本文研究如下Kirchhoff型问题

(1.1) $ \begin{equation} \left\{\begin{array}{ll} { } -\Big(a+b\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x\Big)\Delta u+V(x)u = |u|^{p-2}u+\varepsilon|u|^4u, & x\in{{\Bbb R}} ^3, \\ u>0, \, \, u\in H^1({{\Bbb R}} ^3) \end{array}\right. \end{equation} $

正基态解的存在性, 其中$ a>0 $, $ b>0 $, $ 4<p<6 $, $ \varepsilon>0 $. 此外, $ V(x) $是一个非负函数且满足

$ (V_0):\quad V(x)\in L_{\rm loc}^{\frac{3}{2}}({{\Bbb R}} ^3), \quad\lim\limits_{|x|\rightarrow \infty}V(x) = V_{\infty}, \quad V(x)\geq V_0>0\, \, \, {\rm a.e.}\, \, x\in {{\Bbb R}} ^3. $

问题(1.1) 与下面方程

(1.2) $ \begin{equation} \rho\frac{\partial^{2}u}{\partial t^{2}}-\bigg(\frac{p_{0}}{h}+\frac{E}{2L}\int_{0}^{L}\bigg|\frac{\partial u}{\partial x}\bigg|^{2}{\rm d}x\bigg)\frac{\partial^{2}u}{\partial x^{2}} = f(u) \end{equation} $

对应的稳态相关. 1983年, 作为经典D'Alembert波动方程的延伸, Kirchhoff^[1]在研究拉伸弦的横向振动, 特别是考虑到横向振动引起的弦的长度变化时首次提出方程(1.2), 其中$ u $表示变量, 且当$ b $与弦的内在属性相关时, $ h $是外力, $ a $是初始应变量, 如Young's模.

近年来, 采用非线性分析的工具和变分方法, 许多学者对下列非线性Kirchhoff型方程

(1.3) $ \begin{equation} \left\{\begin{array}{ll} { } -\Big(a+b\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x\Big)\Delta u+V(x)u = f(x, u), & { x\in{{\Bbb R}} ^3 , }\\ u\in H^1({{\Bbb R}} ^3) \end{array}\right. \end{equation} $

进行了大量的研究, 建立了基态解, 束缚态解和半经典态解等的存在性和多重性. 例如: 当$ V(x): = C $时, Xu和Chen^[2]证明了方程(1.3) 具有径向基态解, 其中$ f(x, u) $满足临界Berestycki-Lions型条件. 当$ C = 0 $时, 问题(1.3) 可简化为如下方程

(1.4) $ \begin{equation} \left\{\begin{array}{ll} { } -\big(a+b\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x\Big)\Delta u = f(x, u), & { x\in{{\Bbb R}} ^3 , }\\ u\in H^1({{\Bbb R}} ^3), \end{array}\right. \end{equation} $

Liu和Guo^[3]研究了临界增长中具有一般非线性的问题(1.4) 的基态解的存在性. 当(1.4) 式中$ f(x, u) = g(x)|u|^{2^{\ast}-2}u+\lambda h(x)|u|^{q-1}u $时, Li^[4]通过Nehari法和变分法得到了问题(1.4) 的正基态解的存在性. 有关问题(1.4) 基态解的更多结果, 参看文献[5-7].

关于基态解的研究方面, 最近, Li和Ye^[8]假设$ V(x) $满足以下假设.

(ⅰ) $ V(x)\in{\cal C}({{\Bbb R}} ^3, {{\Bbb R}} ) $弱可微且满足$ (\nabla V(x), x)\in L^{\infty}({{\Bbb R}} ^3)\cup L^{\frac{3}{2}}({{\Bbb R}} ^3) $和

$ V(x)-2(\nabla V(x), x)\geq0\, \, \, {\rm a.e.}\, \, x\in{{\Bbb R}} ^3; $

(ⅱ) 对任意的$ x\in{{\Bbb R}} ^3 $, $ V(x)\leq\liminf\limits_{|y|\rightarrow \infty}V(y): = V_{\infty}<+\infty $且其在Lebesgue正测度的子集中是严格的;

(ⅲ) 存在$ \bar{C}>0 $使得

$ \bar{C} = \inf\limits_{u\in H^1({{\Bbb R}} ^3\backslash\{0\})}\frac{\int_{{{\Bbb R}} ^3}|\nabla u|^2+V(x)|u|^2}{\int_{{{\Bbb R}} ^3}|u|^2}>0, $

采用单调性技巧, Pohozaev-Nehari流形和全局紧性引理证明了当$ f(x, u) = |u|^{p-1}u $, $ 2<p<5 $时问题(1.3) 正基态解的存在性. Wu^[9]利用Pohozaev流形证明了问题(1.3) 存在正基态解. 当$ f(x, u) $在无穷远是次临界且在原点附近是超线性的, $ V(x) $满足与上面(ⅰ) 和(ⅱ) 相似的一些条件时, Guo^[10]用变分方法证明了问题(1.3) 存在正基态解. 当$ f(x, u) = K(x)|u|^4u+g(x, u) $且$ V(x) $满足渐近周期条件时, 作者们在文献[11]中证明了问题(1.3) 存在正基态解. 当$ f(x, u) $在无穷远是次临界, 在原点是超线性的且满足Berestycki-Lions条件和位势$ V(x) $满足与上面(ⅰ)–(ⅲ) 类似的一些条件时, Liu和Guo^[12]利用Jeanjean建立的抽象临界点定理和一个新全局紧性引理证明了问题(1.3) 至少存在一个基态解. 随后, Tang和Chen^[13]对位势$ V(x)\in{\cal C}({{\Bbb R}} ^3, [0, \infty)) $提出一些更强的条件, 他们证明了问题(1.3) 存在一个Nehari-Pohozaev型的基态解. 后来, Chen和Tang^[14]又证明了问题(1.3) 存在一个基态解, 其中$ f(x, u) $满足一般的Berestycki-Lions假设和位势$ V(x)\in{\cal C}({{\Bbb R}} ^3, [0, \infty)) $满足类似文献[13]的条件. Ye^[15]证明了问题(1.3) 存在正基态解, 其中$ f(x, u) = a(x)f(u)+u^5 $且$ V(x) $在无穷远处满足指数阶衰减. 当位势$ V(x) $有一个井位势, Sun和Wu^[16]得到了一类如下Kirchhoff型问题基态解的存在性

$ \left\{\begin{array}{ll} { } -\Big(a\int_{{{\Bbb R}} ^N}|\nabla u|^2{\rm d}x+b\Big)\Delta u+\lambda V(x)u = f(x, u), \, \, \, x\in{{\Bbb R}} ^N, \\ u\in H^{1}({{\Bbb R}} ^{N}). \end{array}\right. $

关于问题(1.3) 基态解的更多结果, 参看文献[17-26].

受上述文献的启发, 本文考虑具有小临界扰动项的问题(1.1) 基态解的存在性. 与上述文献相比, 我们只需求$ V(x)\in L_{\rm loc}^{\frac{3}{2}}({{\Bbb R}} ^3) $或$ V(x) $可能在局部区域比$ V_{\infty} $大. 这是本文主要结果的新奇之处, 其方法是基于约束极小化方法. 主要的困难在于非局部项$ \int_{{{\Bbb R}} ^3}|\nabla u|^2\, {\rm d}x\Delta u $的出现, 由于$ {{\Bbb R}} ^3 $的无界性以及带有临界扰动项的非线性而缺乏紧性. 此外, 由于$ V(x) $非径向对称, 故不能将问题限制在径向对称Sobolev空间$ H_r^1({{\Bbb R}} ^3) $中, 其中$ H_r^1({{\Bbb R}} ^3)\hookrightarrow L^q({{\Bbb R}} ^3) $$ (2<s<6) $是紧的. 为克服这些困难, 必须进行更仔细的分析. 特别地, 对于序列$ \{u_n\}\subset H^1({{\Bbb R}} ^3) $且$ u_n $在$ H^1({{\Bbb R}} ^3) $弱收敛于$ u $, 将仔细分析$ \int_{{{\Bbb R}} ^3}|\nabla u_n|^2\, {\rm d}x $和$ \int_{{{\Bbb R}} ^3}|\nabla u|^2\, {\rm d}x $的不同来恢复紧性.

若$ \varepsilon = 0 $, 则问题(1.1) 有一个正基态解, 显然, 当$ \varepsilon $趋于0时, 对于方程(1.1) 来说, 我们期望这种结果不会改变. 本文将试图证明该现象. 本文的主要结果如下.

定理1.1  假设$ V(x) $满足$ (V_0) $和

(1.5) $ \begin{equation} V(x)\leq V_{\infty}\quad {\rm a.e.}\ \, \, x\in{{\Bbb R}} ^3, \end{equation} $

那么存在$ \varepsilon_0>0 $, 对任意的$ \varepsilon\in(0, \varepsilon_{0}) $, 问题(1.1) 有一个正基态解.

当位势$ V(x) $不满足(1.5) 式时, 通过考虑下列极限问题的正基态解

(1.6) $ \begin{equation} \left\{\begin{array}{ll} { } -\Big(a+b\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x\Big)\Delta u+V_{\infty}u = |u|^{p-2}u, \, \, \, x\in{{\Bbb R}} ^3, \\ u\in H^1({{\Bbb R}} ^3) \end{array}\right. \end{equation} $

证明问题(1.1) 的基态解的存在性. 事实上, 根据文献[18, 27], 在相差平移的情形下, 方程(1.6) 存在唯一的正的径向对称解, 记为$ w $. 现陈述如下结果.

定理1.2 假设$ V(x) $满足$ (V_0) $, 如果存在$ z\in{{\Bbb R}} ^3 $满足

(1.7) $ \begin{equation} \int_{{{\Bbb R}} ^3}V(x)|w_{z}|^{2}\, {\rm d}x<\int_{{{\Bbb R}} ^3} V_{\infty}w^2\, {\rm d}x, \end{equation} $

其中$ w_z(x): = w(x-z) $, 且$ w $是问题(1.6) 的正径向基态解. 那么对任意小的$ \varepsilon $, 问题(1.1) 存在正的基态解.

注意到当$ V(x)\equiv V_{\infty} $时, 那么对任意$ z\in{{\Bbb R}} ^3 $, (1.7) 式成立.

此篇论文的结构如下. 在第2节, 将给出一些符号且回忆了一些学过的知识. 在第3节, 将给出定理1.1和定理1.2的证明.

不失一般性, 假设$ V_{\infty} = 1 $. 在下文中, 将使用以下符号.

$ \bullet $$ H^1({{\Bbb R}} ^3) $是Sobolev空间, 其内积和范数如下

$ (u, v): = \int_{{{\Bbb R}} ^3}(a\nabla u\nabla v+uv){\rm d}x, \quad\|u\|^2: = \int_{{{\Bbb R}} ^3}(a|\nabla u|^2+u^2){\rm d}x. $

同时, 在这里我们将引入一种等价范数

$ \|u\|_V^2: = \int_{{{\Bbb R}} ^3}(a|\nabla u|^2+V(x)u^2){\rm d}x. $

事实上

$ \int_{{{\Bbb R}} ^3}(a|\nabla u|^2+u^2){\rm d}x\leq C\int_{{{\Bbb R}} ^3}(a|\nabla u|^2+V(x)u^2){\rm d}x $

显然成立. 反之, 根据$ (V_0) $可得, 存在$ R>0 $使得当$ |x|\geq R $时有$ V(x)<2V_{\infty} $, 从而有

$ \begin{eqnarray*} \int_{{{\Bbb R}} ^3}(a|\nabla u|^2+V(x)u^2){\rm d}x& = &\int_{{{\Bbb R}} ^3}a|\nabla u|^2{\rm d}x+\int_{|x|\geq R}V(x)u^2{\rm d}x+\int_{|x|\leq R}V(x)u^2{\rm d}x\\ &\leq&\int_{{{\Bbb R}} ^3}a|\nabla u|^2{\rm d}x+2\int_{|x|\geq R}V_{\infty}u^2{\rm d}x \\ &&+\int_{|x|\leq R}\big((V(x))^{\frac{3}{2}}{\rm d}x\big)^\frac{2}{3}\Big(\int_{{{\Bbb R}} ^3}u^6{\rm d}x\Big)^\frac{1}{3}\\ &\leq& C\int_{{{\Bbb R}} ^3}(a|\nabla u|^2+u^2){\rm d}x. \end{eqnarray*} $

$ \bullet $$ {\cal D}^{1, 2}({{\Bbb R}} ^3) $是通常的Sobolev空间, 其标准范数为$ \|u\|_{{\cal D}}: = (\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x)^{\frac{1}{2}} $.

