数学杂志  2018, Vol. 38 Issue (5): 943-950   PDF    
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乔花玲
吴玉梅
一类半线性分数Laplacian方程多解的存在性问题
乔花玲, 吴玉梅    
西安财经学院统计学院, 陕西 西安 710061
摘要:本文研究了一类半线性分数Laplacian方程$\left\{\begin{array}{l@{\quad \qquad} r} (-\Delta)^{s}u=f(x,u), \quad x\in \Omega,\\ u=0,\qquad\qquad\qquad x\in R^{n}\setminus\Omega\end{array}\right.$在原点附近无穷多解的存在性问题.利用改进的Clark's定理,获得了方程对应的泛函有收敛于零的临界点序列的结果,推广了关于整数阶半线性方程多解的存在性结果.
关键词分数Laplacian算子    临界点    无穷多解    Clark's定理    
THE EXSITENCE OF MULTIPLE SOLUTIONS FOR A CLASS OF SEMILINEAR FRACTIONAL LAPLACIAN EQUATIONS
QIAO Hua-ling, WU Yu-mei    
School of Statistics, Xi'an University of Finance and Economics, Xi'an 710061, China
Abstract: In this paper, we study the existence of infinitely many solutions near the origin for a class semilinear fractional Laplacian equtions $\left\{\begin{array}{l@{\quad \qquad} r} (-\Delta)^{s}u=f(x,u), \quad x\in \Omega,\\ u=0,\qquad\qquad\qquad x\in R^{n}\setminus\Omega\end{array}\right.$ By improved Clark's theorem, we obtain the result that the corresponding functional of the equation has a critical sequence that converges to zero. The results of the existence of multiple solutions for integral order semilinear equations are generalized.
Key words: fractional Laplacian     critical     infinitely many solutions     Clark's theorem    
1 引言

分数Laplacian算子$(-\Delta)^s$的定义如下:

$ \widehat {{{( - \Delta )}^s}}u(\xi ) = |\xi {|^{2s}}\hat u(\xi ), $

其中^表示$u$的Fourier变换.如果$u$充分光滑, 它也可以是下面的奇异积分

$ {(-\Delta)^{s}u(x)={\large C_{ n, s}\, P.V.}\int_{R^{n}}\frac{u(x)-u(y)}{|x-y|^{n+2s}}\, dy={C_{ n, s}}\lim\limits_{\varepsilon\to 0^{+}}\int_{R^{n}\setminus B_{\varepsilon}(x)}\frac{u(x)-u(y)}{|x-y|^{n+2s}}\, dy, } $

其中$s\in(0, 1), P.V.$表示主值, ${\large C_{ n, s}}=\displaystyle\frac{\Gamma(\frac{n}{2}+s)}{\pi^{2s+\frac{n}{2}}\cdot\Gamma(-s)}$.有关分数Laplacian算子$(-\Delta)^s$的更详细的内容可以看参考文献[1].

分数Laplacian算子$(-\Delta)^s$及更一般的拟微分算子已经有经典的泛函分析方法研究成果, 该算子在连续介质力学、相变现象、种群动态、对策论的研究中常出现, 它是Lévy过程的随机稳定的无穷小生成元[2-4].由于分数空间和非局部方程在很多科学领域有着重要应用, 在过去的几年里对涉及分数算子问题的研究兴趣仍不断高涨, 如障碍问题[5]、优化与金融[6, 7]、共形几何与极小曲面[8-10]、材料学[11]、反常扩散[12-14]等方面应用.

$n\geq2, s\in(0, 1)$, 方程

$ \begin{equation} \left\{\begin{array}{l@{\quad \qquad} r} (-\Delta)^{s}u=f(x, u), \quad x\in \Omega, \\ u=0, \qquad\qquad\qquad x\in R^{n}\setminus\Omega\end{array}\right. \end{equation} $ (1.1)

已经得到广泛的关注.Rox-Oton和Serra[15]建立与(1.1)式相联系的Pohozaev恒等式, 并证明了在有界光滑的星型域上, 若非线性项$f(u)$是局部的Lipschitz函数, 有不等式

$ \begin{equation*} {\frac{n-2s}{2n}}uf(u)\geq\int_{0}^{u}f(t)dt, \;\;\quad\forall u\in R \end{equation*} $

成立, 则问题(1.1)没有有界正解, 当不等式严格成立时, 问题(1.1)没有非平凡的有界解; 若在$\bar{\Omega}\times R$的任意紧子集上$f(x, u)$是Lipschitz函数, 有不等式

