本文主要研究如下广义Choquard-Pekar方程基态解的存在性

$ \begin{equation} -\Delta u+u+\mu\int_{{\Bbb R}^N}\frac {u^2(y){\rm d}y}{|x-y|^{\lambda}} u(x)=f(u), \ \ {\rm in}\ {\Bbb R}^N, \end{equation} $

(0.1)

其中$N\geq 3, \mu=1, 0 < \lambda < \min (4, N), f(u)=|u|^{p-1}u, 3 < p < \frac{N+2}{N-2}$.

当$N=3, \mu=-1, \lambda=1, f(u)=0$时, (1.1)式是由Pekar^[1]提出来的Choquard-Pekar方程, 它是用来描述量子理论中一个极化子在静止状态时的模型(也可参见文献[2]).对Choquard-Pekar方程早期的研究工作是由Lieb^[3]和Lions^[4]完成的, 他们证明了非平凡解的存在性.目前, 已有一系列关于Choquard-Pekar方程的研究结果^[5-13].当$N\geq 3, 0 < \lambda < \min(4, N), 0 < 2h < \lambda, 0\leq h < 2, V(x)\geq V_0>0$, 且$\lim\limits_{|x|+|y|\rightarrow\infty} {\mu}(x, y)\leq 0$时, 曹道珉^[14]利用集中紧致原理证明得到了下面这类广义Choquard-Pekar方程解的存在性

$ -\Delta u+V(x)u+\int_{{\Bbb R}^N}\frac {\mu(x, y)u^2(y){\rm d}y}{|y|^h|x-y|^{\lambda-2h}} \frac{u(x)}{|x|^h}=0, \ \ {\rm in}\ {\Bbb R}^N. $

受文献[14]的启发, 本文主要研究含超线性项且非局部项前的系数$\mu$为大于零的常数（本文为简单起见考虑$\mu=1>0$）的广义Choquard-Pekar方程(1.1)基态解的存在性.

本文运用Nehari流形和集中紧致原理证明了如下结论.

定理1.1  当$3 < p < \frac{N+2}{N-2}$时, 方程(1.1)存在一个基态解.

本文结构安排如下:第二节, 给出文章的框架和估计中要用到的重要不等式, 并且讨论Nehari流形上临界值的性质; 第三节, 用集中紧致原理证明Nehari流形上极小值点是可达的, 从而证明定理1.1成立.

记$E:=H^1({\Bbb R}^N)$为赋以范数$\|u\|=(\int_{{\Bbb R}^N}(|\nabla u|^2+ |u|^2){\rm d}x)^{\frac{1}{2}}$的索伯列夫空间.方程(1.1)所对应的能量泛函为

$ \begin{eqnarray*} I(u)=\frac{1}{2}\int_{{\Bbb R}^N}(|\nabla u|^2+|u|^2){\rm d}x+\frac{1}{4} \int_{{\Bbb R}^N}\int_{{\Bbb R}^N} \frac{u^2(x)u^2(y)}{|x-y|^{\lambda}}{\rm d}x{\rm d}y- \frac{1}{p+1} \int_{{\Bbb R}^N} |u|^{p+1}{\rm d}x. \end{eqnarray*} $

注意到$I\in C^1(E, {\Bbb R})$, 且$\forall\eta\in C^\infty_0({\Bbb R}^N)$, 有

$ \begin{eqnarray*} \langle I'(u), \eta\rangle& =&\int_{{\Bbb R}^N} (\nabla u\cdot \nabla\eta+u\eta){\rm d}x+ \int_{{\Bbb R}^N}\int_{{\Bbb R}^N} \frac{u^2(x)u(y)\eta(y)}{|x-y|^{\lambda}}{\rm d}x{\rm d}y %\\&& -\int_{{\Bbb R}^N}|u|^{p-1}u\eta {\rm d}x. \end{eqnarray*} $

特别地, 我们有

$ \begin{eqnarray*} \langle I'(u), u\rangle& =&\int_{{\Bbb R}^N} (|\nabla u|^2+|u|^2){\rm d}x+ \int_{{\Bbb R}^N}\int_{{\Bbb R}^N} \frac{u^2(x)u^2(y)}{|x-y|^{\lambda}}{\rm d}x{\rm d}y -\int_{{\Bbb R}^N}|u|^{p+1}{\rm d}x. \end{eqnarray*} $

在以下的证明中, 我们要用到下面的不等式.

