本文中, 我们将对带临界指数增长的椭圆型方程组的解进行研究

$ \left\{\begin{array}{ll} { } -\Delta u+a(x)u=\frac{2\alpha }{\alpha +\beta }u^{\alpha -1}v^\beta +f(x), & x\in \Omega, \\ { } -\Delta v+b(x)v=\frac{2\beta }{\alpha +\beta }u^\alpha v^{\beta -1} +g(x), & x\in \Omega, \\ u>0, v>0, &x\in\Omega, \\ u=v=0, &x\in\partial\Omega \end{array}\right.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \begin{array}{r} (1.1)\\ (1.2)\\ (1.3)\\ (1.4)\end{array} $

其中$ \Omega $是$ {{\Bbb R}} ^N $中一个光滑有界区域, $ N=3, 4, a\geq 2, \beta\geq 2, \alpha +\beta =2^*=\frac{2N}{N-2}, f(x)\geq 0, $$ g(x)\geq 0, $$ f(x), g(x)\in H^{-1}( \Omega )\times H^{-1}(\Omega), a(x)\geq 0, b(x)\geq0. $并且$ a(x), b(x) $满足条件

$ (A_1)\quad a(x), b(x)\in L^\infty( \Omega ). $

$ (A_2) $ 在$ \Omega $中存在一个区域$ \Omega_1 $, 使得任意的$ x\in \Omega_1, $有$ a(x)=b(x)=0. $

Alves, Morais Filho和Souto在文献[1]中证明了一类不带扰动项$ f(x), g(x) $并且一次项系数为常数的方程组解的存在性和非存在性. Han在文献[2]中通过构造Pala-Smale序列和min-max原理研究得到在有界域上一类不带扰动项$ f(x), g(x) $并且一次项系数充分小时方程组高能量解的存在性. Wan和Yang在文献[3]中证明了非齐次半线性椭圆方程多个正解的存在性, 其中有一个解是零附近的局部极少值解且能量小于零, 有两个能量大于零的正解是通过Ljusternik-Schnirelman簇数的方法得到的, 还有一个是高能量的正解. 万优艳在文献[4]中得到了问题(1.1)–(1.4)在零附近的局部极小值点, 该极小值点是问题(1.1)–(1.4)的一个能量小于零的正解.

受以上文献的启发, 我们希望得到问题(1.1)–(1.4)除了文献[4]中能量小于零的局部极小值解以外, 还有多解的存在性. 我们通过Ljusternik-Schnirelman簇数的方法得到问题(1.1)–(1.4)的两个能量大于零的正解. 我们的主要结果如下.

定理1.1  假设$ N\geq 3, \alpha \geq 2, \beta\geq 2, \alpha +\beta =2^*=\frac{2N}{N-2}, f(x)\geq 0, $$ g(x)\geq 0, $$ f(x), g(x)\in L^p( \Omega ), p>\frac{N}{2}, a(x)\geq 0, b(x)\geq 0 $且满足条件$ (A_1) $和$ (A_2), $那么对于充分小的$ \| (f, g)\|_{H^{-1}( \Omega )\times H^{-1} ( \Omega )}, $问题(1.1)–(1.4)存在两个解.

记$ \| (u, v)\| _E=(\int _{ \Omega }|\nabla u |^2+|\nabla v| ^2{\rm d}x)^{\frac{1}{2}} $, 其中$ E=H^1_0( \Omega )\times H^1_0( \Omega ). $对给定的$ a(x), b(x), f(x) $和$ g(x), $我们定义泛函$ I_{f, g} (u, v):E\rightarrow {{\Bbb R}} , $有

$ \begin{eqnarray*} I_{f, g} (u, v)&=&\frac{1}{2}\int _{ \Omega }(|\nabla u |^ 2+| \nabla v|^ 2+a(x)u^2+b(x)v^2){\rm d}x \\ &&-\frac{2}{\alpha +\beta }\int _{ \Omega }u_+^ \alpha v_+ \beta {\rm d}x-\int _{ \Omega }(fu+gv){\rm d}x, \end{eqnarray*} $

$ J_{f, g} (u, v)= \max\limits_{t>0 }I_{f, g} (tu, tv): \Sigma _+\rightarrow {{\Bbb R}} , $

其中$ \Sigma : =\{( u, v ) \in E ; \| (u, v)\| _E=1\}, \Sigma _+:=\{(u, v)\in \Sigma ;u_+\equiv 0, v_+\equiv 0\}. $

为了得到方程组的两个正解, 我们将借助Ljusternik-Schnirelman簇数的知识, 并应用以下两个引理的结论.

引理1.2^{[5, 6]}  设$ M $是Hilbert流形, $ \Psi \in C^1(M, {{\Bbb R}} ), $假设对$ c_0\in {{\Bbb R}} $和$ k\in \mathbb{N}, $有

(1) 任给$ c \leq c_0, \Psi (x) $满足$ (PS) _c $条件,

(2) $ cat(\{x\in M; \Psi (x) \leq c_0\})\geq k. $

则$ \Psi (x) $在$ \{x\in M; \Psi (x)\leq c_0\} $中至少有$ k $个临界点.

引理1.3^[7]  设$ N\geq 1, X $是拓扑空间. 假设存在两个连续映射

$ F:S^{N-1} =\{y\in {{\Bbb R}} ^N;| y| =1\}\rightarrow X, G:X\rightarrow S^{N-1 } $

使得$ G\circ F $和恒等映射: $ S^{N-1} \rightarrow S^{N-1} ;x\mapsto x $同伦, 即存在连续映$ \zeta :[0, 1]\times S^{N-1} \rightarrow S^{N-1} $, 使得

$ \zeta (0, x)=(G\circ F)(x) \qquad \forall A * P x\in S^{N-1}, $

$ \zeta (1, x)=x \qquad \forall A * P x\in S^{N-1}, $

则

$ cat(X)\geq 2 . $

首先, 我们得到一个全局紧的结论: 存在正偶数$ K\in \mathbb{N}, $当且仅当$ c=I_{f, g} (u_0(x), v_0(x))+\frac{k}{N}\frac{S_{\alpha, \beta } }{2})^{\frac{N}{2}}, k=2, 4.\cdots, K $时, $ I_{f, g }(u, v) $和$ J_{f, g} (u, v) $的$ (PS) _c $条件被破坏, 其中$ (u_0(x), v_0(x)) $是$ I_{f, g }(u, v) $的一个临界点.于是, $ \forall \varepsilon >0, $对于充分小的$ \| (f, g)\| _{E^*}, $当$ c<I_{f, g} (u_{\rm loc} (x), v_{\rm loc} (x)) $$ +\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2}} $时, $ I_{f, g} (u, v) $和$ J_{f, g }(u, v) $满足$ (PS) _c $条件, 其中$ (u_{\rm loc} (x), v_{\rm loc} (x)) $是问题(1.1)–(1.4)在零点附近的局部极小解^[4]. 其中, 记$ [J_{f, g}\leq c]=\{(u, v)\in \Sigma _+;J_{f, g} (u, v) \leq c\}. $由引理1.3, 我们得到: 对于充分小的$ \varepsilon >0, [J_{f, g} \leq I_{f, g} (u_{\rm loc} (x), v_{\rm loc} (x)) +\frac{2}{N}\frac{S_{\alpha, \beta } }{2})^{ \frac{N}{2} }-\varepsilon ] $是非空的, 并且

(1.5) $ \begin{align} cat\bigg(\bigg[J_{f, g} \leq I_{f, g} (u_{\rm loc} (x), v_{\rm loc} (x))+\frac{2}{N}(\frac{S_{\alpha, \beta } }{2}) ^{ \frac{N}{2} }-\varepsilon\bigg ]\bigg)\geq 2, \end{align} $

其中$ cat $表示Ljusternik-Schnirelman簇数. 又由引理1.2, 我们得到了方程组的两个正解$ (u_i, v_i), i=1, 2 $的存在性, 并且有$ 0<I_{f, g} (u_i, v_i)<I_{f, g} (u_{\rm loc} (x), v_{\rm loc} (x))+\frac{2}{N}(\frac{S_{\alpha, \beta } }{2})^{ \frac{N}{2}}, i=1, 2. $为了证明(1.5)式成立, 受到文献[1, 3]和[4]的启发, 我们构造映射

$ F:S^{N-1} =\{y\in {{\Bbb R}} ^N;| y| =1\}\rightarrow \bigg [J_{f, g} \leq I_{f, g} (u_{\rm loc} (x), v_{\rm loc} (x))+ \frac{2}{N}(\frac{S_{\alpha, \beta } }{2}) ^{\frac{N}{2} }-\varepsilon \bigg], $

$ G:\bigg[J_{f, g} \leq I_{f, g} (u_{\rm loc} (x), v_{\rm loc} (x))+\frac{2}{N}(\frac{S_{\alpha, \beta } }{2}) ^{\frac{N}{2}} -\varepsilon \bigg]\rightarrow S^{N-1}, $

使得$ F\circ G $与恒等映射同伦. 当$ f(x)\equiv 0, g(x)\equiv 0 $时, $ u_{\rm loc} \equiv 0, v_{\rm loc} \equiv 0 $并且

$ \bigg[J_{0, 0}\leq I_{0, 0} (u_{\rm loc} (x), v_{\rm loc} (x))+\frac{2}{N} (\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2}} \bigg]=\emptyset, $

所以(1.5)式不成立.

本文的第二部分我们将建立文章的基本框架、讨论能量泛函的性质、研究全局紧的结论、得到$ (PS) _c $条件; 第三部分将构建$ F\circ G $与恒等映射同伦; 第四部分将证明两个正解的存在性.

由于在讨论能量泛函的性质时, 我们需要用到能量泛函的二阶导数, 因此本文的结论只对$ N=3, 4 $成立.

记E为Sobolev空间$ H^1_0( \Omega )\times H^1_0( \Omega ) $且这个空间的范数为

$ \| (u, v)\| _E=\bigg(\int _{ \Omega }(|\nabla u |^ 2+| \nabla v| ^ 2){\rm d}x\bigg)^{ \frac{1}{2} }. $

记$ E^* $为$ E $的对偶空间, 即$ E^*=H^{-1} ( \Omega )\times H^{-1} ( \Omega ). $

令

$ S_{\alpha, \beta} ( \Omega ):= \inf\limits_{(u, v)\in E\backslash \{0\} } \frac{\int _{ \Omega }(|\nabla u |^ 2+| \nabla v|^ 2){\rm d}x} {(\int _{ \Omega }| u| ^ \alpha | v|^ \beta {\rm d}x)^{\frac{2}{\alpha +\beta } }}, $

$ S_{\alpha +\beta} ( \Omega ):= \inf\limits_{u\in E\backslash \{0\} } \frac{\int _{ \Omega }|\nabla u |^ 2{\rm d}x} {(\int _{ \Omega }| u| ^ {\alpha +\beta} {\rm d}x)^{\frac{2}{\alpha +\beta } }}. $

由文献[1, 定理5], 我们得到

$ S_{\alpha, \beta} ( \Omega ):=\bigg[(\frac{\alpha }{\beta })^{ \frac{\beta }{\alpha +\beta }} +(\frac{\beta }{\alpha })^{ \frac{\alpha }{\alpha +\beta } }\bigg] S_{\alpha +\beta} ( \Omega ) . $

根据文献[8, 引理1.42], 我们得到$ U(x)=\frac{[N(N-2)] ^{\frac{N-2}{4} }}{[1+| x|^2]^{ \frac{N-2}{2} } } $是$ S_{\alpha +\beta} $在$ {{\Bbb R}} ^N $中的达到函数. 定义

$ U_{\varepsilon, l }(x)=\varepsilon^{ \frac{2-N}{2}} U(\frac{x-l}{\varepsilon }), \quad \forall A * P \varepsilon>0, x\in {{\Bbb R}} ^N, l\in {{\Bbb R}} ^N, $

$ u_{\varepsilon, l }(x)=\xi _l(x)U_{\varepsilon, l }(x), \quad \forall A * P \varepsilon >0, x\in {{\Bbb R}} ^N, l\in \Omega _\eta, $

其中$ \Omega _\eta :=\{x\in \Omega, dist(x, \partial \Omega )>2\eta, \eta >0\}, \xi _l(x)=\xi (x-l), \xi (x)\in C^\infty_0({{\Bbb R}} ^N) $是一个非负函数, 同时$ \xi (x)\equiv 1, \forall A * P x\in B(0, \eta ) ; \xi (x)\equiv 0, A * P x\in B(0, 2\eta )^ c. $根据文献[8]中引理1.46, 我们得到

(2.1) $ \begin{equation} \int _{ \Omega }| \nabla u_\varepsilon |^2{\rm d}x=\int _{{{\Bbb R}} ^N }| \nabla U_\varepsilon |^2{\rm d}x+O(\varepsilon ^{N-2} )=S^{\frac{N}{2}} +O(\varepsilon^ {N-2} ), \end{equation} $

(2.2) $ \begin{equation} \int _{ \Omega }| u_\varepsilon |^2 {\rm d}x=\int _{{{\Bbb R}} ^N }| U_\varepsilon |^2 {\rm d}x+O(\varepsilon ^{N})=S^{\frac{N}{2}} +O(\varepsilon^{N}) . \end{equation} $

若$ f(x)\geq 0, g(x)\geq 0, f(x), g(x)\in H^{-1}( \Omega ) $成立, 且$ a(x), b(x) \geq0, a(x), b(x)\in L^\infty( \Omega ) $也成立, 那么我们定义泛函$ I_{f, g} (u, v):E\rightarrow {{\Bbb R}} , $有

(2.3) $ \begin{eqnarray} I_{f, g} (u, v)&=&\frac{1}{2}\int _{ \Omega }(|\nabla u |^2+|\nabla v|^2+a(x)u^2+b(x)v^2){\rm d}x \\ && -\frac{2}{\alpha +\beta }\int _{ \Omega }u_+^ \alpha v_+^ \beta {\rm d}x -\int _{ \Omega }(fu+gv){\rm d}x . \end{eqnarray} $

记$ I_{0, 0} (u, v):E\rightarrow {{\Bbb R}} , $有

$ I_{0, 0} (u, v)=\frac{1}{2}\int _{ \Omega }(|\nabla u |^2+|\nabla v|^2 +a(x)u^2+b(x)v^2){\rm d}x-\frac{2}{\alpha +\beta }\int _{ \Omega }u_+^ \alpha v_+^ \beta {\rm d}x . $

记$ \Sigma : =\{( u, v ) \in E ; \| (u, v)\| _E=1\}, \Sigma _+:=\{(u, v)\in \Sigma ;u_+\not\equiv 0, v_+\not\equiv 0\}. $定义

$ J_{f, g} (u, v)= \max\limits_{t>0 }I_{f, g} (tu, tv): \Sigma _+\rightarrow {{\Bbb R}} , J_{0, 0} (u, v)=\max\limits_{t>0 }I_{0, 0} (tu, tv): \Sigma _+\rightarrow {{\Bbb R}} . $

引理2.1  假设$ a(x) $和$ b(x) $是非负函数且满足条件$ (A_1), $那么$ \inf\limits_{(u, v)\in \Sigma _+ }J_{0, 0} (u, v)=\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2}} $是不可达的.

