双震
, 孙义静
中国科学院大学数学科学学院, 北京 100049
2018年1月22日 收稿; 2018年4月13日 收修改稿
基金项目: 国家自然科学基金（11571339，11771468）资助
通信作者: 双震, E-mail:
shuangzhen16@mails.ucas.edu.cn 摘要: 研究矩阵型强奇异偏微分方程 $\left\{ \begin{array}{l} - {\rm{div}}(M\left( x \right)\nabla u) = {\rm{ }}f\left( x \right){u^{ - p}} + \lambda {u^q},\;\;\;\;\;\;{\rm{in}}~\Omega ,\\u > 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{in}}~\Omega ,\\u = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{on}}\;\partial \Omega ,\end{array} \right.$ 其中，
$ \Omega \subset {{\mathbb{R}}^{n}}$是有界开集，
M（
x）是定义在Ω上的实对称矩阵，
$ -p <-1, \text{ }0 <q <1, \lambda >0$是参数，
$f\left( x \right)\in {{L}^{1}}\left( \Omega \right), \text{ }f\left( x \right)>0~\ \ \text{a}\text{.e}\text{.}\ \ \text{ in}\ \ ~\Omega $。证明，如果存在
${{u}_{0}}~\in H_{0}^{1}\left( \Omega \right) $满足
$\int_{\Omega }{{}}f\left( x \right){{\left| {{u}_{0}} \right|}^{1-p}}{\rm{d}}x\text{ } <+\infty $，则对任意的
λ>0上述方程都有正
H01-解，即慢速解。我们注意到，对于奇异方程，古典解即
${{C}^{2}}\left( \Omega \right)\cap C\left( {\bar{\Omega }} \right) $解不一定是
$ H_{0}^{1}\left( \Omega \right)$解。
关键词:
H01-解实对称矩阵强奇性
Exsitence of positive solutions for matrix-type partial differential equations with strongly singular nonlinearities
SHUANG Zhen
, SUN Yijing
School of Mathematical Sciences, University of Chinese Academy of Sciences, Beijing 100049, China
Abstract: We investigate the strongly singular partial differential equations of matrix-type, $\left\{ \begin{array}{l} - {\rm{div}}(M\left( x \right)\nabla u) = {\rm{ }}f\left( x \right){u^{ - p}} + \lambda {u^q},\;\;\;\;\;\;{\rm{in}}~\Omega ,\\u > 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{in}}~\Omega ,\\u = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{on}}\;\partial \Omega ,\end{array} \right.$ where Ω is a bound and open set in
$ {{\mathbb{R}}^{n}}$,
M(
x) is a real symmetric matrix on Ω,
$-p <-1, \text{ }0 <q <1, \lambda >0 $ are parameters,
f(
x)∈
L1(Ω),
f(
x)>0 a.e. in Ω. We prove that the above-mentioned equation admits at least one positive
H01-solution when
λ>0 if there exists
${{u}_{0}}~\in H_{0}^{1}\left( \Omega \right) $ such that
$\int_{\Omega }{{}}f\left( x \right){{\left| {{u}_{0}} \right|}^{1-p}}{\rm{d}}x\text{ } <+\infty $. It should be noted that a classical solution, namely, the
$ {{C}^{2}}\left( \Omega \right)\cap C\left( {\bar{\Omega }} \right)$ -solution, is not necessarily a
$ H_{0}^{1}\left( \Omega \right)$ -solution for singular equations.
Keywords:
H01-solutionreal symmetric matrixstrong singularity
本文研究一类具有强奇性的矩阵型偏微分方程。
| $\left\{ {\begin{array}{*{20}{l}}{ - {\mathop{\rm div}\nolimits} \left( {\mathit{\boldsymbol{M}}(x){\nabla _u}} \right) = f(x){u^{ - p}} + \lambda {u^q},}&{{\rm{in }}\;\;\;\Omega ,}\\{u > 0,}&{{\rm{in }}\;\;\;\Omega ,}\\{u = 0,}&{{\rm{on }}\;\;\;\partial \Omega ,}\end{array}} \right.$ | (1) |
其中
$ \Omega \subset {{\mathbb{R}}^{n}}$是有界开集,
M(
x)是Ω上实对称矩阵, 满足存在正常数
α,
β使得
$\mathit{\boldsymbol{M}}\left( x \right)\xi \cdot\xi \ge \alpha {\left| \xi \right|^2}, |{\rm{det}}\;\mathit{\boldsymbol{M}}\left( x \right)|{\rm{ }} \le \beta , \forall \xi \in {{\mathbb{R}}^n}, \forall x \in \Omega , - p < - 1, {\rm{ }}0 < q < 1, {\rm{ }}f\left( x \right) > 0\;\;{\rm{a}}{\rm{.e}}{\rm{.}}\;\;{\rm{in}}\;\;{\rm{ }}\Omega , \lambda > 0 $是参数。
1991年, 美国数学家Lazer和McKenna
[1]研究一类特殊情形
M(
x)≡
I,
λ=0, 即方程
$ - \Delta u = h\left( x \right){u^{ - p}}$。得到如下结果:如果
$h\left( x \right) \in {C^\alpha }(\bar \Omega )\left( {0 <\alpha <1} \right), h\left( x \right) > 0, \forall x \in \bar \Omega $, 那么对任一-
p < 0方程存在唯一解
${u_{ - p}} \in {C^{2 + \alpha }}(\Omega ) \cap C(\bar \Omega ), {\rm{ }}{u_{ - p}} $不属于
${C^1}(\bar \Omega ) $如果-
p < -1,
u-p属于
H01(Ω)当且仅当-
p>-3。其实当-
p < -1时
u-p的梯度在
$\partial \Omega $处爆破, 当-
p≤-3时
u-p的梯度的爆破速度快到没有
L2可积性, 即
${\smallint _\Omega }|\nabla {u_{ - p}}{|^2}{\rm{d}}x = + \infty $.这就好像唯一解
u-p随着-
p→-∞在Ω的边界附近变得越来越陡峭。意大利数学家Boccardo和Orsina
[2]证明当
f(
x)是非负
L1可积函数并且
λ=0时, 对任意的-
p < -1方程存在
Hloc1(Ω)解
u且
${u^{\frac{{1 + p}}{2}}} \in H_0^1(\Omega ) $。关于矩阵型方程的解的存在性问题, 其中矩阵的性质如前文所述, Boccardo和其他数学家做出过大量研究, 详情可见文献[
3-
7]。本文处理问题的主要思想来源于文献[
8-
11]。
1 本文的结论
定理1.1??设Ω是
${{\mathbb{R}}^{n}}$中具有光滑边界的有界开集, 其中
n≥3,
M(
x)是定义在Ω上的实对称矩阵, 满足存在正常数
α,
β使得
$\mathit{\boldsymbol{M}}\left( x \right)\xi \cdot\xi \ge \alpha {\left| \xi \right|^2}, {\rm{ }}|{\rm{det}}\;M\left( x \right)|{\rm{ }} \le \beta , {\rm{ }}\forall \xi \in {{\mathbb{R}}^{n}}, \forall x \in \Omega , - p < - 1, {\rm{ }}0 <q < 1, {\rm{ }}f\left( x \right) > 0\;\;{\rm{a}}{\rm{.e}}{\rm{. }}\;\;{\rm{in}}\;\;{\rm{ }}\Omega . $如果存在
$ {u_0} \in H_0^1\left( \Omega \right)$满足
| $\int_{\Omega} f(x)\left|u_{0}\right|^{1-p} \mathrm{d} x<+\infty,$ | (2) |
那么对每一个
λ>0方程(1)都有
$H_0^1(\Omega ) $-解。
定理1.2 ??设Ω是
${{\mathbb{R}}^{n}}$中包含原点的具有光滑边界的有界开集,
n≥3,
M(
x)是Ω上的实对称矩阵满足定理1.1中的条件, 0 <
μ <
n, -3 < -
p < -1, 则对任意的
λ>0, 在
$f\left( x \right) = {\left| x \right|^{ - \mu }} $的情况下方程(1)存在正解
$ {u_\lambda } \in H_0^1(\Omega )$。
注:在定理1.1和定理1.2中, 要求Ω具有光滑边界, 实际上只要Ω具有锥性质就足够了。因为只需要保证Sobolev嵌入定理成立, 具体可见文献[
12]。
我们称
u是方程(1)的
$H_0^1(\Omega ) $-解, 如果
$ u \in H_0^1(\Omega )$,
u>0 a.e. in Ω, 满足
$\forall \varphi \in H_0^1(\Omega ), {\smallint _\Omega }\mathit{\boldsymbol{M}}\left( x \right)\nabla {\rm{ }}u\cdot\nabla {\rm{ }}\varphi {\rm{d}}x - {\smallint _\Omega }\frac{{f\left( x \right)}}{{{u^p}}}\cdot{\rm{ }}\varphi {\rm{d}}x - \lambda {\smallint _\Omega }{u^q}\varphi {\rm{d}}x = 0. $由于
M(
x)是实对称矩阵
$ (\mathit{\boldsymbol{M}}\left( x \right)\nabla u\cdot\nabla \varphi = {\rm{ }}\mathit{\boldsymbol{M}}\left( x \right)\nabla \varphi \cdot\nabla u)$, 故考虑如下的能量泛函：
| $\begin{array}{l}I(u) = \frac{1}{2}\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla u \cdot \nabla u{\rm{d}}x + \\\;\;\;\;\;\;\;\;\;\;\frac{1}{{p - 1}}\int_\Omega f (x)|u{|^{1 - p}}{\rm{d}}x - \\\;\;\;\;\;\;\;\;\;\;\frac{\lambda }{{1 + q}}\int_\Omega | u{|^{1 + q}}{\rm{d}}x.\end{array}$ | (3) |
