该文考虑如下形式的Kirchhoff椭圆方程

(1.1) $\begin{equation}\label{fc1}-\left(a+b\int_{\mathbb{R} ^3}|\nabla u|^{2}{\rm d}x\right)\bigtriangleup u+V(x)u=\beta|u|^pu+\lambda u, \ \ x\in\mathbb{R} ^3, \end{equation}$

其中$\lambda\in\mathbb{R} \setminus\{0\}, \ \ a, b$为正常数, $\beta$为实参数, $V(x)\in C(\mathbb{R} ^3, \mathbb{R} ^+), \ \ p\in(0, 4).$

当$b = 0$时,上述方程(1.1)即为经典的半线性椭圆方程,特别是$N=2$的情形对应于冷原子物理中的Bose-Einstein凝聚现象相关的Gross-Pitaevskii (GP)方程.这时GP方程的$L^2$范数解有明确的物理意义,即质量约束解.近年来关于GP方程的$L^2$范数约束解备受关注,例如Guo等^[6]讨论GP方程的$L^2$范数约束解关于参数$\beta$的渐近行为;而Zhou等^[1]研究了参数$\beta$固定时GP方程的$L^2$范数约束解随着参数$p$趋向于$2$ (即GP方程的质量临界指标)的渐近行为.另外Zeng等^[2]证明了当$\beta = 1, \ V(x)\equiv 0$时方程(1.1)在约束$\|u\|_{L^{2}}=c$下解的存在性和唯一性.

当$b\neq 0$时,方程(1.1)就从局部型变为非局部型的Kirchhoff椭圆方程.近十多年来,很多学者都十分关注Kirchhoff型椭圆方程的研究.例如Zou等^[3]利用山路引理和Nehari流形方法得到了方程(1.1)在$\lambda=0, \ \ p\in(2, 4)$时基态解的存在性.当$V(x)$为非负常数时, Wu等^[4]研究了方程(1.1)的径向对称解.另外Guo等^[5]讨论了如下形式的$V(x)$

(1.2) $\begin{equation}\label{v}0\not\equiv V(x)\in C(\mathbb{R} ^N, \mathbb{R} ^+), \ \ \inf\limits_{x\in\mathbb{R} ^N}V(x)=0\ \ \mbox{及}\ \ \lim\limits_{|x|\to\infty}V(x)=\infty, \end{equation}$

利用约束变分法结合能量估计讨论了方程(1.1)解的存在性并研究了解的渐近行为.

已有文献的研究的结果表明,为了更加具体地刻画解的爆破点的位置往往需要知道$V(x)$更加具体的形式.例如文献[6]中在假设

$\begin{eqnarray*}V(x)=h(x)\prod\limits_{i=1}^N|x-x_i|^{p_i}, \ \ C<h(x)<\frac{1}{C}, \ \ \mbox{其中}\ \ C>0, \ \ x\in\mathbb{R} ^N\end{eqnarray*}$

的条件下精确分析了GP方程$L^2$范数约束解的渐近行为.而文献[7]中则在假设

$\begin{eqnarray*}V(x)=\left(\sqrt{\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}}-1\right)^2, \ \ \mbox{其中}\ \ a>b>0, \ \ x=(x_1, x_2)\in\mathbb{R} ^2\end{eqnarray*}$

的前提下精细刻画了GP方程$L^2$范数约束解的爆破性质.受以上文献的启发,本文的主要目的就是希望将文献[7]中关于GP方程的相关结果推广到Kirchhoff方程.为此,我们假设$V(x)$为具有椭球型底部的势阱即

(1.3) $\begin{equation}\label{v1}V(x)=\left(\sqrt{\frac{x_1^2}{a_1^2}+\frac{x_2^2}{a_2^2}+\frac{x_3^2}{a_3^2}}-1\right)^2, \ \ \mbox{其中}\ \ a_1>\max\{a_2, a_3\}>0, \ \ x=(x_1, x_2, x_3)\in\mathbb{R} ^3.\end{equation}$

显然势阱$V(x)$底部是一个椭球$\{(x_1, x_2, x_3)\in\mathbb{R} ^3:\frac{x_1^2}{a_1^2}+\frac{x_2^2}{a_2^2}+\frac{x_3^2}{a_3^2}=1\}$.本文将在此类位势下讨论方程(1.1)解的爆破行为,并将更加具体地确定解的爆破点的可能位置.

为了下文讨论的方便,我们记

(1.4) $\begin{equation}\label{v2}f(x)=\sqrt{\frac{x_1^2}{a_1^2}+\frac{x_2^2}{a_2^2}+\frac{x_3^2}{a_3^2}}, \end{equation}$

并令

(1.5) $\begin{equation}\label{v3}Z_1\triangleq\{(\pm a_1, 0, 0)\}, \end{equation}$

则集合$Z_1$为势阱$V(x)$的椭球底部的长轴的两个端点,并且$V(x)=(f(x)-1)^2$.

该文引入如下半线性椭圆方程

(1.6) $\begin{equation}\label{fc2}-\frac{3p}{4}\bigtriangleup u+(1-\frac{p}{4})u-u^{p+1}=0, \ \ \mbox{其中}\ \ 0<p<4, \ \ x\in\mathbb{R} ^3.\end{equation}$

文献[8]已证明了方程(1.6)存在唯一的径向对称正解$\phi_p\in H^1(\mathbb{R} ^3)$,另外$\phi_p$在无穷远处呈指数型衰减,即存在$R>0$,使得当$|x|>R$时有

(1.7) $\begin{equation}\label{fc3}\phi_p(x)<Ce^{-\delta|x|}, \end{equation}$

其中$C, \delta$为正常数.对此$\phi_p$定义

(1.8) $\begin{equation}\label{p1}\beta_p\triangleq\frac{b}{2}||\phi_p||_{L^2}^p, \ \ \mbox{特别地,当}\ \ p={p^*}\triangleq\frac{8}{3}\ \ \mbox{时我们记}\ \ \beta_{p^*}\triangleq\frac{b}{2}||\phi_{p^*}||_{L^2}^{p^*}.\end{equation}$

我们知道为了得到方程(1.1)的解,一个最为经典的方法就是研究如下约束变分问题

(1.9) $\begin{equation}\label{d1}d_\beta(p)=\inf\limits_{u\in S_1} E_p^\beta(u), \end{equation}$