$ \bullet $$ {\cal L}^q({\cal O}) $表示Lebesgue空间, 其中$ 1\leq q\leq\infty $, $ {\cal O}\subseteq{{\Bbb R}} ^3 $是一个可测集. 当$ {\cal O} $是$ {{\Bbb R}} ^3 $的适当可测子集时, $ {\cal L}^q({\cal O}) $上的范数为$ |\cdot|_{{\cal L}^q({\cal O})} $. 当$ {\cal O} = {{\Bbb R}} ^3 $时, 其范数为$ |\cdot|_q $.

$ \bullet $$ c, c_i, C, C_i, \cdots $表示一些正常数.

问题(1.1) 对应的能量泛函是$ I_{\varepsilon}:H^{1}({{\Bbb R}} ^3)\rightarrow {{\Bbb R}} $定义为

$ I_{\varepsilon}(u) = \frac{1}{2}\int_{{{\Bbb R}} ^3}(a|\nabla u|^{2}+V(x)u^{2}){\rm d}x+\frac{b}{4} \Big(\int_{{{\Bbb R}} ^3}|\nabla u|^{2}{\rm d}x\Big)^{2}-\frac{1}{p}\int_{{{\Bbb R}} ^3}|u|^{p}{\rm d}x-\frac{\varepsilon}{6}\int_{{{\Bbb R}} ^3}|u|^6{\rm d}x. $

显然, $ I_{\varepsilon}\in C^1(H^1({{\Bbb R}} ^3), {{\Bbb R}} ) $且$ I_{\varepsilon} $的临界点是问题(1.1) 的弱解. 问题(1.1) 对应的极限方程如下

(2.1) $ \begin{equation} \left\{\begin{array}{ll} { } -\Big(a+b\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x\Big)\Delta u+V_{\infty}u = |u|^{p-2}u+\varepsilon|u|^4u, \, \, \, x\in{{\Bbb R}} ^3, \\ u\in H^1({{\Bbb R}} ^3), \end{array}\right. \end{equation} $

且其对应的泛函为$ I_{\varepsilon, \infty}:H^1({{\Bbb R}} ^3)\rightarrow {{\Bbb R}} $, 定义如下

$ I_{\varepsilon, \infty}(u) = \frac{1}{2}\int_{{{\Bbb R}} ^3}(a|\nabla u|^2+u^2){\rm d}x+\frac{b}{4} \Big(\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x\Big)^2-\frac{1}{p}\int_{{{\Bbb R}} ^3}|u|^p{\rm d}x-\frac{\varepsilon}{6}\int_{{{\Bbb R}} ^3}|u|^6{\rm d}x. $

与$ I_{\varepsilon} $具有相同性质的泛函定义如下

$ I(u) = \frac{1}{2}\int_{{{\Bbb R}} ^3}(a|\nabla u|^2+V(x)u^2){\rm d}x+\frac{b}{4} \Big(\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x\Big)^2-\frac{1}{p}\int_{{{\Bbb R}} ^3}|u|^p{\rm d}x, $

$ I_{\infty}(u) = \frac{1}{2}\int_{{{\Bbb R}} ^3}(a|\nabla u|^2+u^2){\rm d}x+\frac{b}{4} \Big(\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x\Big)^2-\frac{1}{p}\int_{{{\Bbb R}} ^3}|u|^p{\rm d}x. $

定义泛函$ I $, $ I_{\varepsilon} $, $ I_{\infty} $和$ I_{\varepsilon, \infty} $上的Nehari流形如下

$ {\cal N} = \{u\in H_0^1({{\Bbb R}} ^3)\backslash\{0\}:I'(u)[u] = 0\}, \, \, \, \, \, {\cal N}_{\varepsilon} = \{u\in H_0^1({{\Bbb R}} ^3)\backslash\{0\}:I'_{\varepsilon}(u)[u] = 0\}, $

$ {\cal N}_{\infty} = \{u\in H^1({{\Bbb R}} ^3)\backslash\{0\}:I'_{\infty}(u)[u] = 0\}, \, \, \, \, \, {\cal N}_{\varepsilon, \infty} = \{u\in H^1({{\Bbb R}} ^3)\backslash\{0\}:I'_{\varepsilon, \infty}(u)[u] = 0\}. $

定义

(2.2) $ \begin{equation} m: = \inf\limits_{{\cal N}_{\infty}}I_{\infty}, \quad \quad m_{\varepsilon}: = \inf\limits_{N_{\varepsilon, \infty}}I_{\varepsilon, \infty}. \end{equation} $

注意到, 当$ \varepsilon>0 $时有$ m_{\varepsilon}\leq m $. 事实上, 设$ w $满足$ I_{\infty}(w) = m $且存在$ \tau_{\varepsilon}>0 $使得$ \tau_{\varepsilon}w\in {\cal N}_{\varepsilon, \infty} $, 那么

(2.3) $ \begin{equation} m_{\varepsilon}\leq I_{\varepsilon, \infty}(\tau_{\varepsilon}w)\leq I_{\infty}(\tau_{\varepsilon}w)\leq I_{\infty}(w) = m. \end{equation} $

引理2.1  (ⅰ) 存在唯一$ t_u>0 $使得$ t_{u}u\in{\cal N} $, 有

$ I(t_{u}u) = \max\limits_{t>0}I(tu), $

且$ u\longmapsto t_u $是从$ H^1({{\Bbb R}} ^3)\backslash\{0\} $到$ {{\Bbb R}} ^{+} $的连续映射. 若在$ {\cal N}_{\infty} $, $ {\cal N}_{\varepsilon} $和$ {\cal N}_{\varepsilon, \infty} $上分别考虑$ I_{\infty} $, $ I_{\varepsilon} $和$ I_{\varepsilon, \infty} $, 类似的结论成立.

(ⅱ) 对任意的$ u\in{\cal N}_{\varepsilon, \infty} $, 存在正常数$ C>0 $使得$ \|u\|\geq C>0 $, 其中$ C $与$ \varepsilon $无关.

(ⅲ) 对任意的$ \varepsilon\in(0, \varepsilon_0) $, 设$ u $是约束在$ {\cal N}_{\varepsilon, \infty} $上的$ I_{\varepsilon, \infty} $的极小元, 那么存在正常数$ C_1>0 $使得$ |u|_p^p\geq C_1>0 $, 其中$ C_1 $与$ \varepsilon $无关.

证   根据标准的讨论, 可得(ⅰ) 成立.

(ⅱ) 对任意的$ u\in {\cal N}_{\varepsilon, \infty} $, 由Sobolev嵌入定理得

$ 0 = \|u\|^{2}+b|\nabla u|_2^4-|u|_p^p-\varepsilon|u|_6^6\geq\|u\|^2-c_1\|u\|^p-c_2\varepsilon\|u\|^6, $

即$ \|u\|^2\leq c_1\|u\|^p+c_2\varepsilon\|u\|^6 $, 其中$ c_1, c_2>0 $与$ \varepsilon $和$ u $无关. 因此, 对任意的$ \varepsilon>0 $, 有

(2.4) $ \begin{equation} \|u\|\geq C>0, \quad \forall u\in{\cal N}_{\varepsilon, \infty} \end{equation} $

成立, 其中$ C $是与$ \varepsilon $和$ u $无关的正常数.

(ⅲ) 设$ u $是$ I_{\varepsilon, \infty} $约束在$ {\cal N}_{\varepsilon, \infty} $上的极小元, 可得

$ I_{\varepsilon, \infty}(u) = \frac{1}{4}\|u\|^{2}+\frac{p-4}{4p}|u|_{p}^{p}+\frac{1}{12}\varepsilon|u|_{6}^{6} = m_{\varepsilon}, $

由上式和(2.3) 式可得

(2.5) $ \begin{equation} \|u\|^2\leq4m_{\varepsilon}+o(1)\leq4m. \end{equation} $

从(2.4) 式可得$ \|u\| $有正下界, 且从(2.5) 式可得$ \|u\| $有正上界. 因此, 利用Sobolev嵌入定理, 可得

$ |u|^p_p = \|u\|^2+b\|\nabla u\|_2^4-\varepsilon|u|_6^6\geq\|u\|^2-c\varepsilon\|u\|^6\geq C^2-c_1\varepsilon\geq\frac{C^2}{2}\geq C_1, \quad\forall\varepsilon\in(0, \varepsilon_{0}), $

其中$ c, c_1>0 $与$ \varepsilon $和$ u $无关. 证毕.

命题2.1  下列估计成立

(2.6) $ \begin{equation} m_{\varepsilon}\leq\frac{a}{3}\bigg(\frac{b}{2\varepsilon}{\cal S}^3+\sqrt{\frac{b^2}{4\varepsilon^{2}}{\cal S}^6+\frac{a}{\varepsilon}{\cal S}^{3}}\bigg)+\frac{b}{12}\bigg(\frac{b}{2\varepsilon}{\cal S}^3+\sqrt{\frac{b^2}{4\varepsilon^{2}}{\cal S}^6+\frac{a}{\varepsilon}{\cal S}^3}\bigg)^2: = {\cal S}_{a, b}, \, \forall\varepsilon>0, \end{equation} $

其中$ {\cal S} $是最佳Sobolev常数.

证   注意到(2.8) 式中的$ {\cal S}_{a, b} $是下列方程解的基态水平

(2.7) $ \begin{equation} \left\{\begin{array}{ll} { } -\Big(a+b\int_{{{\Bbb R}} ^{3}}|\nabla u|^{2}{\rm d}x\Big)\Delta u = \varepsilon|u|^{4}u, &\, \, \, x\in{{\Bbb R}} ^3, \\ { } \lim\limits_{|x|\rightarrow \infty}u(x) = 0, & \end{array}\right. \end{equation} $

即

(2.8) $ \begin{eqnarray} {\cal S}_{a, b}& = &\frac{a}{3}\bigg(\frac{b}{2\varepsilon}{\cal S}^3+\sqrt{\frac{b^2}{4\varepsilon^{2}}{\cal S}^6 +\frac{a}{\varepsilon}{\cal S}^3}\bigg)+\frac{b}{12}\bigg(\frac{b}{2\varepsilon}{\cal S}^3+\sqrt{\frac{b^2} {4\varepsilon^{2}}{\cal S}^6+\frac{a}{\varepsilon}{\cal S}^3}\bigg)^2{}\\ & = &\min\bigg\{\frac{a}{2}\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x+\frac{b}{4}\Big(\int_{{{\Bbb R}} ^3} |\nabla u|^2{\rm d}x\Big)^2-\frac{\varepsilon}{6}\int_{{{\Bbb R}} ^3}|u|^6{\rm d}x\, :\, u\in{\cal D}^{1, 2}({{\Bbb R}} ^3), {}\\ &&\int_{{{\Bbb R}} ^3}a|\nabla u|^2{\rm d}x+b\Big(\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x\Big)^2 = \varepsilon\int_{{{\Bbb R}} ^3}|u|^6{\rm d}x\bigg\}. \end{eqnarray} $

事实上, 下列问题

(2.9) $ \begin{equation} \left\{\begin{array}{ll} -\Delta u = |u|^{4}u, \, \, \, x\in{{\Bbb R}} ^3, \\ u\in {\cal D}^{1, 2}({{\Bbb R}} ^{3}) \end{array}\right. \end{equation} $

的正解具有如下形式

$ \psi_{\delta, x_{0}} = \frac{(3\delta)^{\frac{1}{4}}}{(\delta+|x-x_{0}|^2)^{\frac{1}{2}}}, \quad\delta>0, \quad x_{0}\in{{\Bbb R}} ^3. $

事实上, 最佳Sobolev常数$ {\cal S} $可通过下式定义

$ {\cal S} = \inf\limits_{u\in{\cal D}^{1, 2}({{\Bbb R}} ^3)\backslash\{0\}}\frac{\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x}{|u|_6^2} = \inf\limits_{u\in{\cal D}^{1, 2}({{\Bbb R}} ^3)\backslash\{0\}, |u|_6 = 1}\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x. $

注意到, 对任意的$ u\in{\cal D}^{1, 2}({{\Bbb R}} ^3)\backslash\{0\} $, 存在唯一的$ t>0 $使得$ tu $满足

