本文考虑加权的退化椭圆系统

(1.1) $ \begin{equation} \left\{\begin{array}{ll} -\Delta_{G}u = \omega(x)v, \\ -\Delta_{G}v = \omega(x)u^q, \end{array}\right. \qquad {{\Bbb R}} ^N = {{\Bbb R}} ^{N_1}\times {{\Bbb R}} ^{N_2}, \end{equation} $

其中$ q > 1 $, $ a, \beta\geq0 $, $ x = (x, y)\in{{\Bbb R}} ^{N_1}\times {{\Bbb R}} ^{N_2} $, $ \omega(x) = \left (1+\| x \|^{2(a+1)}\right)^{\frac{\beta}{2(a+1)}} $.本文建立椭圆系统(1.1)正稳定解的Liouville定理,即证明系统(1.1)正稳定解的非存在性结果.这里Grushin算子(Grushin梯度)定义为

$ \begin{eqnarray*} \Delta_{G}u = \Delta_{x}u+(a+1)^{2}|x|^{2a}\Delta_{y}u\ (\nabla_{G}u = (\nabla_{x}u, (a+1)|x|^{a}\nabla_{y}u)), \end{eqnarray*} $

Grushin距离定义为$ \|x\| = \; (|x|^{2(a+1)}+|y|^{2})^{\frac{1}{2(a+1)}} $.

首先,可以注意到$ a = 0 $的情况, Grushin算子是拉普拉斯算子.则椭圆系统(1.1)是Lane-Emden方程或者Lane-Emden系统,这类方程和系统已经被许多专家讨论. 2007年, Farina^[1]通过巧妙地运用Morse迭代考虑了Lane-Emden方程

(1.2) $ \begin{equation} -\Delta u = |u|^{q-1}u, \quad \Omega\in{{\Bbb R}} ^N, \end{equation} $

给出了该方程有限Morse指标解(正解或者变号解)的完全分类.为了解决双调和方程$ \Delta^2 u = |u|^{q-1}u $稳定解和有限Morse指标解的完整分类, D$ \acute{a} $vila-Dupaigne-Wang-Wei^[2]通过构造单调公式,将方程非平凡解的不存在性转化为非平凡齐次解的不存在性,从而得到相应的结果.

然而,上面所说的两种方法对于一些加权椭圆系统解的定性分析无效. Cowan-Ghoussoub^[3]和Dupaigne等^[4]各自独立的提出的一种方法能处理Lane-Emden系统和加权椭圆系统.如,基于联合使用Bootstrap方法、Souplet不等式^[5]和稳定解的准则, Hajlaoui-Harrabi-Ye^[6]和Hu-Zeng^[7]建立了$ a = 0 $的加权Lane-Emden系统(1.1)稳定解的Liouville型定理.

对于$ a = 1 $的情况,问题(1.2)与Heisenberg Laplace方程

$ -\Delta_{H} u = f(u), \quad (z, t)\in{\Bbb H}^N = {\Bbb C}^n \times {{\Bbb R}} $

密切相关,其中$ \Delta_{H} $是Heisenberg Laplace算子.设$ u $是关于$ z $的径向函数,即$ u = u(|z|, t) $.则$ \Delta_{H} u = \Delta_{z} u+4|z|^2 \frac{\partial ^2u}{\partial t^2} $是$ a = 1 $时的Grushin算子.对于含Heisenberg Laplace算子的椭圆方程的Liouville定理和对称性质已有了很多优秀的成果,参见文献[8-11].

现在考虑$ a > 0 $的一般情况, Grushin算子相应性质已被许多专家讨论^[12-13].最近,运用Kelvin变换结合移动平面法, Monticelli^[14]证明了方程$ -\Delta_{G}u = u^{q} $非负经典解的Liouville定理成立, Yu^[15]用同样的方法得到了上述方程非负弱解的Liouville定理.这里指数满足$ q < \frac{N_{a}+2}{N_{a}-2} $,其中$ N_{a}: = N_{1}+(a+1)N_{2} $称为齐次维数.

近年来,含Grushin算子的椭圆型方程或系统的稳定解与有限Morse指标解已被许多专家研究(参见文献[16-17]).利用Cowan^[18]的方法, Duong等^[16]得到了含Grushin算子的Lane-Emden系统

$ \left\{\begin{array}{ll} -\Delta_G u = v^p, \\ -\Delta_G v = u^q, \end{array}\right.\quad {{\Bbb R}} ^N = {{\Bbb R}} ^{N_1}\times {{\Bbb R}} ^{N_2} $

稳定解的Liouville定理.

受文Montenegro^[19]启发,下面给出稳定解的定义.

定义1.1  设$ \Omega $是$ {{\Bbb R}} ^N $的子集.系统(1.1)的解$ (u, v)\in C^2(\Omega)\times C^2(\Omega) $称为稳定的,如果特征值问题

(1.3) $ \begin{equation} \left\{\begin{array}{ll} \begin{array}{ll} -\Delta_{G} \phi = \omega\psi+\mathfrak{e}\phi, \\ -\Delta_{G} \psi = q\omega u^{q-1}\phi+\mathfrak{e}\psi, \end{array} & \;\ x \in \Omega, \\ \phi = \psi = 0, & \quad \ \partial \Omega \end{array}\right. \end{equation} $

存在第一特征值$ \mathfrak{e} > 0 $和正的光滑特征函数对$ (\phi, \psi) $.

可以注意到, Grushin算子$ \Delta_{G} $是非对称的且在$ \{ 0 \} \times {{\Bbb R}} ^{N_2} $流行上是退化的,从而拉普拉斯算子的研究方法对Grushin算子不适用.本文需要克服由Grushin算子引起的一些新困难(如积分估计和建立解$ (u, v) $比较原理等).受文献[7, 16, 20]的启发,本文将证明加权的退化椭圆系统(1.1)解的Liouville型定理.

定理1.1  设$ u, v $是系统(1.1)的正稳定解.如果$ N_{a}, q > 1 $和$ \beta\geq0 $满足条件

(1.4) $ \begin{equation} N_{a} < 2+\left [4+\beta+\frac{4(\beta+2)}{q-1}\right ]\frac{\kappa_{0}}{2}, \end{equation} $

其中$ \kappa_{0} $是方程

(1.5) $ \begin{equation} {\cal L} (q, \kappa): = \kappa^4-\frac{32q}{q+1}\kappa^2+\frac{32q(q+3)}{(q+1)^2}\kappa-\frac{64q}{(q+1)^2} = 0 \end{equation} $

的最大根.则系统(1.1)没有正稳定解.

