本文研究如下问题

$ \begin{equation} \left \{ \begin{array}{lll} -\triangle u + \frac{u}{\mid x\mid^2}=\mid u \mid ^{2^*-2}u + g(x),&x \in {\Bbb R}^N, \\ u(x)\rightarrow0 (\mid x \mid\rightarrow\infty), &u\in {\cal D}^{1, 2}({\Bbb R}^N) \end{array} \right . \end{equation} $

(1.1)

多解的存在性, 这里$g(x)\geq0, g(x)\not\equiv 0$且$g(x)\in L^{\frac{2N}{N+2}}({\Bbb R}^N).$

这类问题有较强的物理背景和理论研究价值.国内外有许多学者对这类问题开展了有意义的研究, 得到一些有趣的研究结果(参见文献[1-2, 6-7]等).这类问题的研究方法主要是应用变分原理, 带Hardy项的Sobolev不等式和嵌入定理.为了探究问题(1.1)弱解的存在性, 我们可以转化为研究与其相应变分泛函

$ \begin{equation} f(u)=\frac{1}{2}\int \Big(\mid\triangledown u\mid^2+\frac{u^2}{\mid x\mid^2}\Big){\rm d}x-\frac{1}{2^*}\int\mid u\mid^{2^*}{\rm d}x-\int g(x)u{\rm d}x \end{equation} $

(1.2)

临界点的存在性.

受文献[12]的启发, 本文考虑一定条件下泛函$f(u)$存在一个局部极小的临界点以及山路引理形式的另一个临界点, 运用Lions^[10]的集中紧致原理, 证明上述两个临界点的可达性, 从而得到问题(1.1)至少有两个弱解存在性的结果, 即本文的主要结果.

定理1.1  假设$g(x)\in L^{\frac{2N}{N+2}}({\Bbb R}^N)$, $g(x)\geq0$且$g(x)\not\equiv 0$, 则存在常数$C$, 当$\parallel g\parallel_{L^{\frac{2N}{N+2}}({\Bbb R}^N)}\leq C$时, 问题(1.1)至少有两个非平凡解.

在本文中$L^p({\Bbb R}^N)$的范数记为$\mid\cdot\mid_p$, $ {\cal D}^{1, 2}({\Bbb R}^N)$表示$C^\infty({\Bbb R}^N)$在范数$\parallel u\parallel=(\int(\mid\triangledown u\mid^2+\frac{u^2}{\mid x\mid^2}) {\rm d}x)^\frac{1}{2}$下的完备空间.为了证明我们的结果, 我们需要如下几个引理.

引理2.1(Brezis-lieb引理)  $\{u_n\}\subset L^p({\Bbb R}^N), 1\leq p < \infty$.如果$\{u_n\}$在$L^p({\Bbb R}^N)$中有界, 且$u_n\rightarrow u$在${\Bbb R}^N$中几乎处处成立, 则

$ \begin{equation} \lim_{n\rightarrow\infty}(\mid u_n\mid_p^p-\mid u_n-u\mid_p^p)=\mid u\mid_p^p. \end{equation} $

(2.1)

引理2.2(Ekeland变分原理)  ${X}$是一个Banach空间, $\varphi\in{\cal C}^1({X}, R)$具有下界, $v\in{X}$且$\varepsilon, \delta > 0$,

$ \begin{equation} \varphi(v)\leq\inf_X\varphi+\varepsilon, \end{equation} $

(2.2)

则存在$u\in{X}$使得

$ \begin{equation} \varphi(u)\leq\inf_X\varphi+2\varepsilon, \parallel\varphi'(u)\parallel\leq\frac{8\varepsilon}{\delta}, \parallel u-v\parallel \leq2\delta. \end{equation} $

(2.3)

引理2.3  假设$\{u_n\}$是${\cal D}^{1, 2}({\Bbb R}^N)$中一序列满足$ I_g(u_n)\rightarrow c < \frac{1}{N}S^{\frac{N}{2}}, I'_g(u_n)\rightarrow 0$, 那么在$\{u_n\}$中存在一个强收敛的子序列.

