})} $弱收敛到$ T_{k}(u) $.

注 2.3   令$ u $$ \Omega $上的可测函数, 则对任意$ k>0 $, $ T_{k}(u)\in W_0^{1, p}(\Omega) $, 由[4]可知, 存在唯一的可测函数$ v:\Omega\rightarrow \mathbb{R}^{N} $, 使得$ \nabla T_{k}(u) = v\chi_{\{|u|\leq k\}} $对每个$ k>0 $几乎处处成立.

命题 2.4   假设$ u_{n} $是(1.9)的解, 那么由引理2.3, $ u\in L^{\frac{\theta(p-1)+q}{p}}(\Omega) $且对于任意给定的$ k>0 $, 当$ n\rightarrow +\infty $时, $ T_{k}(u_{n}) $$ W_0^{1, p}(\Omega) $中强收敛到$ T_{k}(u) $.

  对任意给定的$ h>k>0 $, 选取$ T_{2k}[u_{n}-T_{h}(u_{n})+T_{k}(u_{n})-T_{k}(u)] $作为(1.9) 的试验函数, 可得

$ \begin{align} &\int_{\Omega}\frac{a(x, \nabla u_{n})\cdot\nabla T_{2k}[u_{n}-T_{h}(u_{n})+T_{k}(u_{n})-T_{k}(u)]}{(1+|u_{n}|)^{\theta (p-1)}} \\ = &-\int_{\Omega}(|u_{n}|^{q-2}u_{n}-f_{n})T_{2k}[u_{n}-T_{h}(u_{n})+T_{k}(u_{n})-T_{k}(u)]. \end{align} $ (2.10)

由(2.3) 可知$ |u_{n}|^{q-2}u_{n} $$ L^{{\frac{\theta(p-1)+q}{q-1}}}(\Omega) $内有界, 又因为$ f_{n} $$ L^{{\frac{\theta(p-1)+q}{q-1}}}(\Omega) $内有界, 故当$ n\rightarrow +\infty $, $ h\rightarrow +\infty $时, 等式右边收敛到$ 0 $. 记$ \epsilon(n, h) $是一个无穷小量, 使得

$ \lim\limits_{h\rightarrow +\infty}\lim\limits_{n\rightarrow +\infty}\epsilon (n, h) = 0, $

那么

$ -\int_{\Omega}(|u_{n}|^{q-2}u_{n}-f_{n})T_{2k}[u_{n}-T_{h}(u_{n})+T_{k}(u_{n})-T_{k}(u)] = \epsilon(n, h). $

$ M = 4k+h $, 若$ |u_{n}|\geq M $, 则$ \nabla {T_{2k}[u_{n}-T_{h}(u_{n})+T_{k}(u_{n})-T_{k}(u)]} = 0 $, 由(2.10) 可得

$ \begin{align*} \epsilon(n, h) = &\int_{\{|u_{n}|<k\}}\frac{a(x, \nabla u_{n})\cdot\nabla T_{2k}[u_{n}-T_{h}(u_{n})+T_{k}(u_{n})-T_{k}(u)]}{(1+|u_{n}|)^{\theta (p-1)}}\\ &+\int_{\{|u_{n}|\geq k\}}\frac{a(x, \nabla u_{n})\cdot\nabla T_{2k}[u_{n}-T_{h}(u_{n})+T_{k}(u_{n})-T_{k}(u)]}{(1+|u_{n}|)^{\theta(p-1)}}. \end{align*} $

又因为当$ |u_{n}|\leq k\; $, $ u_{n} = T_{h}(u_{n}) $; 当$ |u_{n}|\geq k $, $ \nabla T_{k}(u_{n}) = 0 $$ a(x, 0) = 0 $,