证  对于任意的$ (u, v)\in \Sigma _+, $有

$ \begin{eqnarray*} J_{0, 0} (u, v)&=& \max\limits_{t>0 }I_{0, 0} (tu, tv) \\ &=& \max\limits_{t>0 }\bigg\{\frac{t^2}{2}\int _{ \Omega }(|\nabla u|^2+|\nabla v|^2+a(x)u^2+b(x)v^2){\rm d}x-\frac{2t^{2^* }} {\alpha +\beta }\int _{ \Omega }u_+^ \alpha v_+ ^\beta {\rm d}x\bigg\} . \end{eqnarray*} $

若$ \frac{{\rm d}I_{0, 0} (tu, tv)}{{\rm d}t} =0, $我们就得到

$ t_0=\bigg(\frac{1+\int _{ \Omega }(a(x)u^2+b(x)v^2){\rm d}x} {2\int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x}\bigg)^{\frac{1}{2^*-1} }\ \mbox{或}\ t_0 =0. $

由于

$ I_{0, 0} (t_0u, t_0v)=\frac{1}{N}\frac{[1+\int _{ \Omega }(a(x)u^2+b(x)v^2){\rm d}x]^{ \frac{N}{2} }} {(2\int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x)^{ \frac{N-2}{2} }}>I_{0, 0} (0, 0) = 0, $

并且当$ t\rightarrow +\infty $时, 有$ I_{0, 0} (tu, tv)\rightarrow - \infty, $于是

(2.4) $ \begin{equation} J_{0, 0} (u, v)=I_{0, 0} (t_0u, t_0v)=\frac{1}{N} \frac{[1+\int _{ \Omega }(a(x)u^2+b(x)v^2){\rm d}x]^{ \frac{N}{2} }} {(2\int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x)^{ \frac{N-2}{2} } }. \end{equation} $

又由于$ a(x)\geq 0, b(x) \geq0, $并且根据Sobolev不等式, 我们得到

(2.5) $ \begin{eqnarray} J_{0, 0} (u, v)&\geq &\frac{1}{N} \bigg(2\int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x\bigg)^{ -\frac{N-2}{2} } \geq \frac{1}{N}\bigg(2\int _{ \Omega }| u|^ \alpha | v| ^ \beta {\rm d}x\bigg)^{ -\frac{N-2}{2}}\\ &\geq& \frac{1}{N}2^{-\frac{N-2}{2} }S^{\frac{N}{2}} _{\alpha, \beta } =\frac{2}{N}\bigg(\frac{S_{\alpha, \beta} }{2}\bigg)^{ \frac{N}{2}}, \end{eqnarray} $

即有$ \inf\limits_{(u, v)\in \Sigma _+ }J_{0, 0} (u, v) \geq\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2}}. $令$ (u, v)=\frac{(Bu_{\varepsilon, l }, Du_{\varepsilon, l} )} {\| (Bu_{\varepsilon, l}, Du_{\varepsilon, l })\| _E} $, 其中$ B/D=\sqrt{\alpha /\beta }, l\in \Omega $并且$ B(l, 2\eta )\subset \Omega . $由(2.1), (2.2)和(2.4)式我们可以得到

(2.6) $ \begin{eqnarray} J_{0, 0} (u, v)&=& \max\limits_{t>0 }I_{0, 0}\bigg (t\frac{(Bu_{\varepsilon, l}, Du_{\varepsilon, l} )} {\| (Bu_{\varepsilon, l}, Du_{\varepsilon, l} )\| _E}\bigg) \\ &=& \frac{1}{N}\frac{[1+\frac{1}{\| (Bu_{\varepsilon, l}, Du_{\varepsilon, l} )\|^2_E} \int _{ \Omega }(a(x)B^2u^2_{\varepsilon, l} +b(x)D^2u^2_{\varepsilon, l} ){\rm d}x]^{\frac{N}{2}}} {(\frac{2\int _{ \Omega }B^\alpha u^\alpha _{\varepsilon, l} D^\beta u^\beta _{\varepsilon, l} {\rm d}x}{\| (Bu_{\varepsilon, l}, Du_{\varepsilon, l} )\|^{ \alpha +\beta } _E})^{ \frac{N-2}{2} } } \\ &=&\frac{1}{N2^{\frac{N-2}{2}}} \bigg[1+\frac{1}{\| (Bu_{\varepsilon, l}, Du_{\varepsilon, l} )\|^2_E} \int _{ \Omega }(a(x)B^2u^2_{\varepsilon, l} +b(x)D^2u^2_{\varepsilon, l} ){\rm d}x\bigg]^{\frac{N}{2}} \\ &&\times \bigg[(\frac{\alpha }{\beta})^{\frac{\beta }{2^*}} +(\frac{\beta}{\alpha })^{\frac{\alpha }{2^*}}\bigg]^{\frac{N}{2} } (S^{\frac{N}{2}} _{\alpha +\beta} +O(\varepsilon^{ N-2} )) \\ &=&\frac{1}{N2^{\frac{N-2}{2} }} \bigg[1+\frac{1}{\| (Bu_{\varepsilon, l}, Du_{\varepsilon, l} )\|^2_E} \int _{ \Omega }(a(x)B^2u^2_{\varepsilon, l} +b(x)D^2u^2_{\varepsilon, l} ){\rm d}x\bigg]^{\frac{N}{2}}\\ &&\times ( S^{\frac{N}{2}} _{\alpha, \beta}+O(\varepsilon^{ N-2} )) . \end{eqnarray} $

由条件$ (A_1) $和文献[3, (2.11)式], 我们得到

(2.7) $ \begin{eqnarray} \int _{ \Omega }a(x)u^2_{\varepsilon, l} {\rm d}x\leq O(\varepsilon ^{\frac{N}{p'}-N+2 }), \int _{ \Omega }b(x)u^2_{\varepsilon, l} {\rm d}x\leq O(\varepsilon^{ \frac{N}{p'}-N+2} ), \end{eqnarray} $

其中$ 0<\frac{N}{p'}-N+2<N-2. $根据(2.6)式, 我们得到

(2.8) $ \begin{eqnarray} J_{0, 0} (u, v)&\leq &\bigg[\frac{1}{N2^{\frac{N-2}{2}}}S^{\frac{N}{2}} _{\alpha, \beta} +O(\varepsilon ^{N-2} )\bigg](1+O(\varepsilon^{ \frac{N}{p'}-N+2 }))\\ &=&\frac{1}{N2^{\frac{N-2}{2}} }S^{\frac{N}{2}} _{\alpha, \beta} +O(\varepsilon ^{\frac{N}{p'}-N+2 })\\ &=&\frac{2}{N}(\frac{S_{\alpha, \beta}}{2})^{\frac{N}{2} }+O(\varepsilon ^{\frac{N}{p'}-N+2} ), \end{eqnarray} $

即

$ \inf\limits_{(u, v)\in \Sigma _+ }J_{0, 0} (u, v)=\frac{2}{N}(\frac{S_{\alpha, \beta} }{2}) ^{\frac{N}{2} }. $

往证$ \inf\limits_{(u, v)\in \Sigma _+ }J_{0, 0} (u, v) $是不可达的, 用反证法证明. 假设$ (u, v)\in \Sigma _+ $是$ \inf\limits_{(u, v)\in \Sigma _+ }J_{0, 0} (u, v) $的可达点, 通过(2.4)式得到

$ J_{0, 0} (u, v)=\frac{1}{N}\frac{[1+\int _{ \Omega }(a(x)u^2+b(x)v^2{\rm d}x]^{\frac{N}{2}}} {(2\int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x)^{\frac{N-2}{2}} } =\frac{2}{N}(\frac{S_{\alpha, \beta} }{2}) ^{\frac{N}{2} } . $

又通过(2.5)式, $ a(x) $和$ b(x) $非负, 我们得到

$ \begin{eqnarray*} \frac{2}{N}\bigg(\frac{S_{\alpha, \beta} }{2}\bigg)^{ \frac{N}{2}} & \leq &\frac{1}{N}\bigg(2\int _{ \Omega }| u| ^ \alpha | v| ^\beta {\rm d}x\bigg)^{-\frac{N-2}{2}}\leq \frac{1}{N}\bigg(2\int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x\bigg)^{-\frac{N-2}{2} } \\ & \leq& \frac{1}{N}\frac{[1+\int _{ \Omega }(a(x)u^2+b(x)v^2{\rm d}x]^{ \frac{N}{2} }} {(2\int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x)^{\frac{N-2}{2}} } =\frac{2}{N}\bigg(\frac{S_{\alpha, \beta} }{2}\bigg) ^{\frac{N}{2} }, \end{eqnarray*} $

即有

$ S_{\alpha, \beta} =\frac{\| (u, v)\|^2_E}{(\int _{ \Omega }| u| ^ \alpha | v| ^\beta {\rm d}x)^{\frac{2}{\alpha +\beta } } }. $

因为$ S_{\alpha, \beta} $在有界域上不可达, 所以我们得到的结论与之矛盾, 故$ \inf\limits_{(u, v)\in \Sigma _+ }J_{0, 0} (u, v)=\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2} } $是不可达的.

引理2.2  假设$ (f(x), g(x))\in E^*, $并且$ a(x) $和$ b(x) $满足条件$ (A_1), $则

(ⅰ) $ I_{f, g} (u, v)\in C^2(E, {{\Bbb R}} ) $, 对于任意的$ (\varphi, \psi )\in E, $有

(2.9) $ \begin{eqnarray} I'_{f, g} (u.v)(\varphi, \psi )&=&\int _{ \Omega }\nabla u \nabla \varphi {\rm d}x +\int _{ \Omega } \nabla v \nabla \psi {\rm d}x +\int _{ \Omega }a(x)u\varphi {\rm d}x+\int _{ \Omega }b(x)v\psi {\rm d}x \\ &&-\frac{2\alpha }{\alpha +\beta }\int _{ \Omega }u_+ ^{\alpha -1 }v_+ ^\beta \varphi {\rm d}x -\frac{2\beta }{\alpha +\beta }\int _{ \Omega }u_+^ \alpha v_+^{ \beta -1} \psi {\rm d}x -\int _{ \Omega }(f\varphi +g\psi ){\rm d}x, {\qquad} \end{eqnarray} $

(2.10) $ \begin{eqnarray} I''_{f, g} (u, v)((\varphi, \psi ), (\varphi, \psi )) &=&\int _{ \Omega }| \nabla \varphi |^2{\rm d}x+\int _{ \Omega }| \nabla \psi |^2{\rm d}x +\int _{ \Omega }a(x)\varphi^ 2{\rm d}x+\int _{ \Omega }b(x)\psi^ 2{\rm d}x \\ &&-\frac{2\alpha (\alpha -1)}{\alpha +\beta }\int _{ \Omega }u_+ ^{\alpha -2 } v_+ ^\beta \varphi^ 2{\rm d}x-\frac{2\beta (\beta -1)}{\alpha +\beta } \int _{ \Omega }u_+^ \alpha v_+ ^{\beta -2 }\psi^ 2{\rm d}x . \qquad \end{eqnarray} $

(ⅱ) 若$ f(x), g(x), a(x), b(x) $都是非负的, $ (u, v)\in E $是$ I_{f, g} (u, v) $的临界点, 则$ (u(x), v(x)) $是方程组(1.1)–(1.4)的非负解. 若$ u(x)\equiv 0, v(x)\equiv 0 $或$ f(x)\equiv 0, g(x)\equiv 0, $则$ (u(x), v(x)) $是方程组(1.1)–(1.4)的一个正解.

证  (ⅰ) 证明方法见文献[8].

(ⅱ) 假设$ (u, v)\in E $满足$ I'_{f, g} (u, v)=0, $由(2.9)式可知: 对于任意$ (\varphi, \psi )\in E, $有

$ \begin{eqnarray*} &&\int _{ \Omega }\nabla u \nabla \varphi {\rm d}x+\int _{ \Omega } \nabla v \nabla \psi {\rm d}x +\int _{ \Omega }a(x)u\varphi {\rm d}x+\int _{ \Omega }b(x)v\psi {\rm d}x\\ && -\frac{2\alpha }{\alpha +\beta }\int _{ \Omega }u_+^{ \alpha -1} v_+ ^\beta \varphi {\rm d}x -\frac{2\beta }{\alpha +\beta }\int _{ \Omega }u_+ ^\alpha v_+ ^{\beta -1} \psi {\rm d}x -\int _{ \Omega }(f\varphi +g\psi ){\rm d}x=0 . \end{eqnarray*} $

所以$ (u, v) $是

$ \left\{\begin{array}{ll} { } -\Delta u+a(x)u=\frac{2\alpha }{\alpha +\beta }u^{\alpha -1} _+v^\beta _++f(x), \\ { } -\Delta v+b(x)v=\frac{2\beta }{\alpha +\beta }u^\alpha _+v^{\beta -1} _++g(x), \end{array}\right.\quad x\in \Omega $

的一个弱解. 因为$ f(x)\geq 0 $和$ g(x)\geq 0, $上述两个方程两边分别同时乘以$ u_- $和$ v_- $, 然后再积分可以得到

$ \int _{ \Omega }[| \nabla u_ -|^2+a(x)u^2_-]{\rm d}x=\int _{ \Omega }f(x)u_- \leq 0, $

$ \int _{ \Omega }[| \nabla v_-|^2+b(x)v^2_-]{\rm d}x=\int _{ \Omega }g(x)v_- \leq 0 . $

由于$ a(x)\geq 0 $和$ b(x)\geq 0 $, 可以得到

$ u_-=0 \quad {\rm a.e.}\quad \Omega, \quad v_-=0 \quad {\rm a.e.}\quad \Omega. $

所以$ u \geq 0, v \geq 0 $. 如果$ u(x)\not\equiv 0 $和$ v(x)\not\equiv 0 $或者$ f(x)\not\equiv 0 $和$ g(x)\not\equiv 0, $我们通过条件$ (A_1) $和强极值原理得出$ (u(x), v(x)) $在空间$ \Omega $上是正的.

引理2.3  假设$ (f(x), g(x))\in E^*, a(x)\geq 0 $和$ b(x)\geq 0 $且满足条件$ (A_1) $, 有

(ⅰ) $ \forall (u, v)\in \Sigma _+, \forall A * P \varepsilon \in (0, 1), $有

(2.11) $ \begin{eqnarray} (1-\varepsilon )^{ \frac{N}{2} }J_{0, 0} (u, v)-\frac{1}{2\varepsilon }\| (f, g)\|^2_{E^2} \leq J_{f, g} (u, v)\leq (1+\varepsilon )^{ \frac{N}{2} }J_{0, 0} (u, v)+\frac{1}{2\varepsilon }\| (f, g)\|^2_{E^2 } . \end{eqnarray} $

(ⅱ) 存在$ d_0>0, $当$ \| (f, g)\| _{E^*}\leq d_0 $时, 有

$ \inf\limits_{(u, v)\in \Sigma _+ }J_{f, g} (u, v)>0 . $

证  (ⅰ) 因为$ \forall A * P (u, v)\in \Sigma _+ $和$ \forall A * P \varepsilon \in (0, 1), $有

$ \bigg | \int _{ \Omega }(fu+gv){\rm d}x\bigg|\leq \| (f, g)\| _{E^*} \| (u, v)\| _E \leq\frac{\varepsilon }{2}\| (u, v)\|^2_E+\frac{1}{2\varepsilon }\| (f, g)\|^2_{E^*} . $

于是

$ \begin{eqnarray*} &&\frac{1-\varepsilon }{2}\| (u, v)\|^2_E+\frac{1-\varepsilon }{2}\int _{ \Omega } (a(x)u^2+b(x)v^2){\rm d}x-\frac{2}{\alpha +\beta }\int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x-\frac{1}{2\varepsilon } \| (f, g)\|^2_{E^*}\\ &\leq& I_{f, g} (u, v)\\ &\leq& \frac{1+\varepsilon }{2}\| (u, v)\|^2_E+\frac{1+\varepsilon }{2}\int _{ \Omega } (a(x)u^2+b(x)v^2){\rm d}x-\frac{2}{\alpha +\beta } \int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x+\frac{1}{2\varepsilon }\| (f, g)\|^2_{E^*} . \end{eqnarray*} $

因为$ a(x)\geq 0 $和$ b(x) \geq0, $所以$ \forall A * P (u, v)\in \Sigma _+, $有

$ \begin{eqnarray*} &&(1-\varepsilon ) \max\limits_{t>0 }\bigg\{\frac{1}{2}\| (tu, tv)\|^2_E+\frac{1}{2}\int _{ \Omega } [a(x)(tu)^2+b(x)(tv)^2]{\rm d}x-\frac{2}{\alpha +\beta }\int _{ \Omega } \frac{t^{\alpha +\beta }u^\alpha _+v^\beta _+}{1-\varepsilon }{\rm d}x\bigg\}\\ && -\frac{1}{2\varepsilon }\| (f, g)\|^2_{E^*}\\ &\leq &\max\limits_{t>0 }I_{f, g} (tu, tv) \\ &\leq& (1+\varepsilon ) \max\limits_{t>0 }\bigg\{\frac{1}{2}\| (tu, tv)\|^2_E+\frac{1}{2}\int _{ \Omega } [a(x)(tu)^2+b(x)(tv)^2]{\rm d}x-\frac{2}{\alpha +\beta }\int _{ \Omega } \frac{t^{\alpha +\beta} u^\alpha _+v^\beta _+}{1+\varepsilon }{\rm d}x\bigg\}\\ && +\frac{1}{2\varepsilon }\| (f, g)\|^2_{E^*} . \end{eqnarray*} $

令$ S_1=\frac{t}{(1-\varepsilon ) ^{\frac{1}{2^*-2}}}, S_2=\frac{t}{(1+\varepsilon )^{ \frac{1}{2^*-2} }}, $则有

$ \begin{eqnarray*} &&(1-\varepsilon )^{\frac{2^*}{2^*-2}}\max\limits_{s_1>0 } \bigg\{\frac{S^2_1}{2}\| (u, v)\|^2_E+\frac{S^2_1}{2}\int _{ \Omega } [a(x)u^2+b(x)v^2]{\rm d}x-\frac{S^{2^*} _1}{2^*} \int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x\bigg\}\\ && -\frac{1}{2\varepsilon }\| (f, g)\|^2_{E^*}\\ &\leq&\max\limits_{t>0 }I_{f, g} (tu, tv)\\ &\leq& (1+\varepsilon )^{\frac{2^*}{2^*-2}}\max\limits_{s_2>0 } \bigg\{\frac{S^2_2}{2}\| (u, v)\|^2_E+\frac{S^2_2}{2}\int _{ \Omega } [a(x)u^2+b(x)v^2]{\rm d}x-\frac{S^{2^*} _2}{2^*}\int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x\bigg\}\\ && +\frac{1}{2\varepsilon }\| (f, g)\|^2_{E^*} . \end{eqnarray*} $

所以(2.11)式成立.

(ⅱ)通过(2.11)式和引理2.1, 我们知道

$ \begin{eqnarray*} \inf\limits_{(u, v)\in \Sigma _+ }J_{f, g }(u, v) &\geq& (1-\varepsilon )^{ \frac{N}{2}} J_{0, 0} (u, v)-\frac{1}{2\varepsilon }\| (f, g)\|^2_{E^*}\\ &\geq& (1-\varepsilon )^{ \frac{N}{2}}\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2}} -\frac{1}{2\varepsilon }\| (f, g)\|^2_{E^*}. \end{eqnarray*} $

所以, 存在$ d_0>0, $当$ \| (f, g)\| _{E^*}\leq d_0 $时, 有$ \inf\limits_{(u, v)\in \Sigma _+ }J_{f, g} (u, v)>0. $

引理2.4  假设$ (f(x), g(x))\in E^* $, 非负函数$ a(x), b(x) $满足条件$ (A_1), $则

(ⅰ) 对于任意$ (u, v)\in \Sigma _+, $函数$ t\mapsto I_{f, g} (tu, tv) $在$ [0, \infty) $上至多存在两个临界点.