证  由已知条件

$ \begin{equation} I_g(u_n)=\frac{1}{2}\int_{{\Bbb R}^N}(\mid\triangledown u_n\mid^2+\frac{u_n^2}{\mid x\mid^2}){\rm d}x-\frac{1}{2^*}\int_{{\Bbb R}^N}\mid u_n\mid^{2^*}{\rm d}x-\int_{{\Bbb R}^N} g(x)u_n{\rm d}x\rightarrow c, \end{equation} $

(2.4)

以及对任意$\varphi\in{\cal D}^{1, 2}({\Bbb R}^N)$, 有

$ \begin{eqnarray} \langle I'_g(u_n), \varphi\rangle &=&\int_{{\Bbb R}^N}(\triangledown u_n\triangledown \varphi+\frac{u_n \varphi}{\mid x\mid^2}){\rm d}x-\int_{{\Bbb R}^N}\mid u_n\mid^{2^*-2}u_n \varphi {\rm d}x-\int_{{\Bbb R}^N}g(x)\varphi {\rm d}x\\ &=&o(1)\parallel\varphi\parallel . \end{eqnarray} $

(2.5)

首先可得$\{u_n\}$在${\cal D}^{1, 2}({\Bbb R}^N)$中有界.

反证法:若$\{u_n\}$在${\cal D}^{1, 2}({\Bbb R}^N)$中无界, 则当$n\rightarrow\infty $时, $t_n=(\int_{{\Bbb R}^N}\mid\triangledown u_n\mid^2 {\rm d}x)^{\frac{1}{2}}\rightarrow\infty$.记$\omega_n=\frac{u_n}{t_n}$, 因此$(\int_{{\Bbb R}^N}\mid\triangledown \omega_n\mid^2 {\rm d}x)^{\frac{1}{2}}=1$对任意$n$成立.从而存在子列

$ \mbox{ $\omega_n\rightharpoonup\omega$ 在 ${\cal D}^{1, 2}({\Bbb R}^N)$ 中, $\omega_n\stackrel{\rm a.e.}\rightarrow\omega $ 在 ${\Bbb R}^N$ 中.} $

(2.5)式两边同时除以$t_n$, 得到

$ \begin{equation} \int_{{\Bbb R}^N} \Big(\triangledown \omega_n\triangledown \varphi+\frac{\omega_n \varphi}{\mid x\mid^2} \Big){\rm d}x-\int_{{\Bbb R}^N}\mid u_n\mid^{2^*-2}\omega_n \varphi {\rm d}x-\frac{1}{t_n}\int_{{\Bbb R}^N}g(x)\varphi {\rm d}x=o(1). \end{equation} $

(2.6)

我们知道

$ \int_{{\Bbb R}^N}\mid u_n\mid^{2^*-2}\omega_n \varphi {\rm d}x=t_n^{2^*-2}\int_{{\Bbb R}^N}\mid \omega_n\mid^{2^*-2}\omega_n \varphi {\rm d}x. $

特别地, 对任意$\varphi\in{\cal C}_0^\infty({\Bbb R}^N)$, 有

$ \int_{{\Bbb R}^N}\mid \omega_n\mid^{2^*-2}\omega_n \varphi {\rm d}x\rightarrow\int_{{\Bbb R}^N}\mid \omega\mid^{2^*-2}\omega \varphi {\rm d}x. $

(2.6)式两边同时除以$t_n^{2^*-2}$并取极限, 得到

$ \int_{{\Bbb R}^N}\mid \omega\mid^{2^*-2}\omega \varphi {\rm d}x=0. $

由$\varphi$的任意性可知:在${\Bbb R}^N$中$\omega\stackrel{a.e.}=0$.

在(2.4)式两边同时除以$t_n^2$知

$ \begin{equation} \frac{1}{2}\int_{{\Bbb R}^N} \Big(\mid\triangledown \omega_n\mid^2+\frac{\omega_n^2}{\mid x\mid^2}\Big){\rm d}x-\frac{1}{2^*}\int_{{\Bbb R}^N}\mid u_n\mid^{2^*-2}\omega_n^2{\rm d}x-\frac{1}{t_n}\int_{{\Bbb R}^N} g(x)\omega_n{\rm d}x=o(1). \end{equation} $

(2.7)

