本文考虑如下含调和位势和非线性项指数为质量临界情形的$ p $次分数阶Laplace方程

(1.1) $ \begin{equation} (-\Delta)_p^su+|x|^2|u|^{p-2}u-\beta|u|^{p+\frac{sp^2}{N}-2}u = \nu|u|^{p-2}u, \end{equation} $

其中$ 0<s<1 $, $ p\in (1, \infty) $且$ sp<N $, $ |x|^2 $是通常的调和位势, $ \beta>0 $是非线性项参数, $ (-\Delta)_p^s $表示分数阶$ p $-Laplace算子, 定义为

$ (-\Delta)_p^su(x) = 2\lim\limits_{\varepsilon\rightarrow 0}\int_{{{\Bbb R}} ^N\setminus B_\varepsilon(x)}\frac{|u(x)-u(y)|^{p-2}(u(x)-u(y))}{|x-y|^{N+sp}}{\rm d}x{\rm d}y, $

其中$ x, y\in {{\Bbb R}} ^N $, $ B_\varepsilon(x): = \{y\in {{\Bbb R}} ^N:|x-y|<\varepsilon\}. $

如果$ s = 1 $, 方程(1.1)就是一般的$ p $-Laplace方程, 相应方程非平凡解的存在性, 多重性以及解的相关性质研究结果非常丰富, 可参见文献[7-10, 12, 15-16]及其参考文献.其中文献[12]研究了解的存在性及其衰减性.文献[7]研究了正解的对称性.文献[8]研究了$ 1<p<2 $且位势是调和位势时方程(1.1)基态解的存在性, 并证明了非线性项参数趋于临界情形时基态解的集中行为.进一步, 文献[9]研究了一般$ p $情形且位势函数是阱位势时方程(1.1)基态解的存在性, 并证明了非线性项参数趋于临界情形时基态解的集中行为, 但是此时由于极限方程的解不具有唯一性, 因此相对而言基态解的集中行为比较困难.

若$ 0<s<1 $, 方程(1.1)被称为分数阶$ p $-Laplace方程.分数阶Laplace方程由于其在数学物理和其它领域的广泛应用, 近年来受到许多数学研究者的关注, 是国内外研究热点, 特别是当$ p = 2 $时, 关于解的存在性、唯一性、多重性等情形非常多.文献[4]利用约束变分的方法得到了一维情形方程(1.1)基态解的存在性, 并用常微分方程理论得到了基态解的唯一性.进一步, 文献[5]中改进了常微分方程理论得到了高维情形方程(1.1)基态解的存在性和唯一性.而对于一般的$ p $, 解的存在性参见文献[11], 相应的极大值原理和解的对称性参见文献[2], 解的正则性结果参见文献[3].

本文的目的是希望能利用$ L^p $约束极小的方法, 将求解方程(1.1)基态解的存在性转化成求约束在$ L^p $流形上的泛函极小值问题.关于极小解存在性所用的方法来源于文献[8, 15-16].进一步, 关于极小解的渐近行为所使用的方法是能量估计, 来源于文献[10]和^[8].但是由于是$ p $次分数阶Laplace方程, 相关结果较少, 所需的Gagliardo-Nirenberg不等式需要重新推导, 同时极限方程基态解的唯一性不能确定也带来了一定的困难.

研究方程(1.1)基态解的存在性, 可以考虑如下极小化问题

(1.2) $ \begin{equation} m(\beta) = \inf \left\{E_\beta(u):u\in M\right\}, \end{equation} $

其中

$ \begin{eqnarray*} \label{eq1.2} E_\beta(u) = \frac{1}{p}[u]_{W^{s, p}}^p+\frac{1}{p}\int_{{{\Bbb R}} ^N} |x|^2|u|^p{\rm d}x-\frac{\beta}{p+\frac{sp^2}{N}}\int_{{{\Bbb R}} ^N} |u|^{p+\frac{sp^2}{N}}{\rm d}x, \end{eqnarray*} $

$ M: = \left\{\ u\in{\Bbb H}:\int_{{{\Bbb R}} ^N}|u|^p{\rm d}x = 1\right\}. $

$ [u]_{W^{s, p}}^p $表示Gagliardo半范数

$ [u]_{W^{s, p}}^p = \int_{{{\Bbb R}} ^N}\int_{{{\Bbb R}} ^N}\frac{|u(x)-u(y)|^p}{|x-y|^{N+sp}}{\rm d}x{\rm d}y. $

对$ s\in (0, 1) $, $ p\in[1, +\infty) $, 设$ W^{s, p}({{\Bbb R}} ^N): = \{u\in L^p({{\Bbb R}} ^N):\frac{|u(x)-u(y)|}{|x-y|^{\frac{N}{p}+s}}\in L^p({{\Bbb R}} ^N\times {{\Bbb R}} ^N)\} $且具有范数

$ ||u||: = ||u||_{W^{s, p}({{\Bbb R}} ^N)}: = \left(\int_{{{\Bbb R}} ^N}|u|^p{\rm d}x+ \int_{{{\Bbb R}} ^N}\int_{{{\Bbb R}} ^N}\frac{|u(x)-u(y)|^p}{|x-y|^{N+sp}}{\rm d}x{\rm d}y\right)^{\frac{1}{p}}. $

空间$ {\Bbb H} = \{u\in W^{s, p}({{\Bbb R}} ^N): \int_{{{\Bbb R}} ^N} |x|^2|u|^p{\rm d}x<\infty\} $, 很自然的可以在空间$ {\Bbb H} $中讨论方程(1.1)基态解的存在性.

设$ E(u) = \frac{1}{p}[u]_{W^{s, p}}^p+\frac{1}{p}\int_{{{\Bbb R}} ^N} |x|^2|u|^p{\rm d}x-\frac{\beta}{q}\int_{{{\Bbb R}} ^N} |u|^{q}{\rm d}x $, 对任意固定的$ \varphi\in W^{s, p}({{\Bbb R}} ^N) $且$ \int_{{{\Bbb R}} ^N} |\varphi|^p{\rm d}x = 1 $.设$ u_\lambda(x) = \lambda^{\frac{N}{p}}\varphi(\lambda x) $, 则容易计算得

$ E(u_\lambda) = \frac{\lambda^{sp}}{p}[\varphi]_{W^{s, p}}^p+\frac{\lambda^{-2}}{p}\int_{{{\Bbb R}} ^N} |x|^2|\varphi|^p{\rm d}x-\frac{\beta\lambda^{\frac{N(q-p)}{p}}}{q}\int_{{{\Bbb R}} ^N} |\varphi|^{q}{\rm d}x. $

因此当$ \lambda $比较大时, 可以得出$ q<p+\frac{sp^2}{N} $时泛函$ E(u_\lambda) $有下界, 而$ q>p+\frac{sp^2}{N} $时泛函$ E(u_\lambda) $无下界.则$ q = p+\frac{sp^2}{N} $是$ L^p $约束极小存在与否的临界指数, 因此本文直接考虑$ L^p $约束临界即问题(1.2)极小解的存在性.