(ⅱ) 如果$ \| (f, g)\| _{E^*}\leq d_0 $$ ( d_0 $由引理2.3给出), 那么对于任意$ (u, v)\in \Sigma _+, $存在唯一的$ t_{f, g} (u, v)>0, $使得

$ I_{f, g} (t_{f, g} (u, v)u, t_{f, g} (u, v)v)=J_{f, g} (u, v) . $

并且

(2.12) $ \begin{equation} t_{f, g} (u, v)>\bigg(\frac{1+\int _{ \Omega }(a(x)u^2+b(x)v^2){\rm d}x} {2(2^*-1)\int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x}\bigg)^{ \frac{1}{2^*-2}} \geq [2(2^*-1)S^{-\frac{2^*}{2}} _{\alpha, \beta} ]^{ -\frac{1}{2^*-2}} >0, \end{equation} $

(2.13) $ \begin{equation} I''_{f, g} (t_{f, g} (u, v)(u, v))((u, v), (u, v))<0 . \end{equation} $

(ⅲ) 如果$ t\mapsto I_{f, g} (tu, tv) $有一个异于$ t_{f, g} (u, v) $的临界点$ \bar{ t} $, 那么

$ \bar{t}\in [0, (1-\frac{1}{2^*-1})^{ -1} \| (f, g)\| _{E^*} ]. $

证  $ \forall (u, v)\in \Sigma _+, $令

(2.14) $ \begin{eqnarray} g(t)&=&I_{f, g} (tu, tv){}\\ &=&\frac{t^2}{2}+\frac{t^2}{2}\int _{ \Omega } [a(x)u^2+b(x)v^2]{\rm d}x-\frac{2t^{2^*} }{\alpha +\beta }\int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x -t\int _{ \Omega }(fu+gv){\rm d}x. \end{eqnarray} $

(ⅰ) 注意到

(2.15) $ \begin{equation} g'(t)=t+t\int _{ \Omega }[a(x)u^2+b(x)v^2]{\rm d}x-2t^{2^*-1} \int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x-\int _{ \Omega }(fu+gv){\rm d}x, \end{equation} $

(2.16) $ \begin{equation} g''(t)=1+\int _{ \Omega }[a(x)u^2+b(x)v^2]{\rm d}x-2(2^*-1)t^{2^*-2} \int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x, \end{equation} $

所以

(2.17) $ \begin{eqnarray} \left\{\begin{array}{ll} { } g''(t)>0, t<\bigg(\frac{1+\int _{ \Omega }[a(x)u^2+b(x)v^2]{\rm d}x}{2(2^*-1)\int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x}\bigg) ^{\frac{1}{2^*-2}}, \\ { } g''(t)<0, t>\bigg(\frac{1+\int _{ \Omega }[a(x)u^2+b(x)v^2]{\rm d}x}{2(2^*-1)\int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x}\bigg)^{\frac{1}{2^*-2}}. \end{array}\right. \end{eqnarray} $

令$ t_0=(\frac{1+\int _{ \Omega }[a(x)u^2+b(x)v^2]{\rm d}x}{2(2^*-1)\int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x})^{\frac{1}{2^*-2} } $, 那么$ g'(t) $至多有两个零点$ t_1 $和$ t_2 $, 并且

(2.18) $ \begin{eqnarray} 0\leq t_1 \leq t_0\leq t_2 . \end{eqnarray} $

(ⅱ) 已知$ g(0)=0 $且$ t\rightarrow +\infty $时$ g(t)\rightarrow -\infty $. 所以$ \sup\limits_{t>0 }g(t) $是有限的, 通过引理2.3, 我们知道存在$ d_0>0, $当$ \| (f, g)\| _{E^*}\leq d_0, $有$ \inf\limits_{(u, v)\in \Sigma _+ }J_{f, g} (u, v)>0, $即$ \inf\limits_{(u, v)\in \Sigma _+ } \sup\limits_{t>0 }I_{f, g} (tu, tv)>0. $从而$ \sup\limits_{t>0 }g(t)>0. $所以, 存在$ t_{f, g} (u, v)\geq t_0 $使得

$ I_{f, g} (t_{f, g} (u, v)u, t_{f, g} (u, v)v)=J_{f, g} (u, v), $

其中$ t_{f, g} (u, v)\not= t_0, $否则就有$ g'(t_0)=0. $由(2.17)式可知, $ t_0 $是$ g'(t) $的最大值点, 所以对于任意$ t>0, $有$ g'(t)\leq 0, $又由于$ g(0)=0, $于是对于任意$ t>0, $有$ g(t)\leq 0 $. 这与$ \sup\limits_{t>0 }g(t)>0 $相矛盾. 所以(2.12)式成立. 由(2.10)式可知

(2.19) $ \begin{eqnarray} &&I''_{f, g} (tu, tv)((u, v), (u, v))\\ &=&\int _{ \Omega }|\nabla u|^2{\rm d}x+\int _{ \Omega }|\nabla v|^2{\rm d}x +\int _{ \Omega }a(x)u^2{\rm d}x+\int _{ \Omega }b(x)v^2{\rm d}x \\ &&-\frac{2\alpha (\alpha -1)t^{2^*-2 }}{\alpha +\beta }\int _{ \Omega }u_+ ^\alpha v_+ ^\beta {\rm d}x -\frac{2\beta (\beta -1)t^{2^*-2 }}{\alpha +\beta }\int _{ \Omega }u_+ ^\alpha v_+ ^\beta {\rm d}x \\ &=&1+\int _{ \Omega }[a(x)u^2+b(x)v^2]{\rm d}x-2(2^*-1)t^{2^*-2 }\int _{ \Omega }u_+ ^\alpha v_+^ \beta {\rm d}x . \end{eqnarray} $

由(2.16)和(2.19)式可知

(2.20) $ \begin{eqnarray} g''(t)=I''_{f, g} (tu, tv)((u, v), (u, v)) . \end{eqnarray} $

根据(2.12), (2.17)和(2.20)式, 我们可以得出(2.13)式成立. 由(2.12)和(2.18)式可以得出$ t_{f, g} (u, v) $的唯一性.

(ⅲ) 假设$ g(t) $有一个不同于$ t_{f, g} (u, v) $的临界点$ \bar{ t} $, 由(2.12)和(2.18)式可得, $ t_{f, g} (u, v)>t_0\geq\bar{t} $, 即

(2.21) $ \begin{eqnarray} \bar{t} \leq \bigg(\frac{1+\int _{ \Omega }a(x)u^2+b(x)v^2{\rm d}x}{2(2^*-1)\int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x}\bigg)^{ \frac{1}{2^*-2} }. \end{eqnarray} $

因为$ g'(\bar{t})=0, $所以

$ \bar{t}+\bar{t}\int _{ \Omega }[a(x)u^2+b(x)v^2]{\rm d}x-2\bar{t}^{ 2^*-1 } \int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x-\int _{ \Omega }(fu+gv){\rm d}x=0, $

即

$ \begin{eqnarray*} \bar{t} &=& \frac{\int _{ \Omega }(fu+gv){\rm d}x} {1 + \int _{ \Omega }[a(x)u^2+b(x)v^2]{\rm d}x-2\bar{t}^{ 2^*-2 }\int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x} \\ & \leq& \frac{\| (f, g)\| _{E^*} }{1 + \int _{ \Omega }[a(x)u^2+b(x)v^2]{\rm d}x -\frac{2(1 + \int _{ \Omega }[a(x)u^2+b(x)v^2]{\rm d}x) \int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x}{[2(2^*-1)\int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x]}} \\ & =&\frac{\| (f, g)\| _{E^*} }{(1-\frac{1}{2^*-1})(1 + \int _{ \Omega }[a(x)u^2+b(x)v^2]{\rm d}x)} \\ & \leq& (1-\frac{1}{2^*-1})^{-1} \| (f, g)\| _{E^*} . \end{eqnarray*} $

引理2.4证毕.

引理2.5  假设$ (f(x), g(x))\in E^* $, 并且非负函数$ a(x) $和$ b(x) $满足条件$ (A_1), $则存在$ r_1>0 $和$ d_1\in (0, d_0], $使得

(ⅰ) $ I_{f, g} (u, v) $在$ B(r_1)=\{(u, v)=E;\| (u, v)\| _E<r_1\} $上是严格凸的.

(ⅱ) 如果$ \| (f, g)\| _{E^*}\leq d_1, $那么$ \inf\limits_{\| (u, v)\| _E=r_1 }I_{f, g} (u, v)>0. $此外, $ I_{f, g} (u, v) $在$ B(r_1) $上有唯一临界点$ u_{\rm loc} (f, g;x) $且满足

$ (u_{{\rm loc}} ((f, g);x), v_{\rm loc} ((f, g);x))\in B(r_1), $

$ I_{f, g} (u_{\rm loc} ((f, g);x), v_{\rm loc} ((f, g);x))= \inf _{(u, v)\in B(r_1) }I_{f, g} (u, v) . $

证明见文献[4, 引理2].

注2.6  $ (u_{{\rm loc}} (f, g;x), v_{{\rm loc}} (f, g;x)) $是$ I_{f, g} (u, v) $在$ B(r_1) $上的唯一临界点, 则

$ (u_{{\rm loc}} (0, 0;x), v_{{\rm loc}} (0, 0;x))=(0, 0). $

引理2.7  当$ \| (f, g)\| _{E^*} \rightarrow 0 $时, 有$ \| u_{{\rm loc}} (f, g;x)\| _E\rightarrow 0. $

证明见文献[4, 引理3].

引理2.8  令$ d_2= \min \{d_1, (1-\frac{1}{2^*-1})r_1\}>0, $如果$ \| (f, g)\| _{E^*} \leq d_2, $非负函数$ a(x) $和$ b(x) $满足条件$ (A_1) $, 那么

(ⅰ) $ J_{f, g} (u, v)\in C^1( \Sigma _+, {{\Bbb R}} ), $对于任意$ h\in T_{(u, v) } \Sigma _+=\{(\varphi, \psi )\in E;\langle (u, v), (\varphi, \psi )\rangle_E=0\}, $有

(2.22) $ \begin{eqnarray} J'_{f, g} (u, v)h=t_{f, g} (u, v)I'_{f, g} (t_{f, g} (u, v)(u, v))h . \end{eqnarray} $

(ⅱ) $ (u, v)\in \Sigma _+ $是$ J_{f, g} (u, v) $的临界点的充分必要条件是$ t_{f, g} (u, v)(u, v)\in E $是$ I_{f, g} (u, v) $的临界点.

(ⅲ) $ I_{f, g} (u, v) $的临界点集是

(2.23) $ \begin{eqnarray} \{t_{f, g} (u, v)(u, v);(u, v)\in \Sigma _+, J'_{f, g} (u, v)=0\}\cup \{(u_{{\rm loc}} (f, g;x), v_{{\rm loc}} (f, g;x))\} . \end{eqnarray} $

证  (ⅰ) 由(2.9)式, 我们知道: 对任意$ (u, v)\in \Sigma _+, $有

$ I'_{f, g} (tu, tv)(u, v)=t+t\int _{ \Omega }(a(x)u^2+b(x)v^2){\rm d}x-2t^{2^*-1 } \int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x-\int _{ \Omega }(f\varphi +g\psi ){\rm d}x, $

$ I'_{f, g} (tu, tv)(u, v): \Sigma _+\times (0, +\infty)\rightarrow {{\Bbb R}} . $

假设$ Q\in \Sigma _+\times (0, +\infty) $是开子集, $ ((u, v), t_{f, g} (u, v))\in Q, $根据引理2.2可知$ I'_{f, g} (tu, tv)(u, v):Q\rightarrow {{\Bbb R}} $是连续的, 同时有

(1) 根据引理2.4的(ⅱ)可知: 存在唯一的$ t_{f, g} (u, v)>0, $使得

$ J_{f, g} (u, v)=I_{f, g} (t_{f, g} (u, v)(u, v)), I'_{f, g} (t_{f, g} (u, v)(u, v))(u, v)=0 ; $

(2) 由引理2.2, (2.14)和(2.20)式可知: 对于任意$ ((u, v), t)\in Q, $有

$ g''(t)=I''_{f, g} ((tu, tv))((u, v), (u, v)) $

是连续的. 通过(2.13)式可知

$ g''(t_{f, g} (u, v))=I''_{f, g} (t_{f, g} (u, v)(u, v))((u, v), (u, v))<0. $

根据隐函数定理可知: 存在$ (u, v) $的邻域$ U $、$ t_{f, g} (u, v) $的邻域$ W $和唯一的连续映射$ z :U\rightarrow W $满足

$ z (u, v)=t_{f, g} (u, v), $

$ I'_{f, g} (z (u, v)(u, v))(u, v)=0, \quad \forall (u, v)\in U, U\times W\subset Q, $

同时$ z (u, v)=t_{f, g} (u, v)\in C^1( \Sigma _+, (0, +\infty)). $从而

$ J_{f, g} (u, v)=I_{f, g} (t_{f, g} (u, v)(u, v))\in C^1( \Sigma _+, {{\Bbb R}} ) . $

由于

(2.24) $ \begin{equation} I'_{f, g} (t_{f, g} (u, v)(u, v))(u, v)=0, \end{equation} $

于是对于任意$ h\in T_{(u, v)} \Sigma _+=\{(\varphi, \psi )\in E, \langle (\varphi, \psi ), (u, v)\rangle_E=0\}, $有

$ \begin{eqnarray*} J'_{f, g} (u, v)h &=&I'_{f, g} (t_{f, g} (u, v)(u, v))(t_{f, g} (u, v)h +(t'_{f, g} (u, v), h) (u, v)) \\ &=&t_{f, g} (u, v)I'_{f, g} (t_{f, g} (u, v)(u, v))h+(t'_{f, g} (u, v), h)I'_{f, g} (t_{f, g} (u, v)(u, v))(u, v) \\ &=&t_{f, g} (u, v)I'_{f, g} (t_{f, g} (u, v)(u, v))h . \end{eqnarray*} $

(ⅱ) 根据(2.12)式可知$ t_{f, g} (u, v)>0, $又通过(2.22)式可知$ J'_{f, g} (u, v)=0 $成立的充分必要条件是对于任意$ (\varphi, \psi )\in T_{(u, v) } \Sigma _+, $有$ I'_{f, g} (t_{f, g} (u, v)(u, v))(\varphi, \psi )=0. $再结合(2.24)式可知上式等价于$ I'_{f, g} (t_{f, g} (u, v)(u, v))=0. $

(ⅲ) 如果$ (\tilde{u}, \tilde{v})\in E $是$ I_{f, g} (u, v) $的临界点. 令$ (\tilde{u}, \tilde{v})=(tu, tv), $其中$ (u, v)\in \Sigma _+, t>0. $由引理2.4的(ⅲ)可知

$ t=t_{f, g} (u, v)\ \mbox{ 或者}\ t (1-\frac{1}{2^*-1})^{-1} \| (f, g)\| _{E^*} . $

那么$ (\tilde{u}, \tilde{v})\in E $是$ J_{f, g} (u, v) $的临界点或$ \| (u, v)\| _E (1-\frac{1}{2^*-1})^{-1} d_2 \leq r_1. $由引理2.5可知$ (u_{{\rm loc}} (f, g;x), v_{{\rm loc}} (f, g;x)) $是$ I_{f, g} (u, v) $在$ B(r_1) $上的唯一临界点.

引理2.9  假设$ (f(x), g(x))\in E^*, a(x)\geq 0 $和$ b(x) \geq 0 $满足条件$ (A_1), $$ (u_m, v_m)\in E $是$ I_{f, g} $的$ (PS) _c $序列, 则存在实数$ k\in N $, 实数序列$ \{\varepsilon^ j_m\}\subset R, $点列$ \{x^j_m\}\subset \Omega, $其中$ \varepsilon^ j_m\rightarrow 0 (m\rightarrow \infty), dist(x^j_m, \partial \Omega )/\varepsilon ^j_m\rightarrow \infty, 1\leq j\leq k $, 存在方程组(1.1), (1.2)和(1.4)的一个非负解$ (u^0, v^0)\in E $和方程组

$ \left\{\begin{array}{ll} { } -\Delta u=\frac{2\alpha }{\alpha +\beta }u^{\alpha -1} _+v^\beta _+, \\ { } -\Delta v=\frac{2\beta }{\alpha +\beta }u^\alpha _+v^{\beta -1} _+, \end{array}\right. \quad x\in {{\Bbb R}} ^N $

的非平凡解$ (u^j, v^j)\in D^{1, 2} ({{\Bbb R}} ^N)\times D^{1, 2} ({{\Bbb R}} ^N), 1\leq j\leq k, $使$ (u_m, v_m) $的子序列(仍记为$ (u_m, v_m) ) $满足: 当$ m\rightarrow \infty $时, 有

$ \bigg \| u_m-u^0-\sum\limits^k_{j=1}u^j_m\bigg\| _{D^{1, 2} ({{\Bbb R}} ^N) }\rightarrow 0, \quad\; \; \bigg \| v_m-v^0-\sum\limits^k_{j=1}v^j_m\bigg\| _{D^{1, 2} ({{\Bbb R}} ^N)} \rightarrow 0, $

其中$ (u^j_m, v^j_m)=(\varepsilon^ j_m)^{ -\frac{N-2}{2}} (u^j(\frac{x-x^j_m}{\varepsilon^ j_m}), v^j(\frac{x-x^j_m}{\varepsilon^ j_m})), 1 \leq j \leq k, m\in N. $此外

$ I_{f, g} (u_m, v_m )\rightarrow I_{f, g} (u^0, v^0)+\sum\limits^k_{j=1}I^\infty(u^j, v^j), $

其中$ I^\infty(u, v)=\frac{1}{2}\int _{{{\Bbb R}} ^N }(|\nabla u|^2+|\nabla v|^2){\rm d}x-\frac{2}{\alpha +\beta }\int _{{{\Bbb R}} ^N }u^\alpha _+v^\beta _+{\rm d}x, u, v\in D^{1, 2} ({{\Bbb R}} ^N) $.