其中

(1.10) $\begin{equation}\label{d2}E_p^\beta(u)=\frac{a}{2}\int_{\mathbb{R} ^3} |\nabla u|^2{\rm d}x+\frac{b}{4}\left(\int_{\mathbb{R} ^3} |\nabla u|^2{\rm d}x\right)^2-\frac{\beta}{p+2}\int_{\mathbb{R} ^3} |u|^{p+2}{\rm d}x+\frac{1}{2}\int_{\mathbb{R} ^3} V(x)u^2{\rm d}x, \end{equation}$

(1.11) $\begin{equation}\label{d3}S_1=\left\{u\in H:\int_{\mathbb{R} ^3} |u|^2{\rm d}x=1\right\}, \ \H\triangleq\left\{u\in H^1(\mathbb{R} ^3):\int_{\mathbb{R} ^3} V(x)|u|^2{\rm d}x<+\infty\right\}, \end{equation}$

其中$a, b$为正常数, $\beta$为实参数, $0<p<{p^*}\triangleq\frac{8}{3}$.

当$V(x)\equiv0$时,上述约束变分问题我们记为

(1.12) $\begin{equation}\label{d4}\tilde{d}_\beta(p)=\inf\limits_{u\in \tilde{S}_1} \tilde{E}_p^\beta(u), \end{equation}$

其中

(1.13) $\begin{equation}\label{d5}\tilde{E}_p^\beta(u)=\frac{a}{2}\int_{\mathbb{R} ^3} |\nabla u|^2{\rm d}x+\frac{b}{4}\left(\int_{\mathbb{R} ^3} |\nabla u|^2{\rm d}x\right)^2-\frac{\beta}{p+2}\int_{\mathbb{R} ^3} |u|^{p+2}{\rm d}x, \end{equation}$

(1.14) $\begin{equation}\label{d6}\tilde{S}_1=\left\{u\in H^1(\mathbb{R} ^3):\int_{\mathbb{R} ^3} |u|^2{\rm d}x=1\right\}, \end{equation}$

文献[5]已证明了当$\beta>\beta_{p^*}, \ \ 0<p<{p^*}$时,变分问题(1.12)总存在非负的极小元,并且如果还有$V(x)$满足(1.2)式,变分问题(1.9)也存在非负的极小元,并且极小元在$V(x)=0$处爆破.文献[5]的结果概括如下.

定理1.1^{[5, Theorem, 1.5]}  对任意给定的$\beta>\beta_{p^*}$,若$0<p<{p^*}\triangleq\frac{8}{3}$且$V(x)$满足(1.2)式,记极小化问题(1.9)的非负极小元为$u_p\in S_1$.则对任意满足$p\nearrow {p^*}$的序列$\{p\}$,存在$\{y_{\varepsilon_p}\}\subset\mathbb{R} ^3$和$y_0\in\mathbb{R} ^3$使得

(1.15) $\begin{equation}\label{t1.1}\lim\limits_{p\nearrow\ p^*} \varepsilon_p^{\frac{3}{2}}u_p(\varepsilon_px+\varepsilon_py_{\varepsilon_p})=\frac{1}{||\phi_{p^*}||_{L^2}}\phi_{p^*}(|x-y_0|)\ \ \mbox{于}\ \ H^1(\mathbb{R} ^3), \end{equation}$

其中

(1.16) $\begin{equation}\label{t1.2}\varepsilon_p=\left(\frac{\beta p}{\beta_pp^*}\right)^{-\frac{p^*}{4(p^*-p)}}\rightarrow 0\ \ (\mbox{当}\ \ p\nearrow {p^*}\ \ \mbox{时}), \end{equation}$

此外, $\{y_{\varepsilon_p}\}$满足$\varepsilon_py_{\varepsilon_p}\rightarrow z_0$ (当$p\nearrow {p^*}$时),并且$z_0$是$V(x)$的一个全局最小点,即$V(z_0)=0$.

在文献[5]的基础上,为了更加具体地刻画解的爆破位置,该文进一步假设$V(x)$满足(1.3)式,并证明解在$z_0\in Z_1$即椭球的长轴端点处爆破.主要结论如下.

定理1.2  对任意给定的$\beta>\beta_{p^*}$,且$V(x)$满足(1.3)式,任取$(0, {p^*})$中满足$p\nearrow {p^*}$的的一列$\{p\}$,则一定存在$\{y_{\varepsilon_p}\}\subset\mathbb{R} ^3$和$y_0\in\mathbb{R} ^3$使得当$p\nearrow {p^*}$时$\varepsilon_py_{\varepsilon_p}\rightarrow z_0$且$V(z_0)=0$.此外

(ⅰ)对于(1.9)和(1.12)式中定义的$d_\beta(p)$和$\tilde{d}_\beta(p)$必存在$\{p\}$的子列$\{p_k\}\nearrow {p^*}$满足

(1.17) $\begin{equation}\label{t2.1}\lim\limits_{k\rightarrow\infty} \frac{d_\beta(p_k)-\tilde{d}_\beta(p_k)}{\varepsilon_{p_k}^2}=\frac{1}{6a_1^2||\phi_{p^*}||_{L^2}^2}\int_{\mathbb{R} ^3} |z|^2\phi_{p^*}^2(z){\rm d}z, \end{equation}$

(ⅱ)对于$\{y_{\varepsilon_p}\}$和$y_0$我们有

$\begin{eqnarray*}\lim\limits_{p\nearrow\ p^*} \varepsilon_p^{\frac{3}{2}}u_p(\varepsilon_px+\varepsilon_py_{\varepsilon_p})=\frac{1}{||\phi_{p^*}||_{L^2}}\phi_{p^*}(|x-y_0|)\ \ \mbox{于}\ \ H^1(\mathbb{R} ^3), \end{eqnarray*}$

(ⅲ)对于上述子列$\{p_k\}$和$z_0$必有

(1.18) $\begin{equation}\label{t2.2}\lim\limits_{k\rightarrow\infty} \frac{f(\varepsilon_{p_k}y_{\varepsilon_{p_k}})-1}{\varepsilon_{p_k}}=-\bigtriangledown f(z_0)\cdot y_0, \ \ \mbox{其中}\ \ z_0\in Z_1, \end{equation}$

并且

$\varepsilon_{p_k}y_{\varepsilon_{p_k}}\rightarrow z_0\in Z_1\ \ \mbox{当}\ \ k\rightarrow\infty, $

其中

$\varepsilon_p=\left(\frac{\beta p}{\beta_p{p^*}}\right)^{-\frac{p^*}{4({p^*}-p)}}\rightarrow 0\ \ (\mbox{当}\ \ p\nearrow {p^*}\ \ \mbox{时}).$

在这一节中我们会证明一些必要的引理并在下一节证明我们的结论.

