本文考虑了加权椭圆系统

$ \begin{equation}\label{eq:1.1} \left\{\begin{array}{ll} -\Delta u=|x|^{\beta} v, \\ -\Delta v=|x|^{\alpha} |u|^{p-1}u, \end{array}\right. \qquad x \in \Omega, \end{equation} $

(1.1)

其中$\Omega$是${\Bbb R}^N$的一个开子集, $N \ge 5$, $\alpha>-4$, $0 \le \beta <N-4$和$p>1$.

Lane-Emden系统出现在物理、化学和生物等领域, 其著名的Lane-Emden猜想已引起许多专家的关注并获得很多创新性研究成果^[1-3].而描述气象物理星结构的Henon方程$-\Delta u=|x|^{\alpha} u^p$因为权项的出现会致使研究变得更加复杂, 也产生了许多新方法^[4-5].利用单调公式的方法研究椭圆方程和系统已有很多优秀的研究成果, 而构建其单调公式的方法有很多, 如文献[6-13]利用Pohozaev恒等式或直接计算进行构造.但对于二阶椭圆系统(1.1), 利用Pohozaev恒等式建立其单调公式计算相对繁琐不直观^[12].基于此, 受文献[8-9]的思想启发, 本文利用分析的技巧重新构建系统(1.1) 的单调公式.相比于文献[12, 定理2.1], 其计算简便直接.

定义1.1  系统(1.1) 的能量泛函

$ {\cal E}(u)=\int_{\Omega} \left [\frac{1}{2}\frac{|\Delta u|^2}{|x|^{\beta}}-\frac{1}{p+1} |x|^{\alpha} |u|^{p+1} \right]{\rm d}x $

的临界点$u \in C^2(\Omega)$称为系统(1.1) 的稳定解, 如果对任意的$\zeta \in C_0^2 (\Omega)$, 有

$ p\int_{\Omega} |x|^{\alpha}|u|^{p-1}\zeta^2 {\rm d}x \le \int_{\Omega} \frac{|\Delta \zeta|^2}{|x|^{\beta}} {\rm d}x. $

引理1.1^[12]  设$u$是(1.1) 的稳定解, 则对于充分大的$m$和具有关系$0\le \psi \le 1$的任意$\psi \in C_0^4({\Bbb R}^N)$, 有

$ \begin{eqnarray*} \int_{{\Bbb R}^N}\left [\frac{|\Delta u|^2}{|x|^{\beta}} +|x|^{\alpha}|u|^{p+1}\right] \psi^{2m} {\rm d}x &\le& C \int_{{\Bbb R}^N}|x|^{-\frac{2\alpha+\beta p+\beta}{p-1}}{\cal Q}(\psi^m)^{\frac{p+1}{p-1}}{\rm d}x \\[0.15cm] && +C \int_{{\Bbb R}^N}|x|^{-\frac{2\alpha+(\beta+2)(p+1)}{p-1}}{\cal R}(\psi^m)^{\frac{p+1}{p-1}}{\rm d}x, \end{eqnarray*} $

$ \begin{equation}\label{eq:1.2} \int_{B_R(x)}\left [\frac{|\Delta u|^2}{|z|^{\beta}}+|z|^{\alpha}|u|^{p+1}\right]\psi^{2m} {\rm d}z \le CR^{N-4-\beta-\frac{8+2\alpha+2\beta}{p-1}}, \end{equation} $

(1.2)

其中

$ {\cal Q}(\psi^m)= |\nabla \psi|^4+\psi^{2(2-m)}\Big [|\nabla (\Delta \psi^m)\cdot \nabla \psi^m |+|\Delta \psi^m|^2+ \left |\Delta |\nabla \psi^m|^2 \right | \Big], $

$ {\cal R}(\psi^m)= \psi^{2(2-m)} \Big [\left |\Delta \psi^m \right |\left |x \cdot \nabla \psi^m \right |+ \left |\nabla \psi^m \right |^2+\left |x\cdot \nabla (|\nabla \psi^m|^2)\right | \Big]. $

任给定$x\in \Omega$.设$0<r<R$和$B_r(x) \subset B_R(x) \subset \Omega$, 令$u\in C^4(\Omega\backslash \{0\})\cap C^2(\Omega)$, 定义