在(2.6)式中取$\varphi=\omega_n$有

$ \begin{equation} \int_{{\Bbb R}^N}\Big(\mid\triangledown \omega_n\mid^2+\frac{\omega_n^2}{\mid x\mid^2}\Big){\rm d}x-\int_{{\Bbb R}^N}\mid u_n\mid^{2^*-2}\omega_n^2{\rm d}x-\frac{1}{t_n}\int_{{\Bbb R}^N} g(x)\omega_n{\rm d}x=o(1). \end{equation} $

(2.8)

由于在${\cal D}^{1, 2}({\Bbb R}^N)$中$\omega_n\rightharpoonup\omega$, 所以有

$ \int_{{\Bbb R}^N} g(x)\omega_n{\rm d}x\rightarrow 0. $

结合(2.7)和(2.8)式, 可得

$ \begin{equation} \frac{1}{2}+\frac{1}{2}\lim_{n\rightarrow\infty}\int_{{\Bbb R}^N}\frac{\omega_n^2}{\mid x\mid^2}{\rm d}x=\frac{1}{2^*}\lim_{n\rightarrow\infty}\int_{{\Bbb R}^N}\mid u_n\mid^{2^*-2}\omega_n^2{\rm d}x, \end{equation} $

(2.9)

$ \begin{equation} 1+\lim_{n\rightarrow\infty} \int_{{\Bbb R}^N}\frac{\omega_n^2}{\mid x\mid^2}{\rm d}x=\lim_{n\rightarrow\infty}\int_{{\Bbb R}^N}\mid u_n\mid^{2^*-2}\omega_n^2{\rm d}x. \end{equation} $

(2.10)

由(2.9)和(2.10)式, 可得

$ \begin{equation} \Big(\frac{2^*}{2}-1\Big)\Big(1+\lim_{n\rightarrow\infty} \int_{{\Bbb R}^N} \frac{\omega_n^2}{\mid x\mid^2}{\rm d}x\Big)=0. \end{equation} $

(2.11)

(2.11)式显然是矛盾的.因此$\{u_n\}$在${\cal D}^{1, 2}({\Bbb R}^N)$中有界.

由于序列$\{u_n\}$在${\cal D}^{1, 2}({\Bbb R}^N)$中有界, 因此存在$\{u_n\}$的子列(仍记为$\{u_n\}$), 满足:

a) $u_n\rightharpoonup u$, 在${\cal D}^{1, 2}({\Bbb R}^N)$中;

b) $u_n\stackrel{\rm a.e.}{\longrightarrow} u$在${\Bbb R}^N$中.

运用带Hardy项的Sobolev嵌入定理可知, 序列$\{u_n\}$在$L^{2^*}({\Bbb R}^N)$中有界, 所以$u_n^{2^*-1}$在$L^{\frac{2N}{N+2}}({\Bbb R}^N)$中有界, 从而可知存在子列在$L^{\frac{2N}{N+2}}({\Bbb R}^N)$中弱收敛以及在${\Bbb R}^N$中几乎处处收敛.因此知在${\cal D}^{1, 2}({\Bbb R}^N)$中

$ \begin{equation} u_n^{2^*-1}\rightharpoonup u^{2^*-1}. \end{equation} $

(2.12)

由$I'_g(u_n)\rightarrow0$, 则对任意函数$\varphi\in{\cal D}^{1, 2}({\Bbb R}^N)$, 有

$ \begin{equation} \langle I'_g(u_n), \varphi\rangle\rightarrow 0. \end{equation} $

(2.13)

即

$ \begin{equation} \int_{{\Bbb R}^N}\Big(\triangledown u_n\triangledown \varphi+\frac{u_n \varphi}{\mid x\mid^2}\Big){\rm d}x-\int_{{\Bbb R}^N}\mid u_n\mid^{2^*-2}u_n \varphi {\rm d}x-\int_{{\Bbb R}^N}g(x)\varphi {\rm d}x\rightarrow 0. \end{equation} $

(2.14)

所以有

$ \begin{equation} \int_{{\Bbb R}^N}\Big(\triangledown u\triangledown \varphi+\frac{u \varphi}{\mid x\mid^2}\Big){\rm d}x-\int_{{\Bbb R}^N}\mid u \mid ^{2^*-2}u \varphi {\rm d}x-\int_{{\Bbb R}^N}g(x)\varphi {\rm d}x=0. \end{equation} $

(2.15)

