本文考虑如下与时间无关的非线性Schrödinger-Poisson (SP)方程

$ \begin{equation} -\Delta u(x)+V(x)u(x)+\int_{{\Bbb R}^{N}}\frac{u^{2}(y)}{|x-y|}{\rm d}yu(x)=\lambda u(x)+a|u(x)|^{p}u(x), \, \ u(x)\in H^{1}({\Bbb R}^{N}), \end{equation} $

(1.1)

其中空间维数$N\geq2$, 系数$a\in{\Bbb R^{+}}$, 参数$\lambda$表示拉格朗日乘子.另外指标$p$满足:当$N= 2$时, $p>0$; 当$N\geq 3$时, $0 < p < \frac{4}{N-2}$.方程(1.1)对应的约束变分问题为

$ \begin{equation} {\rm e}^{p}(a):=\inf\limits_{\{u\in {\cal H}, \|u\|^{2}_{2}=1\}} E^{p}_{a}(u), \end{equation} $

(1.2)

这里能量泛函$E^{p}_{a}(u)$满足

$ \begin{eqnarray} E^{p}_{a}(u)&:=&\int_{{\Bbb R}^{N}}|\nabla u(x)|^{2}+V(x)u^{2}(x){\rm d}x+\frac{1}{2}\int_{{\Bbb R}^{N}} \int_{{\Bbb R}^{N}}\frac{u^{2}(y)u^{2}(x)}{|x-y|}{\rm d}y{\rm d}x\\ &&-\frac{2a}{(p+2)}\int_{{\Bbb R}^{N}}|u(x)|^{p+2}{\rm d}x, \end{eqnarray} $

(1.3)

位势$V(x)$满足

$(V_1)$ $0\leq V(x)\in L_{loc}^{\infty}({\Bbb R}^{N})$, $\lim\limits_{|x|\rightarrow\infty} V(x) = \infty$并且$\inf\limits_{x\in{\Bbb R}^{N}} V(x)=0$.

(1.2)式中${\cal H}$定义为

$ {\cal H}:=\Big\{u\in H^{1}({\Bbb R}^{N})\Big |\int_{{\Bbb R}^{N}}V(x)u^{2}(x){\rm d}x < \infty\Big\}, $

范数定义为

$ \|u\|_{{\cal H}}=\Big(\int_{{\Bbb R}^{N}} |\nabla u(x)|^{2}+V(x)u^{2}(x){\rm d}x\Big)^{\frac{1}{2}}. $

受郭玉劲及其合作者最近工作的启发^{[5-6, 8]}, 本文研究(1.2)约束极小元的存在性及非存在性.在文献[5-6, 8]中, 作者考虑了当$N=2$时如下不带Poisson项的$L^{2}$ -质量临界(即$ p=2$)情形下的约束变分问题

$ \begin{equation} {\rm e}^{2}(a):=\inf\limits_{\{u\in {\cal H}, \|u\|^{2}_{2}=1\}} E^{2}_{a}(u), \end{equation} $

(1.4)

其中$E^{2}_{a}(u)$满足

$ \begin{equation} E^{2}_{a}(u):=\int_{{\Bbb R}^{2}}|\nabla u(x)|^{2}+V(x)u^{2}(x){\rm d}x-\frac{a}{2}\int_{{\Bbb R}^{2}}|u(x)|^{4}{\rm d}x. \end{equation} $

(1.5)

问题(1.4)极小元的存在性及非存在性结果概述如下:

$\bullet$ 变分问题(1.4)存在极小元当且仅当$0 < a < a^{\ast}:=\|\phi\|^{2}_{2}$.

上述$\phi$ (在平移的意义下)是当$N=2$时下面方程唯一的径向对称正解^{[4, 9]}

$ \begin{equation} -\Delta u(x)+ u(x)-u^{1+\frac{4}{N}}(x)=0, \ \, u\in H^1({\Bbb R}^{N}). \end{equation} $

(1.6)

本文由于受到非局部Poisson项的影响, (1.2)式中${\rm e}^{p}(a)$的泛函结构发生了变化.特别是在$L^{2}$ -质量临界(即$ p=\frac{4}{N}$)情形下, 当$a=a^{\ast}:=\|\phi\|^{\frac{4}{N}}_{2}$时($N\geq2, \ \phi$满足(1.6)式) ${\rm e}^{p}(a)$的结构变化将会影响极小元的存在性.受文献[7]的启示, 运用Ekeland's变分原理^{[10, Theorem 5.1]}, 文章证明了当 $ p=\frac{4}{N}$且$a=a^{\ast}$时${\rm e}^{p}(a)$极小元的存在性.在给出主要结果之前, 下面引入一些必备的知识.特别说明, 文章中$\|u\|_{p}=\big(\int_{{\Bbb R}^{N}}|u(x)|^{p}{\rm d}x\big) ^{\frac{1}{p}}$, 字母$C$代表正常数.