$ \begin{eqnarray*} {\cal M}(r;x, u)&=& r^{\frac{8+2\alpha+2\beta}{p-1}+4+\beta-N} \int_{B_r(x)} \frac{1}{2}\frac{|\Delta u|^2}{|y|^{\beta}}-\frac{1}{p+1}|y|^{\alpha}|u|^{p+1}\\ && +\frac{1+\beta}{2}\frac{4+\alpha+\beta}{p-1}\left [N-2-\frac{4+\alpha+\beta}{p-1} \right]\left (r^{\frac{8+2\alpha+2\beta}{p-1}+1-N}\int_{\partial B_r(x)} u^2 \right )\\ && +\frac{4+\alpha+\beta}{2(p-1)}\left [N-2-\frac{4+\alpha+\beta}{p-1} \right] \frac{\rm d}{{\rm d}r} \left (r^{\frac{8+2\alpha+2\beta}{p-1}+2-N} \int_{\partial B_r(x)} u^2 \right )\\ & &+\frac{r^3}{2}\frac{\rm d}{{\rm d}r} \left [r^{\frac{8+2\alpha+2\beta}{p-1}+1-N}\int_{\partial B_r(x)} \left (\frac{4+\alpha+\beta}{p-1}r^{-1}u+\frac{\partial u}{\partial r}\right )^2 \right] \\ && +\frac{1+\beta}{2}r^{\frac{8+2\alpha+2\beta}{p-1}+3-N} \int_{\partial B_r(x)} \left (|\Delta u|^2-\left |\frac{\partial u}{\partial r}\right |^2 \right ) \\ && +\frac{1}{2}\frac{\rm d}{{\rm d}r} \left [ r^{\frac{8+2\alpha+2\beta}{p-1}+4-N} \int_{\partial B_r(x)} \left (|\Delta u|^2-\left |\frac{\partial u}{\partial r}\right |^2 \right ) \right], \end{eqnarray*} $

则可以证明${\cal M}(r;x, u)$关于$r$是单调的.

定理2.1  假定$N \ge 5$, $p>\frac{N+4+2\alpha+\beta}{N-4-\beta}$, $u\in C^4(\Omega\backslash \{0\}) \cap C^2(\Omega)$是系统(1.1) 的一个稳定解.则${\cal M}(r;x, u)$关于$r \in (0, R)$是非减的.而且

$ \begin{eqnarray*} \frac{\rm d}{{\rm d}r}{\cal M}(r;0, u) \ge C(N, p, \alpha, \beta) r^{-N+2+\frac{8+2\alpha+2\beta}{p-1}} \int_{\partial B_r(x)} \left (\frac{4+\alpha+\beta}{p-1} r^{-1}u+\frac{\partial u}{\partial r} \right )^2 {\rm d}S, \end{eqnarray*} $

其中

$ \rho =N-1-\frac{8+2\alpha+2\beta}{p-1}, \ \varsigma =\frac{4+\alpha+\beta}{p-1}\left [\frac{4+\alpha+\beta}{p-1}+2-N\right], $

$ C(N, p, \alpha, \beta) =2\left [\left (1+\frac{\beta}{2} \right )\rho-\varsigma-\left (1+\frac{\beta}{4} \right )^2 \right] >0. $

证  由方程(1.1) 的变分结构, 定义重尺度能量泛函为

$ \begin{equation}\label{eq:3.1} E(\lambda)=\lambda^{\frac{8+2\alpha+2\beta}{p-1}+4+\beta-N} \int_{B_{\lambda}}\frac{1}{2}\frac{|\Delta u|^2}{|x|^{\beta}}-\frac{1}{p+1}|x|^{\alpha}|u|^{p+1}. \end{equation} $

(2.1)

从引理1.1中的(1.2) 式, 对任意的$\lambda \ge 0$, 有$E(\lambda) \le C$, 从而$E(\lambda)$的定义是有意义的.记

$ u^{\lambda}(x):=\lambda^{\frac{4+\alpha+\beta}{p-1}}u(\lambda x)\;\ \mbox{和}\;\ v^{\lambda}(x):=\lambda^{\frac{2+\alpha+2p+\beta p}{p-1}}v(\lambda x). $

容易验证

$ -\Delta u^{\lambda}(x)=|x|^{\beta}v^{\lambda}(x), ~~ -\Delta v^{\lambda}(x)=|x|^{\alpha}|u^{\lambda}(x)|^{p-1}u^{\lambda}(x), $