引理2.1^[15-16] (Hardy-Littlewood不等式)如果$u\in L^r({\Bbb R}^N), v\in L^s({\Bbb R}^N), $则

$ \begin{eqnarray*} \bigg|\int_{{\Bbb R}^N}\int_{{\Bbb R}^N} \frac{u(x)v(y){\rm d}x{\rm d}y}{|x|^h|y|^l|x-y|^{\lambda-h-l}}\bigg| \leq C|u|_{L^r({\Bbb R}^N)}|v|_{L^s({\Bbb R}^N)}, \end{eqnarray*} $

其中$\lambda, s, r, h, l$满足如下关系式

$ 0<\lambda<N, \alpha+\beta\geq 0, 1<r, \, s <\infty, $

$ 1-\frac{1}{r}-\frac{\lambda}{N}<\frac{h}{N}<1-\frac{1}{r}, \frac{1}{r}+\frac{1}{s}+\frac{\lambda+h+l}{N}=2. $

令$\Sigma =\{u\in E\backslash \{0\}:\langle I'(u), u\rangle =0\}.$我们有下面的结论.

引理2.2  当$p>3$时, $\Sigma$是光滑流形.

证  令$\varphi(u)=\langle I'(u), u\rangle, \forall u\in \Sigma$.则

$ \begin{eqnarray*} \langle \phi'(u), u\rangle& =&2 \int_{{\Bbb R}^N} (|\nabla u|^2+|u|^2){\rm d}x+4 \int_{{\Bbb R}^N} \int_{{\Bbb R}^N} \frac{u^2(x)u^2(y){\rm d}x{\rm d}y}{|x-y|^{\lambda}} \\&& -(p+1)\int_{{\Bbb R}^N} |u|^{p+1}{\rm d}x. \end{eqnarray*} $

因为$u\in \Sigma$且$p>3$, 所以

$ \begin{eqnarray*} \langle \varphi'(u), u\rangle =(1-p) \int_{{\Bbb R}^N} (|\nabla u|^2+|u|^2){\rm d}x+(3-p) \int_{{\Bbb R}^N} \int_{{\Bbb R}^N} \frac{u^2(x)u^2(y){\rm d}x{\rm d}y}{|x-y|^{\lambda}} <0. \end{eqnarray*} $

由隐函数存在定理可得命题成立.

定义

$ C=\inf\limits_{u\in \Sigma}I(u), C^*=\inf\limits_{\gamma\in \Gamma}\max\limits_{ t\in [0, 1]}I(\gamma(t)), C^{**}=\inf\limits_{u\in E\backslash \{0\}}\max\limits_{t\geq 0}I(tu), $

其中$\Gamma =\{\gamma\in C([0, 1], {\mathbb E}): \gamma(0)=0, I(\gamma(1)) < 0\}$.

引理2.3  当$p>3$时, $C=C^*=C^{**}$.

证  首先, 我们证明$C=C^{**}$.事实上, 我们只要能够证明$ \forall u\in E\backslash \{0\}$, 射线$R_t=\{tu, t\geq 0\}$与流形$\Sigma$仅相交于$I(tu), t\geq 0$取得唯一最大值的点$\theta u(\theta>0)$即可.假设

$ g(t)=t^{p-1}D-t^2B-A, \ \ t\in [0, +\infty), $

其中

$ A=\|u\|^2, B=\int_{{\Bbb R}^N} \int_{{\Bbb R}^N} \frac{u^2(x)u^2(y)}{|x-y|^{\lambda}}{\rm d}x{\rm d}y, D=|u|^{p+1}_{p+1}. $

我们可得

$ \left\{\begin{array}{ll}g''(t)<0, &t<t_1, \\ g'(t)>0, &t>t_1, \end{array}\right. $

其中$t_1=\sqrt[^{^{p-3}}]{\frac{2B}{(p-1)(p-2)D}}.$此外, 存在$t_2=0, t_3=\sqrt[^{^{p-3}}]{\frac{2B}{(p-1)D}}.$使得$g'(t_i)=0, i=2, 3, $且$t\geq t_3$时$g(t)$严格单调递增, $t\leq t_3$时$ g(t)$严格单调递减.又因为$g(0)=-A < 0$, 所以射线$R_t$与$\Sigma$仅相交于$I(tu), t\geq 0$取得唯一最大值的点$\theta u (\theta>0).$于是$C=C^{**}$.