证  证明方法见文献[9, 引理3.1]和[2, 引理2.2].

引理2.10  如果$ (f(x), g(x))\in E^*, \| (f, g)\| _{E^*} \leq d_2 $$ (d_2>0 $由引理2.8给出), 非负函数$ a(x) $和$ b(x) $满足条件$ (A_1), $那么

(ⅰ) 当$ dist_E((u, v), \partial \Sigma _+)\equiv \inf \{\| (\varphi, \psi )-(u, v)\| _E; (\varphi, \psi )\in \partial \Sigma _+\}\rightarrow 0 $时, 有$ J_{f, g} (u, v)\rightarrow \infty. $

(ⅱ) 假设$ (u_m, v_m)^\infty _{m=1}\subset\Sigma _+ $满足: 当$ m\rightarrow \infty $时, 有$ J_{f, g} (u_m, v_m)\rightarrow c, $其中$ c\in {{\Bbb R}} , c>0; $

$ \begin{array}{rl} \| J'_{f, g} (u_m, v_m)\| _{T^*_{(u_m, v_m) } \Sigma _+ } & \equiv\sup \{J'_{f, g} (u_m, v_m)(\varphi, \psi );(\varphi, \psi )\in T_{(u_m, v_m)} \Sigma _+, \| (\varphi, \psi )\| _E=1\}\\ &\rightarrow 0, \end{array} $

那么存在$ k\in \mathbb{N}, $实数序列$ \{\varepsilon^ j_m\}\subset {{\Bbb R}} , $点列$ \{x^j_m\}\subset \Omega, $其中$ \varepsilon^ j_m\rightarrow 0 \ (m\rightarrow \infty), $$ dist(x^j_m, \partial \Omega )/\varepsilon^ j_m $$ \rightarrow \infty, $$ 1\leq j\leq k, $存在方程组(1.1), (1.2)和(1.4)的一个非负解$ (u^0, v^0)\in E $和$ I^\infty $的非平凡临界点$ (u^j, v^j), 1\leq j\leq k, $使$ (u_m, v_m) $的子序列(仍记为$ (u_m, v_m) $)满足: 当$ m\rightarrow \infty $时, 有

$ \bigg\| u_m-\frac{u^0-\sum\limits^k_{j=1}u^j_m} {\sqrt{\| u^0+\sum\limits^k_{j=1}u^j_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N) } + \| v^0+\sum\limits^k_{j=1}v^j_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N)} }}\bigg\| _{D^{1, 2} ({{\Bbb R}} ^N) }\rightarrow 0, $

$ \bigg\| v_m-\frac{v^0-\sum\limits^k_{j=1}v^j_m} {\sqrt{\| u^0+\sum\limits^k_{j=1}u^j_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N)} + \| v^0+\sum\limits^k_{j=1}v^j_m\|^2_ {D^{1, 2} ({{\Bbb R}} ^N)} }}\bigg\|_{D^{1, 2} ({{\Bbb R}} ^N) }\rightarrow 0, $

其中$ (u^j_m(x), v^j_m(x))=(\varepsilon^ j_m) ^{-\frac{N-2}{2}} (u^j(\frac{x-x^j_m}{\varepsilon^ j_m}), v^j(\frac{x-x^j_m}{\varepsilon^ j_m})), 1\leq j\leq k, m\in \mathbb{N}. $此外

$ J_{f, g} (u_m, v_m)\rightarrow I_{f, g} (u^0, v^0)+\sum\limits^k_{j=1}I^\infty (u^j, v^j). $

证  (ⅰ) 由引理2.3的(ⅰ)和(2.4)式, 我们有

$ \begin{eqnarray*} J_{f, g} (u, v) &\geq &(1-\varepsilon )^{\frac{N}{2} } J_{0, 0} (u, v)-\frac{1}{2\varepsilon }\| (f, g)\| _{E^*} \\ & =&(1-\varepsilon )^{\frac{N}{2}} \frac{1}{N}\frac{[1+\int _{ \Omega }(a(x)u^2+b(x)v^2){\rm d}x]^{\frac{N}{2} }} {(2\int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x)^{\frac{N-2}{2} }} -\frac{1}{2\varepsilon }\| (f, g)\| _{E^*} \\ &\geq& ( 1-\varepsilon )^{\frac{N}{2}} \frac{1}{N}\bigg(2\int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x\bigg )^{-\frac{N-2}{2}} -\frac{1}{2\varepsilon }\| (f, g)\| _{E^*} . \end{eqnarray*} $

因为$ dist((u, v), \partial \Sigma _+)\rightarrow 0, $所以$ \| (u_+, v_+)\| _E\rightarrow 0, $于是

$ \int _{ \Omega }u^\alpha _+v^\beta _+{\rm d}x \leq \bigg(\int _{ \Omega }u^{2^*} _+{\rm d}x\bigg)^{ \frac{\alpha }{2^*}}\bigg (\int _{ \Omega }v^{2^*} _+{\rm d}x\bigg) ^{\frac{\beta }{2^*}} \rightarrow 0 . $

综上所述可知$ J_{f, g} (u, v)\rightarrow \infty. $

(ⅱ) 根据(2.12)和(2.22)式可知: 当$ m\rightarrow \infty $时, 有

$ \begin{eqnarray*} \| I'_{f, g} (t_{f, g} (u_m, v_m)(u_m, v_m))\| _{E^*} & =&\frac{1}{t_{f, g} (u_m, v_m)}\| J'_{f, g} (u_m, v_m)\|_{T^*_{(u_m, v_m) } \Sigma_+}\\ & \leq &[2(2^*-1)S^{-\frac{2^*}{2}}_{\alpha, \beta}]^{-\frac{1}{2^*-2} } \| J'_{f, g} (u_m, v_m)\|_{T^*_{(u_m, v_m) } \Sigma _+} \\ &\rightarrow & 0 . \end{eqnarray*} $

并且有

$ I_{f, g} (t_{f, g} (u_m, v_m)(u_m, v_m))=J_{f, g} (u_m, v_m)\rightarrow c . $

所以, $ t_{f, g} (u_m, v_m)(u_m, v_m) $是$ I_{f, g} $在$ E $上的$ (PS) _c $序列. 根据引理2.9我们知道: 存在实数$ k\in \mathbb{N}, $实数序列$ \{\varepsilon ^j_m\}\subset {{\Bbb R}} , $点列$ \{x^j_m\}\subset \Omega, $其中$ \varepsilon ^j_m\rightarrow 0\ (m\rightarrow \infty ), $$ dist(x^j_m, \partial \Omega )/\varepsilon^ j_m\rightarrow \infty, 1\leq j\leq k, $有方程组(1.1), (1.2)和(1.4)的一个非负解$ (u^0, v^0)\in E $和$ I^\infty $的非平凡临界点$ (u^j, v^j), 1\leq j\leq k, $使$ \{t_{f, g} (u_m, v_m)(u_m, v_m)\} $的子序列(仍记为$ \{t_{f, g} (u_m, v_m)(u_m, v_m)\} ) $满足: 当$ m\rightarrow \infty $时, 有

(2.25) $ \begin{eqnarray} \begin{array}{ll} { } \Big\| t_{f, g} (u_m, v_m)u_m-u^0-\sum\limits^k_{j=1}u^j_m\Big\| _{D^{1, 2} ({{\Bbb R}} ^N) }\rightarrow 0, \\ { } \Big \| t_{f, g} (u_m, v_m)v_m-v^0-\sum\limits^k_{j=1}v^j_m\Big\| _{D^{1, 2} ({{\Bbb R}} ^N)} \rightarrow 0 . \end{array} \end{eqnarray} $

并且有

(2.26) $ \begin{equation} I_{f, g} (t_{f, g} (u_m, v_m)(u_m, v_m))\rightarrow I_{f, g} (u^0, v^0)+\sum\limits^k_{j=1}I^\infty(u^j) . \end{equation} $

我们断定$ \| t_{f, g} (u_m, v_m)(u_m, v_m)\|_{D^{1, 2} ({{\Bbb R}} ^N)} $是有界的. 事实上, 因为$ (t_{f, g} (u_m, v_m)(u_m, v_m)) $是$ I_{f, g} $的一个$ (PS) _c $序列. 令

$ A=\int _{ \Omega }[|\nabla (t_{f, g} (u_m, v_m)u_m)|^2+|\nabla (t_{f, g} (u_m, v_m)v_m)|^2]{\rm d}x, $

$ B=\int _{ \Omega }[a(x)t^2_{f, g} (u_m, v_m)u^2_m+b(x)t^2_{f, g} (u_m, v_m)v^2_m]{\rm d}x, $

$ D=\int _{ \Omega }(t_{f, g} (u_m, v_m)u_{m_+} )^\alpha (t_{f, g} (u_m, v_m)v_{m_+} )^\beta {\rm d}x, $

$ E=\int _{ \Omega }[ft_{f, g} (u_m, v_m)u_m+gt_{f, g} (u_m, v_m)v_m]{\rm d}x . $

我们有

$ \frac{1}{2}A+\frac{1}{2}B-\frac{2}{\alpha +\beta }D-E=C+O_m(1), $

$ A+B-2D-E=O_m(1)\| t_{f, g} (u_m, v_m)(u_m, v_m)\| _E . $

于是

$ (\frac{1}{2}-\frac{1}{\alpha +\beta })(A+B)-(1-\frac{1}{\alpha +\beta })E=C+O_m(1)\| t_{f, g} (u_m, v_m)(u_m, v_m)\|_E+O_m(1) . $

从而

(2.27) $ \begin{equation} (\frac{1}{2}-\frac{1}{\alpha +\beta })A-[(1-\frac{1}{\alpha +\beta })\| (f, g)\| _{E^*} +O_m(1)] \| t_{f, g} (u_m, v_m)(u_m, v_m)\| _E\leq C+O_m(1) . \end{equation} $

又因为$ \| (f, g)\| _{E^*} \leq d_2 $和(2.27)式, 所以$ \| t_{f, g} (u_m, v_m)(u_m, v_m)\| _{D^{1, 2} ({{\Bbb R}} ^N)} $有界, 根据

$ \| (u_m, v_m)\| _E=1, $

我们可以得到$ \{t_{f, g} (u_m, v_m)\} $也是有界的. 由(2.25)式和$ \| t_{f, g} (u_m, v_m)(u_m, v_m)\| _{D^{1, 2} ({{\Bbb R}} ^N)} $有界, 我们得到

$ \left | t^2_{f, g} (u_m, v_m)\| u_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N) }- \| u^0+\sum\limits^k_{j=1}u^j_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N)}\right | \rightarrow 0, $

$ \left | t^2_{f, g} (u_m, v_m)\| v_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N)} - \| v^0+\sum\limits^k_{j=1}v^j_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N) }\right| \rightarrow 0 . $

于是

$ \left | t^2_{f, g} (u_m, v_m)-\| u^0+\sum\limits^k_{j=1}u^j_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N)} -\| v^0+\sum\limits^k_{j=1}v^j_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N)}\right | \rightarrow 0 . $

因为$ \{t_{f, g} (u_m, v_m)\} $有界, 所以有

(2.28) $ \begin{eqnarray} \left | t_{f, g} (u_m, v_m)-\sqrt{\| u^0+\sum\limits^k_{j=1}u^j_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N) } +\| v^0+\sum\limits^k_{j=1}v^j_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N) }}\right| \rightarrow 0 . \end{eqnarray} $

由(2.25)和(2.28)式, 我们得到

$ \begin{eqnarray*} && \left\| t_{f, g} (u_m, v_m)u_m-u_m\sqrt{ \| u^0+\sum\limits^k_{j=1}u^j_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N)} +\| v^0+\sum\limits^k_{j=1}v^j_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N)} }\right.\\ &&\left.+u_m\sqrt{\| u^0+\sum\limits^k_{j=1}u^j_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N) } +\| v^0+\sum\limits^k_{j=1}v^j_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N)} }-u^0-\sum\limits^k_{j=1}u^j_m\right\| _{D^{1, 2} ({{\Bbb R}} ^N)} \rightarrow 0 . \end{eqnarray*} $

于是,

$ \left \| u_m\sqrt{\| u^0+\sum\limits^k_{j=1}u^j_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N)} +\| v^0+\sum\limits^k_{j=1}v^j_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N) }}-u^0-\sum\limits^k_{j=1}u^j_m\right\| _{D^{1, 2} ({{\Bbb R}} ^N) }\rightarrow 0 . $

所以

$ \left \| u_m-\frac{u^0+\sum\limits^k_{j=1}u^j_m} {\sqrt{\| u^0+\sum\limits^k_{j=1}u^j_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N)} +\| v^0+\sum\limits^k_{j=1}v^j_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N)} }}\right\| _{D^{1, 2} ({{\Bbb R}} ^N) }\rightarrow 0 . $

运用同样的方法可以得到

$ \left\| v_m-\frac{v^0+\sum\limits^k_{j=1}v^j_m}{\sqrt{ \| u^0+\sum\limits^k_{j=1}u^j_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N) } +\| v^0+\sum\limits^k_{j=1}v^j_m\|^2_{D^{1, 2} ({{\Bbb R}} ^N) }}}\right\| _{D^{1, 2} ({{\Bbb R}} ^N) }\rightarrow 0 . $

引理2.10得证.

引理2.11  假设非负函数$ a(x) $和$ b(x) $满足条件$ (A_1), $那么对于任意的$ \varepsilon >0, $存在$ \delta (\varepsilon )\in (0, d_2), $当$ \| (f, g)\| _{E^*}\leq \delta (\varepsilon ) $成立时, 有

(ⅰ) $ \frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2} }-\varepsilon\leq \inf\limits_{(u, v)\in \Sigma _+ }J_{f, g} (u, v) \leq\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2}} +\varepsilon . $

(ⅱ) 当$ c\in (-\infty, I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2}}) $时, $ J_{f, g} (u, v) $满足$ (PS) _c $条件.