且关于$\lambda$求导与关于$x$求导和求积分是可以进行交换的.重尺度变换(2.1) 式可得

$ E(\lambda)=\int_{B_1}\left [\frac{1}{2}\frac{|\Delta u^{\lambda} |^2}{|x|^{\beta}}-\frac{1}{p+1} |x|^{\alpha} |u^{\lambda}|^{p+1} \right]{\rm d}x. $

对充分小的$\epsilon>0$, 定义

$ \widetilde{E}(\lambda)=\displaystyle \int_{B_1\backslash B_{\epsilon}}\left [\frac{1}{2}\frac{|\Delta u^{\lambda} |^2}{|x|^{\beta}}-\frac{1}{p+1} |x|^{\alpha} |u^{\lambda}|^{p+1} \right]{\rm d}x, $

并对$\widetilde{E}(\lambda)$关于$\lambda$求导

$ \begin{eqnarray*} \frac{{\rm d}\widetilde{E}(\lambda)}{{\rm d}\lambda}& =&\int_{B_1\backslash B_{\epsilon}} \frac{\Delta u^{\lambda}}{|x|^{\beta}} \Delta \frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}-|x|^{\alpha}|u^{\lambda}|^{p-1}u^{\lambda}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda} \\ &=&\int_{\partial B_1}\left [\frac{\partial v^{\lambda}}{\partial r}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}-v^{\lambda}\frac{\partial }{\partial r}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\right]-\int_{\partial B_{\epsilon}}\left [\frac{\partial v^{\lambda}}{\partial r}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}-v^{\lambda}\frac{\partial }{\partial r}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\right]. \end{eqnarray*} $

计算可得当$\epsilon \to 0$时, 有$\widetilde{E}(\lambda) \to E(\lambda)$和$\frac{{\rm d}\widetilde{E}(\lambda)}{{\rm d}\lambda} \to \frac{{\rm d} E(\lambda)}{{\rm d}\lambda}$, 即

$ \begin{equation}\label{eq:3.2} \frac{{\rm d}E(\lambda)}{{\rm d}\lambda}=\int_{\partial B_1} \frac{\partial v^{\lambda}}{\partial r}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}-v^{\lambda}\frac{\partial }{\partial r}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}. \end{equation} $

(2.2)

下面, 将$u^{\lambda}$关于变量$r=|x|$的所有导数将全部用关于变量$\lambda$的导数进行表述.从$u^{\lambda}$和$v^{\lambda}$的定义, 分别关于$\lambda$求导可得

$ \begin{equation}\label{eq:3.3} \frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}(x)=\frac{1}{\lambda} \left [\frac{4+\alpha+\beta}{p-1}u^{\lambda}(x)+r\frac{\partial u^{\lambda}}{\partial r}(x)\right] \end{equation} $

(2.3)

和

$ \frac{{\rm d}v^{\lambda}}{{\rm d}\lambda}(x)=\frac{1}{\lambda} \left [\frac{2+\alpha+2p+\beta p}{p-1}v^{\lambda}(x)+r\frac{\partial v^{\lambda}}{\partial r}(x) \right]. $

对(2.3) 式再次关于$\lambda$求导

$ \lambda \frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}+\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}=\frac{4+\alpha+\beta}{p-1}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}+r\frac{\partial }{\partial r}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}(x). $

综上所述, 有

$ r\frac{\partial}{\partial r}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}=\lambda \frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}+\frac{p-5-\alpha-\beta}{p-1}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}(x), $

$ r\frac{\partial v^{\lambda}}{\partial r}=\lambda \frac{{\rm d}v^{\lambda}}{{\rm d}\lambda}-\frac{2+\alpha+2p+2\beta}{p-1}v^{\lambda}(x). $

把上面两个式子代入(2.2) 式可得

$ \begin{eqnarray*} \frac{{\rm d}E(\lambda)}{{\rm d}\lambda}&=&\int_{\partial B_1}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\left (\lambda\frac{{\rm d}v^{\lambda}}{{\rm d}\lambda}-\frac{2+\alpha+2p+\beta p}{p-1}v^{\lambda} \right ) \\ && -v^{\lambda}\left (\lambda \frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}+\frac{p-5-\alpha-\beta}{p-1}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda} \right )\\ & =&-\int_{\partial B_1}\lambda v^{\lambda}\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}+(3+\beta)v^{\lambda}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda} -\lambda\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\frac{{\rm d}v^{\lambda}}{{\rm d}\lambda}. \end{eqnarray*} $