其次, 我们证明$C^*=C^{**}$.显然有$C^{**}\geq C^*$.我们只需证明$C^{**}\leq C^*$. $\forall u\in E\backslash \{0\}$, 令$\bar{t}$为唯一的正实数使得$\bar{t}u\in \Sigma$.于是

$ C^{**}=\inf\limits_{u\in K}I(u), $

其中$K=\{u=\bar{t}u:u\in E, u\neq 0, \bar{t} < \infty\}$.令$\gamma\in \Gamma$是一条道路.如果$\forall \gamma\in \Gamma, \gamma\cap K\neq \phi$, 那么结论成立.如果存在$\gamma\in\Gamma$使得$\gamma(t)\not\in K, \forall t\in [0, 1]$, 那么我们有

$ \begin{eqnarray*} \int_{{\Bbb R}^N} (|\nabla \gamma|^2+\gamma^2){\rm d}x+ \int_{{\Bbb R}^N} \int_{{\Bbb R}^N} \frac{\gamma^2(x)\gamma^2(y)}{|x-y|^{\lambda}}{\rm d}x{\rm d}y> \int_{{\Bbb R}^N} |\gamma|^{p+1}{\rm d}x. \end{eqnarray*} $

于是，由$p>3$, 可得

$ \begin{eqnarray*} I(\gamma)&=&\frac{1}{2} \int_{{\Bbb R}^N} (|\nabla \gamma|^2+\gamma^2){\rm d}x+ \frac{1}{4} \int_{{\Bbb R}^N}\int_{{\Bbb R}^N}\frac{\gamma^2(x)\gamma^2(y)}{|x-y|^{\lambda}}{\rm d}x{\rm d}y -\frac{1}{p+1} \int_{{\Bbb R}^N} |\gamma|^{p+1}{\rm d}x\\ &>&\Big(\frac{1}{2}-\frac{1}{p+1}\Big) \int_{{\Bbb R}^N} (|\nabla \gamma|^2+\gamma^2){\rm d}x+ \Big(\frac{1}{4}-\frac{1}{p+1}\Big) \int_{{\Bbb R}^N}\int_{{\Bbb R}^N}\frac{\gamma^2(x)\gamma^2(y)} {|x-y|^{\lambda}}{\rm d}x{\rm d}y \end{eqnarray*} $

这与$I(\gamma(1)) < 0$矛盾.因此$C^*=C^{**}$.综上所述, 结论成立.

本节中, 我们将运用集中紧致原理^[17-18]来证明在Nehari流形上极小值点是可达的, 从而基态解存在.定义

$ \begin{eqnarray*} \Phi(u)&=&\Big(\frac{1}{2}-\frac{1}{p+1}\Big) \int_{{\Bbb R}^N} (|\nabla u|^2+u^2){\rm d}x+ \Big(\frac{1}{4}-\frac{1}{p+1}\Big) \int_{{\Bbb R}^N}\int_{{\Bbb R}^N}\frac{u^2(x)u^2(y){\rm d}x{\rm d}y}{|x-y|^{\lambda}} \geq 0. \end{eqnarray*} $

选取$\{u_n\}\subset \Sigma$使得$\lim\limits_{n\rightarrow\infty}I(u_n)=C.$因为$u_n\in \Sigma, $所以$\lim\limits_{n\rightarrow\infty}\Phi (u_n)=C.$又由引理2.1, 可得

$ \bigg|\int_{{\Bbb R}^N}\int_{{\Bbb R}^N} \frac{u_n^2(x)u_n^2(y){\rm d}x{\rm d}y}{|x-y|^{\lambda}}\bigg| \leq C|u_n|^4_{L^{2r({\Bbb R}^N)}}, $

其中$r=\frac{2N}{2N-\lambda}$.再由Sobolev嵌入不等式，$\{u_n\}$在$H^1({\Bbb R}^N)$中有界.