即对任意的$\varphi\in{\cal D}^{1, 2}({\Bbb R}^N)$, 有下式成立

$ \langle I'_g(u), \varphi\rangle=0. $

令$\varphi=u$, 则由(2.15)式有

$ \begin{equation} \int_{{\Bbb R}^N} \Big(\mid\triangledown u\mid^2+\frac{u^2}{\mid x\mid^2}\Big){\rm d}x=\int_{{\Bbb R}^N}\mid u\mid^{2^*-1}u{\rm d}x+\int_{{\Bbb R}^N} g(x)u{\rm d}x. \end{equation} $

(2.16)

所以存在适当小的常数$C$, 当$\parallel g\parallel_{L^{\frac{2N}{N+2}} ({\Bbb R}^N)}\leq C$时, 有

$ \begin{eqnarray} I_g(u) &=&\frac{1}{2}\int_{{\Bbb R}^N} \Big(\mid\triangledown u\mid^2+\frac{u^2}{\mid x\mid^2}\Big){\rm d}x-\frac{1}{2^*}\int_{{\Bbb R}^N}\mid u\mid^{2^*}{\rm d}x-\int_{{\Bbb R}^N} g(x)u{\rm d}x\\ &=&\Big (\frac{1}{2}-\frac{1}{2^*}\Big)\int_{{\Bbb R}^N}\mid u\mid^{2^*}{\rm d}x-\frac{1}{2}\int_{{\Bbb R}^N} g(x)u{\rm d}x\geq 0. \end{eqnarray} $

(2.17)

下面, 我们来证明$u_n\longrightarrow u$(在${\cal D}^{1, 2}({\Bbb R}^N)$中强收敛).

记$v_n=u_n-u$, 由引理2.1得

$ \int_{{\Bbb R}^N}\mid\triangledown u_n\mid^2{\rm d}x=\int_{{\Bbb R}^N}\mid\triangledown v_n\mid^2{\rm d}x+\int_{{\Bbb R}^N}\mid\triangledown u\mid^2{\rm d}x+o(1), \\ \begin{equation} \int_{{\Bbb R}^N}\frac{u_n^2}{\mid x\mid^2}{\rm d}x=\int_{{\Bbb R}^N}\frac{v_n^2}{\mid x\mid^2}{\rm d}x+\int_{{\Bbb R}^N}\frac{u^2}{\mid x\mid^2}{\rm d}x+o(1), \end{equation}\\ \int_{{\Bbb R}^N}\mid u_n\mid^{2^*}{\rm d}x=\int_{{\Bbb R}^N}\mid v_n\mid^{2^*}{\rm d}x+\int_{{\Bbb R}^N}\mid u\mid^{2^*}{\rm d}x+o(1). $

(2.18)

由引理条件及(2.18)式知

$ \begin{eqnarray} I_g(u_n)&=&\frac{1}{2}\int_{{\Bbb R}^N} \Big(\mid\triangledown u_n\mid^2+\frac{u_n^2}{\mid x\mid^2}\Big){\rm d}x-\frac{1}{2^*}\int_{{\Bbb R}^N}\mid u_n\mid^{2^*}{\rm d}x-\int_{{\Bbb R}^N} g(x)u{\rm d}x \\ &=&I_g(u)+\frac{1}{2}\int_{{\Bbb R}^N} \Big(\mid\triangledown v_n\mid^2+\frac{v_n^2}{\mid x\mid^2}\Big){\rm d}x-\frac{1}{2^*}\int_{{\Bbb R}^N}\mid v_n\mid^{2^*}{\rm d}x+o(1). \end{eqnarray} $

(2.19)

由(2.13)式可知$\langle I'_g(u_n), u_n\rangle\rightarrow 0$, 再结合(2.18)式子可得

$ \begin{eqnarray*} \langle I'_g(u_n), u_n\rangle &=&\int_{{\Bbb R}^N} \Big(\mid\triangledown u_n\mid^2+\frac{u_n^2}{\mid x\mid^2}\Big){\rm d}x-\int_{{\Bbb R}^N}\mid u_n\mid^{2^*-1}u{\rm d}x-\int_{{\Bbb R}^N} g(x)u{\rm d}x\\ &=&\langle I'_g(u), u\rangle+\int_{{\Bbb R}^N} \Big(\mid\triangledown v_n\mid^2+\frac{v_n^2}{\mid x\mid^2}\Big){\rm d}x-\int_{{\Bbb R}^N}\mid v_n\mid^{2^*}{\rm d}x+o(1)\\ &=&\int_{{\Bbb R}^N}\Big(\mid\triangledown v_n\mid^2+\frac{v_n^2}{\mid x\mid^2}\Big){\rm d}x-\int_{{\Bbb R}^N}\mid v_n\mid^{2^*}{\rm d}x+o(1)\rightarrow 0. \end{eqnarray*} $