首先我们给出下列的Gagliardo-Nirenberg不等式^[9]

(2.1) $\begin{equation}\label{g1}||u||_{L^{p+2}}^{p+2}\leq\frac{p+2}{2||\phi_p||_{L^2}^p}||\bigtriangledown u||_{L^2}^{\frac{3p}{2}}||u||_{L^2}^{2-\frac{p}{2}}, \ \ u\in H^1(\mathbb{R} ^3), \ \ 0<p<4.\end{equation}$

根据文献[9]和Pohozaev恒等式,我们有

(2.2) $\begin{equation}\label{g2}\int_{\mathbb{R} ^3} \phi_p^2{\rm d}x=\int_{\mathbb{R} ^3} |\bigtriangledown\phi_p|^2{\rm d}x, \ \\int_{\mathbb{R} ^3} \phi_p^2{\rm d}x=\frac{2}{p+2}\int_{\mathbb{R} ^3} |\phi_p|^{p+2}{\rm d}x, \ \ 0<p<4.\end{equation}$

文献[5]对$d_\beta(p)$的非负极小元$u_p$和$\tilde{d}_\beta(p)$的非负极小元$\tilde{u}_p$有如下估计.

引理2.1^{[5, Theorem 3.2]}  对任意给定的$\beta>\beta_{p^*}$,若$0<p<{p^*}$且$V(x)$满足(1.3)式,设$u_p$是$d_\beta(p)$的非负极小元, $\tilde{u}_p$是$\tilde{d}_\beta(p)$的非负极小元.令$r_p\triangleq(\frac{\beta p}{\beta_p{p^*}})^{\frac{p^*}{{p^*}-p}}$,则有

(2.3) $\begin{equation}\label{l1.1}\lim\limits_{p\nearrow\ {p^*}}\frac{(\int_{\mathbb{R} ^3} |\nabla u_p|^2{\rm d}x)^2}{r_p}=\lim\limits_{p\nearrow\ {p^*}}\frac{(\int_{\mathbb{R} ^3} |\nabla\tilde{u}_p|^2{\rm d}x)^2}{r_p}=1.\end{equation}$

类似于文献[7],我们对$\bigtriangledown f(z)$有如下估计.

引理2.2  设$V(x), f(x)$和$Z_1$分别由(1.3), (1.4)和(1.5)式给出.对任意$z=(z_1, z_2, z_3)\in\mathbb{R} ^3$,如果$V(z)=0$,则有

(2.4) $\begin{equation}\label{l2.1}|\bigtriangledown f(z)|^2\geq\frac{1}{a_1^2}.\end{equation}$

此外, (2.4)式中等号成立当且仅当$z\in Z_1$.

证  令$z=(z_1, z_2, z_3)\in\mathbb{R} ^3$满足$V(z)=0$.于是

$\begin{eqnarray*}f(z)=\sqrt{\frac{z_1^2}{a_1^2}+\frac{z_2^2}{a_2^2}+\frac{z_3^2}{a_3^2}}=1, \ \ \bigtriangledown f(z)=\bigg(\frac{z_1}{a_1^2}, \frac{z_2}{a_2^2}, \frac{z_3}{a_3^2}\bigg).\end{eqnarray*}$

由$a_1>\max\{a_2, a_3\}>0$及$\frac{z_1^2}{a_1^2}=1-\frac{z_2^2}{a_2^2}-\frac{z_3^2}{a_3^2}$我们有

$\begin{eqnarray*}|\bigtriangledown f(z)|^2&=&\frac{z_1^2}{a_1^4}+\frac{z_2^2}{a_2^4}+\frac{z_3^2}{a_3^4}=\frac{1}{a_1^2}\left(1-\frac{z_2^2}{a_2^2}-\frac{z_3^2}{a_3^2}\right)+\frac{z_2^2}{a_2^4}+\frac{z_3^2}{a_3^4}\\&=&\frac{1}{a_1^2}+\frac{z_2^2}{a_2^2}\left(\frac{1}{a_2^2}-\frac{1}{a_1^2}\right)+\frac{z_3^2}{a_3^2}\left(\frac{1}{a_3^2}-\frac{1}{a_1^2}\right)\geq\frac{1}{a_1^2}.\end{eqnarray*}$

上式等号成立当且仅当$z_2=z_3=0$,此时必有$z_1=\pm a_1$,即$z\in Z_1$.

我们对方程(1.6)的唯一正解$\phi_p\in H^1(\mathbb{R} ^3)$有如下公式.

引理2.3  对任意给定的$y\in\mathbb{R} ^3$和方程(1.6)的唯一正解$\phi_p\in H^1(\mathbb{R} ^3)$有

(2.5) $\begin{equation}\label{l3.1}\int_{\mathbb{R} ^3} (y\cdot x)\phi_p^2(x){\rm d}x=0, \end{equation}$

并且

(2.6) $\begin{equation}\label{l3.2}\int_{\mathbb{R} ^3} |y\cdot x|^2\phi_p^2(x){\rm d}x=\frac{|y|^2}{3}\int_{\mathbb{R} ^3} |x|^2\phi_p^2(x){\rm d}x.\end{equation}$

证  令$y=(y_1, y_2, y_3)\in\mathbb{R} ^3$,对$x=(x_1, x_2, x_3)\in\mathbb{R} ^3$使用极坐标变换

(2.7) $\begin{equation}\label{lp3.1}\left\{\begin{array}{l}x_1=r\cos\theta_1\\x_2=r\sin\theta_1\cos\theta_2\\x_3=r\sin\theta_1\sin\theta_2\end{array}\right.\end{equation}$

其中$r\geq0, \ \ 0\leq\theta_1\leq\pi, \ \ 0\leq\theta_2\leq 2\pi$.雅可比行列式为

(2.8) $\begin{equation}\label{lp3.2}J=\left|\frac{\partial(x_1, x_2, x_3)}{\partial(r, \theta_1, \theta_2)}\right|=r^2\sin\theta_1.\end{equation}$

由(1.7)式知$\int_{\mathbb{R} ^3} (y\cdot x)\phi_p^2(x){\rm d}x, \ \\int_{\mathbb{R} ^3} |y\cdot x|^2\phi_p^2(x){\rm d}x, \ \ \int_{\mathbb{R} ^3}|x|^2\phi_p^2(x){\rm d}x$存在并有界.