定义如下测度

$ \begin{eqnarray*} \rho_n(\Omega)=\Big(\frac{1}{2}-\frac{1}{p+1}\Big) \int_{\Omega} (|\nabla u_n|^2+|u_n|^2){\rm d}x+ \Big(\frac{1}{4}-\frac{1}{p+1}\Big) \int_{\Omega}\int_{\Omega}\frac{u^2_n(x)u^2_n(y){\rm d}x{\rm d}y}{|x-y|^{\lambda}}. \end{eqnarray*} $

由Lions^[17]的集中紧致原理, 有下面三种可能性.

(ⅰ) 消失:$\forall R>0, \lim\limits_{n\rightarrow\infty}\sup\limits_{\xi\in {\Bbb R}^N} \int_{B_R(\xi)}{\rm d}\rho_n=0;$

(ⅱ)二分:存在$\bar{C}\in (0, C)$, 两个序列$\{\xi_n\}$和$\{R_n\}$, 其中$R_n\rightarrow+\infty, n\rightarrow\infty$, 存在两个非负测度$\rho^1_n$和$\rho^2_n$, 使得

$ 0\leq \rho^1_n+\rho^2_n\leq \rho_n, \rho^1_n({\Bbb R}^N)\rightarrow \bar{C}, \rho^2_n({\Bbb R}^N)\rightarrow C-\bar{C}, $

$ {\rm supp}( \rho^1_n)\subset B_{R_n}(\xi_n), {\rm supp}( \rho^2_n)\subset {\Bbb R}^N\backslash B_{2R_n}(\xi_n); $

(ⅲ)紧性:存在$\{\xi_n\}\subset {\Bbb R}^N$满足下面的性质:$\forall \delta>0, $存在${\cal R}=R(\delta)$, 使得

$ \int_{B_R(\xi_n)}{\rm d}\rho_n\geq C-\delta. $

引理3.1  对于测度$\rho_n$, 消失和二分不可能成立.

证   (ⅰ) 首先, 我们证明消失不可能成立.

假设$\rho_n$是消失的.由文献[18]中的引理1.1, 我们有在$L^s({\Bbb R}^N)\ (2 < S < 2^*)$中$u_n\rightarrow 0, n\rightarrow\infty$.由于$\{u_n\}\subset E$, 我们有

$ \begin{equation} 0\leq I(u_n)= -\frac{1}{4} \int_{{\Bbb R}^N}\int_{{\Bbb R}^N}\frac{u^2_n(x)u^2_n(y){\rm d}x{\rm d}y}{|x-y|^{\lambda}} +\Big(\frac{1}{2}-\frac{1}{p+1}\Big) \int_{{\Bbb R}^N} |u_n|^{p+1}{\rm d}x. \end{equation} $

(3.1)

由引理2.1, 我们可得

$ \begin{equation} \bigg|\int_{{\Bbb R}^N}\int_{{\Bbb R}^N} \frac{u_n^2(x)u_n^2(y){\rm d}x{\rm d}y}{|x-y|^{\lambda}}\bigg| \leq C|u_n|^4_{L^{2r({\Bbb R}^N)}}\rightarrow0, n\rightarrow\infty, \end{equation} $

(3.2)

其中$r=\frac{2N}{2N-\lambda}$.

又因为$ \int_{{\Bbb R}^N} |u_n|^{p+1}{\rm d}x \rightarrow0, ~n\rightarrow\infty, $由(3.1)和(3.2)式, 可得

$ 0\leq I(u_n)\rightarrow0, n\rightarrow\infty. $

这与$ I(u_n)\rightarrow C>0, ~n\rightarrow\infty$矛盾.所以$\rho_n$不可能消失.

(ⅱ)其次, 我们证明二分不可能成立.

我们选择一个截断函数$\psi_n\in C^1({\Bbb R}^N)$满足

$ \left\{\begin{array}{ll} \psi_n\equiv 1, &x\in B_{R_n}(\xi_n), \\ \psi_n\equiv 0, &x\in {\Bbb R}^N\backslash B_{2R_n}(\xi_n), \\ 0\leq \psi_n\leq 1, &x\in B_{2R_n(\xi_n)}\backslash B_{R_n}(\xi_n), \end{array}\right. $

且$|\nabla \psi_n|\leq \frac{2}{R_n}$.令$v_n:=\psi_nu_n, w_n:=(1-\psi_n)u_n$.我们有

$ \liminf\limits_{n\rightarrow\infty}I(v_n)\geq\bar{C}, \liminf\limits_{n\rightarrow\infty}I(w_n)\geq C-\bar{C}. $