需要注意, 由于强奇性(-
p < -1), 泛函
I在
$H_0^1(\Omega ) $上有奇性。我们将通过对如下两个约束集合之间的交替运用讨论方程(1)的可解性。
| $\begin{array}{l}{N_1}: = \left\{ {u \in H_0^1(\Omega ):u > 0{\rm{ a}}{\rm{.e}}{\rm{.e}}{\rm{.in }}\Omega {\rm{ and }}} \right.\\\;\;\;\;\;\;\;\;\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla u \cdot \nabla u{\rm{d}}x - \int_\Omega f (x){u^{1 - p}}{\rm{d}}x - \\\;\;\;\;\;\;\;\;\left. {\lambda \int_\Omega {{u^{1 + q}}} {\rm{d}}x \ge 0} \right\},\end{array}$ | (4) |
| $\begin{array}{l}{N_2}: = \left\{ {u \in H_0^1(\Omega ):u > 0{\rm{ a}}{\rm{.e}}{\rm{. in }}\Omega {\rm{ and }}} \right.\\\;\;\;\;\;\;\;\;\;\int_\Omega M (x){\nabla _u} \cdot \nabla u{\rm{d}}x - \int_\Omega f (x){u^{1 - p}}{\rm{d}}x - \\\;\;\;\;\;\;\;\;\;\left. {\lambda \int_\Omega {{u^{1 + q}}} {\rm{d}}x = 0} \right\}.\end{array}$ | (5) |
这里用‖·‖表示
$H_0^1(\Omega ) $中常用的范数, 即
$ \parallel u\parallel = {\left( {{\smallint _\Omega }|\nabla u{|^2}{\rm{d}}x} \right)^{\frac{1}{2}}}$。
2 定理的证明先介绍一些引理。
引理2.1 ??设
M(
x)是定义Ω上实对称矩阵, 满足存在正常数
α,
β使得
$ \mathit{\boldsymbol{M}}\left( x \right)\mathit{\boldsymbol{\xi }}\cdot\mathit{\boldsymbol{\xi }} \ge \alpha {\mathit{\boldsymbol{\xi }}^2}, |{\rm{det}}\;\mathit{\boldsymbol{M}}\left( x \right)|{\rm{ }} \le \beta , \forall \mathit{\boldsymbol{\xi }} \in {{{\mathbb{R}}^{n}}}, \forall x \in \Omega $则
| $\begin{array}{*{20}{c}} {|\mathit{\boldsymbol{M}}(x)\mathit{\boldsymbol{\xi }} \cdot \mathit{\boldsymbol{\eta }}| \leqslant \frac{\beta }{{{\alpha ^{n - 1}}}}|\mathit{\boldsymbol{\xi }}||\mathit{\boldsymbol{\eta }}|,} \\ {\forall x \in \Omega ,\forall \mathit{\boldsymbol{\xi }},\mathit{\boldsymbol{\eta }} \in {\mathbb{R}^n}.} \end{array}$ | (6) |
证明??固定
x∈Ω。因为
M(
x)是实对称矩阵, 所以存在正交矩阵
Q(
x)使得
| $\mathit{\boldsymbol{Q}}{\left( x \right)^{\text{T}}}\mathit{\boldsymbol{M}}\left( x \right)\mathit{\boldsymbol{Q}}\left( x \right) = \left( {\begin{array}{*{20}{c}} {{\lambda _1}\left( x \right)}&{}&{} \\ {}& \ddots &{} \\ {}&{}&{{\lambda _n}\left( x \right)} \end{array}} \right)$ |
其中
$ {\lambda _i}\left( x \right), i = 1, \ldots , n, $是矩阵
M(
x)的特征值, 且
${\lambda _i}\left( x \right) \ge \alpha > 0, i = 1, \ldots , n $, 这是因为
λi(
x)是
M(
x)的特征值, 所以存在
x0≠0, 满足
M(
x)
x0=
λi(
x)
x0, 从而
$ \alpha |{x_0}{|^2} \le \mathit{\boldsymbol{M}}\left( x \right){x_0}\cdot{x_0} = {\lambda _i}\left( x \right){x_0}\cdot{x_0} = {\lambda _i}\left( x \right)|{x_0}{|^2}$, 故
${\lambda _i}\left( x \right) \ge \alpha , i = 1 \ldots n $。
对任意
$ \mathit{\boldsymbol{\xi }}, \mathit{\boldsymbol{\eta }} \in {{{\mathbb{R}}^{n}}}$, 令
$\mathit{\boldsymbol{x}} = {\rm{ }}{\mathit{\boldsymbol{Q}}^{\rm{T}}}\mathit{\boldsymbol{\xi }}, \mathit{\boldsymbol{y}} = \mathit{\boldsymbol{ }}{\mathit{\boldsymbol{Q}}^{\rm{T}}}\mathit{\boldsymbol{\eta }} $, 则
$\mathit{\boldsymbol{M\xi }}\cdot\mathit{\boldsymbol{\eta }} = {\rm{ }}\mathit{\boldsymbol{x}}{\mathit{\boldsymbol{Q}}^{\rm{T}}}\mathit{\boldsymbol{MQy}} = {\lambda _1}{x_1}{y_1} + {\lambda _2}{x_2}{y_2} + \ldots + {\lambda _n}{x_n}{y_n}$, 故
| $\begin{array}{l}\left| {\mathit{\boldsymbol{M\xi }} \cdot \mathit{\boldsymbol{\eta }}} \right| = \left| {\sum\limits_{i = 1}^n {{\lambda _i}{x_i}{y_i}} } \right|\\ = \left| {\left( {\prod\limits_{i = 1}^n {{\lambda _i}} } \right)\left( {\frac{{{\lambda _1}}}{{\Pi {\lambda _i}}}{x_1}{y_1} + \frac{{{\lambda _2}}}{{\Pi {\lambda _i}}}{x_2}{y_2} + \cdots + \frac{{{\lambda _n}}}{{\Pi {\lambda _i}}}{x_n}{y_n}} \right)} \right|\\ = |\det \mathit{\boldsymbol{M}}(x)|\left| {\left( {\frac{{{\lambda _1}}}{{\Pi {\lambda _i}}}{x_1}{y_1} + \frac{{{\lambda _2}}}{{\Pi {\lambda _i}}}{x_2}{y_2} + \cdots + \frac{{{\lambda _n}}}{{\Pi {\lambda _i}}}{x_n}{y_n}} \right)} \right|\\ \le \frac{\beta }{{{\alpha ^{n - 1}}}}\left| \mathit{\boldsymbol{x}} \right|\left| \mathit{\boldsymbol{y}} \right| = \frac{\beta }{{{\alpha ^{n - 1}}}}|\mathit{\boldsymbol{\xi }}||\mathit{\boldsymbol{\eta }}|,\end{array}$ |
因为
$ {\left| \mathit{\boldsymbol{x}} \right|^2} = {\mathit{\boldsymbol{x}}^{\rm{T}}}\mathit{\boldsymbol{x}} = {({\mathit{\boldsymbol{Q}}^{\rm{T}}}\mathit{\boldsymbol{\xi }})^{\rm{T}}}{\mathit{\boldsymbol{Q}}^{\rm{T}}}\mathit{\boldsymbol{\xi }} = {\rm{ }}{\mathit{\boldsymbol{\xi }}^{\rm{T}}}\mathit{\boldsymbol{Q }}{\mathit{\boldsymbol{Q}}^{\rm{T}}}\mathit{\boldsymbol{\xi }} = {\rm{ }}{\mathit{\boldsymbol{\xi }}^{\rm{T}}}\mathit{\boldsymbol{\xi }} = {\rm{ }}{\left| \mathit{\boldsymbol{\xi }} \right|^2}$, 同理|
y|
2=|
η|
2。
引理2.2??在
$H_0^1(\Omega ) $上定义
$ \parallel u{\parallel _1} = {\left( {{\smallint _\Omega }\mathit{\boldsymbol{M}}\left( x \right)\nabla u\cdot\nabla u} \right)^{\frac{1}{2}}}$, 可证‖·‖
1是
$H_0^1(\Omega ) $的范数且与‖·‖范数等价, 从而对偶空间相同, 即
| $\left(H_{0}^{1}(\Omega),\|\cdot\|_{1}\right)^{*}=\left(H_{0}^{1}(\Omega),\|\cdot\|\right)^{*}。$ |
证明??根据
M(
x)的性质和引理2.1, 可以得到
| $\begin{array}{l}0 \le \alpha \int_\Omega | \nabla u{|^2} \le \int_\Omega \mathit{\boldsymbol{M}} (x)\nabla u \cdot \nabla u\\\;\; \le \int_\Omega {\left| {\mathit{\boldsymbol{M}}(x)\nabla u \cdot \nabla u} \right|} \le \frac{\beta }{{{\alpha ^{n - 1}}}}\int_\Omega | \nabla u{|^2} < + \infty .\end{array}$ |
从而‖·‖
1:
$H_0^1(\Omega ) $→[0, +∞)是非负泛函, 而且知道如果‖·‖
1是范数, 则必与范数‖·‖等价。下面证明‖·‖
1是
$H_0^1(\Omega ) $上的范数, 只需验证三角不等式, 其他显然。
| $\begin{array}{l}\left\| {u + v} \right\|_1^2 = \int_\Omega \mathit{\boldsymbol{M}} (x)\nabla (u + v) \cdot \nabla (u + v)\\\;\;\;\;\;\;\;\;\;\;\;\; = \int_\Omega \mathit{\boldsymbol{M}} \nabla u \cdot \nabla u + 2\int_\Omega \mathit{\boldsymbol{M}} \nabla u \cdot \nabla v + \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\int_\Omega \mathit{\boldsymbol{M}} \nabla v \cdot \nabla v,\end{array}$ | (7) |