$ \begin{equation*} {\frac{n-2s}{2}}uf(x, t)\geq nF(x, t)+F_{x}(x, t), \;\;\quad\forall x\in \Omega, t\in R \end{equation*} $

成立, 则问题(1.1)没有有界正解, 当不等式严格成立时, 问题(1.1)没有非平凡的有界解. Fall和Weth [16]给出了在有界星型域$\Omega(0\in\bar{\Omega})$上, 非线性项$f(x, t):\bar{\Omega}\setminus\{0\}\times [0, \infty)\rightarrow R$$\bar{\Omega}\setminus\{0\}$上的每一个子集上关于$t$是一致局部Lipschitz且在某种意义下是超临界的, 则问题(1.1)没有正解, 参见文献[16, 定理1.1].在无界的星型域上, 当$f(x, u)=u^{p}(1\leq p\leq \dfrac{n+2s}{n-2s})$时, 问题(1.1)没有非平凡解.如果$f(x, u)=|u|^{\frac{4s}{n-2s}}u$, Fang[17]研究了带有纯临界非线性项的Laplacian问题, 在空间$D^{s, 2}(R^{n})$上, 得出问题(1.1)有无穷多非径向变号解; Gonzalez[18]等讨论了该方程在双曲空间上层解的存在性、对称性; Servadei[19]等发现具有齐次Dirichlet边界条件非局部微积分算子对应的方程有变分结构, 利用山路定理证明了其非平凡解的存在性, 此结果对一般的分数微积分算子也成立, 作为其特殊情形证明了半线性椭圆问题(1.1)非线性项$f:\Omega\times R\rightarrow R$是Carathéodory函数时, 非平凡解的存在性; 当$s=\frac{1}{2}$时, Cabré[20]等研究了光滑有界域上正解的存在性, 对问题(1.1)非平凡解研究已经有比较完善的结果了, 关于问题(1.1)解的存在性等其它问题的许多结果请参考文献[21-26].对于其是否有无穷解的研究结果为数不多.最近, Liu等[27]研究了如果$f(x, u)$是关于$u$的奇函数且满足一定的增长条件, 则经典的Dirichlet边界值问题

$ \begin{equation} \left\{\begin{array}{l@{\quad \qquad} r} -\Delta u=f(x, u), \quad x\in \Omega, \\ u=0, \qquad\qquad\quad x\in \partial\Omega\end{array}\right. \end{equation} $ (1.2)

有无穷多解$u_{k}$, 当$k\rightarrow\infty$时, 有$\|u_{k}\|_{L^{\infty}}\rightarrow 0$.受此启发, 本文将研究半线性椭圆问题(1.1)多解的存在性.

$\Omega\subset R^{n}$是具有光滑边界的有界域, 记

$ \begin{align*} H^{s}(R^{n})&=\left\{u\in L^{2}(R^{n}):\frac{|u(x)-u(y)|}{|x-y|^\frac{n+2s}{2}}\in L^{2}(R^{n}\times R^{n})\right\}, \\ \|u\|_{H^{s}(R^{n})}&=\left(\int_{R^{2n}}\frac{|u(x)-u(y)|^{2}}{|x-y|^{n+2s}}\, dxdy+\int_{\Omega} |u|^{2}dx \right)^{\frac{1}{2}}. \end{align*} $

定义1.1 若对任意$\varphi\in C^{\infty}(R^{n})$, 有

$ \begin{equation*} \int_{R^{2n}}\frac{(u(x)-u(y))(\varphi(x)-\varphi(y))}{|x-y|^{n+2s}}dxdy=\int_{\Omega}f(x, u)\varphi(x)dx\quad u(x)\in H^{s}(R^{n}) \end{equation*} $

成立, 则称$u(x)$是问题(1.1)的弱解.

定义1.2 设$\Phi\in C^{1}(H^{s}(R^{n}) )$, 称$\Phi$满足PS条件是指:若任意序列$\{u_{k}\}^{\infty}_{k=1}\in H^{s}(R^{n})$满足下列条件

(1) $\{\Phi(u_{k})\}^{\infty}_{k=1}$有界,

(2) 在$H^{s}(R^{n})$$\Phi^{'}(u_{k})\rightarrow 0$有收敛序列.