$ a\int_{{{\Bbb R}} ^3}|\nabla(tu)|^2{\rm d}x+b\Big(\int_{{{\Bbb R}} ^3}|\nabla(tu)|^2{\rm d}x\Big)^2 = \varepsilon\int_{{{\Bbb R}} ^3}|tu|^6{\rm d}x, $

即

(2.10) $ \begin{equation} \varepsilon t^{4}\int_{{{\Bbb R}} ^{3}}|u|^{6}{\rm d}x-bt^{2}\Big(\int_{{{\Bbb R}} ^{3}}|\nabla u|^{2}{\rm d}x\Big)^{2}-a\int_{{{\Bbb R}} ^{3}}|\nabla u|^{2}{\rm d}x = 0, \end{equation} $

通过计算得

$ t^{2} = \frac{b(\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x)^2+\sqrt{b^2(\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x)^4+4a\varepsilon\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x\int_{{{\Bbb R}} ^3}|u|^6{\rm d}x}}{2\varepsilon\int_{{{\Bbb R}} ^3}|u|^6{\rm d}x}. $

因此, 由上式可知

(2.11) $ \begin{eqnarray} &&\frac{a}{2}\int_{{{\Bbb R}} ^3}|\nabla(tu)|^2{\rm d}x+\frac{b}{4} \Big(\int_{{{\Bbb R}} ^3}|\nabla(tu)|^2{\rm d}x\Big)^2-\frac{\varepsilon}{6}\int_{{{\Bbb R}} ^3}|tu|^6{\rm d}x{}\\ & = &\frac{a}{3}t^2\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x+\frac{b}{12}t^4 \Big(\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x\Big)^2{}\\ & = &\frac{a}{3}\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x\frac{b(\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x)^2+\sqrt{b^2(\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x)^4+4a\varepsilon\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x\int_{{{\Bbb R}} ^3}|u|^6{\rm d}x}}{2\varepsilon\int_{{{\Bbb R}} ^3}|u|^6{\rm d}x}{}\\ &&+\frac{b}{12}\Big(\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x\Big)^2\bigg[\frac{b(\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x)^2+\sqrt{b^2(\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x)^4+4a\varepsilon\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x\int_{{{\Bbb R}} ^3}|u|^6{\rm d}x}}{2\varepsilon\int_{{{\Bbb R}} ^3}|u|^6{\rm d}x}\bigg]^2{}\\ & = &\frac{a}{6\varepsilon}\frac{\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x}{(\int_{{{\Bbb R}} ^3}|u|^6{\rm d}x)^{\frac{1}{3}}}\bigg[b\bigg(\frac{\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x}{(\int_{{{\Bbb R}} ^3}|u|^6{\rm d}x)^{\frac{1}{3}}}\bigg)^2+\sqrt{b^2\bigg(\frac{\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x}{(\int_{{{\Bbb R}} ^3}|u|^6{\rm d}x)^{\frac{1}{3}}}\bigg)^4+4a\varepsilon\frac{\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x}{(\int_{{{\Bbb R}} ^3}|u|^6{\rm d}x)^{\frac{1}{3}}}}\bigg]{}\\ &&+\frac{b}{12}\bigg\{\frac{\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x}{2\varepsilon(\int_{{{\Bbb R}} ^3}|u|^6{\rm d}x)^{\frac{1}{3}}}\bigg[b\bigg(\frac{\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x}{(\int_{{{\Bbb R}} ^3}|u|^6{\rm d}x)^{\frac{1}{3}}}\bigg)^2+\sqrt{b^2\bigg(\frac{\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x}{(\int_{{{\Bbb R}} ^3}|u|^6{\rm d}x)^{\frac{1}{3}}}\bigg)^4+4a\varepsilon\frac{\int_{{{\Bbb R}} ^3}|\nabla u|^2{\rm d}x}{(\int_{{{\Bbb R}} ^3}|u|^6{\rm d}x)^{\frac{1}{3}}}}\bigg]\bigg\}^2{}\\ &\geq&\frac{a}{6\varepsilon}{\cal S}(b{\cal S}^2+\sqrt{b^2{\cal S}^4+4a\varepsilon{\cal S}})+\frac{b}{12}\bigg[\frac{{\cal S}}{2\varepsilon}(b{\cal S}^2+\sqrt{b^2{\cal S}^4+4a\varepsilon{\cal S}})\bigg]^2{}\\ & = &\frac{a}{3}\bigg(\frac{b}{2\varepsilon}{\cal S}^3+\sqrt{\frac{b^2}{4\varepsilon^2}{\cal S}^6+\frac{a}{\varepsilon}{\cal S}^3}\bigg)+\frac{b}{12}\bigg(\frac{b}{2\varepsilon}{\cal S}^3+\sqrt{\frac{b^2}{4\varepsilon^2}{\cal S}^6+\frac{a}{\varepsilon}{\cal S}^3}\bigg)^2. \end{eqnarray} $

另一方面, 关于Sobolev常数$ {\cal S} $, 设$ U_0 $是满足$ |U_0|_6 = 1 $的达到$ {\cal S} $的函数, 那么存在唯一的$ t_{U_0}>0 $使得$ t_{U_0}U_0 $满足(2.10) 式. 因此, 通过计算, 有

(2.12) $ \begin{eqnarray} &&\frac{a}{3}t_{U_0}^{2}\int_{{{\Bbb R}} ^3}|\nabla U_0|^2{\rm d}x+\frac{b}{12}t_{U_0}^4 \Big(\int_{{{\Bbb R}} ^3}|\nabla U_0|^2{\rm d}x\Big)^2{}\\ & = &\frac{a}{3}\int_{{{\Bbb R}} ^3}|\nabla U_0|^2{\rm d}x\frac{b(\int_{{{\Bbb R}} ^3}|\nabla U_0|^2{\rm d}x)^2+\sqrt{b^2(\int_{{{\Bbb R}} ^3}|\nabla U_0|^2{\rm d}x)^4+4a\varepsilon\int_{{{\Bbb R}} ^3}|\nabla U_0|^2{\rm d}x}}{2\varepsilon}{}\\ &&+\frac{b}{12}\Big(\int_{{{\Bbb R}} ^3}|\nabla U_0|^2{\rm d}x\Big)^2\bigg[\frac{b(\int_{{{\Bbb R}} ^3}|\nabla U_0|^2{\rm d}x)^2+\sqrt{b^2(\int_{{{\Bbb R}} ^3}|\nabla U_0|^2{\rm d}x)^4+4a\varepsilon\int_{{{\Bbb R}} ^3}|\nabla U_0|^2{\rm d}x}}{2\varepsilon}\bigg]^2{}\\ & = &\frac{a}{3}\bigg[\frac{b}{2\varepsilon} \Big(\int_{{{\Bbb R}} ^3}|\nabla U_0|^2{\rm d}x\Big)^3+\sqrt{\frac{b^2}{4\varepsilon^{2}} \Big(\int_{{{\Bbb R}} ^3}|\nabla U_0|^2{\rm d}x\Big)^6+\frac{a}{\varepsilon}\Big(\int_{{{\Bbb R}} ^3}|\nabla U_0|^2{\rm d}x\Big)^3}\bigg]{}\\ &&+\frac{b}{12}\bigg[\frac{b}{2\varepsilon} \Big(\int_{{{\Bbb R}} ^3}|\nabla U_0|^2{\rm d}x\Big)^3+\sqrt{\frac{b^2}{4\varepsilon^{2}} \Big(\int_{{{\Bbb R}} ^3}|\nabla U_0|^2{\rm d}x\Big)^6+\frac{a}{\varepsilon} \Big(\int_{{{\Bbb R}} ^3}|\nabla U_0|^2{\rm d}x\Big)^3}\bigg]^2{}\\ & = &\frac{a}{3}\bigg(\frac{b}{2\varepsilon}{\cal S}^3+\sqrt{\frac{b^2}{4\varepsilon^{2}}{\cal S}^6+\frac{a}{\varepsilon}{\cal S}^3}\bigg)+\frac{b}{12}\bigg(\frac{b}{2\varepsilon}{\cal S}^3+\sqrt{\frac{b^2}{4\varepsilon^{2}}{\cal S}^6+\frac{a}{\varepsilon}{\cal S}^3}\bigg)^2. \end{eqnarray} $

从而由(2.11) 式和(2.12) 式可得(2.8) 式成立.

现在, 设径向函数$ \bar{U}\in{\cal D}^{1, 2}({{\Bbb R}} ^3) $达到(2.8) 式的函数. 设$ v_n(x) = \zeta(|x|)n^{\frac{1}{2}}\bar{U}(nx) $, $ n\in{\Bbb N} $, 其中$ \zeta\in{\cal C}_0^{\infty}({{\Bbb R}} ^{+}, [0, 1]) $是一个截断函数, 且对$ s\in[0, 1] $有$ \zeta(s) = 1 $. 因此, 存在唯一的$ t_n>0 $满足$ u_n: = t_nv_n\in{\cal N}_{\varepsilon, \infty} $, 即

(2.13) $ \begin{equation} a\|v_n\|^2+bt_n^2|\nabla v_n|_2^4 = t_n^{p-2}|v_n|_p^p+\varepsilon t_n^4|v_n|_6^6. \end{equation} $

由$ v_n $的定义可得

$ \begin{eqnarray*} |v_n-n^{\frac{1}{2}}\bar{U}(nx)|_6^6& = &\int_{{{\Bbb R}} ^3}\Big|\zeta(|x|)n^{\frac{1}{2}}\bar{U}(nx)-n^{\frac{1}{2}}\bar{U}(nx)\Big|^6{\rm d}x\\ & = &n^3\int_{{{\Bbb R}} ^3}\Big|(\zeta(|x|)-1)\bar{U}(nx)\Big|^6{\rm d}x\\ & = &\int_{{{\Bbb R}} ^3}\Big|[\zeta(\frac{|x|}{n})-1]\bar{U}(x)\Big|^6{\rm d}x, \end{eqnarray*} $

利用Lebesgue控制收敛定理可得

(2.14) $ \begin{equation} |v_n-n^{\frac{1}{2}}\bar{U}(nx)|_6\rightarrow0, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, {当 \, n\rightarrow \infty }. \end{equation} $

类似可得

(2.15) $ \begin{equation} |\nabla v_n-\nabla(n^{\frac{1}{2}}\bar{U}(nx))|_2\rightarrow0, \, \, \, {当\, n\rightarrow \infty }, \end{equation} $

(2.16) $ \begin{equation} v_n\rightarrow0\quad {在 \, \, L^2({{\Bbb R}} ^3) }, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, {当 \, n\rightarrow \infty }. \end{equation} $

由(2.16) 式和插值不等式可知, $ v_n\rightarrow0 $在$ L^p({{\Bbb R}} ^3) $. 由(2.15) 式和(2.16) 式得, 当$ n\rightarrow \infty $时有$ \|v_n-n^{\frac{1}{2}}\bar{U}(nx)\|_{H^1}\rightarrow0 $.