首先我们来证明(2.5)式.注意到对任意正整数$n$有

(2.9) $\begin{equation}\label{lp3.3}\int_0^\pi\sin^n\theta\cos\theta {\rm d}\theta=0, \end{equation}$

于是

$\begin{eqnarray*}&&\int_{\mathbb{R} ^3} (y\cdot x)\phi_p^2(x){\rm d}x\\&=&\int_0^\infty r^3\phi_p^2(r) \int_0^\pi \int_0^{2\pi} (y_1\sin\theta_1\cos\theta_1+y_2\sin^2\theta_1\cos\theta_2+y_3\sin^2\theta_1\sin\theta_2){\rm d}\theta_2{\rm d}\theta_1{\rm d}r\\&=&\int_0^\infty r^3\phi_p^2(r) \int_0^\pi 2\pi\cdot (y_1\sin\theta_1\cos\theta_1){\rm d}\theta_1{\rm d}r=0.\end{eqnarray*}$

即证明了(2.5)式.

现在我们来证明(2.6)式.由(2.9)式有

$\int_{\mathbb{R} ^3} 2(y_1y_2x_1x_2)\phi_p^2(x){\rm d}x=\int_0^\infty 2r^4\phi_p^2(r) \int_0^\pi \int_0^{2\pi} (y_1y_2\sin^2\theta_1\cos\theta_1\cos\theta_2){\rm d}\theta_2{\rm d}\theta_1{\rm d}r=0, $

$\int_{\mathbb{R} ^3} 2(y_1y_3x_1x_3)\phi_p^2(x){\rm d}x=\int_0^\infty 2r^4\phi_p^2(r) \int_0^\pi \int_0^{2\pi} (y_1y_3\sin^2\theta_1\cos\theta_1\sin\theta_2){\rm d}\theta_2{\rm d}\theta_1{\rm d}r=0, $

$\int_{\mathbb{R} ^3} 2(y_2y_3x_2x_3)\phi_p^2(x){\rm d}x=\int_0^\infty 2r^4\phi_p^2(r) \int_0^\pi \int_0^{2\pi} (y_2y_3\sin^3\theta_1\sin\theta_2\cos\theta_2){\rm d}\theta_2{\rm d}\theta_1{\rm d}r=0, $

于是

(2.10) $\begin{eqnarray}\label{lp3.4}\int_{\mathbb{R} ^3} |y\cdot x|^2\phi_p^2(x){\rm d}x&=&\int_{\mathbb{R} ^3} (y_1x_1+y_2x_2+y_3x_3)^2\phi_p^2(x){\rm d}x\nonumber\\&=&\int_{\mathbb{R} ^3} (y_1^2x_1^2+y_2^2x_2^2+y_3^2x_3^2)\phi_p^2(x){\rm d}x.\end{eqnarray}$

对$x$做一次坐标轮换,即令$x_1=x_2, \ \ x_2=x_3, \ \ x_3=x_1$,我们有

(2.11) $\begin{equation}\label{lp3.5}\int_{\mathbb{R} ^3} |y\cdot x|^2\phi_p^2(x){\rm d}x=\int_{\mathbb{R} ^3} (y_1^2x_2^2+y_2^2x_3^2+y_3^2x_1^2)\phi_p^2(x){\rm d}x.\end{equation}$

再对$x$做一次坐标轮换,即令$x_1=x_3, \ \ x_2=x_1, \ \ x_3=x_2$,我们有

(2.12) $\begin{equation}\label{lp3.6}\int_{\mathbb{R} ^3} |y\cdot x|^2\phi_p^2(x){\rm d}x=\int_{\mathbb{R} ^3} (y_1^2x_3^2+y_2^2x_1^2+y_3^2x_2^2)\phi_p^2(x){\rm d}x.\end{equation}$

把(2.10), (2.11)式和(2.12)式相加再除以$3$,有

$\begin{eqnarray*}\int_{\mathbb{R} ^3} |y\cdot x|^2\phi_p^2(x){\rm d}x&=&\frac{1}{3}\int_{\mathbb{R} ^3} (\sum\limits_{i, j=1}^3y_i^2x_j^2)\phi_p^2(x){\rm d}x\\&=&\frac{(y_1^2+y_2^2+y_3^2)}{3}\int_{\mathbb{R} ^3} (x_1^2+x_2^2+x_3^2)\phi_p^2(x){\rm d}x\\&=&\frac{|y|^2}{3}\int_{\mathbb{R} ^3}|x|^2\phi_p^2(x){\rm d}x, \end{eqnarray*}$

即证明了(2.6)式.

证  (ⅰ)我们先证

(3.1) $\begin{equation}\label{tp2.1}\lim\limits_{p\nearrow p^*}\frac{d_\beta(p)-\tilde{d}_\beta(p)}{\varepsilon_p^2}\leq\frac{1}{6a_1^2||\phi_{p^*}||_{L^2}^2}\int_{\mathbb{R} ^3} |z|^2\phi_{p^*}^2(z){\rm d}z.\end{equation}$

对$\forall x_0\in Z_1\triangleq\{(\pm a_1, 0, 0)\}, \ \ \forall t>0$,我们令

(3.2) $\begin{equation}\label{tp2.2}u_t(x)\triangleq\frac{t^{\frac{3}{2}}}{||\phi_p||_{L^2}}\cdot\phi_p(t(x-x_0)), \end{equation}$

于是

(3.3) $\begin{equation}\label{tp2.3}\int_{\mathbb{R} ^3} u_t^2(x){\rm d}x=\frac{1}{||\phi_p||_{L^2}^2}\int_{\mathbb{R} ^3} t^3\phi_p^2(t(x-x_0)){\rm d}x=\frac{1}{||\phi_p||_{L^2}^2}\int_{\mathbb{R} ^3} \phi_p^2(z){\rm d}z=1.\end{equation}$