主要结果是

定理1.1 若存在$\delta >0, $使得$f\in C(\bar{\Omega}\times (-\delta, \delta), R), f(x, 0)=0, f$关于$u$是奇函数, 而且存在一个球域$B_{r}(x_{0})\subset\Omega$使得对$x\in B_{r}(x_{0})$, 有

$ \begin{equation}\lim\limits_{|u|\rightarrow 0}\frac{F(x, u)}{|u|^{2}}=+\infty\end{equation} $ (1.3)

一致成立, 则问题(1.1)有无穷多解$u_{k}$, 且当$k\rightarrow\infty$时, $\|u_{k}\|_{L^{\infty}}\rightarrow 0, $其中

$ F(x, u)=\int^{u}_{0}f(x, t)dt, \;\;\quad B_{r}(x_{0})=\{x_{0}\in R^{n}|\, |x-x_{0}|< r \}, r>0. $

本文安排如下:第2部分, 验证PS条件; 第3部分, 利用Clark's定理证明定理1.1, 并作为推广, 给出方程组多解的存在性.

2 验证PS条件

为了证明主要结果, 考虑问题

$ \begin{equation} \left\{\begin{array}{l@{\quad \qquad} r} (-\Delta)^{s}u(x)=\hat{f}(x, u), \quad x\in \Omega\subset R^{n}, \\ u=0, \qquad \qquad\qquad\quad \;\;x\in R^{n}\setminus\Omega, \end{array}\right. \end{equation} $ (2.1)

其中$s\in(0, 1), \hat{f}\in C(\bar{\Omega}\times R, R), \hat{f}$关于$u$是奇函数,

$ \begin{equation*} \hat{f}(x, u)=\left\{\begin{array}{l@{\quad \qquad} r} f(x, u), \quad x\in \bar{\Omega}, |u|<\frac{\delta}{2}, \\ 0, \qquad\quad\; x\in \bar{\Omega}, |u|>\delta.\end{array}\right. \end{equation*} $

(2.1)式相应的泛函为

$ \Phi(u)=\frac{1}{2}\displaystyle\int_{R^{2n}}\frac{|u(x)-u(y)|^{2}}{|x-y|^{n+2s}}dxdy-\displaystyle\int_{\Omega} \hat{F}(x, u)\, dx, \;\;\quad u\in H^{s}(R^{n}), $

其中$\hat{F}(x, u)=\displaystyle\int^{u}_{0}\hat{f}(x, t)dt.$

引理2.1 $\Phi(u)\in C^{1}(H^{s}(R^{n}))$是偶的, 强制的且下有界.

 由假设$\hat{f}\in C(\bar{\Omega}\times R, R), \hat{f}$关于$u$是奇函数, $\hat{F}(x, u)=\displaystyle\int^{u}_{0}\hat{f}(x, t)dt$, 所以$\hat{F}(x, u)\in C^{1}(\bar{\Omega}\times R, R)$, $\hat{F}(x, u)$关于$u$是偶函数, 从而$\Phi(u)\in C^{1}(H^{s}(R^{n}))$是偶的.

由(1.3)式可得存在常数$C_{1}$, 使得$\displaystyle\int_{\Omega} \hat{F}(x, u)dx\leq C_{1}\displaystyle\int_{\Omega} |u|^{2}dx$, 根据空间$H^{s}(R^{n})$上范数的定义有

$ \displaystyle\int_{R^{2n}}\frac{|u(x)-u(y)|^{2}}{|x-y|^{n+2s}}\, dxdy=\|u\|^{2}_{H^{s}(R^{n})}-\displaystyle\int_{\Omega} |u|^{2}dx, $

因此有

$ \begin{eqnarray*} \Phi(u)&\geq &\frac{1}{2}\left(\|u\|^{2}_{H^{s}(R^{n})}-\displaystyle\int_{\Omega} |u|^{2}dx\right)-C_{1}\displaystyle\int_{\Omega}|u|^{2}dx\\ &=&\frac{1}{2}\|u\|^{2}_{H^{s}(R^{n})}-C_{2}\displaystyle\int_{\Omega}|u|^{2}dx =\frac{1}{2}\|u\|^{2}_{H^{s}(R^{n})}-C_{2}\|u\|^{2}_{L^{2}(\Omega)}, \end{eqnarray*} $

$\Phi$是强制的且下有界, 其中$C_{2}=C_{1}+\frac{1}{2}$是适当的常数.

引理2.2 若$\{u_{n}\}$$H^{s}(R^{n})$上弱收敛到$u$, 则在$L^{q}(R^{n})$$\{u_{n}\}$满足PS条件, 其中$2\leq q\leq 2^{*}_{s}=\dfrac{2n}{n-2s}$.

