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 $.这就好像唯一解u_-p随着-p→-∞在Ω的边界附近变得越来越陡峭。意大利数学家Boccardo和Orsina^[2]证明当f(x)是非负L¹可积函数并且λ=0时, 对任意的-p < -1方程存在H_loc¹(Ω)解u且${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 $, 这是因为λ_i(x)是M(x)的特征值, 所以存在x₀≠0, 满足M(x)x₀=λ_i(x)x₀, 从而$ \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} $, 则存在u_i的子列(仍记为u_i), 满足$ {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dI(tu₀)/dt在(0, +∞)内有唯一零点, 记为t(u₀)=t₀, 而且在区间(0, t₀)内dI(tu₀)/dt < 0, 在区间(t₀, +∞)内dI(tu₀)/dt>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u_i→u ($H_0^1(\Omega ) $)且u_i∈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u_i的一个子列成立(仍记为u_i)

$ \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Hölder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

$ \|u\| \geqslant C_{0}. $

(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Hölder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{u_i}有界, 即存在M>0, 使得‖u_i‖≤M, 从而存在u_i的一个子列(仍记为u_i), 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   {u_i}中有一个子列位于N₂内(下面仍把这个子列记为u_i)。

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$ L^1+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}} $的唯一零点, 又由于u_i∈N₂, 所以得到${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t_k→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u_i∈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u_i∈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Hölder不等式, 引理2.1, M(x)的性质以及$H_0^1(\Omega ) $ $ \hookrightarrow$ L^1+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{u_i}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u_i∈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u_i∈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{u_i}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u_i∈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u_i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u_i∈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u_i∈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u_i+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u_i(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u_p>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=u_p, 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。