令$\Omega_n:=B_{2R_n}(\xi_n)\backslash B_{R_n}(\xi_n)$.我们可得

$ \rho_n(\Omega_n)\rightarrow 0, n\rightarrow\infty. $

也就是

$ \begin{eqnarray} \int_{\Omega_n} (|\nabla u_n|^2+u_n^2){\rm d}x\rightarrow 0, n\rightarrow\infty, \end{eqnarray} $

(3.3)

$ \begin{eqnarray} \int_{\Omega_n} \int_{{\Bbb R}^N} \frac{u^2_n(x)u^2_n(y){\rm d}x{\rm d}y} {|x-y|^{\lambda}}\rightarrow 0, n\rightarrow\infty. \end{eqnarray} $

(3.4)

于是

$ \int_{\Omega_n} (|\nabla v_n|^2+v_n^2){\rm d}x\rightarrow 0, n\rightarrow\infty, $

$ \int_{\Omega_n} (|\nabla w_n|^2+w_n^2){\rm d}x\rightarrow 0, n\rightarrow\infty. $

从而

$ \begin{eqnarray} \int_{{\Bbb R}^N} (|\nabla u_n|^2+u_n^2){\rm d}x= \int_{{\Bbb R}^N} (|\nabla v_n|^2+v_n^2){\rm d}x+ \int_{{\Bbb R}^N} (|\nabla w_n|^2+w_n^2){\rm d}x+o_n(1), \end{eqnarray} $

(3.5)

$ \begin{eqnarray} \int_{{\Bbb R}^N} |u_n|^{p+1}{\rm d}x= \int_{{\Bbb R}^N} |v_n|^{p+1}{\rm d}x +\int_{{\Bbb R}^N} |w_n|^{p+1}{\rm d}x+o_n(1). \end{eqnarray} $

(3.6)

根据(3.4)式, 我们有

$ \begin{eqnarray} \int_{{\Bbb R}^N}\int_{{\Bbb R}^N}\frac{u^2_n(x)u^2_n(y){\rm d}x{\rm d}y}{|x-y|^{\lambda}} &=&\int_{{\Bbb R}^N}\int_{{\Bbb R}^N}\frac{v^2_n(x)v^2_n(y){\rm d}x{\rm d}y}{|x-y|^{\lambda}} +\int_{{\Bbb R}^N}\int_{{\Bbb R}^N}\frac{w^2_n(x)w^2_n(y){\rm d}x{\rm d}y}{|x-y|^{\lambda}} \nonumber\\ &&+2\int_{B_{R_n}}\int_{B^c_{2R_n}} \frac{u^2_n(x)u^2_n(y){\rm d}x{\rm d}y}{|x-y|^{\lambda}}+o_n(1) \nonumber\\ &\geq& \int_{{\Bbb R}^N}\int_{{\Bbb R}^N}\frac{v^2_n(x)v^2_n(y){\rm d}x{\rm d}y}{|x-y|^{\lambda}} +\int_{{\Bbb R}^N}\int_{{\Bbb R}^N}\frac{w^2_n(x)w^2_n(y){\rm d}x{\rm d}y}{|x-y|^{\lambda}}. \end{eqnarray} $

(3.7)

由(3.5)-(3.7)式, 我们得

$ I(u_n)\geq I(v_n)+I(w_n)+o_n(1). $

因此

$ C=\lim\limits_{n\rightarrow\infty}I(u_n)\geq \liminf\limits_{n\rightarrow\infty} I(v_n)+\liminf\limits_{n\rightarrow\infty} I(w_n) \geq\bar{C}+(C-\bar{C})=C. $

于是

$ \begin{eqnarray} \lim\limits_{n\rightarrow\infty}I(v_n)=\bar{C}, \lim\limits_{n\rightarrow\infty}I(w_n)=C-\bar{C}. \end{eqnarray} $

(3.8)

因为$u_n\in \Sigma, $又由(3.5)-(3.7)式, 我们可得

$ \begin{eqnarray} 0=\varphi(u_n)=\langle I'(u_n), u_n\rangle \geq \varphi(v_n)+\varphi(w_n)+o_n(1). \end{eqnarray} $