 由$\Phi^{'}$的定义有

$ \begin{eqnarray} \langle\Phi^{'}(u_{n}), u_{n}-u\rangle&=&\langle\Phi^{'}(u_{n})-\Phi^{'}(u), u_{n}-u\rangle=o(1), \;\;\quad n\rightarrow\infty, \end{eqnarray} $ (2.2)
$ \begin{eqnarray}\langle\Phi^{'}(u_{n})-\Phi^{'}(u), u_{n}-u\rangle&=&\displaystyle\int_{R^{2n}}\frac{|u_{n}(x)-u(x)-u_{n}(y)+u(y)|^{2}}{|x-y|^{n+2s}}dxdy\nonumber\\ &&-\displaystyle\int_{\Omega} \left[f(x, u_{n}(x))-f(x, u(x))\right]\left(u_{n}(x)-u(x)\right)dx. \end{eqnarray} $ (2.3)

由Hölder不等式有

$ \begin{eqnarray*}&&\displaystyle\int_{\Omega} \displaystyle|\left[f(x, u_{n}(x))-f(x, u(x))\right]\left(u_{n}(x)-u(x)\right)|dx \\ &\leq&\left[\displaystyle\int_{\Omega}|f(x, u_{n}(x))-f(x, u(x))|^{2}dx\right]^{\frac{1}{2}} \cdot\left[\displaystyle\int_{\Omega}\left|u_{n}(x)-u(x)\right|^{2}dx\right]^{\frac{1}{2}}.\nonumber \end{eqnarray*} $

$L^{q}(R^{n})$$u_{n}(x)\rightarrow u(x)$, 映射$t|\rightarrow f(x, t)$关于$t$连续, $t\in R$, 所以

$ f(x, u_{n}(x))\rightarrow f(x, u(x))\;\;\quad (n\rightarrow\infty), $

$\Omega$上a.e.成立.

由假设$f(x, u(x))$有界及控制收敛定理知

$ \displaystyle\int_{\Omega} |f(x, u_{n}(x))|dx\rightarrow\displaystyle\int_{\Omega} |f(x, u(x))|dx. $

再由引理2.1知$\Phi(u)$是强制的, 故序列$\{u_{n}\}$是有界的, 因此有

$ \begin{equation} \displaystyle\int_{\Omega} \left[f(x, u_{n}(x))-f(x, u(x))\right]\left(u_{n}(x)-u(x)\right)dx\rightarrow 0. \end{equation} $ (2.4)

联立(2.2)-(2.4)式可得

$ \begin{eqnarray*} &&\displaystyle\int_{R^{2n}}\frac{|u_{n}(x)-u(x)-u_{n}(y)+u(y)|^{2}}{|x-y|^{n+2s}}\, dxdy=o(1), \\ &&\|u_{n}(x)-u(x)\|^{2}_{H^{s}(R^{n})}\\&=&\displaystyle\int_{R^{2n}}\frac{|u_{n}(x)-u(x)-u_{n}(y)+u(y)|^{2}}{|x-y|^{n+2s}}dxdy +\displaystyle\int_{\Omega}\left|u_{n}(x)-u(x)\right|^{2}dx\nonumber =o(1).\nonumber \end{eqnarray*} $

因此$\Phi(u)$满足PS条件.

3 定理1.1的证明

为了证明定理1.1, 需要如下关键定理.

定理3.1 (见文献[27])设$X$是Banach空间, $\Phi\in C^{1}(X, R), \Phi$是偶泛函, 下有界且$\Phi(0)=0, $满足PS条件.如果对任意$k\in N$, 存在$X$$k$ -维子空间$X^{k}$$\rho_{k}>0$使得$\sup\limits_{\scriptscriptstyle X^{k}\bigcap S_{\rho_{k}}}\Phi<0$, 其中$S_{\rho}=\{u\in X|\;\|u\|=\rho\}$, 那么下面的结论至少有一个成立:

1) 存在临界点序列$\{u_{k}\}$满足对任意的$k$, 当$k\rightarrow\infty$时, $\|u_{k}\|\rightarrow 0, \Phi(u_{k})<0;$

2) 存在$r>0$使得对任意$0<a<r$, 存在临界点$u$使得$\|u\|=a, \Phi(u)=0$.