$\bullet$  Gagliardo-Nirenberg不等式^[11]:当$N\geq 2$且$0 < p < \frac{4}{N-2}$时下面不等式成立

$ \begin{equation} \int_{{\Bbb R}^{N}} |u(x)|^{2+p}{\rm d}x \leq C \Big(\int_{{\Bbb R}^{N}}|\nabla u(x)|^2{\rm d}x \Big)^{\frac{pN}{4}}\\ \Big(\int_{{\Bbb R}^{N}}|u(x)|^{2}{\rm d}x \Big)^{1+\frac{p}{2}-\frac{pN}{4}}, \, \ u(x)\in H^1({\Bbb R}^{N}). \end{equation} $

(1.7)

尤其当$p=\frac{4}{N}$时

$ \begin{equation} \int_{{\Bbb R}^{N}} |u(x)|^{2+\frac{4}{N}}{\rm d}x \leq \frac{N+2}{N\|\phi\|_2^{\frac{4}{N}}}\int_{{\Bbb R}^{N}}|\nabla u(x)|^2{\rm d}x \Big(\int_{{\Bbb R}^{N}}|u(x)|^{2}{\rm d}x \Big)^{\frac{2}{N}}, \, \ u(x)\in H^1({\Bbb R}^{N}). \end{equation} $

(1.8)

(1.8)式等号成立当且仅当$u(x)=\phi(x)$, 其中$\phi$满足(1.6)式.由(1.6)和(1.8)式得到

$ \begin{equation} \frac{N}{2}\int_{{\Bbb R}^{N}}|\phi(x)|^2{\rm d}x = \int_{{\Bbb R}^{N}}|\nabla \phi(x)|^2{\rm d}x = \frac{N}{N+2}\int_{{\Bbb R}^{N}} |\phi(x)|^{2+\frac{4}{N}}{\rm d}x. \end{equation} $

(1.9)

由文献[4, 命题4.1]知$\phi(x)$还满足:当$|x|\to \infty$时

$ \begin{equation} |\nabla \phi(x)|, \phi(|x|) = O(|x|^{-\frac{N-1}{2}}{\rm e}^{-|x|}), \end{equation} $

(1.10)

另外, 当空间维数$N\geq2$时, 运用Hardy-Littewood-Sobolev不等式, 类似文献[3, 13]的方法可以推出

$ \begin{equation} \int_{{\Bbb R}^{N}}\int_{{\Bbb R}^{N}}\frac{u^{2}(y)u^{2}(x)}{|x-y|}{\rm d}y{\rm d}x\leq C\Big(\int_{{\Bbb R}^{N}}|\nabla u(x)|^{2}{\rm d}x\Big)^{\frac{1}{2}} \Big(\int_{{\Bbb R}^{N}}|u(x)|^{2}{\rm d}x\Big)^{\frac{3}{2}}. \end{equation} $

(1.11)

下面给出紧嵌入引理^{[1, Theorem 2.1]}:

引理1.1  假设 $V(x)$满足条件$(V_1)$, 则对任意$r\in [2, 2^{\ast})$, $2^{\ast}=\frac{2N}{N-2}$, 下列空间

$ \begin{equation} {\cal H}\hookrightarrow L^{r}({\Bbb R}^{N}) \end{equation} $

(1.12)

是紧的.

基于上述事实, 本文结果如下.

定理1.1  假设$N\geq2$, $V(x)$满足条件$(V_1)$, $\phi(x)$ (在平移的意义下)是方程(1.6)唯一的径向对称正解, 令$a^{\ast}:=\|\phi\|^{\frac{4}{N}}_{2}$, 则以下结论成立

(ⅰ) 对任意给定的系数$a>0$, 如果$0 < p < \frac{4}{N}$, 那么问题(1.2)至少存在一个极小元.

(ⅱ) 对任意给定的系数$a>0$, 如果$p>\frac{4}{N}$, 那么问题(1.2)不存在极小元.

(ⅲ) 当$p=\frac{4}{N}$时, 问题(1.2)存在极小元当且仅当$0 < a\leq a^{\ast}$.

注1.1  解决问题(1.2)极小元存在性的主要困难在于, 当$p=\frac{4}{N}$且$a=a^{\ast}$时, 极小化序列的有界性不再容易得到, 文献[5-6, 8]的方法也不再适用于这种情况.受文献[7]的启示, 本文运用Ekeland's变分原理^{[10, Theorem 5.1]} 证明了极小化序列实际上是有界的, 并且在 ${\cal H}$中存在一个强收敛子序列.还需要提及的是, 在文献[5-6, 8]中作者还分析了位势阱$V(x)$在不同形状下, 当$p=\frac{4}{N}$且$a$从下方接近$a^{\ast}$时, 极小元的爆破行为.而在本文, 当$p=\frac{4}{N}$且$a=a^{\ast}$时, 定理1.1给出了极小元的存在性, 所以极小元不可能发生爆破, 这也是本文只研究极小元存在性的原因.

这一部分主要是对定理1.1进行证明.下面对指标$p$分类, 进而讨论问题(1.2)约束极小元的存在性和非存在性.

情形1  在定理1.1的假设下, 对任意给定的$a>0$, 如果$0 < p < \frac{4}{N}$, 那么问题(1.2)至少存在一个极小元.