(3.9)

由引理2.3, $\forall n\geq 1$, 存在$\theta_n>0$使得$\theta_nv_n\in \Sigma$, 于是

$ \begin{eqnarray} \theta^2_n\int_{{\Bbb R}^N} (|\nabla v_n|^2+v_n^2){\rm d}x+ \theta^4_n\int_{{\Bbb R}^N} \int_{{\Bbb R}^N} \frac{v^2_n(x)v^2_n(y){\rm d}x{\rm d}y}{|x-y|^{\lambda}} =\theta^{p+1}_n\int_{{\Bbb R}^N} | v_n|^{p+1}{\rm d}x. \end{eqnarray} $

(3.10)

下面分三种情形讨论.

情形1   存在$\{v_n\}$的子列, 仍记作$\{v_n\}$, 使得$\varphi(v_n)\leq 0$.

由(3.10)式, 我们有

$ \begin{eqnarray} \varphi(v_n)=(1-\theta^{1-p}_n)\int_{{\Bbb R}^N} (|\nabla v_n|^2+v^2_n){\rm d}x+ (1-\theta^{3-p}_n)\int_{{\Bbb R}^N} \int_{{\Bbb R}^N} \frac{v^2_n(x)v^2_n(y){\rm d}x{\rm d}y}{|x-y|^{\lambda}}. \end{eqnarray} $

(3.11)

由于$p>3, \theta_n\leq 1$.于是, $\forall n\geq 1$, 因为$\theta_nv_n\in \Sigma$, 所以我们有

$ C\leq I(\theta_nv_n)=\Phi(\theta_nv_n)\leq \Phi(v_n)\rightarrow \bar{C}<C, n\rightarrow\infty. $

矛盾.

情形2   存在$\{w_n\}$的子列, 仍记作$\{w_n\}$, 使得$\varphi(w_n)\leq 0$.与情形1类似的讨论, 我们也可得出矛盾.

情形3  存在$\{v_n\}$和$\{w_n\}$的子列, 仍分别记作$\{v_n\}$和$\{w_n\}$, 使得$\varphi(v_n)> 0$且$\varphi(w_n)> 0$.由(3.9)式, 我们有$\varphi(v_n)=o_n(1) $且$\varphi(w_n)=o_n(1) $.如果$\theta_n\leq 1+o_n(1) $, 我们分下面三种情况讨论

(a) 当$\limsup\limits_{n\rightarrow \infty} \theta_n>1$时, 由(3.11)式可得当$n$较大时$\|v_n\|=0$.这与(3.8)式和$\bar{C}>0$矛盾;

(b) 当$\limsup\limits_{n\rightarrow \infty} \theta_n < 1$时, 与情形1类似, 我们可以得到矛盾;

(c) 当$\limsup\limits_{n\rightarrow \infty} \theta_n=1$时, 如果存在子列使得每个元素都严格大于$1$, 则类似于(a).如果存在子列使得每个元素都严格小于$1$, 则类似于(b).如果$\theta_n\equiv 1$, 则$v_n\in \Sigma$.所以$I(v_n)= \Phi(v_n)\rightarrow\bar{C} < C$.这与$C$的定义矛盾.

于是$ \lim\limits_{n\rightarrow\infty}\theta_n=\theta_0>1$.从而

$ \begin{eqnarray*} o_n(1)&=&\varphi(v_n)\\ &=&(1-\theta^{1-p}_n)\int_{{\Bbb R}^N} (|\nabla v_n|^2+v^2_n){\rm d}x+ (1-\theta^{3-p}_n)\int_{{\Bbb R}^N} \int_{{\Bbb R}^N} \frac{v^2_n(x)v^2_n(y){\rm d}x{\rm d}y}{|x-y|^{\lambda}}. \end{eqnarray*} $

因此, 在$H^1({\Bbb R}^N)$中$v_n\rightarrow 0, n\rightarrow\infty$.这与(3.8)式矛盾.

综上所述, 二分不可能发生.

