20世纪50年代,Alexandrov创立了移动平面法并用其研究了 常平均曲率曲面嵌入问题的唯一性^[1]. 其后,经过Serrin^[2],Gidas,Ni 和Nirenberg^[3],Caffarelli,Gidas和 Spruck^[4],Li^[5],Chen和 Li^[6,7],Damascelli,Pacella和Ramaswamy^[8] 等学者的进一步发展和完善, 移动平面法已成为研究非线性椭圆方程解的对称性,单调性, 非存在性以及先验估计的有力工具.

Monge-Ampere方程是典型的完全非线性方程,利用移动平面方法 Delanoe^[9]证明了单位球上的一类 Monge-Ampere方程凸解的径向对称性. 对于一般的完全非线性椭圆方程Dirichlet边值问题 $$\begin{equation} \label{jinhu1} \left\{ \begin{array}{ll} F(x,u,u_{11},\cdots,u_{NN})=0,~~& x\in S=(-1,1)\times \Omega,\\ u=0,& x\in \partial S, \end{array} \right.(1.1) \end{equation}$$ 其中$\Omega$是${\Bbb R}^{N-1}$中的单位球. Li^[10]证明了: 若$u\in C^{2, \alpha}(\overline{S})$是问题(1.1)的正解,且(i) $F$在$x_1$ 和$\{p_{12},\cdots,p_{1N}\}$ 方向上是对称的; (ii) $F$在$x_1$方向上的左半区域是递增的, 即$F_{x_1}(x_1,y,u,$ $u_{11},\cdots,u_{NN})\geq0$,$-1<x_1<0$. 那么$u$在$x_1$方向上是对称的,并且只有一个极大值. 其后,Li^[11] 将该结果推广到无界区域上.

对于半线性椭圆方程组,前人也做了大量的工作,如文献^[]. 其中,Busca和Sirakov^[12]研究了半线性椭圆方程组 $$\begin{equation} \label{jinhu2} \left\{ \begin{array}{ll} \Delta u+g(u,v)=0,~~& x\in {\Bbb R}^N,\\ \Delta v+f(u,v)=0,& x\in {\Bbb R}^N, \end{array} \right.(1.2) \end{equation}$$ 具有衰减性质正解的对称性. 假设$(u,v)$是方程组(1.2)的正解,当$|x|\rightarrow\infty$时, $u(x)\rightarrow0$,$v(x)\rightarrow0$. 若$f$,$g$满足

(i)~ $f$,$g \in C^1([0,\infty)\times [0,\ \ \infty),{\Bbb R}),\ (u,v)\in [0,\infty)\times [0,\infty)$;

(ii)~ $\frac{\partial g}{\partial v}(u,v)\geq0$,$\frac{\partial f}{\partial u}(u,v)\geq0$;

(iii)~ $\frac{\partial g}{\partial u}(0,0)<0,\ \frac{\partial f}{\partial v}(0,0)<0$; (iv)~ $\det A>0,\ A=\left( \begin{array}{cc} \frac{\partial g}{\partial u} & \frac{\partial g}{\partial v} \\ \frac{\partial f}{\partial u} & \frac{\partial g}{\partial v} \\ \end{array} \right)(0,0)$. \\ 那么存在两点$x_0$,$x_1\in {\Bbb R}^N$,使得$u(x)=u(|x-x_0|),\ v(x)=v(|x-x_1|)$,并且对于$\forall r_1=|x-x_0|$ 和$\forall r_2=|x-x_1|$,分别有$\frac{{\rm d}u}{{\rm d}r_1}<0$和$\frac{{\rm d}v}{{\rm d}r_2}<0$成立.

本文通过结合上面介绍的完全非线性椭圆方程与半线性椭圆方程组解的对称性研究办法, 考虑如下完全非线性椭圆方程组解的对称性 $$\begin{equation} \label{jinhu3} \left\{ \begin{array}{lll} F(x,u,u_{11},\cdots,u_{NN})+g(u,v)=0,~~ & x\in S=(-1,1)\times \Omega,\\ G(x,v,v_{11},\cdots,v_{NN})+f(u,v)=0,& x\in S=(-1,1)\times \Omega,\\ u(x)=0,v(x)=0,&x\in \partial S, \end{array} \right.(1.3) \end{equation}$$ 其中,$\Omega$是${\Bbb R}^{N-1}$中的单位球, $F$和$G$满足一致椭圆型条件.

定义1.1 (一致椭圆型条件^[18]) 设偏微分算子 $$L=\sum_{i,j=1}^Na^{ij}(x)D_{ij}+\sum_{i=1}^Nb^i(x)D_i+c(x),a^{ij}=a^{ji},$$ 其中,$a_{ij}(x),b_i(x),c(x)$为定义在开集$\Omega\subset {\Bbb R}^N$上的函数. 若存在的正数$\theta$,使得 $$\sum_{i,j=1}^Na^{ij}(x)\xi_i\xi_j\ge \theta {\left| \xi \right|^2},\ \forall x \in \Omega,\ \xi \in {\Bbb R}^N,$$ 则称$L$是一致椭圆型算子.

对于方程组(1.3)中的$F$满足一致椭圆型条件是指存在正数$\theta$,使得 $$\begin{equation} \label{def:uelli} \sum\limits_{i,j=1}^N {{F_{{p_{ij}}}}\left( {x,u,p_{11},\cdots,p_{NN}} \right)} {\xi _i}{\xi _j} \ge \theta{\left| \xi \right|^2}(1.4) \end{equation}$$ 对所有的$\xi\in {\Bbb R}^N$均成立.

本文的主要定理如下.

{\heiti\bf 定理1.2} 设$F$,$G\in C^1(S,[0,\infty),{\Bbb R}^\frac{N\times(N+1)}{2})$,$f$,$g\in C^1([0,\infty)\times[0,\infty))$,当满足如下条件

(i) $${F_p}\left({-1,y,0,p_{11},\cdots,p_{NN}} \right) + \frac{{\partial g}}{{\partial u}}\left( {0,0} \right) < - {c_1},~c_1>0,~\forall p_{ij}\in {\Bbb R},\forall y\in \Omega;$$ $${G_s}\left({-1,y,0,s_{11},\cdots,s_{NN}} \right) + \frac{{\partial f}}{{\partial v}}\left( {0,0} \right) < - {c_2},~c_2>0,~\forall s_{ij}\in {\Bbb R},\forall y\in \Omega.$$

(ii)~ $F$关于$x_1$和$\left\{ {{p_{12}},\cdots ,{p_{1N}}} \right\}$对称,$G$关于$x_1$和$\left\{ {{s_{12}},\cdots ,{s_{1N}}} \right\}$对称,并且对于$\forall (x_1,y)\in S,\ x_1<0,\ \forall u,v>0,\ \forall {p_{ij}},{s_{ij}} \in {\Bbb R}$,$i=1,\cdots,N$, $j=i,\cdots,N$,有下式成立 $${F_{{x_1}}}\left( {{x_1},y,u,p_{11},\cdots,p_{NN}} \right) \ge 0;\ {G_{{x_1}}}\left( {{x_1},y,v,s_{11},\cdots,s_{NN}} \right) \ge 0.$$