定理1.1的证明 由引理2.1和引理2.2可知$\Phi(u)\in C^{1}(H^{s}(R^{n}))$是偶泛函, 下有界, 满足PS条件.由定理$1$的假设$f(x, 0)=0$$\hat{F}(x, u)$的定义, 有$\Phi(0)=0$. $\forall K>0, \exists\, \delta=\delta(K)>0$, 使得如果$u\in C^{\infty}_{0}(B_{r}(x_{0})), |u|_{\infty}<\delta, $

$ \begin{eqnarray*} \hat{F}(x, u(x))&\geq& K|u(x)|^{2}, \\ \:\Phi(u)&=&\frac{1}{2}\displaystyle\int_{R^{2n}}\frac{|u(x)-u(y)|^{2}}{|x-y|^{n+2s}}\;dxdy-\displaystyle\int_{\Omega} \hat{F}(x, u(x))dx\nonumber\\ &\leq&\frac{1}{2}\displaystyle\int_{R^{2n}}\frac{|u(x)-u(y)|^{2}}{|x-y|^{n+2s}}\;dxdy-K\displaystyle\int_{\Omega}|u(x)|^{2}dx\nonumber\\ &\leq&\|u(x)\|^{2}_{H^{s}(R^{n})}-K\|u(x)\|^{2}_{L^{2}(\Omega)}.\nonumber \end{eqnarray*} $

这表明$\forall k\in N, $如果$X^{k}$$C^{\infty}_{0}(B_{r}(x_{0}))$$k$ -维子空间, $\rho_{k}>0$充分小, 则有$\sup\limits_{\scriptscriptstyle X^{k}\bigcap S_{\rho_{k}}}\Phi<0, $ $S_{\rho_{k}}=\{u(x)\in H^{s}(R^{n})|\;\|u(x)\|=\rho_{k}\}$, 由定理3.1知$\Phi$有非平凡临界点列$\{u_{k}(x)\}$满足$\forall\, k$$\Phi(u_{k}(x))< 0$, 当$k\rightarrow\infty$时, $\|u_{k}(x)\|\rightarrow 0$.

最后, 来证明当$k\rightarrow\infty$时, 有$\|u_{k}(x)\|_{L^{\infty}}\rightarrow 0 $, 即当$k$充分大时, $u_{k}(x)$也是问题(1.1)的解.记$2^{*}_{s}=\dfrac{2n}{n-2s}$, 假设$1<p<n$, 当$p\geq n$时, 可以类似证明结论成立.若$u$是(1.1)式的解, $\alpha>0$, 设$M>0$, 记$u^{M}(x)=\max\{-M, \min\{u(x), M\}\}$.给(2.1)式两边同乘以$|u^{M}|^{\alpha}u^{M}$可以得到

$ \frac{2^{2}}{(\alpha+2)^{2}}\displaystyle\int_{R^{n}}|\nabla|u^{M}|^{\frac{\alpha}{2}+1}|^{2}\leq C\displaystyle\int_{R^{n}}|u^{M}|^{\alpha+1}. $

再由分数Sobolev-Hardy不等式[28], 则有

$ \begin{equation} \|u^{M}\|_{L^{\frac{(\alpha+2)n}{n-2s}}(R^{n})}\leq C_{1}(\alpha+2)^{\frac{2}{\alpha+2}}\|u^{M}\|^{\frac{\alpha+1}{\alpha+2}}_{L^{\alpha+1}(R^{n})}, \end{equation} $ (3.1)

其中$C_{1}>1$是常数, 且与$u$$\alpha$无关.令$\alpha_{0}=2^{*}_{s}-1, \alpha_{k}=\dfrac{(\alpha_{k-1}+2)n}{n-2s}-1$, 即$\alpha_{k}=\dfrac{(2^{*}_{s}/{2})^{k+1}-1}{(2^{*}_{s}/{2})-1}\alpha_{0}, k=1, 2, 3, \cdots$, 对$\alpha_{k}$重复利用不等式(3.1)可得

$ {u^M}{_{{L^{{\alpha _{k + 1}} + 1}}({R^n})}} \le \exp (\sum\limits_{i = 0}^k {\frac{{2\ln ({C_1}({\alpha _i} + 2))}}{{{\alpha _i} + 2}}} ){u^M}_{{L^{2_s^*}}({R^n})}^{{\nu _k}}, $

其中$\nu_{k}=\prod\limits^{k}_{i=0}\frac{\alpha_{i}+1}{\alpha_{i}+2}$.令$M\rightarrow+\infty$, 则$k\rightarrow\infty$, 因此有下面的不等式成立

$ {\left\| {{u^M}} \right\|_{{L^\infty }({R^n})}} \le \exp (\sum\limits_{i = 0}^\infty {\frac{{2\ln ({C_1}({\alpha _i} + 2))}}{{{\alpha _i} + 2}}} )\left\| {{u^M}} \right\|_{{L^{2_s^*}}({R^n})}^\nu , $

其中$\nu=\prod\limits^{\infty}_{i=0}\frac{\alpha_{i}+1}{\alpha_{i}+2}\in (0, 1), \exp(\sum\limits^{\infty}_{i=0}\frac{2\ln(C_{1}(\alpha_{i}+2))}{\alpha_{i}+2})$是正数, 因此当$k\rightarrow\infty$时, $\|u_{k}(x)\|_{L^{\infty}}\rightarrow 0$, 结论得证.