证  (ⅰ) 根据引理2.1, 我们得到$ \inf\limits_{(u, v)\in \Sigma _+ }J_{0, 0} (u, v)=\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2}} $, 又由引理2.3的(ⅰ)可知: 对于任意的$ \sigma \in (0, 1), $有

$ \begin{eqnarray*} && \inf\limits_{(u, v)\in \Sigma _+ }J_{0, 0} (u, v)\\ & \leq&\inf\limits_{(u, v)\in \Sigma _+ }J_{f, g} (u, v) +[(1-\sigma )^{ -\frac{N}{2}} -1] \inf\limits_{(u, v)\in \Sigma _+ } J_{f, g} (u, v)+(1-\sigma )^{ -\frac{N}{2} }\frac{1}{2\sigma }\| (f, g)\|^2_{E^*}. \end{eqnarray*} $

由引理2.3可知, 当$ \| (f, g)\|^2_{E^*}\leq 1 $时, 有

$ ( \frac{1}{2})^{ \frac{N}{2}} \frac{2}{N} (\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2}} -1\leq \inf\limits_{(u, v)\in \Sigma _+ }J_{f, g} (u, v)\leq( \frac{3}{2})^{ \frac{N}{2}} \frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2} }+1 . $

即如果$ \| (f, g)\| _{E^*}\leq 1 $成立, 则$ \inf\limits_{(u, v)\in \Sigma _+ }J_{f, g} (u, v) $有界. 取$ \sigma _1>0 $是充分小的, 当$ 0<\sigma <\sigma _1 $且$ \| (f, g)\| _{E^*}\leq 1 $时, 有$ [(1-\sigma )^{ -\frac{N}{2} }-1]\inf\limits _{(u, v)\in \Sigma _+ }J_{f, g} (u, v)<\frac{\varepsilon }{2}. $对于任意固定的$ \sigma (0<\sigma <\sigma _1), $存在$ \bar{d}(\varepsilon )\in (0, \min \{1, d_2\}], $当$ \| (f, g)\| _{E^*} \leq\bar{d}(\varepsilon ) $时, 有$ (1-\sigma )^{ -\frac{N}{2}} \frac{1}{2\sigma }\| (f, g)\|^2_{E^*} <\frac{\varepsilon }{2}. $于是, 当$ \| (f, g)\| _{E^*}\leq \bar{d}(\varepsilon ) $时, 有

$ \frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2} }\leq\inf\limits_{(u, v)\in \Sigma _+ }J_{f, g} (u, v)+\varepsilon, $

即$ \frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2} }-\varepsilon \leq\inf\limits_{(u, v)\in \Sigma _+ }J_{f, g} (u, v). $

另一方面, 由引理2.3可知: 对于任意$ \sigma \in (0, 1), $有

$ \begin{eqnarray*} && \inf\limits_{(u, v)\in \Sigma _+ }J_{f, g} (u, v)+[(1+\sigma )^{ -\frac{N}{2}} -1] \inf\limits_{(u, v)\in \Sigma _+ }J_{f, g} (u, v)-(1+\sigma )^{ -\frac{N}{2} } \frac{1}{2\sigma }\| (f, g)\| _{E^*}\\ &&\leq \inf\limits_{(u, v)\in \Sigma _+ }J_{0, 0} (u, v) . \end{eqnarray*} $

因为当$ \| (f, g)\| _{E^*} \leq 1 $时, $ \inf\limits_{(u, v)\in \Sigma _+ }J_{f, g} (u, v) $是有界的, 所以可令$ \sigma _2>0 $充分小, 使得

$ -[(1+\sigma )^{ -\frac{N}{2} }-1] \inf\limits_{(u, v)\in \Sigma _+ }J_{f, g} (u, v)<\frac{\varepsilon }{2} . $

对于任意固定的$ \sigma (0<\sigma <\sigma _2), $存在$ \hat{d}(\varepsilon )\in (0, \min \{d_2, 1\}], $使得当$ \| (f, g)\| _{E^*} \leq\hat{d}(\varepsilon ) $时, 有$ (1+\sigma )^{ -\frac{N}{2} }\frac{1}{2\sigma }\| (f, g)\|^2_{E^*} <\frac{\varepsilon }{2}, $所以$ \inf\limits_{(u, v)\in \Sigma _+ }J_{f, g} (u, v)\leq \frac{2}{N}(\frac{S_{\alpha, \beta} }{2}) ^{\frac{N}{2}} +\varepsilon. $因此, 对于任意的$ \varepsilon >0, $存在$ d(\varepsilon )= \min \{\bar{d}(\varepsilon ), \hat{d}(\varepsilon )\}, $使得当$ \| (f, g)\| _{E^*} \leq d(\varepsilon ) $时, 有

$ \frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2} }-\varepsilon \leq\inf\limits_{(u, v)\in \Sigma _+ }J_{f, g} (u, v) \leq \frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2}} +\varepsilon . $

(ⅱ) 注意到对于任意的$ u, v\in D^{1, 2} ({{\Bbb R}} ^N) $且满足

$ \left\{\begin{array}{ll} { } -\Delta u=\frac{2\alpha }{\alpha +\beta }u^{\alpha -1 }_+v^\beta _+, \quad & x\in {{\Bbb R}} ^N, \\ { } -\Delta v=\frac{2\beta }{\alpha +\beta }u^\alpha _+v^{\beta -1 }_+, & x\in {{\Bbb R}} ^N, \\ { } u\not\equiv 0, v\not\equiv 0, & x\in {{\Bbb R}} ^N , \end{array}\right.\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\; \begin{array}{r} (2.29)\\ (2.30)\\ (2.31) \end{array} $

有

$ I^\infty(u, v)=\frac{1}{2}\int _{{{\Bbb R}} ^N }(|\nabla u|^2+|\nabla v|^2){\rm d}x-\frac{2}{\alpha +\beta } \int _{{{\Bbb R}} ^N }u^\alpha _+v^\beta _+{\rm d}x \geq \frac{2}{N}(\frac{S_{\alpha, \beta} }{2}) ^{\frac{N}{2} }. $

事实上, 将(2.29)和(2.30)式两边分别同时乘以$ u_- $和$ v_-, $再积分可以得到

$ \int _{{{\Bbb R}} ^N }|\nabla u _-|^2{\rm d}x=0, {\qquad} \int _{{{\Bbb R}} ^N }|\nabla v_-|^2{\rm d}x=0. $

于是$ u \geq 0 $和$ v\geq 0. $再将(2.29)和(2.30)两边分别同时乘以$ u_+ $和$ v_+, $在积分可以得到

$ \int _{{{\Bbb R}} ^N }|\nabla u _+|^2{\rm d}x=\frac{2\alpha }{\alpha +\beta }\int _{{{\Bbb R}} ^N }u^\alpha _+v^\beta _+{\rm d}x, $

$ \int _{{{\Bbb R}} ^N }|\nabla v_+|^2{\rm d}x=\frac{2\beta }{\alpha +\beta }\int _{{{\Bbb R}} ^N }u^\alpha _+v^\beta _+{\rm d}x . $

因为$ \int _{{{\Bbb R}} ^N }(| \nabla u _+|^2+| \nabla v_+|^2){\rm d}x\geq S_{\alpha, \beta} (\int _{{{\Bbb R}} ^N }u^\alpha _+v^\beta _+{\rm d}x)^{ \frac{2}{\alpha +\beta } }, $所以有$ \int _{{{\Bbb R}} ^N }u^\alpha _+v^\beta _+{\rm d}x \geq (\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2} }. $从而有

$ \begin{eqnarray*} I^\infty(u, v)&=&\int _{{{\Bbb R}} ^N }u^\alpha _+v^\beta _+{\rm d}x-\frac{2}{\alpha +\beta }\int _{{{\Bbb R}} ^N }u^\alpha _+v^\beta _+{\rm d}x +\frac{1}{2}\int _{{{\Bbb R}} ^N }(|\nabla u _-|^2+|\nabla v_-|^2){\rm d}x \\ &=& \frac{2}{N}\int _{{{\Bbb R}} ^N }u^\alpha _+v^\beta _+{\rm d}x \\ &\geq& \frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2} }. \end{eqnarray*} $

由引理2.10, 我们知道: 当$ m\rightarrow \infty $时, 有$ c+o(1)=J_{f, g} (u_m, v_m)\rightarrow I_{f, g} (u^0, v^0) +\sum\limits^k_{j=1}I^\infty(u^j, v^j). $由于$ c<I_{f, g} (u_{{\rm loc}} (f, g;x), v_{{\rm loc}} (f, g;x))+\frac{2}{N} (\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2}}, $又由引理2.3的(ⅱ), 引理2.8的(ⅲ)和$ I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} ) \leq 0 $可以得到

$ I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )=\inf \{I_{f, g} (u^0, v^0);u^0, v^0\ \mbox{是} I_{f, g} (u, v) \mbox{在} E \mbox{上的临界点\}. } $

所以, 由引理2.10可以得到: 在$ E $中, 当$ m\rightarrow \infty $时, 有

$ u_m\rightarrow \frac{u^0}{\sqrt{\| u^0\|^2_{D^{1, 2} ({{\Bbb R}} ^N) }+\| v^0\|^2_{D^{1, 2} ({{\Bbb R}} ^N) }}}, $

$ v_m\rightarrow \frac{v^0}{\sqrt{\| u^0\|^2_{D^{1, 2} ({{\Bbb R}} ^N)} +\| v^0\|^2_{D^{1, 2} ({{\Bbb R}} ^N) }} }. $

引理2.11得证.

记$ [J_{f, g}\leq c]=\{(u, v)\in \Sigma _+;J_{f, g} (u, v)\leq c\}. $我们将构造下面这两个映射

$ F:S^{N-1} \rightarrow [J_{f, g} \leq I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N} (\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2}} -\varepsilon ], $

$ G:[J_{f, g} \leq I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2} } -\varepsilon ]\rightarrow S^{N-1}, $

使$ F\circ G $和恒等映射同伦.

我们先构造映射$ F $. 记$ \Omega ^ \eta _1:=\{x\in \Omega _1;dist(x, \partial \Omega _1)>2\eta \}. $

引理3.1  假设$ f(x) \geq 0, g(x)\geq 0, f(x)\not\equiv 0, g(x)\not\equiv 0, \| (f, g)\| _{E^*} \leq d_2 $$ ( d_2 $由引理2.8给出), $ f(x), g(x)\in L^p( \Omega ), p>\frac{N}{2}, $非负函数$ a(x), b(x) $满足条件$ (A_1) $和$ (A_2), $那么存在$ \varepsilon _0>0, $使得对于任意$ 0<\varepsilon <\varepsilon _0, t\geq 0, B/D=\sqrt{\alpha /\beta } $和$ a.e\ e\in \Omega^ \eta _1 $有

(3.1) $ \begin{eqnarray} I_{f, g} (u_{{\rm loc}} +tBu_{\varepsilon, e}, v_{{\rm loc}} +tDu_{\varepsilon, e} )<I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} ) +\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2} } . \end{eqnarray} $

证

(3.2) $ \begin{eqnarray} && I_{f, g} (u_{{\rm loc}} +tBu_{\varepsilon, e}, v_{{\rm loc}} +tDu_{\varepsilon, e} )\\ &=& \frac{1}{2}\int _{ \Omega }|\nabla (u_{{\rm loc}} +tBu_{\varepsilon, e} )|^2{\rm d}x+\frac{1}{2}\int _{ \Omega }|\nabla (v_{{\rm loc}} +tDu_{\varepsilon, e} )|^2{\rm d}x \\ &&+ \frac{1}{2}\int _{ \Omega }a(x)(u_{{\rm loc}} +tBu_{\varepsilon, e} )^2{\rm d}x+\frac{1}{2}\int _{ \Omega }b(x) (v_{{\rm loc}} +tDu_{\varepsilon, e} )^2{\rm d}x \\ &&-\frac{2}{\alpha +\beta }\int _{ \Omega }(u_{{\rm loc}} +tBu_{\varepsilon, e} )^\alpha (v_{{\rm loc}} +tDu_{\varepsilon, e} )^\beta {\rm d}x \\ &&-\int _{ \Omega }f(x)(u_{{\rm loc}} +tBu_{\varepsilon, e} ){\rm d}x-\int _{ \Omega }g(x)(v_{{\rm loc}} +tDu_{\varepsilon, e} ){\rm d}x . \end{eqnarray} $

因为$ I'_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )=0, $所以

(3.3) $ \begin{eqnarray} &&\langle I'_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} ), (tBu_{\varepsilon, e}, tDu_{\varepsilon, e} )\rangle\\ &=&tB\int _{ \Omega } \nabla u _{{\rm loc}} \nabla u _{\varepsilon, e} {\rm d}x +tD\int _{ \Omega }\nabla v_{{\rm loc}} \nabla u _{\varepsilon, e} {\rm d}x+tB\int _{ \Omega }a(x)u_{{\rm loc}} u_{\varepsilon, e} {\rm d}x \\ && +tD\int _{ \Omega }b(x)v_{{\rm loc}} u_{\varepsilon, e} {\rm d}x+\frac{2\alpha tB}{\alpha +\beta } \int _{ \Omega }u^{\alpha -1 }_{{\rm loc}} v^\beta _{{\rm loc}} u_{\varepsilon, e} {\rm d}x +\frac{2\beta tD}{\alpha +\beta }\int _{ \Omega }u^\alpha _{{\rm loc}} v^{\beta -1} _{{\rm loc}} u_{\varepsilon, e} {\rm d}x\\ && -\int _{ \Omega }tBf(x)u_{\varepsilon, e} {\rm d}x-\int _{ \Omega }tDg(x)u_{\varepsilon, e} {\rm d}x=0 . \end{eqnarray} $

将(3.3)式代入(3.2)式中, 就有

$ \begin{eqnarray*} I_{f, g} (u_{{\rm loc}} +tBu_{\varepsilon, e}, v_{{\rm loc}} +tDu_{\varepsilon, e} ) &=&I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+I_{0, 0} (tBu_{\varepsilon, e}, tDv_{\varepsilon, e} ) \\ &&-\frac{2}{\alpha +\beta }\bigg[\int _{ \Omega }(u_{{\rm loc}} +tBu_{\varepsilon, e} )^\alpha ( v_{{\rm loc}} +tDv_{\varepsilon, e} )^\beta {\rm d}x \\ &&-\int _{ \Omega }u^\alpha _{{\rm loc}} v^\beta _{{\rm loc}} {\rm d}x -t^{\alpha +\beta } B^\alpha D^\beta \int _{ \Omega }u^{\alpha +\beta } _{\varepsilon, e} {\rm d}x\\ &&-\alpha tB\int _{ \Omega }u^{\alpha -1} _{{\rm loc}} v^\beta _{{\rm loc}} u_{\varepsilon, e} {\rm d}x -\beta tD\int _{ \Omega }u^\alpha _{{\rm loc}} v^{\beta -1 }_{{\rm loc}} u_{\varepsilon, e} {\rm d}x\bigg]. \end{eqnarray*} $

根据(2.1)式, (2.2)式和类似于(2.4)式的证明, 我们可以得到

$ \begin{eqnarray*} &&\max\limits_{t>0 }I_{0, 0} (tBu_{\varepsilon, e}, tDu_{\varepsilon, e} ) \\ &=&\frac{1}{N}\frac{[(B^2+D^2)\int _{ \Omega }|\nabla u _{\varepsilon, e} |^2{\rm d}x +\int _{ \Omega }(B^2a(x)+D^2b(x))u^2_{\varepsilon, e} {\rm d}x]^{\frac{N}{2} }} {(2B^\alpha D^\beta \int _{ \Omega }u^{\alpha +\beta } _{\varepsilon, e} {\rm d}x)^{\frac{N-2}{2} }}\\ &= &\frac{1}{N}\frac{[(B^2+D^2)S^{\frac{N}{2}} +O(\varepsilon^{ N-2 }) +\int _{ \Omega }(B^2a(x)+D^2b(x))u^2_{\varepsilon, e} {\rm d}x]^{\frac{N}{2} }} {(2B^\alpha D^\beta S^{\frac{N}{2} }+O(\varepsilon ^N))^{ \frac{N-2}{2} }}\\ & =& \frac{1}{N}[(B^2+D^2)S^{\frac{N}{2}}]^\frac{N}{2} \bigg [1+O(\varepsilon ^{N-2 })+(B^2+D^2)^{-1} S^{-\frac{N}{2}} \int _{ \Omega }(B^2a(x)+D^2b(x))u^2_{\varepsilon, e} {\rm d}x\bigg]^\frac{N}{2}\\ &&\times [(2B^\alpha D^\beta S^{-\frac{N}{2}} )^{\frac{N-2}{2}} +O(\varepsilon ^N)]\\ & = &\frac{2}{N}\bigg[\frac{1}{2}((\frac{\alpha }{\beta})^{\frac{\beta }{\alpha +\beta } }+ (\frac{\beta}{\alpha })^{\frac{\alpha }{\alpha +\beta }} )S\bigg]^{\frac{N}{2}} \bigg [1+O(\varepsilon^{ N-2} )+(B^2+D^2)^{-1} S^{-\frac{N}{2}} \int _{ \Omega }(B^2a(x)\\ &&+D^2b(x))u^2_{\varepsilon, e} {\rm d}x\bigg]^{\frac{N}{2}} + O(\varepsilon^ N)\\ &=&\frac{2}{N}(\frac{S_{\alpha, \beta}}{2})^{\frac{N}{2}} \bigg[1+(B^2+D^2)^{-1} S^{-\frac{N}{2} }\int _{ \Omega }(B^2a(x)+D^2b(x))u^2_{\varepsilon, e} {\rm d}x\bigg]^{\frac{N}{2} } +O(\varepsilon^{ N-2} ) . \end{eqnarray*} $

由$ a(x)\geq 0, b(x) \geq 0 $和条件$ (A_2) $, 我们得到: 对于任意$ e\in \Omega ^ \eta _1, $有

$ \int _{ \Omega }a(x)u^2_{\varepsilon, e} {\rm d}x=\int _{ \Omega }b(x)u^2_{\varepsilon, e} {\rm d}x=0 . $

于是, 对于任意$ e\in \Omega ^ \eta _1, $有

$ I_{0, 0} (tBu_{\varepsilon, e}, tDv_{\varepsilon, e}\leq \frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2}} +O(\varepsilon^{ N-2} ) . $

因此

(3.4) $ \begin{eqnarray} && I_{f, g} (u_{{\rm loc}} +tBu_{\varepsilon, e}, v_{{\rm loc}} +tDu_{\varepsilon, e} )\\ &\leq& I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta}}{2})^{\frac{N}{2}} -\frac{2}{\alpha +\beta }\bigg[\int _{ \Omega }(u_{{\rm loc}} +tBu_{\varepsilon, e} ) ^ \alpha ( v_{{\rm loc}} +tDu_{\varepsilon, e} )^\beta {\rm d}x \\ &&-\int _{ \Omega }u^\alpha _{{\rm loc}} v^\beta _{{\rm loc}} {\rm d}x -t^{\alpha +\beta } B^\alpha D^\beta \int _{ \Omega }u^{\alpha +\beta }_{\varepsilon, e} {\rm d}x -\alpha tB\int _{ \Omega }u^{\alpha -1} _{{\rm loc}} v^\beta _{{\rm loc}} u_{\varepsilon, e} {\rm d}x\\ && -\beta tD\int _{ \Omega }u^\alpha _{{\rm loc}} v^{\beta -1 }_{{\rm loc}} u_{\varepsilon, e} {\rm d}x\bigg] + O(\varepsilon^ {N-2 }) . \end{eqnarray} $

因为$ t\rightarrow 0 $时, 有

$ I_{f, g} (u_{{\rm loc}} +tBu_{\varepsilon, e}, v_{{\rm loc}} +tDu_{\varepsilon, e} )\rightarrow I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )<0, $