注1.1  (1)使用稳定解准则(2.1)和Grushin算子性质,本文建立了系统(1.1)解的精确积分估计.

(2)通过发展文献[16, 21]中的方法,本文建立了一个新的比较原理,以及利用Bootstrap方法给出解的$ L^2 $ -估计,从而可以证明定理1.1.

推论1.1  设$ u $是双Grushin算子椭圆方程

$ \Delta_G^2 u = u^q, \qquad {{\Bbb R}} ^N = {{\Bbb R}} ^{N_1}\times {{\Bbb R}} ^{N_2} $

的正稳定解.如果$ N_{a}, q > 1 $满足下面条件

$ N_a <2+ \frac{2(q+1)}{q-1} \kappa_0, $

其中$ \kappa_{0} $是如同定理1.1,则方程没有正稳定解.

本文$ C $总表示一般的正常数,与其它变量无关. $ B_R(x) $表示$ {{\Bbb R}} ^N $上以$ x $为球心$ R $为半径的开球.另记$ B_{R} = B_{R}(0) $.

首先建立下面稳定解的两个准则.

引理2.1  设$ (u, v) $是椭圆系统(1.1)的非负稳定解,则下面两个不等式成立

(2.1) $ \begin{equation} q\int_{\Omega} \omega u^{q-1}\zeta^{2} {\rm d}x \le \int_{\Omega}\frac{|\Delta_{G}\zeta|^{2}}{\omega} {\rm d}{\bf x}, \quad \forall \zeta \in C_0^2 (\Omega) \end{equation} $

和

(2.2) $ \begin{equation} \sqrt{q}\int_{\Omega} \omega u^{\frac{q-1}{2}}\zeta^{2} {\rm d}x \le \int_{\Omega}|\nabla_{G}\zeta|^{2}{\rm d}x, \quad \forall \zeta \in C_0^1(\Omega). \end{equation} $

证  用$ \frac{\zeta^{2}}{\phi} $乘以方程(1.3)的第二个方程,其中$ \zeta \in C_0^2 (\Omega) $,并分部积分可得

$ q\int_{\Omega} \omega u^{q-1}\phi \frac{\zeta^{2}}{\phi} {\rm d}x \le \int_{\Omega}-\Delta_{G}\psi \frac{\zeta^{2}}{\phi} {\rm d}x \leq\int_{\Omega}\frac{\Delta_{G}\phi}{\omega}\cdot\Delta_{G}(\zeta^2 \phi^{-1}){\rm d}x. $

简单计算可得

$ \begin{eqnarray*} \Delta_{G}\phi\Delta_{G}(\zeta^2 \phi^{-1}& = & - \Big (\zeta\phi^{-1}\Delta_{G}\phi-\Delta_{G}\zeta \Big )^{2}+|\Delta_{G}\zeta|^{2} -2\phi^{-1}(\omega\psi+\mathfrak{e}\phi)(\nabla_{G}\zeta-\zeta\phi^{-1}\nabla_{G}\phi)^{2}\\ &\le & |\Delta_{G}\zeta|^{2}. \end{eqnarray*} $

因此有

$ \begin{eqnarray*} q\int_{\Omega} \omega u^{q-1}\zeta^{2} {\rm d}{ x} \le \int_{\Omega}\frac{|\Delta_{G}\zeta|^{2}}{\omega} {\rm d}{ x}, \qquad \forall \zeta \in C_0^2 (\Omega). \end{eqnarray*} $

下面证明(2.2)式.由稳定解的定义可得

$ -\frac{\Delta_{G}\psi}{\psi}\geq q\omega u^{q-1}\frac{\phi}{\psi} $

和

$ -\frac{\Delta_{G}\phi}{\phi}\geq\frac{\omega\psi}{\phi}. $

分别用$ \zeta^{2} $和$ \eta^{2} $乘上面两个不等式,其中$ \zeta, \eta \in C_0^1 (\Omega) $,积分可得

$ q\int_{\Omega} \omega u^{q-1}\frac{\phi}{\psi}\zeta^{2} {\rm d}x \le -\int_{\Omega}\frac{\Delta_{G}\psi}{\psi}\zeta^{2}{\rm d}x $

和

$ \int_{\Omega} \omega\frac{\psi}{\phi}\eta^{2} {\rm d}x \le -\int_{\Omega}\frac{\Delta_{G}\phi}{\phi}\eta^{2}{\rm d}x . $

直接计算有

$ \begin{eqnarray*} \int_{\Omega} \left (-\frac{\Delta_{G}\psi}{\psi}\zeta^{2}-|\nabla_{G}\zeta|^{2} \right ){\rm d}x = -\int_{\Omega}\Big [ \zeta\psi^{-1}\nabla_{G}\psi-\nabla_{G}\zeta \Big ]^{2}{\rm d}x\le 0. \end{eqnarray*} $

即

$ \begin{eqnarray*} q\int_{\Omega} \omega u^{q-1}\frac{\phi}{\psi}\zeta^{2} {\rm d}x \le \int_{\Omega}|\nabla_{G}\zeta|^{2}{\rm d}x. \end{eqnarray*} $

类似可得

$ \begin{eqnarray*} \int_{\Omega}\omega\frac{\psi}{\phi}\eta^{2}{\rm d}x\leq\int_{\Omega}|\nabla_{G}\eta|^{2}{\rm d}x . \end{eqnarray*} $

则使用不等式$ a^{2}+b^{2}\geq2ab $且取$ \zeta = \eta $,有

$ \begin{eqnarray*} \sqrt{q}\int_{{{\Bbb R}} ^N} \omega u^{\frac{q-1}{2}}\zeta^{2} {\rm d}x \le \int_{{{\Bbb R}} ^N}|\nabla_{G}\zeta|^{2}{\rm d}x, \qquad \forall \zeta \in C_0^1 (\Omega). \end{eqnarray*} $

结论得证.

为了方便,记$ {\cal D} (x): = 1+\|x\|^{2(a+1)} $.