定义$ \beta^* = |Q|_p^{\frac{sp^2}{N}} $, 其中$ Q(x) $是方程

(1.3) $ \begin{equation} (-\Delta)_p^s\varphi+\frac{sp}{N}|\varphi|^{p-2}\varphi = |\varphi|^{\frac{sp^2}{N}-2}\varphi, \ \; \; x\in{{\Bbb R}} ^N \end{equation} $

的径向基态解.针对问题(1.2)极小解的存在性, 本文结果如下：

定理1.1   (1)如果$ 0\leq \beta<\beta^* $, 则问题(1.2)至少存在一个非负极小解;

(2) 如果$ \beta \geq \beta^* $, 则问题(1.2)不存在极小解;

(3) 如果$ \beta<\beta^* $, 则$ m(\beta)>0 $, $ \lim\limits_{\beta\rightarrow \beta^*}m(\beta) = m(\beta^*) = 0 $.

利用细致的能量估计, 进一步可得基态解的渐近行为如下:

定理1.2   假设$ u_\beta $是$ \beta<\beta^* $时极小化问题(1.2)的非负极小解, 设

$ \mu = \left(\frac{2\int_{{{\Bbb R}} ^N} |x|^2|Q|^p{\rm d}x}{sp}\right)^{\frac{1}{2+sp}}, $

则任意满足当$ n\rightarrow \infty $时$ \beta_n\rightarrow \beta^* $的序列$ \{\beta_n\} $存在子列(仍记作$ \{\beta_n\} $), 使得

$ \lim\limits_{n\rightarrow \infty}(\beta^*-\beta_n)^{\frac{N}{p(sp+2)}}u_{\beta_n}((\beta^*-\beta_n)^{\frac{1}{sp+2}}x) = \left(\frac{\mu}{(\beta^*)^{\frac{2+N}{sp(2+sp)}}}\right)^{\frac{N}{p}}Q\left((\beta^*)^{\frac{sp-N}{sp(2+sp)}}\mu x\right), $

$ \lim\limits_{n\rightarrow \infty}\frac{m(\beta_n)}{(\beta^*-\beta_n)^{-\frac{2}{sp+2}}} = \frac{1}{(p\beta^*)^{\frac{2+N}{2+sp}}}\left[\left(\frac{2}{sp}\right)^{\frac{sp}{2+sp}}+ \left(\frac{sp}{2}\right)^{\frac{2}{2+sp}}\right]\left(\int_{{{\Bbb R}} ^N} |x|^2|Q|^p{\rm d}x\right)^{\frac{sp}{2+sp}}. $

本文中, 不加特殊说明的情形下, $ C $表示常数, $ [u]_{W^{s, p}}^p $表示Gagliardo半范数, $ |u|_r $表示$ L^r({{\Bbb R}} ^N) $中的范数.

相似文献[1]中紧性引理的证明, 可以得到如下嵌入紧性定理.

引理2.1   假设$ 2<p<q< p^* $, 则嵌入$ {\Bbb H}\hookrightarrow L^q({{\Bbb R}} ^N) $是紧的.

为了得到极小解的存在性, 需要相应的Gagliardo-Nirenberg不等式, 相似文献[14]中的证明, 结合文献[13]中的嵌入不等式, 可得如下Gagliardo-Nirenberg不等式.

引理2.2   设$ p\in [1, \infty) $, $ s\in (0, 1) $, $ sp<N $.则对任意$ u\in W^{s, p}({{\Bbb R}} ^N) $, 有

(2.1) $ \begin{equation} \int_{{{\Bbb R}} ^N}|u|^{p+\frac{sp^2}{N}}{\rm d}x\leq \frac{N+sp}{N|Q(x)|_p^{\frac{sp^2}{N}}}\left( \int_{{{\Bbb R}} ^N}|u|^p{\rm d}x\right)^{\frac{sp}{N}}[u]_{W^{s, p}}^p, \end{equation} $

其中函数$ Q(x) $是方程(1.3)的基态解, 且(2.1)式中等号成立当且仅当$ u $是基态解$ Q(x) $的伸缩平移.

证   由文献[13], 对$ p^* = \frac{Np}{N-ps} $, 存在一个正的常数$ C = C(n, p, s) $, 使得

(2.2) $ \begin{equation} |u|_{p*}^p\leq C\int_{{{\Bbb R}} ^N}\int_{{{\Bbb R}} ^N}\frac{|u(x)-u(y)|^p}{|x-y|^{N+sp}}{\rm d}x{\rm d}y. \end{equation} $

上式结合Hölder不等式可得

$ \int_{{{\Bbb R}} ^N}|u|^{p+\frac{sp^2}{N}}{\rm d}x\leq C(N, q, s)\left( \int_{{{\Bbb R}} ^N}|u|^p{\rm d}x\right)^{\frac{sp}{N}}\int_{{{\Bbb R}} ^N}\int_{{{\Bbb R}} ^N}\frac{|u(x)-u(y)|^p}{|x-y|^{N+sp}}{\rm d}x{\rm d}y. $

令

$ J_{s, p}(u) = \frac{\left( \int_{{{\Bbb R}} ^N}|u|^p{\rm d}x\right)^{\frac{sp}{N}}\int_{{{\Bbb R}} ^N}\int_{{{\Bbb R}} ^N}\frac{|u(x)-u(y)|^p}{|x-y|^{N+sp}}{\rm d}x{\rm d}y}{\int_{{{\Bbb R}} ^N}|u|^{p+\frac{sp^2}{N}}{\rm d}x}. $

