本文研究含Hardy型势和临界指数的退化椭圆型方程

(1.1) $ \begin{equation} -(\Delta _x + |x|^{2\alpha }\Delta _y) u -\mu \frac{\psi ^2 u}{d(z)^2} = \frac{\psi ^s |u|^{2^*(s)-2}u}{d(z)^s}, \quad (x, y)\in{{\Bbb R}} ^{m}\times{{\Bbb R}} ^{n} \end{equation} $

非平凡解的存在性和渐近性问题, 其中$ \alpha>0 $, $ -(\Delta_{x}+|x|^{2\alpha}\Delta_{y}) $是Grushin算子, $ 0\leq \mu<\mu_{G} $, $ 0\leq s<2 $. 当$ x\neq0 $时, 该算子是椭圆型的且在流形$ \{0\}\times {{\Bbb R}} ^{n} $上是退化的; 当$ \alpha $是非负整数时, 该算子是Hörmander型算子; 当$ \alpha = 0 $时, Grushin算子就变化为经典的Laplace算子; 当$ \alpha = 1 $时, Grushin算子就变化为Heisenberg群上的次Laplace算子. 另外, 从Grushin算子的定义可以看出: 当$ \alpha>0 $时方程(1.1)在$ x $方向平移不是不变的, 从而与$ \alpha = 0 $的情形相比这是一个新的困难.

设

$ X_{i} = \frac{\partial }{\partial x_i}, \quad i = 1, 2, \cdots , m, \quad X_{m+j} = |x|^\alpha \frac{\partial }{\partial y_j}, \quad j = 1, 2, \cdots , n, $

则$ \nabla _\alpha = (X_{1}, X_{2}, \cdots , X_{N}) $表示Grushin梯度, 其中$ N = m+n $. 设$ h = (h_{1}, h_{2}, \cdots, h_{N})\in C^{1}({{\Bbb R}} ^{N}, {{\Bbb R}} ^{N}) $, 则Grushin向量场诱导的Grushin散度可表示为

$ {\rm{div}}_{\alpha}(h) = \sum\limits_{i = 1}^{N}X_{i}h_{i}. $

因此, Grushin算子表示如下

$ (\Delta_{x}+|x|^{2\alpha}\Delta_{y}) = {\rm{div}}_{\alpha}\nabla_{\alpha} = \sum\limits_{i = 1}^{N}X_{i}^{2}. $

对任意的$ z = (x, y)\in{{\Bbb R}} ^{m}\times{{\Bbb R}} ^{n} $, 定义$ d(z) = (|x|^{2(\alpha +1)}+(\alpha +1)^2|y|^2)^{\frac{1}{2(\alpha +1)}}, $则$ d(z) $称之为Grushin算子的自然度量. 在伸缩变换$ \delta _\lambda (x, y) = (\lambda x, \lambda ^{\alpha +1} y) $ ($ \lambda >0 $)下, $ d(z) $关于$ \lambda $是一次齐次的. 直接计算可知: $ d(z)^{2-Q} $是Grushin算子的基本解, 即$ d(z)^{2-Q} $满足如下方程的解$ (\Delta_{x}+|x|^{2\alpha}\Delta_{y})w(z) = 0, \quad z\neq 0. $

设$ D^{1, 2}_{\alpha}({{\Bbb R}} ^{N}) $为函数空间$ C^{\infty}_{0}({{\Bbb R}} ^{N}) $在范数$ \|\cdot\|_{D^{1, 2}_{\alpha}({{\Bbb R}} ^{N})} $下的完备化空间

$ \|u\|_{D^{1, 2}_{\alpha}({{\Bbb R}} ^{N})} = \bigg(\int_{{{\Bbb R}} ^{N}}|\nabla_{\alpha}u|^{2}{\rm d}z\bigg)^{\frac{1}{2}}. $

设$ I_{\mu}:D^{1, 2}_{\alpha}({{\Bbb R}} ^{N})\to {{\Bbb R}} $为方程(1.1)对应的能量泛函

$ I_{\mu}(u) = \frac{1}{2}\int_{{{\Bbb R}} ^{N}}( |\nabla_{\alpha} u|^2 - \mu \frac{\psi^{2}|u|^2}{d(z)^{2}}) {\rm d}z- \frac{1}{2^*(s)}\int_{{{\Bbb R}} ^{N}} \frac{\psi^{s}|u|^{2^*(s)}}{d(z)^{s}}{\rm d}z. $

则$ I_{\mu}\in C^{1}(D^{1, 2}_{\alpha} ({{\Bbb R}} ^{N}), {{\Bbb R}} ) $. 若对任意的$ v\in D^{1, 2}_{\alpha} ({{\Bbb R}} ^{N}) $, 有

$ \langle I'_{\mu}(u), v\rangle = \int_{{{\Bbb R}} ^{N}}\bigg( \nabla_{\alpha} u \nabla_{\alpha} v - \mu\frac{\psi^{2} uv}{d(z)^{2}}- \frac{ \psi^{s} |u|^{2^*(s)-2}uv}{d(z)^{s}}\bigg){\rm d}z = 0, $

则称$ u\in D^{1, 2}_{\alpha} ({{\Bbb R}} ^{N}) $为方程(1.1)的解.

为了运用变分方法处理方程(1.1), 我们需要如下形式的Hardy不等式^[1]

(1.2) $ \begin{equation} \int _{{{\Bbb R}} ^{N}} |\nabla _{\alpha} u|^{2} {\rm d}z \geq \mu_{G} \int _{{{\Bbb R}} ^{N}} \frac{\psi ^2 |u|^2}{d(z)^2}{\rm d}z, \quad \, \forall u\in C_0^{\infty }({{\Bbb R}} ^{N}), \end{equation} $

其中$ \psi: = |\nabla_{\alpha} d(z)| = (\frac{|x|}{d(z)})^{\alpha} $, $ \mu_{G} = (\frac{Q-2}{2})^{2} $是最佳常数, 且该常数是不可达的. 相关内容可参见文献[2-4]等. 结合(1.2)式与Sobolev不等式

$ \bigg(\int _{{{\Bbb R}} ^{N}} |u|^{2^*}{\rm d}z\bigg)^{\frac{2}{2^*}}\leq C_{m, n, \alpha} \int _{{{\Bbb R}} ^{N}} |\nabla _{\alpha} u|^{2} {\rm d}z, \quad \forall u\in C_0^{\infty }({{\Bbb R}} ^{N}), $

其中$ C_{m, n, \alpha} $是仅依赖$ m $, $ n $, $ \alpha $的正常数, $ 2^* = \frac{2Q}{Q-2} $是Sobolev临界指数, 可得如下的Hardy-Sobolev不等式

(1.3) $ \begin{equation} \bigg( \int _{{{\Bbb R}} ^N} \frac{\psi^s |u| ^{2^*(s)}}{d(z)^s} {\rm d}z \bigg) ^{\frac{2}{2^*(s)}} \leq C_{m, n, \alpha, s}\int _{{{\Bbb R}} ^N} |\nabla _{\alpha} u|^2 {\rm d}z, \quad \forall u\in D_{\alpha}^{1, 2} ({{\Bbb R}} ^N), \end{equation} $

其中$ C_{m, n, \alpha, s} $是仅依赖于$ m $, $ n $, $ \alpha $, $ s $的正常数, $ 0\leq s<2 $, $ 2^*(s): = \frac{2(Q-s)}{Q-2} $是Sobolev嵌入$ D^{1, 2}_{\alpha}({{\Bbb R}} ^{N})\hookrightarrow L^{2^*(s)}({{\Bbb R}} ^{N}, \frac{\psi^{s}}{d(z)^{s}}{\rm d}z) $临界指数. 不等式(1.3)的证明过程请见引理2.1的证明.

当$ \mu< \mu_G $时, 定义范数

$ \|u\| : = \bigg( \int_{{{\Bbb R}} ^N} ( |\nabla_{\alpha} u|^2 - \mu \frac{\psi^{2}|u|^2}{d(z)^{2}}) {\rm d}z \bigg)^\frac{1}{2}. $

由(1.2)式可知

(1.4) $ \begin{equation} (1-\frac{\mu_+}{\mu_G}) \|u\|^2_{D^{1, 2}_{\alpha}({{\Bbb R}} ^N)} \le \|u\|^2 \le (1+\frac{\mu_-}{\mu_G}) \|u\|^2_{D^{1, 2}_{\alpha}({{\Bbb R}} ^N)}, \quad \forall u\in D^{1, 2}_{\alpha}({{\Bbb R}} ^{N}), \end{equation} $

其中$ \mu_+ = \rm{max} \{\mu, 0\} $ and $ \mu_- = - \rm{max} \{\mu, 0\} $. 因此, 不等式(1.4)表明范数$ \| \cdot \| $与$ \| \cdot\|_{D^{1, 2}_{\alpha}({{\Bbb R}} ^N)} $是等价的. 结合(1.3)–(1.4)式, 存在常数$ C>0 $使得

$ \begin{eqnarray*} \label{eq1-6} C\bigg(\int_{{{\Bbb R}} ^N}\frac{\psi^{s} |u|^{2^*(s)}}{d(z)^{s}}{\rm d}z\bigg)^\frac{2}{{2^*(s)}} \leq \int_{{{\Bbb R}} ^N}\bigg( |\nabla_{\alpha}u|^2 - \mu \frac{\psi^{2} |u|^{2}}{d(z)^{2}}\bigg){\rm d}z, \quad \forall u \in D^{1, 2}_{\alpha} ({{\Bbb R}} ^N). \end{eqnarray*} $

从而, 从上述不等式出发可以定义如下的最佳常数

$ \begin{eqnarray*} \Lambda_{\mu, s}({{\Bbb R}} ^N) = \inf\limits_{u \in D^{1, 2}_{\alpha} ({{\Bbb R}} ^{N})\setminus \{0\}} \frac{\int_{{{\Bbb R}} ^{N}} |\nabla_{\alpha} u|^2 {\rm d}z - \mu \int_{{{\Bbb R}} ^{N}} \frac{\psi^{2}|u|^2}{d(z)^{2}} {\rm d}z} {(\int_{{{\Bbb R}} ^{N}} \frac{\psi^{s}|u|^{2^*(s)}}{d(z)^{s}}{\rm d}z)^\frac{2}{2^*(s)}}. \end{eqnarray*} $

显然该常数的可达性与方程(1.1)解的存在性是密切相关的.