证  假设$u\in {\cal H}$并且$\|u\|^{2}_{2}=1$.由(1.3)和(1.7)式计算得

$ \begin{eqnarray} E^{p}_{a}(u)&\geq& \int_{{\Bbb R}^{N}}\Big(|\nabla u(x)|^{2}+V(x)u^{2}(x)\Big){\rm d}x \\ &&+\frac{1}{2}\int_{{\Bbb R}^{N}} \int_{{\Bbb R}^{N}}\frac{u^{2}(y)u^{2}(x)}{|x-y|}{\rm d}y{\rm d}x-C\Big(\int_{{\Bbb R}^{N}}|\nabla u(x)|^{2}{\rm d}x\Big)^{\frac{pN}{4}} \\ &\geq& \int_{{\Bbb R}^{N}}|\nabla u(x)|^{2}{\rm d}x-C\Big(\int_{{\Bbb R}^{N}}|\nabla u(x)|^{2}{\rm d}x\Big)^{\frac{pN}{4}}. \end{eqnarray} $

(2.1)

因为$0 < p < \frac{4}{N}$, 由(2.1)式知$E^{p}_{a}(u)$是下方有界的.假设$\{u_{n}\}\subset {\cal H}$是${\rm e}^{p}(a)$的一个极小化序列且满足

$ \begin{equation} \lim\limits_{n\rightarrow\infty}E^{p}_{a}(u_{n})={\rm e}^{p}(a) \ \hbox{且}\ \|u_{n}\|^{2}_{2}=1. \end{equation} $

(2.2)

由(2.1)式知对任意$a>0$, $\int_{{\Bbb R}^{N}}\big(|\nabla u_{n}(x)|^{2}+V(x)u_{n}^{2}(x)\big){\rm d}x$关于$n$一致有界.运用引理1.1, 选取合适的子列(仍用$\{u_{n}\}$表示), 则存在$u_{0}\in {\cal H}$使得

$ \begin{equation} u_{n} \stackrel{n}\rightarrow u_{0}\ \hbox{在}\, L^{r}({\Bbb R}^{N})\, \hbox{中成立, }\, \, r\in [2, 2^{\ast}). \end{equation} $

(2.3)

利用(1.11)和(2.3)式可得

$ \begin{eqnarray} &&\lim\limits_{n\rightarrow\infty}\int_{{\Bbb R}^{N}} |u_{n}(x)|^{2}{\rm d}x= \int_{{\Bbb R}^{N}} |u_{0}(x)|^{2}{\rm d}x=1, \\ &&\lim\limits_{n\rightarrow\infty}\int_{{\Bbb R}^{N}} |u_{n}(x)|^{p+2}{\rm d}x=\int_{{\Bbb R}^{N}} |u_{0}(x)|^{p+2}{\rm d}x, \\ &&\lim\limits_{n\rightarrow\infty}\int_{{\Bbb R}^{N}} \int_{{\Bbb R}^{N}}\frac{u_{n}^{2}(y)u_{n}^{2}(x)}{|x-y|}{\rm d}y{\rm d}x = \int_{{\Bbb R}^{N}} \int_{{\Bbb R}^{N}}\frac{u_{0}^{2}(y)u_{0}^{2}(x)}{|x-y|}{\rm d}y{\rm d}x. \end{eqnarray} $

(2.4)

由(2.4)式推出$E^{p}_{a}(u_{n})$是弱下半连续的, 且满足

$ \begin{equation} {\rm e}^{p}(a)=\lim\limits_{n\rightarrow\infty} E^{p}_{a}(u_{n})\geq\liminf\limits_{n\rightarrow\infty} E^{p}_{a}(u_{n})\geq E^{p}_{a}(u_{0})={\rm e}^{p}(a). \end{equation} $

(2.5)

从(2.5)式获知

$ u_{n} \stackrel{n}\rightarrow u_{0}\ \hbox{在}\ {\cal H}\ \hbox{ 中成立, }~~\hbox{其中}\ u_{0}\ \hbox{满足}\ E^{p}_{a}(u_{0})={\rm e}^{p}(a). $

因此$u_{0}$是问题(1.2)的极小元.

情形2  在定理1.1的假设下, 对任意给定的$a>0$, 如果$p>\frac{4}{N}$, 那么问题(1.2)不存在极小元.

证  受文献[5]启发, 只需证明${\rm e}^{p}(a)=-\infty$即可.选取试验函数

$ \begin{equation} u_{\tau}(x)=A_{\tau}\frac{\tau^{\frac{N}{2}}}{\| \phi\|_2}\varphi(x-x_0) \phi\big(\tau|x-x_0|\big), \end{equation} $

(2.6)

其中$\tau>0$, $x_0\in {\Bbb R}^{N}$满足$V(x_{0})=0$, $\phi(x)$满足(1.6)式; $0\leq \varphi \in C_0^\infty({\Bbb R}^{N})$满足:当$|x|\leq 1$时, $\varphi(x)=1$, 当$|x|\geq 2$时, $\varphi(x)=0$; $A_{\tau}>0$的选取决定$\int_{{\Bbb R}^{N}}u_{\tau}(x)^2 {\rm d}x =1$.利用(1.10)和(1.11)式, 当$\tau\rightarrow\infty$时计算得

$ \begin{equation} 1\le A_{\tau}^2 \leq 1+C{\rm e}^{-2\tau}, \end{equation} $

(2.7)