设$ u_{\lambda, \mu}(x) = \mu u(\lambda x) $, 通过计算可得

$ \begin{eqnarray*} J_{s, p}(u_{\lambda, \mu}) = \frac{\lambda^{sp-N}\mu^p\int_{{{\Bbb R}} ^N}\int_{{{\Bbb R}} ^N}\frac{|u(x)-u(y)|^p}{|x-y|^{N+sp}}{\rm d}x{\rm d}y \left( \lambda^{-N}\mu^p\int_{{{\Bbb R}} ^N}|u|^p{\rm d}x\right)^{\frac{sp}{N}}}{\lambda^{-N}\mu^{p+\frac{sp^2}{N}}\int_{{{\Bbb R}} ^N}|u|^{p+\frac{sp^2}{N}}{\rm d}x} = J_{s, p}(u), \end{eqnarray*} $

即$ J_{s, p}(u) $保持伸缩不变.进一步易知对任意的$ u\in W^{s, p}({{\Bbb R}} ^N) $有$ J_{s, p}(u)>0 $.可知存在一个非负极小化序列$ \{u_m\}\subset W^{s, p}({{\Bbb R}} ^N)\cap L^{p+\frac{sp^2}{N}}({{\Bbb R}} ^N) $, 使得

$ \alpha = \lim\limits_{ m \rightarrow\infty}J_{s, p}(u_m) = \inf\limits_{u\in W^{s, p}({{\Bbb R}} ^N)}J_{s, p}(u)<+\infty. $

取

$ \lambda_m = \frac{|u_m|_p^{\frac{1}{s}}}{[u_m]_{W^{s, p}}^{\frac{1}{s}}}, \ \mu_m = \frac{|u_m|_p^{\frac{N}{sp}-1}}{[u_m]_{W^{s, p}}^{\frac{N}{sp}}}, \ \ v_m = u_m^{\lambda_m, \mu_m}(x) = \mu_mu_m(\lambda_m x), $

则

$ \int_{{{\Bbb R}} ^N}|v_m(x)|^p{\rm d}x = \mu_m^p\lambda_m^{-N}\int_{{{\Bbb R}} ^N}|u_m|^p{\rm d}x = 1, $

$ \int_{{{\Bbb R}} ^N}\int_{{{\Bbb R}} ^N}\frac{|v_m(x)-v_m(y)|^p}{|x-y|^{N+sp}}{\rm d}x{\rm d}y = \mu_m^p\lambda_m^{sp-N}\int_{{{\Bbb R}} ^N}\int_{{{\Bbb R}} ^N}\frac{|u_m(x)-u_m(y)|^p}{|x-y|^{N+sp}}{\rm d}x{\rm d}y = 1, $

$ \int_{{{\Bbb R}} ^N}|v_m(x)|^{p+\frac{sp^2}{N}}{\rm d}x = \mu_m^{p+\frac{sp^2}{N}}\lambda_m^{-N}\int_{{{\Bbb R}} ^N}|u_m|^{p+\frac{sp^2}{N}}{\rm d}x = \frac{|u_m|_p^{-\frac{sp^2}{N}}}{[u_m]_{W^{s, p}}^{p}}|u_m|_{p+\frac{sp^2}{N}}^{p+\frac{sp^2}{N}}, $

$ J_{s, p}(v_m) = \frac{|u_m|_p^{\frac{sp^2}{N}}[u_m]_{W^{s, p}}^{p}}{|u_m|_{p+\frac{sp^2}{N}}^{p+\frac{sp^2}{N}}} = J_{s, q}(u_m). $

上面公式显示

(2.3) $ \begin{equation} \alpha = \lim\limits_{ m \rightarrow\infty}J_{s, p}(u_m) = J_{s, q}(v_m) = \lim\limits_{ m \rightarrow\infty}\inf\limits_{u\in W^{s, p}({{\Bbb R}} ^N)}J_{s, p}(u). \end{equation} $

利用文献[6]中关于$ \int_{{{\Bbb R}} ^N}\int_{{{\Bbb R}} ^N}\frac{|u(x)-u(y)|^p}{|x-y|^{N+sp}}{\rm d}x{\rm d}y $的对称递减重排不等式, 记$ v_m $的对称递减重排为$ v_m^* $.则$ v_m^* $满足

(ⅰ) $ v_m^*\geq 0 $, $ \forall x\in {{\Bbb R}} ^N $;

(ⅱ) $ v_m^* $是径向对称函数;

(ⅲ) 对任意$ r\in [1, \infty] $, 若$ v_m\in L^r({{\Bbb R}} ^N) $, 则$ |v_m^*|_r = |v_m|_r $;

(ⅳ) 若$ v_m\in W^{s, p}({{\Bbb R}} ^N) $, 则

$ \int_{{{\Bbb R}} ^N}\int_{{{\Bbb R}} ^N}\frac{|v_m(x)-v_m(y)|^p}{|x-y|^{N+sp}}{\rm d}x{\rm d}y\geq \int_{{{\Bbb R}} ^N}\int_{{{\Bbb R}} ^N}\frac{|v_m^*(x)-v_m^*(y)|^p}{|x-y|^{N+sp}}{\rm d}x{\rm d}y. $

进一步可得

$ J_{s, p}(v_m^*)\leq J_{s, p}(v_m). $

这意味着序列$ \{v_m^*\} $也是一个极小化序列且$ v_m^* = v_m^*(|x|) $关于$ |x| $是径向单调递减的.类似文献[12], 可得

(2.4) $ \begin{equation} |v_m^*|\leq |x|^{-\frac{N}{p}}, \end{equation} $

$ |v_m^*|_p = 1 $.利用条件(ⅲ)和(ⅳ), 可得

$ |v_m^*|_p = |v_m|_p = 1, \ \int_{{{\Bbb R}} ^N}\int_{{{\Bbb R}} ^N}\frac{|v_m^*(x)-v_m^*(y)|^p}{|x-y|^{N+sp}}{\rm d}x{\rm d}y\leq 1, $

即$ \{v_m^*\} $在空间$ W^{s, p}({{\Bbb R}} ^N) $中有界.则存在序列$ \{v_m^*\} $的一个子列, 仍记作$ \{v_m^*\} $, 存在$ v\in W^{s, p}({{\Bbb R}} ^N) $, 使得

$ v_m^*\rightharpoonup v, \ \mbox{in } W^{s, p}({{\Bbb R}} ^N);\ \ v_m^*\rightharpoonup v, \ \mbox{in } L^{t}({{\Bbb R}} ^N), \ p<t<p^*. $