| $\begin{array}{*{20}{r}}{{{\left( {{{\left\| u \right\|}_1} + {{\left\| v \right\|}_1}} \right)}^2} = \int_\Omega \mathit{\boldsymbol{M}} \nabla u \cdot \nabla u + \int_\Omega \mathit{\boldsymbol{M}} \nabla v \cdot \nabla v + }\\{2{{\left( {\int_\Omega \mathit{\boldsymbol{M}} \nabla u \cdot \nabla u} \right)}^{\frac{1}{2}}}{{\left( {\int_\Omega \mathit{\boldsymbol{M}} \nabla v \cdot \nabla v} \right)}^{\frac{1}{2}}}.}\end{array}$ |
由于
M(
x)是正定矩阵, 从而存在实可逆矩阵
M1(
x)使得
$\mathit{\boldsymbol{M}} = \mathit{\boldsymbol{M}}_1^{\rm{T}}{\mathit{\boldsymbol{M}}_1} $, 从而
| $\begin{array}{l}\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla u \cdot \nabla v \le \int_\Omega {\left| {{\mathit{\boldsymbol{M}}_1}(x)\nabla u \cdot {\mathit{\boldsymbol{M}}_1}(x)\nabla v} \right|} \\\;\;\;\; \le \int_\Omega {\left| {{\mathit{\boldsymbol{M}}_1}(x)\nabla u} \right|} \cdot \left| {{\mathit{\boldsymbol{M}}_1}(x)\nabla v} \right|\\\;\;\;\; \le {\left( {\int_\Omega {{{\left| {{\mathit{\boldsymbol{M}}_1}(x)\nabla u} \right|}^2}} } \right)^{\frac{1}{2}}}{\left( {\int_\Omega {{{\left| {{\mathit{\boldsymbol{M}}_1}(x)\nabla v} \right|}^2}} } \right)^{\frac{1}{2}}}\\\;\;\;\; = {\left( {\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla u \cdot \nabla u} \right)^{\frac{1}{2}}}{\left( {\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla v \cdot \nabla v} \right)^{\frac{1}{2}}}.\end{array}$ | (8) |
结合式(7), 可推得
| $\|u+v\|_{1} \leqslant\|u\|_{1}+\|v\|_{1} .$ |
故‖·‖
1是
$H_0^1(\Omega ) $的范数, 与范数‖·‖等价。所以有
| $\left(H_{0}^{1}(\Omega),\|\cdot\|_{1}\right)^{*}=\left(H_{0}^{1}(\Omega),\|\cdot\|\right)^{*}.$ |
引理2.3 ??
N1是闭集。
证明??设
$ {u_i} \to u(H_0^1(\Omega ))$且
${u_i} \in {N_1} $, 则存在
ui的子列(仍记为
ui), 满足
$ {u_i}\left( x \right) \to u\left( x \right)\;\;{\rm{ a}}{\rm{.e}}{\rm{. }}\;\;{\rm{ in}}\;\;{\rm{ }}\Omega .$由于
${u_i} > 0\;\;{\rm{ a}}{\rm{.e}}{\rm{. }}\;\;{\rm{ in}}\;\;{\rm{ }}\Omega $, 所以
$u \ge 0\;\;{\rm{ a}}{\rm{.e}}{\rm{. }}\;\;{\rm{ in}}\;\;\Omega $。由于
$ {u_i} \in {N_1}, {\rm{ }}f\left( x \right) > 0\;\;{\rm{ a}}{\rm{.e}}{\rm{. }}\;\;{\rm{ in}}\;\Omega , \lambda > 0$以及引理2.1, 可知
| $\begin{array}{l}\int_\Omega f (x)u_i^{1 - p}{\rm{d}}x \le \int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla u\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \le \frac{\beta }{{{\alpha ^{n - 1}}}}\int_\Omega {{{\left| {\nabla {u_i}} \right|}^2}} {\rm{d}}x,\end{array}$ |
从而
| $\begin{array}{l}\mathop {\lim \inf }\limits_{i \to \infty } \int_\Omega f (x)u_i^{1 - p}{\rm{d}}x \le \mathop {\lim \inf }\limits_{i \to \infty } \frac{\beta }{{{\alpha ^{n - 1}}}}\int_\Omega {{{\left| {\nabla {u_i}} \right|}^2}} {\rm{d}}x\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \frac{\beta }{{{\alpha ^{n - 1}}}}\int_\Omega {{{\left| {\nabla u} \right|}^2}{\rm{d}}x} < + \infty .\end{array}$ |
根据Fatou引理知
$f\left( x \right){u^{1 - p}} = \mathop {{\rm{lim }}\;{\rm{inf}}}\limits_{i \to \infty } {\rm{ }}f\left( x \right)u_i^{1 - p} $是Ω内可积函数
$ (u > 0\;\;{\rm{ a}}{\rm{.e}}{\rm{. }}\;\;{\rm{ in}}\;{\rm{ }}\Omega )$, 而且
| $\begin{array}{l}\int_\Omega f (x){u^{1 - p}}{\rm{d}}x \le \mathop {\lim \inf }\limits_{i \to \infty } \int_\Omega f (x)u_i^{1 - p}{\rm{d}}x\\ \le \mathop {\lim \inf }\limits_{i \to \infty } \left[ {\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla {u_i}{\rm{d}}x - \frac{\lambda }{{1 + q}}\int_\Omega {u_i^{1 + q}} {\rm{d}}x} \right].\end{array}$ | (9) |
因为
$H_0^1(\Omega ) \to {L^{1 + q}}(\Omega )$是紧嵌入, 所以
${\smallint _\Omega }u_i^{1 + q}{\rm{d}}x \to {\rm{ }}{\smallint _\Omega }{u^{1 + q}}{\rm{d}}x $。根据引理2.2可知
$ {\smallint _\Omega }\mathit{\boldsymbol{M}}\left( x \right)\nabla {u_i}\cdot\nabla {u_i}{\rm{d}}x \to {\rm{ }}{\smallint _\Omega }\mathit{\boldsymbol{M}}\left( x \right)\nabla u\cdot\nabla u{\rm{d}}x$。再由式(9), 可得
| $\begin{array}{l}\int_\Omega f (x){u^{1 - p}}{\rm{d}}x \le \int_\Omega \mathit{\boldsymbol{M}} (x)\nabla u \cdot \nabla u{\rm{d}}x - \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{\lambda }{{1 + q}}\int_\Omega {{u^{1 + q}}} {\rm{d}}x,\end{array}$ |
所以
u∈
N1。因此
N1是闭集。
引理2.4 ??设
u0∈
$H_0^1(\Omega ) $, 满足
| $\int_\Omega f (x){\left| {{u_0}} \right|^{1 - p}}{\rm{d}}x < + \infty ,$ |
则存在唯一的
${t_0} = t({u_0}) > 0 $, 使得
1)
$ I(t({u_0}){u_0}) \le I(t{u_0})\;\;{\rm{ }}\forall t > 0$, 即在
t(
u0)点达到最小值；
2)
t(
u0)
u0∈
N2。
证明 ??因为
${\smallint _\Omega }f\left( x \right)|{u_0}{|^{1 - p}}{\rm{d}}x < + \infty , {\rm{ }} - p < - 1 $, 所以
$ {u_0} \ne 0$, 从而有
${\smallint _\Omega }\mathit{\boldsymbol{M}}\left( x \right)\nabla {u_0}\cdot\nabla {u_0} \ge \alpha {\smallint _\Omega }|\nabla {u_0}{|^2} > 0 $。又由于
f(
x)>0, 可知
$ {\smallint _\Omega }f\left( x \right)|{u_0}{|^{1 - p}} > 0$。
| $\begin{array}{l}\frac{{{\rm{d}}I\left( {t{u_0}} \right)}}{{{\rm{d}}t}} = t\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_0} \cdot \nabla {u_0} - {t^{ - p}}\int_\Omega f (x){\left| {{u_0}} \right|^{1 - p}} - \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\lambda {t^q}\int_\Omega {{{\left| {{u_0}} \right|}^{1 + q}}} ,\end{array}$ | (10) |
容易看出d
I(
tu0)/d
t在(0, +∞)内有唯一零点, 记为
t(
u0)=
t0, 而且在区间(0,
t0)内d
I(
tu0)/d
t < 0, 在区间(
t0, +∞)内d
I(
tu0)/d
t>0, 说明I(
tu0),
$ \forall $t>0在(0,
t0]单调递减, 在[
t0, +∞)单调递增, 所以
I(
tu0), t >0在
t0处取得最小值, 从而式(1)成立。另外还可得到
| $\begin{array}{l}t_0^2\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_0} \cdot \nabla {u_0}{\rm{d}}x - t_0^{1 - p}\int_\Omega f (x){\left| {{u_0}} \right|^{1 - p}}{\rm{d}}x - \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\lambda t_0^{1 + q}\int_\Omega {{{\left| {{u_0}} \right|}^{1 + q}}} {\rm{d}}x = 0,\end{array}$ | (11) |
所以
t(
u0)
u0∈
N2。
引理2.5??泛函
I在
N1中下半连续。
证明??设
ui→
u (
$H_0^1(\Omega ) $)且
ui∈
N1, 要证
| $I(u) \le \mathop {\lim }\limits_{i \to \infty } \inf I\left( {{u_i}} \right).$ | (12) |
同引理2.3中的证明，由Fatou引理可知对
ui的一个子列成立(仍记为
ui)
| $\frac{1}{{p - 1}}\int_\Omega f (x){u^{1 - p}}{\rm{d}}x \le \mathop {\lim \inf }\limits_{i \to \infty } \frac{1}{{p - 1}}\int_\Omega f (x)u_i^{1 - p}{\rm{d}}x.$ |
通过
$ H_0^1(\Omega )\hookrightarrow {\rm{ }}{L^{1 + q}}(\Omega ) $是紧嵌入以及引理2.2即得式(12).