推论3.1 若存在$\delta > 0$, 使得$F\in C^{1}(\bar{\Omega}\times B_{\delta}(0), R), F(x, 0)=0, F$关于$u$是偶函数, 而且存在一个球域$B_{r}(x_{0})\subset\Omega$, 使得对$x\in B_{r}(x_{0})$, 有

$ \begin{equation}\lim\limits_{|u|\rightarrow 0}\frac{F(x, u)}{|u|^{2}}=+\infty\end{equation} $ (3.2)

一致成立, 则方程组

$ \begin{equation} \left\{\begin{array}{l@{\quad \qquad} r} (-\Delta)^{s}u=F_{u}(x, u), \quad x\in \Omega, \\ u=0, \qquad x\in R^{n}\setminus\Omega\end{array}\right. \end{equation} $ (3.3)

有无穷多解$u_{k}$, 且当$k\rightarrow\infty$时, $\|u_{k}\|_{L^{\infty}}\rightarrow 0$, 其中$\Omega\subset R^{n}$是有光滑边界的有界域,

$ B_{r}(x_{0})=\left\{x\in R^{n}||x-x_{0}|< r\right\}, r>0, \;\; \quad B_{\delta}(0)=\left\{x\in R^{m}||x|< \delta\right\}, \delta>0, $

$u=(u_{1}, u_{2}, \cdots, u_{m})$$m$维向量函数, $(-\Delta)^{s}u=((-\Delta)^{s}u_{1}, (-\Delta)^{s}u_{2}, \cdots, (-\Delta)^{s}u_{m})$.

 考虑方程组

$ \begin{equation} \left\{\begin{array}{l@{\quad \qquad} r} (-\Delta)^{s}u=\hat{F}_{u}(x, u), \quad x\in \Omega, \\ u=0, \qquad x\in R^{n}\setminus\Omega, \end{array}\right. \end{equation} $ (3.4)

其中$s\in(0, 1), \hat{F}\in C^{1}(\bar{\Omega}\times R^{m}, R)$使得$\hat{F}$关于$u$是偶函数,

$ \begin{equation*} \hat{F}(x, u)=\left\{\begin{array}{l@{\quad \qquad} r} F(x, u), \quad x\in \bar{\Omega}, &|u|<\frac{\delta}{2}, \\ 0, \qquad\qquad x\in\bar{\Omega}, &|u|>\delta.\end{array}\right. \end{equation*} $

(3.4)式相应的泛函为

$ \Phi(u)=\frac{1}{2}\displaystyle\int_{R^{2n}}\frac{|u(x)-u(y)|^{2}}{|x-y|^{n+2s}}dxdy-\displaystyle\int_{\Omega}\hat{F}(x, u)dx, \;\;\quad u\in H^{s}(R^{n}, R^{m}), $

$\Phi (u) \in {C^1}({H^s}({R^n},{R^m})$是偶的, 强制的, 下有界, 且满足PS条件. $\forall k\in N$, 如果$X^{k}$$C^{\infty}_{0}(B_{r}(x_{0}), R^{m})$$k$ -维子空间, $\rho_{k}>0$充分小, 则有$\sup\limits_{\scriptscriptstyle X^{k}\bigcap S_{\rho_{k}}}\Phi<0, $ $S_{\rho}=\{u(x)\in H^{s}(R^{n}, R^{m})|\|u(x)\|=\rho\}$, 由定理$3.1$$\Phi$有非平凡临界点列$\{u_{k}(x)\}$满足对任意$k$$\Phi(u_{k})< 0$, 当$k\rightarrow \infty$时, $\|u_{k}(x)\|\rightarrow 0$.

与定理1.1的证明类似, 可以证明当$k\rightarrow\infty$时, 有$\|u_{k}(x)\|_{L^{\infty}}\rightarrow0$, 即对于$k$充分大时, $u_{k}(x)$仍然是方程组(3.3)的解.

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