奇异非线性问题因为与大量的数学物理问题有着密切的联系而一直受到人们的广泛关注, 近年来有许多数学学者对此类问题进行了研究, 主要讨论这类问题解的存在性和多解性, 如文献[5-11]研究了Laplacian和$ p $-Laplacian算子的相关问题; 有关分数阶Laplacian算子方程的结果可以参考文献[12-16]等. 由于Grushin型算子是退化的, 与之相关的含Hardy型势和临界增长方程相关问题的研究结果还不是很丰富, 本文作者已有了部分结果^[23-24]}. 利用上述文献的思路与方法, 结合Ekeland变分原理, 本文首先研究含临界Sobolev-Hardy指数和奇性项的退化椭圆方程解的存在性. 得出了如下的结论.

定理 1.1    若$ 0\leq s<2 $, $ 0<\mu<\mu_{G} $. 则常数$ \Lambda_{\mu, s}({{\Bbb R}} ^N) $是可达的.

此外, 在研究含Hardy型势的Laplace算子临界指数增长问题时, 讨论其相应的极限方程基态解的存在性和渐近性质是研究这类问题的基础. 确切的说, 在运用Brezis-Nirenberg分析技巧解决此类问题时, 我们常需要用极限方程的基态解作为试验函数来估算临界指数方程相应的能量泛函的水平上界, 此时极限方程解的表达式显得尤为重要. 但有些非线性算子的极限方程基态解并没有具体的表达式, 此时就需要估算该解在奇性点和无穷远点的渐近性质. 本文利用Grushin向量场算子的Moser迭代和Kelvin变换技巧, 运用与经典Laplace方程相类似的研究技巧, 得到了方程(1.1)的解在原点和无穷远点的渐近性质, 从而将Laplace算子方程相类似的性质推广到Grushin型退化算子方程中, 为研究Grushin算子相关的奇异问题奠定了基础. 具体结论表述如下.

定理 1.2    设$ 0\leq s<2 $, $ 0\leq\mu<\mu_{G} $. 若$ u\in D^{1, 2}_{\alpha}({{\Bbb R}} ^{N}) $是方程(1.1)的非平凡解, 则当$ d(z)\to 0 $时, 有

(1.5) $ \begin{equation} u(z) = O\bigg(\frac{1}{d(z)^{\frac{Q-2}{2}-\sqrt{(\frac{Q-2}{2})^{2}-\mu}}}\bigg); \end{equation} $

当$ d(z)\to +\infty $时, 有

(1.6) $ \begin{equation} u(z) = O\bigg(\frac{1}{d(z)^{\frac{Q-2}{2}+\sqrt{(\frac{Q-2}{2})^{2}-\mu}}}\bigg). \end{equation} $

本文结构如下: 在第2节中先给出Grushin向量场Sobolev-Hardy不等式的证明, 然后完成定理1.1的证明. 在第3节中用Moser迭代技巧和Kelvin变换方法给出定理1.2的证明.

本节首先给出(3)式的证明.

引理 2.1    假设$ 0\leq s<2 $. 则存在常数$ C_{1}>0 $使得

(2.1) $ \begin{equation} \bigg(\int_{{{\Bbb R}} ^{N}} \frac{\psi^{s}|u|^{2^*(s)}}{d(z)^{s}}{\rm d}z\bigg)^\frac{2}{{2^*(s)}} \leq C_{1} \int_{{{\Bbb R}} ^N} |\nabla_{\alpha}u|^2 {\rm d}z, \quad \forall u \in D^{1, 2}_{\alpha}({{\Bbb R}} ^N) . \end{equation} $

进一步, 若$ \mu< \mu_{G} $, 则存在常数$ C_{2}>0 $使得

(2.2) $ \begin{equation} C_{2}\bigg(\int_{{{\Bbb R}} ^N} \frac{\psi^{s}|u|^{2^*(s)}}{d(z)^{s}}{\rm d}z\bigg)^\frac{2}{{2^*(s)}} \leq \int_{{{\Bbb R}} ^N} |\nabla_{\alpha}u|^2 {\rm d}z - \mu \int_{{{\Bbb R}} ^N}\frac{\psi^{2} |u|^{2}}{d(z)^{2}}{\rm d}z, \quad \forall u \in D^{1, 2}_{\alpha} ({{\Bbb R}} ^N). \end{equation} $

证    当$ s = 0 $, 不等式(2.1)就是Grushin向量场Sobolev不等式. 下面考虑$ 0 <s< 2 $. 利用Hölder不等式, Hardy不等式和Sobolev不等式可得

$ \begin{eqnarray*} \int_{{{\Bbb R}} ^N} \frac{\psi^{s}|u|^{2^*(s)}}{d(z)^{s}}{\rm d}z & = &\int_{{{\Bbb R}} ^N} \frac{\psi^{s}|u|^s}{d(z)^{s}} |u|^{2^*(s)-s}{\rm d}z \\ & \leq& \bigg(\int_{{{\Bbb R}} ^N} \frac{\psi^{2} |u|^2}{d(z)^{2}}{\rm d}z\bigg)^\frac{s}{2} \bigg(\int_{{{\Bbb R}} ^N} |u|^{(2^*(s)-s) \frac{2}{2-s} }{\rm d}z\bigg)^\frac{2-s}{2} \\ & = &\bigg(\int_{{{\Bbb R}} ^N} \frac{\psi^{2} |u|^2}{d(z)^{2}}{\rm d}z\bigg)^\frac{s}{2} \bigg(\int_{{{\Bbb R}} ^N} |u|^{2^*}{\rm d}z\bigg)^\frac{2-s}{2} \\ &\leq& \bigg(\frac{1}{\mu_{G}}\int_{{{\Bbb R}} ^N} |\nabla_{\alpha}u|^2 {\rm d}z\bigg)^\frac{s}{2} \bigg(C_{m, n, \alpha}\int_{{{\Bbb R}} ^N} |\nabla_{\alpha}u|^2 {\rm d}z\bigg)^{\frac{2^*}{2}.\frac{2-s}{2}}\\ & = &\mu_{G}^{-\frac{s}{2}}(C_{m, n, \alpha})^{\frac{Q(2-s)}{2(Q-2)}} \bigg(\int_{{{\Bbb R}} ^N} |\nabla_{\alpha}u|^2 {\rm d}z\bigg)^ \frac{2^*(s)}{2}. \end{eqnarray*} $

从而(2.1)式得证.

从$ \mu_{G} $的定义可以看出, 对任意的$ u \in D^{1, 2}_{\alpha}({{\Bbb R}} ^N), $有

$ \frac{\int_{{{\Bbb R}} ^N} |\nabla_{\alpha}u|^2 {\rm d}z- \mu \int_{{{\Bbb R}} ^N} \frac{\psi^{2}|u|^{2}}{d(z)^{2}}{\rm d}z} { (\int_ {{{\Bbb R}} ^N} \frac{\psi^{s}|u|^{2^*(s)}}{d(z)^s}{\rm d}z)^\frac{2}{2^*(s)} } \geq (1- \frac{\mu}{\mu_{G} }) \frac{ \int_{{{\Bbb R}} ^N} |\nabla_{\alpha}u|^2 {\rm d}z} { (\int_ {{{\Bbb R}} ^N} \frac{\psi^{s}|u|^{2^*(s)}}{d(z)^s}{\rm d}z)^\frac{2}{2^*(s)} }\geq C_{2}>0. $

从而, 当$ \mu< \mu_G $时, 有(2.2)式成立. 引理2.1得证.

定理1.1的证明    对任意的$ 0\leq \mu<\mu_{G} $, $ 0\leq s<2 $. 考察最佳Sobolev-Hardy常数$ \Lambda_{\mu, s}({{\Bbb R}} ^N) $达到函数的存在性等价于在空间$ D^{1, 2}_{\alpha}({{\Bbb R}} ^{N}) $上研究如下函数

$ I(u) = \frac{\int_{{{\Bbb R}} ^{N}} |\nabla_{\alpha} u|^2 {\rm d}z - \mu \int_{{{\Bbb R}} ^{N}} \frac{\psi^{2}|u|^2}{d(z)^{2}} {\rm d}z} {(\int_{{{\Bbb R}} ^{N}} \frac{\psi^{s}|u|^{2^*(s)}}{d(z)^{s}}{\rm d}z)^\frac{2}{2^*(s)}} $

极小值的存在性问题.

下面分两种情形来考虑.

情形1 $ 0<s <2 $. 对函数$ I(u) $运用Ekeland变分原理^[17]可知: 存在极小序列$ \{u_k\}_k\subset D^{1, 2}_{\alpha} ({{\Bbb R}} ^{N}) $使得

(2.3) $ \begin{equation} \lim\limits_{k\to \infty}\int_ {{{\Bbb R}} ^N} \frac{\psi^{s}|u_k|^ {2^*(s)}}{d(z)^s}{\rm d}z = 1, \end{equation} $

(2.4) $ \begin{equation} \lim\limits_{k\to \infty}I(u_k) = \Lambda_{\mu, s}({{\Bbb R}} ^N), \end{equation} $

以及在$ D^{1, 2}_{\alpha} ({{\Bbb R}} ^{N}) $的对偶空间$ (D^{1, 2}_{\alpha} ({{\Bbb R}} ^{N}))' $中有

(2.5) $ \begin{equation} \lim\limits_{k\to \infty}I'(u_k) = 0. \end{equation} $

定义函数$ J $, $ K:D^{1, 2}_{\alpha} ({{\Bbb R}} ^{N})\to {{\Bbb R}} $如下

$ J(u) = \frac{1}{2} \int_{{{\Bbb R}} ^{N}}\bigg( |\nabla_{\alpha} u|^2 - \mu\frac{\psi^{2}|u|^{2}}{d(z)^{2}}\bigg){\rm d}z, {\quad} K(u) = \frac{1}{2^*(s)}\int_{{{\Bbb R}} ^{N}}\frac{ \psi^{s}|u|^{2^*(s)}}{d(z)^{s}} {\rm d}z. $

则由(2.4)式和(2.5)式可得

(2.6) $ \begin{equation} \lim\limits_{k\to \infty}J(u_k) = \frac{1}{2} \Lambda_{\mu, s}({{\Bbb R}} ^N), \qquad \lim\limits_{k\to \infty}[J'(u_k) - \Lambda_{\mu, s}({{\Bbb R}} ^N) K'(u_k)] = 0. \end{equation} $