引理2.6??存在常数
C0>0，使得
$ \forall u \in {N_1}, \parallel u\parallel \ge {C_0}$。
证明 ??设
u∈
N1, 则
| $\frac{\beta }{{{\alpha ^{n - 1}}}}{\left\| u \right\|^2} \ge \int_\Omega {\left| {\mathit{\boldsymbol{M}}(x)\nabla u \cdot \nabla u} \right|} \ge \int_\Omega f (x){u^{1 - p}}.$ |
由反向H?lder不等式和Poincaré不等式可以得到
| $\begin{array}{*{20}{c}}{\int_\Omega f (x){u^{1 - p}}{\rm{d}}x \ge {{\left( {\int_\Omega f {{(x)}^{\frac{1}{p}}}} \right)}^p}{{\left( {\int_\Omega u } \right)}^{1 - p}},}\\{\int_\Omega u {\rm{d}}x \le {{\left( {\int_\Omega {{u^2}} {\rm{d}}x} \right)}^{\frac{1}{2}}}|\Omega {|^{\frac{1}{2}}},}\\{{{\left( {\int_\Omega u {\rm{d}}x} \right)}^{1 - p}} \ge C{{\left\| u \right\|}^{1 - p}}.}\end{array}$ |
从而可以得到
(这里
C0与Ω,
β,
α,
p,
f(
x)有关。)
引理2.7??定义
$J\left( u \right) = {\smallint _\Omega }\mathit{\boldsymbol{M}}\left( x \right)\nabla u\cdot\nabla \varphi {\rm{d}}x $, 则
$J\left( u \right), \forall u \in H_0^1(\Omega )$是有界线性泛函。
证明??根据引理2.1和H?lder不等式, 可得
$\mathit{\boldsymbol{M}}\left( x \right)\nabla u\cdot\nabla \varphi $可积。
J(
u)显然是线性的, 下面只需证明有界性。
由于
M(
x)是正定矩阵, 所以存在可逆矩阵
M1(
x), 使得
$\mathit{\boldsymbol{M}} = \mathit{\boldsymbol{M}}_1^{\rm{T}}{\mathit{\boldsymbol{M}}_1} $成立, 从而有
| $\begin{array}{l}\left| {J(u)} \right| = \left| {\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla u \cdot \nabla \varphi } \right|\\\;\;\;\;\;\;\;\;\; \le \int_\Omega {\left| {{\mathit{\boldsymbol{M}}_1}(x)\nabla u \cdot {\mathit{\boldsymbol{M}}_1}(x)\nabla \varphi } \right|} \\\;\;\;\;\;\;\;\;\; \le \int_\Omega {\left| {{\mathit{\boldsymbol{M}}_1}(x)\nabla u} \right|} \cdot \left| {{\mathit{\boldsymbol{M}}_1}(x)\nabla \varphi } \right|\\\;\;\;\;\;\;\;\;\; \le {\left( {\int_\Omega {{{\left| {{\mathit{\boldsymbol{M}}_1}(x)\nabla u} \right|}^2}} } \right)^{\frac{1}{2}}}{\left( {\int_\Omega {{{\left| {{\mathit{\boldsymbol{M}}_1}(x)\nabla \varphi } \right|}^2}} } \right)^{\frac{1}{2}}}\\\;\;\;\;\;\;\;\;\; = {\left( {\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla u \cdot \nabla u} \right)^{\frac{1}{2}}}{\left( {\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla \varphi \cdot \nabla \varphi } \right)^{\frac{1}{2}}}\\\;\;\;\;\;\;\;\;\; \le \frac{\beta }{{{\alpha ^{n - 1}}}}\left\| u \right\|\left\| \varphi \right\|.\end{array}$ |
接下来开始定理1.1的证明。
定理1.1的证明??由引理2.3, 2.4和2.5, 可知
N1是闭集,
N1非空,
I在
N1上有定义而且下半连续。因为
$\forall u \in {N_1} $有
| $\begin{array}{l}I(u) \ge \frac{1}{2}\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla u \cdot \nabla u{\rm{d}}x - \frac{\lambda }{{1 + q}}\int_\Omega {{u^{1 + q}}} {\rm{d}}x\\\;\;\;\;\;\; \ge \frac{\alpha }{2}{\left\| u \right\|^2} - C{\left\| u \right\|^{1 + q}},\end{array}$ | (13) |
所以
I在
N1上有下界。从而根据Ekeland变分原理, 可取最优化极小值序列, 即存在序列
$\{ {u_i}\} \subset {N_1} $, 使得
| $\begin{array}{l}\left. 1 \right)I\left( {{u_i}} \right) < \mathop {\inf }\limits_{{N_1}} I + \frac{1}{i};\\\left. 2 \right)I\left( w \right) \ge I\left( {{u_i}} \right) - \frac{1}{i}\left\| {w - {u_i}} \right\|,\forall w \in {N_1}.\end{array}$ | (14) |
由于
I在
N1上强制, 可知{
ui}有界, 即存在
M>0, 使得‖
ui‖≤
M, 从而存在
ui的一个子列(仍记为
ui), 存在
u*∈
$H_0^1(\Omega ) $, 成立
| $u_{i} \rightarrow u^{*} \quad \text { in } \quad H_{0}^{1}(\Omega),$ | (15) |
| $u_{i} \rightarrow u^{*} \quad \text { in } \quad L^{r}(\Omega), \forall r \in\left[1, \frac{2 n}{n-2}\right),$ | (16) |
| $u_{i}(x) \rightarrow u^{*}(x) \quad \text { a.e.in } \Omega.$ | (17) |
下面证明
$\mathop {{\rm{inf}}}\limits_{{N_1}} I = \mathop {{\rm{inf}}}\limits_{{N_2}} I $。根据引理2.1可知,
$\forall u \in {N_1} $有
| $\begin{array}{l}\int_\Omega f (x)|u{|^{1 - p}} \le \int_\Omega {\left| {\mathit{\boldsymbol{M}}(x)\nabla u \cdot \nabla u} \right|} \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \le \frac{\beta }{{{\alpha ^{n - 1}}}}{\left\| u \right\|^2} < + \infty .\end{array}$ |
从而由引理2.4可知, 存在唯一的正实数
t(
u), 使得
t(
u)
u∈
N2而且
I(
t(
u)
u)≤
I(
u), 进而
$ I\left( u \right) \ge \mathop {{\rm{inf}}}\limits_{{N_2}} I$。所以
$ \mathop {{\rm{inf}}}\limits_{{N_1}} I \ge \mathop {{\rm{inf}}}\limits_{{N_2}} I$。又因为
$ {N_2} \subset {N_1}$, 故
$\mathop {{\rm{inf}}}\limits_{{N_2}} I \ge \mathop {{\rm{inf}}}\limits_{{N_1}} I $, 从而知
| $\mathop {\inf }\limits_{{N_1}} I = \mathop {\inf }\limits_{{N_2}} I.$ |
下面需要分两种情况讨论。
情况1 ??{
ui}中有一个子列位于
N2内(下面仍把这个子列记为
ui)。
固定
$\varphi \in H_0^1(\Omega ), \varphi \ge 0 $。设
t≥0, 因为
| $\int_\Omega f (x){\left( {{u_i} + t\varphi } \right)^{1 - p}}\int_\Omega f (x)u_i^{1 - p} < + \infty $ |
根据引理2.4，可知存在唯一的正实数, 记为
${f_{i, \varphi }}\left( t \right) $, 满足
| $\left\{ {\begin{array}{*{20}{l}}{I\left[ {{f_{i,\varphi }}(t)\left( {{u_i} + t\varphi } \right)} \right] \le I\left[ {\theta \left( {{u_i} + t\varphi } \right)} \right],\forall \theta > 0,}\\{{f_{i,\varphi }}(t)\left( {{u_i} + t\varphi } \right) \in {N_2}.}\end{array}} \right.$ |
下面证明
${f_{i, \varphi }}\left( t \right) $在
t≥0上连续。记
| $g(t): = \int_\Omega \mathit{\boldsymbol{M}} (x)\nabla \left( {{u_i} + t\varphi } \right) \cdot \nabla \left( {{u_i} + t\varphi } \right){\rm{d}}x,\forall t \ge 0,$ |
| $h(t): = \int_\Omega f (x){\left| {{u_i} + t\varphi } \right|^{1 - p}}{\rm{d}}x,\forall t \ge 0,$ |
| $k(t): = \int_\Omega {{{\left| {{u_i} + t\varphi } \right|}^{1 + q}}} {\rm{d}}x,\forall t \ge 0.$ |
根据引理2.2, 控制收敛定理以及
$H_0^1(\Omega ) $ $ \hookrightarrow$ L1+q(Ω), 可以证明
$ g\left( t \right), h\left( t \right), k\left( t \right)$都是连续的。由
${f_{i, \varphi }}\left( t \right)({u_i} + t\varphi ) \in {N_2} $, 可得
| $f_{i, \varphi}^{1-q}(t) g(t)=f_{i, \varphi}^{-p-q}(t) h(t)+\lambda k(t),$ | (18) |
任取
t0∈[0, +∞), 考虑函数
| $f(x): = {x^{1 - q}}g\left( {{t_0}} \right) - \frac{1}{{{x^{p + q}}}}h\left( {{t_0}} \right) + \lambda k\left( {{t_0}} \right).$ |