由(2.2)式有

(3.4) $\begin{eqnarray}\label{tp2.4}\int_{\mathbb{R} ^3} |\bigtriangledown u_t(x)|^2{\rm d}x&=&\frac{1}{||\phi_p||_{L^2}^2}\int_{\mathbb{R} ^3} t^3|\bigtriangledown\phi_p(t(x-x_0))|^2{\rm d}x=\frac{1}{||\phi_p||_{L^2}^2}\int_{\mathbb{R} ^3} t^2|\bigtriangledown\phi_p(z)|^2{\rm d}z\nonumber\\&=&\frac{1}{||\phi_p||_{L^2}^2}\int_{\mathbb{R} ^3} t^2\phi_p^2(z){\rm d}z=t^2.\end{eqnarray}$

再由(2.2)式和$\beta_p\triangleq\frac{b}{2}||\phi_p||_{L^2}^p$可得

(3.5) $\begin{eqnarray}\label{tp2.5}\int_{\mathbb{R} ^3} u_t^{p+2}(x){\rm d}x&=&\frac{1}{||\phi_p||_{L^2}^{p+2}}\int_{\mathbb{R} ^3} t^{\frac{3}{2}(p+2)}\phi_p^{p+2}(t(x-x_0)){\rm d}x=\frac{1}{||\phi_p||_{L^2}^{p+2}}\int_{\mathbb{R} ^3} t^{\frac{3p}{2}}\phi_p^{p+2}(z){\rm d}z\nonumber\\&=&\frac{p+2}{2||\phi_p||_{L^2}^p}t^{\frac{3p}{2}}=\frac{b(p+2)}{4\beta_p}t^{\frac{4p}{p^*}}.\end{eqnarray}$

由(1.7)式知$\int_{\mathbb{R} ^3} |x|^2\phi_p^2(x){\rm d}x$存在并有界.又$x_0\in Z_1$,由引理2.2知$|\bigtriangledown f(x_0)|^2=\frac{1}{a_1^2}$.于是由(1.3)及(2.6)式有

(3.6) $\begin{eqnarray}\label{tp2.6}\lim\limits_{t\rightarrow\infty}t^2\int_{\mathbb{R} ^3} V(x)u_t^2(x){\rm d}x&=&\lim\limits_{t\rightarrow\infty}\frac{t^2}{||\phi_p||_{L^2}^2}\int_{\mathbb{R} ^3} t^3V(x)\phi_p^2(t(x-x_0)){\rm d}x\nonumber\\&=&\lim\limits_{t\rightarrow\infty}\frac{t^2}{||\phi_p||_{L^2}^2}\int_{\mathbb{R} ^3} V\left(\frac{z}{t}+x_0\right)\phi_p^2(z){\rm d}z\nonumber\\&=&\lim\limits_{t\rightarrow\infty}\frac{1}{||\phi_p||_{L^2}^2}\int_{\mathbb{R} ^3}\left(\frac{f\left(\frac{z}{t}+x_0\right)-f(x_0)}{1/t}\right)^2\phi_p^2(z){\rm d}z\nonumber\\&=&\frac{1}{||\phi_p||_{L^2}^2}\int_{\mathbb{R} ^3} (\bigtriangledown f(x_0)\cdot z)^2\phi_p^2(z){\rm d}z\nonumber\\&=&\frac{|\bigtriangledown f(x_0)|^2}{3||\phi_p||_{L^2}^2}\int_{\mathbb{R} ^3} |z|^2\phi_p^2(z){\rm d}z\nonumber\\&=&\frac{1}{3a_1^2||\phi_p||_{L^2}^2}\int_{\mathbb{R} ^3} |z|^2\phi_p^2(z){\rm d}z<+\infty.\end{eqnarray}$

由(3.3)式及(3.6)式有$u_t\in S_1$,故结合(3.4), (3.5)式以及定义(1.10)有

(3.7) $\begin{eqnarray}\label{tp2.7}E_p^\beta(u_t)=\frac{a}{2}t^2+\frac{b}{4}t^4-\frac{b\beta}{4\beta_p}(t^4)^{\frac{p}{p^*}}+\frac{1}{2}\int_{\mathbb{R} ^3} V(x)u_t^2(x){\rm d}x.\end{eqnarray}$

设$u_p\in S_1$是(1.9)式的极小元, $\tilde{u}_p\in\tilde{S}_1$是(1.12)式的极小元.则由$||\tilde{u}_p||_{L^2}=1$及(2.1)式有

(3.8) $\begin{eqnarray}\label{tp2.8}||\tilde{u}_p||_{L^{p+2}}^{p+2}&\leq&\frac{p+2}{2||\phi_p||_{L^2}^p}||\bigtriangledown\tilde{u}_p||_{L^2}^{\frac{3p}{2}}||\tilde{u}_p||_{L^2}^{2-\frac{p}{2}}=\frac{p+2}{2||\phi_p||_{L^2}^p}||\bigtriangledown\tilde{u}_p||_{L^2}^{\frac{3p}{2}}\nonumber\\&=&\frac{p+2}{2||\phi_p||_{L^2}^p}\left(\int_{\mathbb{R} ^3} |\nabla\tilde{u}_p|^2{\rm d}x\right)^{\frac{3p}{4}}=\frac{p+2}{2||\phi_p||_{L^2}^p}\left(\int_{\mathbb{R} ^3} |\nabla\tilde{u}_p|^2{\rm d}x\right)^{\frac{2p}{{p^*}}}.\end{eqnarray}$

于是由(3.7)和(3.8)式有

$\begin{eqnarray*}d_\beta(p)-\tilde{d}_\beta(p)&=&E_p^\beta(u_p)-\tilde{E}_p^\beta(\tilde{u}_p)\leq E_p^\beta(u_t)-\tilde{E}_p^\beta(\tilde{u}_p)\\&\leq&\frac{a}{2}t^2+\frac{b}{4}t^4-\frac{b\beta}{4\beta_p}(t^4)^{\frac{p}{p^*}}+\frac{1}{2}\int_{\mathbb{R} ^3} V(x)u_t^2(x){\rm d}x\\&&-\frac{a}{2}\int_{\mathbb{R} ^3} |\nabla\tilde{u}_p|^2{\rm d}x-\frac{b}{4}\left(\int_{\mathbb{R} ^3} |\nabla\tilde{u}_p|^2{\rm d}x\right)^2+\frac{b\beta}{4\beta_p}\left(\int_{\mathbb{R} ^3} |\nabla\tilde{u}_p|^2{\rm d}x\right)^{\frac{2p}{{p^*}}}.\end{eqnarray*}$