定义集中函数

$ {\cal Q}(r) = \int_{B_{d}(0, r)}\frac{\psi^{s}|u_k|^ {2^*(s)}}{d(z)^s}{\rm d}z, \quad r> 0, $

其中$ B_{d}(0, r): = \{z\in {{\Bbb R}} ^{m}\times{{\Bbb R}} ^{n}:d(z)<r\} $表示在度量$ d(z) $下以原点$ 0 $为圆心, $ r>0 $为半径的开球. 因为对任意的$ k \in \mathbb{N} $有$ \int_ {{{\Bbb R}} ^{N}}\frac{\psi^{s}|u_k|^ {2^*(s)}}{d(z)^s}{\rm d}z = 1 $, 利用连续性可知: 存在$ r_k>0 $使得对任意的$ k \in \mathbb{N} $有

$ {\cal Q}(r_k) = \int_{B_{d}(0, {r_k})}\frac{\psi^{s}|u_k|^ {2^*(s)}}{d(z)^s}{\rm d}z = \frac{1}{2}. $

令$ v_k(z): = r_k^{\frac{Q-2}{2}} u_k(\delta_{r_{k}}(z)) $ ($ k \in \mathbb{N} $). 对任意的$ \lambda>0 $, 简单计算可得

$ d\delta_{\lambda}(z) = \lambda^{Q} {\rm d}z, \qquad d(\delta_{\lambda}(z)) = \lambda d(z), $

$ \psi(z) = |\nabla_{\alpha}d(z)| = \bigg(\frac{|x|}{d(z)}\bigg)^{\alpha} = \bigg(\frac{\lambda|x|}{\lambda d(z)}\bigg)^{\alpha} = \psi(\delta_{\lambda}(z)). $

因此

$ \begin{eqnarray*} &&\int_{{{\Bbb R}} ^N} |\nabla_{\alpha} v_k|^2 {\rm d}z - \mu \int_{{{\Bbb R}} ^N} \frac{\psi^{2}|v_k(z)|^{2}}{d(z)^{2}}{\rm d}z\\ & = & r_{k}^{Q}\int_{{{\Bbb R}} ^N} |\nabla_{\alpha} u_k(\delta_{r_{k}}(z))|^2 \frac{{\rm d}\delta_{r_{k}}(z)}{r_{k}^{Q}} - \mu r_{k}^{Q-2}\int_{{{\Bbb R}} ^N}\psi(\delta_{r_{k}}(z))^{2} \frac{|u_k(\delta_{r_{k}}(z))|^{2}} {\frac{d(\delta_{r_{k}}(z))^{2}}{r_{k}^{2}}} \frac{{\rm d}\delta_{r_{k}}(z)}{r_{k}^{Q}}\\ & = &\int_{{{\Bbb R}} ^N} |\nabla_{\alpha} u_k(\delta_{r_{k}}(z))|^2 {\rm d}\delta_{r_{k}}(z) - \mu \int_{{{\Bbb R}} ^N}\psi(\delta_{r_{k}}(z))^{2} \frac{|u_k(\delta_{r_{k}}(z))|^{2}} {d(\delta_{r_{k}}(z))^{2}}{\rm d}\delta_{r_{k}}(z)\\ & = & \int_{{{\Bbb R}} ^N} |\nabla_{\alpha} u_k|^2 {\rm d}z - \mu \int_{{{\Bbb R}} ^N}\frac{\psi^{2} |u_k|^{2}}{d(z)^{2}}{\rm d}z, \end{eqnarray*} $

以及

$ \begin{eqnarray*} \int_ {{{\Bbb R}} ^N}\frac{ \psi(z)^{s}|v_k(z)|^ {2^*(s)}}{d(z)^s}{\rm d}z & = &\int_ {{{\Bbb R}} ^N} \psi(\delta_{r_{k}}(z))^{s} \frac{r_{k}^{\frac{Q-2}{2}\frac{2(Q-s)}{Q-2}}|u_k(\delta_{r_{k}}(z))|^{2^*(s)}} {r_{k}^{-s}d(\delta_{r_{k}}(z))^s}\frac{{\rm d}\delta_{r_{k}}(z)}{r_{k}^{Q}}\\ & = &\int_ {{{\Bbb R}} ^N} \frac{\psi(z)^{s}|u_k(z)|^ {2^*(s)}}{d(z)^s}{\rm d}z. \end{eqnarray*} $

从而, 结合上述两式与(2.3)–(2.4)式可得

(2.7) $ \begin{equation} \lim\limits_{k \to \infty} \int_{{{\Bbb R}} ^N} \bigg( |\nabla_{\alpha} v_k|^2 - \mu\frac{\psi^{2}|v_k|^{2}}{d(z)^{2}} \bigg){\rm d}z = \Lambda_{\mu, s}({{\Bbb R}} ^N); \end{equation} $

(2.8) $ \begin{equation} \lim\limits_{k \to \infty}\int_ {{{\Bbb R}} ^N} \frac{\psi^{s}|v_k|^ {2^*(s)}}{d(z)^s}{\rm d}z = 1. \end{equation} $

同理, 运用上述计算过程可得

(2.9) $ \begin{equation} \int_{B_{d}(0, 1)}\frac{\psi^{s} |v_k|^ {2^*(s)}}{d(z)^s} {\rm d}z = \frac{1}{2}, \quad \forall k \in \mathbb{N}. \end{equation} $

由(1.4)和(2.7)式可知$ \{\|v_k\|_{D^{1, 2}_{\alpha} ({{\Bbb R}} ^{N})} \} _{k \in \mathbb{N}} $是有界的. 所以, 存在子列, 仍记为$ \{v_k\} $, 使得

(2.10) $ \begin{equation} \left\{\begin{array}{llll} & v_k \rightharpoonup v & \rm{在 D^{1, 2}_{ \mathsf{ α}}({{\Bbb R}} ^N) 中}, \\ &v_k \to v& \rm{在 L_{\rm loc}^{q}({{\Bbb R}} ^N) 中}, \, \forall q\in [1, 2^*), \\ &v_k(z) \to v(z) &\rm{几乎处处于 {{\Bbb R}} ^N. } \end{array}\right. \end{equation} $

下证$ v\not\equiv0 $. 不妨假设$ v\equiv0 $, 则由(2.10)式可得

(2.11) $ \begin{equation} v_k \rightharpoonup 0\, \, {于 D^{1, 2}_{ \mathsf{ α}}({{\Bbb R}} ^{N}) 中}; \quad v_k \to 0\, \, {于 L_{\rm loc}^{q}({{\Bbb R}} ^N) (1\le q<2^*) 中}. \end{equation} $

取截断函数$ \eta \in C_0^{\infty}({{\Bbb R}} ^{N}) $, 当$ z\in B_{d}(0, 1) $时有$ 0 \le \eta \le 1 $; 当$ z\in B_{d}(0, \frac{1}{2}) $时有$ \eta\equiv 1 $. 在(2.6)式中取试验函数$ \eta^2 v_k(z) $, 则得

(2.12) $ \begin{eqnarray} &&\int_{{{\Bbb R}} ^{N}}\nabla_{\alpha} v_k \nabla_{\alpha} (\eta^2 v_k ) {\rm d}z - \mu \int_{{{\Bbb R}} ^{N}} \frac{\psi^{2}\, v_k (\eta^2 v_k) }{d(z)^{2}}{\rm d}z\\ & = & \Lambda_{\mu, s}({{\Bbb R}} ^N) \int_ {{{\Bbb R}} ^N}\frac{\psi^{s} \, |v_k|^{2^*(s)-2}v_{k} (\eta^2 v_k ) }{d(z)^s}{\rm d}z+o(1). \end{eqnarray} $

利用(2.11)式, 有

$ \begin{eqnarray*} \label{eq2-1-13} \int_{{{\Bbb R}} ^N} \nabla_{\alpha} v_k \cdot \nabla_{\alpha}(\eta^2 v_k) {\rm d}z & = &\int_{{{\Bbb R}} ^N}(| \nabla_{\alpha}(\eta v_k)|^2-|v_k \nabla_{\alpha} \eta|^2){\rm d}z\\ & = &\int_{{{\Bbb R}} ^N} | \nabla_{\alpha}(\eta v_k)|^2 {\rm d}z -\int_{{{\Bbb R}} ^N} |v_k \nabla_{\alpha} \eta|^2 {\rm d}z \\ & = &\int_{{{\Bbb R}} ^N} | \nabla_{\alpha}(\eta v_k)|^2 {\rm d}z - \int_{\rm{supp}(|\nabla_{\alpha} \eta|)}|\nabla_{\alpha} \eta|^2 |v_k |^2 {\rm d}z\\ & = &\int_{{{\Bbb R}} ^N} | \nabla_{\alpha}(\eta v_k)|^2 {\rm d}z+o(1). \end{eqnarray*} $

结合上式与(2.8), (2.9)和(2.12)式, 可得

(2.13) $ \begin{eqnarray} \|\eta v_k\|^2 & = & \int_{{{\Bbb R}} ^N} |\nabla_{\alpha} (\eta v_k)|^2 {\rm d}z - \mu\int_{{{\Bbb R}} ^N} \frac{\psi^{2} |\eta v_k|^2 }{d(z)^{2}}{\rm d}z\\ & = &\Lambda_{\mu, s}({{\Bbb R}} ^N)\int_ {{{\Bbb R}} ^N} \frac{\psi^{s}| v_k|^{2^*(s)-2} (|\eta v_k|^2) }{d(z)^s}{\rm d}z+o(1) \\ & \leq& \Lambda_{\mu, s}({{\Bbb R}} ^N) \int_{B_{d}(0, 1)} \frac{\psi^{s}|v_k|^{2^*(s)}}{d(z)^s} {\rm d}z+o(1) \\ & = & \frac{\Lambda_{\mu, s}({{\Bbb R}} ^N)}{2^{1-\frac{2}{2^*(s)}}} \bigg( \int_{B_{d}(0, 1)}\frac{\psi^{s} |v_k|^{2^*(s)}}{d(z)^s} {\rm d}z \bigg)^\frac{2}{2^*(s)}+o(1). \end{eqnarray} $