假设$ \int_{{\Bbb R}^N}\Big(\mid\triangledown v_n\mid^2+\frac{v_n^2}{\mid x\mid^2}\Big){\rm d}x\rightarrow b, $则

$ \begin{equation} \int_{{\Bbb R}^N}\mid v_n\mid^{2^*}{\rm d}x \rightarrow b, \end{equation} $

(2.20)

以及$S$的定义

$ \begin{equation} S\Big(\int_{{\Bbb R}^N}\mid v_n\mid^{2^*}{\rm d}x\Big)^{\frac{2}{2^*}}\leq\int_{{\Bbb R}^N} \Big(\mid\triangledown v_n\mid^2+\frac{v_n^2}{\mid x\mid^2}\Big){\rm d}x =\int_{{\Bbb R}^N}\mid v_n\mid^{2^*}{\rm d}x+o(1), \end{equation} $

(2.21)

可得$b\geq S b^{\frac{2}{2^*}}.$

如果$b=0$, 则定理得证; 如果$b\neq0$, 由(2.17)及(2.19)式可得

$ \frac{1}{N}S^{\frac{N}{2}}=\Big(\frac{1}{2}-\frac{1}{2^*}\Big)S^{\frac{N}{2}} \leq\Big(\frac{1}{2}-\frac{1}{2^*}\Big)b\leq c < \frac{1}{N}S^{\frac{N}{2}}. $

上式显然是矛盾的, 所以$b=0$, 序列$\{u_n\}$在${\cal D}^{1, 2}({\Bbb R}^N)$中有强收敛子列.

在这一部分, 我们将给出主要结果的证明.

定义3.1  定义

$ \overline{B_R}=\{u\in{\cal D}^{1, 2}({\Bbb R}^N)|\parallel u\parallel\leq R\}, I_0=\inf_{u\in\overline{B_R}}I_g(u). $

定理3.1  假设$ u\in\overline{B_R}, g(x)\in L^{\frac{2N}{N+2}}({\Bbb R}^N)$, $g(x)\geq 0, g(x)\not\equiv 0$, $\parallel g\parallel_{L^{\frac{2N}{N+2}}({\Bbb R}^N)}\leq C$, 其中$C$和$R$适当小, 则存在$u_0\in\overline{B_R}, $使得$I_0=I_g(u_0)$和$I'_g(u_0)=0.$即$u_0$是问题(1.1)的解.

证  对任意的$u\in{\cal D}^{1, 2}({\Bbb R}^N), u > 0, $

$ \begin{equation}I_g(tu)=\frac{t^2}{2}\int \Big(\mid\triangledown u\mid^2+\frac{u^2}{\mid x\mid^2}\Big){\rm d}x-\frac{t^{2^*}}{2^*}\int\mid u\mid^{2^*}{\rm d}x-t\int g(x)u{\rm d}x, \end{equation} $

(3.1)

$ \begin{equation} \frac{{\rm d}I_g(tu)}{{\rm d}t}=t \int\Big(\mid\triangledown u\mid^2+\frac{u^2}{\mid x\mid^2}\Big) -t^{2^*-1}\int\mid u\mid^{2^*}{\rm d}x-\int g(x)u{\rm d}x. \end{equation} $

(3.2)

如果$0 < t < \varepsilon$, 显然有$\frac{{\rm d}I_g(tu)}{{\rm d}t} < 0, $所以当$t\in(0, \varepsilon)$时, $I_g(tu)$是减函数.由$I_g(0)=0, \ I_g(tu)\in C^1$, 对于充分小的$R$, 当$u\in\overline{B_R}$时, 有

$ \begin{equation} I_g(u) < 0, I_0=\inf_{u\in\overline{B_R}}I_g(u) < 0. \end{equation} $

(3.3)