(iii)~ 对于$u$,$v \in \left[{0,\infty } \right) \times \left[{0,\infty } \right)$, $\frac{{\partial g}}{{\partial v}} \left( {u,v} \right) \ge 0$, $\frac{{\partial f}}{{\partial u}}\left({u,v} \right) \ge 0.$

(iv)~ $\forall p_{ij},s_{ij}\in {\Bbb R}$,$\forall y,y'\in \Omega$, $$\left| \begin{array}{*{20}{cc}} F_p\left(-1,y,0,p_{11},\cdots,p_{NN} \right)+ \frac{\partial g}{\partial u}\left(0,0\right) ~~& \frac{\partial g}{\partial v}(0,0) \\[3mm] \frac{\partial f}{\partial u}\left(0,0\right) ~~& G_s\left(-1,y',0 ,s_{11},\cdots,s_{NN} \right)+\frac{\partial f}{\partial v}\left(0,0\right)\\ \end{array} \right| > 0.$$

那么对于方程组(1.3)的任意$C^2$正解$(u,v)$必须在$x_1$方向上对称, 且只有一个极大值,即对于$(x_1,y)\in S,\ - 1 <x_1< 0$,有 $$u(x_1,y) = u(-x_1,y ),\ u_1(x_1,y)>0,$$ $$v(x_1,y)= v(-x_1,y),\ v_1(x_1,y)>0.$$

上述完全非线性方程组的一个特殊情形为实椭圆型Monge-Ampere方程组. 在几何中可用于两个耦合曲面的预定高斯曲率问题^[19,20].

利用上述定理1.2,对于实Monge-Ampere方程组 $$\begin{equation} \label{corollary1} \left\{ \begin{array}{ll} -\det \left(D^2u \right) + g\left(u,v\right) = 0,~~& x \in S =(-1,1) \times (-1,1),\\ -\det \left(D^2v \right) + f\left(u,v \right) = 0,& x \in S =(-1,1) \times (-1,1),\\ u(x)=v(x)=0,& x\in \partial S,\\ u(x)>0,v(x)>0,& x\in S, \end{array} \right.(1.5) \end{equation}$$ 可以得到如下推论.

推论1.3 设

(i')~ $\frac{\partial g}{\partial u}(0,0)<-c_1$, $\frac{\partial f}{\partial v}(0,0)<-c_2$,$c_1,c_2>0$; (ii')~ 对于任意的$u,v\geq 0$有 $\frac{\partial g}{\partial v}(u,v) \geq0$, $\frac{\partial f}{\partial u}(u,v)\geq0$;

(iii')~ $\left| \begin{array}{cc} \frac{\partial g}{\partial u}\left(0,0\right) & \frac{\partial g}{\partial v}(0,0) \\ \frac{\partial f}{\partial u}\left(0,0\right) & \frac{\partial f}{\partial v}\left(0,0\right)\\ \end{array} \right| > 0.$ \\ 若$u,v\in {C^2}\left( S \right) \cap C\left( {\bar S} \right)$ 为方程组(1.5)的正的严格凹解, 则 $$u(x_1,y) = u(-x_1,y ),~~ u_1(x_1,y)>0,~~ \forall x_1<0,y\in (-1,1);$$ $$v(x_1,y)= v(-x_1,y),~~ v_1(x_1,y)>0,~~ \forall x_1<0,y\in (-1,1).$$

本节对要用到的引理进行引用和证明并为定理的证明做一些准备.

取$T_\lambda=\{x\in S|x_1=\lambda,\ -1 < \lambda \leq 0\}$ 是一个与$x_1$轴垂直的平面, $${\Sigma _\lambda }= \left\{ {x \in S\left| { - 1 < {x_1} < \lambda } \right.} \right\}$$ 是$S$中在平面$T_\lambda$左边的部分. 对于$-1<\lambda\leq0$, 定义$x^\lambda=(2\lambda-x_1,y)$是$(x_1,y)$关于$T_\lambda$的对称点. 我们比较解$u$和$v$在$\Sigma_\lambda$上某点的值与其对称点的值. 对于$x \in {\Sigma _\lambda }$,记 $${u^\lambda }\left( x \right) = u\left( {{x^\lambda }} \right),~~ {v^\lambda }\left( x \right) = v\left( {{x^\lambda }} \right);$$ $$\omega \left( {x,\lambda } \right) = {u^\lambda }\left( x \right) - u\left( x \right), ~~\varphi \left( {x,\lambda } \right) = {v^\lambda }\left( x \right) - v\left( x \right).$$

这些函数也可看成是定义在 $${Q_\mu } = \left\{ {\left( {{x_1},y,\lambda } \right) \in S \times [- 1,0]\left| { - 1 < {x_1} < \lambda < \mu } \right.} \right\}$$ 上的关于$(x,\lambda)$的函数. 注意到,我们只需证明解$u$,$v$关于$T_0$对称, 即$\omega(x,0)\equiv0$和$\varphi(x,0)\equiv0$.

记 $$\Sigma_{\lambda}^{\omega^-}=\{x\in \Sigma_\lambda|\omega(x,\lambda)<0\},~~ \Sigma_{\lambda}^{\varphi^-}=\{x\in \Sigma_\lambda|\varphi(x,\lambda)<0\}.$$

引理2.1 若 $F$,$G\in C^1(S,[0,\infty),{\Bbb R}^\frac{N\times(N+1)}{2})$满足定理1.2中条件(i), 则存在正数$\delta_1$,使得当$-1<\lambda<-1+\delta_1$时, 对任意$x\in \Sigma_\lambda^{\omega^-}$,有 $$\int_0^1 F_p\left(x,u(x)+t\omega(x),u_{ij}^{\lambda}(x)\right){\rm d}t+ \frac{\partial g}{\partial u}\left(\xi,v(x)\right)<0,$$ 其中,$\xi$为介于$u^{\lambda}(x)$和$u(x)$之间的任意值; 存在正数$\delta_2$, 使得当$-1<\lambda<-1+\delta_2$时,对任意$x\in \Sigma_\lambda^{\varphi^-}$,有 $$\int_0^1 G_s \left(x,v(x)+t\varphi(x),v_{ij}^{\lambda}(x)\right){\rm d}t + \frac{\partial f}{\partial v}\left(u(x),\eta \right) < 0,$$ 其中,$\eta$可取$v^{\lambda}(x)$和$v(x)$之间的任意值.