$ \begin{equation} \int_{{\Bbb R}^{N}}|\nabla u_\tau(x)|^2{\rm d}x \leq\frac{A_{\tau }^2\tau^2}{\| \phi\|^2_2}\int_{{\Bbb R}^{N}} |\nabla \phi(x)|^{2}{\rm d}x +C {\rm e}^{-2\tau}, \end{equation} $

(2.8)

$ \begin{equation} \int_{{\Bbb R}^{N}}|u_\tau(x)|^{2+p}{\rm d}x \geq\frac{A_{\tau}^{2+p}\tau^\frac{Np}{2}}{\| \phi\|^{2+p}_2}\int_{{\Bbb R}^{N}} |\phi(x)|^{2+p}{\rm d}x -C {\rm e}^{-(1+p)\tau}, \end{equation} $

(2.9)

$ \begin{equation} \int_{{\Bbb R}^{N}}\int_{{\Bbb R}^{N}}\frac{u^{2}_{\tau}(y)u^{2}_{\tau}(x)}{|x-y|}{\rm d}y{\rm d}x \leq C\Big(\int_{{\Bbb R}^{N}}|\nabla u_\tau(x)|^2{\rm d}x\Big)^{\frac{1}{2}}\leq C\tau\Big(\int_{{\Bbb R}^{N}} |\nabla \phi(x)|^{2}{\rm d}x\Big)^{\frac{1}{2}}. \end{equation} $

(2.10)

与此同时, 由于$x\to V(x)\varphi^{2}\big(\frac{x-x_0}{R}\big)$是有界的并有紧支集, 可以得到

$ \begin{equation} \int_{{\Bbb R}^{N}}V(x)u^2_\tau(x) {\rm d}x \leq \frac{A_{\tau }^2}{\| \phi\|^2_2} \bigg[\int_{0\leq|x|\leq\sqrt{\tau}}V \Big(\frac{x}{\tau}+x_0\Big)\phi^2(x){\rm d}x +C {\rm e}^{-\sqrt{\tau}}\bigg]. \end{equation} $

(2.11)

联合(2.7)-(2.11)式可知

$ \begin{eqnarray} {\rm e}^{p}(a)&\leq&\lim\limits_{\tau\to\infty} E^{p}_{a}(u_\tau)\\ &\leq&\lim\limits_{\tau\rightarrow\infty}\Big[\frac{A_{\tau }^2\tau^2}{\| \phi\|^2_2}\int_{{\Bbb R}^{N}} |\nabla \phi(x)|^{2}{\rm d}x-\frac{2a}{p+2}\frac{A_{\tau}^{2+p}\tau^\frac{Np}{2}}{\| \phi\|^{2+p}_2}\int_{{\Bbb R}^{N}} |\phi(x)|^{2+p}{\rm d}x \\ &&+C\tau\Big(\int_{{\Bbb R}^{N}} |\nabla \phi(x)|^{2}{\rm d}x\Big)^{\frac{1}{2}}+C{\rm e}^{-\sqrt{\tau}}\Big]+V(x_{0})=-\infty. \end{eqnarray} $

(2.12)

故问题(1.2)不存在极小元.

情形3  在定理1.1的假设下, 当$p=\frac{4}{N}$时, 分以下三步证明问题(1.2)极小元存在性和非存在性.

步骤1  当$0 < a < a^{\ast}$时, 则问题(1.2)存在极小元.

证  假设$u\in {\cal H}$且$\|u\|^{2}_{2}=1$.利用(1.8)式, 类似(2.1)式的计算可得

$ \begin{eqnarray} E^{p}_{a}(u)&\geq &\int_{{\Bbb R}^{N}}|\nabla u(x)|^{2}+V(x)u^{2}(x){\rm d}x \\ &&+\frac{1}{2}\int_{{\Bbb R}^{N}} \int_{{\Bbb R}^{N}}\frac{u^{2}(y)u^{2}(x)}{|x-y|}{\rm d}y{\rm d}x-\frac{a}{a^{\ast}}\int_{{\Bbb R}^{N}}|\nabla u(x)|^{2}{\rm d}x\\ &\geq& \Big(\frac{a^{\ast}-a}{a^{\ast}}\Big)\int_{{\Bbb R}^{N}}|\nabla u(x)|^{2}{\rm d}x+\int_{{\Bbb R}^{N}}V(x)u^{2}(x){\rm d}x\\ &&+\frac{1}{2}\int_{{\Bbb R}^{N}}\int_{{\Bbb R}^{N}}\frac{u^{2}(y)u^{2}(x)}{|x-y|}{\rm d}y{\rm d}x. \end{eqnarray} $

(2.13)

当$a < a^{\ast}$时, 由(2.13)式可知$E^{p}_{a}(u)\geq0$.假设$\{u_{n}\}\subset {\cal H}$是${\rm e}^{p}(a)$的一个极小化序列, 由(2.13)式知$\int_{{\Bbb R}^{N}}|\nabla u_{n}(x)|^{2}+V(x)u_{n}^{2}(x){\rm d}x$关于$n$一致有界.接下来的证明与情形1完全相同, 不再详述.

步骤2  当$a>a^{\ast}$时, 则问题(1.2)不存在极小元.