引理2.2  对任意的$ \zeta, \eta \in C_0^4 ({{\Bbb R}} ^N) $,有下面不等式成立

(2.3) $ \begin{eqnarray} \int_{{{\Bbb R}} ^N} \frac{\Delta_{G} \zeta\Delta_{G}(\zeta\eta^{2})}{\omega} {\rm d}x & = & \int_{{{\Bbb R}} ^N}\frac{[\Delta_{G}(\zeta\eta)]^{2}}{\omega} {\rm d}x \\ && +\int_{{{\Bbb R}} ^N}\frac{[-4(\nabla_{G}\zeta\cdot\nabla_{G}\eta)^{2}+2\zeta\Delta_{G}\zeta|\nabla_{G}\eta|^{2}]}{\omega}{\rm d}x \\ && +\int_{{{\Bbb R}} ^N}\frac{\zeta^{2}}{\omega} \Big [2\nabla_{G}(\Delta_{G}\eta)\cdot\nabla_{G}\eta+(\Delta_{G}\eta)^{2} \Big ]{\rm d}x \\ & & -2\beta \int_{{{\Bbb R}} ^N}\frac{\zeta^{2}\Delta_{G}\eta|x|^{a}}{{\cal D}(x) \omega} \left (\nabla_{G}\eta, (|x|^a x, y) \right ){\rm d}x \end{eqnarray} $

和

(2.4) $ \begin{eqnarray} 2 \int_{{{\Bbb R}} ^N} \frac{|\nabla_{G}\zeta|^2|\nabla_{G}\eta|^2}{\omega} {\rm d}x & = & 2\int_{{{\Bbb R}} ^N}\frac{\zeta(-\Delta_{G}\zeta)|\nabla_{G}\eta|^{2}}{\omega}{\rm d}x +\int_{{{\Bbb R}} ^N}\frac{\zeta^{2}\Delta_{G}(|\nabla_{G}\eta|^{2})}{\omega}{\rm d}x \\ && -2\beta \int_{{{\Bbb R}} ^N}\frac{\zeta^{2}|x|^{a}}{{\cal D}(x) \omega} \left (\nabla_{G}(|\nabla_{G}\eta|^{2}), (|x|^{a}x, y) \right ){\rm d}{ x} \\ & & +\beta \int_{{{\Bbb R}} ^N}\frac{\zeta^{2}|\nabla_{G}\eta|^{2}|x|^{2a}}{{\cal D} (x) \omega} \left [(\beta+2a+2)\frac{\|x\|^{2(a+1)}}{{\cal D} (x)} -(2a+N_a) \right ]{\rm d}x.{}\\ \end{eqnarray} $

证  简单计算可得

$ \begin{eqnarray*} \Delta_{G}\zeta\Delta_{G}(\zeta\eta^{2})& = & [\Delta_{G}(\zeta\eta)]^{2}-4(\nabla_{G}\zeta\cdot\nabla_{G}\eta)^{2} -\zeta^{2}(\Delta_{G}\eta)^{2} \\ && +2\zeta\Delta_{G}\zeta|\nabla_{G}\eta|^{2}-4\zeta\Delta_{G}\eta\nabla_{G}\zeta\cdot\nabla_{G}\eta, \end{eqnarray*} $

$ \nabla_{G}(\omega^{-1}) = -\beta \left (1+\|x\|^{2(a+1)} \right )^{-\frac{\beta}{2(a+1)}-1}\left (|x|^{2a} x, |x|^a y \right ) $

和

$ \begin{eqnarray*} -4\int_{{{\Bbb R}} ^N}\frac{\zeta\Delta_{G}\eta\nabla_{G}\zeta\cdot\nabla_{G}\eta}{\omega} {\rm d}x & = &2\int_{{{\Bbb R}} ^N}\frac{\zeta^{2}}{\omega}\Big [\nabla_{G}(\Delta_{G}\eta)\cdot\nabla_{G}\eta+(\Delta_{G}\eta)^{2} \Big ] {\rm d}x\\ && +2\int_{{{\Bbb R}} ^N}\zeta^{2}\Delta_{G}\eta\nabla_{G}\eta\cdot\nabla_{G}(\omega^{-1}){\rm d}x. \end{eqnarray*} $

因此,综合上面三个等式可得等式(2.3)成立.

另一方面,容易验证

$ \Delta_{G}(\omega^{-1}) = \beta \left (1+\|x\|^{2(a+1)} \right )^{-\frac{\beta}{2(a+1)}-1} |x|^{2a}\left [(\beta+2a+2)\frac{\|x\|^{2(a+1)}}{{\cal D}(x)}-2a-N_a \right ] $

和

$ \begin{eqnarray*} \int_{{{\Bbb R}} ^N} \frac{\Delta_{G}(\zeta^{2})|\nabla_{G}\eta|^{2}}{\omega} {\rm d}x & = &\int_{{{\Bbb R}} ^N}\frac{\zeta^{2}\Delta_{G}(|\nabla_{G}\eta|^{2})}{\omega}{\rm d}x +2\int_{{{\Bbb R}} ^N}\zeta^{2}\nabla_{G}(|\nabla_{G}\eta|^{2})\cdot\nabla_{G}(\omega^{-1}){\rm d}x \\ && +\int_{{{\Bbb R}} ^N}\zeta^{2}|\nabla_{G}\eta|^{2}\Delta_{G}(\omega^{-1}){\rm d}x. \end{eqnarray*} $

联合等式$ \frac{1}{2}\Delta_{G}(\zeta^{2}) = |\nabla_{G}\zeta|^{2}+\zeta\Delta_{G}\zeta $可得(2.4)式成立.

采用类似于文献[7, Lemma 2.4]和[20, Lemma 2.3]的方法,可以给出系统(1.1)稳定解的精确积分估计.