由于$ 2^*(s) < 2^* $, 利用(2.10)式可得

$ \int_{B_{d}(0, 1)} |v_k|^ {2^*(s)} {\rm d}z \to 0\quad (k \to \infty). $

因此, 结合上式及Hölder不等式, 计算可得

$ \begin{eqnarray*} \label{eq2-1-15} &&\bigg( \int_{B_{d}(0, 1)}\frac{\psi^{s} |v_k|^ {2^*(s)}}{d(z)^s} {\rm d}z \bigg)^\frac{1}{2^*(s)}\\ & = & \bigg( \int_{B_{d}(0, 1)} \frac{\psi^{s} |\eta v_k +(1- \eta) v_k |^ {2^*(s)}}{d(z)^s} {\rm d}z\bigg)^\frac{1}{2^*(s)}\\ &\leq& \bigg( \int_{B_{d}(0, 1)}\frac{\psi^{s} |\eta v_k|^{2^*(s)}}{d(z)^s} {\rm d}z \bigg)^\frac{1}{2^*(s)} + \bigg( \int_{B_{d}(0, 1)} \frac{\psi^{s}|(1- \eta )v_k|^ {2^*(s)}}{d(z)^s} {\rm d}z\bigg)^\frac{1}{2^*(s)}\\ & \leq &\bigg( \int_{{{\Bbb R}} ^N} \frac{\psi^{s}|\eta v_k|^ {2^*(s)}}{d(z)^s} {\rm d}z \bigg)^\frac{1}{2^*(s)} + C \bigg( \int_{B_{d}(0, 1)}|v_k|^{2^*(s)} {\rm d}z \bigg)^\frac{1}{2^*(s)}\\ & = &\bigg( \int_{{{\Bbb R}} ^N} \frac{\psi^{s}|\eta v_k|^ {2^*(s)}}{d(z)^s} {\rm d}z \bigg)^\frac{1}{2^*(s)} +o(1). \end{eqnarray*} $

所以

(2.14) $ \begin{equation} \bigg( \int_{B_{d}(0, 1)} \frac{\psi^{s}|v_k|^ {2^*(s)}}{d(z)^s} {\rm d}z \bigg)^\frac{2}{2^*(s)} \leq \bigg( \int_{{{\Bbb R}} ^N} \frac{\psi^{s}|\eta v_k|^ {2^*(s)}}{d(z)^s} {\rm d}z \bigg)^\frac{2}{2^*(s)} + o(1). \end{equation} $

因此, 将(2.14)式带入(2.13)式, 得

(2.15) $ \begin{equation} \|\eta v_k\|^2 \leq\frac{\Lambda_{\mu, s}({{\Bbb R}} ^N) }{2^{1-\frac{2}{2^*(s)}}} \bigg(\int_ {{{\Bbb R}} ^N}\frac{\psi^{s} |\eta v_k|^{2^*(s)}}{d(z)^s}{\rm d}z\bigg)^\frac{2}{2^*(s)}+o(1). \end{equation} $

另一方面, 由最佳常数$ \Lambda_{\mu, s}({{\Bbb R}} ^N) $的定义和(2.15) 式可知

$ \begin{eqnarray*} \label{eq2-1-18} \Lambda_{\mu, s}({{\Bbb R}} ^N) \bigg(\int_ {{{\Bbb R}} ^N} \frac{\psi^{s}|\eta v_k|^{2^*(s)}}{d(z)^s}{\rm d}z\bigg)^\frac{2}{2^*(s)} &\leq& \|\eta v_k\|^2 \\ &\leq& \frac{\Lambda_{\mu, s}({{\Bbb R}} ^N) }{2^{1-\frac{2}{2^*(s)}}} \bigg(\int_ {{{\Bbb R}} ^N} \frac{\psi^{s}|\eta v_k|^{2^*(s)}}{d(z)^s}{\rm d}z\bigg)^\frac{2}{2^*(s)}+o(1). \end{eqnarray*} $

从而, 由$ 2^{1-\frac{2}{2^*(s)}}>1 $可得

$ \lim\limits_{k\to \infty}\int_ {{{\Bbb R}} ^N} \frac{\psi^{s}|\eta v_k|^{2^*(s)}}{d(z)^s}{\rm d}z = \lim\limits_{k\to \infty}\int_{B_{d}(0, 1)} \frac{\psi^{s}|v_k|^{2^*(s)}}{d(z)^s} {\rm d}z = 0. $

与(2.9)式矛盾.故$ v\not\equiv 0 $.

下证$ v_k $在$ {{\Bbb R}} ^{N} $中弱收敛于$ v $, 且使得$ \int_ {{{\Bbb R}} ^{N}}\frac{\psi^{s} |v|^ {2^*(s)}}{d(z)^s}{\rm d}z = 1 $.

令$ \theta_k = v_k-v $, 由Brezis-Lieb引理^[18]可得

(2.16) $ \begin{equation} \int_{{{\Bbb R}} ^{N}} \frac{\psi^{s} |v_k|^{2^*(s)}}{d(z)^{s}}{\rm d}z = \int_{{{\Bbb R}} ^{N}} \frac{\psi^{s} |v|^{2^*(s)}}{d(z)^{s}}{\rm d}z + \int_{{{\Bbb R}} ^{N}} \frac{\psi^{s} |\theta_k|^{2^*(s)}}{d(z)^{s}}{\rm d}z+o(1), \end{equation} $

以及

(2.17) $ \begin{equation} \|v_k\|^2 = \|v\|^2+\|\theta_k\|^2 + o(1). \end{equation} $

由(2.6), (2.16)和(2.17)式和常数$ \Lambda_{\mu, s}({{\Bbb R}} ^N) $的定义可得

(2.18) $ \begin{eqnarray} o(1) & = &\|v_k\|^2 - \Lambda_{\mu, s}({{\Bbb R}} ^N) \int_{{{\Bbb R}} ^N} \frac{\psi^{s}|v_k|^{2^*(s)}}{d(z)^{s}}{\rm d}z \\& = &\|v\|^2 - \Lambda_{\mu, s}({{\Bbb R}} ^N) \int_{{{\Bbb R}} ^N} \frac{\psi^{s}|v|^{2^*(s)}}{d(z)^{s}}{\rm d}z + \|\theta_k\|^2 - \Lambda_{\mu, s}({{\Bbb R}} ^N) \int_{{{\Bbb R}} ^N}\frac{\psi^{s} |\theta_k|^{2^*(s)}}{d(z)^{s}}{\rm d}z + o(1) \\ & \geq& \Lambda_{\mu, s}({{\Bbb R}} ^N) \left[ \bigg(\int_{{{\Bbb R}} ^N} \frac{\psi^{s}|v|^{2^*(s)}}{d(z)^{s}}{\rm d}z\bigg)^{\frac{2}{2^*(s)}} - \int_{{{\Bbb R}} ^N} \frac{\psi^{s}|v|^{2^*(s)}}{d(z)^{s}}{\rm d}z \right] \\ &&+ \Lambda_{\mu, s}({{\Bbb R}} ^N) \left[ \bigg(\int_{{{\Bbb R}} ^N} \frac{\psi^{s}|\theta_k|^{2^*(s)}}{d(z)^{s}}{\rm d}z\bigg)^{\frac{2}{2^*(s)}} - \int_{{{\Bbb R}} ^N}\frac{\psi^{s}|\theta_k|^{2^*(s)}}{d(z)^{s}}{\rm d}z \right]+o(1) \\ & = &\Lambda_{\mu, s}({{\Bbb R}} ^N) (A+B), \end{eqnarray} $

其中

$ A: = \bigg(\int_{{{\Bbb R}} ^N} \frac{\psi^{s}|v|^{2^*(s)}}{d(z)^{s}}{\rm d}z\bigg)^{\frac{2}{2^*(s)}} - \int_{{{\Bbb R}} ^N} \frac{\psi^{s}|v|^{2^*(s)}}{d(z)^{s}}{\rm d}z, $

$ B: = \bigg(\int_{{{\Bbb R}} ^N} \frac{\psi^{s}|\theta_k|^{2^*(s)}}{d(z)^{s}}{\rm d}z\bigg)^{\frac{2}{2^*(s)}} - \int_{{{\Bbb R}} ^N} \frac{\psi^{s}|\theta_k|^{2^*(s)}}{d(z)^{s}}{\rm d}z. $

结合$ \lim\limits_{k\to +\infty}\int_{{{\Bbb R}} ^{N}} \frac{\psi^{s}|v_{k}|^{2^*(s)}}{d(z)^{s}}{\rm d}z = 1 $及(2.16)式可得

$ 0\leq \int_{{{\Bbb R}} ^{N}} \frac{\psi^{s}|v|^{2^*(s)}}{d(z)^{s}}{\rm d}z\leq 1, \qquad 0\leq \int_{{{\Bbb R}} ^{N}} \frac{\psi^{s}|\theta_k|^{2^*(s)}}{d(z)^{s}}{\rm d}z\leq 1. $

从而$ A\geq 0 $, $ B\geq 0 $, 且$ A+B = o(1) $. 因此, $ A = 0 $, $ B = o(1) $. 又因为$ v \not\equiv 0 $, 所以$ \int_{{{\Bbb R}} ^{N}}\frac{\psi^{s} |v|^{2^*(s)}}{d(z)^{s}}{\rm d}z = 1 $. 同时上述过程亦可推出

$ \int_{{{\Bbb R}} ^{N}}\bigg(|\nabla_{\alpha} v|^2 - \mu \frac{\psi^{2} |v|^{2}}{d(z)^{2}}\bigg){\rm d}z = \Lambda_{\mu, s}({{\Bbb R}} ^N). $

所以, 对任意的$ s \in (0, 2) $, 存在最佳常数$ \Lambda_{\mu, s}({{\Bbb R}} ^N) $的达到函数$ v $. 又因为$ |v| $亦满足上述过程, 所以不放假设$ v $是非负的. 即对任意的$ s \in (0, 2) $存在非平凡, 非负达到函数$ v $.

情形2 $ s = 0 $. 由文献[19]可知当$ s = 0 $, $ \mu = 0 $时$ \Lambda_{0, 0}({{\Bbb R}} ^N) $是可达的. 因此, 下面考虑当$ s = 0 $且$ 0<\mu <\mu_{G} $时$ \Lambda_{\mu, 0}({{\Bbb R}} ^N) $是可达的.