因为$\lambda\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda} =\frac{4+\alpha+\beta}{p-1}u^{\lambda}+r\frac{\partial u^{\lambda}}{\partial r}$, 则

$ \lambda \frac{\partial }{\partial r}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}= \frac{p+3+\alpha+\beta}{p-1}\frac{\partial u^{\lambda}}{\partial r}+r\frac{\partial^2 u^{\lambda}}{\partial r^2}, $

从而在$\partial B_1$上意味着

$ \begin{eqnarray*} \frac{\partial^2 u^{\lambda}}{\partial r^2}&= &\lambda \frac{\partial }{\partial r}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}-\frac{p+3+\alpha+\beta}{p-1}\frac{\partial u^{\lambda}}{\partial r}\\ &=&\lambda^2 \frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}+\frac{p-5-\alpha-\beta}{p-1} \lambda \frac{{\rm d}u^{\lambda}}{{\rm d}\lambda} \\ && -\frac{p+3+\alpha+\beta}{p-1}\left ( \lambda \frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}-\frac{4+\alpha+\beta}{p-1}u^{\lambda} \right )\\ & =& \lambda^2\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}-\frac{8+2\alpha+2\beta}{p-1}\lambda \frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}+\left [1+\frac{4+\alpha+\beta}{p-1} \right]\frac{4+\alpha+\beta}{p-1}u^{\lambda}. \end{eqnarray*} $

对于$r=|x|$和$\theta=\frac{x}{|x|} \in {\Bbb S}^{N-1}$, 使用球面坐标有$u^{\lambda}(x)=u^{\lambda}(r, \theta)$, 则在$\partial B_1$上可得

$ \begin{eqnarray*} -v^{\lambda}&=& -|x|^{\beta}v^{\lambda}=\Delta u^{\lambda}= \frac{\partial^2 u^{\lambda}}{\partial r^2}+\frac{N-1}{r}\frac{\partial u^{\lambda}}{\partial r}+\frac{1}{r^2}\Delta_{\theta} u^{\lambda}\\ &= &\lambda^2\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}-\frac{8+2\alpha+2\beta}{p-1}\lambda \frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}+\left [1+\frac{4+\alpha+\beta}{p-1} \right]\frac{4+\alpha+\beta}{p-1}u^{\lambda}\\ &&+(N-1)\lambda \frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}-(N-1)\frac{4+\alpha+\beta}{p-1}u^{\lambda}+\Delta_{\theta}u^{\lambda}\\ &=&\lambda^2 \frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}+\rho \lambda \frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}+\varsigma u^{\lambda} +\Delta_{\theta}u^{\lambda}, \end{eqnarray*} $

其中

$ \rho: =N-1-\frac{8+2\alpha+2\beta}{p-1}, $

$ \varsigma: =\frac{4+\alpha+\beta}{p-1}\left [\frac{4+\alpha+\beta}{p-1}+2-N\right]. $

因此

$ \begin{eqnarray*} \frac{{\rm d}E(\lambda)}{{\rm d}\lambda}&= &\int_{\partial B_1} \lambda \left ( \lambda^2\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}+\rho \lambda\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}+\varsigma u^{\lambda} \right )\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}\\ &&+(3+\beta)\left (\lambda^2\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}+\rho\lambda \frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}+\varsigma u^{\lambda} \right )\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\\ && -\lambda \frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\frac{\rm d}{{\rm d}\lambda} \left (\lambda^2\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}+\rho \lambda\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}+\varsigma u^{\lambda} \right )\\ && +\int_{\partial B_1}\lambda \Delta_{\theta}u^{\lambda}\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}+(3+\beta)\Delta_{\theta}u^{\lambda}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda} -\lambda\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\Delta_{\theta}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\\ &:= &\ell_1+\ell_2. \end{eqnarray*} $