证 由于$F\in C^1(S,[0,\infty),{\Bbb R}^\frac{N\times(N+1)}{2})$且满足(i),因此存在$\epsilon>0$,使得当$-1<\lambda<-1+\epsilon$时,对任意$p_{ij}\in {\Bbb R}^\frac{N\times(N+1)}{2}$,$p>0,s>0,\xi>0$,$0<p<\epsilon$,$\xi+s<\epsilon$,及任意$x\in \Sigma_\lambda$, 有 $$F_p\left( {x,p,{p_{ij}}} \right) + \frac{{\partial g}}{{\partial u}}\left( {\xi,s} \right) < -c_1/2<0.$$ 由于当$x_1=-1$时,$u(x)=0$,$v(x)=0$,因此对上述$\epsilon>0$,存在$\delta_1>0 (\delta_1<\epsilon)$,使得 $-1< \lambda <-1+\delta_1$时有 $u(x)+v(x)<\epsilon$. 于是对于$x\in \Sigma_\lambda^{\omega^-}$, $0<u^\lambda(x)<u(x)$,此时有 $u(x)+t\omega(x)=(1-t)u(x)+tu^{\lambda}(x)<u(x)<\epsilon$,$\forall t\in (0,1)$. 对于任意$\xi\in (u^{\lambda}(x),u(x))$,有 $\xi+v(x)<u(x)+v(x)<\epsilon$. 故有,对$t\in (0,1)$ $$F_p\left(x,u(x)+t\omega(x),u_{ij}^{\lambda}(x)\right)+ \frac{\partial g}{\partial u}\left(\xi,v(x)\right)<-c_1/2.$$ 两边对$t$在$(0,1)$上积分得 $$\int_0^1 F_p\left(x,u(x)+t\omega(x),u_{ij}^{\lambda}(x)\right){\rm d}t+ \frac{\partial g}{\partial u}\left(\xi,v(x)\right)<0.$$

对于$G$的结果,证明类似.

引理2.2 设$F$,$G\in C^1(S,[0,\infty),{\Bbb R}^\frac{N\times(N+1)}{2})$,$f$,$g\in C^1([0,\infty)\times[0,\infty))$,满足定理1.2中的条件(iv), 则存在$\delta_3>0$,使得$-1<\lambda<-1+\delta_3$时, 对任意$x^*,x^{**}\in \Sigma_\lambda^{\omega^-}\cap \Sigma_{\lambda}^{\varphi^-}$,有 $$\alpha(\lambda)\delta(\lambda)-\beta(\lambda)\gamma(\lambda)>0,$$ 其中, $$\alpha(\lambda)=\int_0^1F_p(x^*,u(x^*)+t\omega(x^*),u_{ij}^{\lambda}(x^*)){\rm d}t+\frac{\partial g}{\partial u}(\xi_1,v(x^*)),$$ $$\delta(\lambda)=\int_0^1G_p(x^{**},v(x^{**})+t\varphi(x^{**}), v_{ij}^{\lambda}(x^{**})){\rm d}t+\frac{\partial f}{\partial v}(u(x^{**}),\eta_2),$$ $$\beta(\lambda)= \frac{\partial g}{\partial v}(u^\lambda(x^*),\eta_1),\gamma(\lambda)=\frac{\partial f}{\partial u}(\xi_2,v(x^{**}));$$ $\xi_1$可取介于$u(x^*)$和$u^{\lambda}(x^{*})$之间的任意值, $\xi_2$可取介于$u(x^{**})$和$u^{\lambda}(x^{**})$之间的任意值, $\eta_1$可取介于$v(x^*)$和$v^{\lambda}(x^{*})$之间的任意值, $\eta_2$可取介于$v(x^{**})$和$v^{\lambda}(x^{**})$之间的任意值.

证 记 $$\begin{eqnarray*} & & A(x,\tilde{x},p,p',\tilde{p},p_{ij},s,s',\tilde{s},s_{ij},\xi_1,\xi_2,\eta_1,\eta_2)\\ & =& \left| \begin{array}{cc} F_p(x,p,p_{ij})+\frac{\partial g}{\partial u}(\xi_1,s)~~ & \frac{\partial g}{\partial v}(p',\eta_1)\\[3mm] \frac{\partial f}{\partial u}(\xi_2,\tilde{s})~~ & G_p(\tilde{x},s',s_{ij})+\frac{\partial f}{\partial v}(\tilde{p},\eta_2) \end{array} \right|. \end{eqnarray*}$$ 由(iv)可知,存在$\epsilon>0$,使得当$-1<\lambda<<-1+\epsilon$时, 对任意$x,\tilde{x}\in \Sigma_{\lambda}$,$p>0$,$s>0$,$\tilde{p}>0$, $\tilde{s}>0$,$\xi_i>0$,$\eta_i>0$ $(i=1,2)$, $p+s<\epsilon$,$\tilde{p}+\tilde{s}<\epsilon$,$0<p'<p<\epsilon$, $0<s'<s<\epsilon$,$\xi_1+s<\epsilon$,$\xi_2+\tilde{s}<\epsilon$, $\eta_1+p<\epsilon$,$\eta_2+\tilde{p}<\epsilon$,有 $$A(x,\tilde{x},p,p',\tilde{p},p_{ij},s,s',\tilde{s},s_{ij}, \xi_1,\xi_2,\eta_1,\eta_2)>0,~~\forall p_{ij},s_{ij}\in {\Bbb R}.$$ 对于上述$\epsilon$,存在$\delta_3>0$,使得$-1<\lambda<-1+\delta_3$时,对任意$x^*,x^{**}\in \Sigma_\lambda^{\omega^-}\cap \Sigma_{\lambda}^{\varphi^-}$,有 $u(x^*)+v(x^*)<\epsilon$,$u(x^{**})+v(x^{**})<\epsilon$,且$0<u^\lambda(x^*)<u(x^*)<\epsilon$,$v^{\lambda}(x^{**})<v(x^{**})$,$0<v(x^{**})+t_2\varphi(x^{**})=t_2v^{\lambda}(x^{**})+(1-t_2)v(x^{**})<v(x^{**})<\epsilon$,$\xi_1+v(x^*)<u(x^*)+v(x^*)<\epsilon$,$\xi_2+v(x^{**})<u(x^{**})+v(x^{**})<\epsilon$,$\eta_1+u(x^*)<v(x^*)+u(x^*)<\epsilon$,$\eta_2+u(x^{**})<v(x^{**})+u(x^{**})<\epsilon$. 于是对于 $\forall t_1,t_2\in (0,1)$, $$\begin{eqnarray} \nonumber 0&<& A(x^*,x^{**},u(x^*)+t_1\omega(x^*),u^\lambda(x^*),u(x^{**}),u_{ij}^{\lambda}(x^*),\\ \nonumber & & v(x^{*}),v(x^{**})+t_2\varphi(x^{**}),v(x^{**}),v_{ij}^{\lambda}(x^{**}),\xi_1,\xi_2,\eta_1,\eta_2)\\ \nonumber &=&\left(F_p(x^*,u(x^*)+t_1\omega(x^*),u_{ij}^{\lambda}(x^*))+\frac{\partial g}{\partial u}(\xi_1,v(x^*))\right)\\ \nonumber & & \cdot \left(G_p(x^{**},v(x^{**})+t_2\varphi(x^{**}),v_{ij}^{\lambda}(x^{**}))+\frac{\partial f}{\partial v}(u(x^{**}),\eta_2)\right) \\ \label{eq:a} & & -\frac{\partial g}{\partial v}(u^\lambda(x^*),\eta_1)\cdot\frac{\partial f}{\partial u}(\xi_2,v(x^{**})).(2.1) \end{eqnarray}$$ 注意到当$-1<\lambda<-1+\delta_3$时,对任意$x^*,x^{**}\in \Sigma_\lambda^{\omega^-}\cap \Sigma_{\lambda}^{\varphi^-}$,有 $$\begin{eqnarray*} & & \alpha(\lambda)\delta(\lambda)-\beta(\lambda)\gamma(\lambda)\\ &=& \left(\int_0^1F_p(x^*,u(x^*)+t\omega(x^*),u_{ij}^{\lambda}(x^*)){\rm d}t+\frac{\partial g}{\partial u}(\xi_1,v(x^*))\right)\\ & & \cdot \left(\int_0^1G_p(x^{**},v(x^{**})+t\varphi(x^{**}),v_{ij}^{\lambda}(x^{**})){\rm d}t+\frac{\partial f}{\partial v}(u(x^{**}),\eta_2)\right)\\ & & -\frac{\partial g}{\partial v}(u^\lambda(x^*),\eta_1)\cdot \frac{\partial f}{\partial u}(\xi_2,v(x^{**}))\\ &=& \int_0^1\int_0^1\left[\left(F_p(x^*,u(x^*)+t_1\omega(x^*),u_{ij}^{\lambda}(x^*))+\frac{\partial g}{\partial u}(\xi_1,v(x^*))\right)\right.\\ & & \qquad \qquad \cdot \left(G_p(x^{**},v(x^{**})+t_2\varphi(x^{**}),v_{ij}^{\lambda}(x^{**})){\rm d}t +\frac{\partial f}{\partial v}(u(x^{**}),\eta_2)\right)\\ & & \qquad \qquad \left.-\frac{\partial g}{\partial v}(u^\lambda(x^*), \eta_1)\cdot \frac{\partial f}{\partial u}(\xi_2,v(x^{**}))\right]{\rm d}t_1{\rm d}t_2\\ &>& 0. \end{eqnarray*}$$