证  类似情形2方法, 选取和(2.6)式相同的试验函数, 由(1.9)式以及(2.7)-(2.11)式推出

$ \begin{eqnarray} {\rm e}^{p}(a)&\leq&\lim\limits_{\tau\to\infty}E^{p}_{a}(u_\tau)\\ &\leq&\lim\limits_{\tau\rightarrow\infty}\Big[\frac{A_{\tau }^2\tau^2}{\| \phi\|^2_2}\int_{{\Bbb R}^{N}} |\nabla \phi(x)|^{2}{\rm d}x-\frac{aN}{N+2}\frac{A_{\tau}^{2+\frac{4}{N}}\tau^2}{\| \phi\|^{2+\frac{4}{N}}_2}\int_{{\Bbb R}^{N}} |\phi(x)|^{2+\frac{4}{N}}{\rm d}x \\ &&+C\tau\Big(\int_{{\Bbb R}^{N}} |\nabla \phi(x)|^{2}{\rm d}x\Big)^{\frac{1}{2}}+C{\rm e}^{-\sqrt{\tau}}\Big]+V(x_{0})\\ &=&\lim\limits_{\tau\rightarrow\infty}\Big[\frac{A_{\tau }^2\tau^2}{\| \phi\|^2_2}\Big(\frac{a^{\ast}-a}{a^{\ast}}\Big)\int_{{\Bbb R}^{N}} |\nabla \phi(x)|^{2}{\rm d}x+C\tau\Big(\int_{{\Bbb R}^{N}} |\nabla \phi(x)|^{2}{\rm d}x\Big)^{\frac{1}{2}}\\ &&+C{\rm e}^{-\sqrt{\tau}}\Big]+V(x_{0})=-\infty. \end{eqnarray} $

(2.14)

因此问题(1.2)不存在极小元.

步骤3  当$p=\frac{4}{N}$且$a=a^{\ast}$时, 则问题(1.2)存在极小元.

证  令$a=a^{\ast}$, 由问题(2.13)可知

$ \begin{equation} E^{\frac{4}{N}}_{a^{\ast}}(u)\geq\int_{{\Bbb R}^{N}}V(x)u^{2}(x){\rm d}x+\frac{1}{2}\int_{{\Bbb R}^{N}}\int_{{\Bbb R}^{N}}\frac{u^{2}(y)u^{2}(x)}{|x-y|}{\rm d}y {\rm d}x\geq 0. \end{equation} $

(2.15)

尽管上式得到$E^{\frac{4}{N}}_{a^{\ast}}(u)$下方有界, 但仅通过(2.15)式不足以说明所选取的极小化序列在${\cal H}$中是一致有界的.受文献[7]的启发, 下面将运用Ekeland's变分原理来证明极小化序列的一致有界性.假设极小化序列$\{u_k\}\subset {\cal H} $满足

$ \lim\limits_{k\to \infty} E^{\frac{4}{N}}_{a^{\ast}}(u_k) = {\rm e}^{\frac{4}{N}}(a^{\ast})\ \mbox{ 且 }\ \|u_k\|_2^2=1. $

若能获知$\int_{{\Bbb R}^{N}}|\nabla u_k(x)|^2{\rm d}x $关于$k$一致有界, 类似于情形1可完成证明.下证

$ \int_{{\Bbb R}^{N}}|\nabla u_k(x)|^2 {\rm d}x \leq C \, \, \mbox{一致的关于}\, \, k. $

利用反证法, 假设$\int_{{\Bbb R}^{N}}|\nabla u_k(x)|^2{\rm d}x \stackrel{k}\rightarrow \infty$, 获得矛盾即可.

首先, 定义度量空间

$ {\cal K}:=\Big\{u(x)\in {\cal H}, \|u\|_2^2=1 \Big\}, $

度量定义成$d(u, v):=\|u-v\|_{{\cal K}}$, 其中$u, v\in {\cal H}$.容易验证$({\cal K}, d)$是一个完备的度量空间.通过Ekeland's变分原理^{[10, Theorem 5.1]}, 如果 ${\rm e}^{\frac{4}{N}}(a^{\ast})$下方有界, 那么存在一个极小化序列$\{u_{k}\}\subset {\cal K}$满足

$ \begin{equation} {\rm e}^{\frac{4}{N}}(a^{\ast})\leq E^{\frac{4}{N}}_{a^{\ast}}(u_{k})\leq {\rm e}^{\frac{4}{N}}(a^{\ast})+\frac{1}{k}, \end{equation} $

(2.16)

$ \begin{equation} E^{\frac{4}{N}}_{a^{\ast}}(v)\geq E_{a^{\ast}}^{\frac{4}{N}}(u_{k})-\frac{1}{k}\|u_{k}-v\|_{{\cal H}} ~~ \hbox{对任意}v \in {\cal K}{成立.} \end{equation} $

(2.17)

假设$\int_{{\Bbb R}^{N}}|\nabla u_k(x)|^2{\rm d}x \stackrel{k}\rightarrow \infty$,

$ \begin{equation} \hbox{令}\ \varepsilon_k^{-2}:=\int_{{\Bbb R}^{N}}|\nabla u_k(x)|^2{\rm d}x, \ \hbox{ 那么}\ \varepsilon_k \stackrel{k}\rightarrow 0. \end{equation} $