$ \begin{align*} \epsilon(n, h) = &\int_{\{|u_{n}|\leq k\}}\frac{a(x, \nabla T_{M}(u_{n}))\cdot\nabla[T_{k}(u_{n})-T_{k}(u)]}{(1+|u_{n}|)^{\theta(p-1)}}\\ &+\int_{\{|u_{n}|\geq k\}}\frac{a(x, \nabla T_{k}(u_{n}))\cdot\nabla[u_{n}-T_{h}(u_{n})-T_{k}(u)]}{(1+|u_{n}|)^{\theta(p-1)}}\\ = &\int _{\Omega} \frac{a(x, \nabla T_{k}(u_{n}))\cdot \nabla[T_{k}(u_{n})-T_{k}(u)]}{(1+|u_{n}|)^{\theta(p-1)}} +\int _{\{|u_{n}|\geq k\}}\frac{a(x, \nabla T_{M}(u_{n}))\cdot\nabla [u_{n}-T_{h}(u_{n})]}{(1+|u_{n}|)^{\theta(p-1)}}\\ &-\int_{\{|u_{n}|\geq k\}}\frac{a(x, \nabla T_{M}(u_{n}))\cdot\nabla T_{k}(u)}{(1+|u_{n}|)^{\theta(p-1)}}, \end{align*} $

由右式第二项是非负的, 可得

$ \begin{align*} \epsilon(n, h)\geq&\int_{\Omega}\frac{[a(x, \nabla T_{k}(u_{n}))-a(x, \nabla T_{k}(u))]\cdot\nabla [T_{k}(u_{n})-T_{k}(u)]}{(1+k)^{\theta(p-1)}}\\ &+\int_{\Omega} \frac{{a(x, \nabla T_{k}(u))}\cdot \nabla[T_{k}(u_{n})-T_{k}(u)]}{(1+|u_{n}|)^{\theta(p-1)}} -\int_{\{|u_{n}|\geq k\}}\frac{a(x, \nabla T_{M}(u_{n}))\cdot\nabla T_{k}(u)}{(1+|u_{n}|)^{\theta(p-1)}}\\ \doteq& A_{n}+B_{n}-C_{n}. \end{align*} $

由引理2.3, $ T_{k}(u_{n}) $$ W_0^{1, p}(\Omega) $中弱收敛到$ T_{k}(u) $, 由假设(1.5), $ \frac{a(x, \nabla T_{k}(u))}{(1+|u_{n}|)^{\theta(p-1)}} $$ (L^{p}(\Omega))^{N} $中有界, 故

$ \lim\limits_{n\rightarrow +\infty}B_{n} = \lim\limits_{n\rightarrow +\infty}\int _{\Omega} \frac{{a(x, \nabla T_{k}(u))}\cdot \nabla[T_{k}(u_{n})-T_{k}(u)]}{(1+|u_{n}|)^{\theta(p-1)}} = 0. $

最后一部分可以写成

$ C_{n} = \int _{\Omega}\frac{a(x, \nabla T_{M}(u_{n}))\cdot \nabla T_{k}(u)\chi_{|u_{n}|\geq k}}{(1+|u_{n}|)^{\theta(p-1)}}. $

因为$ M $为定值, $ a(x, \nabla T_{M}(u_{n})) $$ (L^{\frac{p}{p-1}}(\Omega))^{N} $中有界. 因此, 存在$ \sigma\in(L^{\frac{p}{p-1}}(\Omega))^{N} $且有子序列$ {a(x, \nabla T_{M}(u_{nj}))} $, 使得当$ j\rightarrow +\infty $时, $ a(x, \nabla T_{M}(u_{nj})) $$ (L^{\frac{p}{p-1}}(\Omega))^{N} $中弱收敛到$ \sigma $. 另一方面, 当$ n\rightarrow +\infty $时, $ \frac{\nabla T_{k}(u)\chi_{|u_{n}|\geq k}}{(1+|u_{n}|)^{\theta(p-1)}} $$ (L^{p}(\Omega))^{N} $中强收敛到$ 0 $, 故

$ \lim\limits_{n\rightarrow +\infty}C_{n} = \lim\limits_{n\rightarrow +\infty}\int _{\{|u_{n}|\geq k\}}\frac{a(x, \nabla T_{M}(u_{n}))\cdot \nabla T_{k}(u)}{(1+|u_{n}|)^{\theta(p-1)}} = 0. $