定理1.2的证明要用到角点区域的Hopf引理,其证明见文献^[3].

引理2.3 (角点区域的Hopf引理,参见文献[10,Lemma A.1]或^[3])

设$\Omega$是${\Bbb R}^N$中的区域,原点在其边界上, 并且在原点附近的边界包含两个横向分割的$C^2$超曲面$\rho=0$, $\sigma=0$. 在$\Omega$中有$\rho<0$,$\sigma<0$. 并设$u\in C^2$是微分不等式 $Lu = \sum\limits_{i,j} {{a_{ij}}\left( x \right)} {u_{ij}}\left( x \right) + \sum\limits_i {{b_i}} \left( x \right){u_i}\left( x \right) + c\left( x \right)u\left( x \right) \le 0$ 在$\overline{\Omega}$中的正解. 其中$a_{ij}\in C(\overline{\Omega})$,$L$是一致椭圆的. 原点处记$\mu=\frac{{{a_{ij}}{\rho _i}{\sigma _j}}}{{\sqrt {{a_{ij}}{\rho _i}{\rho _j}} \sqrt {{a_{ij}}{\sigma _i}{\sigma _j}} }}$,则$-1<\mu<1$. 令$\theta_0=\arccos (-\mu)$. 如果在原点附近的$\overline{\Omega}$内,$u\in C^{k, \alpha}$,$k+\alpha>\frac{\pi}{\theta_0}$,则 $\left(\frac{\partial }{\partial s}\right)^ju ,\ j = 0,\cdots ,k$ 中至少有一个在原点处严格正, 这里$s$是指向$\Omega$内的任意的一个方向.

本节给出定理1.2及推论1.3的证明.

先对方程组做一些处理. 由条件(ii),$F$和$G$关于$x_1$和$\left\{ {{u_{12}},\cdots,{u_{1N}}} \right\}$对称, 可知$u^\lambda$和$v^\lambda$同样满足方程组(1.3),即 $$\begin{equation} \label{jinhu4} \left\{ \begin{array}{ll} F(x^\lambda,u^\lambda,u^\lambda_{ij})+g(u^\lambda,v^\lambda)=0,\ x \in {\Sigma _\lambda },\\ G(x^\lambda,u^\lambda,u^\lambda_{ij})+f(u^\lambda,v^\lambda)=0,\ x \in {\Sigma _\lambda }. \end{array}(3.1) \right. \end{equation}$$ 方程组(1.3)和(3.1)分别对应相减得 $$\begin{equation} \label{jinhu5} F(x^\lambda,u^\lambda,u_{ij}^\lambda) - F\left( {x,u,{u_{ij}}} \right) + g\left( {{u^\lambda },{v^\lambda }} \right) - g\left( {u,v} \right) = 0,(3.2) \end{equation}$$ $$\begin{equation} \label{jinhu6} G(x^\lambda,v^\lambda,v_{ij}^\lambda) - G\left( {x,v,{v_{ij}}} \right) + f\left( {{u^\lambda },{v^\lambda }} \right) - f\left( {u,v} \right) = 0.(3.3) \end{equation}$$ 利用中值定理处理(3.2)和(3.3)式得到 $$\begin{eqnarray} \label{jinhu7} \nonumber &&\int_0^1 {{F_{{x_1}}}} \left( {{x_1} + t\left( {2\lambda - 2{x_1}} \right),y,{u^\lambda },u_{ij}^\lambda } \right)\left( {2\lambda - 2{x_1}}\right){\rm d}t\\ \nonumber&&+ \left[{\int_0^1 {{F_p}\left( {x,u + t\omega ,u_{ij}^\lambda } \right){\rm d}t + \frac{{\partial g}}{{\partial u}}\left( {{\xi _1}\left( {x,\lambda } \right),v} \right)} } \right]\omega\\ \nonumber&&+ \sum\limits_{i,j} {\int_0^1 {{F_{{p_{ij}}}}} } \left( {x,u,{u_{ij}} + t{\omega _{ij}}} \right){\rm d}t\cdot \omega _{ij}\\ &=& - \frac{{\partial g}}{{\partial v}}\left( {{u^\lambda },{\eta _1}\left( {x,\lambda } \right)} \right)\varphi.(3.4) \end{eqnarray}$$ $$\begin{eqnarray} \label{jinhu8} \nonumber &&\int_0^1 {{G_{{x_1}}}} \left( {{x_1} + t\left( {2\lambda - 2{x_1}} \right),y,{v^\lambda },v_{ij}^\lambda } \right)\left( {2\lambda - 2{x_1}}\right){\rm d}t\\ \nonumber&&+ \left[{\int_0^1 {{G_p}\left( {x,v + t\varphi ,v_{ij}^\lambda } \right){\rm d}t + \frac{{\partial f}}{{\partial v}}\left( u,{{\eta _2}\left( {x,\lambda } \right)} \right)} } \right]\varphi\\ \nonumber&&+ \sum\limits_{i,j} {\int_0^1 {{G_{{p_{ij}}}}} } \left( {x,v,{v_{ij}} + t{\varphi_{ij}}} \right){\rm d}t\cdot \varphi_{ij}\\ &=& - \frac{{\partial f}}{{\partial u}}\left( {{\xi_2}\left( {x,\lambda }\right),{v^\lambda }} \right)\omega,(3.5) \end{eqnarray}$$ 其中, $${\xi _i}\left( {x,\lambda } \right) \in \left( {\min \left\{ {u\left( x \right),{u^\lambda }\left( x \right)} \right\},\max \left\{ {u\left( x \right),{u^\lambda }\left( x \right)} \right\}} \right),$$ $${\eta _i}\left( {x,\lambda } \right) \in \left( {\min \left\{ {v\left( x \right),{v^\lambda }\left( x \right)} \right\},\max \left\{ {v\left( x \right),{v^\lambda }\left( x \right)} \right\}} \right),\ i=1,2;$$ $$F_{p_{ij}}\left(x,u,u_{ij}+t\omega _{ij}\right) = F_{p_{ij}}\left(x,u,u_{11},\cdots ,u_{ij - 1},u_{ij} + t{\omega _{ij}},u_{ij + 1}^\lambda ,\cdots ,u_{NN}^\lambda \right),$$ $$G_{p_{ij}}\left(x,v,v_{ij}+ t\varphi _{ij} \right) = {F_{{p_{ij}}}}\left( {x,v,{v_{11}},\cdots ,{v_{ij - 1}},{v_{ij}} + t{\varphi _{ij}},v_{ij + 1}^\lambda ,\cdots ,v_{NN}^\lambda } \right),$$ $$\ i=1,2,\cdots,N,\ j= i,i+1,\cdots ,N-1;$$ $$F_{p_{iN}}\left(x,u,u_{iN}+t\omega _{iN}\right) = F_{p_{iN}}\left(x,u,u_{11},\cdots ,u_{iN - 1},u_{iN} + t{\omega _{iN}},u_{i+1 i+1}^\lambda ,\cdots ,u_{NN}^\lambda \right),$$ $$G_{p_{iN}}\left(x,v,v_{iN}+ t\varphi _{iN} \right) = {G_{{p_{iN}}}}\left( {x,v,{v_{11}},\cdots ,{v_{iN - 1}},{v_{iN}} + t{\varphi _{iN}},v_{i+1i+1}^\lambda ,\cdots ,v_{NN}^\lambda } \right),$$ $$i=1,2,\cdots,N.$$