能够得到
f(
x)在(0, +∞)内严格递增且在(0, +∞)内有唯一零点, 即
$ {f_{i, \varphi }}\left( {{t_0}} \right)$。从而
${f_{i, \varphi }}\left( t \right) $在
t≥0上连续。
因为从引理2.4的证明中可以看出
${f_{i, \varphi }}\left( 0 \right) $是
$\frac{{{\rm{d}}I(t{u_i})}}{{{\rm{d}}t}} $的唯一零点, 又由于
ui∈
N2, 所以得到
${f_{i, \varphi }}\left( 0 \right) $=1。
定义| $f_{i,\varphi }^\prime (0): = \mathop {\lim }\limits_{t \to {0^ + }} \frac{{{f_{i,\varphi }}(t) - 1}}{t} \in [ - \infty , + \infty ]$ |
如果极限不存在, 可取
tk→0
+, 记
$ {{f'}_{i, \varphi }}\left( 0 \right): = \mathop {{\rm{lim}}}\limits_{k \to \infty } \frac{{{f_{i, \varphi }}({t_k}) - 1}}{{{t_k}}}$, 在[-∞, +∞]内取值。
下面证明
$ {{f'}_{i, \varphi }}\left( 0 \right)$有界。由于
${f_{i, \varphi }}\left( t \right)({u_i} + t\varphi ) \in {N_2} $和
ui∈
N2, 可得
| $\begin{array}{l}0 = f_{i,\varphi }^2\left( t \right)\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla \left( {{u_i} + t\varphi } \right) \cdot \nabla \left( {{u_i} + t\varphi } \right)} - \\\;\;\;\;\;f_{i,\varphi }^{1 - p}\left( t \right)\int_\Omega {f\left( x \right){{\left( {{u_i} + t\varphi } \right)}^{1 - p}}} - \\\;\;\;\;\;\lambda f_{i,\varphi }^{1 + q}\left( t \right)\int_\Omega {{{\left( {{u_i} + t\varphi } \right)}^{1 + q}}} - \\\;\;\;\;\;\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla {u_i} + \int_\Omega f (x)u_i^{1 - p} + \lambda \int_\Omega {u_i^{1 + q}} \\\;\;\;\;\; = \left( {f_{i,\varphi }^2(t) - 1} \right)\int_0 \mathit{\boldsymbol{M}} (x)\nabla \left( {{u_i} + t\varphi } \right) \cdot \nabla \left( {{u_i} + t\varphi } \right) + \\\;\;\;\;\;2t\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla \varphi + {t^2}\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla \varphi \cdot \nabla \varphi - \\\;\;\;\;\;\left( {f_{i,\varphi }^{1 - p}(t) - 1} \right)\int_\Omega f (x){\left( {{u_i} + t\varphi } \right)^{1 - p}} - \\\;\;\;\;\;\int_\Omega f (x)\left( {{{\left( {{u_i} + t\varphi } \right)}^{1 - p}} - u_i^{1 - p}} \right) - \\\;\;\;\;\;\lambda \left( {f_{i,\varphi }^{1 + q}(t) - 1} \right)\int_\Omega {{{\left( {{u_i} + t\varphi } \right)}^{1 + q}}} - \\\;\;\;\;\;\lambda \int_\Omega {{{\left( {{u_i} + t\varphi } \right)}^{1 + q}}} - u_i^{1 + q}{\rm{d}}x.\end{array}$ |
从而利用
$ - p < - 1, {\rm{ }}{u_i} > 0, {\rm{ }}t > 0, {\rm{ }}\varphi \ge 0, {\rm{ }}f > 0$, 知
| $\begin{array}{l}0 \ge \frac{{{f_{i,\varphi }}\left( t \right) - 1}}{t}\\\left\{ {\left( {{f_{i,\varphi }}\left( t \right) + 1} \right)\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla \left( {{u_i} + t\varphi } \right) \cdot \nabla \left( {{u_i} + t\varphi } \right) - } \right.\\(1 - p){(1 + o(1))^{ - p}}\int_\Omega f (x){\left( {{u_i} + t\varphi } \right)^{1 - p}} - \\\left. {\lambda (1 + q){{(1 + o(1))}^q}\int_\Omega {{{\left( {{u_i} + t\varphi } \right)}^{1 + q}}} } \right\} + \\2\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla \varphi + t\int_\Omega {\bf{M}} (x){\nabla _\varphi } \cdot {\nabla _\varphi } - \\\lambda \int_\Omega {\frac{{{{\left( {{u_i} + t\varphi } \right)}^{1 + q}} - u_i^{1 + q}}}{t}} {\rm{d}}x,\end{array}$ | (19) |
其中
o(1)表示当
t→0
+时的无穷小。利用控制收敛定理可以证明当
t→0
+时,
| $\int_{\Omega} \frac{\left(u_{i}+t \varphi\right)^{1+q}-u_{i}^{1+q}}{t} \mathrm{d} x \rightarrow(1+q) \int_{\Omega} u_{i}^{q} \varphi \mathrm{d} x.$ |
利用前面4个函数的连续性以及
ui∈
N2, 在式(19)两边令
t→0
+可得到
| $\begin{array}{*{20}{l}}{\lambda (1 + q)\int_\Omega {u_i^q} \varphi - 2\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla \varphi \ge f_{i,\varphi }^\prime (0)}\\{\left[ {(1 - q)\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla {u_i} + (p + q)\int_\Omega f (x)u_i^{1 - p}} \right].}\end{array}$ | (20) |
利用H?lder不等式, 引理2.1,
M(
x)的性质以及
$H_0^1(\Omega ) $ $ \hookrightarrow$L1+q(Ω), 可以得到一系列不等式,
| $\left\{ \begin{array}{l}\int_\Omega {u_i^q} \varphi {\rm{d}}x \le \left\| {{u_i}} \right\|_{{L^{1 + q(\Omega )}}}^q{\left\| \varphi \right\|_{{L^{1 + q}}(\Omega )}}\\\;\;\;\; \le C{\left\| {{u_i}} \right\|^q}\left\| \varphi \right\| \le C\left\| \varphi \right\|,\\\left| {\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla \varphi } \right| \le \frac{\beta }{{{\alpha ^{n - 1}}}}\left\| {{u_i}} \right\|\left\| \varphi \right\|\\\;\;\; \le C\left\| \varphi \right\|,\\\left( {1 - q} \right)\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla {u_i} + \left( {p + q} \right)\int_\Omega {f\left( x \right)u_i^{1 - p}} \\\;\;\;\; \le \left( {1 - q} \right)\alpha C_0^2,\end{array} \right.$ |
其中还用到引理2.6和{
ui}是有界的。结合式(20)得到
$ {{f'}_{i, \varphi }}\left( 0 \right)$有一致上界, 即存在
${{C}_{1}}\in \mathbb{R} $, 使得
$ {{f'}_{i, \varphi }}\left( 0 \right)$≤
C1对任意的正整数
i成立。
根据式(14)中的第2个结论以及
${f_{i, \varphi }}\left( t \right)({u_i} + t\varphi ) \in {N_2}$, 有
| $\begin{array}{l}I\left( {{u_i}} \right) \le I\left( {{f_{i,\varphi }}(t)\left( {{u_i} + t\varphi } \right)} \right) + \\\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{i}\left\| {{u_i} - {f_{i,\varphi }}(t)\left( {{u_i} + t\varphi } \right)} \right\|.\end{array}$ |
代入计算, 因为
ui∈
N2, 得到
| $\begin{array}{l}\left( {\frac{1}{2} - \frac{1}{{1 - p}}} \right)\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla {u_i} + \\\lambda \left( {\frac{1}{{1 - p}} - \frac{1}{{1 + q}}} \right)\int_\Omega {u_i^{1 + q}} \le \\\left( {\frac{1}{2} - \frac{1}{{1 - p}}} \right)f_{i,\varphi }^2(t)\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla \left( {{u_i} + t\varphi } \right) \cdot \\\nabla \left( {{u_i} + t\varphi } \right) + \lambda \left( {\frac{1}{{1 - p}} - \frac{1}{{1 + q}}} \right)f_{i,\varphi }^{1 + q}(t) \cdot \\\int_\Omega {{{\left( {{u_i} + t\varphi } \right)}^{1 + q}}} + \frac{1}{i}\left\| {\left( {1 - {f_{i,\varphi }}(t)} \right){u_i} - t{f_{i,\varphi }}(t)\varphi } \right\|.\end{array}$ |