引理2.3  设$ (u, v) $是系统(1.1)的一个非负稳定解.则对任意$ R > 0 $,有

$ \int_{B_R\times B_{R^{a+1}}}\omega \left (v^{2}+u^{q+1}\right ) {\rm d}x\leq CR^{N_a-4-\beta-\frac{4(\beta+2)}{q-1}}. $

证  设$ \xi_i \in C_c^{\infty}({{\Bbb R}}, [0, 1]) $, $ i = 1, 2 $使得在$ [-1, 1] $上$ \xi_i = 1 $,在$ [-2^{1+(i-1)a}, 2^{1+(i-1)a}] $外$ \xi_i = 0 $.对任意的$ R > 0 $,定义$ \varphi_R(x, y) = \xi_1(\frac{|x|}{R}) \xi_2(\frac{|y|}{R^{a+1}}) $.则存在与$ R $无关的常数$ C > 0 $,使得

$ \begin{eqnarray*} &&|\nabla_{x}\varphi_{R}|\leq CR^{-1}, \quad \quad |\nabla_{y}\varphi_{R}|\leq CR^{-(a+1)}, \\ &&|\Delta_{x}\varphi_{R}|\leq CR^{-2}, \quad \quad |\Delta_{y}\varphi_{R}|\leq CR^{-2(a+1)}. \end{eqnarray*} $

用$ u\varphi^{2}_{R} $乘以系统(1.1)的第二个方程并分部积分可得

$ \int_{{{\Bbb R}} ^N}\omega u^{q+1}\varphi_{R}^{2}{\rm d}x = \int_{{{\Bbb R}} ^N}\frac{\Delta_{G}u\Delta_{G}(u\varphi_{R}^{2})}{\omega}{\rm d}x. $

把测试函数$ \zeta = u\varphi_{R} $插入(2.1)式中可得

$ q\int_{{{\Bbb R}} ^N}\omega u^{q+1}\varphi^{2}_{R}{\rm d}x\leq \int_{{{\Bbb R}} ^N}\frac{[\Delta_{G}(u\varphi_{R}]^{2}}{\omega}{\rm d}x. $

则联合(2.3)和(2.4)式,有

$ \begin{eqnarray*} (q-1)\int_{{{\Bbb R}} ^N} \omega u^{q+1}\varphi_R^{2}{\rm d}x &\le& 6 \int_{{{\Bbb R}} ^N}\frac{u(-\Delta_{G}u) |\nabla_{G}\varphi_R|^2}{\omega}{\rm d}x \\ && + \int_{{{\Bbb R}} ^N}\frac{u^{2}}{\omega} \Big [2\Delta_{G}(|\nabla_{G}\varphi_R|^{2})-2\nabla_{G}(\Delta_{G}\varphi_R)\cdot\nabla_{G}\varphi_R \Big ]{\rm d}x \\ && +2\beta \int_{{{\Bbb R}} ^N}\frac{u^{2}|x|^{a}}{\omega {\cal D} (x)} \Big ( \Delta_{G}\varphi_R \nabla_{G}\varphi_R-2\left (\nabla_{G}(|\nabla_{G}\varphi_R|^2), (|x|^a x, y) \right ) \Big ) {\rm d}x \\ & &+2\beta \int_{{{\Bbb R}} ^N}\frac{u^{2}|\nabla_{G}\varphi_R|^{2}|x|^{2a}}{\omega {\cal D} (x)} \left [(\beta+2a+2)\frac{\|x\|^{2(a+1)}}{{\cal D} (x)} -(2a+N_a) \right ]{\rm d}x. \end{eqnarray*} $

因为$ \Delta_{G}(u\varphi_R) = -\omega v\varphi_R+2\nabla_{G}u\cdot\nabla_{G}\varphi_R+u\Delta_{G}\varphi_R $,联合上面不等式和引理 可得

(2.5) $ \begin{eqnarray} \int_{{{\Bbb R}} ^N}\omega \left ( v^{2}+u^{q+1}\right )\varphi_R^{2}{\rm d}x &\le& C\int_{{{\Bbb R}} ^N}uv|\nabla_{G}\varphi_R|^{2}{\rm d}x \\ && +C\int_{{{\Bbb R}} ^N}\frac{u^{2}}{\omega}\Big [|\Delta_{G}(|\nabla_{G}\varphi_R|^{2})| +|\nabla_{G}(\Delta_{G}\varphi_R)\nabla_{G}\varphi_R | \Big ]{\rm d}x \\ && +C\int_{{{\Bbb R}} ^N}\frac{u^{2}|x|^{a}}{\omega {\cal D}(x)} \left (|\nabla_{G}(|\nabla_{G}\varphi_R|^{2})|+|\Delta_{G}\varphi_R \nabla_{G}\varphi_R |\right ) \left | (|x|^{a}x , y)\right | {\rm d}x \\ & &+C\int_{{{\Bbb R}} ^N}\frac{u^{2}|\nabla_{G}\varphi_R|^{2}|x|^{2a}}{\omega {\cal D} (x)}\left [\frac{\|x\|^{2(a+1)}}{{\cal D} (x)} +C \right ]{\rm d}x. \end{eqnarray} $

在不等式(2.5)中,用$ \varphi_R^m $代替$ \varphi_R $, $ m $充分大.容易验证

$ \int_{{{\Bbb R}} ^N} uv \varphi_R^{2(m-1)}|\nabla_{G}\varphi_R|^{2}{\rm d}x \leq\frac{1}{2C}\int_{{{\Bbb R}} ^N}\omega v^{2}\varphi_R^{2m}{\rm d}x +C\int_{{{\Bbb R}} ^N}\frac{u^{2}}{\omega}\varphi_R^{2(m-1)}|\nabla_{G} \varphi_R|^{4}{\rm d}x. $

则联合上面两个不等式,有

$ \int_{{{\Bbb R}} ^N}\omega \left (v^{2}+u^{q+1} \right )\varphi_R^{2m}{\rm d}x \le C\int_{{{\Bbb R}} ^N}u^2 {\cal F}(\varphi_R, m, x) \omega^{-1} {\rm d}x, $

其中

$ \begin{eqnarray*} {\cal F} (\varphi_R, m, x)& = & \varphi_R^{2(m-1)}|\nabla_{G}\varphi_R|^{4}+|\Delta_{G}(|\nabla_{G}\varphi_R^m|^{2})| +|\nabla_{G}(\Delta_{G}\varphi_R^m)\nabla_{G}\varphi_R^m | \\ && +\frac{|x|^{a}}{1+\|x\|^{2(a+1)}}\Big [\left (|\nabla_{G}(|\nabla_{G}\varphi_R^m|^{2})|+|\Delta_{G}\varphi_R^m \nabla_{G}\varphi_R^m |\right ) \left | (|x|^a x, y)\right | \\ && +|x|^{a}|\nabla_{G}\varphi_R^m|^2 \Big ] +\frac{|\nabla_{G}\varphi_R^m|^{2}|x|^{2a}}{(1+\|x\|^{2(a+1)})^2}\|x\|^{2(a+1)}. \end{eqnarray*} $

应用Hölder不等式可得

$ \int_{B_R\times B_{R^{a+1}}}\omega \left (v^{2}+u^{q+1} \right ){\rm d}x\leq CR^{N_a-4-\beta-\frac{4(\beta+2)}{q-1}}. $

结论成立.