定理1.2的证明分四步.

第一步 证明存在$-1<\lambda^*\leq0$, 使得当$-1<\lambda<\lambda^*\leq0$时, $$\omega(x,\lambda)\geq 0, \varphi(x,\lambda)\geq 0,\forall x\in \Sigma_\lambda.$$

用反证法证明. 假设上述不等式不成立,则对于每一个$\lambda \in (-1,0)$, 都存在点$x_\lambda \in \Sigma_\lambda$,使得 $\omega(x_\lambda,\lambda)<0$或$\varphi(x_\lambda,\lambda)<0$. 不妨设$\omega(x_\lambda,\lambda)<0$. 由于所考虑的函数$u$在区域$S$的边界上等于$0$,在区域$S$的内部严格大于零, 因此当$x\in \partial \Sigma_\lambda$时, $\omega(x,\lambda)\geq 0$. 所以,$\omega(x,\lambda)$在$\Sigma_\lambda$的内部能够取得负的最小值. 记 $$\omega \left( {{x_0}},\lambda\right) = \mathop {\min }\limits_{x \in \overline{\Sigma _\lambda }} \omega \left( x ,\lambda \right) < 0,\ {x_0} = \left( {x_0^1,x_0^2,\cdots x_0^N} \right),$$ 则有 $$\nabla _x\omega \left(x_0,\lambda \right) = 0,\ \left\{ \omega _{ij}\left(x_0,\lambda \right) \right\} \geq 0.$$

根据引理2.1,引理2.2, 取$\delta = \min \left\{\delta _1,\delta_2,\delta_3,\frac{1}{2} \right\}$,当 $-1<\lambda<-1+\delta$时,有 $$\alpha \left( \lambda \right) := \int_0^1 {{F_p}\left( {{x_0},u(x_0) + t\omega \left( {{x_0}} \right),u_{ij}^\lambda \left( {{x_0}} \right)} \right){\rm d}t + \frac{{\partial g}}{{\partial u}}\left( {{\xi _1}\left( {{x_0},\lambda } \right),v\left( {{x_0}} \right)} \right)} < 0.$$ 根据条件(ii),则有 $$\int_0^1 {{F_{x_1}}} \left( {x_0^1 + t\left( {2\lambda - x_0^1} \right)},x_0^2,\cdots x_0^N,u + t\omega \left( {{x_0}} \right),\cdots \right){\rm d}t\cdot \left( {2\lambda - 2x_0^1} \right) \ge 0.$$ 根据$F$的一致椭圆性知 $$\sum\limits_{i,j} {\int_0^1 {{F_{{p_{ij}}}}\left( {x,u,{u_{11}},\cdots ,{u_{ij - 1}},{u_{ij}} + t{\omega _{ij}},u_{ij + 1}^\lambda ,\cdots ,u_{NN}^\lambda } \right)} }{\rm d}t\cdot {\omega _{ij}}\left( {{x_0}} \right) \ge 0.$$ 于是由(3.4)式得 $$\begin{eqnarray} \label{jinhu9} \nonumber&&\left[\int_0^1 F_p\left( x,u\left(x_0 \right)+t\omega\left(x_0 \right),u_{ij}^\lambda \right){\rm d}t + \frac{\partial g}{\partial u}\left(\xi _1\left( x_0,\lambda \right),v\left(x_0\right) \right) \right]\omega \left(x_0 \right) \\ &\le&-\frac{\partial g}{\partial v}\left( {{u^\lambda }\left( {{x_0}} \right),{\eta _1}\left( {{x_0},\lambda } \right)} \right)\varphi \left( {{x_0}} \right).(3.6) \end{eqnarray}$$ 注意到上式左端严格大于零(见引理2.1),又由条件(iii)知,$- \frac{{\partial g}}{{\partial v}}\left( {{u^\lambda }\left( {{x_0}} \right),{\eta _1}\left( {{x_0},\lambda } \right)} \right) < 0$, 得$\varphi(x_0,\lambda)<0.$ 同样由于$\varphi(x)\geq 0$,$x\in \partial \Sigma_\lambda$,于是 $\varphi(x)$在$\Sigma_\lambda$内部取到负的最小值,即存在$x_1\in\Sigma_\lambda$, 使得$\varphi \left( {{x_1}} \right) = \mathop {\min }\limits_{x \in \overline{\Sigma _\lambda }} \varphi \left( x \right) < 0.$