从而有
| $\begin{array}{l}\frac{{{f_{i,\varphi }}(t) - 1}}{t}\left\{ {\left( {\frac{{ - 1}}{2} + \frac{1}{{1 - p}}} \right)\left( {{f_{i,\varphi }}(t) + 1} \right)} \right.\\\;\;\;\;\;\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla \left( {{u_i} + t\varphi } \right) \cdot \nabla \left( {{u_i} + t\varphi } \right) + \\\;\;\;\;\;\left. {\lambda \left( {1 + \frac{{1 + q}}{{p - 1}}} \right){{(1 + o(1))}^q}\int_\Omega {{{\left( {{u_i} + t\varphi } \right)}^{1 + q}}} } \right\} - \\\;\;\;\;\;\frac{1}{i}\left| {\frac{{{f_{i,\varphi }}(t) - 1}}{t}} \right|\left\| {{u_i}} \right\| \le \left( {\frac{1}{2} + \frac{1}{{p - 1}}} \right)\\\;\;\;\;\;\left\{ {2\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla \varphi + t\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla \varphi \cdot \nabla \varphi } \right\} + \\\;\;\;\;\;\frac{1}{i}{f_{i,\varphi }}(t)\left\| \varphi \right\|.\end{array}$ |
在上式中令
t→0
+,
| $\begin{array}{l}f_{i,\varphi }^\prime (0)\left\{ { - 2\left( {\frac{1}{2} + \frac{1}{{p - 1}}} \right)\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla {u_i} + } \right.\\\;\;\;\;\left. {\lambda \left( {1 + \frac{{1 + q}}{{p - 1}}} \right)\int_\Omega {u_i^{1 + q}} - \frac{1}{i}{\mathop{\rm sgn}} \left[ {f_{i,\varphi }^\prime (0)} \right]\left\| {{u_i}} \right\|} \right\}\\\;\;\;\; \le 2\left( {\frac{1}{2} + \frac{1}{{p - 1}}} \right)\int_\Omega M (x)\nabla {u_i} \cdot \nabla \varphi + \frac{{\left\| \varphi \right\|}}{i}.\end{array}$ | (21) |
由引理2.6和
ui∈
N2可以得到,
| $\begin{array}{l} - 2\left( {\frac{1}{2} + \frac{1}{{p - 1}}} \right)\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla {u_i} + \\\;\;\;\;\;\lambda \left( {1 + \frac{{1 + q}}{{p - 1}}} \right)\int_\Omega {u_i^{1 + q}} = \\\;\;\;\;\; - \left( {\frac{{1 - q}}{{p - 1}}\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla {u_i} + \frac{{p + q}}{{p - 1}}\int_\Omega f (x)u_i^{1 - p}} \right)\\\;\;\;\;\; \le - \frac{{1 - q}}{{p - 1}}\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla {u_i}\\\;\;\;\;\; \le - \frac{{1 - q}}{{p - 1}}\alpha {\left\| {{u_i}} \right\|^2} \le - \frac{{1 - q}}{{p - 1}}\alpha C_0^2.\end{array}$ |
由于{
ui}有界, 所以
| $\frac{1}{i}{\mathop{\rm sgn}} \left[ {f_{i,\varphi }^\prime (0)} \right]\left\| {{u_i}} \right\| \to 0(i \to \infty ).$ |
从而可知存在
$ N\in {{\mathbb{N}}^{*}}$, 当
i>
N时有
| $\begin{array}{l} - 2\left( {\frac{1}{2} + \frac{1}{{p - 1}}} \right)\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla {u_i} + \\\;\;\;\;\lambda \left( {1 + \frac{{1 + q}}{{p - 1}}} \right)\int_\Omega {u_i^{1 + q}} - \frac{1}{i}{\mathop{\rm sgn}} \left[ {f_{i,\varphi }^\prime (0)} \right]\left\| {{u_i}} \right\|\\\;\;\;\; \le - \frac{{(1 - q)\alpha C_0^2}}{{2(p - 1)}} < 0\end{array}$ | (22) |
结合式(21)和式(22)以及
$ |{\smallint _\Omega }\mathit{\boldsymbol{M}}\left( x \right)\nabla {u_i}\cdot\nabla \varphi | \le C\parallel \varphi \parallel $, 可以得到存在
C2∈
${\mathbb{R}} $满足
| $C_{2} \leqslant f_{i, \varphi}^{\prime}(0) \leqslant C_{1}, \forall i>N$ |
根据式(14)中的第2个结论以及
${f_{i, \varphi }}\left( t \right)({u_i} + t\varphi ) \in {N_2}$, 有
| $\begin{array}{l}\frac{1}{{p - 1}}\int_\Omega {\frac{{f(x)u_i^{1 - p} - f(x){{\left( {{u_i} + t\varphi } \right)}^{1 - p}}}}{t}} {\rm{d}}x + \\\frac{1}{{p - 1}}\frac{{\left( {{f_{i,\varphi }}(t) - 1} \right)(p - 1){{(1 + o(1))}^{ - p}}}}{t}\\\int_\Omega f (x){\left( {{u_i} + t\varphi } \right)^{1 - p}} \le \frac{1}{2}\frac{{{f_{i,\varphi }}(t) - 1}}{t}\left( {{f_{i,\varphi }}(t) + 1} \right)\\\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla \left( {{u_i} + t\varphi } \right) \cdot \nabla \left( {{u_i} + t\varphi } \right) + \\\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla \varphi + \frac{1}{2}t\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla \varphi \cdot \nabla \varphi - \\\frac{\lambda }{{1 + q}}\frac{{{f_{i,\varphi }}\left( t \right) - 1}}{t}\left( {1 + q} \right){\left( {1 + o\left( 1 \right)} \right)^q} \cdot \\\int_\Omega {{{\left( {{u_i} + t\varphi } \right)}^{1 + q}}} - \frac{\lambda }{{1 + q}}\int_\Omega {\frac{{{{\left( {{u_i} + t\varphi } \right)}^{1 + q}} - u_i^{1 + q}}}{t}} + \\\frac{1}{i}\frac{{\left| {{f_{i,\varphi }}\left( t \right) - 1} \right|}}{t}\left\| {{u_i}} \right\| + \frac{1}{i}{f_{i,\varphi }}\left( t \right)\left\| \varphi \right\|,\end{array}$ |
其中
o(1)表示当
t→0
+时的无穷小。在上式中令
t→0
+取下极限, 因为
ui∈
N2, 可知
| $\begin{array}{l}\frac{1}{{p - 1}}\mathop {\lim \inf }\limits_{t \to {0^ + }} \int_\Omega {\frac{{f\left( x \right)u_i^{1 - p} - f\left( x \right){{\left( {{u_i} + t\varphi } \right)}^{1 - p}}}}{t}{\rm{d}}x} \le \\\;\;\;\;\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla \varphi - \lambda \int_\Omega {u_i^q\varphi } + \\\;\;\;\;\frac{{\left| {{{f'}_{i,\varphi }}\left( 0 \right)} \right|\left\| {{u_i}} \right\| + \left\| \varphi \right\|}}{i}.\end{array}$ | (23) |
另一方面，由Fatou引理可以得到
| $\begin{array}{l}(p - 1)\int_\Omega f (x)u_i^{ - p}\varphi \\\;\;\;\;\;\;\; \le \mathop {\lim \inf }\limits_{t \to {0^ + }} \int_\Omega {\frac{{f(x)u_i^{1 - p} - f(x){{\left( {{u_i} + t\varphi } \right)}^{1 - p}}}}{t}} {\rm{d}}x.\end{array}$ | (24) |
由式(17)用Fatou引理, 并结合式(23)和式(24), 可得
| $\begin{array}{*{20}{c}}{\int_\Omega f (x){{\left( {{u^*}} \right)}^{ - p}}\varphi \le \mathop {\lim \inf }\limits_{i \to \infty } \left\{ {\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla \varphi - } \right.}\\{\left. {\lambda \int_\Omega {u_i^q} \varphi + \frac{{\left| {f_{i,\varphi }^\prime (0)} \right|\left\| {{u_i}} \right\| + \left\| \varphi \right\|}}{i}} \right\}.}\end{array}$ | (25) |