令$t=t_p\triangleq\left(\int_{\mathbb{R} ^3} |\nabla\tilde{u}_p|^2{\rm d}x\right)^{\frac{1}{2}}$,则可知

$\begin{eqnarray*}\frac{d_\beta(p)-\tilde{d}_\beta(p)}{\varepsilon_p^2}\leq\frac{1}{2\varepsilon_p^2}\int_{\mathbb{R} ^3} V(x)u_t^2(x){\rm d}x.\end{eqnarray*}$

由引理2.1及$t^2=\int_{\mathbb{R} ^3} |\nabla\tilde{u}_p|^2{\rm d}x, \ \r_p\triangleq(\frac{\beta p}{\beta_pp^*})^{\frac{p^*}{p^*-p}}=\frac{1}{\varepsilon_p^4}$知$\mathop {\lim }\limits_{p \nearrow {p^*}} \varepsilon_p^2\cdot t^2=1$.又因当$t\rightarrow\infty$$ (p\nearrow {p^*})$时有$\varepsilon_p\rightarrow 0$$ (p\nearrow {p^*})$于是

$\begin{eqnarray*}\lim\limits_{p\nearrow\ {p^*}} \frac{d_\beta(p)-\tilde{d}_\beta(p)}{\varepsilon_p^2}&\leq&\lim\limits_{p\nearrow\ {p^*}} \frac{1}{2\varepsilon_p^2\cdot t^2}\cdot\lim\limits_{t\rightarrow\infty} t^2\int_{\mathbb{R} ^3} V(x)u_t^2(x){\rm d}x\\&=&\frac{1}{6a_1^2||\phi_{p^*}||_{L^2}^2}\int_{\mathbb{R} ^3} |z|^2\phi_{p^*}^2(z){\rm d}z.\end{eqnarray*}$

即证明了(3.1)式.

下证(1.17)式.由定理1.1,对任意给定的$\beta>\beta_{p^*}$,若$0<p<{p^*}$且$V(x)$满足(1.3)式,则方程(1.9)存在非负的极小元$u_p\in S_1$.并且对任意满足$p\nearrow {p^*}$的可达元序列$\{u_p\}$,存在$\{y_{\varepsilon_p}\}\subset\mathbb{R} ^3$和$y_0\in\mathbb{R} ^3$使得当$p\nearrow {p^*}$时有

$\varepsilon_py_{\varepsilon_p}\rightarrow z_0, \ \mbox{并且}\ \V(z_0)=0\ \ \mbox{即}\ \ f(z_0)=1.$

对$\varepsilon_p=\left(\frac{\beta p}{\beta_p{p^*}}\right)^{-\frac{p^*}{4({p^*}-p)}}$,我们定义

(3.9) $\begin{equation}\label{tp2.9}w_p(x)\triangleq \varepsilon_p^{\frac{3}{2}}u_p(\varepsilon_px+\varepsilon_py_{\varepsilon_p}), \end{equation}$

以及

(3.10) $\begin{equation}\label{tp2.10}\mu_p\triangleq\frac{1}{\varepsilon_p^2}\int_{\mathbb{R} ^3} V(\varepsilon_px+\varepsilon_py_{\varepsilon_p})w_p^2(x){\rm d}x, \end{equation}$

则

(3.11) $\begin{eqnarray}\label{tp2.11}\mu_p&=&\frac{1}{\varepsilon_p^2}\int_{\mathbb{R} ^3} V(\varepsilon_px+\varepsilon_py_{\varepsilon_p})w_p^2(x){\rm d}x=\frac{1}{\varepsilon_p^2}\int_{\mathbb{R} ^3} V(\varepsilon_px+\varepsilon_py_{\varepsilon_p})\varepsilon_p^3u_p^2(\varepsilon_px+\varepsilon_py_{\varepsilon_p}){\rm d}x\nonumber\\&=&\frac{1}{\varepsilon_p^2}\int_{\mathbb{R} ^3} V(z)u_p^2(z){\rm d}z.\end{eqnarray}$

设$u_p\in S_1$是(1.9)式的极小元, $\tilde{u}_p\in\tilde{S}_1$是(1.12)式的极小元.我们有

$d_\beta(p)-\tilde{d}_\beta(p)=E_p^\beta(u_p)-\tilde{E}_p^\beta(\tilde{u}_p)\geq E_p^\beta(u_p)-\tilde{E}_p^\beta(u_p)=\frac{1}{2}\int_{\mathbb{R} ^3} V(z)u_p^2(z){\rm d}z.$

于是

(3.12) $\begin{equation}\label{tp2.12}\frac{d_\beta(p)-\tilde{d}_\beta(p)}{\varepsilon_p^2}\geq\frac{1}{2\varepsilon_p^2}\int_{\mathbb{R} ^3} V(z)u_p^2(z){\rm d}z=\frac{1}{2}\mu_p, \end{equation}$

结合(3.1)式,我们知道$\{\mu_p\}$关于$p$一致有界.由定理1.1知在$H^1(\mathbb{R} ^3)$范数意义下有

$\lim\limits_{p\nearrow\ {p^*}}w_p=\frac{1}{||\phi_{p^*}||_{L^2}}\phi_{p^*}(|x-y_0|), $

其中$y_0\in\mathbb{R} ^3$,则在$L^2(\mathbb{R} ^3)$范数意义下有

$\lim\limits_{p\nearrow\ {p^*}}w_p=\frac{1}{||\phi_{p^*}||_{L^2}}\phi_{p^*}(|x-y_0|).$

于是对任意满足$p_k\nearrow {p^*}(k\rightarrow\infty)$的序列$\{p_k\}$,存在子列,我们依然记为$\{p_k\}$,使得

(3.13) $\begin{equation}\label{tp2.13}w_{p_k}\mathop{\rightarrow}\limits^{k}\frac{1}{||\phi_{p^*}||_{L^2}}\phi_{p^*}(|x-y_0|)\ \ \mbox{在}\ \ \mathbb{R} ^3\ \ \mbox{中几乎收敛}.\end{equation}$

下面,我们用反证法证明$\left\{\frac{f(\varepsilon_{p_k}y_{\varepsilon_{p_k}})-f(z_0)}{\varepsilon_{p_k}}\right\}$有界.