由(2.4)式, 可进一步推导得$ v_m^*\rightarrow v, \ \mbox{in } L^{t}({{\Bbb R}} ^N), \ p<t<p^*. $即

$ \alpha\leq J_{s, p}(v)\leq \frac{1}{|v|_{p+\frac{sp^2}{N}}^{p+\frac{sp^2}{N}}} = \lim\limits_{m\rightarrow\infty}J_{s, p}(v_m^*) = \alpha. $

从而$ |v|_p = [v]_{W_{s, p}}^p = 1 $且在空间$ W^{s, p}({{\Bbb R}} ^N) $上$ v_m^*\rightarrow v $.进一步可知函数$ v $满足如下Euler-Lagrange方程

$ \frac{{\rm d}J_{s, p}(v+t\eta)}{{\rm d}t}\Big|_{t = 0} = 0, \mbox{ for all } \eta \in C_0^\infty({{\Bbb R}} ^N). $

代入$ \alpha = \frac{1}{|v|_{p+\frac{sp^2}{N}}^{p+\frac{sp^2}{N}}} $, $ |v|_p = |v| = 1 $, 有

$ (-\Delta)_p^sv+\frac{sp}{N}|v|^{p-2}v = \frac{\alpha (N+sp)}{N}|v|^{{p+\frac{sp^2}{N}}-2}v, \ \ \mbox{ in } {{\Bbb R}} ^N. $

令$ v = \left(\frac{N}{\alpha (N+sp)}\right)^{\frac{1}{\frac{sp^2}{N}}}\varphi $, 可得函数$ \varphi $满足方程

(2.5) $ \begin{equation} (-\Delta)_p^s\varphi+\frac{sp}{N}|\varphi|^{p-2}\varphi = |\varphi|^{q-2}\varphi, \ \ \mbox{ in } {{\Bbb R}} ^N. \end{equation} $

假设函数$ Q(x) $是方程(2.5)的一个基态解, 则

$ \alpha = \frac{N}{N+sp}\left(\int_{{{\Bbb R}} ^N}|Q(x)|^p{\rm d}x\right)^{\frac{sp}{N}}. $

进一步, 利用Pohozaev恒等式和方程(2.5), 通过计算可得

(2.6) $ \begin{equation} [Q]_{W^{s, p}}^p = \int_{{{\Bbb R}} ^N}|Q|^p{\rm d}x = \frac{N}{N+sp}\int_{{{\Bbb R}} ^N}|Q|^{p+\frac{sp^2}{N}}{\rm d}x. \end{equation} $

引理2.2得证.

引理2.3   设$ 0\leq \beta<\beta^* $, $ u_\beta $是极小化问题(1.2)的解, 则存在常数$ \gamma>0 $, 使得

$ \gamma(\beta^*-\beta)^{\frac{2}{2+sp}}\leq m(\beta)\leq C(s, p, N)(\beta^*-\beta)^{\frac{2}{sp+2}}, $

其中$ C(s, p, N) = \frac{1}{p\beta^*\frac{N+2}{sp+2}} \left[\big(\frac{2}{sp}\big)^{\frac{sp}{sp+2}}+\big(\frac{sp}{2}\big)^{\frac{sp}{sp+2}}\right]\left(\int_{{{\Bbb R}} ^N} |x|^2|Q|^p{\rm d}x\right)^{\frac{sp}{sp+2}} $.

证   由引理2.2中的Gagliardo-Nirenberg不等式结合Young不等式可知

$ \begin{eqnarray*} m(\beta) = E_\beta(u_\beta)&\geq &\frac{N(\beta^*-\beta)}{p(N+sp)}\int_{{{\Bbb R}} ^N} |u_\beta|^{p+\frac{sp^2}{N}}{\rm d}x+\frac{1}{p}\int_{{{\Bbb R}} ^N} |x|^2|u_\beta|^p{\rm d}x\\ &\geq& \frac{c}{p}+\frac{N(\beta^*-\beta)}{p(N+sp)}\int_{{{\Bbb R}} ^N} |u_\beta|^{p+\frac{sp^2}{N}}{\rm d}x-\frac{1}{p}\int_{{{\Bbb R}} ^N} (c-|x|^2)_+|u_\beta|^p{\rm d}x\\ &\geq &\frac{c}{p}-\frac{sp}{p(N+sp)(\beta^*-\beta)^{\frac{N}{sp}}}\int_{{{\Bbb R}} ^N} (c-|x|^2)_+^{\frac{N+sp}{sp}}{\rm d}x\\ &\geq &\frac{c}{p}-Cc^{\frac{N+sp}{sp}+\frac{N}{2}}. \end{eqnarray*} $

设$ \alpha>0 $充分小, 取$ c = \alpha(\beta^*-\beta)^{\frac{2}{2+sp}} $, 则由上述表达式可知存在$ \gamma>0 $, 使得

(2.7) $ \begin{equation} m(\beta)\geq \gamma(\beta^*-\beta)^{\frac{2}{2+sp}}. \end{equation} $

设$ u_{\lambda}(x) = \frac{\lambda^{\frac{N}{p}}}{|Q|_{p}}Q(\lambda x) $, 则

$ \int_{{{\Bbb R}} ^N} |x|^2|u_\lambda|^p{\rm d}x = \lambda^{-2}(\beta^*)^{-\frac{N}{sp}}\int_{{{\Bbb R}} ^N} |x|^2|Q|^p{\rm d}x, $

$ E_\beta(u_\lambda) = \left(1-\frac{\beta}{\beta^*}\right)\frac{1}{p}\lambda^{sp}+\frac{1}{p}\lambda^{-2}(\beta^*)^{-\frac{N}{sp}}\int_{{{\Bbb R}} ^N} |x|^2|Q|^p{\rm d}x. $

令$ \lambda = \Big( \frac{2(\beta^*)^{1-\frac{N}{sp}}}{sp(\beta^*-\beta)}\int_{{{\Bbb R}} ^N} |x|^2|Q|^p{\rm d}x\Big)^{\frac{1}{sp+2}} $, 可得

$ \begin{eqnarray*} m(\beta)\leq E_\beta(u_\lambda)& = &\frac{(\beta^*-\beta)^{\frac{2}{sp+2}}}{p\beta^*\frac{N+2}{sp+2}} \left[\left(\frac{2}{sp}\right)^{\frac{sp}{sp+2}}+\left(\frac{sp}{2}\right)^{\frac{sp}{sp+2}}\right]\left(\int_{{{\Bbb R}} ^N} |x|^2|Q|^p{\rm d}x\right)^{\frac{sp}{sp+2}}\\ & = &C(s, p, N)(\beta^*-\beta)^{\frac{2}{sp+2}}, \end{eqnarray*} $

其中$ C(s, p, N) = \frac{1}{p\beta^*\frac{N+2}{sp+2}} \left[\big(\frac{2}{sp}\big)^{\frac{sp}{sp+2}}+\big(\frac{sp}{2}\big)^{\frac{sp}{sp+2}}\right]\left(\int_{{{\Bbb R}} ^N} |x|^2|Q|^p{\rm d}x\right)^{\frac{sp}{sp+2}} $.引理2.3得证.