(2.18)

定义函数

$ \begin{equation} w_k(x):=\varepsilon_k^{\frac{N}{2}}u_k({\varepsilon_k x}), \end{equation} $

(2.19)

由(1.11), (2.18)和(2.19)式推出

$ \begin{equation} \int_{{\Bbb R}^{N}}|\nabla w_{k}(x)|^{2}{\rm d}x=1,   \int_{{\Bbb R}^{N}} \int_{{\Bbb R}^{N}}\frac{w_{k}^{2}(y)w_{k}^{2}(x)}{|x-y|}{\rm d}y{\rm d}x\leq C. \end{equation} $

(2.20)

再由(2.15)及(2.16)式得到

$ \begin{equation} \int_{{\Bbb R}^{N}}V(x)u_{k}^{2}(x){\rm d}x=\int_{{\Bbb R}^{N}}V(\epsilon_{k}x)w_{k}^{2}(x){\rm d}x\leq {\rm e}^{\frac{4}{N}}(a^{\ast})+\frac{1}{k}. \end{equation} $

(2.21)

观察(1.2)式, 代入(2.19)式得

$ \begin{eqnarray} E_{a^{\ast}}^{\frac{4}{N}}(u_{k})& =& \frac{1}{\epsilon^{2}_{k}}\int_{{\Bbb R}^{N}}|\nabla w_{k}(x)|^{2}{\rm d}x+\int_{{\Bbb R}^{N}}V(\epsilon_{k}x)w_{k}^{2}(x){\rm d}x \\ &&+\frac{1}{2\epsilon_{k}}\int_{{\Bbb R}^{N}} \int_{{\Bbb R}^{N}}\frac{w_{k}^{2}(y)w_{k}^{2}(x)}{|x-y|}{\rm d}y{\rm d}x-\frac{Na^{\ast}}{(N+2)\epsilon^{2}_{k}}\int_{{\Bbb R}^{N}}|w_{k}(x)|^{\frac{4}{N}+2}{\rm d}x. \end{eqnarray} $

(2.22)

联合(2.15)-(2.16)式以及(2.20)-(2.22)式分别得到

$ \begin{equation} \lim\limits_{k\rightarrow\infty}\epsilon^{2}_{k} {\rm e}^{\frac{4}{N}}(a^{\ast})\leq\lim\limits_{k\rightarrow\infty}\epsilon^{2}_{k} E_{a^{\ast}}^{\frac{4}{N}}(u_{k})=0, \quad\lim\limits_{k\rightarrow\infty}\int_{{\Bbb R}^{N}}|w_{k}(x)|^{\frac{4}{N}+2}{\rm d}x\rightarrow \frac{N+2}{Na^{\ast}}. \end{equation} $

(2.23)

接着需证明存在序列$\{y_k\}\subset{\Bbb R}^{N}$和正常数$R$, $\eta$满足

$ \begin{equation} \liminf\limits_{k\to\infty} \int_{B_R(y_k)} |w_k(x)|^2 {\rm d}x \geq\eta>0. \end{equation} $

(2.24)

假设(2.24)式不成立, 则对任意$R>0$, 存在子序列$\{ w_k\}$ (仍用$\{ w_k\}$表示)使得

$ \begin{equation} \lim\limits_{k\to\infty}\sup\limits_{x\in{\Bbb R}^{N}}\int_{B_{R}(y)}| w_k(x)|^2 {\rm d}x = 0. \end{equation} $

(2.25)

通过消失引理^{[12, Lemma 1.21]}可知$w_{k} \stackrel{n}\rightarrow 0$在$L^r({\Bbb R}^{N})$中成立, $r \in(2, 2^*)$.显然与(2.23)式矛盾, 故(2.24)式成立.

另外, 还需要引入下面引理.

引理2.1  假设$w_k$形如(2.19)式, 则在子序列意义下$w_k \stackrel{k} \rightharpoonup w_0$在$ {\cal K}$中成立, 且$w_0$满足

$ \begin{equation} -\Delta w_0(x)+\frac{2}{N}w_0(x)=a^{\ast}w_0^{1+\frac{4}{N}}(x). \end{equation} $

(2.26)

证  对任意的$\varphi(x)\in C^\infty_0({\Bbb R}^{N})\cap{\cal H}$, 定义

$ \tilde{\varphi}(x):=\varphi\Big(\frac{x}{\varepsilon_k}\Big) $

和

$ f(\tau, \sigma):=\frac{1}{2}\int_{{\Bbb R}^N}|u_k(x)+\tau u_k(x)+\sigma\tilde{\varphi}(x)|^2 {\rm d}x-\frac{1}{2}. $

因此$f(\tau, \sigma)$满足

$ f(0, 0)=0, \, \, \frac{\partial f(0, 0)}{\partial\tau}=\int_{{\Bbb R}^{N}}| u_k(x)|^2{\rm d}x=1 $

和

$ \frac{\partial f(0, 0)}{\partial\sigma}=\int_{{\Bbb R} ^N}u_k(x)\tilde{\varphi}(x) {\rm d}x. $