$ t\rightarrow +\infty $时, 有

$ I_{f, g} (u_{{\rm loc}} +tBu_{\varepsilon, e}, v_{{\rm loc}} +tDu_{\varepsilon, e} )\rightarrow -\infty, $

所以存在$ 0<m<M, $使对于任意的$ t\in [0, m]\cup [M, \infty), $有

$ I_{f, g} (u_{{\rm loc}} +tBu_{\varepsilon, e}, v_{{\rm loc}} +tDu_{\varepsilon, e} ) \leq 0 . $

于是, 当$ t\in [0, m]\cup [M, \infty) $时, 有(3.1)式成立. 下面, 我们研究$ t\in [m, M] $的情形. 因为对于任意的$ p\geq 2, a \geq 0, b\geq 0, $有$ (a+b)^p\geq a^p+b^p+pa^{p-1} b, $那么就有

(3.5) $ \begin{eqnarray} &&(u_{{\rm loc}} +tBu_{\varepsilon, e} )^\alpha (v_{{\rm loc}} +tDu_{\varepsilon, e} )^\beta\\ &\geq &(u^\alpha _{{\rm loc}} +t^\alpha B^\alpha u^\alpha _{\varepsilon, e} +\alpha tBu^{\alpha -1} _{{\rm loc}} u_{\varepsilon, e} )(v^\beta _{{\rm loc}} +t^\beta D^\beta u^\beta _{\varepsilon, e} +\beta tDv^{\beta -1} _{{\rm loc}} u_{\varepsilon, e} )\\ &=&u^\alpha _{{\rm loc}} v^\beta _{{\rm loc}} +t^\beta D^\beta u^\alpha _{{\rm loc}} u^\beta _{\varepsilon, e} +\beta tDu^\alpha _{{\rm loc}} v^{\beta -1} _{{\rm loc}} u_{\varepsilon, e} +t^\alpha B^\alpha u^\alpha _{\varepsilon, e} v^\beta _{{\rm loc}} \\ && +t^{\alpha +\beta }B^\alpha D^\beta u^{\alpha +\beta } _{\varepsilon, e} +t^{\alpha +1 }B^\alpha \beta Dv^{\beta -1} _{{\rm loc}} u^{\alpha +1} _{\varepsilon, e} +\alpha tBu^{\alpha -1} _{{\rm loc}} v^\beta _{{\rm loc}} u_{\varepsilon, e} \\ && +t^{\beta +1} \alpha BD^\beta u^{\alpha -1} _{{\rm loc}} u^{\beta +1} _{\varepsilon, e} +t^\alpha \alpha B\beta Du^{\alpha -1} _{{\rm loc}} v^{\beta -1} _{{\rm loc}} u^2_{\varepsilon, e} . \end{eqnarray} $

因为$ f(x), g(x)\in L^p( \Omega ), p>\frac{N}{2}, $由引理2.2的(ⅱ)可知$ u_{{\rm loc}} \in C^0(\bar{\Omega}). $于是, 我们得到$ \min\limits_{x\in B(e, 2\eta ) }u_{{\rm loc}} >0, $$ \min\limits_{x\in B(e, 2\eta ) }v_{{\rm loc}} >0. $从而

(3.6) $ \begin{eqnarray} t^\beta D^\beta \int _{ \Omega }u^\alpha _{{\rm loc}} u^\beta _{\varepsilon, e} {\rm d}x& \geq& c\int _{ \Omega }u^\alpha _{{\rm loc}} \xi^ \beta (x-e) \bigg[\frac{(N(N-2))^{\frac{N-2}{4}}\varepsilon^{-\frac{N-2}{2}}} {(1+|\frac{x-e}{\varepsilon}|^2)^{\frac{N-2}{2}}}\bigg]^\beta {\rm d}x \\ &\geq& c\varepsilon ^{\frac{2N-(N-2)\beta }{2} }\min\limits_{x\in B(e, 2\eta ) }u_{{\rm loc}} \int _{B(e, \eta ) }\frac{1}{(1+| \frac{x-e}{\varepsilon }|^2) ^{\frac{N-2}{2}\beta} \varepsilon^ N}{\rm d}x \\ &=& c\varepsilon^{ \frac{2N-(N-2)\beta }{2}}\min\limits_{x\in B(e, 2\eta ) }u_{{\rm loc}} \int _{B(0, \eta /\varepsilon ) } \frac{1}{(1+| z|^2)^{\frac{N-2}{2}\beta }}{\rm d}z \\ &=& c\varepsilon^{\frac{2N-(N-2)\beta }{2}}\min\limits_{x\in B(e, 2\eta ) }u_{{\rm loc}} \int _{B(0, \eta ) } \frac{1}{(1+| z|^2)^{ \frac{N-2}{2}\beta} }{\rm d}z \\ &=& O(\varepsilon^{ \frac{2N-(N-2)\beta }{2} }) . \end{eqnarray} $

类似地, 我们有

(3.7) $ \begin{equation} \int _{ \Omega }t^\alpha B^\alpha u^\alpha _{\varepsilon, e} v^\beta _{{\rm loc}} {\rm d}x\geq O(\varepsilon^{ \frac{2N-(N-2)\alpha }{2}} ), \end{equation} $

(3.8) $ \begin{equation} \int _{ \Omega }t^{\alpha +1 }B^\alpha \beta Dv^{\beta -1} _{{\rm loc}} u^{\alpha +1} _{\varepsilon, e} {\rm d}x \geq O(\varepsilon^{ \frac{2N-(N-2)(\alpha +1)}{2}} ), \end{equation} $

(3.9) $ \begin{equation} \int _{ \Omega }t^{\beta +1} \alpha BD^\beta u^{\alpha -1 }_{{\rm loc}} u^{\beta +1} _{\varepsilon, e} {\rm d}x\geq O(\varepsilon^{ \frac{2N-(N-2)(\beta +1)}{2}} ), \end{equation} $

(3.10) $ \begin{equation} \int _{ \Omega }t^2\alpha B\beta Du^{\alpha -1} _{{\rm loc}} v^{\beta -1} _{{\rm loc}} u^2_{\varepsilon, e} {\rm d}x \geq O(\varepsilon^ 2), \end{equation} $

其中$ \frac{2N-(N-2)(\alpha +1)}{2}>\frac{2N-(N-2)(\alpha +\beta )}{2}=0 $和$ \frac{2N-(N-2)(\beta +1)}{2}>\frac{2N-(N-2)(\alpha +\beta )}{2}=0. $将(3.5)–(3.10)式代入(3.4)式中, 得到

$ \begin{eqnarray*} I_{f, g} (u_{{\rm loc}} +tBu_{\varepsilon, e}, v_{{\rm loc}} +tDu_{\varepsilon, e} ) & \leq& I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta}}{2})^{\frac{N}{2}} \\ &&+ O(\varepsilon^{ N-2 })-O(\varepsilon^{ \frac{2N-(N-2)\beta }{2} }) -O(\varepsilon^{ \frac{2N-(N-2)\alpha }{2}} )\\ &&-O(\varepsilon^{\frac{2N-(N-2)(\alpha +1)}{2} })-O(\varepsilon^{\frac{2N-(N-2)(\beta +1)}{2}} )-O(\varepsilon^2) . \end{eqnarray*} $

如果$ N\not= 3, \alpha \not= 3, \beta \not= 3, $有$ N-2> \min \{\frac{2N-(N-2)(\alpha +1)}{2}, \frac{2N-(N-2)(\beta +1)}{2}\}. $

如果$ N=3, \alpha =3, \beta =3, $因为$ (a+b) ^3=a^3+3a^2b+3ab^2+b^3, $所以

$ I_{f, g} (u_{{\rm loc}} +tBu_{\varepsilon, e}, v_{{\rm loc}} +tDu_{\varepsilon, e} ) \leq I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2}} +O(\varepsilon )-O(\varepsilon^{ \frac{1}{2}} ) . $

于是, 存在$ \varepsilon _0>0, $使对于任意的$ 0<\varepsilon <\varepsilon _0, $$ t\geq 0 $及$ e\in \Omega ^ \eta _1, $有

$ I_{f, g} (u_{{\rm loc}} +tBu_{\varepsilon, e}, v_{{\rm loc}} +tDu_{\varepsilon, e} ) <I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2} } . $

引理3.1得证.

注3.2  当$ f(x)\equiv 0, g(x)\equiv 0 $时, (3.1)式不成立. 事实上, 当$ f(x)\equiv 0, g(x)\equiv 0 $时, $ u_{{\rm loc}} \equiv 0, v_{{\rm loc}} \equiv 0 $且

$ \begin{eqnarray*} I_{f, g} (u_{{\rm loc}} +Bt_{f, g} (u_{\varepsilon, e} )u_{\varepsilon, e}, v_{{\rm loc}} +Dt_{f, g} (u_{\varepsilon, e} )u_{\varepsilon, e} ) & =&I_{0, 0} (Bt_{f, g} (u_{\varepsilon, e} )u_{\varepsilon, e}, Dt_{f, g} (u_{\varepsilon, e} ) u_{\varepsilon, e} )\\ &>&\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2} } . \end{eqnarray*} $

引理3.3  假设$ (f(x), g(x))\in E^*, \| (f(x), g(x))\| _{E^*} \leq d_2, $非负函数$ a(x) $和$ b(x) $满足条件$ (A_1) $和$ (A_2), $那么存在$ d_3\in (0, d_2] $及$ \varepsilon _1<\varepsilon _0 $$ ( \varepsilon _0 $由引理3.1给出), 使得对于任意$ \| (f, g)\| _{E^*} \leq d_3, 0<\varepsilon \leq \varepsilon _1 $及$ a.e\ e\in \Omega ^ \eta _1, $在$ (\frac{B^2+D^2}{2B^\alpha D^\beta })^{\frac{N-2}{4}} $的一个邻域里存在唯一的$ s=s(f, g, \varepsilon, e)>0 $满足

(3.11) $ \begin{eqnarray} \| (u_{{\rm loc}} +sBu_{\varepsilon, e}, v_{{\rm loc}} +sDu_{\varepsilon, e} )\| _E =t_{f, g}\bigg (\frac{(u_{{\rm loc}} +sBu_{\varepsilon, e}, v_{{\rm loc}} +sDu_{\varepsilon, e} )} {\| (u_{{\rm loc}} +sBu_{\varepsilon, e}, v_{{\rm loc}} +sDu_{\varepsilon, e} )\| _E}\bigg). \end{eqnarray} $

此外, 映射

$ \varepsilon \in \{\varepsilon \in {{\Bbb R}} ;0<\varepsilon <\varepsilon _1\}\rightarrow s(f, g, \varepsilon, e)\in (0, \infty) $

是连续的.

证  这里我们要运用到隐函数定理. 令

$ \begin{eqnarray*} \Phi(s, f, g, \varepsilon, e)&=&\langle I'_{f, g} (u_{{\rm loc}} +sBu_{\varepsilon, e}, v_{{\rm loc}} +sDu_{\varepsilon, e} ), (u_{{\rm loc}} +sBu_{\varepsilon, e}, v_{{\rm loc}} +sDu_{\varepsilon, e} )\rangle \\ &=&\int _{ \Omega }|\nabla (u_{{\rm loc}} +sBu_{\varepsilon, e} )|^2{\rm d}x +\int _{ \Omega }|\nabla (v_{{\rm loc}} +sDu_{\varepsilon, e} )|^2{\rm d}x\\ &&+\int _{ \Omega }a(x)(u_{{\rm loc}} +sBu_{\varepsilon, e} )^2{\rm d}x + \int _{ \Omega }b(x)(v_{{\rm loc}} +sDu_{\varepsilon, e} )^2{\rm d}x \\ &&-2\int _{ \Omega }(u_{{\rm loc}} +sBu_{\varepsilon, e} )^\alpha (v_{{\rm loc}} +sDu_{\varepsilon, e} )^\beta {\rm d}x -\int _{ \Omega }f(u_{{\rm loc}} +sBu_{\varepsilon, e} ){\rm d}x\\ &&-\int _{ \Omega }g(v_{{\rm loc}} +sDu_{\varepsilon, e} ){\rm d}x . \end{eqnarray*} $

于是(3.11)式成立的充分必要条件是$ \Phi( s, f, g, \varepsilon, e)=0. $事实上, 我们令

$ \| (u_{{\rm loc}} +sBu_{\varepsilon, e}, v_{{\rm loc}} +sDu_{\varepsilon, e} )\| _E=h(s, f, g, \varepsilon, e). $

我们有

$ \begin{eqnarray*} && I_{f, g} \bigg(t\frac{(u_{{\rm loc}} +sBu_{\varepsilon, e}, v_{{\rm loc}} +sDu_{\varepsilon, e} )} {\| (u_{{\rm loc}} +sBu_{\varepsilon, e}, v_{{\rm loc}} +sDu_{\varepsilon, e} )\| _E}\bigg)\\ & =&\frac{t^2h^{-2} }{2}\int _{ \Omega }| \nabla (u_{{\rm loc}} +sBu_{\varepsilon, e} )|^2{\rm d}x +\frac{t^2h^{-2} }{2}\int _{ \Omega }| \nabla (v_{{\rm loc}} +sDu_{\varepsilon, e} )|^2{\rm d}x \\ &&+ \frac{t^2h^{-2} }{2}\int _{ \Omega }[a(x)(u_{{\rm loc}} +sBu_{\varepsilon, e} )^2 +b(x)(v_{{\rm loc}} +sDu_{\varepsilon, e} )^2]{\rm d}x\\ &&-2\frac{t^{\alpha +\beta }h^{-(\alpha +\beta )} }{2^*}\int _{ \Omega } (u_{{\rm loc}} +sBu_{\varepsilon, e} )^\alpha (v_{{\rm loc}} +sDu_{\varepsilon, e} )^ \beta {\rm d}x\\ &&-th^{-1} \int _{ \Omega }(u_{{\rm loc}} +sBu_{\varepsilon, e}f ){\rm d}x-th^{-1} \int _{ \Omega }(v_{{\rm loc}} +sDu_{\varepsilon, e} )g{\rm d}x, \end{eqnarray*} $

(3.12) $ \begin{eqnarray} && \frac{\rm d}{{\rm d}t}\bigg(I_{f, g}\bigg (t\frac{(u_{{\rm loc}} +sBu_{\varepsilon, e}, v_{{\rm loc}} +sDu_{\varepsilon, e} )} {\| (u_{{\rm loc}} +sBu_{\varepsilon, e}, v_{{\rm loc}} +sDu_{\varepsilon, e} )\| _E}\bigg)\bigg)\\ &=&th^{-2} \int _{ \Omega }| \nabla (u_{{\rm loc}} +sBu_{\varepsilon, e} )|^2{\rm d}x +th^{-2} \int _{ \Omega }|\nabla (v_{{\rm loc}} +sDu_{\varepsilon, e} )|^2{\rm d}x\\ &&+th^{-2} \int _{ \Omega }[a(x)(u_{{\rm loc}} +sBu_{\varepsilon, e} )^2+b(x)(v_{{\rm loc}} +sDu_{\varepsilon, e} )^2]{\rm d}x\\ &&-2t^{2^*-1 }h^{-2^*} \int _{ \Omega }(u_{{\rm loc}} +sBu_{\varepsilon, e} )^ \alpha (v_{{\rm loc}} +sDu_{\varepsilon, e} ) ^\beta {\rm d}x\\ && -h^{-1} \int _{ \Omega }(u_{{\rm loc}} +sBu_{\varepsilon, e} )f{\rm d}x-h^{-1 } \int _{ \Omega }(v_{{\rm loc}} +sDu_{\varepsilon, e} )g{\rm d}x . \end{eqnarray} $

一方面, 假设(3.11)式是成立的, 就有

$ \frac{\rm d}{{\rm d}t}\bigg(I_{f, g} \bigg(t_{f, g}\bigg( \frac{(u_{{\rm loc}} +sBu_{\varepsilon, e}, v_{{\rm loc}} +sDu_{\varepsilon, e} )} {\| (u_{{\rm loc}} +sBu_{\varepsilon, e}, v_{{\rm loc}} +sDu_{\varepsilon, e} )\| _E}\bigg) \frac{(u_{{\rm loc}} +sBu_{\varepsilon, e}, v_{{\rm loc}} +sDu_{\varepsilon, e} )} {\| (u_{{\rm loc}} +sBu_{\varepsilon, e}, v_{{\rm loc}} +sDu_{\varepsilon, e} )\| _E}\bigg)\bigg)=0 . $

即$ \frac{\rm d}{{\rm d}t}(I_{f, g} (u_{{\rm loc}} +sBu_{\varepsilon, e}, v_{{\rm loc}} +sDu_{\varepsilon, e} ))=0. $由(3.12)式得到: $ h=t. $于是

$ \| (u_{{\rm loc}} +sBu_{\varepsilon, e}, v_{{\rm loc}} +sDu_{\varepsilon, e} )\|^{-1}_E \Phi (s, f, g, \varepsilon, e)=0 . $

故$ \Phi(s, f, g, \varepsilon, e)=0. $

另一方面, 假设$ \Phi(s, f, g, \varepsilon, e)=0 $是成立的. 于是

$ \begin{eqnarray*} &&h\int _{ \Omega }\frac{| \nabla (u_{{\rm loc}} +sBu_{\varepsilon, e} )|^2 +|\nabla (v_{{\rm loc}} +sDu_{\varepsilon, e} )|^2}{h^2}{\rm d}x\\ &&+h\int _{ \Omega }\frac{[a(x)(u_{{\rm loc}} +sBu_{\varepsilon, e} )^ 2+b(x)(v_{{\rm loc}} +sDu_{\varepsilon, e} ) ^2]}{h^2}{\rm d}x \\ &&-2h^{2^*-1 }\int _{ \Omega }\frac{(u_{{\rm loc}} +sBu_{\varepsilon, e} )^\alpha (v_{{\rm loc}} +sDu_{\varepsilon, e} )^\beta }{h^{2^*} }{\rm d}x \\ &&-\int _{ \Omega }f\frac{u_{{\rm loc}} +sBu_{\varepsilon, e} }{h}{\rm d}x -\int _{ \Omega }g\frac{v_{{\rm loc}} +sDu_{\varepsilon, e} }{h}{\rm d}x =0. \end{eqnarray*} $

由(3.12)式和引理2.4的(ⅱ)可知：如果有$ \| (f, g)\| _{E^*}\leq d_0, $对于任意的$ (u, v)\in \Sigma _+, $存在唯一的$ t_{f, g} (u, v)>0, $使$ I_{f, g} (t_{f, g} (u, v)(u, v))=J_{f, g} (u, v) $成立. 因此(3.11)式成立.