受文献[16, 21]的启发,我们建立新的比较原理.

引理2.4  设$ (u, v) $是系统(1.1)的正经典解,则不等式成立

$ \frac{u^{q+1}}{q+1}\leq \frac{v^{2}}{2}. $

证  设$ \delta = \sqrt{\frac{2}{q+1}} $, $ \Psi = \delta u^{\frac{q+1}{2}}-v $.直接计算可得

(2.6) $ \begin{eqnarray} \Delta_{G} \Psi = \frac{q-1}{2\delta}u^{\frac{q-3}{2}}|\nabla_{G}u|^{2}+\omega u^{q}-\delta^{-1}u^{\frac{q-1}{2}}\omega v \ge \delta^{-1}\omega u^{\frac{q-1}{2}}\Psi. \end{eqnarray} $

假设矛盾,则可设$ { } M = \sup_{{{\Bbb R}} ^{N}}\Psi, 0 < M\leq+\infty $.下面仅考虑两种情况.

情况1  $ \Psi $的上确界在无穷远处可以取到.

取截断函数$ \xi\in C_{0}^{2}({{\Bbb R}} ^N, [0, 1]) $,并设$ \varphi(x, y) = \xi^{m}(x, y) $,这里$ m > 0 $由后面确定.根据$ \nabla\xi $和$ \Delta\xi $的有界性意味着

$ |\Delta \varphi|\leq C \varphi^{\frac{m+2}{m}}, \qquad \varphi^{-1}|\nabla \varphi |^{2}\leq C\varphi^{\frac{m+2}{m}}. $

记$ \varphi_{R}(x, y): = \varphi \left (\frac{x}{R}, \frac{y}{R^{a+1}} \right) $和$ \Psi_{R}: = \varphi_{R}\Psi $,则

$ \begin{eqnarray*} \begin{array}{cc} \sup\limits_{{{\Bbb R}} ^{n}}\Psi_R (x, y) = \max\limits_{{{\Bbb R}} ^{n}} \Psi_R (x, y) \longrightarrow M \;\ \mbox{as} \;\ R \longrightarrow +\infty. \end{array} \end{eqnarray*} $

设$ (x_{R}, y_{R}) $是最大值点,即$ \Psi_{R}(x_{R}, y_{R}) = \max\limits_{{{\Bbb R}} ^{n}}{\Psi_{R}}(x, y) $.显然

$ \nabla_{G}\Psi_R (x_{R}, y_{R}) = 0, \; \;\ \Delta_{G}\Psi_R (x_{R}, y_{R})\le 0. $

下面所有的估计式都在点$ (x_{R}, y_{R}) $取得.直接计算可得

$ 0 = \nabla_{G}\Psi_{R}(x_R, y_R) = \nabla_{G}\varphi_{R}\Psi+\varphi_R\nabla_{G}\Psi \Longrightarrow \nabla_{G}\Psi = -\varphi_{R}^{-1}\nabla_{G}\varphi_R \Psi. $

不等式$ \Delta_{G}\Psi_{R}(x_{R}, y_{R})\leq0 $表明

$ \begin{eqnarray*} \Delta_{G}\varphi_{R}\Psi+2\nabla_{G}\varphi_{R}\cdot\nabla_{G}\Psi+\varphi_{R} \Delta_G \Psi \le 0. \end{eqnarray*} $

联合上面两个不等式发现

$ \varphi_{R}\Delta_{G}\Psi \le \Big [ 2\varphi_{R}^{-1}|\nabla_{G}\varphi_{R}|^{2}-\Delta_{G}\varphi_{R} \Big ] \Psi \le CR^{-2}\varphi_{R}^{\frac{m-2}{m}}\Psi. $

因此联合不等式(2.6),有

$ \begin{eqnarray*} \delta^{-1}\omega u^{\frac{q-1}{2}} \varphi_R \Psi\leq CR^{-2}\varphi_{R}^{\frac{m-2}{m}}\Psi. \end{eqnarray*} $

选择$ m = \frac{2q+2}{q-1} $,则

$ \delta^{-1}\omega \left [\delta^{-1}(\Psi_R+v\varphi_R)\right ]^{\frac{q-1}{q+1}}\leq CR^{-2}. $

这是一个矛盾.

情况2  存在$ (x_0, y_0) $使得$ M = \sup\limits_{{{\Bbb R}} ^{N}}\Psi = \Psi(x_0, y_0) > 0 $.

从不等式(2.6)容易推出

$ \begin{eqnarray*} \Delta_{G}\Psi(x_0, y_0) \ge \delta^{-1}\omega u^{\frac{q-1}{2}}\Psi(x_0, y_0)>0. \end{eqnarray*} $

因此,至少存在一个指数$ i $,使得$ \frac{\partial^{2}}{\partial x_{i}^{2}}\Psi(x_0, y_0) > 0 $或$ \frac{\partial^{2}}{\partial y_{i}^{2}}\Psi(x_0, y_0) > 0 $.显然这矛盾于$ \Psi(x_0, y_0) = \sup\limits_{{{\Bbb R}} ^{N}} \Psi $.

引理2.5^[16]  设$ \Phi $是光滑函数, $ k_{a} = \frac{N_{a}}{N_{a}-2} $.则有

(2.7) $ \begin{eqnarray} \left (\int_{B_{R}\times B_{R^{a+1}}}\Phi^{2k_{a}}{\rm d}x \right )^{\frac{1}{k_{a}}} &\le & CR^{2+N_{a}\left (\frac{1}{k_{a}}-1 \right )}\int_{B_{2R} \times {B_{(2R)^{a+1}}}}|\nabla_{G}\Phi|^{2}{\rm d}x \\ && +CR^{N_{a}\left (\frac{1}{k_{a}}-1 \right )}\int_{B_{2R} \times {B_{(2R)^{a+1}}}}\Phi^{2}{\rm d}x. \end{eqnarray} $

证  设$ \varphi\in C_c^{2}({{\Bbb R}} ^{N}, [0, 1]) $是截断函数,使得在$ B_{1}\times B_{1} $上有$ \varphi = 1 $,在$ B_{2}\times B_{2}^{a+1} $外$ \varphi = 0 $.使用Sobolev不等式^[15],则