断言当$ n\rightarrow \infty $时, $ t_n\rightarrow1 $. 事实上, 从(2.13) 式可知, $ \{t_n\} $有界. 由(2.14)–(2.16) 式可得

$ a|\nabla(n^{\frac{1}{2}}\bar{U}(nx))|_2^2+bt_n^2|\nabla(n^{\frac{1}{2}}\bar{U}(nx))|_2^4 = o(1)+\varepsilon t_n^4|n^{\frac{1}{2}}\bar{U}(nx)|_6^6, $

即

(2.17) $ \begin{equation} a|\nabla\bar{U}|_2^2+bt_n^2|\nabla\bar{U}|_2^4 = \varepsilon t_n^4|\bar{U}|_6^6+o(1). \end{equation} $

由$ \bar{U} $的定义和(2.17) 式, 可推出

$ a|\nabla\bar{U}|_2^2+t_n^2(\varepsilon|\bar{U}|_6^6-a|\nabla\bar{U}|_2^2) = \varepsilon t_n^4|\bar{U}|_6^6+o(1), $

即

(2.18) $ \begin{equation} (1-t_n^2)a|\nabla\bar{U}|_2^2 = \varepsilon(t_n^2-1)t_n^2|\bar{U}|_6^6+o(1), \end{equation} $

从而推出了$ t_n\rightarrow1 $. 因此, 有

$ \begin{eqnarray*} &&I_{\varepsilon, \infty}(u_n)-\bigg(\frac{a}{2}\int_{{{\Bbb R}} ^3}|\nabla(n^{\frac{1}{2}}\bar{U}(nx))|^2{\rm d}x+\frac{b}{4} \Big(\int_{{{\Bbb R}} ^3}|\nabla(n^{\frac{1}{2}}\bar{U}(nx))|^2{\rm d}x\Big)^2-\frac{\varepsilon}{6}\int_{{{\Bbb R}} ^3}|n^{\frac{1}{2}}\bar{U}(nx)|^6{\rm d}x\bigg)\\ & = &\frac{a}{2}\int_{{{\Bbb R}} ^3}|\nabla u_n|^2{\rm d}x+\frac{1}{2}\int_{{{\Bbb R}} ^3}|u_n|^2{\rm d}x+\frac{b}{4} \Big(\int_{{{\Bbb R}} ^3}|\nabla u_n|^2{\rm d}x\Big)^2-\frac{1}{p}\int_{{{\Bbb R}} ^3}|u_n|^p{\rm d}x-\frac{\varepsilon}{6}\int_{{{\Bbb R}} ^3}|u_n|^6{\rm d}x\\ &&-\bigg(\frac{a}{2}\int_{{{\Bbb R}} ^3}|\nabla(n^{\frac{1}{2}}\bar{U}(nx))|^2{\rm d}x+\frac{b}{4} \Big(\int_{{{\Bbb R}} ^3}|\nabla(n^{\frac{1}{2}}\bar{U}(nx))|^2{\rm d}x\Big)^2-\frac{\varepsilon}{6}\int_{{{\Bbb R}} ^3}|n^{\frac{1}{2}}\bar{U}(nx)|^6{\rm d}x\bigg)\\ & = &\frac{1}{2}\int_{{{\Bbb R}} ^3}|u_n|^2{\rm d}x-\frac{1}{p}\int_{{{\Bbb R}} ^3}|u_n|^p{\rm d}x\rightarrow0, \quad {当\; n\rightarrow \infty }. \end{eqnarray*} $

注意到

$ \begin{eqnarray*} &&\frac{a}{2}\int_{{{\Bbb R}} ^3}|\nabla(n^{\frac{1}{2}}\bar{U}(nx))|^2{\rm d}x+\frac{b}{4} \Big(\int_{{{\Bbb R}} ^3}|\nabla(n^{\frac{1}{2}}\bar{U}(nx))|^2{\rm d}x\Big)^2-\frac{\varepsilon}{6}\int_{{{\Bbb R}} ^3}|n^{\frac{1}{2}}\bar{U}(nx)|^6{\rm d}x\\ & = &\frac{a}{3}\bigg(\frac{b}{2\varepsilon}{\cal S}^3+\sqrt{\frac{b^2}{4\varepsilon^{2}}{\cal S}^6+ \frac{a}{\varepsilon}{\cal S}^{3}}\bigg)+\frac{b}{12}\bigg(\frac{b}{2\varepsilon}{\cal S}^3+ \sqrt{\frac{b^2}{4\varepsilon^{2}}{\cal S}^6+\frac{a}{\varepsilon}{\cal S}^3}\bigg)^2\\ & = &{\cal S}_{a, b} \quad\quad \forall n\in{\Bbb N}, \end{eqnarray*} $

因此, 由$ m_{\varepsilon} $的定义可得出(2.6) 式.

推论2.1 对充分小的$ \varepsilon>0 $, 有下面估计成立

(2.19) $ \begin{equation} m_{\varepsilon}<{\cal S}_{a, b}. \end{equation} $

证   由(2.6)式可知, $ m_{\varepsilon}\leq{\cal S}_{a, b} $. 假设$ m_{\varepsilon} = {\cal S}_{a, b} $成立, 由命题2.1可得, 当$ u_n = t_nv_n(x) = t_n\zeta(|x|)n^{\frac{1}{2}}\bar{U}(nx)\in{\cal N}_{\varepsilon, \infty} $时, 其中$ \bar{U} $是方程(2.8) 的一个极小元, 可得$ t_n\rightarrow1 $, $ v_n\rightarrow0 $在$ L^p({{\Bbb R}} ^3) $, 其中$ 2\leq p<6 $和$ I_{\varepsilon, \infty}(u_n)\rightarrow {\cal S}_{a, b} $.

利用引理2.1的(ⅲ) 可得, $ |t_nv_n|_p^p\geq C_1>0 $. 从$ t_n\rightarrow1 $可知, $ |v_n|_p^p\geq C_1>0 $, 与$ v_n\rightarrow0 $在$ L^p({{\Bbb R}} ^3) $矛盾. 因此(2.19) 式成立. 证毕.

对于极限问题(2.1), 有下面的存在性结果.

定理2.1  存在$ \varepsilon_{0}>0 $, 对任意的$ \varepsilon\in(0, \varepsilon_{0}) $, 问题(2.1) 存在正径向对称基态解$ w_{\varepsilon} $.

证   令

$ H_r^1({{\Bbb R}} ^3) = \{u\in H^1({{\Bbb R}} ^3):\ u\ {是径向对称的} \}, \ {\cal N}_r = {\cal N}_{\varepsilon, \infty}\cap H_r^1({{\Bbb R}} ^3), $

易证$ m_{\varepsilon} = \inf\limits_{u\in{\cal N}_r}I_{\varepsilon, \infty}(u) $. 设$ \{u_n\}_n $是$ {\cal N}_r $的一个极小化序列且满足

$ a\|u_n\|_{{\cal D}}^2+|u_n|^2_2+b|\nabla u_n|_2^4 = |u_n|^p_p+\varepsilon|u_n|^6_6, $

$ I_{\varepsilon, \infty}(u_n) = \frac{a}{4}\|u_n\|_{{\cal D}}^2+\frac{1}{4}|u_n|^2_2+\frac{p-4}{4p}|u_n|^p_p+\frac{\varepsilon}{12}|u_n|^6_6 = m_{\varepsilon}+o(1). $

又$ m_{\varepsilon}\leq m $和$ p>4 $, 这导出了$ \{u_n\}_n $在$ H_r^1({{\Bbb R}} ^3) $上有界. 因此, 通过选取子列, 仍记为$ \{u_n\}_n $, 假设存在$ w_{\varepsilon}\in H_r^1({{\Bbb R}} ^3) $使得

(2.20) $ \begin{equation} \left\{ \begin{array}{ll} u_n \rightharpoonup w_{\varepsilon}, & {在 H_r^1({{\Bbb R}} ^3) \, 和\, L^6({{\Bbb R}} ^3) , } \\ u_n \rightarrow w_{\varepsilon}, & {在 L^q({{\Bbb R}} ^3), 2<q<6 , }\\ u_n \rightarrow w_{\varepsilon}, & { {\rm a.e.}\ \, \, x\in{{\Bbb R}} ^3 .} \end{array} \right. \end{equation} $

断言当$ n\rightarrow \infty $时, $ I'_{\varepsilon, \infty}(u_n)\rightarrow0 $. 令$ G(u) = I_{\varepsilon, \infty}' (u)[u]+1, \forall u\in{\cal N}_r = \{u\in H_r^1({{\Bbb R}} ^3)\backslash\{0\}:G(u) = 1\} $, 显然由$ p>2 $可得$ G(u)\in{\cal C}^{2}(H^1({{\Bbb R}} ^3), {{\Bbb R}} ) $. 通过引理2.1 (ⅱ), 可得

(2.21) $ \begin{equation} \langle G' (u), u\rangle = 2\|u\|^2+4b|\nabla u|_2^4-p|u|_p^p-6\varepsilon|u|_6^6 = -2\|u\|^2+(4-p)|u|_p^p-2\varepsilon|u|_6^6\leq-2C^2<0, \end{equation} $

这意味着对任意的$ u\in{\cal N}_{\varepsilon, \infty} $有$ G'(u)\neq0 $. 由于$ I_{\varepsilon, \infty}(u_n)\rightarrow m_{\varepsilon} $, 利用Ekeland变分原理(参见文献[28, 定理8.5])可得: 对于$ \delta = \frac{1}{\sqrt{n}} $, 存在$ v_n\in{\cal N}_{\varepsilon, \infty} $使得

$ \|u_n-v_n\|\leq\frac{2}{\sqrt{n}}, \quad \min\limits_{\lambda\in{{\Bbb R}} }\|I_{\varepsilon, \infty}' (v_n)-\lambda G' (v_n)\|\leq\frac{8}{\sqrt{n}}. $

从而, 利用$ I_{\varepsilon, \infty}' $和$ G' $的连续性, 不妨将$ v_n $替换为$ u_n $, 易得

(2.22) $ \begin{equation} I'_{\varepsilon, \infty}(u_n)[v] = \lambda_{n}G'(u_n)[v]+o(1)\|v\|, \quad \forall v\in H_r^1({{\Bbb R}} ^3), \end{equation} $

其中$ \lambda_{n}\in{{\Bbb R}} $. 选取$ v = u_n\in{\cal N}_r $, 可得

(2.23) $ \begin{equation} 0 = I'_{\varepsilon, \infty}(u_n)[u_n] = \lambda_{n}G'(u_n)[u_n]+o(1)\|u_n\|. \end{equation} $

由$ \{u_n\} $的有界性和(2.21) 式可得$ \lambda_{n} = o(1) $. 因此, 由(2.22) 式可知

(2.24) $ \begin{eqnarray} &&a\int_{{{\Bbb R}} ^3}\nabla u_n\cdot\nabla v{\rm d}x+\int_{{{\Bbb R}} ^3}u_n v{\rm d}x+b\int_{{{\Bbb R}} ^3}|\nabla u_n|^2{\rm d}x\int_{{{\Bbb R}} ^3}\nabla u_n\cdot\nabla v{\rm d}x -\int_{{{\Bbb R}} ^3}|u_n|^{p-2}u_nv{\rm d}x{}\\ &&-\varepsilon\int_{{{\Bbb R}} ^3}|u_n|^4u_nv{\rm d}x = o(1)\|v\|, \quad \forall v\in H_r^1({{\Bbb R}} ^3), \end{eqnarray} $

断言成立.

令$ w_n: = u_n-w_{\varepsilon} $, 欲证当$ n\rightarrow \infty $时, 有$ \|w_n\|_{{\cal D}}\rightarrow0 $. 存在一个直到子序列, 仍记为$ w_n $, 假设$ \lim\limits_{n\rightarrow \infty}\|w_n\|_{{\cal D}} = l_1\geq0 $.

由(2.20) 式, Brezis-Lieb引理和$ u_n\in {\cal N}_r $可推导出

$ a\|u_n\|_{{\cal D}}^2+|u_n|^2_2+b\|u_n\|_{{\cal D}}^4-|u_n|^p_p-\varepsilon|u_n|^6_6 = o(1), $

$ \|u_n\|^2 = \|w_{\varepsilon}\|^2+\|w_n\|^2+o(1), $

$ \|u_n\|_{{\cal D}}^4 = \|w_{\varepsilon}\|_{{\cal D}}^4+\|w_n\|_{{\cal D}}^4+2\|w_{\varepsilon}\|_{{\cal D}}^2\|w_n\|_{{\cal D}}^2+o(1), $

$ |u_n|^6_6 = |w_n|^6_6+|w_{\varepsilon}|^6_6+o(1), \quad |u_n|_p^p\rightarrow |w_{\varepsilon}|_p^p. $

因此

(2.25) $ \begin{eqnarray} && a\|w_n\|_{{\cal D}}^2+|w_n|^2_2+a\|w_{\varepsilon}\|_{{\cal D}}^2+|w_{\varepsilon}|^2_2+b\|w_n\|_{{\cal D}}^4 +b\|w_{\varepsilon}\|_{{\cal D}}^4 {}\\ &&+2bl_1^2\|w_{\varepsilon}\|_{{\cal D}}^2 -\varepsilon|w_{\varepsilon}|^6_6-\varepsilon|w_n|^6_6-|w_{\varepsilon}|^p_p = o(1). \end{eqnarray} $

利用(2.24) 式, 有

(2.26) $ \begin{equation} a\|w_{\varepsilon}\|_{{\cal D}}^2+|w_{\varepsilon}|^2_2+bl_1^2\|w_{\varepsilon}\|_{{\cal D}}^2+b\|w_{\varepsilon}\|_{{\cal D}}^4-\varepsilon|w_{\varepsilon}|^6_6-|w_{\varepsilon}|^p_p = 0. \end{equation} $

由(2.25) 式和(2.26) 式可得

(2.27) $ \begin{equation} a\|w_n\|_{{\cal D}}^2+|w_n|^2_2+b\|w_n\|_{{\cal D}}^4+bl_1^2\|w_{\varepsilon}\|_{{\cal D}}^2-\varepsilon|w_n|^6_6 = o(1). \end{equation} $

若当$ n\rightarrow \infty $时, $ |w_n|_6\rightarrow0 $. 由(2.27) 式得$ w_n\rightarrow0 $在$ H_r^1({{\Bbb R}} ^3) $, 故$ l_1 = 0 $.