运用隐函数定理, 存在常数$\delta_k>0$和函数$\tau(\sigma)\in C^1((-\delta_k, \delta_k), {\Bbb R})$满足

$ \tau(0)=0, \, \, \tau'(0)=-\int_{{\Bbb R}^N}u_k(x)\tilde{\varphi}(x) {\rm d}x $

和

$ f(\tau(\sigma), \sigma)=f(0, 0)=0. $

因此推出

$ (u_k+\tau(\sigma) u_k+\sigma\tilde{\varphi})\in {\cal K}, $

其中$\sigma\in(-\delta_k, \delta_k).$

由(2.17)式得

$ E^{\frac{4}{N}}_{a^{\ast}}(u_k+\tau(\sigma) u_k+\sigma\tilde{\varphi})-E^{\frac{4}{N}}_{a^{\ast}}(u_k)\geq -\frac{1}{k}\|\tau(\sigma) u_k+\sigma\tilde{\varphi}\|_{{\cal H}}. $

取极限$\sigma\to0^{+}$, $\sigma\to0^{-}$得

$ \begin{equation} |\langle E^{\frac{4}{N}}_{a^{\ast}}(u_k), \tau'(0) u_k+\sigma\tilde{\varphi}\rangle|\leq\frac{1}{k}\|\tau'(0) u_k+\sigma\tilde{\varphi}\|_{{\cal H}}. \end{equation} $

(2.27)

通过计算得

$ \begin{eqnarray} &&\Big|\int_{{\Bbb R}^N}\nabla u_k(x)\nabla \tilde{\varphi}(x) {\rm d}x +\int_{{\Bbb R}^N}V(x)u_k(x)\tilde{\varphi}(x) {\rm d}x \\ &&+\frac{1}{2}\int_{{\Bbb R}^N}\int_{{\Bbb R}^N}\frac{u_k^{2}(y)}{|x-y|}{\rm d}yu_k(x)\tilde{\varphi}(x) {\rm d}x -a^{\ast}\int_{{\Bbb R}^N}u_k^{1+\frac{4}{N}}(x)\tilde{\varphi}(x) {\rm d}x \\ &&+\tau'(0)\Big[\int_{{\Bbb R}^N}|\nabla u_k(x)|^2 {\rm d}x +\int_{{\Bbb R}^N}V(x)u^2_k(x) {\rm d}x \\ && +\frac{1}{2}\int_{{\Bbb R}^N}\int_{{\Bbb R}^N}\frac{u_k^{2}(x)u_k^{2}(y)}{|x-y|}{\rm d}y {\rm d}x -a^{\ast}\int_{{\Bbb R}^N}u_k^{2+\frac{4}{N}}(x) {\rm d}x \Big]\Big| \\ &\leq&\frac{1}{k}\|\tau'(0)u_k+\tilde{\varphi}\|_{{\cal H}}. \end{eqnarray} $

(2.28)

由(2.16)式及(2.20)-(2.24)式推出

$ \tau'(0)=-\int_{{\Bbb R}^N}u_k(x)\tilde{\varphi}(x) {\rm d}x =-\varepsilon_k^{\frac{N}{2}}\int_{{\Bbb R}^N}w_k(x)\varphi(x) {\rm d}x, \\ \|\tau'(0)u_k+\tilde{\varphi}\|_{{\cal H}}\leq |\tau'(0)|\|u_k\|_{{\cal H}}+\|\tilde{\varphi}\|_{{\cal H}} \leq C \varepsilon_k^{\frac{N}{2}-1}, $

(2.29)

以及

$ \begin{eqnarray} &&\int_{{\Bbb R}^N}|\nabla u_k(x)|^2 {\rm d}x +\int_{{\Bbb R}^N}V(x)u^2_k(x) {\rm d}x\\ && +\frac{1}{2}\int_{{\Bbb R}^N}\int_{{\Bbb R}^N}\frac{u_k^{2}(x)u_k^{2}(y)}{|x-y|}{\rm d}y {\rm d}x -a^{\ast}\int_{{\Bbb R}^N}u_k^{2+\frac{4}{N}}(x) {\rm d}x\\ &=&\varepsilon_k^{-2}\bigg[ \varepsilon_k^{2}E^{\frac{4}{N}}_{a^{\ast}}(u_k)-\frac{2a^{\ast}}{N+2}\int_{{\Bbb R}^N}w_k^{2+\frac{4}{N}}(x){\rm d}x \Big] \\ &=&-\Big(\frac{2}{N}+o(1)\bigg)\varepsilon_k^{-2},\quad \hbox{当}\ k\rightarrow\infty. \end{eqnarray} $

(2.30)

在(2.28)式中, 用$w_k(x)$代替$u_k(x)$, 再由(2.29)及(2.30)式计算得, 当$ k\rightarrow\infty$时

$ \begin{eqnarray} &&\Big|\int_{{\Bbb R}^N}\nabla w_k(x)\nabla \varphi(x) {\rm d}x +\varepsilon_k^2\int_{{\Bbb R}^N}V(\varepsilon_kx)w_k(x)\varphi(x) {\rm d}x \\ &&+\frac{\varepsilon_k}{2}\int_{{\Bbb R}^N}\int_{{\Bbb R}^N}\frac{w_k^{2}(y)}{|x-y|}{\rm d}yw_k(x)\varphi(x) {\rm d}x -a^{\ast}\int_{{\Bbb R}^N}w_k^{1+\frac{4}{N}}(x)\varphi(x) {\rm d}x \\ &&+\Big(\frac{2}{N}+o(1)\Big)\int_{{\Bbb R}^N}w_k(x)\varphi(x) {\rm d}x \Big| \\ &\leq& \frac{1}{k}C\varepsilon_k, \end{eqnarray} $

(2.31)

由(2.31)式可知$w_0$满足(2.26)式.