由(2.1), (2.2)和(2.7)式, 我们得到: 当$ \varepsilon \rightarrow 0+ $时, 有

$ \begin{eqnarray*} &&\Phi\bigg(\bigg(\frac{B^2+D^2}{2B^\alpha D^\beta }\bigg)^{\frac{N-2}{4} }, 0, 0, \varepsilon, e\bigg) \\ &=&\bigg( \frac{B^2+D^2}{2B^\alpha D^\beta }\bigg)^{\frac{N-2}{2}} (B^2+D^2)\int _{ \Omega }|\nabla u_{\varepsilon, e} |^2{\rm d}x \\ &&+\bigg(\frac{B^2+D^2}{2B^\alpha D^\beta }\bigg)^{\frac{N-2}{2} }\int _{ \Omega }[B^2a(x)+D^2b(x)]u^2_{\varepsilon, e} {\rm d}x - 2 \bigg(\frac{B^2+D^2}{2B^\alpha D^\beta }\bigg)^{\frac{N}{2}} B^\alpha D^\beta \int _{ \Omega }u^{2^*} _{\varepsilon, e} {\rm d}x \\ & \rightarrow &0, \\ && \frac{\partial }{\partial s}\bigg|_{s=(\frac{B^2+D^2}{2B^\alpha D^\beta })^{\frac{N-2}{4}}} \Phi(s, 0, 0, \varepsilon, e)\\ &=&\bigg[2s(B^2+D^2)\int _{ \Omega }| \nabla u_{\varepsilon, e} |^2{\rm d}x+2s\int _{ \Omega }[B^2a(x)+D^2b(x)\bigg]u^2_{\varepsilon, e} {\rm d}x \\ &&-2\cdot 2^*s^{2^*-1 }B^\alpha D^\beta \int _{ \Omega }u^{2^*} _{\varepsilon, e} {\rm d}x\bigg|_ {s=(\frac{B^2+D^2}{2B^\alpha D^\beta })^{ \frac{N-2}{4}}}\\ & \rightarrow & -(B^2+D^2)^{\frac{N+2}{4}} (2B^\alpha D^\beta )^{\frac{2-N}{4}} (2^*-2)S^{\frac{N}{2} }<0. \end{eqnarray*} $

根据隐函数定理可知: 在$ (\frac{B^2+D^2}{2B^\alpha D^\beta })^{\frac{N-2}{4} } $的一个邻域里存在唯一的映射$ s=s(f, g, \varepsilon, e) $, 使$ \Phi (s, f, g, \varepsilon, e)=0 $并且$ \varepsilon \mapsto s(f, g, \varepsilon, e) $是连续的.

定义映射$ F_\rho (y):S^{N-1} =\{y\in {{\Bbb R}} ^N;| y| =1\}\rightarrow \Sigma _+ $

$ F_\rho (y)=\frac{(u_{{\rm loc}} +s(f, g, \varepsilon, \rho y)Bu_{\varepsilon, \rho y} (x), v_{{\rm loc}} +s(f, g, \varepsilon, \rho y)Du_{\varepsilon, \rho y} (x))} {\| (u_{{\rm loc}} +s(f, g, \varepsilon, \rho y)Bu_{\varepsilon, \rho y} (x), v_{{\rm loc}} +s(f, g, \varepsilon, \rho y)Du_{\varepsilon, \rho y} (x))\| _E}, $

其中$ e=\rho y\in \Omega^ \eta _1\backslash \{0\}=\{x\in \Omega _1;dist(x, \partial \Omega _1)>2\eta \} \backslash\{0\} $由引理3.1给出.

引理3.4  假设$ f(x)\geq 0, g(x)\geq 0, f(x), g(x)\in L^p( \Omega ), p>\frac{N}{2}, $$ 0<\| (f(x), g(x))\| _{E^*} \leq d_3 $$ (d_3 $由引理3.3给出), 非负函数$ a(x) $和$ b(x) $满足条件$ (A_1) $和$ (A_2) $. 那么对于任意的$ 0<\varepsilon <\varepsilon _1 $$ ( \varepsilon _1 $由引理3.3给出), 有

$ J_{f, g} (F_\rho (y))<I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2} }. $

证  由$ J_{f, g} (u, v)=I_{f, g} (t_{f, g} (u, v)(u, v)) $和引理3.3可知

$ \begin{eqnarray*} J_{f, g} (F_\rho(y)) &=&J_{f, g} \bigg(\frac{(u_{{\rm loc}} +s(f, g, \varepsilon, \rho y)Bu_{\varepsilon, \rho y }(x), v_{{\rm loc}} +s(f, g, \varepsilon, \rho y)Du_{\varepsilon, \rho y }(x))} {\| (u_{{\rm loc}} +s(f, g, \varepsilon, \rho y)Bu_{\varepsilon, \rho y }(x), v_{{\rm loc}} +s(f, g, \varepsilon, \rho y)Du_{\varepsilon, \rho y} (x))\| _E}\bigg)\\ &=& I_{f, g} (u_{{\rm loc}} +s(f, g, \varepsilon, \rho y)Bu_{\varepsilon, \rho y} (x), v_{{\rm loc}} +s(f, g, \varepsilon, \rho y)Du_{\varepsilon, \rho y }(x)) . \end{eqnarray*} $

根据引理3.1, 有

$ J_{f, g} (F_\rho (y))<I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N} (\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2}} . $

引理3.4得证.

引理3.5  假设$ f(x)\geq 0, g(x)\geq 0, f(x), g(x)\in L^p( \Omega ), p>\frac{N}{2}, $$ 0<\| (f(x), g(x))\| _{E^*}\leq d_3, $非负函数$ a(x) $和$ b(x) $满足条件$ (A_1) $和$ (A_2). $那么对于任意的$ 0<\varepsilon <\varepsilon _1, $存在$ \delta _1(x)>0, $使得

$ F_\rho (S^{N-1} )\subset\bigg [J_{f, g} \leq I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N} (\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2}} -\delta _1(\varepsilon )\bigg] . $

证  因为$ S^{N-1} $是紧的, $ F_\rho (y) $是连续的, 所以$ F_\rho (S^{N-1} ) $也是紧的. 由引理3.4, 对于任意的$ (u, v)\in F_\rho (S^{N-1} ), $有

$ J_{f, g} (u, v)<I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2} }. $

于是, 存在$ \delta (u, v)>0, $使得

(3.13) $ \begin{eqnarray} J_{f, g} (u, v) \leq I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2}} -\delta (u, v) . \end{eqnarray} $

取$ \delta (u, v)=\frac{1}{2}(I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2}} -J_{f, g} (u, v)) $. 根据引理2.8的(ⅰ)可得$ \delta (u, v)\in C^1( \Sigma _+, {{\Bbb R}} ). $因为$ F_\rho (S^{N-1} ) $是紧的, 所以存在$ (u_0, v_0)\in F_\rho (S^{N-1} ), $使得

$ \begin{eqnarray*} \min\limits_{(u, v)\in F_\rho (S^{N-1} ) }\delta (u, v) &=&\min\limits_{(u, v)\in F_\rho (S^{N-1} ) } \frac{1}{2}\bigg(I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2} } -J_{f, g} ( u, v )\bigg)\\ &= &\frac{1}{2}\bigg(I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2} } -J_{f, g} (u_0, v_0 )\bigg) >0 . \end{eqnarray*} $

由(3.13)式可知

$ \begin{eqnarray*} J_{f, g} (u, v)&\leq &I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2}} -\min\limits_{(u, v)\in F_\rho (y) }\delta (u, v) \\ &= &I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2}} -\frac{1}{2}\bigg(I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2}} -J_{f, g} (u_0, v_0)\bigg). \end{eqnarray*} $

取$ \delta _1(\varepsilon )=\frac{1}{2}(I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta} }{2}) ^{ \frac{N}{2}} -J_{f, g} (u_0, v_0)). $那么, 对于$ 0<\| (f(x), g(x))\| _{E^*}\leq d_3, $非负函数$ a(x) $和$ b(x) $满足条件$ (A_1) $和$ (A_2), $$ 0<\varepsilon <\varepsilon _1, $存在$ \delta _1(\varepsilon )>0, $使得

$ F_\rho (S^{N-1} )\subset \bigg[J_{f, g} \leq I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N} (\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2} }-\delta _1(\varepsilon )\bigg] . $

引理3.5得证.

其次, 我们构造映射$ G $.

引理3.6  假设非负函数$ a(x) $和$ b(x) $满足条件$ (A_1), $那么存在$ \delta _0>0, $当$ J_{0, 0} (u, v)\leq \frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2}}+\delta _0 $时, 有

$ \int _{ \Omega }\frac{x}{| x| }(|\nabla u|^2+|\nabla v|^2){\rm d}x\not= 0. $

证  由引理2.1可知: $ \inf\limits_{(u, v)\in \Sigma _+ }J_{0, 0} (u, v)=\frac{2}{N}(\frac{S_{\alpha, \beta} }{2}) ^{ \frac{N}{2}} $是不可达的. 由引理2.10的(ⅱ)可知: 对于任意的$ R\ge 0 $, 存在$ \sigma =\sigma (R)>0, $若$ (u, v)\in \Sigma _+ $满足

$ J_{0, 0} (u, v) \leq \frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2}} +\sigma, $

$ \| J'_{0, 0} (u, v)\| _{T^*_{(u, v)} \Sigma _+ }\leq \sigma, $

那么, 存在$ y\in \Omega $ (不妨设$ y\not= 0 ), $$ 0<\varepsilon <\frac{1}{R} $满足$ \frac{\varepsilon }{dist(y, \partial \Omega )}<\frac{1}{R}, $使得

$ \left\| u-\frac{(\frac{B^2+D^2}{2B^\alpha D^\beta })^{\frac{N-2}{4} }BU_{\varepsilon, y }} {\sqrt{\|(\frac{B^2+D^2}{2B^\alpha D^\beta })^{\frac{N-2}{4}} BU_{\varepsilon, y} \|^2_{D^{1, 2} ({{\Bbb R}} ^N)} +\| (\frac{B^2+D^2}{2B^\alpha D^\beta })^{\frac{N-2}{4}} DU_{\varepsilon, y }\|^2_{D^{1, 2} ({{\Bbb R}} ^N) }}}\right\|_{D^{1, 2} ({{\Bbb R}} ^N) } \leq \frac{1}{R}, $

$ \left\|v-\frac{(\frac{B^2+D^2}{2B^\alpha D^\beta })^{\frac{N-2}{4}} DU_{\varepsilon, y} } {\sqrt{\|(\frac{B^2+D^2}{2B^\alpha D^\beta })^{\frac{N-2}{4}} BU_{\varepsilon, y} \|^2_{D^{1, 2} ({{\Bbb R}} ^N)} +\|( \frac{B^2+D^2}{2B^\alpha D^\beta })^{\frac{N-2}{4} }DU_{\varepsilon, y} \|^2_{D^{1, 2} ({{\Bbb R}} ^N) }}}\right\| _{D^{1, 2} ({{\Bbb R}} ^N) }\leq \frac{1}{R} . $

即

(3.14) $ \begin{equation} \left\| u-\frac{B}{\sqrt{B^2+D^2}}\frac{U_{\varepsilon, y} }{\| U_{\varepsilon, y} \| _{D^{1, 2} ({{\Bbb R}} ^N) }}\right\| _{D^{1, 2} ({{\Bbb R}} ^N) }\leq \frac{1}{R}, \end{equation} $

(3.15) $ \begin{equation} \left \| v-\frac{D}{\sqrt{B^2+D^2}}\frac{U_{\varepsilon, y} }{\| U_{\varepsilon, y} \| _{D^{1, 2} ({{\Bbb R}} ^N)} }\right\| _{D^{1, 2} ({{\Bbb R}} ^N)}\leq \frac{1}{R} . \end{equation} $

假设$ (u, v)\in \Sigma _+ $且有$ J_{0, 0} (u, v)\leq \frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2} }+\delta _0. $由Ekeland变分原理, 存在$ (\tilde{u}, \tilde{v})\in \Sigma _+, $使得$ \| (\tilde{u}, \tilde{v})-(u, v)\| _E\leq \sqrt{\delta _0}, \| J'_{0, 0} (\tilde{u}, \tilde{v})\| _{T^*_{(u, v)} \Sigma _+ }\leq \sqrt{\delta _0}, J_{0, 0} (\tilde{u}, \tilde{v}) \leq \frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2}} +\delta _0. $取$ \delta _0\leq \{\sigma (2R^2), \sigma (2R), \frac{1}{4R^2}\}. $根据(3.14)和(3.15)式有: 存在$ y\in \Omega $ (不妨设$ y\not= 0 ), $$ 0<\varepsilon <\frac{1}{R}, $使得

$ \begin{eqnarray*} && \bigg \| u-\frac{B}{\sqrt{B^2+D^2}}\frac{U_{\varepsilon, y} }{\| U_{\varepsilon, y} \| _{D^{1, 2} ({{\Bbb R}} ^N) }}\bigg\| _ {D^{1, 2} ({{\Bbb R}} ^N)} \\ &\leq &\| u-\tilde{u}\| _{D^{1, 2} ({{\Bbb R}} ^N) }+\bigg\|\tilde{ u } -\frac{B}{\sqrt{B^2+D^2}} \frac{U_{\varepsilon, y} }{\| U_{\varepsilon, y} \| _{D^{1, 2} ({{\Bbb R}} ^N)} }\bigg\| _{D^{1, 2} ({{\Bbb R}} ^N)} \\ & \leq& \sqrt{\delta _0}+\frac{1}{2R}\leq \frac{1}{R} . \end{eqnarray*} $

类似地, 我们有

$ \bigg\| v-\frac{D}{\sqrt{B^2+D^2}}\frac{U_{\varepsilon, y} } {\| U_{\varepsilon, y} \| _{D^{1, 2} ({{\Bbb R}} ^N) }}\bigg\| _{D^{1, 2} ({{\Bbb R}} ^N)}\leq \frac{1}{R} . $