计算$\ell_1$得

$ \begin{eqnarray*} \ell_1&=&\int_{\partial B_1} \lambda \left ( \lambda^2\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}+\rho \lambda\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}+\varsigma u^{\lambda} \right )\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}\\[0.08cm] &&+(3+\beta)\left (\lambda^2\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}+\rho\lambda \frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}+\varsigma u^{\lambda} \right )\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\\[0.08cm] && -\lambda \frac{{\rm d}u^{\lambda}}{{\rm d}\lambda} \left (\lambda^2\frac{d^3u^{\lambda}}{{\rm d}\lambda^3}+(2+\rho) \lambda\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}+(\rho+\varsigma)\frac{{\rm d} u^{\lambda}}{{\rm d}\lambda} \right )\\[0.08cm] &=&\int_{\partial B_1}\lambda^3 \left (\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}\right )^2+(1+\beta)\lambda^2\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda} +\varsigma \lambda u^{\lambda}\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}\\[0.08cm] && +(3+\beta)\varsigma u^{\lambda}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}+\Big [(2+\beta)\rho-\varsigma\Big]\lambda \left (\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\right )^2 -\lambda^3 \frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\frac{d^3 u^{\lambda}}{{\rm d}\lambda^3}. \end{eqnarray*} $

注意到

$ -\lambda^3\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\frac{{\rm d}^3u^{\lambda}} {{\rm d}\lambda^3}= \frac{\rm d}{{\rm d}\lambda}\left (-\frac{\lambda^3}{2}\frac{\rm d}{{\rm d}\lambda}\left (\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\right )^2 \right )+3\lambda^2\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}+\lambda^3\left (\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}\right )^2, $

$ \lambda u^{\lambda}\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}= \frac{{\rm d}^2}{{\rm d}\lambda^2}\left (\frac{\lambda \left (u^{\lambda}\right )^2}{2} \right ) -2u^{\lambda}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}-\lambda \left (\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda} \right )^2. $

因此有

$ \begin{eqnarray*} \ell_1&=& \int_{\partial B_1}2\lambda \left [\lambda^2\left (\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2} \right )^2 +\left (2+\frac{\beta}{2}\right )\lambda \frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}+\left (\Big (1+\frac{\beta}{2}\Big )\rho-\varsigma \right ) \left (\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\right )^2\right]\\[0.08cm] && +\frac{1+\beta}{2}\varsigma \frac{{\rm d}\left (u^{\lambda}\right )^2}{{\rm d}\lambda}+\frac{\varsigma}{2}\frac{{\rm d}^2}{{\rm d}\lambda^2}\left [\lambda \left (u^{\lambda}\right )^2\right]-\frac{1}{2}\frac{\rm d}{{\rm d}\lambda} \left [\lambda^3\frac{\rm d}{{\rm d}\lambda}\left (\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\right )^2 \right]\\[0.08cm] &\ge&\int_{\partial B_1}\frac{1+\beta}{2}\varsigma \frac{d\left (u^{\lambda}\right )^2}{{\rm d}\lambda}+\frac{\varsigma}{2}\frac{{\rm d}^2}{{\rm d}\lambda^2}\left [\lambda \Big (u^{\lambda}\Big )^2\right]-\frac{1}{2}\frac{\rm d}{{\rm d}\lambda} \left [\lambda^3\frac{\rm d}{{\rm d}\lambda}\left (\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\right )^2 \right]. \end{eqnarray*} $

选择$p>\frac{N+4+2\alpha+\beta}{N-4-\beta}$, 从而有

$ \begin{eqnarray*} \left (1+\frac{\beta}{2}\right )\rho-\varsigma&=&\left (1+\frac{\beta}{2} \right )\left (N-1-\frac{8+2\alpha+2\beta}{p-1}\right ) -\frac{4+\alpha+\beta}{p-1}\left (\frac{4+\alpha+\beta}{p-1}+2-N \right )\\ & >&\left (1+\frac{\beta}{4} \right )^2. \end{eqnarray*} $

则可得

$ \begin{eqnarray}\label{eq:3.7} && 2\lambda^3\left (\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2} \right )^2+(4+\beta)\lambda^2\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda} +((2+\beta)\rho-2\varsigma)\lambda \left (\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\right )^2 \nonumber \\ & =&2\lambda\left (\lambda\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}+\left (1+\frac{\beta}{4}\right )\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda} \right )^2 +\left [(2+\beta)\rho-2\varsigma-2\left (1+\frac{\beta}{4}\right )^2 \right]\lambda \left (\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\right )^2 \\ & \ge& 0. \end{eqnarray} $