由引理2.1知 $$\delta \left( \lambda \right) := \int_0^1 {{G_p}\left( {{x_1},v(x_1) + t{\varphi(x_1) },v_{ij}^\lambda \left( {{x_1}} \right)} \right){\rm d}t + \frac{{\partial f}}{{\partial v}}\left( {u\left( {{x_1}} \right),{\eta _2}\left( {{x_1},\lambda } \right)} \right)} < 0.$$ 且有 $$\begin{eqnarray} \label{jinhu10} \nonumber&&\left[{\int_0^1 {{G_p}\left( {{x_1},v(x_1) + t{\varphi(x_1)},v_{ij}^\lambda \left( {{x_1}} \right)} \right){\rm d}t + \frac{{\partial f}}{{\partial v}}\left( {u\left( {{x_1}} \right),{\eta _2}\left( {{x_1},\lambda } \right)} \right)} } \right]\varphi \left( {{x_1}} \right)\\ &\le&-\frac{{\partial f}}{{\partial u}}\left( {{\xi _2}\left( {{x_1},\lambda } \right),{v^\lambda }\left( {{x_1}} \right)} \right)\omega \left( {{x_1}} \right).(3.7) \end{eqnarray}$$ 并且有$\omega(x_1)<0$.

记 $$\beta \left( \lambda \right) = \frac{{\partial g}}{{\partial v}}\left( {{u^\lambda }\left( {{x_0}} \right),{\eta _1}\left( {{x_0},\lambda } \right)} \right) >0,\ \gamma \left( \lambda \right) = \frac{{\partial f}}{{\partial u}}\left( {{\xi _2}\left( {{x_1},\lambda } \right),{v^\lambda }\left( {{x_1}} \right)} \right)> 0.$$

改写(3.6)和(3.7)式我们得到 $$\begin{eqnarray} \nonumber \omega\left( {{x_0}} \right) &\ge& -\frac{{\beta \left( \lambda \right)}}{{\alpha \left( \lambda \right)}}\varphi \left( {{x_0}} \right) \ge -\frac{{\beta \left( \lambda \right)}}{{\alpha \left( \lambda \right)}}\varphi \left( {{x_1}} \right)\\ \nonumber &\ge& \frac{{\beta \left( \lambda \right)\gamma \left( \lambda \right)}}{{\alpha \left( \lambda \right)\delta \left( \lambda \right)}}\omega \left( {{x_1}} \right) \ge \frac{{\beta \left( \lambda \right)\gamma \left( \lambda \right)}}{{\alpha \left( \lambda \right)\delta \left( \lambda \right)}}\omega \left( {{x_0}} \right), \end{eqnarray}$$ 于是得 $$\begin{eqnarray} \label{jinhu11} \alpha \left( \lambda \right)\delta \left( \lambda \right) - \beta \left( \lambda \right)\gamma \left( \lambda \right) \le 0.(3.8) \end{eqnarray}$$

另一方面,由于此时$x_0,x_1\in \Sigma_{\lambda}^{\omega^-}\cap \Sigma_{\lambda}^{\varphi^-}$,利用引理2.2知 $\alpha \left( \lambda \right)\delta \left( \lambda \right) - \beta \left( \lambda \right)\gamma \left( \lambda \right)>0$. 这与(3.8)式矛盾. 因此假设错误.

故存在$\lambda^*\in (-1,0]$,使得$\lambda\in (-1,\lambda^*]$时,有 $$\begin{eqnarray} \label{jinhu13} \omega(x,\lambda) \geq0,\ \varphi(x,\lambda)\geq0,~~\forall x \in {\Sigma _\lambda }(3.9) \end{eqnarray}$$ 均成立 (第一步完成).

第二步 连续地增加$\lambda$的值, 只要不等式(3.9)成立, 就继续向右移动平面$T_\lambda$. 我们要证明如此移动下去, 平面$T_\lambda$不会在触到原点之前停下来. 更为具体地说,令 $$\bar \mu = \sup \{ {\mu| {\omega ( {x,\lambda } ) \ge 0,~\varphi ( {x,\lambda } ) \ge 0,~ \forall x \in {\Sigma _\lambda }},~ \forall \lambda\leq \mu}\} ,$$ 则必有$\overline{\mu}\geq0$.

用反证法证明. 假设$\overline{\mu}<0$. 由$\overline{\mu}$的定义知, 当$\lambda<\overline{\mu}$时,有 $\omega(x,\lambda) \geq0$和 $\varphi(x,\lambda)\geq0$,$x\in{\Sigma _{ \lambda }}$. 由于所考虑的函数对于$\lambda$ 都是连续的, 并且根据条件(iii)和(3.9)式,对于所有的$x \in {\Sigma _{\bar \mu }}$,有 $$\begin{eqnarray} \label{jinhu14} \nonumber & & \int_0^1F_{x_1} \left(x_1+t\left(2\bar\mu-2x_1 \right),y,u^{\lambda},u_{ij}^\lambda \right){\rm d}t\cdot \left(2\bar\mu-2x_1\right)\\ \nonumber &&+ \sum\limits_{i,j}\int_0^1 F_{p_{ij}} \left(x,u,u_{ij}+t\omega_{ij}\left(x,\bar\mu\right)\right){\rm d}t\cdot \omega _{ij}\left(x,\bar \mu\right)\\ \nonumber&&+ \left[\int_0^1 F_p\left(x,u+t\omega\left(x,\bar\mu\right),u_{ij}^\lambda \right){\rm d}t+\frac{\partial g}{\partial u}\left(\xi _1\left(x,\bar\mu\right),v\right)\right]\omega\left(x,\bar\mu\right)\\ &=& -\frac{{\partial g}}{{\partial v}}\left( {{u^{\bar \mu }},{\eta _1}\left( {x,\bar \mu } \right)} \right)\varphi \left( {x,\bar \mu } \right)\le 0,(3.10)(3.10) \end{eqnarray}$$ $$\begin{eqnarray} \label{jinhu15} \nonumber&&\int_0^1 {{G_{{x_1}}}} \left( {{x_1} + t\left( {2\bar \mu - 2{x_1}} \right)},y,v^\lambda,v_{ij}^\lambda \right){\rm d}t\cdot \left( {2\bar \mu - 2{x_1}} \right)\\ \nonumber&&+ \sum\limits_{i,j} {\int_0^1 {{G_{{p_{ij}}}}} } \left(x,v,{{v_{ij}} + t{\varphi _{ij}}\left( {x,\bar \mu } \right)} \right){\rm d}t\cdot {\varphi _{ij}}\left( {x,\bar \mu } \right) \\ \nonumber&&+ \left[{\int_0^1 {{G_p}\left(x,{v + t\varphi \left( {x,\bar \mu }\right)},v_{ij}^\lambda \right){\rm d}t + \frac{{\partial f}}{{\partial v}}\left( {u,{\eta _2}\left( {x,\bar \mu } \right)} \right)} } \right]\varphi \left( {x,\bar \mu } \right) \\ &= & - \frac{{\partial f}}{{\partial u}}\left( {{\xi _2}\left( {x,\bar \mu } \right),{v^{\bar \mu }}} \right)\omega \left( {x,\bar \mu } \right)\le 0.(3.11) \end{eqnarray}$$