于是

$ \begin{eqnarray*} && \bigg | \int _{ \Omega }\frac{x}{| x| }\bigg[|\nabla u|^2+|\nabla v|^2-\frac{|\nabla U_{\varepsilon, y} |^2} {\| U_{\varepsilon, y} \|^2_{D^{1, 2} ({{\Bbb R}} ^N)} }\bigg]{\rm d}x\bigg| \\ &\leq& \int _{ \Omega }\bigg| |\nabla u|^2+|\nabla v|^2-\frac{|\nabla U_{\varepsilon, y} |^2} {\| U_{\varepsilon, y} \|^2_{D^{1, 2} ({{\Bbb R}} ^N) }}\bigg| {\rm d}x\\ & \leq &\int _{{{\Bbb R}} ^N }\bigg| |\nabla u|^2-\frac{B^2}{B^2+D^2}\frac{| \nabla U _{\varepsilon, y} |^2} {\| U_{\varepsilon, y} \|^2_{D^{1, 2} ({{\Bbb R}} ^N)} }\bigg| {\rm d}x +\int _{{{\Bbb R}} ^N }\bigg| |\nabla v|^2-\frac{D^2}{B^2+D^2}\frac{|\nabla U _{\varepsilon, y} |^2} {\| U_{\varepsilon, y} \|^2_{D^{1, 2} ({{\Bbb R}} ^N)} }\bigg| {\rm d}x\\ & =&\int _{{{\Bbb R}} ^N }\bigg| |\nabla u | -\frac{B}{\sqrt{B^2+D^2}} \frac{| \nabla U _{\varepsilon, y} | }{\| U_{\varepsilon, y} \| _{D^{1, 2} ({{\Bbb R}} ^N) }} \bigg |\bigg (| \nabla u | +\frac{B}{\sqrt{B^2+D^2}}\frac{| \nabla U _{\varepsilon, y} | } {\| U_{\varepsilon, y} \| _{D^{1, 2} ({{\Bbb R}} ^N) }}\bigg){\rm d}x \\ &&+\int _{{{\Bbb R}} ^N }\bigg| |\nabla v| -\frac{D}{\sqrt{B^2+D^2}}\frac{|\nabla U _{\varepsilon, y} | } {\| U_{\varepsilon, y} \| _{D^{1, 2} ({{\Bbb R}} ^N) }}\bigg| \bigg(|\nabla v| +\frac{D}{\sqrt{B^2+D^2}} \frac{|\nabla U _{\varepsilon, y} | }{\| U_{\varepsilon, y} \| _{D^{1, 2} ({{\Bbb R}} ^N)} }\bigg){\rm d}x \\ &\leq& \int _{{{\Bbb R}} ^N } \bigg|\nabla \bigg(u-\frac{B}{\sqrt{B^2+D^2}}\frac{ U_{\varepsilon, y} } {\| U_{\varepsilon, y} \| _{D^{1, 2} ({{\Bbb R}} ^N) }}\bigg)\bigg| \bigg (| \nabla u | +\frac{B}{\sqrt{B^2+D^2}} \frac{|\nabla U _{\varepsilon, y} | }{\| U_{\varepsilon, y} \| _{D^{1, 2} ({{\Bbb R}} ^N) }}\bigg){\rm d}x\\ && +\int _{{{\Bbb R}} ^N }\bigg| \nabla \bigg(v-\frac{D}{\sqrt{B^2+D^2}}\frac{ U_{\varepsilon, y} } {\| U_{\varepsilon, y} \| _{D^{1, 2} ({{\Bbb R}} ^N) }}\bigg)\bigg| \bigg (|\nabla v| +\frac{D}{\sqrt{B^2+D^2}} \frac{| \nabla U _{\varepsilon, y} | }{\| U_{\varepsilon, y} \| _{D^{1, 2} ({{\Bbb R}} ^N)} }\bigg){\rm d}x \\ &\leq &\bigg(\int _{{{\Bbb R}} ^N }\bigg| \nabla \bigg(u-\frac{B}{\sqrt{B^2+D^2}}\frac{U_{\varepsilon, y} } {\| U_{\varepsilon, y} \| _{D^{1, 2} ({{\Bbb R}} ^N) }}\bigg)\bigg|^2{\rm d}x\bigg)^{\frac{1}{2}}\\ &&\times \bigg (\int _{{{\Bbb R}} ^N } \bigg(|\nabla u | +\frac{B}{\sqrt{B^2+D^2}} \frac{| \nabla U _{\varepsilon, y} | }{\| U_{\varepsilon, y} \| _{D^{1, 2} ({{\Bbb R}} ^N) }} \bigg)^2{\rm d}x \bigg)^{\frac{1}{2}}\\ &&+\bigg(\int _{{{\Bbb R}} ^N } \bigg| \nabla \bigg(v-\frac{D}{\sqrt{B^2+D^2}}\frac{U_{\varepsilon, y} } {\| U_{\varepsilon, y} \| _{D^{1, 2} ({{\Bbb R}} ^N) }}\bigg)\bigg|^2{\rm d}x\bigg)^{\frac{1}{2}}\\ &&\times \bigg (\int _{{{\Bbb R}} ^N } \bigg(|\nabla v| +\frac{D}{\sqrt{B^2+D^2}}\frac{|\nabla U _{\varepsilon, y} | } {\| U_{\varepsilon, y} \| _{D^{1, 2} ({{\Bbb R}} ^N)} }\bigg)^2{\rm d}x\bigg )^{\frac{1}{2}}\\ & \leq& 4\bigg(\int _{{{\Bbb R}} ^N } \bigg| \nabla\bigg (u-\frac{B}{\sqrt{B^2+D^2}}\frac{U_{\varepsilon, y} } {\| U_{\varepsilon, y} \| _{D^{1, 2} ({{\Bbb R}} ^N) }}\bigg)\bigg|^2{\rm d}x\bigg)^{\frac{1}{2}}\\ && +4\bigg(\int _{{{\Bbb R}} ^N }\bigg|\nabla\bigg (v-\frac{D}{\sqrt{B^2+D^2}}\frac{U_{\varepsilon, y} } {\| U_{\varepsilon, y} \| _{D^{1, 2} ({{\Bbb R}} ^N) }}\bigg)\bigg|^2{\rm d}x\bigg)^{\frac{1}{2}}\\ & \leq &\frac{8}{R} . \end{eqnarray*} $

因此

$ \bigg| \int _{ \Omega }\frac{x}{| x| }\frac{| \nabla U _{\varepsilon, y} |^2} {\| U_{\varepsilon, y} \|^2_{D^{1, 2} ({{\Bbb R}} ^N) }}{\rm d}x\bigg| -\frac{8}{R} \leq \bigg | \int _{ \Omega }\frac{x}{| x| }(|\nabla u|^2+|\nabla v|^2){\rm d}x\bigg| . $

由文献[3, 引理3.2.6], 我们得到

$ \bigg | \int _{ \Omega }\frac{x}{| x| }\frac{|\nabla U _{\varepsilon, y} |^2} {\| U_{\varepsilon, y} \|^2_{D^{1, 2} ({{\Bbb R}} ^N) }}{\rm d}x\bigg|\geq 1-c\varepsilon^{(w-1)N+3w-2} -c\varepsilon^{(1-w)(N-2) }-c(\frac{1}{R})^{ N-2 }, $

其中$ \frac{N+2}{N+3}<w<1. $从而

$ 1-c\varepsilon^{ (w-1)N+3w-2} -c\varepsilon^{ (1-w)(N-2) } -c(\frac{1}{R})^{ N-2} -\frac{8}{R} \leq \bigg | \int _{ \Omega }\frac{x}{| x| }(|\nabla u|^2+|\nabla v|^2){\rm d}x\bigg|, $

其中$ \frac{N+2}{N+3}<w<1. $故结论成立.

引理3.7  假设$ (f(x), g(x))\in E^*, \| (f, g)\| _{E^*} \leq d_3 $$ ( d_3 $由引理3.3给出), 非负函数$ a(x) $和$ b(x) $满足条件$ (A_1). $那么存在$ d_4\in (0, d_3], $当$ \| (f, g)\| _{E^*} \leq d_4 $时, 有

$ \bigg [J_{f, g} <I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N} (\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2}} \bigg]\subset \bigg[J_{0, 0} <\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2}} +\delta _0\bigg], $

其中$ \delta _0 $由引理3.6给出.

证  由引理2.3的(ⅰ)可知: 对于任意的$ \varepsilon \in (0, 1) $和$ (u, v)\in \Sigma _+, $有

$ J_{0, 0} (u, v)\leq (1-\varepsilon ) ^{-\frac{N}{2}} \bigg (J_{f, g} (u, v)+\frac{1}{2\varepsilon }\| (f, g)\| _{E^*} \bigg), $

注意到$ I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )\leq 0 $, 则有

(3.16) $ \begin{eqnarray} J_{0, 0} (u, v)&<&(1-\varepsilon )^{-\frac{N}{2} }\bigg(I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+ \frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2}} +\frac{1}{2\varepsilon }\| (f, g)\| _{E^*} \bigg){}\\ &= &\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2} }+(1-\varepsilon )^{-\frac{N}{2}} \frac{1}{2\varepsilon }\| (f, g)\| _{E^*} +[(1-\varepsilon )^{-\frac{N}{2}} -1]\frac{2}{N} (\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2} }\\ &&+ (1-\varepsilon )^{-\frac{N}{2}} I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )\\ &\leq& \frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2}} +(1-\varepsilon )^{-\frac{N}{2}} \frac{1}{2\varepsilon }\| (f, g)\| _{E^*} +[(1-\varepsilon )^{-\frac{N}{2}} -1] \frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2} } . \end{eqnarray} $

存在$ \varepsilon _3>0, $当$ 0<\varepsilon <\varepsilon _3 $时, 有

(3.17) $ \begin{eqnarray} [(1-\varepsilon )^{ -\frac{N}{2}} -1]\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2}} <\frac{\delta _0}{2} . \end{eqnarray} $

于是, 对任意$ \varepsilon >0 $满足$ 0<\varepsilon <\varepsilon _3, $取$ d_4\in (0, d_3], $使得当$ \| (f, g)\| _{E^*}\leq d_4 $时, 有

(3.18) $ \begin{eqnarray} (1-\varepsilon )^{ -\frac{N}{2}} \frac{1}{2\varepsilon }\| (f, g)\|^2_{E^*} <\frac{\delta _0}{2} . \end{eqnarray} $

由(3.16)–(3.18)式可得结论是成立的.

定义映射

$ G:\bigg[J_{f, g} <I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2} }\bigg]\rightarrow S^{N-1}, $

$ G(u, v)=\int _{ \Omega }\frac{x}{| x| }[|\nabla u|^2+|\nabla v|^2]{\rm d}x/ \bigg | \int _{ \Omega }\frac{x}{| x| }[|\nabla u|^2+|\nabla v|^2]{\rm d}x\bigg| . $

由引理3.6和3.7可知: 当$ \| (f, g)\| _{E^*} \leq d_4 $时, 对于任意的$ (u, v)\in [J_{f, g} <I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{\frac{N}{2} }], $有

$ \int _{ \Omega }\frac{x}{| x| }[|\nabla u|^2+|\nabla v|^2]{\rm d}x\not= 0. $

所以$ G(u, v) $是有意义的.

引理3.8  假设$ f(x) \geq 0, g(x) \geq0, $$ f(x), g(x)\in L^p( \Omega ), p>\frac{N}{2}, $$ (f(x), g(x))\not\equiv 0, $非负函数$ a(x) $和$ b(x) $满足条件$ (A_1) $和$ (A_2) $. 那么, 对于充分小的$ \varepsilon >0, $充分小的$ \| (f, g)\| _{E^*} $和$ a.e\ \rho y\in \Omega ^ \eta _1\backslash \{0\}=\{x\in \Omega _1;dist(x, \partial \Omega _1)>2\eta \}\backslash \{0\}, $有

$ G\circ F_\rho :S^{N-1} \rightarrow S^{N-1} ;y\mapsto G(F_\rho (y)) $

与恒等映射同伦.

证  假设$ \gamma (\theta ), \theta \in [\theta _1, \theta _2] $是$ S^{N-1} $上$ G(\frac{(Bu_{\varepsilon, \rho y }, Du_{\varepsilon, \rho y} )} {\| (Bu_{\varepsilon, \rho y }, Du_{\varepsilon, \rho y} )\| _E}) $和$ G(F_\rho (y)) $之间的正规测地线, 其中$ \gamma (\theta _1)=G(\frac{(Bu_{\varepsilon, \rho y}, Du_{\varepsilon, \rho y })} {\| (Bu_{\varepsilon, \rho y}, Du_{\varepsilon, \rho y })\| _E}), $$ \gamma (\theta _2)=G(F_\rho (y)). $

令

$ \bar{ \gamma }(\theta _1)=\gamma (2(\theta _1-\theta _2)\theta +\theta _2), \quad \theta \in [0, \frac{1}{2}] . $

定义

$ \zeta (\theta, y):[0, 1]\times S^{N-1} \rightarrow S^{N-1}, $

$ \zeta (\theta, y)=\left\{\begin{array}{ll} \bar{\gamma}(\theta ), & { } \theta \in [0, \frac{1}{2}), \\ { } G\bigg(\frac{(Bu_{2(1-\theta )\varepsilon, \rho y }(x), Du_{2(1-\theta )\varepsilon, \rho y }(x))}{ \| (Bu_{2(1-\theta )\varepsilon, \rho y }(x), Du_{2(1-\theta )\varepsilon, \rho y }(x))\| _E}\bigg), \quad & { } \theta \in [ \frac{1}{2}, 1), \\ y, &\theta =1. \end{array}\right. $

首先, $ \zeta (\theta, y) $是有意义的. 事实上, 由于非负函数$ a(x) $和$ b(x) $满足条件$ (A_1) $, 由(2.8)式, 我们得到: 对于任意的$ e\in \Omega^ \eta _1, $当$ \varepsilon \rightarrow 0+ $时, 有

$ J_{0, 0} \bigg(\frac{(Bu_{\varepsilon, e}, Du_{\varepsilon, e} )} {\| (Bu_{\varepsilon, e}, Du_{\varepsilon, e} )\| _E}\bigg) \leq \frac{2}{N}(\frac{S_{\alpha, \beta} }{2}) ^{ \frac{N}{2}} + o(1) . $

于是, 对于充分小的$ \varepsilon >0 $和任意的$ e\in \Omega ^\eta _1, $有

$ J_{0, 0} \bigg(\frac{(Bu_{\varepsilon, e}, Du_{\varepsilon, e} )}{\| (Bu_{\varepsilon, e}, Du_{\varepsilon, e} )\| _E}\bigg) \leq \frac{2}{N}(\frac{S_{\alpha, \beta} }{2})^{ \frac{N}{2} } + \delta_0, $

其中$ \delta_0 $由引理3.6给出, 所以$ \zeta (\theta, y) $是有意义的.

其次, 我们将证明$ \zeta (\theta, y)\in C([0, 1]\times S^{N-1}, S^{N-1} ), $事实上

$ \lim _{\theta \rightarrow \frac{1}{2} }\bar{\gamma }(\theta )=\gamma (\theta _1)= G\bigg(\frac{(Bu_{\varepsilon, \rho y }, Du_{\varepsilon, \rho y })} {\| (Bu_{\varepsilon, \rho y}, Du_{\varepsilon, \rho y} )\| _E}\bigg)=\zeta (\frac{1}{2}, y) . $

我们断言

(3.19) $ \begin{eqnarray} \lim _{\theta \rightarrow 1 }G \bigg(\frac{(Bu_{2(1-\theta )\varepsilon, \rho y} (x), Du_{2(1-\theta )\varepsilon, \rho y} (x))} {\| (Bu_{2(1-\theta )\varepsilon, \rho y }(x), Du_{2(1-\theta )\varepsilon, \rho y} (x))\| _E}\bigg)=y . \end{eqnarray} $

由文献[3]中的引理3.2.8可知: 对于任意的$ e\in \Omega ^\eta _1\setminus \{0\}, $当$ \varepsilon \rightarrow 0+ $时, 有

$ \bigg | \int _{ \Omega }\bigg(\frac{x}{| x| }-\frac{e}{| e| }\bigg)\frac{|\nabla u _{\varepsilon, e} |^2} {\| u_{\varepsilon, e} \|^2_E}{\rm d}x\bigg| =o(1) . $

于是, 对于任意的$ e\in \Omega ^ \eta _1\backslash \{0\}, $当$ \varepsilon \rightarrow 0+ $时, 有

$ \int _{ \Omega }\frac{x}{| x| }\frac{| \nabla u_{\varepsilon, e} |^2}{\| u_{\varepsilon, e} \|^2_E}{\rm d}x=\frac{e}{| e| }+o(1) . $

因此, 对于任意$ \rho y\in \Omega ^\eta _1\backslash\{0\}, | y| =1, $有

$ \lim _{\theta \rightarrow 1 }G \bigg(\frac{(Bu_{2(1-\theta )\varepsilon, \rho y }(x), Du_{2(1-\theta )\varepsilon, \rho y} (x))} {\| (Bu_{2(1-\theta )\varepsilon, \rho y} (x), Du_{2(1-\theta )\varepsilon, \rho y }(x))\| _E}\bigg) = \lim _{\theta \rightarrow 1 }\frac{\int _{ \Omega }\frac{x}{| x| }\frac{| \nabla u_{2(1-\theta )\varepsilon, \rho y} |^2} {\| u_{2(1-\theta )\varepsilon, \rho y} \|^2_E}{\rm d}x} {| \int _{ \Omega }\frac{x}{| x| }\frac{| \nabla u_{2(1-\theta )\varepsilon, \rho y }|^2}{\| u_{2(1-\theta )\varepsilon, \rho y } \|^2_E}{\rm d}x| } = \frac{\frac{\rho y}{| \rho y| }}{| \frac{\rho y}{| \rho y| }| } = y . $

所以(3.19)式成立. 根据$ \zeta (\theta, y) $的定义可得: 对于任意的$ y\in S^{N-1}, $当$ 0<\varepsilon <\varepsilon _2 $时, 有

$ \gamma (0, y)=G(F_\rho (y)), \; \; \; \; \gamma (1, y)=y . $

所以$ G\circ F_\rho $与恒等映射同伦.

定理1.1的证明  由引理2.11的(ⅱ), 引理3.8, 引理1.2和引理1.3可知$ J_{f, g} (u, v) $在

$ [J_{f, g} (u, v) <I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta} }{2}) ^{ \frac{N}{2}} ] $

中至少有两个临界点$ (\bar{u}_1, \bar{v}_1), (\bar{u}_2, \bar{v}_2). $令

$ (u_1, v_1)=t_{f, g} (\bar{u}_1, \bar{v}_1)(\bar{u}_1, \bar{v}_1 ), $

$ (u_2, v_2)=t_{f, g} (\bar{u}_2, \bar{v}_2)(\bar{u}_2, \bar{v}_2). $

根据引理2.3的(ⅱ), 我们可得

$ 0<I_{f, g} (u_i, v_i)<I_{f, g} (u_{{\rm loc}}, v_{{\rm loc}} )+\frac{2}{N}(\frac{S_{\alpha, \beta} }{2}) ^{\frac{N}{2}}, \quad i=1, 2. $

通过引理2.2的(ⅱ)和引理2.8的(ⅱ)我们得到$ (u_1, v_1), (u_2, v_2) $是问题(1.1)–(1.4)的两个能量大于零的正解.