$ \begin{eqnarray*} \left (\int_{B_{1}\times B_{1}}\Phi^{2k_{a}}{\rm d}x \right )^{\frac{1}{2k_{a}}} & \le & \left (\int_{B_{2}\times B_{2^{a+1}}}(\Phi\varphi)^{2k_{a}}{\rm d}x \right )^{\frac{1}{2k_{a}}}\\ & \le & C \left (\int_{B_{2}\times B_{2^{a+1}}}|\nabla_{G}(\Phi\varphi)|^{2}{\rm d}x \right )^{\frac{1}{2}}\\ &\le & C\left (\int_{B_{2}\times B_{2^{a+1}}} \Big [|\nabla_{G}\Phi|^{2}+\Phi^{2} \Big ]{\rm d}x \right )^{\frac{1}{2}}. \end{eqnarray*} $

利用Rescaling方法意味着不等式(2.7)成立.

引理3.1  设$ (u, v) $是系统(1.1)的正稳定解,则对任意的$ \kappa\geq2 $和$ R > 0 $,有

$ \begin{eqnarray*} \int_{B_R\times B_{R^{a+1}}}\omega u^{q}v^{\kappa -1}{\rm d} x\leq\frac{C}{R^{2}}\int_{B_{2R}\times B_{(2R)^{a+1}}}v^{\kappa}{\rm d}x, \end{eqnarray*} $

其中$ k $满足

$ \begin{eqnarray*} {\cal L} (q, \kappa): = \kappa^4-\frac{32q}{q+1}\kappa^2+\frac{32q(q+3)}{(q+1)^2}\kappa-\frac{64q}{(q+1)^2}<0. \end{eqnarray*} $

证  设$ \xi_{i}\in C_c^2({{\Bbb R}}, [0, 1]) $, $ i = 1, 2 $,使得在$ [-1, 1] $上$ \xi_{i} = 1 $,在$ [-2^{1+(i-1)a}, 2^{1+(i-1)a}] $外$ \xi_{i} = 0 $.对任意的$ R > 0 $,定义$ \varphi_{R}(x, y) = \xi_{1}(\frac{|x|}{R})\xi_{2}(\frac{|y|}{R^{a+1}}) $.则存在与$ R $无关的常数$ C > 0 $,使得

$ \begin{eqnarray*} &&|\nabla_{x}\varphi_{R}|\leq CR^{-1}, \; \ \; \ |\nabla_{y}\varphi_{R}|\leq CR^{-(a+1)}, \\ &&|\Delta_{x}\varphi_{R}|\leq CR^{-2}, \; \ \; \ |\Delta_{y}\varphi_{R}|\leq CR^{-2(a+1)}. \end{eqnarray*} $

在(2.2)式中取测试函数$ \zeta = u^{\frac{\lambda+1}{2}}\varphi_R $,从而有

$ \begin{eqnarray*} \sqrt{q}\int_{{{\Bbb R}} ^N}\omega u^{\frac{q-1}{2}}u^{\lambda+1}\varphi_R^{2}{\rm d}x &\le & \int_{{{\Bbb R}} ^N}u^{\lambda+1}|\nabla_{G}\varphi_R|^{2}{\rm d}x +\int_{{{\Bbb R}} ^N}\left |\nabla_{G}\left (u^{\frac{\lambda+1}{2}} \right ) \right |^{2}\varphi_R^{2}{\rm d}x \\ && +(\lambda+1)\int_{{{\Bbb R}} ^N}u^{\lambda}\varphi_R\nabla_{G}u\cdot\nabla_{G}\varphi_R {\rm d}x. \end{eqnarray*} $

直接计算

$ (\lambda+1)\int_{{{\Bbb R}} ^N}u^{\lambda}\varphi_R\nabla_{G}u\cdot\nabla_{G}\varphi_R {\rm d}x = -\frac{1}{2}\int_{{{\Bbb R}} ^N}u^{\lambda+1}\Delta_{G}(\varphi_R^{2}){\rm d}x $

和

$ \begin{eqnarray*} \int_{{{\Bbb R}} ^N} \left |\nabla_{G} \left (u^{\frac{\lambda+1}{2}} \right ) \right |^{2} \varphi_R^{2}{\rm d}x & = &\frac{(\lambda+1)^{2}}{4}\int_{{{\Bbb R}} ^N}u^{\lambda-1}|\nabla_{G}u|^{2}\varphi_R^{2}{\rm d}x\\ & = & -\frac{(\lambda+1)^{2}}{4\lambda}\int_{{{\Bbb R}} ^N}u^{\lambda}\varphi_R^{2}\Delta_{G}u{\rm d}x -\frac{\lambda+1}{4\lambda}\int_{{{\Bbb R}} ^N}\nabla_{G}(u^{\lambda+1})\cdot\nabla_{G}(\varphi_R^{2}){\rm d}x\\ & = & \frac{(\lambda+1)^{2}}{4\lambda}\int_{{{\Bbb R}} ^N}\omega u^{\lambda}v\varphi_R^{2}{\rm d}x +\frac{\lambda+1}{4\lambda}\int_{{{\Bbb R}} ^N}u^{\lambda+1}\Delta_{G}(\varphi_R^{2}){\rm d}x. \end{eqnarray*} $

因此

$ \begin{eqnarray*} \sqrt{q}\int_{{{\Bbb R}} ^N}\omega u^{\frac{q-1}{2}}u^{\lambda+1}\varphi_R^{2}{\rm d}x & \le &\frac{(\lambda+1)^2}{4\lambda}\int_{{{\Bbb R}} ^N}\omega u^{\lambda}v\varphi_R^{2}{\rm d}x \\ && +C \int_{{{\Bbb R}} ^N}u^{\lambda+1}\Big [|\nabla_{G}\varphi_R|^{2}+ |\Delta_{G}(\varphi_R^{2})| \Big ]{\rm d}x. \end{eqnarray*} $

