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数学物理学报, 2018, 38(5): 893-902 doi:

论文

变号深阱位势分数阶Schrödinger方程非平凡解的存在性和集中性

王文波,1, 李全清,2

Existence and Concentration of Nontrivial Solutions for the Fractional Schrödinger Equations with Sign-Changing Steep Well Potential

Wang Wenbo,1, Li Quanqing,2

通讯作者: 李全清, E-mail: shili06171987@126.com

收稿日期: 2017-07-28  

基金资助: 云南省应用基础研究青年项目和红河学院科研基金博士专项项目.  XJ17B11

Received: 2017-07-28  

Fund supported: the Yunnan Province Applied Basic Research for Youths and Honghe University Doctoral Research Program.  XJ17B11

作者简介 About authors

王文波,E-mail:wenbowangmath@163.com , E-mail:wenbowangmath@163.com

摘要

考虑分数阶Schrödinger方程

\begin{equation}\renewcommand{\theequation}{$P_{\lambda}$} (-\Delta)^{s}u+\lambda V(x)u+V_{0}(x)u=P(x)|u|^{p-2}u+Q(x)|u|^{q-2}u, ~~~~~x\in {\Bbb R}^{N}\;\;\;\;\;\;\;\left( {{P_\lambda }} \right) \end{equation}

非平凡解的存在性和集中性,其中\lambda>0, s\in(0, 1), N>2s, 2<q<p<2_{s}^{\ast} (2_{s}^{\ast}=\frac{2N}{N-2s}), P\in L^{\infty}有正的下界, Q\in L^{\infty}可正可负或变号, V是深势阱位势, V_{0}\in L^{\infty}.\lambda充分大时,此方程存在非平凡解.进一步,如果V(x)\geq0,其解序列拥有某种集中现象.特别地,对于解的存在性, V允许变号.

关键词: 分数阶Schrödinger方程 ; 势阱位势 ; 变号位势 ; 集中性

Abstract

Consider the following fractional Schrödinger equation

\begin{equation}\renewcommand{\theequation}{$P_{\lambda}$} (-\Delta)^{s}u+\lambda V(x)u+V_{0}(x)u=P(x)|u|^{p-2}u+Q(x)|u|^{q-2}u, ~~~~~x\in {\Bbb R}^{N}, \end{equation}

where \lambda>0, s\in(0, 1), N>2s, 2<q<p<2_{s}^{\ast} (2_{s}^{\ast}=\frac{2N}{N-2s}), P\in L^{\infty} is positive, Q\in L^{\infty} may be positive, sign-changing or negative, V is steep well potential, and V_{0}\in L^{\infty}. When \lambda is large, the existence of nontrivial solutions is obtained via variational methods. Furthermore, if V(x)\geq0, concentration results are also obtained. In particular, the potential V is allowed to be sign-changing for the existence.

Keywords: Fractional Schrödinger equations ; Steep well potential ; Sign-changing potential ; Concentration

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本文引用格式

王文波, 李全清. 变号深阱位势分数阶Schrödinger方程非平凡解的存在性和集中性. 数学物理学报[J], 2018, 38(5): 893-902 doi:

Wang Wenbo, Li Quanqing. Existence and Concentration of Nontrivial Solutions for the Fractional Schrödinger Equations with Sign-Changing Steep Well Potential. Acta Mathematica Scientia[J], 2018, 38(5): 893-902 doi:

1 引言主和要结果

该文致力于研究如下的分数阶Schrödinger方程

\begin{equation} (-\Delta)^{s}u+\lambda V(x)u+V_{0}(x)u=P(x)|u|^{p-2}u+Q(x)|u|^{q-2}u, ~~~~~x\in {\Bbb R}^{N}, \end{equation}
(1.1)

其中\lambda是正参数, 0<s<1, N>2s, 2<q<p<2_{s}^{\ast} (2_{s}^{\ast}=\frac{2N}{N-2s}), (-\Delta)^{s}u是分数阶拉普拉斯算子.分数阶拉普拉斯可以理解为稳定的Lévy扩散过程的无穷维算子,其也出现在等离子体中的不规则的扩散,火焰的扩散,液体的化学反应,种群动态.更多的物理背景,请参看文献[20, 23].

最近,分数阶Schrödinger方程

\begin{equation}(-\Delta)^{s}u+ V(x)u= f(x, u), ~~~~~x\in {\Bbb R}^{N}, \end{equation}
(1.2)

引起了许多物理学家和数学家的兴趣.对于存在性,唯一性,正则性,渐进衰减性请参看文献[3, 7-8, 11, 14, 27, 28].另外,文献[2, 9, 26]研究了有界区域的分数阶方程.文献[9]研究了结点解.通过s -调和延拓,文献[8]把非局部问题转化为局部问题.