如果$t > \varepsilon$,

$ \begin{eqnarray*} \frac{{\rm d}I_g(tu)}{{\rm d}t}&=&t \int \Big(\mid\triangledown u\mid^2+\frac{u^2}{\mid x\mid^2}\Big)-t^{2^*-1}\int\mid u\mid^{2^*}{\rm d}x-\int g(x)u{\rm d}x\\ &=&t\parallel u\parallel^2-t^{2^*-1}\mid u\mid^{2^*}_{2^*}-\int g(x)u{\rm d}x\\ &\geq &t\parallel u\parallel^2-t^{2^*-1}\mid u\mid^{2^*}_{2^*}-\mid g\mid_\frac{2N}{N+2}\mid u\mid_{2^*}\\ &\geq &t\parallel u\parallel^2-C_1t^{2^*-1} \parallel u \parallel^{2^*}-C_2\parallel u \parallel. \end{eqnarray*} $

记$\parallel u \parallel=A$, 有

$ \begin{equation} g(t)=t\parallel u\parallel^2-C_1t^{2^*-1} \parallel u \parallel^{2^*}-C_2\parallel u \parallel=A^2t-C_1A^{2^*}t^{2^*-1}-C_2A, \end{equation} $

(3.4)

令$ g'(t)=A^2-C_1(2^*-1)A^{2^*}t^{2^*-2}=0$, 可得

$ \begin{equation} t_0=[C_1(2^*-1)]^{\frac{-1}{2^*-2}}A^{-1}, \end{equation} $

(3.5)

$ \begin{eqnarray*} g(t_0)&=&A^2[C_1(2^*-1)]^{\frac{-1}{2^*-2}}A^{-1}-C_1A^{2^*}[C_1(2^*-1)]^\frac{1-2^*}{2^*-2}A^{1-2^*}-C_2A\\ &=&A \Big\{[C_1(2^*-1)]^{\frac{-1}{2^*-2}}-C_1[C_1(2^*-1)]^{\frac{1-2^*}{2^*-2}}-C_2\Big\}\\ &=&A\Big \{[C_1(2^*-1)]^{\frac{-1}{2^*-2}}\Big(1-\frac{1}{2^*-1}\Big)-C_2\Big\}. \end{eqnarray*} $

由$1-\frac{1}{2^*-1}>0, C_2$适当小, 我们有$g(t_0)>0.$所以当$t=t_0$时, $I(tu)$是增函数.

上述表明, 对于$u\in\overline{B_R}, u>0, $函数$I(tu)$存在局部极小值.

由引理2.2, 对于充分小的$R$, 我们可以得到$I_g(tu)$的一个$(PS)_c$序列:

$ \{u_n\}\subset{\cal D}^{1, 2}({\Bbb R}^N), I_g(u_n)\rightarrow I_0, I'_g(u_n)\rightarrow 0 $

由$I_0=\inf\limits_{u\in\overline{B_R}}I_g(u) < 0 < \frac{1}{N}S^\frac{N}{2}$, 利用引理2.3可知:存在$u_0\in\overline{B_R}, $使得$I_0=I_g(u_0)$和$I'_g(u_0)=0.$即$u_0$是问题(1.1)的解.

定理3.2  假设$ u\in{\cal D}^{1, 2}({\Bbb R}^N)$, $ g(x)\in L^{\frac{2N}{N+2}}({\Bbb R}^N)$, $g(x)\geq0, g(x)\not\equiv 0$, $\parallel g\parallel_{L^{\frac{2N}{N+2}}({\Bbb R}^N)}\leq C$, 其中$C$适当小, 由山路引理, 问题(1.1)存在另一个解.

证  令$v$是$S$的极小可达函数, 其中$ S=\inf\limits_{u\in{\cal D}^{1, 2}({\Bbb R}^N)}\frac{\parallel u\parallel^2}{\mid u\mid_{2^*}^{2}}.$

$ \begin{eqnarray*} \max_{t\geq0}I_g(tv)&=&\max_{t\geq0} \Big(\frac{t^2}{2}\int \mid \triangledown v\mid^2+\frac{v^2}{\mid x\mid^2}-\frac{t^{2^*}}{2^*}\int\mid v\mid^{2^*}-t\int g(x)v \Big)\\ & < &\max_{t\geq0}\Big(\frac{t^2}{2}\int \mid \triangledown v\mid^2+\frac{v^2}{\mid x\mid^2}-\frac{t^{2^*}}{2^*}\int\mid v\mid^{2^*}\Big ). \end{eqnarray*} $