定理1.1的证明  由引理3.1可得紧性成立.也就是, 存在$\{\xi_n\}\subset {\Bbb R}^N$满足下面的性质:$\forall \delta >0, $存在$R=R(\delta)$, 使得

$ \begin{equation} \int_{B_R(\xi_n)}{\rm d}\rho_n\geq C-\delta. \end{equation} $

(3.12)

定义$\bar{u}_n(\cdot):=u_n(\cdot -\xi_n)\in H^1({\Bbb R}^N)$.由于$u_n\in \Sigma$, 由积分对平移的不变性, 我们有$\bar{u}_n\in \Sigma$且$\lim\limits_{n\rightarrow \infty}I(\bar u_n)=C$.根据(3.12)式, 我们可得:$\forall \delta>0$, 存在$R(\delta)>0$, 使得对$\forall n\geq 1$, 有

$ \begin{equation} \|\bar{u}\|_{H^1(B^c_R)}<\delta. \end{equation} $

(3.13)

因为$\{u_n\}$在$H^1({\Bbb R}^N)$中有界, 所以$\{\bar{u}_n\}$在$H^1({\Bbb R}^N)$中也有界.于是存在$u\in H^1({\Bbb R}^N)$, 使得当$n\rightarrow\infty$时, 有

$ \begin{equation} \mbox{在$ H^1({\Bbb R}^N)$中, }\ \bar{u}_n\rightharpoonup u, \end{equation} $

(3.14)

$ \begin{equation} \mbox{在$ L^s_{loc}({\Bbb R}^N)$中, }\ \bar{u}_n\rightarrow u, \ \forall 2\leq s< \frac{2N}{N-2}, \end{equation} $

(3.15)

$ \begin{equation} \mbox{在${\Bbb R}^N$中, }\ \bar{u}_n\mathop{\longrightarrow}\limits^{\rm a.e.} u. \end{equation} $

(3.16)

根据(3.13)-(3.15)式, 取$S\in [2, \frac{2N}{N-2}), \forall\delta>0$, 存在$R>0$使得对充分大的$n$, 由Sobolev嵌入定理, 有

$ \begin{eqnarray*} \|\bar{u}_n-u\|_{L^s({\Bbb R}^N)}&\leq & \|\bar{u}_n-u\|_{L^s(B_R)}+\|\bar{u}_n-u\|_{L^s(B_R^c)}\\ &\leq& \delta +c(\|\bar{u}_n\|_{H^1(B_R^c)}+ \|u\|_{H^1(B_R^c)})\\ &\leq &(1+2c)\delta. \end{eqnarray*} $

因此, 当$s\in [2, \frac{2N}{N-2})$时, 在$L^s({\Bbb R}^N)$中, 有

$ \begin{eqnarray} \bar{u}_n\rightarrow u, u\rightarrow\infty. \end{eqnarray} $

(3.17)

于是, 我们有

$ \begin{eqnarray} \lim\limits_{n\rightarrow\infty} \int_{{\Bbb R}^N}\int_{{\Bbb R}^N}\frac{\bar{u}^2_n(x)\bar{u}^2_n(y){\rm d}x{\rm d}y} {|x-y|^{\lambda}} =\int_{{\Bbb R}^N}\int_{{\Bbb R}^N}\frac{u^2(x)u^2(y){\rm d}x{\rm d}y}{|x-y|^{\lambda}}. \end{eqnarray} $

(3.18)

由(3.14)式, 得$\liminf\limits_{n\rightarrow \infty} \|\bar{u}_n\|\geq \|u\|$.如果$\liminf\limits_{n\rightarrow \infty} \|\bar{u}_n\| > \|u\|$, 则存在$\theta\in (0, 1)$使得$\theta u\in \Sigma$.于是

$ \begin{eqnarray*} C\leq I(\theta u)=\Phi(\theta u)<\Phi(u)\leq \liminf\limits_{n\rightarrow \infty}\Phi(\bar{u}_n)\leq C, \end{eqnarray*} $

矛盾.故$\liminf\limits_{n\rightarrow \infty} \|\bar{u}_n\| = \|u\|$.从而存在子列(仍记作$\bar{u}_n$)使得

$ \begin{eqnarray} \|\bar{u}_n\|\rightarrow\|u\|, n\rightarrow\infty. \end{eqnarray} $

(3.19)

因此在$H^1({\Bbb R}^N)$中$\bar{u}_n$强收敛于$u$.又由积分的平移不变性和(3.17)-(3.19)式, 有$u\in \Sigma$且$I(u)=C$.从而$u$是方程(1.1)的基态解.