取$ \mu_{1}: = \frac{4\lambda\sqrt{q}}{(\lambda+1)^{2}} $,即

$ \begin{eqnarray*} \int_{{{\Bbb R}} ^N}\omega u^{\frac{q-1}{2}} u^{\lambda+1}\varphi_R^{2}{\rm d}x \le \frac{1}{\mu_{1}}\int_{{{\Bbb R}} ^N}\omega u^{\lambda}v\varphi_R^{2}{\rm d}x +\frac{C}{R^{2}}\int_{B_{2R} \times B_{(2R)^{a+1}}}u^{\lambda+1}{\rm d}x. \end{eqnarray*} $

又在(2.2)式中插入$ \zeta = v^{\frac{\tau+1}{2}}\varphi_R $, $ \tau\geq1 $,则可得

$ \begin{eqnarray*} \sqrt{q} \int_{{{\Bbb R}} ^N} \omega u^{\frac{q-1}{2}}v^{\tau+1}\varphi_R^{2}{\rm d}x & \le& \int_{{{\Bbb R}} ^N}u^{\tau+1}|\nabla_{G}\varphi_R|^{2}{\rm d}x +\int_{{{\Bbb R}} ^N} \left |\nabla_{G}\left (v^{\frac{\tau+1}{2}}\right ) \right |^{2}\varphi_R^{2}{\rm d}x \\ &&-\frac{1}{2}\int_{{{\Bbb R}} ^N}v^{\tau+1}\Delta_{G}(\varphi_R^{2}){\rm d}x. \end{eqnarray*} $

类似的选择$ \mu_{2}: = \frac{4\tau\sqrt{q}}{(\tau+1)^{2}} $,有

$ \begin{eqnarray*} \int_{{{\Bbb R}} ^N}\omega u^{\frac{q-1}{2}}v^{\tau+1}\varphi_R^{2}{\rm d}x \le \frac{1}{\mu_{2}}\int_{{{\Bbb R}} ^N}\omega u^{q}v^{\tau}\varphi_R^{2}{\rm d}x +\frac{C}{R^{2}}\int_{B_{2R} \times B_{(2R)^{a+1}}}v^{\tau+1}{\rm d}x. \end{eqnarray*} $

记

$ {\cal J}_{1}: = \int_{{{\Bbb R}} ^N}\omega u^{\frac{q-1}{2}}u^{\lambda+1}\varphi_{R}^{2}{\rm d}x, $

$ {\cal J}_{2}: = \int_{{{\Bbb R}} ^N}\omega u^{\frac{q-1}{2}}v^{\tau+1}\varphi_{R}^{2}{\rm d}x. $

因此

$ \begin{eqnarray*} {\cal J}_1+\mu_{2}^{\tau+1}{\cal J}_2 &\le &\frac{1}{\mu_{1}}\int_{{{\Bbb R}} ^N}\omega u^{\lambda}v\varphi_R^{2}{\rm d}{ x}+\mu_{2}^{\tau}\int_{{{\Bbb R}} ^N}\omega u^{q}v^{\tau}\varphi_R^{2}{\rm d}x\\ & & +\frac{C}{R^{2}} \int_{B_{2R} \times B_{(2R)^{a+1}}}(u^{\lambda+1}+v^{\tau+1})\varphi_R^{2}{\rm d}{ x}. \end{eqnarray*} $

选择$ \lambda $和$ \tau $满足关系式

$ \begin{eqnarray*} 2\lambda = (q+1)\tau+q-1 \Longleftrightarrow \lambda+1 = \frac{(q+1)(\tau+1)}{2}. \end{eqnarray*} $

使用Young不等式可得

$ \begin{eqnarray*} \frac{1}{\mu_{1}}\int_{{{\Bbb R}} ^N}\omega u^{\lambda}v\varphi_R^{2}{\rm d}x \le \frac{\tau}{\tau+1}\int_{{{\Bbb R}} ^N}\omega u^{\frac{q-1}{2}}u^{\lambda+1}\varphi_R^{2}{\rm d}x +\frac{1}{\mu_{1}^{\tau+1}(\tau+1)}\int_{{{\Bbb R}} ^N}\omega u^{\frac{q-1}{2}}v^{\tau+1}\varphi_R^{2}{\rm d}x. \end{eqnarray*} $

类似的计算,有

$ \begin{eqnarray*} \mu_{2}^{\tau}\int_{{{\Bbb R}} ^N}\omega u^{q}v^{\tau}\varphi_R^{2}{\rm d}x \le \frac{1}{\tau+1}\int_{{{\Bbb R}} ^N}\omega u^{\frac{q-1}{2}}u^{\lambda+1}\varphi_R^{2}{\rm d}x +\frac{\mu_{2}^{\tau+1}\tau}{\tau+1}\int_{{{\Bbb R}} ^N}\omega u^{\frac{q-1}{2}}v^{\tau+1}\varphi_R^{2}{\rm d}x. \end{eqnarray*} $

则再结合上面三个不等式可知

$ \begin{eqnarray*} {\cal J}_1+\mu_{2}^{\tau+1}{\cal J}_2 &\le & \frac{\tau}{\tau+1}{\cal J}_1+\frac{1}{\tau+1}{\cal J}_1 +\frac{1}{{\mu_{1}^{\tau+1}}(\tau+1)}{\cal J}_2 +\frac{\mu_2^{\tau+1}\tau}{\tau+1}{\cal J}_2\\ && +\frac{C}{R^{2}}\int_{B_{2R} \times B_{(2R)^{a+1}}} (u^{\lambda+1}+v^{\tau+1}){\rm d}x. \end{eqnarray*} $

即

$ \begin{eqnarray*} \frac{(\mu_1 \mu_2)^{\tau+1}-1}{\mu_1^{\tau+1}(\tau+1)}{\cal J}_2 \le \frac{C}{R^{2}} \int_{B_{2R} \times B_{(2R)^{a+1}}} (u^{\lambda+1}+v^{\tau+1}){\rm d}x. \end{eqnarray*} $

如果$ \mu_1 \mu_2 > 1 $,则从引理2.4发现

$ \begin{eqnarray*} \int_{B_R \times B_{R^{a+1}}} \omega u^{\frac{q-1}{2}}v^{\tau+1}{\rm d} x\leq\frac{C}{R^{2}} \int_{B_{2R} \times B_{(2R)^{a+1}}}v^{\tau+1}{\rm d}x. \end{eqnarray*} $