通过上述事实, 接下来完成步骤3的证明.首先需证

$ \begin{equation} \int_{{\Bbb R}^N}|w_0(x)|^2 {\rm d}x = 1. \end{equation} $

(2.32)

通过Pohozaev's恒等式^{[2, Lemma 8.1.2]}, 对(2.26)式的任意解$w_0$满足

$ \begin{eqnarray*} \int_{{\Bbb R}^{N}}|w_0(x)|^2{\rm d}x=\int_{{\Bbb R}^{N}}| \nabla w_0(x)|^2{\rm d}x=\frac{Na^{\ast}}{N+2}\int_{{\Bbb R}^{N}}|w_0(x)|^{2+\frac{4}{N}}{\rm d}x. \end{eqnarray*} $

再利用Gagliardo-Nirenberg不等式(1.8)可知

$ \begin{eqnarray} \frac{Na^{\ast}}{N+2}&\leq&\frac{\int_{{\Bbb R}^{N}}|\nabla w_{0}(x)|^2{\rm d}x \Big(\int_{{\Bbb R}^{N}}|w_{0}(x)|^{2}{\rm d}x \Big)^{\frac{2}{N}}} {\int_{{\Bbb R}^{N}} |w_{0}(x)|^{2+\frac{4}{N}}{\rm d}x} \\ &=&\frac{Na^{\ast}}{N+2}\Big(\int_{{\Bbb R}^{N}}|w_{0}(x)|^{2}{\rm d}x \Big)^{\frac{2}{N}}. \end{eqnarray} $

(2.33)

故

$ 1\leq \int_{{\Bbb R}^{N}}|w_0(x)|^2{\rm d}x, $

联合事实

$ \int_{{\Bbb R}^{N}}|w_0(x)|^2{\rm d}x\leq \liminf\limits_{n\to\infty} \int_{{\Bbb R}^{N}}|w_k(x)|^2{\rm d}x=1, $

说明(2.32)式成立.由(2.24)式知$w_0\not\equiv0$, 另外(2.33)式实际取等号, 因此可以知道$w_0$是Gagliardo-Nirenberg不等式(1.8)最佳达到元.通过保范性, 可以知道$w_{n} \stackrel{n}\rightarrow w_{0}$在$L^{2}({\Bbb R}^{2})$中成立.利用$\{w_{n}\}$在$H^{1}({\Bbb R}^{2})$的一致有界性可知$w_{n} \stackrel{n}\rightarrow w_{0}$在$L^{q}({\Bbb R}^{2})$中成立, 其中$2\leq q < 2^{\ast}$.利用(1.11)式和上述事实可知

$ \begin{eqnarray} &&\lim\limits_{n\rightarrow\infty}\int_{{\Bbb R}^{N}} |w_{k}(x)|^{2}{\rm d}x = \int_{{\Bbb R}^{N}} |w_{0}(x)|^{2}{\rm d}x=1, \\ &&\lim\limits_{n\rightarrow\infty}\int_{{\Bbb R}^{N}} |w_{k}(x)|^{\frac{4}{N}+2}{\rm d}x= \int_{{\Bbb R}^{N}} |w_{0}(x)|^{\frac{4}{N}+2}{\rm d}x, \\ &&\lim\limits_{n\rightarrow\infty}\int_{{\Bbb R}^{N}} \int_{{\Bbb R}^{N}}\frac{w_{k}^{2}(y)w_{k}^{2}(x)}{|x-y|}{\rm d}y{\rm d}x = \int_{{\Bbb R}^{N}} \int_{{\Bbb R}^{N}}\frac{w_{0}^{2}(y)w_{0}^{2}(x)}{|x-y|}{\rm d}y{\rm d}x. \end{eqnarray} $

(2.34)

再由(2.15)及(2.19)式推出

$ \begin{equation} {\rm e}^{\frac{4}{N}}(a^*)=\lim\limits_{k\to\infty}E^{\frac{4}{N}}_{a^{\ast}}(u_k) \geq\lim\limits_{k\to\infty}\frac{1}{\varepsilon_k}\int_{{\Bbb R}^{N}} \int_{{\Bbb R}^{N}}\frac{w_{k}^{2}(y)w_{k}^{2}(x)}{|x-y|}{\rm d}y{\rm d}x \geq\infty. \end{equation} $

(2.35)

显然(2.35)式与${\rm e}^{\frac{4}{N}}(a^{\ast})$有界性矛盾.而上述矛盾的根源就是, $\|\nabla u_k\|_2^2$关于$k$实际上是一致有界的.接下来的过程和情形1类似, 故完成了定理1.1的证明.