引理2.4   若$ u_\beta $是极小化问题(1.2)的一个极小解, 则存在一个与$ \beta $无关的正常数$ A $, 使得当$ \beta\rightarrow\beta^* $时有

$ 0<\frac{1}{A}(\beta^*-\beta)^{-\frac{sp}{sp+2}}\leq \int_{{{\Bbb R}} ^N} |u_\beta|^{p+\frac{sp^2}{N}}{\rm d}x\leq A (\beta^*-\beta)^{-\frac{sp}{sp+2}}. $

证   利用引理2.2中的Gagliardo-Nirenberg不等式有

$ m(\beta) = E_\beta(u_\beta)\geq \frac{N(\beta^*-\beta)}{p(N+sp)}\int_{{{\Bbb R}} ^N} |u_\beta|^{p+\frac{sp^2}{N}}{\rm d}x. $

上式结合引理2.3可得

$ \int_{{{\Bbb R}} ^N} |u_\beta|^{p+\frac{sp^2}{N}}{\rm d}x\leq \frac{p(N+sp)}{N(\beta^*-\beta)}C(s, p, N)(\beta^*-\beta)^{\frac{2}{sp+2}} = A(\beta^*-\beta)^{-\frac{sp}{sp+2}}, $

其中$ A = \frac{p(N+sp)}{N}C(s, p, N) $.现假设$ 0<\delta<\beta $, 可得

$ \begin{eqnarray*} m(\delta)\leq E_\delta(u_\beta)& = &\frac{1}{p}[u_\beta]_{W^{s, p}}^p+\frac{1}{p}\int_{{{\Bbb R}} ^N} |x|^2|u_\beta|^p{\rm d}x-\frac{\delta N}{p(N+sp)}\int_{{{\Bbb R}} ^N} |u_\beta|^{p+\frac{sp^2}{N}}{\rm d}x\\ & = &E_\beta(u_\beta)+\frac{ N(\beta-\delta)}{p(N+sp)}\int_{{{\Bbb R}} ^N} |u_\beta|^{p+\frac{sp^2}{N}}{\rm d}x\\ & = &m(\beta)+\frac{ N(\beta-\delta)}{p(N+sp)}\int_{{{\Bbb R}} ^N} |u_\beta|^{p+\frac{sp^2}{N}}{\rm d}x, \end{eqnarray*} $

即

$ \frac{m(\delta)-m(\beta)}{\beta-\delta}\leq \frac{ N}{p(N+sp)}\int_{{{\Bbb R}} ^N} |u_\beta|^{p+\frac{sp^2}{N}}{\rm d}x. $

结合引理2.3可得

$ \frac{ N}{p(N+sp)}\int_{{{\Bbb R}} ^N} |u_\beta|^{p+\frac{sp^2}{N}}{\rm d}x\geq \frac{m(\delta)-m(\beta)}{\beta-\delta}\geq (\beta^*-\beta)^{-\frac{sp}{sp+2}}\frac{\gamma(1+\theta)^{\frac{2}{sp+2}}-C(s, p, N)}{\theta}, $

其中$ \theta = \frac{\beta-\delta}{\beta^*-\beta} $.当$ \beta $充分靠近$ \beta^* $时, $ \theta $足够大, 使得$ \frac{\gamma(1+\theta)^{\frac{2}{sp+2}}-C(s, p, N)}{\theta}>0 $.引理2.4得证.

定理1.1的证明   对任意的$ u\in M $, 由引理2.2中的Gagliardo-Nirenberg不等式有

$ \begin{eqnarray*} E_\beta(u)& = &\frac{1}{p}[u]_{W^{s, p}}^p+\frac{1}{p}\int_{{{\Bbb R}} ^N} |x|^2|u|^p{\rm d}x-\frac{\beta}{p+\frac{sp^2}{N}}\int_{{{\Bbb R}} ^N} |u|^{p+\frac{sp^2}{N}}{\rm d}x\\ &\geq & \frac{1}{p}[u]_{W^{s, p}}^p+\frac{1}{p}\int_{{{\Bbb R}} ^N} |x|^2|u|^p{\rm d}x-\frac{\beta}{p+\frac{sp^2}{N}}\frac{N+sp}{N|Q|_p^{\frac{sp^2}{N}}}[u]_{W^{s, p}}^p\\ & = &\bigg(1-\frac{\beta}{|Q|_p^{\frac{sp^2}{N}}}\bigg)\frac{1}{p}[u]_{W^{s, p}}^p+\frac{1}{p}\int_{{{\Bbb R}} ^N} |x|^2|u|^p{\rm d}x. \end{eqnarray*} $

由$ \beta^* $的定义有

(3.1) $ \begin{equation} E_\beta(u)\geq \left(1-\frac{\beta}{\beta^*}\right)\frac{1}{p}[u]_{W^{s, p}}^p+\frac{1}{p}\int_{{{\Bbb R}} ^N} |x|^2|u|^p{\rm d}x. \end{equation} $

(1) 当$ \beta<\beta^* $时.由(3.1)式知$ E_\beta(u)\geq 0 $, 因此泛函$ E_\beta(u) $有下界.