因此

$ \epsilon(n, h)\geq A_{n} = \int _{\Omega} \frac{[a(x, \nabla T_{k}(u_{n}))-a(x, \nabla T_{k}(u))]\cdot \nabla [T_{k}(u_{n})-T_{k}(u)]}{(1+k)^{\theta(p-1)}}, $

由假设(1.6),

$ \lim\limits_{n\rightarrow +\infty}\int _{\Omega}[a(x, \nabla T_{k}(u_{n}))-a(x, \nabla T_{k}(u))]\cdot\nabla[T_{k}(u_{n})-T_{k}(u)] = 0. $

再由([5, 6]), 可得$ T_{k}(u_{n}) $$ W_0^{1, p}(\Omega) $中强收敛到$ T_{k}(u) $.

推论 2.5   取引理$ 2.3 $$ u_{n} $$ u $, 那么$ \nabla u_{n} $$ \Omega $中几乎处处收敛到$ \nabla u $, $ \nabla u $的定义如注2.3所述.

引理 2.6   取引理$ 2.3 $中的$ u_{n} $$ u $, 那么$ \frac{|\nabla u_{n}|^{p-1}}{(1+|u_{n}|)^{\theta(p-1)}} $$ (L^{1}(\Omega))^{N} $中强收敛到$ \frac{|\nabla u|^{p-1}}{(1+|u|)^{\theta(p-1)}} $, 并且$ u_{n} $$ L^{\frac{\theta(p-1)+q}{p}}(\Omega) $中强收敛到$ u $.

  我们先证$ \frac{|\nabla u_{n}|^{p-1}}{(1+|u_{n}|)^{\theta(p-1)}} $收敛到$ \frac{|\nabla u|^{p-1}}{(1+|u|)^{\theta(p-1)}} $, 令$ \epsilon>0, \; k>0 $且充分大, 使得

$ [\int_{\{|u_{n}|\geq k\}}|f|^{\frac{\theta(p-1)+q}{p(q-1)}}]^{\frac{q-1}{\theta(p-1)+q}}<\epsilon, $

定义$ E $是一个可测集, 易知

$ \begin{align*} \int_{E}\frac{|\nabla u_{n}|^{p-1}}{(1+|u_{n}|)^{\theta(p-1)}}& = \int_{E\cap{\{|u_{n}|\leq k\}}}\frac{|\nabla u_{n}|^{p-1}}{(1+|u_{n}|)^{\theta(p-1)}}+\int_{E\cap{\{|u_{n}|\geq k\}}}\frac{|\nabla u_{n}|^{p-1}}{(1+|u_{n}|)^{\theta(p-1)}}\\ & = \int_{E}\frac{|\nabla T_{k}(u_{n})|^{p-1}}{(1+|\nabla T_{k}(u_{n})|)^{\theta(p-1)}}+\int_{E\cap{\{|u_{n}|\geq k\}}}\frac{|\nabla u_{n}|^{p-1}}{(1+|u_{n}|)^{\theta(p-1)}}, \end{align*} $