设$ \{u_k\}_{k\in \mathbb{N}} \subset D^{1, 2}_{\alpha}({{\Bbb R}} ^{N}) \setminus \{ 0 \} $是$ \Lambda_{\mu, 0}({{\Bbb R}} ^N) $的极小序列, 即

(2.19) $ \begin{equation} \lim\limits_{k \to \infty}\int_ {{{\Bbb R}} ^N} |u_k|^{2^*}{\rm d}z = 1, \end{equation} $

(2.20) $ \begin{equation} \lim\limits_{k \to \infty} \int_{{{\Bbb R}} ^N} \bigg( |\nabla_{\alpha} u_k|^2 - \mu \frac{\psi^{2}|u_k|^{2}}{d(z)^{2}}\bigg){\rm d}z = \Lambda_{\mu, 0}({{\Bbb R}} ^N). \end{equation} $

所以序列$ \{u_k\} _{k \in \mathbb{N}} $在$ D^{1, 2}_{\alpha}({{\Bbb R}} ^{N}) $中是有界的. 因此, 存在$ \{u_k\} _{k \in \mathbb{N}} $的子序列, 仍记为$ \{u_k\} _{k \in \mathbb{N}} $, 使得$ u_k \rightharpoonup u $于$ D^{1, 2}_{\alpha} ({{\Bbb R}} ^{N}) $, 并且

(2.21) $ \begin{equation} \| u_k\|^2_{D^{1, 2}_{\alpha}({{\Bbb R}} ^{N})} = \| u_k - u\|^2_{D^{1, 2}_{\alpha}({{\Bbb R}} ^{N})} +\| u\|^2_{D^{1, 2}_{\alpha}({{\Bbb R}} ^{N})}+o(1), \end{equation} $

(2.22) $ \begin{equation} \int_{{{\Bbb R}} ^{N}} \frac{\psi^{2}|u_{k}|^{2}}{d(z)^{2}}{\rm d}z = \int_{{{\Bbb R}} ^{N}} \frac{\psi^{2}|u_k-u|^{2}}{d(z)^{2}}{\rm d}z +\int_{{{\Bbb R}} ^{N}} \frac{\psi^{2}|u|^{2}}{d(z)^{2}}{\rm d}z+ o(1) = \int_{{{\Bbb R}} ^{N}} \frac{\psi^{2}|u|^{2}}{d(z)^{2}}{\rm d}z+ o(1), \end{equation} $

以及

(2.23) $ \begin{equation} \int_ {{{\Bbb R}} ^{N}} |u_k|^{2^*}{\rm d}z = \int_ {{{\Bbb R}} ^{N}} |u_k-u|^{2^*}{\rm d}z+\int_ {{{\Bbb R}} ^{N}} |u|^{2^*}{\rm d}z+o(1). \end{equation} $

从而, 利用(2.20)及(2.21)–(2.23)式可得

$ \begin{eqnarray*} \label{eq2-1-24*} \Lambda_{\mu, 0}({{\Bbb R}} ^N) & = & \int_{{{\Bbb R}} ^N} |\nabla_{\alpha} u_k|^2 {\rm d}z- \mu \int_{{{\Bbb R}} ^{N}}\frac{\psi^{2} |u_k|^{2}}{d(z)^{2}}{\rm d}z+ o(1) \\ & \geq& \| u_k - u \|^2_{D^{1, 2}_{\alpha}} +\| u\|^2_{D^{1, 2}_{\alpha}} - \mu \int_{{{\Bbb R}} ^{N}} \frac{\psi^{2} |u|^{2}}{d(z)^{2}}{\rm d}z +o(1)\\ &\geq& \Lambda_{0, 0}({{\Bbb R}} ^N) \bigg(\int_ {{{\Bbb R}} ^{N}} |u_k - u|^{2^*}{\rm d}z\bigg)^\frac{2}{2^*}+ \Lambda_{\mu, 0}({{\Bbb R}} ^N) \bigg(\int_ {{{\Bbb R}} ^{N}} |u|^{2^*}{\rm d}z\bigg)^\frac{2}{2^*}+o(1)\\ &\geq &\Lambda_{0, 0}({{\Bbb R}} ^N) \int_ {{{\Bbb R}} ^{N}} |u_k - u|^{2^*}{\rm d}z \qquad \bigg(\rm{因为}\int_ {{{\Bbb R}} ^{N}} |u_{k}-u|^{2^*}{\rm d}z\in[0, 1]\bigg)\\ &&+\Lambda_{\mu, 0}({{\Bbb R}} ^N) \int_ {{{\Bbb R}} ^{N}} |u|^{2^*}{\rm d}z+o(1) \qquad\bigg(\rm{因为}\int_ {{{\Bbb R}} ^{N}} |u|^{2^*}{\rm d}z\in[0, 1]\bigg)\\ & = & \Lambda_{0, 0}({{\Bbb R}} ^N)\int_ {{{\Bbb R}} ^{N}} |u_k - u|^{2^*}{\rm d}z + \Lambda_{\mu, 0}({{\Bbb R}} ^N) \bigg(\int_ {{{\Bbb R}} ^{N}} |u_{k} |^{2^*}{\rm d}z- \int_ {{{\Bbb R}} ^{N}} |u_{k}-u |^{2^*}{\rm d}z\bigg)+o(1)\\ & = & \big(\Lambda_{0, 0}({{\Bbb R}} ^N) - \Lambda_{\mu, 0}({{\Bbb R}} ^N)\big) \int_ {{{\Bbb R}} ^N} |u_k - u|^ {2^*}{\rm d}z+\Lambda_{\mu, 0}({{\Bbb R}} ^N)+o(1). \end{eqnarray*} $

即

(2.24) $ \begin{equation} \big(\Lambda_{0, 0}({{\Bbb R}} ^N) - \Lambda_{\mu, 0}({{\Bbb R}} ^N)\big) \int_ {{{\Bbb R}} ^N} |u_k-u|^ {2^*}{\rm d}z+o(1)\leq 0. \end{equation} $

又因为对任意的$ \mu>0 $, 总是有$ \Lambda_{\mu, 0}({{\Bbb R}} ^N)< \Lambda_{0, 0}({{\Bbb R}} ^N) $. 所以, (2.24)式表明$ u_k $强收敛于$ u $在$ L^{2^*}({{\Bbb R}} ^N) $中, 从而$ \int_ {{{\Bbb R}} ^N} |u|^{2^*}{\rm d}z = 1 $. 由于$ I $是下半连续的, 所以$ u $是$ \Lambda_{\mu, 0}({{\Bbb R}} ^N) $的极小函数. 类似于情形1可得: 当$ s = 0 $且$ 0<\mu <\mu_{G} $时, 存在$ \Lambda_{\mu, 0}({{\Bbb R}} ^N) $的非平凡, 非负达到函数. 定理1.1证明完毕.

为了讨论方程(1.1)的解在原点和无穷远点的渐近性质, 引入如下加权形式的Sobolev-Hardy不等式, 参见文献[20, 定理6.1]或[21].设$ 0\leq s\leq 2 $, $ \gamma> \frac{2-Q}{2} $, 则对任意的$ u\in C_0^\infty ({{\Bbb R}} ^N) $, 存在仅依赖于$ s $, $ Q $, $ \gamma $的常数$ S>0 $, 使得

(3.1) $ \begin{equation} \bigg(\int_{{{\Bbb R}} ^N} \psi^s \frac{| d(z)^{\gamma} u| ^{2^*(s)}}{d(z)^s}{\rm d}z\bigg) ^{\frac{2}{2^*(s)}}\leq S\int _{{{\Bbb R}} ^N} |d(z)^{\gamma}\nabla _\alpha u|^2 {\rm d}z, \end{equation} $

其中$ 2^*(s) = \frac{2(Q-s)}{Q-2} $. 显然: 当$ \gamma = 0 $时, (3.1) 式就变成Sobolev-Hardy不等式(1.3).

为了研究引理3.1, 引入Banach空间$ D^{1, 2}_{\alpha, \theta}({{\Bbb R}} ^{N}) = D^{1, 2}_{\alpha}({{\Bbb R}} ^{N}, d(z)^{-2\theta}{\rm d}z) $, 其范数定义如下

$ \|u\|_{D^{1, 2}_{\alpha, \theta}} = \bigg(\int_{{{\Bbb R}} ^{N}}|\nabla_{\alpha}u|^{2}d(z)^{-2\theta}{\rm d}z\bigg)^{\frac{1}{2}} = \bigg(\int_{{{\Bbb R}} ^{N}}|d(z)^{-\theta}\nabla_{\alpha}u|^{2}{\rm d}z\bigg)^{\frac{1}{2}}. $

引理 3.1    假设$ 0<s<2 $, $ 0<\theta<\frac{Q-2}{2} $. 若$ v\in D^{1, 2}_{\alpha, \theta}({{\Bbb R}} ^{N}) $是方程

(3.2) $ \begin{equation} -\rm{div}_{\alpha}(d(z)^{-2\theta}\nabla_{\alpha}v) = \psi^{s}\frac{v^{2^*(s)-1}}{d(z)^{s+\theta 2^*(s)}}, \quad \forall z\in {{\Bbb R}} ^{N} \end{equation} $

的非负非平凡弱解, 则有$ v\in L^{\infty}({{\Bbb R}} ^{N}) $. 特别地, 存在常数$ R>0 $和$ C>0 $使得

$ v(z)\geq C>0, \quad \forall z\in B_{d}(0, R). $

证    对任意的$ \beta>1 $和$ L>0 $, 定义函数

$ v_{L} = \min\{|v|\, , \, L\}, \qquad \phi(z) = v_{L}^{2(\beta-1)}v. $

则由$ \nabla_{\alpha}\phi = v_{L}^{2(\beta-1)}\nabla_{\alpha}v+2(\beta-1)v_{L}^{2(\beta-1)-1}\nabla_{\alpha}v_{L}v $可得

(3.3) $ \begin{eqnarray} &&\int_{{{\Bbb R}} ^{N}}d(z)^{-2\theta}\langle\nabla_{\alpha}v, \nabla_{\alpha}\phi \rangle {\rm d}z\\ & = &\int _{{{\Bbb R}} ^{N}}d(z)^{-2\theta} v_{L}^{2(\beta-1)}|\nabla_{\alpha}v|^{2}{\rm d}z+2(\beta-1)\int _{{{\Bbb R}} ^{N}}d(z)^{-2\theta} v_{L}^{2(\beta-1)-1}v\langle\nabla_{\alpha}v, \nabla_{\alpha}v_{L}\rangle {\rm d}z\\ & = &\int _{{{\Bbb R}} ^{N}}d(z)^{-2\theta } v_{L}^{2(\beta-1)}|\nabla_{\alpha}v|^{2}{\rm d}z+2(\beta-1)\int _{|v|<L}d(z)^{-2\theta } v^{2(\beta-1)}|\nabla_{\alpha}v|^{2}{\rm d}z\\ &\geq& \int _{{{\Bbb R}} ^{N}}d(z)^{-2\theta } v_{L}^{2(\beta-1)}|\nabla_{\alpha}v|^{2}{\rm d}z. \end{eqnarray} $