设$ \{u_n\}\subset M $是问题(1.2)的一个非负极小化序列, 由定义有

$ \{u_n\}\subset{\Bbb H}, \ \ \ \int_{{{\Bbb R}} ^N} |u_n|^p{\rm d}x = 1, \ \ \ \lim\limits_{n\rightarrow \infty}E_\beta(u_n) = m(\beta). $

结合(3.1)式可得

$ 0(1)+m(\beta) = E_\beta(u_n)\geq \left(1-\frac{\beta}{\beta^*}\right)\frac{1}{p}[u_n]_{W^{s, p}}^p+\frac{1}{p}\int_{{{\Bbb R}} ^N} |x|^2|u_n|^p{\rm d}x. $

由此可知

$ [u_n]_{W^{s, p}}^p\leq C, \ \ \ \int_{{{\Bbb R}} ^N} |x|^2|u_n|^p{\rm d}x\leq C. $

由引理2.1, 存在序列$ \{u_n\} $的子序列, 仍记作$ \{u_n\} $, 存在$ u\in {\Bbb H} $, 使得在空间$ {\Bbb H} $中$ u_n\rightharpoonup u $, 在空间$ L^q({{\Bbb R}} ^N) (p\leq q<p^*) $中$ u_n\rightarrow u $.因此

$ \int_{{{\Bbb R}} ^N} |u|^p{\rm d}x = \lim\limits_{n\rightarrow \infty}\int_{{{\Bbb R}} ^N} |u_n|^p{\rm d}x = 1, $

进一步可知$ u\geq 0 $.而后由范数的弱下半连续性可知

$ E_\beta(u) = m(\beta), $

即$ u\in {\Bbb H} $为$ m(\beta) $的极小解.

(2) 当$ \beta>\beta^* $时.对常数$ \lambda $, 设

(3.2) $ \begin{equation} u_{\lambda}(x) = \frac{\lambda^{\frac{N}{p}}}{|Q|_{p}}Q(\lambda x). \end{equation} $

通过计算并利用(2.6)式可知

$ \int_{{{\Bbb R}} ^N} |u_\lambda|^p{\rm d}x = 1, \ \ \ \int_{{{\Bbb R}} ^N} |u_\lambda|^{p+\frac{sp^2}{N}}{\rm d}x = \frac{(N+sp)\lambda^{sp}}{N|Q|_p^{\frac{sp^2}{N}}}, $

$ [u_\lambda]_{W^{s, p}}^p = \lambda^{sp}, \ \ \ \int_{{{\Bbb R}} ^N} |x|^2|u_\lambda|^p{\rm d}x = \frac{1}{\lambda^2|Q|_p^p}\int_{{{\Bbb R}} ^N} |x|^2|Q|^p{\rm d}x. $

进一步

$ m(\beta)\leq \lim\limits_{\lambda\rightarrow \infty}E_\beta(u_\lambda) = \lim\limits_{\lambda\rightarrow \infty}\left[\left(1-\frac{\beta}{\beta^*}\right)\frac{1}{p}\lambda^{sp}+\frac{1}{\lambda^2|Q|_p^p}\int_{{{\Bbb R}} ^N} |x|^2|Q|^p{\rm d}x\right] = -\infty. $

这意味着当$ \beta>\beta^* $时问题(1.2)不存在极小解.

若$ \beta = \beta^* $.此时

$ m(\beta^*)\leq \lim\limits_{\lambda\rightarrow \infty}E_\beta(u_\lambda) = \lim\limits _{\lambda\rightarrow \infty}\frac{1}{\lambda^2|Q|_p^p}\int_{{{\Bbb R}} ^N} |x|^2|Q|^p{\rm d}x = 0. $

而由(3.1)式有$ E_\beta(u)\geq \int_{{{\Bbb R}} ^N} |x|^2|u|^p{\rm d}x\geq 0, $即$ m(\beta^*)\geq 0 $.因此$ m(\beta^*) = 0 $.此时若$ m(\beta^*) $存在极小元$ u_1\in M $, 不妨设$ u_1 $是非负的, 有

$ 0 = m(\beta^*)\geq \int_{{{\Bbb R}} ^N} |x|^2|u_1|^p{\rm d}x\geq 0, $

即

(3.3) $ \begin{equation} \int_{{{\Bbb R}} ^N} |x|^2|u_1|^p{\rm d}x = 0. \end{equation} $

故$ u_1 = 0, $ a.e. $ x\in {{\Bbb R}} ^N $, 这与$ u_1\in M $矛盾.因此当$ \beta = \beta^* $时极小化问题(1.2)不存在极小解.

(3) 的证明.由(3.1)式可知当$ \beta<\beta^* $时, $ m(\beta)\geq 0 $, 进而可得$ m(\beta)>0 $.类似于(2)中(3.2)式的取法并令$ \lambda = (\beta^*-\beta)^{-\frac{1}{sp+1}} $, 可得当$ \beta\rightarrow \beta^* $时, 有

$ E_\beta(u_\lambda) = \frac{(\beta^*-\beta)^{\frac{1}{sp+1}}}{p\beta^*}+\frac{(\beta^*-\beta)^{\frac{2}{sp+1}}}{|Q|_p^p}\int_{{{\Bbb R}} ^N} |x|^2|Q|^p{\rm d}x\rightarrow 0. $

因此可得$ \lim\limits_{\beta\rightarrow \beta^*}m(\beta) = m(\beta^*) = 0. $

定理1.2的证明   设$ u_\beta $是极小化问题(1.2)的极小解, 则

$ m(\beta) = E_\beta(u_\beta)\geq \frac{\beta^*-\beta}{\beta^*p}[u_\beta]_{W^{s, p}}^p+\frac{1}{p}\int_{{{\Bbb R}} ^N} |x|^2|u_\beta|^p{\rm d}x. $

设$ \varepsilon_\beta = (\beta^*-\beta)^{\frac{1}{sp+2}} $, 因此可得

$ [u_\beta]_{W^{s, p}}^p\leq \frac{\beta^*p}{\beta^*-\beta}m(\beta)\leq \frac{\beta^*p}{\beta^*-\beta}C(s, p, N)(\beta^*-\beta)^{\frac{2}{sp+2}} = p\beta^*C(s, p, N)\varepsilon_\beta^{-sp}, $

$ \int_{{{\Bbb R}} ^N} |x|^2|u_\beta|^p{\rm d}x\leq pm(\beta)\leq pC(s, p, N)\varepsilon_\beta^{2}. $

设$ v_\beta(x) = \varepsilon_\beta^{\frac{N}{p}} u_\beta(\varepsilon_\beta x) $, 通过计算可得

(3.4) $ \begin{equation} [v_\beta]_{W^{s, p}}^p = \varepsilon^{sp}[u_\beta]_{W^{s, p}}^p\leq \varepsilon^{sp} p\beta^*C(s, p, N)\varepsilon_\beta^{-sp} = p\beta^*C(s, p, N), \end{equation} $