由积分连续性以及(2.8), 可得

$ \begin{align*} &\int_{E\cap{\{|u_{n}|\geq k\}}}\frac{|\nabla u_{n}|^{p-1}}{(1+|u_{n}|)^{\theta(p-1)}}\\ \leq&\int_{E\cap{\{|u_{n}|\geq k\}}}\frac{|\nabla u_{n}|^{p-1}}{(1+|u_{n}|)^{\frac{\theta(p-1)^{3}+q(p-1)^{2}}{p^{2}}}}(1+|u_{n}|)^{\frac{\theta(p-1)^{3}+q(p-1)^{2}-\theta p^{2}(p-1)}{p^{2}}}\\ \leq&[\int_{E\cap{|u_{n}|\geq k}}\frac{|\nabla u_{n}|^{p}}{(1+|u_{n}|)^{\frac{\theta(p-1)^{2}+q(p-1)}{p}}}]^{\frac{p-1}{p}}[\int_{E\cap{\{|u_{n}|\geq k\}}}(1+|u_{n}|)^{\frac{\theta(p-1)^{3}+q(p-1)^{2}-\theta p^{2}(p-1)}{p}}]^{\frac{1}{p}}\\ \leq&[\int_{E\cap{\{|u_{n}|\geq k\}}}\frac{|\nabla u_{n}|^{p}}{(1+|u_{n}|)^{\frac{\theta(p-1)^{2}+q(p-1)}{p}}}]^{\frac{p-1}{p}}[\int_{E }(1+|u_{n}|)^{\frac{\theta(p-1)^{3}+q(p-1)^{2}-\theta p^{2}(p-1)}{p}}]^{\frac{1}{p}}\\ \leq& C[\int_{\{|u_{n}|\geq k\}}|f|^{\frac{\theta(p-1)+q}{p(q-1)}}]^{\frac{q-1}{\theta(p-1)+q}\cdot\frac{p-1}{p}}\leq C[\int_{\{|u_{n}|\geq k\}}|f|^{\frac{\theta(p-1)+q}{p(q-1)}}]^{\frac{q-1}{\theta(p-1)+q}}<C\epsilon. \end{align*} $

根据Hölder不等式和(2.6), 可得

$ \int_{E}\frac{|\nabla T_{k}(u_{n})|^{p-1}}{(1+|\nabla T_{k}(u_{n})|)^{\theta(p-1)}}\leq\int_{E}|\nabla T_{k}(u_{n})|^{p-1}\leq C[{\rm{meas}}(E)]^{\frac{1}{p}}k^{\frac{p-1}{p}}(1+k)^{\frac{\theta(p-1)^{2}}{p}}(\int_{\Omega}|f|)^{\frac{p-1}{p}}, $

因此, 有

$ \int_{E}\frac{|\nabla u_{n}|^{p-1}}{(1+|u_{n}|)^{\theta(p-1)}}\leq C[\rm{meas}(E)]^{\frac{1}{p}}k^{\frac{p-1}{p}}(1+k)^{\frac{\theta(p-1)^{2}}{p}}(\int_{\Omega}|f|)^{\frac{p-1}{p}}+C\epsilon. $

固定$ k $, 选取meas($ E $) 充分小, 有$ \int_{E}\frac{|\nabla u_{n}|^{p-1}}{(1+|u_{n}|)^{\theta(p-1)}}\leq C\epsilon, $其中, 常数$ C $不依赖于$ n $$ \epsilon $. 由Vitali收敛定理, $ \frac{|\nabla u_{n}|^{p-1}}{(1+|u_{n}|)^{\theta(p-1)}} $$ (L^{1}(\Omega))^{N} $中强收敛到$ \frac{|\nabla u|^{p-1}}{(1+|u|)^{\theta(p-1)}} $.

对于第二个收敛性, 由(2.3),

$ \begin{align*} \int_{E}|u_{n}|^{\frac{\theta(p-1)+q}{p}}&\leq \int_{E\cap{\{|u_{n}|\leq k\}}}|u_{n}|^{\frac{\theta(p-1)+q}{p}}+\int_{E\cap{\{|u_{n}|\geq k\}}}|u_{n}|^{\frac{\theta(p-1)+q}{p}}\\ &\leq k^{\frac{\theta(p-1)+q}{p}}{\rm{meas}}(E)+\int_{{\{|u_{n}|\geq k\}}}|f|^{\frac{\theta(p-1)+q}{p(q-1)}}. \end{align*} $

选取合适的$ k $, 使右式第二项积分充分小, 并且让$ E $的测度充分小, 使得第一项积分充分小. 因为$ u_{n} $几乎处处收敛到$ u $, 由Vitali收敛定理, $ u_{n} $$ L^{\frac{\theta(p-1)+q}{p}}(\Omega) $中强收敛到$ u $.