若$ \lim\limits_{n\rightarrow \infty}|w_n|_6\neq0 $, 假设$ l_2 = \lim\limits_{n\rightarrow \infty}|w_n|^6_6>0 $.

从(2.27) 式和$ {\cal S}l_2^{\frac{1}{3}}\leq l_1^2 $, 可推导出$ al_1^2+bl_1^4\leq\varepsilon l_2 $且$ \varepsilon l_2\geq al_1^2+bl_1^4\geq a{\cal S}l_2^{\frac{1}{3}}+b{\cal S}^{2}l_2^{\frac{2}{3}}, $这意味着$ \varepsilon l_2^{\frac{2}{3}}-b{\cal S}^2l_2^{\frac{1}{3}}-a{\cal S}\geq0, $即

(2.28) $ \begin{equation} l_2^{\frac{1}{3}}\geq\frac{b{\cal S}^2+\sqrt{b^2{\cal S}^4+4\varepsilon a{\cal S}}}{2\varepsilon}. \end{equation} $

由(2.26) 式可得

$ \begin{eqnarray*} m_{\varepsilon}+o(1)& = &I_{\varepsilon, \infty}(u_n)-\frac{1}{6}\langle I'_{\varepsilon, \infty}(u_n), u_n\rangle\\ & = &\frac{a}{3}\|u_n\|_{{\cal D}}^2+\frac{1}{3}|u_n|^2_2+\frac{b}{12}\|u_n\|^4_{{\cal D}}+\frac{p-6}{6p}|u_n|^p_p\\ & = &\frac{a}{3}(\|w_{\varepsilon}\|_{{\cal D}}^2+\|w_n\|_{{\cal D}}^2)+\frac{1}{3}(|w_n|^2_2+|w_{\varepsilon}|^2_2)+\frac{b}{12}(\|w_n\|_{{\cal D}}^4+\|w_{\varepsilon}\|_{{\cal D}}^4+2l_1^2\|w_{\varepsilon}\|_{{\cal D}}^2)\\ &&+\frac{p-6}{6p}|w_{\varepsilon}|^p_p+o(1)\\ &\geq&\frac{a}{3}\|w_n\|_{{\cal D}}^2+\frac{b}{12}\|w_n\|_{{\cal D}}^4+\frac{1}{3}(a\|w_{\varepsilon}\|_{{\cal D}}^2+|w_{\varepsilon}|^2_2)+\frac{b}{12}\|w_{\varepsilon}\|_{{\cal D}}^4+\frac{b}{6}l_1^2\|w_{\varepsilon}\|_{{\cal D}}^2\\ &&+\frac{p-6}{6p}|w_{\varepsilon}|^p_p+o(1)\\ &\geq&\frac{a}{3}\|w_n\|_{{\cal D}}^2+\frac{b}{12}\|w_n\|_{{\cal D}}^4+\frac{p-4}{4p}|w_{\varepsilon}|^p_p+o(1)\\ &\geq&\frac{a}{3}l_1^2+\frac{b}{12}l_1^4, \end{eqnarray*} $

结合(2.28) 式, 有

$ \begin{eqnarray*} m_{\varepsilon}&\geq&\frac{a}{3}{\cal S}l_2^{\frac{1}{3}}+\frac{b}{12}{\cal S}^2l_2^{\frac{2}{3}}\\ & \geq&\frac{a}{3}{\cal S}\frac{b{\cal S}^2+\sqrt{b^2{\cal S}^4+4\varepsilon a{\cal S}}}{2\varepsilon}+\frac{b}{12}{\cal S}^2 \bigg(\frac{b{\cal S}^{2}+\sqrt{b^2{\cal S}^4+4\varepsilon a{\cal S}}}{2\varepsilon}\bigg)^2\\ & = &\frac{a}{3}\bigg(\frac{b}{2\varepsilon}{\cal S}^3+\sqrt{\frac{b^2}{4\varepsilon^{2}}{\cal S}^6+\frac{a}{\varepsilon}{\cal S}^3}\bigg)+\frac{b}{12}\bigg(\frac{b}{2\varepsilon}{\cal S}^3+\sqrt{\frac{b^2}{4\varepsilon^{2}}{\cal S}^6+\frac{a}{\varepsilon}{\cal S}^3}\bigg)^2\\ & = &{\cal S}_{a, b}. \end{eqnarray*} $

这与(2.19) 式矛盾. 因此可得

(2.29) $ \begin{equation} u_n\rightarrow w_{\varepsilon}\, \, {在}\, \, H_r^1({{\Bbb R}} ^3). \end{equation} $

由(2.24) 和(2.29) 式, 可知$ w_{\varepsilon}\in H_r^1({{\Bbb R}} ^3) $是极小化函数且使得$ w_{\varepsilon} $是下面方程的解

(2.30) $ \begin{equation} -(a+b|\nabla u|_2^2)\triangle u+u = |u|^{p-2}u+\varepsilon|u|^4u, \, \, \, x\in{{\Bbb R}} ^3. \end{equation} $

为证明$ w_{\varepsilon} $是正的, 注意到$ |w_{\varepsilon}| $也是$ I_{\varepsilon, \infty} $约束在$ {\cal N}_{\varepsilon, \infty} $上的极小元, 故可假设$ w_{\varepsilon}\geq0 $. 另外, 由$ w_{\varepsilon} $是(2.30) 的解, 由强极值原理可得$ w_{\varepsilon}>0 $.

引理2.2

$ \lim\limits_{\varepsilon\rightarrow0}m_{\varepsilon} = m. $

证   设$ w_{\varepsilon} $是$ m_{\varepsilon} $的极小元, 其中$ \varepsilon\in(0, \varepsilon_0) $和$ w_{\varepsilon} $是定理2.1得到的解. 选取$ t_{\varepsilon}>0 $使得$ t_{\varepsilon}w_{\varepsilon}\in{\cal N}_{\infty} $, 即

(2.31) $ \begin{equation} \|w_{\varepsilon}\|^2+bt_{\varepsilon}^2|\nabla w_{\varepsilon}|_2^4 = t_{\varepsilon}^{p-2}|w_{\varepsilon}|_p^p. \end{equation} $

由引理2.1(ⅲ) 的证明, 易得$ \|w_{\varepsilon}\|^2\leq4m $, 这推出了$ |\nabla w_{\varepsilon}|_2 $有界. 另外, 由引理2.1(ⅲ) 可知$ |w_{\varepsilon}|_p^p\geq C_1>0 $. 因此从(2.31) 式可推出

(2.32) $ \begin{equation} 4m\geq t_{\varepsilon}^{p-2}|w_{\varepsilon}|_p^p-bt_{\varepsilon}^2|\nabla w_{\varepsilon}|_2^4\geq C_1t_{\varepsilon}^{p-2}-bt_{\varepsilon}^2\|w_{\varepsilon}\|^4\geq C_1t_{\varepsilon}^{p-2}-16bm^2t_{\varepsilon}^2. \end{equation} $

由$ 4<p<6 $, 可得$ \{t_{\varepsilon}\} $有界. 因此, 由(2.3) 式可得

(2.33) $ \begin{eqnarray} m_{\varepsilon}\leq m&\leq& I_{\infty}(t_{\varepsilon}w_{\varepsilon}) = I_{\varepsilon, \infty}(t_{\varepsilon}w_{\varepsilon})+\frac{\varepsilon}{6}\int_{{{\Bbb R}} ^3}|t_{\varepsilon}w_{\varepsilon}|^6{\rm d}x {}\\ & \leq& I_{\varepsilon, \infty}(w_{\varepsilon})+\frac{\varepsilon}{6}\int_{{{\Bbb R}} ^3}|t_{\varepsilon}w_{\varepsilon}|^6{\rm d}x = m_{\varepsilon}+C \varepsilon. \end{eqnarray} $

上式令$ \varepsilon\rightarrow0 $, 可得结论成立.

本节主要通过分析在水平$ c $上的$ (PS)_c $序列来证明问题(1.1) 基态解的存在性, 其中$ c $小于极限问题(2.1) 的基态能量值.

引理3.1  设$ \{u_n\}_n $是$ I_{\varepsilon} $的$ (PS)_c $序列, 那么$ \{u_n\}_n $在$ H^1({{\Bbb R}} ^3) $上有界且$ c\geq0 $.

证   由$ \{u_n\}_n $是$ I_{\varepsilon} $的$ (PS)_c $序列, 可得

(3.1) $ \begin{equation} c+o(1)\|u_n\|_{V}\geq I_{\varepsilon}(u_{n})-\frac{1}{4}I'_{\varepsilon}(u_n)[u_n] = \frac{1}{4}\|u_{n}\|_{V}^{2}+\frac{p-4}{4p}|u_{n}|_{p}^{p}+\frac{\varepsilon}{12}|u_{n}|_{6}^{6}, \end{equation} $

这蕴含了$ \|u_n\|_V^2\leq4c+o(1)\|u_n\|_{V}. $因此, $ \{u_n\}_n $在$ H^1({{\Bbb R}} ^3) $有界. 由$ 4<p<6 $和(3.1) 式知$ c\geq0 $.

命题3.1  假设$ V(x) $满足条件$ (V_0) $和(1.5). 设$ \{u_n\}_n $是$ I_{\varepsilon} $约束在$ {\cal N}_{\varepsilon} $的$ (PS)_c $序列. 若对任意的$ \varepsilon\in(0, \varepsilon_0) $, 有$ c<m_{\varepsilon} $, 则$ \{u_n\}_n $在$ H^1({{\Bbb R}} ^3) $中是相对紧的.

证   设$ \{u_n\}_n $是$ I_{\varepsilon} $约束在$ {\cal N}_{\varepsilon} $上的$ (PS)_c $序列, 由引理3.1知, $ \{u_n\}_n $在$ H^1({{\Bbb R}} ^3) $中有界. 令$ \rho_n(x): = V(x)|u_n|^2+|u_n|^p+\varepsilon|u_n|^6\in L^1({{\Bbb R}} ^3) $, 则$ \{\rho_n\} $在$ L^1({{\Bbb R}} ^3) $中有界. 因此, 通过选择子序列, 可假设$ |\rho_n|_1\rightarrow l, $当$ n\rightarrow \infty, $于是得到$ l>0 $, 否则, 有$ I_{\varepsilon}(u_n)\rightarrow0 $, 这与$ I_{\varepsilon}(u_n)\rightarrow c $相矛盾. 事实上, 由$ \{u_n\}_n $是$ (PS)_c $序列且在$ H^1({{\Bbb R}} ^3) $中有界, 可得

$ I_{\varepsilon}'(u_n)[u_n] = o(1)\|u_n\|_{V} = o(1), $

这意味着$ \|u_n\|_V\rightarrow0 $, 因而有$ I_{\varepsilon}(u_n)\rightarrow0 $.

现在, 利用文献[29] 中的集中紧性原理去建立序列$ {\rho_n} $的紧性.