(2.4)

再计算$\ell_2$.在$\partial B_1$上分部积分可得

$ \begin{eqnarray*} \ell_2&=&\int_{\partial B_1}\lambda \Delta_{\theta}u^{\lambda}\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}+(3+\beta)\Delta_{\theta}u^{\lambda}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda} -\lambda\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\Delta_{\theta}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}\\[0.08cm] &=&\int_{\partial B_1}-\lambda \nabla_{\theta}u^{\lambda}\nabla_{\theta}\frac{{\rm d}^2u^{\lambda}}{{\rm d}\lambda^2}-(3+\beta)\nabla_{\theta}u^{\lambda} \nabla_{\theta}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda}+\lambda \left |\nabla_{\theta}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda} \right |^2\\[0.08cm] &=&-\frac{\lambda}{2}\frac{{\rm d}^2}{{\rm d}\lambda^2}\int_{\partial B_1}\left |\nabla_{\theta}u^{\lambda} \right |^2-\frac{3+\beta}{2}\frac{\rm d}{{\rm d}\lambda}\int_{\partial B_1} |\nabla_{\theta}u^{\lambda}|^2+2\lambda \int_{\partial B_1}\left |\nabla_{\theta}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda} \right |^2\\[0.08cm] &=& -\frac{1}{2}\frac{{\rm d}^2}{{\rm d}\lambda^2}\left [\lambda \int_{\partial B_1} \left |\nabla_{\theta}u^{\lambda} \right |^2 \right] -\frac{1+\beta}{2}\frac{\rm d}{{\rm d}\lambda} \int_{\partial B_1}\left |\nabla_{\theta}u^{\lambda} \right |^2 +2\lambda \int_{\partial B_1}\left |\nabla_{\theta}\frac{{\rm d}u^{\lambda}}{{\rm d}\lambda} \right |^2\\[0.08cm] &\ge&-\frac{1}{2}\frac{{\rm d}^2}{{\rm d}\lambda^2}\left [\lambda \int_{\partial B_1} \left |\nabla_{\theta}u^{\lambda} \right |^2 \right] -\frac{1+\beta}{2}\frac{\rm d}{{\rm d}\lambda} \int_{\partial B_1}\left |\nabla_{\theta}u^{\lambda} \right |^2. \end{eqnarray*} $

另一方面, 对上面所有的项应用重尺度变换可得

$ \begin{matrix} \int_{\partial {{B}_{1}}}{\frac{\text{d}}{\text{d}\lambda }}{{\left( {{u}^{\lambda }} \right)}^{2}}=\frac{\text{d}}{\text{d}\lambda }\left( {{\lambda }^{\frac{8+2\alpha +\beta }{p-1}+1-N}}\int_{\partial {{B}_{\lambda }}}{{{u}^{2}}} \right), \\ \int_{\partial {{B}_{1}}}{\frac{{{\text{d}}^{2}}}{\text{d}{{\lambda }^{2}}}}\left[ \lambda {{\left( {{u}^{\lambda }} \right)}^{2}} \right]=\frac{{{\text{d}}^{2}}}{\text{d}{{\lambda }^{2}}}\left( {{\lambda }^{\frac{8+2\alpha +\beta }{p-1}+2-N}}\int_{\partial {{B}_{\lambda }}}{{{u}^{2}}} \right), \\ \int_{\partial {{B}_{1}}}{{{\left( \frac{\text{d}{{u}^{\lambda }}}{\text{d}\lambda } \right)}^{2}}}={{\lambda }^{\frac{8+2\alpha +2\beta }{p-1}+1-N}}\int_{\partial {{B}_{\lambda }}}{{{\left( \frac{4+\alpha +\beta }{p-1}{{\lambda }^{-1}}u+\frac{\partial u}{\partial r} \right)}^{2}}}, \\ \int_{\partial {{B}_{1}}}{\frac{\text{d}}{\text{d}\lambda }}\left[ {{\lambda }^{3}}\frac{\text{d}}{\text{d}\lambda }{{\left( \frac{\text{d}{{u}^{\lambda }}}{\text{d}\lambda } \right)}^{2}} \right]=\frac{\text{d}}{\text{d}\lambda }\left[ {{\lambda }^{3}}\frac{\text{d}}{\text{d}\lambda }\left( {{\lambda }^{\frac{8+2\alpha +\beta }{p-1}+1-N}}\int_{\partial {{B}_{\lambda }}}{{{\left( \frac{4+\alpha +\beta }{p-1}{{\lambda }^{-1}}u+\frac{\partial u}{\partial r} \right)}^{2}}} \right) \right], \\ \frac{{{\text{d}}^{2}}}{\text{d}{{\lambda }^{2}}}\left( \lambda \int_{\partial {{B}_{1}}}{{{\left| {{\nabla }_{\theta }}{{u}^{\lambda }} \right|}^{2}}} \right)=\frac{{{\text{d}}^{2}}}{\text{d}{{\lambda }^{2}}}\left[ {{\lambda }^{\frac{8+2\alpha +2\beta }{p-1}+4-N}}\int_{\partial {{B}_{\lambda }}}{\left( |\nabla u{{|}^{2}}-{{\left| \frac{\partial u}{\partial r} \right|}^{2}} \right)} \right], \\ \ \ \frac{\text{d}}{\text{d}\lambda }\left( \int_{\partial {{B}_{1}}}{{{\left| {{\nabla }_{\theta }}{{u}^{\lambda }} \right|}^{2}}} \right)=\frac{\text{d}}{\text{d}\lambda }\left[ {{\lambda }^{\frac{8+2\alpha +2\beta }{p-1}+3-N}}\int_{\partial {{B}_{\lambda }}}{\left( |\nabla u{{|}^{2}}-{{\left| \frac{\partial u}{\partial r} \right|}^{2}} \right)} \right]. \\ \end{matrix} $