根据已知的一些结论,
$ {{f'}_{i, \varphi }}\left( 0 \right)$和
ui的有界性, 引理2.7和式(15), 并再次运用式(17)和Fatou引理, 由式(25)可知
| $\begin{array}{l}\int_\Omega f (x){\left( {{u^*}} \right)^{ - p}}\varphi \le \int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u^*} \cdot {\nabla _\varphi } - \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\lambda \int_\Omega {{{\left( {{u^*}} \right)}^q}} \varphi .\end{array}$ | (26) |
注意式(26)是对任意的
$\varphi \in H_0^1(\Omega ) $且
$\varphi \ge 0$成立的, 所以有
| $\begin{array}{*{20}{c}}{\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u^*} \cdot \nabla {u^*} - \int_\Omega f (x){{\left( {{u^*}} \right)}^{1 - p}} - }\\{\lambda \int_\Omega {{{\left( {{u^*}} \right)}^{1 + q}}} \ge 0,}\end{array}$ | (27) |
从而
$ {\smallint _\Omega }f\left( x \right){({u^*})^{1 - p}}{\rm{d}}x < + \infty $, 故
$ {u^*} > 0\;{\rm{a}}{\rm{.e}}{\rm{.}}\;{\rm{in}}\;{\rm{ }}\Omega $。所以
u*∈
N1。
由式(14)中第1个结论和式(16)可得
| $\begin{array}{l}\mathop {\inf }\limits_{{N_1}} I = \mathop {\lim }\limits_{i \to \infty } I\left( {{u_i}} \right) = \mathop {\lim \;inf}\limits_{i \to \infty } I\left( {{u_i}} \right) \ge \\\;\;\;\;\;\;\;\;\;\;\mathop {\lim \inf }\limits_{i \to \infty } \frac{1}{2}\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla {u_i} + \\\;\;\;\;\;\;\;\;\;\;\mathop {\lim \inf }\limits_{i \to \infty } \frac{1}{{p - 1}}\int_\Omega f (x)u_i^{1 - p} - \frac{\lambda }{{1 + q}}\int_\Omega {{{\left( {{u^*}} \right)}^{1 + q}}} \\\;\;\;\;\;\;\;\;\;\; \ge \frac{1}{2}\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u^ * } \cdot \nabla {u^ * } + \\\;\;\;\;\;\;\;\;\;\;\frac{1}{{p - 1}}\int_\Omega {f\left( x \right){{\left( {{u^ * }} \right)}^{1 - p}}} - \\\;\;\;\;\;\;\;\;\;\;\frac{\lambda }{{1 + q}}\int_\Omega {{{\left( {{u^ * }} \right)}^{1 + q}}} = I\left( {{u^ * }} \right)\\\;\;\;\;\;\;\;\;\;\; \ge I\left( {t\left( {{u^ * }} \right){u^ * }} \right) \ge \mathop {\inf }\limits_{{N_2}} I = \mathop {\inf }\limits_{{N_1}} I.\end{array}$ | (28) |
式(28)中还用到了引理2.4和下面两个结论。
1) 由式(15)和引理2.2可知
| ${u_i}\;?\;{u^*}{\rm{ in }}\left( {H_0^1(\Omega ),{{\left\| \cdot \right\|}_1}} \right);$ |
2) 由式(17)和Fatou引理可知
| $\int_\Omega f (x){\left( {{u^*}} \right)^{1 - p}} \le \mathop {\lim \inf }\limits_{i \to \infty } \int_\Omega {f(x)u_i^{1 - p}} .$ |
由式(28)可以得到
$I({u^*}) = I(t({u^*}){u^*}) $, 根据引理2.4中的最小值的性质可知
t(
u*)=1, 从而
u*∈
N2。
情况2??对充分大的
i有
ui∈
N1\
N2。
固定
φ∈
$H_0^1(\Omega ) $满足
φ≥0。仍然可以定义
$g\left( t \right), h\left( t \right), k\left( t \right) $(如同情况1中一样), 它们依然是连续的。式(24)仍然成立。由控制收敛定理还是可以得到当
t→0
+时,
| $\int_\Omega {\frac{{{{\left( {{u_i} + t\varphi } \right)}^{1 + q}} - u_i^{1 + q}}}{t}} {\rm{d}}x \to (1 + q)\int_\Omega {u_i^q} \varphi {\rm{d}}x.$ |
因为
ui∈
N1\
N2, 所以
| $\begin{array}{l}\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla {u_i} - \int_\Omega f (x)u_i^{1 - p} - \\\;\;\;\;\;\;\;\lambda \int_\Omega {u_i^{1 + q}} > 0.\end{array}$ |
由
$ g\left( t \right), h\left( t \right), k\left( t \right)$的连续性可知, 当
t充分小时成立
| $\begin{array}{*{20}{l}}{\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla \left( {{u_i} + t\varphi } \right) \cdot \nabla \left( {{u_i} + t\varphi } \right) - }\\{\;\;\;\;\int_\Omega f (x){{\left( {{u_i} + t\varphi } \right)}^{1 - p}} - \lambda \int_\Omega {{{\left( {{u_i} + t\varphi } \right)}^{1 + q}}} > 0,}\end{array}$ |
即得
ui+
tφ∈
N1\
N2.
因为
$ I\left( w \right) \ge I({u_i}) - \frac{1}{i}\parallel w - {u_i}\parallel , \forall w \in {N_1}$, 所以
| $I\left(u_{i}\right) \leqslant I\left(u_{i}+t \varphi\right)+\frac{1}{i}\left\|u_{i}-\left(u_{i}+t \varphi\right)\right\|.$ |
代入计算, 两边同除以
t得
| $\begin{array}{l}0 \le \int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla \varphi + \frac{1}{2}t\int_\Omega M (x)\nabla \varphi \cdot \nabla \varphi + \\\;\;\;\;\;\frac{1}{{p - 1}}\int_\Omega {\frac{{f\left( x \right){{\left( {{u_i} + t\varphi } \right)}^{1 - p}} - f\left( x \right)u_i^{1 - p}}}{t}} - \\\;\;\;\;\;\frac{\lambda }{{1 + q}}\int_\Omega {\frac{{{{\left( {{u_i} + t\varphi } \right)}^{1 + q}} - u_i^{1 + q}}}{t}} + \frac{{\left\| \varphi \right\|}}{i}.\end{array}$ |
令
t→0
+取上极限(通过引理2.7和式(15)), 得到
| $\begin{array}{l}0 \le \int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u_i} \cdot \nabla \varphi - \int_\Omega f (x)u_i^{ - p}\varphi - \\\;\;\;\;\;\lambda \int_\Omega {u_i^{ - q}\varphi } + \frac{{\left\| \varphi \right\|}}{i}.\end{array}$ | (29) |
另外由
ui(
x)→
u*(
x) a.e.in Ω, 利用Fatou引理可得
| $\int_\Omega f (x){\left( {{u^*}} \right)^{ - p}}\varphi \le \mathop {\lim \inf }\limits_{i \to \infty } \int_\Omega {f(x)u_i^{ - p}\varphi } ,$ | (30) |
| $\int_\Omega {{{\left( {{u^*}} \right)}^q}} \varphi \le \mathop {\lim \inf }\limits_{i \to \infty } \int_\Omega {u_i^q} \varphi .$ | (31) |
结合式(29), 式(30)和式(31)可知
| $\begin{array}{l}\int_\Omega f (x){\left( {{u^*}} \right)^{ - p}}\varphi \le \int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u^ * } \cdot \nabla \varphi - \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\lambda \int_\Omega {{{\left( {{u^*}} \right)}^q}} \varphi ,\end{array}$ |
所以
| $\begin{array}{*{20}{c}}{\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u^*} \cdot \nabla {u^*} - \int_\Omega f (x){{\left( {{u^*}} \right)}^{1 - p}} - }\\{\lambda \int_\Omega {{{\left( {{u^*}} \right)}^{1 + q}}} \ge 0.}\end{array}$ |
从而
$ {\smallint _\Omega }f\left( x \right){({u^*})^{1 - p}}{\rm{d}}x < + \infty $, 故
u*>0 a.e.in Ω, 所以
u*∈
N1。重复情况1中的步骤, 此时也成立
| $\mathop {\inf }\limits_{{N_1}} I \ge I\left( {{u^*}} \right) \ge I\left( {t\left( {{u^*}} \right){u^*}} \right) \ge \mathop {\inf }\limits_{{N_2}} I = \mathop {\inf }\limits_{{N_1}} I,$ |
所以
u*∈
N2.