根据条件(ii)和$\overline{\mu}\geq x_1$, (3.10)和(3.11)式的第一项非负,于是对于所有的$x \in {\Sigma _{\bar \mu }}$,有 $$\begin{eqnarray} \label{jinhu16} \nonumber& &- \sum\limits_{i,j}\int_0^1 F_{p_{ij}} \left(x,u,u_{ij}+t\omega_{ij}\left(x,\bar\mu\right)\right){\rm d}t\cdot \omega _{ij}\left(x,\bar \mu\right)\\ & & -\left[\int_0^1 F_p\left(x,u+t\omega\left(x,\bar\mu\right),u_{ij}^\lambda \right){\rm d}t+\frac{\partial g}{\partial u}\left(\xi _1\left(x,\bar\mu\right),v\right)\right]\omega\left(x,\bar\mu\right)\geq 0,(3.13) \end{eqnarray}$$ $$\begin{eqnarray} \label{jinhu17} \nonumber &&- \sum\limits_{i,j} {\int_0^1 {{G_{{p_{ij}}}}} } \left(x,v,{{v_{ij}} + t{\varphi _{ij}}\left( {x,\bar \mu } \right)} \right){\rm d}t\cdot {\varphi _{ij}}\left( {x,\bar \mu } \right) \\ \nonumber& & -\left[{\int_0^1 {{G_p}\left(x,{v + t\varphi \left( {x,\bar \mu }\right)},v_{ij}^\lambda \right){\rm d}t + \frac{{\partial f}}{{\partial v}}\left( {u,{\eta _2}\left( {x,\bar \mu } \right)} \right)} } \right]\varphi \left( {x,\bar \mu } \right)\geq 0.(3.12) \end{eqnarray}$$

需要注意的是,$\omega(x,\overline{\mu})\equiv0$是不成立的. 假设$\omega(x,\overline{\mu})\equiv0$,取边界上的点$u(-1,y)=0$, 又$\overline{\mu}<0$, 则对称点$(2\overline{\mu}+1,y)$在$S$内部且$u(2\overline{\mu}+1,y)=0$. 这和$u$是正解矛盾.

因为$\overline{\mu}<0$ 是使(3.9)式成立的最大区间的右端点,所以对任意满足$\lambda^k<0$,${\lambda ^k} \searrow \bar \mu$的单调减序列$\{\lambda_k\}$,存在 ${x^k} \in \overline{\Sigma _{\lambda ^k}}$,使得$\omega \left( {{x^k},\ {\lambda ^k}} \right) < 0.$ 由于${x_k} \in S$(有界区域),必存在收敛子列,不妨设$x^k$即为收敛子列, 并且就是函数$\omega(x,\lambda^k)$在区域$\overline{\Sigma _{\lambda ^k}}$中的最小值点, 即 $$\begin{equation} \label{jinhu18} \left\{ \begin{array}{l} \omega (x^k,\lambda^k) = \min \limits_{x \in \overline{\Sigma _{\lambda ^k}}} \omega(x,\lambda^k)< 0,\\ {x^k} \to \bar x = (\bar{x}_1,\bar{x}_2,\cdots,\bar{x}_N) \in \bar \Sigma_{\bar\mu},\;k \to \infty . \end{array} \right.(3.13) \end{equation}$$

由于$x^k$是函数$\omega(x,\lambda^k)$在区域${\bar \Sigma _{{\lambda ^k}}}$内部的最小值点,因此有 $${\nabla _x}\omega \left( {{x^k},{\lambda ^k}} \right) = 0,\ \left\{ {{\omega _{ij}}\left( {{x^k},{\lambda ^k}} \right)} \right\} \ge 0.$$ 取$k\rightarrow\infty$极限,则得 $$\begin{eqnarray} \label{jinhu19} \omega \left( {\bar x,\bar \mu } \right) \le 0,\ {\nabla _x}\omega \left( {\bar x,\bar \mu } \right) = 0,\ \left\{ {{\omega _{ij}}\left( {\bar x,\bar \mu } \right)} \right\} \ge 0.(3.14) \end{eqnarray}$$

又由连续性可知,$\omega \left( {\bar x,\bar \mu } \right)\geq0$. 于是$\omega \left( {\bar x,\bar \mu } \right)=0$.

则$\overline{x}$的位置有三种情况: (a) 位于${\Sigma _{\bar \mu }}$的内部; (b) $\overline{x}$属于$T_{\overline{\mu}}$或${\bar \Sigma _{\bar \mu }} \cap \partial S$; (c) $\bar x \in {T_{\bar \mu }} \cap \partial S$. 以下对这三种情况分别讨论.

(a)~ $\overline{x}$位于${\Sigma _{\bar \mu }}$的内部

对于${\Sigma _{\bar \mu }}$的内部的任意一点$a$, 可以选择一个边界光滑的区域$A$ ($\overline{A}$ 包含于${\Sigma _{\bar \mu }}$的内部),使得$a$和$\overline{x}$都位于$A$的内部. 在$A$上应用强极值原理(原定理条件在边界上函数等于0,此处在$\partial A$上,$\omega \left( {\bar x,\bar \mu } \right)\geq0$,定理仍能够成立),得到在$A$上$\omega \left( {\bar x,\bar \mu } \right)\equiv0$. $\omega \left( {\bar x,\bar \mu } \right)$在$a$点处等于0,由$a$的任意性得,在${\Sigma _{\bar \mu }}$的内部$\omega \left( {\bar x,\bar \mu } \right)\equiv0$. 由于边界条件$u(x)=0$,${\Sigma _{\bar \mu }}$的内部$u(x)>0$, 所以在$\partial {\Sigma _{\bar \mu }}$上$\omega \left( {\bar x,\bar \mu } \right)>0$. 由函数的连续性可得出矛盾.

(b)~ $\overline{x}$属于$T_{\overline{\mu}}$或${\bar \Sigma _{\bar \mu }} \cap \partial S$

取$r>0$充分小,可使某一个以$\overline{x}$为边界点的球${B_r} \subset {\Sigma _{\bar \mu }}$. 因为$\omega \left( {x,\bar \mu } \right) > 0,\ \forall x \in {B_r}\left( {\bar x} \right),\ \omega \left( {\bar x,\bar \mu } \right) = 0$,应用经典的Hopf引理知 $\frac{{\partial\omega }}{{\partial n}}\left( {\bar x,\bar \mu } \right)\ne 0$, 所以${\nabla _x}\omega \left( {\bar x,\bar \mu } \right)\ne 0$, 与(3.14)式矛盾.

(c)~ $\bar x \in {T_{\bar \mu }} \cap \partial S$.