(3.5) $ \begin{equation} \int_{{{\Bbb R}} ^N} |x|^2|v_\beta|^p{\rm d}x = \varepsilon_\beta^{-2}\int_{{{\Bbb R}} ^N} |x|^2|u_\beta|^p{\rm d}x\leq pC(s, p, N). \end{equation} $

再由引理2.4可得

$ 0\leq \frac{1}{A}\leq \int_{{{\Bbb R}} ^N} |v_\beta|^{p+\frac{sp^2}{N}}{\rm d}x = \varepsilon_\beta^{sp}\int_{{{\Bbb R}} ^N} |u_\beta|^{p+\frac{sp^2}{N}}{\rm d}x\leq A. $

取序列$ \{\beta_n\} $, 且当$ n\rightarrow \infty $时$ \beta_n\rightarrow \beta^* $.由(3.4)和(3.5)式容易得序列$ \{v_{\beta_n}\} $在空间$ {\Bbb H} $中有界, 因此存在序列$ \{v_{\beta_n}\} $的一个子序列, 仍记作$ \{v_{\beta_n}\} $, 存在$ v\in {\Bbb H} $, 使得在空间$ {\Bbb H} $中$ v_{\beta_n}\rightharpoonup v $, 在空间$ W^{s, p}({{\Bbb R}} ^N) $中$ v_{\beta_n}\rightharpoonup v $, 在空间$ L^q({{\Bbb R}} ^N)(p\leq q<p^*) $中$ v_{\beta_n}\rightarrow v $.

另一方面, 由于$ u_{\beta_n} $是极小化问题(1.2)的非负极小解, 从而$ u_{\beta_n} $满足方程

(3.6) $ \begin{equation} (-\Delta)_p^su_{\beta_n}+|x|^2|u_{\beta_n}|^{p-2}u_{\beta_n}-\beta_n|u_{\beta_n}|^{p+\frac{sp^2}{N}-2}u_{\beta_n} = \mu_{\beta_n}|u_{\beta_n}|^{p-2}u_{\beta_n}, \end{equation} $

其中Lagrange乘子

$ \mu_{\beta_n} = pm(\beta_n)-\frac{sp\beta_n}{N+sp}\int_{{{\Bbb R}} ^N} |u_{\beta_n}|^{p+\frac{sp^2}{N}}{\rm d}x. $

上式结合引理2.3和引理2.4可得存在一个常数$ \nu>0 $, 使得

$ \lim\limits_{n\rightarrow \infty}\varepsilon_{\beta_n}^{sp}\mu_{\beta_n} = -\nu. $

进一步, 由定义可知

$ \begin{eqnarray*} (-\Delta)_p^sv_{\beta_n}& = &2\lim\limits_{\delta\rightarrow 0}\int_{{{\Bbb R}} ^N\setminus B_\delta(x)}\frac{|v_{\beta_n}(x) -v_{\beta_n}(y)|^{p-2}(v_{\beta_n}(x)-v_{\beta_n}(y))}{|x-y|^{N+sp}}{\rm d}y\\ & = &\epsilon_{\beta_n}^{N+sp-\frac{N}{p}}(-\Delta)_p^su_{\beta_n}, \end{eqnarray*} $

因而函数$ v_{\beta_n} $满足方程

(3.7) $ \begin{equation} (-\Delta)_p^sv_{\beta_n}+\varepsilon_{\beta_n}^{sp}|\varepsilon_{\beta_n}x|^2|v_{\beta_n}|^{p-2}v_{\beta_n} -\beta_n|v_{\beta_n}|^{p+\frac{sp^2}{N}-2}v_{\beta_n} = \mu_{\beta_n}\varepsilon_{\beta_n}^{sp}|v_{\beta_n}|^{p-2}v_{\beta_n}. \end{equation} $

对每个$ \varphi \in W^{s, p}({{\Bbb R}} ^N) $, 定义一个泛函$ T:W^{s, p}({{\Bbb R}} ^N)\rightarrow (W^{s, p})' $为

$ \langle T(\varphi), \psi\rangle = \int_{{{\Bbb R}} ^N}\int_{{{\Bbb R}} ^N}\frac{|\varphi(x)-\varphi(y)|^{p-2}(\varphi(x)-\varphi(y))(\psi(x)-\psi(y))}{|x-y|^{N+sp}}{\rm d}x{\rm d}y. $

由方程(3.7)可知

$ \lim\limits_{n\rightarrow \infty}\langle T(v_{\beta_n}), v_{\beta_n}-v\rangle = 0. $

而由$ v_{\beta_n} $在空间$ W^{s, p}({{\Bbb R}} ^N) $中的弱收敛性可知

(3.8) $ \begin{equation} \lim\limits_{n\rightarrow \infty}\langle T(v_{\beta_n})-T(v), v_{\beta_n}-v\rangle = 0. \end{equation} $

进一步, 利用下述不等式:对任意的$ \xi, \eta\in {{\Bbb R}} ^N $, 有

$ (|\xi|^{p-2}\xi-|\eta|^{p-2}\eta)(\xi-\eta)\geq C_p|\xi-\eta|^p, \ \ p\geq 2, $

$ (|\xi|^{p-2}\xi-|\eta|^{p-2}\eta)(\xi-\eta)\geq \tilde{C}_p\frac{|\xi-\eta|^2}{(|\xi|+|\eta|)^{2-p}}, \ \ 1<p< 2, $

其中$ C_p, \tilde{C}_p $是仅依赖于$ p $的常数, 由(3.8)式有

$ \lim\limits_{n\rightarrow \infty}\int_{{{\Bbb R}} ^N}\int_{{{\Bbb R}} ^N}\frac{|v_{\beta_n}(x)-v_{\beta_n}(y)-v(x)+v(y)|^{p}}{|x-y|^{N+sp}}{\rm d}x{\rm d}y = 0. $

因此在空间$ W^{s, p}({{\Bbb R}} ^N) $中有$ v_{\beta_n} \rightarrow v $.且$ v $满足

(3.9) $ \begin{equation} (-\Delta)_p^sv-\beta^*v^{p+\frac{sp^2}{N}-1} = -\nu v^{p-1}. \end{equation} $

进而由文献[2]中极大值原理可得$ v>0 $.