假设$\left\{\frac{f(\varepsilon_{p_k}y_{\varepsilon_{p_k}})-f(z_0)}{\varepsilon_{p_k}}\right\}$无界,则存在$\{p_k\}$的子列,我们依然记为$\{p_k\}$,使得

$\left|\frac{f(\varepsilon_{p_k}y_{\varepsilon_{p_k}})-f(z_0)}{\varepsilon_{p_k}}\right|\rightarrow\infty\ (k\rightarrow\infty).$

注意到$\varepsilon_{p_k}y_{\varepsilon_{p_k}}\rightarrow z_0\ (k\rightarrow\infty)$且$f(z_0)=1$,由(1.3)式, (3.13)式及Fatou引理,对任意$M>0$我们有

$\begin{eqnarray*}\liminf\limits_{k\rightarrow\infty} \mu_{p_k}&=&\liminf\limits_{k\rightarrow\infty} \frac{1}{\varepsilon_{p_k}^2}\int_{\mathbb{R} ^3} V(\varepsilon_{p_k}x+\varepsilon_{p_k}y_{\varepsilon_{p_k}})w_{p_k}^2(x){\rm d}x\\&=&\liminf\limits_{k\rightarrow\infty} \int_{\mathbb{R} ^3}\left(\frac{f(\varepsilon_{p_k}x+\varepsilon_{p_k}y_{\varepsilon_{p_k}})-f(z_0)}{\varepsilon_{p_k}}\right)^2w_{p_k}^2(x){\rm d}x\\&=&\liminf\limits_{k\rightarrow\infty} \int_{\mathbb{R} ^3}\left(\frac{f(\varepsilon_{p_k}x+\varepsilon_{p_k}y_{\varepsilon_{p_k}})-f(\varepsilon_{p_k}y_{\varepsilon_{p_k}})}{\varepsilon_{p_k}}+\frac{f(\varepsilon_{p_k}y_{\varepsilon_{p_k}})-f(z_0)}{\varepsilon_{p_k}}\right)^2w_{p_k}^2(x){\rm d}x\\&\geq&\int_{\mathbb{R} ^3} \liminf\limits_{k\rightarrow\infty}\left(\frac{f(\varepsilon_{p_k}x+\varepsilon_{p_k}y_{\varepsilon_{p_k}})-f(\varepsilon_{p_k}y_{\varepsilon_{p_k}})}{\varepsilon_{p_k}}+\frac{f(\varepsilon_{p_k}y_{\varepsilon_{p_k}})-f(z_0)}{\varepsilon_{p_k}}\right)^2w_{p_k}^2(x){\rm d}x\\&=&\int_{\mathbb{R} ^3} \left(\bigtriangledown f(z_0)\cdot x+\liminf\limits_{k\rightarrow\infty}\frac{f(\varepsilon_{p_k}y_{\varepsilon_{p_k}})-f(z_0)}{\varepsilon_{p_k}}\right)^2w_{p_k}^2(x){\rm d}x\\&=&\frac{1}{||\phi_{p^*}||_{L^2}^2}\int_{\mathbb{R} ^3}\left(\bigtriangledown f(z_0)\cdot x+\liminf\limits_{k\rightarrow\infty}\frac{f(\varepsilon_{p_k}y_{\varepsilon_{p_k}})-f(z_0)}{\varepsilon_{p_k}}\right)^2\phi_{p^*}^2(|x-y_0|){\rm d}x\\&\geq& M.\end{eqnarray*}$

这与$\{\mu_p\}$有界相矛盾,因此$\left\{\frac{f(\varepsilon_{p_k}y_{\varepsilon_{p_k}})-f(z_0)}{\varepsilon_{p_k}}\right\}$有界.于是一定存在实数$\alpha\in\mathbb{R} $及$\{p_k\}$的子列(仍记为$\{p_k\}$),使得$\frac{f(\varepsilon_{p_k}y_{\varepsilon_{p_k}})-f(z_0)}{\varepsilon_{p_k}}\rightarrow\alpha \ (k\rightarrow\infty)$.

令$K=\bigtriangledown f(z_0)\cdot y_0+\alpha$,则$K$是一个实数.由(1.7)式知$\int_{\mathbb{R} ^3} (\bigtriangledown f(z_0)\cdot z+K)^2\phi_{p^*}^2(z){\rm d}z$存在并有界.注意到$\varepsilon_{p_k}y_{\varepsilon_{p_k}}\rightarrow z_0\ (k\rightarrow\infty), \ \ f(z_0)=1$,由(1.3)式, (3.13)式及Fatou引理有