记$A=\parallel v\parallel^2, B=\mid v\mid_{2^*}^{2^*}$,

$ \begin{equation} g(t)=\frac{t^2}{2}\int \mid \triangledown v\mid^2+ \frac{v^2}{\mid x\mid^2}-\frac{t^{2^*}}{2^*}\int\mid v\mid^{2^*}=\frac{t^2}{2}\parallel v\parallel^2-\frac{t^{2^*}}{2^*}\mid v\mid_{2^*}^{2^*}=\frac{t^2}{2}A-\frac{t^{2^*}}{2^*}B. \end{equation} $

(3.6)

令$g'(t)=At-Bt^{2^*-1}=0, $可得

$ \begin{equation} t_0=\Big(\frac{A}{B}\Big)^{\frac{1}{2^*-2}}, \end{equation} $

(3.7)

$ \begin{eqnarray*} g(t_0)&=&\frac{1}{2}A\Big(\frac{A}{B}\Big)^{\frac{2}{2^*-2}}-\frac{B}{2^*} \Big(\frac{A}{B}\Big)^{\frac{2^*}{2^*-2}}= \Big(\frac{1}{2}-\frac{1}{2^*}\Big)A^\frac{2^*}{2^*-2}/ B^\frac{2}{2^*-2}\\ &=&\Big(\frac{1}{2}-\frac{1}{2^*}\Big)[A/B^{\frac{2}{2^*}}]^{\frac{2^*}{2^*-2}} =\frac{1}{N}\Big(\frac{\parallel v\parallel^2}{\mid v\mid_{2^*}^{2^*}}\Big)^{N/2} =\frac{1}{N}S^{N/2}, \end{eqnarray*} $

所以有

$ \begin{equation} \max_{t\geq0}I_g(tv) < \frac{1}{N}S^{N/2}. \end{equation} $

(3.8)

$ \begin{eqnarray*} I_g(u)&=&\frac{1}{2}\int \Big(\mid\triangledown u\mid^2+\frac{u^2}{\mid x\mid^2}\Big){\rm d}x-\frac{1}{2^*}\int\mid u\mid^{2^*}{\rm d}x-\int g(x)u{\rm d}x\\ &\geq&\frac{1}{2}\parallel u \parallel^2-\frac{C_1}{2^*}\parallel u \parallel^{2^*}-C_2\parallel u \parallel^{2^*}. \end{eqnarray*} $

对于适当小的$C_2$, 存在$ r > 0 $, 使得$b=\inf\limits_{\parallel u \parallel=r}I_g(u) > 0=I_g(0);$并存在$t_0 > 0, $使得$\parallel t_0v\parallel > r$且$I_g(t_0v) < 0.$

定义3.2

$ c=\inf_{\gamma\in\Gamma}\max_{t\in[0, 1]}I_g(\gamma(t)), $

$ \Gamma:=\{\gamma\in{\cal C}([0, 1], {\cal D}^{1, 2}({\Bbb R}^N)); \gamma(0)=0, \gamma(1)=t_0v\}. $

由引理2.3和山路引理可知, 泛函$ I_g(u)$有一个临界值$c\in[b, c^*), $且问题

$ \begin{equation}\label{2.1} \left \{ \begin{array}{lll} triangle u + \frac{u}{\mid x\mid^2}=\mid u \mid ^{2^*-2}u + g(x),&x \in {\Bbb R}^N \setminus \{0\}, \\ [2mm] u\in {\cal D}^{1, 2}({\Bbb R}^N) \end{array} \right . \end{equation} $

(3.9)

有一个非平凡解$u$.

定理1.1的证明  首先由定理3.1证得$u_0$是$B_R$中泛函$I_g(u)$的局部极小点, 再由$I_g(u_0) < \frac{1}{N}S^{N/2}$, 运用引理2.3可知$u_0$是问题(1.1)的一个弱解.然后, 由定理3.2知泛函$I_g(u)$满足山路引理的几何条件且$I_g(u) < \frac{1}{N}S^{N/2}$, 由山路引理及引理2.3可得问题(1.1)的另一个弱解.即定理1.1的结论成立.