(1) (消失性)   若$ {\rho_n} $消失, 则$ \{u_n^2\} $也消失, 即存在$ R>0 $使得

$ \lim\limits_{n\rightarrow \infty}\sup\limits_{y\in{{\Bbb R}} ^3}\int_{B_R(y)}|u_n|^2{\rm d}x = 0. $

利用消失引理[28, 引理1.21]可得, 对于$ 2<r<6 $, 有$ u_n\rightarrow0 $在$ L^r({{\Bbb R}} ^3) $. 于是

$ |u_n|_p^p\rightarrow0, \, \, \, \ $

这意味着

(3.2) $ \begin{equation} I_{\varepsilon}(u_n) = \frac{1}{2}\|u_n\|_{V}^2+\frac{b}{4}\|u_n\|_{{\cal D}}^4-\frac{\varepsilon}{6}|u_n|_6^6+o(1) \end{equation} $

和

(3.3) $ \begin{equation} \|u_n\|_{V}^2+b\|u_n\|_{{\cal D}}^4-\varepsilon|u_n|_6^6 = o(1). \end{equation} $

因此, 可假设

$ \|u_n\|_{{\cal D}}^2\rightarrow t_1>0, \, \, \, \, \, \, |u_n|_6^6\rightarrow t_2>0, \, \, {当 \, n\rightarrow \infty }. $

然后由(3.3) 式和$ {\cal S}t_{2}^{\frac{1}{3}}\leq t_{1}^{2} $, 可推导出下式

$ \varepsilon t_{2}\geq at_{1}^{2}+bt_{1}^{4}+o(1)\geq a{\cal S}t_{2}^{\frac{1}{3}}+b{\cal S}^{2}t_{2}^{\frac{2}{3}}, $

即

$ t_2^{\frac{1}{3}}\geq\frac{b{\cal S}^2+\sqrt{b^2{\cal S}^4+4\varepsilon a{\cal S}}}{2\varepsilon}. $

另一方面, 有

$ \begin{eqnarray*} c+o(1)& = &I_{\varepsilon}(u_n)-\frac{1}{6}\langle I'_{\varepsilon}(u_n), u_n\rangle\\ & = &\frac{1}{2}\|u_n\|_{V}^2+\frac{b}{4}\|u_n\|_{{\cal D}}^4-\frac{\varepsilon}{6}|u_n|_6^6-\frac{1}{6}\|u_n\|_{V}^2-\frac{b}{6}\|u_n\|_{{\cal D}}^4+\frac{\varepsilon}{6}|u_n|_6^6+o(1)\\ &\geq&\frac{a}{3}\|u_n\|_{{\cal D}}^2+\frac{b}{12}\|u_n\|_{{\cal D}}^4 = \frac{a}{3}t_1^2+\frac{b}{12}t_1^4+o(1)\\ &\geq&\frac{a}{3}{\cal S}\frac{b{\cal S}^{2}+\sqrt{b^{2}{\cal S}^{4}+4\varepsilon a{\cal S}}}{2\varepsilon}+\frac{b}{12}{\cal S}^2 \bigg(\frac{b{\cal S}^{2}+\sqrt{b^{2}{\cal S}^{4}+4\varepsilon a{\cal S}}}{2\varepsilon}\bigg)^{2}\\ & = &\frac{a}{3}\bigg(\frac{b}{2\varepsilon}{\cal S}^{3}+\sqrt{\frac{b^{2}}{4\varepsilon^{2}}{\cal S}^{6}+\frac{a}{\varepsilon}{\cal S}^{3}}\bigg)+\frac{b}{12}\bigg(\frac{b}{2\varepsilon}{\cal S}^{3}+\sqrt{\frac{b^{2}}{4\varepsilon^{2}}{\cal S}^{6}+\frac{a}{\varepsilon}{\cal S}^{3}}\bigg)^{2}\\ & = &{\cal S}_{a, b}>m_{\varepsilon}, \end{eqnarray*} $

这与$ c<m_{\varepsilon} $矛盾. 因此, 消失性不会发生.

(2) (二分性)   证明二分性不会发生. 用反证法, 假设存在$ \alpha\in(0, l) $和$ \{x_n\}\subset{{\Bbb R}} ^3 $使得对任意$ r>R_{\varepsilon} $和$ r'>R_{\varepsilon} $, 有

$ \liminf\limits_{n\rightarrow \infty}\int_{B_r(x_n)}\rho_n{\rm d}x\geq\alpha-\varepsilon, \quad\liminf\limits_{n\rightarrow \infty}\int_{B_{r'}^c(x_n)}\rho_n{\rm d}x\geq(l-\alpha)-\varepsilon. $

因此, 这里存在$ \varepsilon_{n}\rightarrow0 $, $ r_n\rightarrow +\infty $和$ r'_n = 4r_n $, 所以

(3.4) $ \begin{equation} \int_{B_{r_n}(x_n)}\rho_n{\rm d}x\geq\alpha-\varepsilon_{n}, \, \, \, \int_{B_{r'_n}^c(x_n)}\rho_n{\rm d}x\geq(l-\alpha)-\varepsilon_{n}, \end{equation} $

这意味着

(3.5) $ \begin{equation} \int_{B_{4r_n}(x_n)\backslash B_{r_n}(x_n)}\rho_{n}{\rm d}x = \int_{B_{r'_n}(x_n)}\rho_{n}{\rm d}x-\int_{B_{r_n}(x_n)}\rho_{n}{\rm d}x\leq\alpha+\varepsilon_{n}-(\alpha-\varepsilon_{n}) = 2\varepsilon_{n}. \end{equation} $

令$ \varphi_{0}\in{\cal C}^{\infty}({{\Bbb R}} _{+}) $是一个截断函数且满足$ 0\leq\varphi_{0}\leq1 $, 当$ 2\leq s\leq3 $时$ \varphi_{0}(s) = 1 $, 当$ s\leq1 $或$ s\geq4 $时$ \varphi_{0}(s) = 0 $和$ |\varphi'_{0}(s)|\leq2 $. 假设$ \varphi_{n}(x) = \varphi_{0}(\frac{|x-x_n|}{r_n}) $, 可得

$ \begin{eqnarray*} && \int_{{{\Bbb R}} ^3}\Big(|\nabla(\varphi_{n}u_n)|^2+V(x)|\varphi_{n}u_n|^2\Big){\rm d}x\\ & = &\int_{B_{4r_n}(x_n)\backslash B_{r_n}(x_n)}\Big(|\nabla\varphi_{n}\cdot u_n+\varphi_{n}\cdot\nabla u_n|^2+ \varphi_{n}^2V(x)|u_n|^2\Big){\rm d}x\\ &\leq&\int_{B_{4r_n}(x_n)\backslash B_{r_n}(x_n)}\Big(|\nabla\varphi_{n}|^2|u_n|^2+2\nabla\varphi_{n}\cdot\nabla u_n\cdot u_n+|\nabla u_n|^2+V(x)|u_n|^2\Big){\rm d}x, \end{eqnarray*} $

其中

$ \begin{eqnarray*} && \int_{B_{4r_n}(x_n)\backslash B_{r_n}(x_n)}V(x)|u_n|^2{\rm d}x\\ & = &\bigg(\int_{B_{4r_n}(x_n)\backslash B_{r_n}(x_n)}(V(x))^{\frac{3}{2}}{\rm d}x\bigg)^{\frac{2}{3}}\bigg(\int_{B_{4r_n}(x_n)\backslash B_{r_n}(x_n)}u_n^6{\rm d}x\bigg)^{\frac{1}{3}}\\ &\leq &C\int_{B_{4r_n}(x_n)\backslash B_{r_n}(x_n)}u_n^2{\rm d}x\leq C_1, \end{eqnarray*} $

$ \bigg(\int_{B_{4r_n}(x_n)\backslash B_{r_n}(x_n)}|\nabla\varphi_{n}|^2|u_n|^2\bigg)^{\frac{1}{2}}\leq\frac{2}{r_n}\bigg(\int_{B_{4r_n}(x_n)\backslash B_{r_n}(x_n)}|u_n|^2\bigg)^{\frac{1}{2}}\leq C. $

由$ \varphi_{n} $的定义和Hölder不等式, 可推导出

(3.6) $ \begin{eqnarray} &&\int_{B_{4r_n}(x_n)\backslash B_{3r_n}(x_n)}u_n\nabla u_n\nabla\varphi_{n}{\rm d}x{}\\ &\leq&\frac{2}{r_n}\int_{B_{4r_n}(x_n)\backslash B_{3r_n}(x_n)}|u_n||\nabla u_n|{\rm d}x{}\\ &\leq&\frac{2}{r_n}\bigg(\int_{B_{4r_n}(x_n)\backslash B_{3r_n}(x_n)}|u_n|^2{\rm d}x\bigg)^{\frac{1}{2}}\bigg(\int_{B_{4r_n}(x_n)\backslash B_{3r_n}(x_n)}|\nabla u_n|^2{\rm d}x\bigg)^{\frac{1}{2}}{}\\ &\leq&\frac{C}{r_n} = o(1). \end{eqnarray} $

类似地, 有

(3.7) $ \begin{equation} \int_{B_{2r_n}(x_n)\backslash B_{r_n}(x_n)}u_n\nabla u_n\nabla\varphi_{n}{\rm d}x = o(1). \end{equation} $

因此, 很容易得到$ \varphi_{n}u_n\in H^1({{\Bbb R}} ^3) $和

(3.8) $ \begin{equation} \langle I'_{\varepsilon}(u_n), \varphi_{n}u_n\rangle\rightarrow0, \, \, {当}\, \, n\rightarrow \infty. \end{equation} $

令$ A: = \lim\limits_{n\rightarrow \infty}\|u_n\|_{{\cal D}}^2>0 $, 断言

(3.9) $ \begin{eqnarray} && a\int_{B_{3r_n}(x_n)\backslash B_{2r_n}(x_n)}|\nabla u_n|^2{\rm d}x+bA\int_{B_{3r_n}(x_n)\backslash B_{2r_n}(x_n)}|\nabla u_n|^2{\rm d}x {}\\ &&+\int_{B_{3r_n}(x_n)\backslash B_{2r_n}(x_n)}V(x)|u_n|^2{\rm d}x = \int_{B_{3r_n}(x_n)\backslash B_{2r_n}(x_n)}(|u_n|^p+\varepsilon|u_n|^6){\rm d}x+o(1). \end{eqnarray} $

事实上, 由(3.5) 式可得

(3.10) $ \begin{equation} \int_{{{\Bbb R}} ^3}\varepsilon|u_n|^6\varphi_{n}{\rm d}x = \int_{B_{3r_n}(x_n)\backslash B_{2r_n}(x_n)}\varepsilon|u_n|^6{\rm d}x+o(1) = o(1), \end{equation} $

(3.11) $ \begin{equation} \int_{{{\Bbb R}} ^3}|u_n|^p\varphi_{n}{\rm d}x = \int_{B_{3r_n}(x_n)\backslash B_{2r_n}(x_n)}|u_n|^p{\rm d}x+o(1) = o(1), \end{equation} $

(3.12) $ \begin{equation} \int_{{{\Bbb R}} ^3}V(x)|u_n|^2\varphi_{n}{\rm d}x = \int_{B_{3r_n}(x_n)\backslash B_{2r_n}(x_n)}V(x)|u_n|^2{\rm d}x+o(1) = o(1). \end{equation} $

因此, 由(3.8) 式和(3.10)–(3.12) 式可得

(3.13) $ \begin{equation} \int_{{{\Bbb R}} ^3}\nabla u_n\cdot\nabla(\varphi_{n}u_n){\rm d}x = o(1). \end{equation} $

因为

$ \begin{eqnarray*} \int_{{{\Bbb R}} ^3}\nabla u_n\nabla(\varphi_{n}u_n){\rm d}x & = &\int_{B_{4r_n}(x_n)\backslash B_{r_n}(x_n)}\nabla u_n\nabla(\varphi_{n}u_n){\rm d}x\\ & = &\int_{B_{4r_n}(x_n)\backslash B_{r_n}(x_n)}|\nabla u_n|^2\varphi_{n}{\rm d}x+\int_{B_{4r_n}(x_n)\backslash B_{r_n}(x_n)}u_n\nabla u_n\nabla\varphi_{n}{\rm d}x\\ & = &\int_{B_{3r_n}(x_n)\backslash B_{2r_n}(x_n)}|\nabla u_n|^2{\rm d}x+\int_{B_{4r_n}(x_n)\backslash B_{3r_n}(x_n)}|\nabla u_n|^2\varphi_{n}{\rm d}x\\ &&+\int_{B_{2r_n}(x_n)\backslash B_{r_n}(x_n)}|\nabla u_n|^2\varphi_{n}{\rm d}x+\int_{B_{4r_n}(x_n)\backslash B_{3r_n}(x_n)}u_n\nabla u_n\nabla\varphi_{n}{\rm d}x\\ &&+\int_{B_{2r_n}(x_n)\backslash B_{r_n}(x_n)}u_n\nabla u_n\nabla\varphi_{n}{\rm d}x, \end{eqnarray*} $

由(3.6), (3.7) 和(3.13) 式可得

$ \int_{B_{4r_n}(x_n)\backslash B_{3r_n}(x_n)}|\nabla u_n|^2\varphi_{n}{\rm d}x = o(1), \, \, \, \, \, \, \, \, \, \, \int_{B_{2r_n}(x_n)\backslash B_{r_n}(x_n)}|\nabla u_n|^2\varphi_{n}{\rm d}x = o(1). $

从而证明了(3.9) 式.