$\begin{eqnarray*}\liminf\limits_{k\rightarrow\infty} \mu_{p_k}&=&\liminf\limits_{k\rightarrow\infty} \frac{1}{\varepsilon_{p_k}^2}\int_{\mathbb{R} ^3} V(\varepsilon_{p_k}x+\varepsilon_{p_k}y_{\varepsilon_{p_k}})w_{p_k}^2(x){\rm d}x\\&=&\liminf\limits_{k\rightarrow\infty} \int_{\mathbb{R} ^3}\left(\frac{f(\varepsilon_{p_k}x+\varepsilon_{p_k}y_{\varepsilon_{p_k}})-f(z_0)}{\varepsilon_{p_k}}\right)^2w_{p_k}^2(x){\rm d}x\\&=&\liminf\limits_{k\rightarrow\infty} \int_{\mathbb{R} ^3}\left(\frac{f(\varepsilon_{p_k}x+\varepsilon_{p_k}y_{\varepsilon_{p_k}})-f(\varepsilon_{p_k}y_{\varepsilon_{p_k}})}{\varepsilon_{p_k}}+\frac{f(\varepsilon_{p_k}y_{\varepsilon_{p_k}})-f(z_0)}{\varepsilon_{p_k}}\right)^2w_{p_k}^2(x){\rm d}x\\&\geq&\int_{\mathbb{R} ^3} \liminf\limits_{k\rightarrow\infty}\left(\frac{f(\varepsilon_{p_k}x+\varepsilon_{p_k}y_{\varepsilon_{p_k}})-f(\varepsilon_{p_k}y_{\varepsilon_{p_k}})}{\varepsilon_{p_k}}+\frac{f(\varepsilon_{p_k}y_{\varepsilon_{p_k}})-f(z_0)}{\varepsilon_{p_k}}\right)^2w_{p_k}^2(x){\rm d}x\\&=&\frac{1}{||\phi_{p^*}||_{L^2}^2}\int_{\mathbb{R} ^3}(\bigtriangledown f(z_0)\cdot x+\alpha)^2\phi_{p^*}^2(|x-y_0|){\rm d}x\\&=&\frac{1}{||\phi_{p^*}||_{L^2}^2}\int_{\mathbb{R} ^3}[\bigtriangledown f(z_0)\cdot z+(\bigtriangledown f(z_0)\cdot y_0+\alpha)]^2\phi_{p^*}^2(z){\rm d}z\\&=&\frac{1}{||\phi_{p^*}||_{L^2}^2}\int_{\mathbb{R} ^3}(\bigtriangledown f(z_0)\cdot z+K)^2\phi_{p^*}^2(z){\rm d}z.\end{eqnarray*}$

由引理2.3有

(3.14) $\begin{eqnarray}\label{tp2.14}\liminf\limits_{k\rightarrow\infty} \mu_{p_k}&\geq&\frac{1}{||\phi_{p^*}||_{L^2}^2}\int_{\mathbb{R} ^3}(\bigtriangledown f(z_0)\cdot z+K)^2\phi_{p^*}^2(z){\rm d}z\nonumber\\&=&\frac{1}{||\phi_{p^*}||_{L^2}^2}\int_{\mathbb{R} ^3}[(\bigtriangledown f(z_0)\cdot z)^2+2K\bigtriangledown f(z_0)\cdot z+K^2]\phi_{p^*}^2(z){\rm d}z\nonumber\\&=&\frac{|\bigtriangledown f(z_0)|^2}{3||\phi_{p^*}||_{L^2}^2}\int_{\mathbb{R} ^N} |z|^2\phi_{p^*}^2(z){\rm d}z+\frac{1}{||\phi_{p^*}||_{L^2}^2}\int_{\mathbb{R} ^3} K^2\phi_{p^*}^2(z){\rm d}z\nonumber\\&\geq&\frac{|\bigtriangledown f(z_0)|^2}{3||\phi_{p^*}||_{L^2}^2}\int_{\mathbb{R} ^3} |z|^2\phi_{p^*}^2(z){\rm d}z, \end{eqnarray}$

由引理2.2有$|\bigtriangledown f(z_0)|^2\geq\frac{1}{a_1^2}$,于是

(3.15) $\begin{equation}\label{tp2.15}\liminf\limits_{k\rightarrow\infty} \mu_{p_k}\geq\frac{|\bigtriangledown f(z_0)|^2}{3||\phi_{p^*}||_{L^2}^2}\int_{\mathbb{R} ^3} |z|^2\phi_{p^*}^2(z){\rm d}z\geq\frac{1}{3a_1^2||\phi_{p^*}||_{L^2}^2}\int_{\mathbb{R} ^3} |z|^2\phi_{p^*}^2(z){\rm d}z, \end{equation}$

结合(3.12)和(3.15)式有

(3.16) $\begin{equation}\label{tp2.16}\liminf\limits_{k\rightarrow\infty} \frac{d_\beta(p_k)-\tilde{d}_\beta(p_k)}{\varepsilon_{p_k}^2}\geq\liminf\limits_{k\rightarrow\infty} \frac{1}{2}\mu_{p_k}\geq\frac{1}{6a_1^2||\phi_{p^*}||_{L^2}^2}\int_{\mathbb{R} ^3} |z|^2\phi_{p^*}^2(z){\rm d}z, \end{equation}$

因此由(3.1)式有

(3.17) $\begin{equation}\label{tp2.17}\liminf\limits_{k\rightarrow\infty} \frac{d_\beta(p_k)-\tilde{d}_\beta(p_k)}{\varepsilon_{p_k}^2}=\frac{1}{6a_1^2||\phi_{p^*}||_{L^2}^2}\int_{\mathbb{R} ^3}| z|^2\phi_{p^*}^2(z){\rm d}z.\end{equation}$

于是(1.17)式成立.

(ⅱ)这是定理1.1的直接推论,本文重点是给出$\{\varepsilon_py_{\varepsilon_p}\}$的极限点更为具体的位置,即我们的结论(ⅲ).

(ⅲ)最后,结合(3.16)和(3.17)式我们知道

(3.18) $\begin{equation}\label{tp2.18}\liminf\limits_{k\rightarrow\infty} \mu_{p_k}=\frac{1}{3a_1^2||\phi_{p^*}||_{L^2}^2}\int_{\mathbb{R} ^3} |z|^2\phi_{p^*}^2(z){\rm d}z.\end{equation}$

为使(3.18)式成立,则(3.14)和(3.15)式中的不等式必须是等式. (3.14)式中的不等式是等式当且仅当$K=0$,即$\alpha=-\bigtriangledown f(z_0)\cdot y_0$,于是

$\lim\limits_{k\rightarrow\infty} \frac{f(\varepsilon_{p_k}y_{\varepsilon_{p_k}})-f(z_0)}{\varepsilon_{p_k}}=\alpha=-\bigtriangledown f(z_0)\cdot y_0, $

即

$\lim\limits_{k\rightarrow\infty} \frac{f(\varepsilon_{p_k}y_{\varepsilon_{p_k}})-1}{\varepsilon_{p_k}}=-\bigtriangledown f(z_0)\cdot y_0, $

于是(1.18)式成立. (3.15)式中的不等式是等式当且仅当$|\bigtriangledown f(z_0)|^2=\frac{1}{a_1^2}$,由引理2.2知$z_0\in Z_1$,即

$\varepsilon_{p_k}y_{\varepsilon_{p_k}}\rightarrow z_0\in Z_1\ \ \mbox{当}\ \ k\rightarrow\infty.$

至此定理1.2证毕.