定理1.1的证明  对任意$ \varphi\in W_0^{1, \infty}(\Omega) $, 等式(1.9) 两边同时与$ \varphi $作用, 有

$ \int_{\Omega}\frac{a(x, \nabla u_{n})\cdot\nabla\varphi}{(1+|u_{n}|)^{\theta (p-1)}}+\int_{\Omega}|u_{n}|^{q-2}u_{n}\varphi = \int_{\Omega}f_{n}\varphi, $

$ n\rightarrow +\infty $时, 易知序列$ {|u_{n}|^{q-1}} $$ {f_{n}} $$ L^{\frac{\theta(p-1)+q}{p(q-1)}}(\Omega) $中分别收敛到$ |u|^{q-1} $$ f $, 显然在$ L^{1}(\Omega) $中也是成立的. 由$ \frac{|\nabla u_{n}|^{p-1}}{(1+|u_{n}\mid)^{\theta(p-1)}} $$ (L^{1}(\Omega))^{N} $中强收敛到$ \frac{|\nabla u|^{p-1}}{(1+|u|)^{\theta(p-1)}} $可得

$ \lim\limits_{n\rightarrow +\infty}\int_{\Omega}\frac{a(x, \nabla u_{n})\cdot\nabla\varphi}{(1+|u_{n}|)^{\theta (p-1)}} = \int_{\Omega}\frac{a(x, \nabla u)\cdot\nabla\varphi}{(1+|u|)^{\theta(p-1)}}. $

再利用Lebesgue控制收敛定理, 有

$ \int_{\Omega}\frac{a(x, \nabla u)\cdot\nabla\varphi}{(1+|u|)^{\theta(p-1)}}+\int_{\Omega}|u|^{q-2}u\varphi = \int_{\Omega}f\varphi, \; \; \varphi\in W_0^{1, \infty}(\Omega), $

$ u $满足(1.7).

3 定理1.3的证明

本节我们讨论逼近解的唯一性, 令$ g\in L^{\frac{\theta(p-1)+q}{p(q-1)}}(\Omega) $, $ g_{n} $$ L^{\infty}(\Omega) $中的函数列, $ g_{n} $$ L^{\frac{\theta(p-1)+q}{p(q-1)}}(\Omega) $中收敛到$ g $$ |g_{n}|\leq|g| $, 若$ z\in W_0^{1, 1}(\Omega)\cap L^{\frac{\theta(p-1)+q}{p}}(\Omega) $是方程

$ \left\{\begin{array}{l} -{\rm{div}}(\frac{a(x, \nabla z)}{(1+|z|)^{\theta(p-1)}})+|z|^{q-2}z = g \; \; \; &x\in\Omega, \\ z = 0 \; \; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; &x\in\partial\Omega, \end{array} \right. $ (3.1)

的一个逼近解, 由定义1.2, 一定存在$ z_{n}\in W_0^{1, 1}(\Omega)\cap L^{\infty}(\Omega) $是逼近方程

$ \left\{\begin{array} -{\rm{div}}(\frac{a(x, \nabla z_{n})}{(1+|z_{n}|)^{\theta(p-1)}})+|z_{n}|^{q-2}z_{n} = g_{n} \; \; \; &x\in\Omega, \\ \; \; \; \; \; \; \; \;z_{n} = 0 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; &x\in\partial\Omega, \end{array} \right. $ (3.2)

的一个弱解. 接下来, 我们将证明问题(1.3) 的逼近解是唯一的, 首先证明一个引理.

引理 3.1   假设$ u_{n} $$ z_{n} $分别是(1.9) 和(3.2) 的解序列, $ u $$ z $分别为其对应的逼近解, 那么

$ \begin{align} \int_{\Omega}|u-z|^{q-1}\leq\int_{\Omega}|f-g|, \end{align} $ (3.3)

且当$ f\leq g $$ \Omega $中几乎处处成立, 可得$ u\leq z $$ \Omega $中几乎处处成立.