此外, 在(3.2)式两边乘上$ \phi $并积分得

(3.4) $ \begin{equation} \int_{{{\Bbb R}} ^{N}}d(z)^{-2\theta}\langle\nabla_{\alpha}v, \nabla_{\alpha}\phi \rangle {\rm d}z = \int _{{{\Bbb R}} ^{N}}\frac{\psi^{s}v^{2^*(s)-1}\phi}{d(z)^{s+\theta2^{*}(s)}} {\rm d}z = \int _{{{\Bbb R}} ^{N}}\frac{\psi^{s}v^{2^*(s)}}{d(z)^{s+\theta2^{*}(s)}}v_{L}^{2(\beta-1)} {\rm d}z. \end{equation} $

由假设条件$ \theta<\frac{Q-2}{2} $可得$ -\theta>\frac{2-Q}{2} $, 所以在不等式(3.1)式中令$ \gamma = -\theta $, 则得到如下加权形式的Sobolev-Hardy不等式

(3.5) $ \begin{equation} \bigg( \int _{{{\Bbb R}} ^{N}}\frac{\psi^{s}| u| ^{2^*(s)}}{d(z)^{s+\theta 2^*(s)}} {\rm d}z \bigg) ^{\frac{2}{2^*(s)}} \leq S\int _{{{\Bbb R}} ^{N}}\frac{| \nabla _\alpha u|^2}{d(z)^{2\theta}} {\rm d}z, \quad \forall u\in D_\alpha^{1, 2}( {{\Bbb R}} ^{N}, d^{-2\theta }{\rm d}z). \end{equation} $

令$ w_{L} = v v_{L}^{\beta-1} $, 则$ \nabla_{\alpha}w_{L} = v_{L}^{\beta-1}\nabla_{\alpha}v+ (\beta-1)v\, v_{L}^{\beta-2}\nabla_{\alpha}v_{L} $, 将其带入(3.5)式, 可得

(3.6) $ \begin{eqnarray} &&\bigg( \int _{{{\Bbb R}} ^{N}}\frac{\psi^{s}|w_{L}|^{2^*(s)}} {d(z)^{s+\theta 2^*(s)}}{\rm d}z\bigg) ^{\frac{2}{2^*(s)}}\\ &\leq& S \int _{{{\Bbb R}} ^{N}}d^{-2\theta }|v_{L}^{\beta-1}\nabla_{\alpha}v+ (\beta-1)v\, v_{L}^{\beta-2}\nabla_{\alpha}v_{L}|^{2}{\rm d}z\\ &\leq &2S \bigg( \int _{{{\Bbb R}} ^{N}}d^{-2\theta }|v_{L}^{\beta-1}\nabla_{\alpha}v|^{2}{\rm d}z+ (\beta-1)^{2}\int _{{{\Bbb R}} ^{N}}d^{-2\theta }|v\, v_{L}^{\beta-2}\nabla_{\alpha}v_{L}|^{2}{\rm d}z\bigg)\\ &\leq& 2S \bigg( \int _{{{\Bbb R}} ^{N}}d^{-2\theta }v_{L}^{2(\beta-1)}|\nabla_{\alpha}v|^{2}{\rm d}z+ (\beta-1)^{2}\int _{{{\Bbb R}} ^{N}}d^{-2\theta }v_{L}^{2(\beta-1)}|\nabla_{\alpha}v|^{2}{\rm d}z\bigg)\\ & = &2S(1+(\beta-1)^{2}) \int _{{{\Bbb R}} ^{N}}d^{-2\theta }v_{L}^{2(\beta-1)}|\nabla_{\alpha}v|^{2}{\rm d}z\\ & = &2S\beta^{2}\bigg(\frac{1}{\beta^2}+(\frac{\beta-1}{\beta})^{2}\bigg) \int _{{{\Bbb R}} ^{N}}d^{-2\theta }v_{L}^{2(\beta-1)}|\nabla_{\alpha}v|^{2}{\rm d}z. \end{eqnarray} $

又因为$ \beta>1 $, 所以$ \frac{1}{\beta^2}+(\frac{\beta-1}{\beta})^{2}<2 $. 从而由(3.3), (3.4)和(3.6)式可得

(3.7) $ \begin{eqnarray} \bigg( \int _{{{\Bbb R}} ^{N}}\frac{\psi^{s}|w_{L}|^{2^*(s)}} {d(z)^{s+\theta 2^*(s)}}{\rm d}z\bigg) ^{\frac{2}{2^*(s)}} &\leq& 4S\beta^{2} \int _{{{\Bbb R}} ^{N}}d^{-2\theta }v_{L}^{2(\beta-1)}|\nabla_{\alpha}v|^{2}{\rm d}z\\ &\leq &4S\beta^{2}\int_{{{\Bbb R}} ^{N}}d(z)^{-2\theta}\langle\nabla_{\alpha}v, \nabla_{\alpha}\phi \rangle {\rm d}z\\ & = &4S\beta^{2}\int _{{{\Bbb R}} ^{N}}\frac{\psi^{s}v^{2^*(s)}v_{L}^{2(\beta-1)}}{d(z)^{s+\theta2^{*}(s)}} {\rm d}z\\ & = &4S\beta^{2}\int _{{{\Bbb R}} ^{N}}\frac{\psi^{s} v^{2^*(s)-2}w_{L}^{2}} {d(z)^{s+\theta2^{*}(s)}} {\rm d}z. \end{eqnarray} $

对任意的$ M>0 $, 由Hölder不等式可得

(3.8) $ \begin{eqnarray} \int _{{{\Bbb R}} ^{N}} \frac{\psi^{s} v^{2^*(s)-2}w_{L}^{2}}{d(z)^{s+\theta2^{*}(s)}} {\rm d}z & = &\int _{v(z)\leq M}\frac{\psi^{s} v^{2^*(s)-2}w_{L}^{2}}{d(z)^{s+\theta2^{*}(s)}} {\rm d}z+ \int _{v(z)>M}\frac{\psi^{s} v^{2^*(s)-2}w_{L}^{2}}{d(z)^{s+\theta2^{*}(s)}} {\rm d}z\\ &\leq& M^{2^*(s)-2}\int _{v(z)\leq M}\frac{\psi^{s}w_{L}^{2}}{d(z)^{s+\theta2^{*}(s)}} {\rm d}z\\ &&+ \int _{v(z)>M} \frac{\psi^{s\cdot\frac{2}{2^*(s)}}w_{L}^{2}}{d(z)^{(s+\theta2^{*}(s))\frac{2}{2^*(s)}}} \cdot \frac{\psi^{s\cdot \frac{2^*(s)-2}{2^*(s)}}v^{2^*(s)-2}} {d(z)^{(s+\theta2^{*}(s))\frac{2^*(s)-2}{2^*(s)}}} {\rm d}z\\ &\leq& M^{2^*(s)-2}\int _{v(z)\leq M} \frac{\psi^{s} w_{L}^{2}}{d(z)^{s+\theta2^{*}(s)}} {\rm d}z\\ &&+ \bigg(\int _{v(z)>M} \frac{\psi^{s}|w_{L}|^{2^*(s)}}{d(z)^{s+\theta2^{*}(s)}}{\rm d}z\bigg)^{\frac{2}{2^*(s)}} \bigg(\int _{v(z)>M}\frac{\psi^{s} |v|^{2^*(s)}} {d(z)^{s+\theta2^{*}(s)}} {\rm d}z\bigg)^{\frac{2^*(s)-2}{2^*(s)}} \\ &\leq& M^{2^*(s)-2}\int _{{{\Bbb R}} ^{N}} \frac{\psi^{s} w_{L}^{2}}{d(z)^{s+\theta2^{*}(s)}} {\rm d}z\\ &&+ \bigg(\int _{{{\Bbb R}} ^{N}} \frac{\psi^{s}|w_{L}|^{2^*(s)}}{d(z)^{s+\theta2^{*}(s)}}{\rm d}z\bigg)^{\frac{2}{2^*(s)}} \bigg(\int _{v(z)>M}\frac{\psi^{s}|v|^{2^*(s)}} {d(z)^{s+\theta2^{*}(s)}} {\rm d}z\bigg)^{\frac{2^*(s)-2}{2^*(s)}}. \end{eqnarray} $

从而结合(3.7)和(3.8)式可得

(3.9) $ \begin{eqnarray} && \bigg( \int _{{{\Bbb R}} ^{N}}\frac{\psi^{s}|w_{L}|^{2^*(s)}} {d(z)^{s+\theta 2^*(s)}}{\rm d}z\bigg) ^{\frac{2}{2^*(s)}} \leq 4S \beta^{2}M^{2^*(s)-2}\int _{{{\Bbb R}} ^{N}} \frac{\psi^{s} w_{L}^{2}}{d(z)^{s+\theta2^{*}(s)}} {\rm d}z \\ && +4S \beta^{2} \bigg(\int _{{{\Bbb R}} ^{N}} \frac{\psi^{s}|w_{L}|^{2^*(s)}}{d(z)^{s+\theta2^{*}(s)}}{\rm d}z\bigg)^{\frac{2}{2^*(s)}} \bigg(\int _{v(z)>M}\frac{\psi^{s}|v|^{2^*(s)}} {d(z)^{s+\theta2^{*}(s)}} {\rm d}z\bigg)^{\frac{2^*(s)-2}{2^*(s)}}. \end{eqnarray} $

因为$ v\in D^{1, 2}_{\alpha, \theta}({{\Bbb R}} ^{N}) $, 所以$ \int _{{{\Bbb R}} ^{N}}\frac{\psi^{s}|v|^{2^*(s)}}{d(z)^{s+\theta2^{*}(s)}} {\rm d}z<\infty $, 则存在充分大的$ M_{0}>1 $使得