无论是情况1还是情况2都得到这样的结果,
| $\left\{ {\begin{array}{*{20}{l}}{{u^*} \in {N_2},{u^*} > 0,}\\{I\left( {{u^*}} \right) = \mathop {\inf }\limits_{{N_1}} I,}\\{\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u^ * } \cdot \nabla \varphi - \int_\Omega f (x){{\left( {{u^*}} \right)}^{ - p}}\varphi - }\\{\lambda \int_\Omega {{{\left( {{u^*}} \right)}^q}} \varphi \ge 0,\forall \varphi \ge 0.}\end{array}} \right.$ | (32) |
下面证明
u*是方程(1)的解。固定
ψ∈
$H_0^1(\Omega ) $和
t>0, (
u*+
tψ)
+∈
$H_0^1(\Omega ) $且(
u*+
tψ)
+≥0。由式(32)可知
| $\begin{array}{l}0 \le \int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u^ * } \cdot \nabla {\left( {{u^*} + t\psi } \right)^ + } - \\\;\;\;\;\;\int_\Omega f (x){\left( {{u^*}} \right)^{ - p}}{\left( {{u^*} + t\psi } \right)^ + } - \\\;\;\;\;\;\lambda \int_\Omega {{{\left( {{u^*}} \right)}^q}} {\left( {{u^*} + t\psi } \right)^ + }\\\;\;\; = \int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u^ * } \cdot \nabla \left( {{u^*} + t\psi } \right) - \\\;\;\;\;\;\int_\Omega f (x){\left( {{u^*}} \right)^{ - p}}\left( {{u^*} + t\psi } \right) - \\\;\;\;\;\;\lambda \int_\Omega {{{\left( {{u^*}} \right)}^q}} \left( {{u^*} + t\psi } \right) - \\\;\;\;\;\;\int_{\left\{ {{u^*} + t\psi < 0} \right\}} {\mathit{\boldsymbol{M}}(x)} \nabla {u^*} \cdot \nabla \left( {{u^*} + t\psi } \right) + \\\;\;\;\;\;\int_{\left\{ {{u^*} + t\psi < 0} \right\}} f (x){\left( {{u^*}} \right)^{ - p}}\left( {{u^*} + t\psi } \right) + \\\;\;\;\;\;\lambda \int_{\left\{ {{u^*} + t\psi < 0} \right\}} {{{\left( {{u^*}} \right)}^q}} \left( {{u^*} + t\psi } \right)\\\;\;\;\;\; \le t\left\{ {\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u^ * } \cdot \nabla \psi - \int_\Omega f (x){{\left( {{u^*}} \right)}^{ - p}}\psi - } \right.\\\;\;\;\;\;\;\;\left. {\lambda \int_\Omega {{{\left( {{u^*}} \right)}^q}} \psi } \right\} - t\int_{\left\{ {{u^*} + t\psi < 0} \right\}} {\mathit{\boldsymbol{M}}(x)\nabla {u^*} \cdot \nabla \psi } .\end{array}$ |
两边除以
t, 令
t→0
+, 其中meas{
u*+
tψ < 0}→0, 可知
| $\begin{array}{*{20}{c}}{\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u^*} \cdot \nabla \psi - \int_\Omega f (x){{\left( {{u^*}} \right)}^{ - p}}\psi - }\\{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\lambda \int_\Omega {{{\left( {{u^*}} \right)}^q}} \psi \ge 0.}\end{array}$ | (33) |
从而
| $\begin{array}{l}\int_\Omega \mathit{\boldsymbol{M}} (x)\nabla {u^*} \cdot \nabla \psi - \int_\Omega f (x){\left( {{u^*}} \right)^{ - p}}\psi - \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\lambda \int_\Omega {{{\left( {{u^*}} \right)}^q}} \psi = 0,\forall \psi \in H_0^1(\Omega ),\end{array}$ | (34) |
即说明
u*是方程(1)的弱解。
定理1.2是运用定理1.1的结论的一个例子。
定理1.2的证明??设
φ1是-Δ在Dirichlet边界条件下的第一特征向量, 即
| $\left\{ \begin{array}{l} - \Delta {\varphi _1} = {\lambda _1}{\varphi _1}\;\;\;{\rm{in}}\;\;\Omega \\{\varphi _1} = 0\;\;{\rm{on}}\;\;\partial \Omega \end{array} \right.$ |
而且在Ω内
φ1>0, 其中
λ1表示-Δ在Dirichlet边界条件下的第一特征值。
根据Lazer和McKenna(文献[
1], Lemma, Theorem 1和2)可知,
$ {\smallint _\Omega }\varphi _1^l\left( x \right){\rm{d}}x < + \infty \Leftrightarrow l > - 1$, 且对任意的-3 < -
p < -1, 存在相应函数
${u_p} \in H_0^1(\Omega ) \cap {C^2}(\Omega ) \cap C(\bar \Omega ) $, 在Ω内
up>0, 存在常数
d0,
d1>0, 使得
| ${d_0}\varphi _1^{\frac{2}{{1 + p}}}(x) \le {u_p}(x) \le {d_1}\varphi _1^{\frac{2}{{1 + p}}}(x),\forall x \in \Omega .$ |
令
u0=
up, 下面验证
$ {\smallint _\Omega }{\left| x \right|^{ - \mu }}u_0^{1 - p}{\rm{d}}x < + \infty $。因为0∈Ω, 所以可取一个小球
${B_r}\left( 0 \right) \subset \Omega $, 从而存在
C1,
C2>0, 使得
$ 0 < {C_1} \le {u_0}\left( x \right) \le {C_2}, \forall x \in {B_r}\left( 0 \right)$。
| $\begin{array}{*{20}{l}}{{{\int_\Omega {\left| x \right|} }^{ - \mu }}u_0^{1 - p}{\rm{d}}x = {{\int_{{B_r}} {\left| x \right|} }^{ - \mu }}u_0^{1 - p}{\rm{d}}x + }\\{\;\;{\smallint _{\Omega \backslash {B_r}}}|x{|^{ - \mu }}u_0^{1 - p}{\rm{d}}x \le C_1^{1 - p}{{\int_{{B_r}} {\left| x \right|} }^{ - \mu }}{\rm{d}}x + }\\{\;\;\;{r^{ - \mu }}{\smallint _{\Omega \backslash {B_r}}}u_0^{1 - p}{\rm{d}}x \le C_1^{1 - p}\int_{{B_r}} | x{|^{ - \mu }}{\rm{d}}x + }\\{\;\;\;{r^{ - \mu }}d_0^{1 - p}{\smallint _{\Omega \backslash {B_r}}}\varphi _1^{\frac{{2(1 - p)}}{{1 + p}}}(x){\rm{d}}x \le C_1^{1 - p}\int_{{B_r}} | {{\left. x \right|}^{ - \mu }}{\rm{d}}x + }\\{\;\;\;{r^{ - \mu }}d_0^{1 - p}\int_\Omega {\varphi _1^{\frac{{2(1 - p)}}{{1 + p}}}} (x){\rm{d}}x < + \infty ,}\end{array}$ | (35) |
其中用到, 因为-
n < -
μ < 0, 所以
$ {\smallint _{{B_r}}}{\left| x \right|^{ - \mu }}{\rm{d}}x < + \infty $, 由于
$1 < p < 3 $, 所以
$ \frac{{2\left( {1 - p} \right)}}{{1 + p}} > - 1$, 故
| $\int_\Omega {\varphi _1^{\frac{{2(1 - p)}}{{1 + p}}}} (x){\rm{d}}x < + \infty .$ |
根据定理1.1即得要证结论。
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