因此有

$ \begin{eqnarray*} \frac{{\rm d}E(\lambda)}{{\rm d}\lambda}& \ge&\frac{1+\beta}{2}\varsigma \frac{\rm d}{{\rm d}\lambda} \left (\lambda^{\frac{8+2\alpha+\beta}{p-1}+1-N}\int_{\partial B_{\lambda}} u^2 \right ) +\frac{\varsigma}{2}\frac{{\rm d}^2}{{\rm d}\lambda^2} \left ( \lambda^{\frac{8+2\alpha+\beta}{p-1}+2-N}\int_{\partial B_{\lambda}} u^2 \right )\\ & &-\frac{1}{2}\frac{\rm d}{{\rm d}\lambda} \left [\lambda^3 \frac{\rm d}{{\rm d}\lambda} \left ( \lambda^{\frac{8+2\alpha+\beta}{p-1}+1-N}\int_{\partial B_{\lambda}} \left (\frac{4+\alpha+\beta}{p-1} \lambda^{-1} u +\frac{\partial u}{\partial r} \right )^2\right ) \right] \\ && -\frac{1+\beta}{2} \frac{\rm d}{{\rm d}\lambda} \left [\lambda^{ \frac{8+2\alpha+2\beta}{p-1}+3-N}\int_{\partial B_{\lambda}}\left (|\nabla u|^2-\left |\frac{\partial u}{\partial r} \right |^2 \right ) \right]\\[0.08cm] && - \frac{1}{2}\frac{{\rm d}^2}{{\rm d}\lambda^2} \left [\lambda^{ \frac{8+2\alpha+2\beta}{p-1}+4-N}\int_{\partial B_{\lambda}}\left (|\nabla u|^2-\left |\frac{\partial u}{\partial r} \right |^2 \right ) \right]. \end{eqnarray*} $

证毕.

注2.1  (ⅰ) 从(2.4) 式可以选取

$ \begin{eqnarray*} C(N, p, \alpha, \beta)=\left [(2+\beta)\rho-2\varsigma-2\left (1+\frac{\beta}{4} \right )^2 \right]>0. \end{eqnarray*} $

特别, 如果$p=\frac{N+4+2\alpha+\beta}{N-4-\beta}$, 则

$ C(N, p, \alpha, \beta)=2+2\beta+\frac{7}{16}\beta^2+ \frac{1}{4}(N-4-\beta)(N-\beta) >0. $

因此, 定理2.1的结论对$p=\frac{N+4+2\alpha+\beta}{N-4-\beta}$也成立.

(ⅱ) 相比于文献[12, 定理2.1]利用Pohozaev恒等式构造单调公式, 本文给出的证明方法其计算简便直接.