(3.10) $ \begin{equation} \bigg(\int _{v(z)>M_{0}}\psi^{s}\frac{v^{2^*(s)}}{d(z)^{s+\theta2^{*}(s)}} {\rm d}z\bigg)^{\frac{2-s}{Q-s}} \leq \frac{1}{8S\beta^{2}}. \end{equation} $

从而由(3.9)和(3.10)式可得

(3.11) $ \begin{equation} \bigg(\int _{{{\Bbb R}} ^{N}} \frac{\psi^{s}|w_{L}|^{2^*(s)}}{d(z)^{s+\theta2^{*}(s)}}{\rm d}z\bigg)^{\frac{2}{2^*(s)}} \leq 8S\beta^{2}M^{2^*(s)-2}_{0}\int _{{{\Bbb R}} ^{N}} \frac{\psi^{s} w_{L}^{2}}{d(z)^{s+\theta2^{*}(s)}} {\rm d}z. \end{equation} $

令$ C_{1} = 8SM^{2^*(s)-2}_{0} $. 在(3.11)式中令$ L\to \infty $, 并利用Fatou引理可得

(3.12) $ \begin{equation} \bigg(\int _{{{\Bbb R}} ^{N}} \frac{\psi^{s} v^{\beta 2^*(s)}}{d(z)^{s+\theta2^{*}(s)}}{\rm d}z\bigg)^{\frac{2}{2^*(s)}} \leq C_{1}\beta^{2}\int _{{{\Bbb R}} ^{N}} \frac{\psi^{s} v^{2 \beta}}{d(z)^{s+\theta2^{*}(s)}} {\rm d}z<+\infty. \end{equation} $

在(3.12)式中取$ \beta = \beta_{1} = \frac{2^*(s)}{2}>1 $, 即得

(3.13) $ \begin{equation} \bigg(\int _{{{\Bbb R}} ^{N}} \frac{\psi^{s} v^{\beta_{1} 2^*(s)}}{d(z)^{s+\theta2^{*}(s)}}{\rm d}z\bigg)^{\frac{2}{2^*(s)}} \leq C_{1}\beta_{1}^{2}\int _{{{\Bbb R}} ^{N}} \frac{\psi^{s} v^{2^*(s)}}{d(z)^{s+\theta2^{*}(s)}} {\rm d}z<+\infty. \end{equation} $

取序列$ \{\beta_{k}\}_{k = 2}^{\infty} $, 其满足如下条件

(3.14) $ \begin{equation} 2\beta_{k+1}+2^*(s)-2 = \beta_{k}\, 2^*(s), \qquad \forall k\geq 1. \end{equation} $

显然: 当$ \beta_{k}>1 $时, 必有$ \beta_{k+1}>1 $. 从而用$ \beta_{k+1} $替换(3.7)式中的$ \beta $, 可得

(3.15) $ \begin{equation} \bigg( \int _{{{\Bbb R}} ^{N}}\frac{\psi^{s}|v v_{L}^{\beta_{k+1}-1}|^{2^*(s)}} {d(z)^{s+\theta 2^*(s)}}{\rm d}z\bigg) ^{\frac{2}{2^*(s)}} \leq 4S\beta^{2}_{k+1}\int _{{{\Bbb R}} ^{N}} \frac{\psi^{s} v^{2^*(s)-2}(vv_{L}^{(\beta_{k+1}-1)})^{2}} {d(z)^{s+\theta2^{*}(s)}} {\rm d}z. \end{equation} $

在(3.15)式中令$ L\to +\infty $, 利用Fatou引理可得

$ \begin{eqnarray*} \label{eq3-15} \bigg( \int _{{{\Bbb R}} ^{N}}\frac{\psi^{s} v^{\beta_{k+1}2^*(s)}} {d(z)^{s+\theta 2^*(s)}}{\rm d}z\bigg) ^{\frac{2}{2^*(s)}} &\leq &4S\beta^{2}_{k+1}\int _{{{\Bbb R}} ^{N}}\frac{\psi^{s}v^{2^*(s)-2+2\beta_{k+1}}} {d(z)^{s+\theta2^{*}(s)}} {\rm d}z\\ & = &4S\beta^{2}_{k+1}\int _{{{\Bbb R}} ^{N}}\frac{\psi^{s} v^{\beta_{k}2^*(s)}} {d(z)^{s+\theta2^{*}(s)}} {\rm d}z, \quad \forall k\geq 1. \end{eqnarray*} $

从而

(3.16) $ \begin{equation} \bigg( \int _{{{\Bbb R}} ^{N}}\psi^{s}\frac{v^{\beta_{k+1}2^*(s)}} {d(z)^{s+\theta 2^*(s)}}{\rm d}z\bigg) ^{\frac{1}{2^*(s)(\beta_{k+1}-1)}} \leq (4S\beta^{2}_{k+1})^{\frac{1}{2(\beta_{k+1}-1)}} \bigg(\int _{{{\Bbb R}} ^{N}}\psi^{s}\frac{v^{\beta_{k}2^*(s)}} {d(z)^{s+\theta2^{*}(s)}} {\rm d}z\bigg)^{\frac{1}{2^*(s)(\beta_{k}-1)}}. \end{equation} $

对任意的$ k\in \mathbb{N} $, 令

$ {\cal A}_{k} = \bigg(\int _{{{\Bbb R}} ^{N}}\frac{\psi^{s}v^{\beta_{k}2^*(s)}} {d(z)^{s+\theta2^{*}(s)}} {\rm d}z\bigg)^{\frac{1}{2^*(s)(\beta_{k}-1)}}, \qquad C_{k} = (4S\beta^{2}_{k+1})^{\frac{1}{2(\beta_{k+1}-1)}}. $

则由(3.16)可得

$ {\cal A}_{k+1}\leq C_{k}{\cal A}_{k}, \qquad \forall k\geq 1. $

从而

(3.17) $ \begin{eqnarray} \ln {\cal A}_{k+1} &\leq& \ln (C_{k}{\cal A}_{k}) = \ln C_{k}+\ln{\cal A}_{k} \leq \ln C_{k}+\ln C_{k-1}+\ln{\cal A}_{k-1}\\ &\leq& \sum\limits_{i = 1}^{k}\ln C_{i}+\ln{\cal A}_{1}\\ & = &\sum\limits_{i = 1}^{k}\frac{1}{2(\beta_{i+1}-1)}(\ln 4S+2\ln \beta_{i+1})+\ln{\cal A}_{1}\\ & = &\sum\limits_{i = 1}^{k}\bigg(\frac{\ln 4S}{2(\beta_{i+1}-1)}+\frac{\ln \beta_{i+1}}{\beta_{i+1}-1}\bigg)+\ln{\cal A}_{1}. \end{eqnarray} $

又因为序列$ \{\beta_{i}\} $满足$ 2\beta_{i+1}+2^*(s)-2 = \beta_{i}\, 2^*(s) $, 即$ \beta_{i+1}-1 = \frac{2^*(s)}{2}(\beta_{i}-1) $, 从而

$ \beta_{i+1}-1 = \frac{2^*(s)}{2}(\beta_{i}-1) = \bigg(\frac{2^*(s)}{2}\bigg)^{2}(\beta_{i-1}-1) = \bigg(\frac{2^*(s)}{2}\bigg)^{i}(\beta_{1}-1). $

由此可得

(3.18) $ \begin{eqnarray} &&\sum\limits_{i = 1}^{k}\bigg(\frac{\ln 4S}{2(\beta_{i+1}-1)}+\frac{\ln \beta_{i+1}}{\beta_{i+1}-1}\bigg)\\ & = &\sum\limits_{i = 1}^{k}\bigg(\frac{\ln 4S}{2(\beta_{1}-1)}\bigg(\frac{2}{2^*(s)}\bigg)^{i}+\frac{\ln \left[\big(\frac{2^*(s)}{2}\big)^{i}(\beta_{1}-1)+1\right]}{\beta_{1}-1}\bigg(\frac{2}{2^*(s)}\bigg)^{i}\bigg)\\ &\leq&\sum\limits_{i = 1}^{k}\bigg(\frac{\ln 4S}{2(\beta_{1}-1)}\bigg(\frac{2}{2^*(s)}\bigg)^{i}+\frac{\ln \big(\frac{2^*(s)}{2}\big)^{i}\beta_{1}}{\beta_{1}-1}\bigg(\frac{2}{2^*(s)}\bigg)^{i}\bigg)\\ &\leq& C\sum\limits_{i = 1}^{k}\bigg(\bigg(\frac{2}{2^*(s)}\bigg)^{i}+i\bigg(\frac{2}{2^*(s)}\bigg)^{i}\bigg). \end{eqnarray} $

由于级数$ \sum\limits_{i = 1}^{\infty}\big(\frac{2}{2^*(s)}\big)^{i} $, $ \sum\limits_{i = 1}^{\infty}i\big(\frac{2}{2^*(s)}\big)^{i} $是收敛的, 从而由(3.17)–(3.18)式可知存在常数$ C>0 $, 使得$ \ln {\cal A}_{k+1}\leq C<+\infty $, 即

(3.19) $ \begin{equation} \bigg(\int _{{{\Bbb R}} ^{N}}\frac{\psi^{s} v^{\beta_{k}2^*(s)}} {d(z)^{s+\theta2^{*}(s)}} {\rm d}z\bigg)^{\frac{1}{2^*(s)(\beta_{k}-1)}} = {\cal A}_{k}\leq e^C<+\infty. \end{equation} $

所以, 对任意的$ R>1 $, 利用(3.19)式以及函数$ \psi $的有界性可以推出

(3.20) $ \begin{eqnarray} \bigg(\int _{d(z)<R}v^{\beta_{k}2^*(s)}{\rm d}z\bigg)^{\frac{1}{2^*(s)\beta_{k}}} &\leq& c R^{\frac{s+\theta 2^*(s)}{2^*(s)\beta_{k}}} \bigg(\int _{d(z)<R}\frac{\psi^{s} v^{\beta_{k}2^*(s)}}{d(z)^{s+\theta2^{*}(s)}} {\rm d}z\bigg) ^{\frac{1}{2^*(s)(\beta_{k}-1)}\frac{\beta_{k}-1}{\beta_{k}}}\\ &\leq &c R^{\frac{s+\theta 2^*(s)}{2^*(s)\beta_{k}}} e^{\frac{C(\beta_{k}-1)}{\beta_{k}}}<+\infty. \end{eqnarray} $

根据序列$ \{\beta_{i}\} $满足的条件可知$ \mathop {\lim }\limits_{i \to \infty }\beta_{i} = \infty $, 因此, 从(3.20)式中可知存在某个常数$ C>0 $使得$ \|v\|_{L^{\infty}({{\Bbb R}} ^{N})}\leq C<+\infty $. 引理3.1得证.