  由(1.9)–(3.2), 可得

$ \begin{align} -{\rm{div}}[\frac{a(x, \nabla u_{n})}{(1+|u_{n}|)^{\theta(p-1)}}-\frac{a(x, \nabla z_{n})}{(1+|z_{n}|)^{\theta(p-1)}}]+|u_{n}|^{q-2}u_{n}-|z_{n}|^{q-2}z_{n} = f_{n}-g_{n}. \end{align} $ (3.4)

选取$ T_{h}(u_{n}-z_{n}) $作为试验函数, 有

$ \begin{align*} &\int_{\Omega}[\frac{a(x, \nabla u_{n})}{(1+|u_{n}|)^{\theta(p-1)}}-\frac{a(x, \nabla z_{n})}{(1+|z_{n}|)^{\theta(p-1)}}]\cdot\nabla T_{h}(u_{n}-z_{n})\\ &+\int_{\Omega}(|u_{n}|^{q-2}u_{n}-|z_{n}|^{q-2}z_{n})T_{h}(u_{n}-z_{n}) = \int_{\Omega}(f_{n}-g_{n})T_{h}(u_{n}-z_{n}). \end{align*} $

上述等式可以换成如下写法

$ \begin{align*} &\int_{\Omega}\frac{[a(x, \nabla u_{n})-a(x, \nabla z_{n})]\cdot\nabla T_{h}(u_{n}-z_{n})}{(1+|u_{n}|)^{\theta(p-1)}} +\int_{\Omega}(|u_{n}|^{q-2}u_{n}-|z_{n}|^{q-2}z_{n})T_{h}(u_{n}-z_{n})\\ = &\int_{\Omega}(f_{n}-g_{n})T_{h}(u_{n}-z_{n}) -\int_{\Omega}[\frac{1}{(1+|u_{n}|)^{\theta(p-1)}}-\frac{1}{(1+|z_{n}|)^{\theta(p-1)}}]a(x, \nabla z_{n})\cdot\nabla T_{h}(u_{n}-z_{n}). \end{align*} $

由假设(1.6), 等式左边第一项非负, 对右式最后一项使用Lagrange中值定理并利用函数$ t\mapsto\frac{1}{(1+|t|)^{\theta(p-1)}} $导数的绝对有界性, 故

$ \begin{align*} &\int_{\Omega}(|u_{n}|^{q-2}u_{n}-|z_{n}|^{q-2}z_{n})\frac{T_{h}(u_{n}-z_{n})}{h}\\ \leq&\int_{\Omega}|(f_{n}-g_{n})|\frac{T_{h}(u_{n}-z_{n})}{h}+\theta(p-1)\int_{\Omega}|a(x, \nabla z_{n})||\nabla T_{h}(u_{n}-z_{n})| \end{align*} $

对每个固定的$ n, \; u_{n} $$ z_{n} $均属于$ W_0^{1, p}(\Omega) $$ a(x, \xi) $满足假设(1.5). 当$ h\rightarrow0 $时, 有

$ \int_{\Omega}|u_{n}-z_{n}|^{q-1}\leq\int_{\Omega}||u_{n}|^{q-2}u_{n}-|z_{n}|^{q-2}z_{n}|\leq\int_{\Omega}|(f_{n}-g_{n})|, $

对不等式两边取极限, (3.3) 得证.

再选取$ T_{h}(u_{n}-z_{n})^{+} $作为试验函数, 并用同样的处理方法得

$ \int_{\Omega}(u_{n}-z_{n})_{+}^{q-1}\leq\int_{\{u_{n}\geq z_{n}\}}(f_{n}-g_{n}), $

因此, 若假设$ f\leq g $几乎处处成立, 当$ n\rightarrow +\infty $时, 再取极限. 则有

$ \int_{\Omega}(u-z)_{+}^{q-1}\leq\int_{\{u\geq z\}}(f-g)\leq0. $

定理1.3的证明  设$ {u, v}\in W_0^{1, 1}(\Omega)\cap L^{\frac{\theta(p-1)+q}{p}}(\Omega) $是问题(1.3) 的逼近解, 由定义1.2可知一定存在$ {u_{n}} , {v_{n}}\in W_0^{1, p}(\Omega)\cap L^{\infty}(\Omega) $分别是逼近方程(1.9) 的解序列. 则由引理3.1可知,

$ \begin{align} \int_{\Omega}|u-v|^{q-1}\leq\int_{\Omega}|f-f| = 0 \end{align} $ (3.5)

即问题(1.3) 的逼近解是唯一的. 证毕.

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