由$ u_{\beta_n} $和$ v_{\beta_n} $的定义, 可以计算能量得

(3.10) $ \begin{eqnarray} m(\beta_n)& = &E_{\beta_n}(u_{\beta_n}) = \frac{1}{p}[u_{\beta_n}]_{W^{s, p}}^p+\frac{1}{p}\int_{{{\Bbb R}} ^N} |x|^2|u_{\beta_n}|^p{\rm d}x-\frac{ N\beta_n}{p(N+sp)}\int_{{{\Bbb R}} ^N} |u_{\beta_n}|^{p+\frac{sp^2}{N}}{\rm d}x\\ & = &\frac{1}{p\varepsilon_{\beta_n}^{sp}}[v_{\beta_n}]_{W^{s, p}}^p+\frac{\varepsilon_{\beta_n}^{2}}{p}\int_{{{\Bbb R}} ^N} |x|^2|v_{\beta_n}|^p{\rm d}x-\frac{ N\beta_n}{p(N+sp)\varepsilon_{\beta_n}^{sp}}\int_{{{\Bbb R}} ^N} |v_{\beta_n}|^{p+\frac{sp^2}{N}}{\rm d}x\\ &\geq& \frac{\varepsilon_{\beta_n}^{2}}{p}\int_{{{\Bbb R}} ^N} |x|^2|v_{\beta_n}|^p{\rm d}x+\frac{N\varepsilon_{\beta_n}^2}{p(N+sp)}\int_{{{\Bbb R}} ^N} |v_{\beta_n}|^{p+\frac{sp^2}{N}}{\rm d}x. \end{eqnarray} $

设

$ v = \left(\frac{N\nu}{sp\beta^*}\right)^{\frac{N}{sp^2}}\tilde{Q}\left(\left(\frac{N\nu}{sp}\right)^{\frac{1}{sp}}(x)\right). $

将$ v $代入方程(3.9)并计算可得$ \tilde{Q} $是方程(1.3)的解, 由于$ v $满足方程(3.9), 结合Pohozaev恒等式可知$ \tilde{Q} $是方程(1.3)的基态解.因此存在$ x_0\in {{\Bbb R}} ^N $, 使得$ \tilde{Q} = Q(x+x_0) $, 这意味着

(3.11) $ \begin{equation} v = \left(\frac{N\nu}{sp\beta^*}\right)^{\frac{N}{sp^2}}Q\left(\left(\frac{N\nu}{sp}\right)^{\frac{1}{sp}}(x+x_0)\right). \end{equation} $

由(3.11)和(2.6)式, 且$ Q $是径向对称函数, 可得

(3.12) $ \begin{eqnarray} &&\liminf\limits_{n\rightarrow \infty}\int_{{{\Bbb R}} ^N} |x|^2|v_{\beta_n}|^p{\rm d}x\geq\int_{{{\Bbb R}} ^N} |x|^2|v|^p{\rm d}x \end{eqnarray} $

(3.13) $ \begin{eqnarray} & = &(\beta^*)^{-\frac{N}{sp}}\left(\frac{N\nu}{sp}\right)^{-\frac{2}{sp}}\int_{{{\Bbb R}} ^N} |x-\left(\frac{N\nu}{sp}\right)^{\frac{1}{sp}}x_0|^2|Q|^p{\rm d}x\\ &\geq& (\beta^*)^{-\frac{N}{sp}}\left(\frac{N\nu}{sp}\right)^{-\frac{2}{sp}}\int_{{{\Bbb R}} ^N} |x|^2|Q|^p{\rm d}x, \end{eqnarray} $

(3.14) $ \begin{equation} \lim\limits_{n\rightarrow \infty}\int_{{{\Bbb R}} ^N} |v_{\beta_n}|^{p+\frac{sp^2}{N}}{\rm d}x = \int_{{{\Bbb R}} ^N} |v|^{p+\frac{sp^2}{N}}{\rm d}x = \frac{\nu(N+sp)}{sp\beta^*}. \end{equation} $

结合(3.10), (3.12)和(3.14)式可得

$ \begin{eqnarray*} \lim\limits_{n\rightarrow \infty}\frac{m(\beta_n)}{(\beta^*-\beta_n)^{-\frac{2}{sp+2}}}&\geq&\frac{1}{p}\left( \frac{\nu N}{sp\beta^*}+ (\beta^*)^{-\frac{N}{sp}}\left(\frac{N\nu}{sp}\right)^{-\frac{2}{sp}}\int_{{{\Bbb R}} ^N} |x|^2|Q|^p{\rm d}x\right)\\ &\geq& \frac{1}{(p\beta^*)^{\frac{2+N}{2+sp}}}\left[\left(\frac{2}{sp}\right)^{\frac{sp}{2+sp}}+ \left(\frac{sp}{2}\right)^{\frac{2}{2+sp}}\right]\left(\int_{{{\Bbb R}} ^N} |x|^2|Q|^p{\rm d}x\right)^{\frac{sp}{2+sp}}. \end{eqnarray*} $

上式能取等号当且仅当$ x_0 = 0 $且

$ \nu = \frac{2^\frac{sp}{2+sp}}{N}(sp)^{\frac{2}{2+sp}}(\beta^*)^{\frac{sp-N}{2+sp}}\left(\int_{{{\Bbb R}} ^N} |x|^2|Q|^p{\rm d}x\right)^{\frac{sp}{2+sp}}. $

从而由引理2.3可得

$ \lim\limits_{n\rightarrow \infty}\frac{m(\beta_n)}{(\beta^*-\beta_n)^{-\frac{2}{sp+2}}} = \frac{1}{(p\beta^*)^{\frac{2+N}{2+sp}}}\left[\left(\frac{2}{sp}\right)^{\frac{sp}{2+sp}}+ \left(\frac{sp}{2}\right)^{\frac{2}{2+sp}}\right]\left(\int_{{{\Bbb R}} ^N} |x|^2|Q|^p{\rm d}x\right)^{\frac{sp}{2+sp}}. $

设$ \mu = \left(\frac{2\int_{{{\Bbb R}} ^N} |x|^2|Q|^p{\rm d}x}{sp}\right)^{\frac{1}{2+sp}} $, 将$ \nu $和$ x_0 $代入表达式(3.11)可得

$ v = \left(\frac{\mu}{(\beta^*)^{\frac{2+N}{sp(2+sp)}}}\right)^{\frac{N}{p}}Q\left((\beta^*)^{\frac{sp-N}{sp(2+sp)}}\mu x\right). $

定理1.2的证毕.