定理1.2的证明    设$ u\in D^{1, 2}_{\alpha}({{\Bbb R}} ^{N}) $为方程(1.1)的解. 令

$ v: = d(z)^{\beta }u, $

其中$ \beta = \frac{Q-2}{2}-\sqrt{(\frac{Q-2}{2})^{2}-\mu} $. 由于$ 0\leq \mu<(\frac{Q-2}{2})^{2} $, 则$ 0<\beta <\frac{Q-2}{2} $, 且$ \beta $是方程

$ \beta ^2-\beta (Q-2) + \mu = 0 $

的解. 计算可得

(3.21) $ \begin{eqnarray} (\Delta_{x}+|x|^{2\alpha}\Delta_{y})u & = &(\Delta_{x}+|x|^{2\alpha}\Delta_{y})(d(z)^{-\beta}v)\\ & = &v[(\Delta_{x}+|x|^{2\alpha}\Delta_{y})d(z)^{-\beta}] +d(z)^{-\beta}[(\Delta_{x}+|x|^{2\alpha}\Delta_{y})v]\\ && +2\langle \nabla_{\alpha} d(z)^{-\beta}, \nabla_{\alpha} v\rangle, \end{eqnarray} $

其中

(3.22) $ \begin{eqnarray} (\Delta_{x}+|x|^{2\alpha}\Delta_{y})d(z)^{-\beta} & = &\psi^{2}[\beta(\beta+1)d(z)^{-\beta-2}+\frac{Q-1}{d(z)}(-\beta)d(z)^{-\beta-1}]\\ & = &\psi^{2}[\beta(\beta+1)d(z)^{-\beta-2}-\beta(Q-1)d(z)^{-\beta-2}]\\ & = &\psi^{2}[\beta^{2}-\beta(Q-2)]d(z)^{-\beta-2}. \end{eqnarray} $

从而将$ u = d(z)^{-\beta}v $带入方程(1.1)的左边, 由(3.21)–(3.22)式可得

(3.23) $ \begin{eqnarray} &&-(\Delta_{x}+|x|^{2\alpha}\Delta_{y})u-\mu\frac{\psi^{2}u}{d(z)^{2}}\\ & = &-v\psi^{2}[\beta^{2}-\beta(Q-2)]d(z)^{-\beta-2} -d(z)^{-\beta}[(\Delta_{x}+|x|^{2\alpha}\Delta_{y})v]\\ && -2\langle \nabla_{\alpha} d(z)^{-\beta}, \nabla_{\alpha} v\rangle -\mu\frac{\psi^{2}}{d(z)^{2}}d(z)^{-\beta}v\\ & = &v\psi^{2}[\beta^{2}-\beta(Q-2)+\mu]d(z)^{-\beta-2} -d(z)^{-\beta}[(\Delta_{x}+|x|^{2\alpha}\Delta_{y})v] -2\langle \nabla_{\alpha} d(z)^{-\beta}, \nabla_{\alpha} v\rangle\\ & = &-d(z)^{-\beta}[(\Delta_{x}+|x|^{2\alpha}\Delta_{y})v] -2\langle \nabla_{\alpha} d(z)^{-\beta}, \nabla_{\alpha} v\rangle. \end{eqnarray} $

将$ u = d(z)^{-\beta}v $带入方程(1.1)的右边

(3.24) $ \begin{equation} \frac{\psi^{s}|u|^{2^*(s)-2}u}{d(z)^{s}} = \frac{\psi^{s}|d(z)^{-\beta}v|^{2^*(s)-2}d(z)^{-\beta}v}{d(z)^{s}} = \frac{\psi^{s} |v|^{2^*(s)-2}v}{d(z)^{s+\beta(2^*(s)-1)}}. \end{equation} $

结合(3.23)和(3.24)式可推出

$ -d(z)^{-\beta}[(\Delta_{x}+|x|^{2\alpha}\Delta_{y})v] -2\langle \nabla_{\alpha} d(z)^{-\beta}, \nabla_{\alpha} v\rangle = \frac{\psi^{s}|v|^{2^*(s)-2}v}{d(z)^{s+\beta(2^*(s)-1)}}. $

在上式的两边分别乘以$ d(z)^{-\beta} $, 则有

$ -\rm{div}_{\alpha}(d(z)^{-2\beta}\nabla_{\alpha}v) = \frac{\psi^{s}|v|^{2^*(s)-2}v}{d(z)^{s+\beta 2^{*}(s)}}. $

即$ v $是方程(3.2)的非平凡解. 由引理3.1知: 存在常数$ C>1 $使得对任意的$ R>0 $有$ C^{-1}\leq v(z)\leq C $, $ \forall z\in B_{d}(0, R) $. 从而

(3.25) $ \begin{equation} \frac{C^{-1}}{d(z)^{\beta}}\leq u(z)\leq \frac{C}{d(z)^{\beta}}, \quad \forall z\in B_{d(z)}(0, R), \end{equation} $

其中$ \beta = \frac{Q-2}{2}-\sqrt{(\frac{Q-2}{2})^{2}-\mu} $. 从而定理1.2中的(1.5)式得证.

下面证明(1.6)式, 即寻求解在无穷远点处的渐近性质.

定义Grushin几何下的球面反演变换$ \sigma:{{\Bbb R}} ^{N}\to {{\Bbb R}} ^{N} $为$ \sigma (z) = \delta _{d(z)^{-2}}(z), \ \forall z\in{{\Bbb R}} ^{N}\backslash\{ 0\}. $反演$ \sigma $在下述意义下是保形映射, 即对任意的$ u\in C^1({{{\Bbb R}} }^N) $, $ z\in{{\Bbb R}} ^{N}\backslash\{ 0\} $, 有

(3.26) $ \begin{equation} |\nabla _\alpha (u\circ \sigma )(z)|^2 = |J_\sigma (z)|^{\frac{2}{Q}}|(\nabla _\alpha u)(\sigma (z))|^2, \end{equation} $

其中$ J_{\sigma} = \rm{det}\frac{\partial \sigma}{\partial z} $是$ \sigma $的Jacobian行列式, 并且满足$ |J_{\sigma }(z)| = d(z)^{-2Q} = \Gamma (z)^{\frac{2Q}{Q-2}} $, 这里的$ \Gamma (z) = \frac{1}{d(z)^{Q-2}} $是Grushin算子$ \Delta_{x}+|x|^{2\alpha}\Delta_{y} $的基本解. 由此, 保形映射可以诱导出如下的Kelvin型变换. 设$ u: {{\Bbb R}} ^N \to{{\Bbb R}} $, 其Kelvin型变换$ w: {{\Bbb R}} ^N\backslash \{0\}\to{{\Bbb R}} $定义为

(3.27) $ \begin{equation} w(z) = \Gamma (z)u(\sigma (z)) = \frac{1}{d(z)^{Q-2}}u(\delta_{d(z)^{-2}}(z)), \quad \forall z\in{{\Bbb R}} ^{N}\backslash\{ 0\}. \end{equation} $

利用文献[22, 定理2.5]中的计算方法可以证明: 若$ u $是方程(1.1)的非平凡解, 则对任意的$ \phi \in C_0^\infty ({{\Bbb R}} ^{N} ) $, $ z\in {{\Bbb R}} ^{N} \backslash\{0\} $, 有下式成立

(3.28) $ \begin{equation} \int _{{{\Bbb R}} ^{N}} \langle \nabla _{\alpha} w, \nabla _{\alpha} \phi \rangle {\rm d}z -\mu \int _{{{\Bbb R}} ^{N}} \frac{\psi(z)^2 w \phi }{d(z)^2}{\rm d}z = \int _{{{\Bbb R}} ^{N}} \frac{\psi^{s} |w|^{2^*(s)-2}w\phi }{d(z)^{s}}{\rm d}z, \end{equation} $

即$ w $亦是方程(1.1)的非平凡解. 从而, 由(3.25)式可知: 存在$ C>0 $和$ R>0 $使得

(3.29) $ \begin{equation} \frac{C^{-1}}{d(z)^{\frac{Q-2}{2}-\sqrt{(\frac{Q-2}{2})^{2}-\mu}}}\leq w(z)\leq \frac{C}{d(z)^{\frac{Q-2}{2}-\sqrt{(\frac{Q-2}{2})^{2}-\mu}}}. \quad \forall z\in B_{d(z)}(0, R)\backslash\{0\}, \end{equation} $

即对任意的$ z\in B_{d(z)}(0, R)\backslash\{0\} $, 有

(3.30) $ \begin{equation} \frac{C^{-1}}{d(z)^{-\frac{Q-2}{2}-\sqrt{(\frac{Q-2}{2})^{2}-\mu}}}\leq u(\delta_{d(z)^{-2}}(z))\leq \frac{C}{d(z)^{-\frac{Q-2}{2}-\sqrt{(\frac{Q-2}{2})^{2}-\mu}}}. \end{equation} $

令$ \xi = \delta_{d(z)^{-2}}(z) $, 则$ d(\xi) = d(\delta_{d(z)^{-2}}(z)) = \frac{1}{d(z)} $, 从而当$ d(z)\to 0 $时有$ d(\xi)\to +\infty $. 所以, 当$ d(\xi)\to+\infty $时, 由(3.30)式可以得出

(3.31) $ \begin{equation} \frac{C^{-1}}{d(\xi)^{\frac{Q-2}{2}+\sqrt{(\frac{Q-2}{2})^{2}-\mu}}}\leq u(\xi)\leq \frac{C}{d(\xi)^{\frac{Q-2}{2}+\sqrt{(\frac{Q-2}{2})^{2}-\mu}}}. \end{equation} $

所以, 当$ d(\xi)\to+\infty $时, 有

(3.32) $ \begin{equation} u(\xi) = O\bigg(\frac{1}{d(\xi)^{\frac{Q-2}{2}+\sqrt{(\frac{Q-2}{2})^{2}-\mu}}}\bigg). \end{equation} $

即(1.6)式得证. 定理1.2证明完毕.