考虑另一截断函数$ \eta $, 满足$ 0\leq\eta\leq1 $, 当$ s\leq2 $时$ \eta(s) = 1 $, 当$ s\geq3 $时$ \eta(s) = 0 $且$ |\eta'(s)|\leq2 $. 令$ U_n(x) = \eta\big(\frac{|x-x_n|}{r_n}\big)u_n(x), $$ W_n(x) = \Big(1-\eta\big(\frac{|x-x_n|}{r_n}\big)\Big)u_n(x), $然后由(3.4) 式可得

$ \int_{{{\Bbb R}} ^3}(V(x)|U_n|^2+|U_n|^p+\varepsilon|U_n|^6){\rm d}x\geq\alpha-\varepsilon_{n}, $

$ \int_{{{\Bbb R}} ^3}(V(x)|W_n|^2+|W_n|^p+\varepsilon|W_n|^6){\rm d}x\geq(l-\alpha)-\varepsilon_{n}. $

从(3.9) 式和$ \eta $的定义可得

$ \begin{eqnarray*} \int_{{{\Bbb R}} ^3}\nabla U_n\nabla W_n{\rm d}x& = &\int_{B_{3r_n}(x_n)\backslash B_{2r_n}(x_n)}\nabla(\eta u_n)\nabla[(1-\eta)u_n]{\rm d}x\\ & = &\int_{B_{3r_n}(x_n)\backslash B_{2r_n}(x_n)}(\nabla\eta\cdot u_n+\eta\cdot\nabla u_n)[\nabla(1-\eta)\cdot u_n+(1-\eta)\cdot\nabla u_n]{\rm d}x\\ & = &\int_{B_{3r_n}(x_n)\backslash B_{2r_n}(x_n)}\Big(\eta(1-\eta)|\nabla u_n|^2-|\nabla\eta|^2|u_n|^2 +(1-2\eta)u_n\nabla u_n\nabla\eta\Big){\rm d}x\\ & = &o(1), \end{eqnarray*} $

这意味着

(3.14) $ \begin{equation} \int_{{{\Bbb R}} ^3}|\nabla u_n|^2{\rm d}x = \int_{{{\Bbb R}} ^3}|\nabla U_n|^2{\rm d}x+\int_{{{\Bbb R}} ^3}|\nabla W_n|^2{\rm d}x+o(1). \end{equation} $

类似地, 由(3.12) 式可得

(3.15) $ \begin{equation} \int_{{{\Bbb R}} ^3}V(x)|u_n|^2{\rm d}x = \int_{{{\Bbb R}} ^3}V(x)|U_n|^2{\rm d}x+\int_{{{\Bbb R}} ^3}V(x)|W_n|^2{\rm d}x+o(1). \end{equation} $

现在区分以下两种情况.

情形1   若$ \{x_n\} $有界, 由$ (V_0) $得, 对任意的$ \varepsilon>0 $, 存在$ R_0>0 $使得

(3.16) $ \begin{equation} |V(x)-V_{\infty}|\leq \varepsilon, \, \, \, \, \forall|x|\geq R_0. \end{equation} $

由$ \{x_n\}\subset{{\Bbb R}} ^3 $的有界性可得, 存在$ R'>0 $使得$ |x_n|\leq R' $. 因此, 当$ n $充分大时, 有$ {{\Bbb R}} ^3\backslash B_{r_n}(x_n)\subset{{\Bbb R}} ^3\backslash B_{r_n-R'}(0)\subset {{\Bbb R}} ^3\backslash B_{R_0}(0) $. 由(3.16) 式和假设(1.5) 可推导出

$ \begin{eqnarray*} \bigg|\int_{{{\Bbb R}} ^3}\big(V(x)-V_{\infty}\big)|W_n|^2{\rm d}x\bigg| & = &\int_{|x-x_n|>2r_n}|V(x)-V_{\infty}||W_n|^2{\rm d}x\\ &\leq&\varepsilon\int_{|x-x_n|>2r_n}|W_n|^2{\rm d}x\leq\varepsilon C, \end{eqnarray*} $

且由$ \varepsilon $的任意性可导出

(3.17) $ \begin{equation} \int_{{{\Bbb R}} ^3}\big(V( x)-V_{\infty}\big)|W_n|^2{\rm d}x = o(1). \end{equation} $

令

$ I_A(W_n) = \frac{1}{2}\|W_n\|^2+\frac{b}{2}A\|W_n\|_{{\cal D}}^2-\frac{1}{p}|W_n|_p^p-\frac{\varepsilon}{6}|W_n|_6^6, $

此时可断言

(3.18) $ \begin{equation} \langle I'_A(W_n), W_n\rangle = \langle I'_{\varepsilon}(u_n), W_n\rangle+o(1) = o(1). \end{equation} $

事实上, 由(3.9) 式可得, 当$ n $充分大时有

(3.19) $ \begin{eqnarray} &&\int_{{{\Bbb R}} ^3}\Big(|\nabla W_n|^2-\nabla u_n\nabla W_n\Big){\rm d}x{}\\ & = &\int_{{{\Bbb R}} ^3}\Big(\big|\nabla[(1-\eta)u_n]\big|^2-\nabla u_n\nabla[(1-\eta)u_n]\Big){\rm d}x{}\\ & = &\int_{{{\Bbb R}} ^3}\Big(|\nabla\eta|^2|u_n|^2-(1-2\eta)u_n\cdot\nabla u_n\cdot\nabla\eta+\eta(\eta-1)|\nabla u_n|^2\Big){\rm d}x{}\\ &\leq&\int_{{{\Bbb R}} ^3}\Big||\nabla\eta|^2|u_n|^2+(2\eta-1)u_n\cdot\nabla u_n\cdot\nabla\eta-\eta(1-\eta)|\nabla u_n|^2\Big|{\rm d}x{}\\ &\leq&\frac{C}{{r_n}^2}+\frac{C}{r_n}+o(1) = o(1). \end{eqnarray} $

从(3.9) 式得, 当$ n $充分大时, 有

(3.20) $ \begin{eqnarray} \bigg|\int_{{{\Bbb R}} ^3}\Big(|W_n|^6-|u_n|^4u_nW_n\Big){\rm d}x\bigg| & = &\bigg|\int_{{{\Bbb R}} ^3}\big((1-\eta)^6|u_n|^6-|u_n|^4u_n(1-\eta)u_n\big){\rm d}x\bigg|{}\\ &\leq&\int_{{{\Bbb R}} ^3}(1-\eta)\big|(1-\eta)^{5}-1\big||u_n|^6{\rm d}x{}\\ & = &\int_{B_{3r_n}(x_n)\backslash B_{2r_n}(x_n)}(1-\eta)\big|(1-\eta)^{5}-1\big||u_n|^6{\rm d}x{}\\ & = &o(1). \end{eqnarray} $

类似地, 有

(3.21) $ \begin{equation} \bigg|\int_{{{\Bbb R}} ^3}\Big(|W_n|^p-|u_n|^{p-2}u_nW_n\Big){\rm d}x\bigg| = o(1), \end{equation} $

且根据(3.12) 和(3.17) 式, 可得

(3.22) $ \begin{eqnarray} &&\bigg|\int_{{{\Bbb R}} ^3}\Big(V_{\infty}|W_n|^2-V(x)u_nW_n\Big){\rm d}x\bigg|{}\\ &\leq&\int_{{{\Bbb R}} ^3}\Big(|V(x)-V_{\infty}||W_n|^2+V(x)(1-\eta)\eta|u_n|^2\Big){\rm d}x{}\\ & = &\int_{{{\Bbb R}} ^3}|V(x)-V_{\infty}||W_n|^2{\rm d}x+\int_{B_{3r_n}(x_n)\backslash B_{2r_n}(x_n)}V(x)(1-\eta)\eta|u_n|^2{\rm d}x{}\\ & = &o(1). \end{eqnarray} $

因此, 由(3.19)–(3.22) 式可推出(3.18) 式. 因为$ \|W_n\|_{{\cal D}}^2<\|u_n\|_{{\cal D}}^2 $, 存在$ t_n<1 $使得$ t_nW_n\in{\cal N}_{\varepsilon, \infty} $和

$ \begin{eqnarray*} m_{\varepsilon}&\leq& I_{\varepsilon, \infty}(t_nW_n) = I_{\varepsilon, \infty}(t_nW_n)-\frac{1}{p}\langle I'_{\varepsilon, \infty}(t_nW_n), t_nW_n\rangle\\ & = &(\frac{1}{2}-\frac{1}{p})\|t_nW_n\|^2+(\frac{1}{4}-\frac{1}{p})b\|t_nW_n\|_{{\cal D}}^4+(\frac{1}{p}-\frac{1}{6})\varepsilon|t_nW_n|_6^6\\ &<&(\frac{1}{2}-\frac{1}{p})\|W_n\|^2+(\frac{1}{4}-\frac{1}{p})b\|W_n\|_{{\cal D}}^4+(\frac{1}{p}-\frac{1}{6})\varepsilon|W_n|_6^6\\ &<&(\frac{1}{2}-\frac{1}{p})\|u_n\|^2+(\frac{1}{4}-\frac{1}{p})b\|u_n\|_{{\cal D}}^4+(\frac{1}{p}-\frac{1}{6})\varepsilon|u_n|_6^6\\ & = &I_{\varepsilon}(u_n)-\frac{1}{p}\langle I'_{\varepsilon, \infty}(u_n), u_n\rangle\\ & = &c+o(1)<m_{\varepsilon}, \end{eqnarray*} $

这就产生了矛盾.

情形2   若$ \{x_n\} $无界, 选取子列, 仍记为$ \{x_n\} $, 使得$ |x_n|\geq 2r_n $. 那么有$ B_{2r_n}(x_n)\subset {{\Bbb R}} ^3\backslash B_{r_n}(0)\subset{{\Bbb R}} ^3\backslash B_{R_0}(0) $. 类似于(3.17) 式的证明, 可推出

$ \int_{{{\Bbb R}} ^3}\big(V(x)-V_{\infty}\big)|U_n|^2{\rm d}x = o(1). $

类似于$ \{x_n\} $有界的情形, 可得出矛盾$ m_{\varepsilon}<m_{\varepsilon} $. 因此, 二分性不会发生.

据以上讨论, 序列$ \{\rho_n\} $是紧的, 即存在$ \{x_n\}\subset{{\Bbb R}} ^3 $, $ \widetilde{R}>0 $使得对任意的$ \hat{\delta}>0 $有

(3.23) $ \begin{equation} \int_{B_{\widetilde{R}}^{c}(x_n)}\rho_n(x){\rm d}x<\hat{\delta}. \end{equation} $

断言序列$ \{x_n\} $一定有界. 否则, 选取$ r_n $使得当$ r_n\rightarrow +\infty $时, 有$ |x_n|\geq r_n\geq \widetilde{R}+R_0 $. 当$ n $充分大时, 有$ B_{\widetilde{R}}(x_n)\subset {{\Bbb R}} ^3\backslash B_{r_n-\widetilde{R}}(0)\subset{{\Bbb R}} ^3\backslash B_{R_0}(0) $. 由(3.16) 和(3.23) 式得

$ \begin{eqnarray*} \bigg|\int_{{{\Bbb R}} ^3}\big(V(x)-V_{\infty}\big)|u_n|^2{\rm d}x\bigg| &\leq&\int_{B_{\widetilde{R}}(x_n)}|V(x)-V_{\infty}||u_n|^2{\rm d}x+\int_{B_{\widetilde{R}}^{c}(x_n)}|V(x)-V_{\infty}||u_n|^2{\rm d}x \\ & \leq&\varepsilon C+o(1). \end{eqnarray*} $

因此, $ I_{\varepsilon}(u_n) = I_{\varepsilon, \infty}(u_n)+o(1) $和$ o(1) = \langle I'_{\varepsilon}(u_n), u_n\rangle = \langle I'_{\varepsilon, \infty}(u_n), u_n\rangle +o(1) $. 于是存在$ t_n\rightarrow1 $使得$ t_nu_n\in{\cal N}_{\varepsilon, \infty} $. 因此, 当$ n $充分大时, 有

$ c = I_{\varepsilon}(u_n)+o(1) = I_{\varepsilon, \infty}(u_n)+o(1) = I_{\varepsilon, \infty}(t_nu_n)+o(1)\geq m_{\varepsilon}+o(1), $

这与条件$ c<m_{\varepsilon} $相矛盾. 因此断言得证.