将$\omega \left( {\bar x,\bar \mu } \right) = 0$代入(3.12)式得 $$\begin{eqnarray} \int_0^1 \sum\limits_{i,j} F_{p_{ij}} \left(\bar x,u,u_{ij}+t\omega_{ij}\left(\bar x,\bar\mu\right)\right)\cdot \omega _{ij}\left(\bar x,\bar \mu\right){\rm d}t \le 0.(3.15) \end{eqnarray}$$ 由$F$满足一致椭圆条件,则有 $$\int_0^1 \sum\limits_{i,j} F_{p_{ij}} \left(\bar x,u,u_{ij}+t\omega_{ij}\left(\bar x,\bar\mu\right)\right)\cdot \omega _{ij}\left(\bar x,\bar \mu\right){\rm d}t \ge 0.$$ 因此有 $$\begin{equation} \label{jinhu20} \int_0^1 \sum\limits_{i,j} F_{p_{ij}} \left(\bar x,u,u_{ij}+t\omega_{ij}\left(\bar x,\bar\mu\right)\right)\cdot \omega _{ij}\left(\bar x,\bar \mu\right){\rm d}t = 0.(3.16) \end{equation}$$ 由于$\{F_{p_{ij}}\}$是对称正定矩阵, $\{\omega_{ij}(\bar x,\overline{\mu})\}$是对称半正定矩阵, 则当且仅当$\{\omega_{ij}(\bar x,\overline{\mu})\}=0$时(3.16)式成立, 即$\omega_{ij}(\bar x,\overline{\mu})=0$, $i,j=1,2,\cdots,N$.

下面我们在$\overline{x}$处运用角点区域的Hopf引理(见引理2.3). 在$\overline{x}$处,$\partial \Sigma_{\overline{\mu}}$ 包含两个横向相交的$C^2$的超曲面$\rho=0,\rho=x_1-\bar\mu$和$\sigma=0,\sigma=x_2^2+\cdots+x_N^2-1$. 记 $$a_{ij}\left( \bar x,\bar \mu \right) = \int_{0}^1F_{p_{ij}} \left(\bar x,u,u_{ij}+t\omega_{ij}\left(\bar x,\bar\mu\right)\right){\rm d}t.$$ 下面我们证明$\sum\limits_{i,j=1}^Na_{ij}(\overline{x},\overline{\mu})\rho_i(\overline{x})\sigma_j(\overline{x})=0$. 即 $$\sum_{j=2}^Na_{1j}(\overline{x},\overline{\mu}) x_j=0.$$

由条件(ii),如果${u_{12}} = \cdots = {u_{1N}} = 0$, 则${F_{{p_{1j}}}}\left( {x,u,{u_{ij}}} \right) = 0$,$j = 2,\cdots ,N$. 注意到,由于$\omega_{ij}(\bar x,\bar \mu)=0$, $$a_{ij}\left( \bar x,\bar \mu \right) = \int_{0}^1F_{p_{ij}} \left(\bar x,u,u_{ij}\right){\rm d}t=F_{p_{ij}} \left(\bar x,u,u_{ij}\right).$$ 因此, 我们只需验证 $${u_{1j}}\left( {\bar x} \right) = u_{1j}^\lambda \left( {\bar x,\bar \mu } \right) = 0,\ j=2,\cdots,N.$$

由于$\overline{x}\in T_{\overline{\mu}}$,所以 $${u_{1j }}\left( {\bar x} \right) = - u_{1j }^\lambda \left( {\bar x,\bar \mu } \right) = - \frac{1}{2}{\omega _{1j }}\left( {\left( {\bar x,\bar \mu } \right)} \right) = 0,\ j = 2,3,\cdots N.$$ 所以在$\overline{x}$处 $\sum\limits_{i,j=1}^Na_{ij}(\overline{x},\overline{\mu})\rho_i(\overline{x})\sigma_j(\overline{x})=0$.

由于$u$是$C^2$的,引理2.3中$\theta_0=\frac{\pi}{2}$, 则可得对于任意指向$S$内部的方向$s$,$\frac{\partial }{{\partial s}}\omega \ne 0$或者${\left( {\frac{\partial }{{\partial s}}} \right)^2}\omega \ne 0.$ 取$s = -\frac{1}{\sqrt{2}}(1,x_2,\cdots,x_N)$.

利用(3.14)式和$\omega_{ij}(\bar{x},\bar{\mu})=0$,直接计算可知,在$\bar{x}$处有 $$\frac{\partial \omega}{\partial s}(\bar{x})=-\frac{1}{\sqrt{2}}(\omega_1+x_2\omega_2+\cdots+x_N\omega_N)|_{\bar{x}}=0;$$ $$\begin{eqnarray*} \frac{\partial^2 \omega}{\partial s^2}(\bar{x})&=& \frac{1}{2}(\omega_{11}+\omega_{12}\bar x_2+\cdots+\omega_{1N}\bar x_N +\omega_{12}\bar x_2+\omega_{2}\bar x_2+\omega_{22}\bar x_2^2+\cdots+\omega_{2N}\bar x_2\bar x_N\\ & & +\cdots+\omega_{1N}\bar x_N+\omega_{2N}\bar x_2\bar x_N+\cdots+\omega_{NN}\bar x_N^2+\omega_N\bar x_N)|_{\bar{x}}\\ &=& 0. \end{eqnarray*}$$ 而此与引理2.3矛盾.

综上所述,假设错误. 由此得$\overline{\mu}=0$.

并且由$\overline{\mu}$的定义知 $$u(x_1,y)\leq u(-x_1,y),\ v(x_1,y)\leq v(-x_1,y),\ x_1<0.$$ (第二步完成).

第三步 我们可以从右侧开始移动平面,由于$F$和$G$在$x_1$方向上是对称的, 能够同样推出$\overline{\mu}=0$. 于是得 $$u(x_1,y)\geq u(-x_1,y),\ v(x_1,y)\geq v(-x_1,y),\ x_1<0.$$

所以对于$(x_1,y)\in S,\ -1<x_1<0$, $$u(x_1,y)=u(-x_1,y),\ v(x_1,y)=v(-x_1,y).$$ (第三步完成).

第四步 证明单调性. 由以上证明过程知,当$\lambda<0$时, $\omega(x,\lambda)\geq 0$和$\varphi(x,\lambda)\geq 0$. 又由于$\omega(x,\lambda)>0$,$x_1=-1$,$\omega(x,\lambda)$不恒等于0. 利用Hopf引理,在$T_\lambda$去掉$\partial S$的边界处, $\frac{{\partial \omega \left( {x,\lambda } \right)}}{{\partial {x_1}}} < 0,$ 又$\frac{{\partial \omega \left( {x,\lambda } \right)}}{{\partial {x_1}}} = - 2{u_1}\left( {{x_1},y} \right).$ 所以, 当$x_1<0,y\in \Omega$时,有$u_1(x_1,y)>0$,$v_1(x_1,y)>0$ (第四步完成). 定理1.2证毕.

下面证明推论1.3.

证 注意到由于$u,v$为严格凹解,则$D^2u$,$D^2v$为严格负定矩阵. 因此对于二维情形下$-\det(D^2u)$和$-\det(D^2v)$满足椭圆型条件 (1.4). 不难验证满足推论1.3条件的方程组(1.5) 满足定理1.2的所有条件, 直接应用定理1.2即可.