下面,我们取$ \kappa = \tau+1 $并结合引理2.4可得

$ \begin{eqnarray*} \int_{B_R \times B_{R^{a+1}}} \omega u^{q}v^{\kappa-1}{\rm d}x & = &\int_{B_R \times B_{R^{a+1}}} \omega u^{\frac{q-1}{2}}u^{\frac{q+1}{2}}v^{\kappa -1}{\rm d} x\\ & \le &C \int_{B_R \times B_{R^{a+1}}} \omega u^{\frac{q-1}{2}}v^{\kappa} {\rm d}x \\ & \le & \frac{C}{R^{2}}\int_{B_{2R} \times B_{(2R)^{a+1}}} v^{\kappa} {\rm d}x. \end{eqnarray*} $

另一方面,因为$ \mu_{1} = \frac{4\lambda\sqrt{q}}{(\lambda+1)^{2}} $, $ \mu_{2} = \frac{4\tau\sqrt{q}}{(\tau+1)^{2}} $, $ 2\lambda = (q+1)\tau+q-1 $, $ \lambda+1 = \frac{(q+1)(\tau+1)}{2} $和$ \kappa = \tau+1 $,有

$ \mu_1\mu_2>1 \Longleftrightarrow 8q \Big [(q+1)(\kappa-1)+q-1 \Big ](\kappa -1) >\frac{(q+1)^2}{4} \kappa^4. $

直接计算表明

$ {\cal L} (q, \kappa): = \kappa^4-\frac{32q}{q+1} \kappa^2+\frac{32q(q+3)}{(q+1)^2}\kappa -\frac{64q}{(q+1)^2}<0. $

结论得证.

注3.1  可以注意到,当$ \kappa\in(2s_{0}^{-}, 2s_{0}^{+}) $, $ q > 1 $时,引理3.1成立(参见文献[7, Remark 3.2]),其中

$ \begin{eqnarray*} &&s_0^-: = \sqrt{\frac{2q}{q+1}}-\sqrt{\frac{2q}{q+1}-\sqrt{\frac{2q}{q+1}}}, \\ &&s_0^+: = \sqrt{\frac{2q}{q+1}}+\sqrt{\frac{2q}{q+1}-\sqrt{\frac{2q}{q+1}}}. \end{eqnarray*} $

而且,由(1.5)式定义的$ {\cal L} $在区间$ [2s_0^+, +\infty) $上有唯一根$ \kappa_0 $,使得对任意的$ \kappa \in [2s_0^+, \kappa_0) $,有$ {\cal L}(q, \kappa) < 0 $.

定理1.1的证明  用$ v^{2s-1} \varphi_R^2 $乘以系统(1.1)的第二个方程并分部积分可得

$ \begin{eqnarray*} \int_{{{\Bbb R}} ^N} |\nabla_G v|^2 v^{2s-2} \varphi_R^2 {\rm d}x& = &\frac{1}{2s-1} \int_{{{\Bbb R}} ^N} \omega u^q v^{2s-1} \varphi_R^2 {\rm d}x +\frac{1}{2s(2s-1)}\int_{{{\Bbb R}} ^N} v^{2s} \Delta_G (\varphi_R^2) {\rm d}x. \end{eqnarray*} $

这里$ \varphi_{R} $的定义与引理3.1相同.取$ \Phi = v^s $,其中$ 2s $满足不等式(1.5),则引理3.1意味着

$ \begin{eqnarray*} \int_{B_R \times B_{R^{a+1}}} |\nabla_G \Phi |^2 {\rm d}x & = &s^2 \int_{B_R \times B_{R^{a+1}}} |\nabla_G v|^2 v^{2s-2} {\rm d}x \\ & \le & \frac{s^2}{2s-1} \int_{B_{2R} \times B_{(2R)^{a+1}}} \omega u^q v^{2s-1} \varphi_R^2 {\rm d} x \\ & & +\frac{s}{2(2s-1)}\int_{B_{2R} \times B_{(2R)^{a+1}}} v^{2s} \Delta_G (\varphi_R^2) {\rm d}x \\ & \le & \frac{C}{R^2} \int_{B_{2R} \times B_{(2R)^{a+1}}} v^{2s} {\rm d}x, \end{eqnarray*} $

联合引理 可得

(3.1) $ \begin{equation} \left (\int_{B_R \times B_{R^{a+1}}} v^{2sk_a} {\rm d}x \right )^{\frac{1}{k_a}} \le CR^{N_a \left (\frac{1}{k_a}-1 \right )} \int_{B_{2R} \times B_{(2R)^{a+1}}} v^{2s} {\rm d}x. \end{equation} $

设$ m $是满足$ 2\kappa_{a}^{m-1} < \kappa < 2\kappa_{a}^{m} $的非负整数.定义一个递增迭代序列

$ s_0 = 1, \; s_1 = k_a, \; s_2 = k_a^2, \cdots, s_m = k_a^m. $

取$ V = v^2 $,迭代不等式(3.1)可得

$ \left (\int_{B_R \times B_{R^{a+1}}} V^{k_a^m} {\rm d}x \right )^{\frac{1}{k_a^m}} \le CR^{N_a \left (\frac{1}{k_a^m}-1 \right )} \int_{B_{2^mR} \times B_{(2^mR)^{a+1}}} V {\rm d}x. $

现在取$ \gamma = 2k_a^m $和$ \gamma < \frac{N_a}{N_a-2} \kappa_0 $,则

$ \begin{eqnarray*} \left (\int_{B_R \times B_{R^{a+1}}} v^{\gamma} {\rm d}x\right )^{\frac{2}{\gamma}}& \le & CR^{N_a \left (\frac{2}{\gamma}-1 \right )} \int_{B_{2^mR} \times B_{(2^mR)^{a+1}}} v^2 {\rm d}x \\ &\le & CR^{N_a\left (\frac{2}{\gamma}-1 \right )+N_a-4-\beta-\frac{4(\beta+2)}{q-1}}. \end{eqnarray*} $

容易看出

$ N_a\left (\frac{2}{\gamma}-1 \right )+N_a-4-\beta-\frac{4(\beta+2)}{q-1}<0 $

等价于

$ N_a<2+\left [4+\beta+\frac{4(\beta+2)}{q-1} \right ] \frac{\kappa_0}{2}. $

因此从不等式(1.4)可得当$ R \to +\infty $时, $ \|v\|_{L^{\gamma}({{\Bbb R}} ^N)} = 0 $,即,在$ {{\Bbb R}} ^N $中$ v \equiv 0 $,这是矛盾.结论得证.