因为$ \{u_n\}_{n} $在$ H^1({{\Bbb R}} ^3) $中有界, 通过取子列, 仍记为$ \{u_n\}_n $, 可假设存在$ \bar{u}\in H^1({{\Bbb R}} ^3) $使得

(3.24) $ \begin{equation} u_n\rightharpoonup \bar{u}\, \, {在}\, \, H^1({{\Bbb R}} ^3), \quad u_n\rightarrow \bar{u}\, \, {在}\, \, L_{\rm loc}^r({{\Bbb R}} ^3)\, (2\leq r<6), \quad u_n\rightarrow \bar{u}\, \, {\rm a.e.}\ \, x\in{{\Bbb R}} ^3. \end{equation} $

因此, 从(3.23) 式可推导出当$ n\rightarrow \infty $时

(3.25) $ \begin{equation} \int_{{{\Bbb R}} ^3}V(x)|u_n|^2{\rm d}x\rightarrow \int_{{{\Bbb R}} ^3}V(x)|\bar{u}|^2{\rm d}x, \, \, \, \, \, \, \int_{{{\Bbb R}} ^3}|u_n|^p{\rm d}x\rightarrow \int_{{{\Bbb R}} ^3}|\bar{u}|^p{\rm d}x. \end{equation} $

令$ v_n: = u_n-\bar{u} $, 欲证当$ n\rightarrow \infty $时, 有$ \|v_n\|_{{\cal D}}\rightarrow0 $. 通过取子列, 仍记为$ v_n $, 假设$ \lim\limits_{n\rightarrow \infty}\|v_{n}\|_{{\cal D}} = \alpha_{1}\geq0 $. 由$ I'_{\varepsilon}(u_n)\rightarrow0 $, (3.24) 式和Brezis-Lieb引理, 可推导出

$ \|u_n\|_V^2+b\|u_n\|_{{\cal D}}^4-|u_n|^p_p-\varepsilon|u_n|^6_6 = o(1), $

$ \|u_n\|_{{\cal D}}^{2} = \|\bar{u}\|_{{\cal D}}^{2}+\|v_n\|_{{\cal D}}^{2}+o(1), $

$ \|u_n\|_{{\cal D}}^{4} = \|\bar{u}\|_{{\cal D}}^{4}+\|v_n\|_{{\cal D}}^{4}+2\|\bar{u}\|_{{\cal D}}^{2}\|v_n\|_{{\cal D}}^{2}+o(1), $

$ |u_n|^6_6 = |v_n|^6_6+|\bar{u}|^6_6+o(1). $

从而有

(3.26) $ \begin{equation} \|v_n\|_V^2+\|\bar{u}\|_V^2+b\|v_n\|_{{\cal D}}^4+b\|\bar{u}\|_{{\cal D}}^4+2b\alpha_1^2\|\bar{u}\|_{{\cal D}}^2 -\varepsilon|\bar{u}|^6_6-\varepsilon|v_n|^6_6-|\bar{u}|^p_p = o(1). \end{equation} $

此外, 有

(3.27) $ \begin{equation} \lim\limits_{n\rightarrow \infty}\langle I'_{\varepsilon}(u_n), \bar{u}\rangle = \|\bar{u}\|_V^2+b\alpha_1^2\|\bar{u}\|_{{\cal D}}^2+b\|\bar{u}\|_{{\cal D}}^4-\varepsilon|\bar{u}|^6_6-|\bar{u}|^p_p = 0. \end{equation} $

由(3.26) 和(3.27) 式可得

(3.28) $ \begin{equation} \|v_n\|_V^2+b\|v_n\|_{{\cal D}}^4+b\alpha_1^2\|\bar{u}\|_{{\cal D}}^2-\varepsilon|v_n|^6_6 = o(1). \end{equation} $

若当$ n\rightarrow \infty $时有$ |v_n|_6\rightarrow0 $, 通过(3.28) 式可知$ v_n\rightarrow0 $在$ H^1({{\Bbb R}} ^3) $, 从而有$ \alpha_1 = 0 $.

若$ \lim\limits_{n\rightarrow \infty}|v_n|_6\neq0 $, 假设$ \alpha_{2} = \lim\limits_{n\rightarrow \infty}|v_n|^6_6>0 $.

由(3.28) 式和$ {\cal S}\alpha_2^{\frac{1}{3}}\leq \alpha_1^2 $, 可推导出$ a\alpha_1^2+b\alpha_1^4\leq\varepsilon \alpha_2 $, 从而引出下式

$ \varepsilon \alpha_{2}\geq a\alpha_{1}^2+b\alpha_{1}^4\geq a{\cal S}\alpha_{2}^{\frac{1}{3}}+b{\cal S}^{2}\alpha_{2}^{\frac{2}{3}}, $

这意味着

$ \varepsilon \alpha_{2}^{\frac{2}{3}}-b{\cal S}^{2}\alpha_{2}^{\frac{1}{3}}-a{\cal S}\geq0, $

即

(3.29) $ \begin{equation} \alpha_{2}^{\frac{1}{3}}\geq\frac{b{\cal S}^{2}+\sqrt{b^{2}{\cal S}^{4}+4\varepsilon a{\cal S}}}{2\varepsilon}. \end{equation} $

由(3.27) 式, 有

$ \begin{eqnarray*} c+o(1)& = &I_{\varepsilon}(u_n)-\frac{1}{6}\langle I'_{\varepsilon}(u_n), u_n\rangle = \frac{1}{3}\|u_n\|_V^2+\frac{b}{12}\|u_n\|^4_{{\cal D}}+\frac{p-6}{6p}|u_n|^p_p\\ & = &\frac{1}{3}(\|\bar{u}\|_V^2+\|v_n\|_V^2)+\frac{b}{12}(\|v_n\|_{{\cal D}}^4+\|\bar{u}\|_{{\cal D}}^4+2\alpha_1^2\|\bar{u}\|_{{\cal D}}^2)+\frac{p-6}{6p}|\bar{u}|^p_p+o(1)\\ &\geq&\frac{a}{3}\|v_n\|_{{\cal D}}^2+\frac{b}{12}\|v_n\|_{{\cal D}}^4+\frac{1}{3}\|\bar{u}\|_V^2+\frac{b}{12}\|\bar{u}\|_{{\cal D}}^4+\frac{b}{6}\alpha_1^2\|\bar{u}\|_{{\cal D}}^2+\frac{p-6}{6p}|\bar{u}|^p_p+o(1)\\ &\geq&\frac{a}{3}\|v_n\|_{{\cal D}}^2+\frac{b}{12}\|v_n\|_{{\cal D}}^4+\frac{p-4}{4p}|\bar{u}|^p_p+o(1)\\ &\geq&\frac{a}{3}\alpha_1^2+\frac{b}{12}\alpha_1^4, \end{eqnarray*} $

结合(3.29) 式, 有

$ \begin{eqnarray*} c&\geq&\frac{a}{3}{\cal S}\alpha_{2}^{\frac{1}{3}}+\frac{b}{12}{\cal S}^2\alpha_{2}^ {\frac{2}{3}}\geq\frac{a}{3}{\cal S}\frac{b{\cal S}^{2}+\sqrt{b^{2}{\cal S}^4+4\varepsilon a{\cal S}} }{2\varepsilon}+\frac{b}{12}{\cal S}^2\bigg(\frac{b{\cal S}^2+\sqrt{b^2{\cal S}^4+4\varepsilon a{\cal S}}}{2\varepsilon}\bigg)^2\\ & = &\frac{a}{3}\bigg(\frac{b}{2\varepsilon}{\cal S}^3+\sqrt{\frac{b^2}{4\varepsilon^{2}}{\cal S}^6+\frac{a}{\varepsilon}{\cal S}^3}\bigg)+\frac{b}{12}\bigg(\frac{b}{2\varepsilon}{\cal S}^3+\sqrt{\frac{b^2}{4\varepsilon^{2}}{\cal S}^6+\frac{a}{\varepsilon}{\cal S}^3}\bigg)^2\\ & = &{\cal S}_{a, b}>m_{\varepsilon}, \end{eqnarray*} $

这与$ c<m_{\varepsilon} $相矛盾. 因此可得当$ n\rightarrow \infty $时, 有$ u_n\rightarrow \bar{u} $在$ H^1({{\Bbb R}} ^3) $. 证毕.

定理1.1的证明   假设$ V(x)\not\equiv1 $. 现在, 断言

(3.30) $ \begin{equation} \inf\limits_{{\cal N}_{\varepsilon}}I_{\varepsilon}<m_{\varepsilon}. \end{equation} $

事实上, 定理2.1中的方程(2.1) 的基态解$ w_{\varepsilon} $, 令$ t>0 $使得$ tw_{\varepsilon}\in{\cal N}_{\varepsilon} $, 则有

$ \inf\limits_{{\cal N}_{\varepsilon}}I_{\varepsilon}\leq I_{\varepsilon}(tw_{\varepsilon})<I_{\varepsilon, \infty}(tw_{\varepsilon})\leq I_{\varepsilon, \infty}(w_{\varepsilon}) = m_{\varepsilon}. $

采用定理2.1的证明方法(产生出$ PS $序列), 利用命题3.1, 可得泛函$ I_{\varepsilon} $在$ {\cal N}_{\varepsilon} $上存在极小元$ u\in H^1({{\Bbb R}} ^3) $. 类似于定理2.1的讨论, 可证得$ u $是正的. 证毕.

定理1.2的证明  设$ z\in{{\Bbb R}} ^3 $使得(1.7) 式成立, 且$ t_{\varepsilon}>0 $使得$ t_{\varepsilon}w_z\in{\cal N}_{\varepsilon} $. 为证明问题(1.1) 存在基态解, 只需证明断言: 对于充分小$ \varepsilon>0 $, 成立

(3.31) $ \begin{equation} I_{\varepsilon}(t_{\varepsilon}w_z)<m_{\varepsilon}. \end{equation} $

事实上, 若(3.31) 式成立, 则$ \inf\limits_{{\cal N}_{\varepsilon}}I_{\varepsilon}<m_{\varepsilon} $. 类似于定理1.1的证明, 可证得问题(1.1) 存在基态解.

为证明(3.31) 式, 令$ w_z\in{\cal N} $, 即$ \|w_z\|_V^2+b|\nabla w_z|_2^4 = |w_z|_p^p $. 断言由(1.7) 式可得

(3.32) $ \begin{equation} I(w_{z})<m. \end{equation} $

现在, 由于

(3.33) $ \begin{equation} I(w_z) = \frac{1}{2}\|w_z\|_V^2+\frac{b}{4}|\nabla w_z|_2^4-\frac{1}{p}|w_z|_p^p = \frac{1}{4}\|w_z\|_V^2+\frac{p-4}{4p}|w_z|_p^p. \end{equation} $

注意到, 由(1.7) 式和$ w $是方程(1.6) 的基态解可得

(3.34) $ \begin{equation} I_{\infty}(w) = \frac{1}{4}\|w\|^2+\frac{p-4}{4p}|w|_p^p = m. \end{equation} $

由(1.7), (3.33)和(3.34)式, 可推得(3.32) 成立.

最后, 断言当$ \varepsilon\rightarrow0 $时, 有$ t_{\varepsilon}\rightarrow 1 $. 事实上, 由$ t_{\varepsilon}w_z\in{\cal N}_{\varepsilon} $, 即

(3.35) $ \begin{equation} \|w_z\|_V^2+bt_{\varepsilon}^2|\nabla w_z|_2^4 = t_{\varepsilon}^{p-2}|w_z|_p^p+\varepsilon t_{\varepsilon}^{4}|w_z|_6^6, \end{equation} $

由上式易推得$ \{t_{\varepsilon}\} $有界. 由$ w_z\in{\cal N} $和(3.35) 式, 有

$ (|w_z|_p^p-b|\nabla w_z|_2^4)+bt_{\varepsilon}^2|\nabla w_z|_2^4 = t_{\varepsilon}^{p-2}|w_z|_p^p+o(1), $

即

$ (t_{\varepsilon}^{2}-1)b|\nabla w_z|_2^4 = (t_{\varepsilon}^{p-2}-1)|w_z|_p^p+o(1), $

上式蕴含了$ t_{\varepsilon}\rightarrow1 $.

由引理2.2可得: 当$ \varepsilon\rightarrow0 $时, 有$ m_{\varepsilon}\rightarrow m $. 因此, 对于充分小的$ \varepsilon $, 由(3.32) 式可推出(3.31) 式成立. 证毕.