许多学者也在研究如下问题的半经典解

\begin{equation}\label{eq1.2}\varepsilon^{2s}(-\Delta)^{s}u+ V(x)u=f(x, u), ~~~~~x\in {\Bbb R}^{N}.\end{equation}
(1.3)

在量子力学中,当\varepsilon\rightarrow0,方程(1.3)的解的存在性和集中现象很重要,见文献[1, 6, 10, 13, 15-18, 28-29].如果f(x, u)=|u|^{p-2}u, \varepsilon^{2s}=\lambda^{-1}, v=\lambda ^{\frac{-1}{p-2}}u,方程(1.3)同如下方程等价

\begin{eqnarray*}(-\Delta)^{s}v+ \lambda V(x)v= f(x, v), ~~~~~x\in {\Bbb R}^{N}.\end{eqnarray*}

然而,对于一般的f (如f(x, u)=|u|^{p-2}u+|u|^{q-2}u, p\neq q),情况是不同的.受到上面提到文献的启发,结合文献[12],本文的主要目标是:研究当位势V变号, \lambda充分大时,问题(1.1)解的存在性和集中性.据我们所知,还没有文献研究方程(1.1), V变号,解的存在性和集中性.

为了陈述我们的结果,对V_{0}, V, PQ,我们做如下假设

(V_{0})V_{0}\in L^{\infty}({\Bbb R}^{N})并且V_{0}:=\inf\limits_{x\in{\Bbb R}^{N}} V_{0}(x)>0.

(V_{1})V \in C ({\Bbb R}^{N}, {\Bbb R})并且V下方有界.

(V_{2})存在b>0使得非空集合\{x\in{\Bbb R}^{N}~ :~ V(x)< b\}测度有限.

(V_{3})\Omega=int V^{-1}(0)非空,边界光滑并且\overline{\Omega}=V^{-1}(0).

(P)P(x)\in L^{\infty}({\Bbb R}^{N}), P_{0}:=\inf\limits_{x\in{\Bbb R}^{N}} P(x)>0.

(Q)Q(x)\in L^{\infty}({\Bbb R}^{N}), Q变号或是负的并且\{x\in{\Bbb R}^{N}~ :~ Q(x)\geq0\}测度有限.

定理1.1  假设(V_{0})-(V_{3}), (P)(Q)成立, 2<q<p<2_{s}^{\ast}并且V变号,则当\lambda充分大并且\|V_{0}\|_{L^{\infty}}\leq C(\lambda)时,问题(1.1)至少有一个非平凡解u_{\lambda}.

定理1.2  假设(V_{0})-(V_{3}), (P)(Q)成立, 2<q<p<2_{s}^{\ast}并且V(x)\geq0,则当\lambda充分大并且\|V_{0}\|_{L^{\infty}}\leq C(\lambda)时,问题(1.1)至少有一个非平凡解u_{\lambda}.此外,在H^{s}({\Bbb R}^{N})中, u_{\lambda}\mathop{\longrightarrow}\limits^{\lambda\rightarrow\infty} \bar{u},其中\bar{u}是如下方程的非平凡解

\begin{equation}\label{eq1.6}\left\{\begin{array}{ll} (-\Delta)^{s}u+V_{0}(x)u=P(x)|u|^{p-2}u+Q(x)|u|^{q-2}u, &x\in \Omega, \\ u=0, &x\in \partial\Omega. \end{array}\right.\end{equation}
(1.4)

注1.1  如果Q满足Q(x)\in L^{\infty}({\Bbb R}^{N}), Q_{0}:=\inf\limits_{x\in{\Bbb R}^{N}} Q(x)>0.定理1.1和1.2仍然成立.

位势V(x)满足(V_{1})-(V_{3})被文献[4]中的作者首次称为是深阱位势的.他们研究了非线性Schrödinger方程.我们也可参看文献[12, 32].在文献[19]中,作者研究了深阱位势的Schrödinger-Poisson方程.最近,文献[33]研究了带有变号深势阱位势的Schrödinger-Poisson方程.文献[20]和[34]研究了深阱位势的分数阶Schrödinger方程.但他们要求V是正的.本文结果不同于文献[33],我们需要重新建立分数阶Schrödinger方程的特征值问题.并且在研究过程中遇到的困难是空间的嵌入缺乏紧性.我们通过证明临界值c_{\lambda}关于\lambda一致有界去恢复紧性.我们需要更精细的估计(参看引理3.4).这篇文章最初的想法是研究当V变号,解的集中现象,然而失败了.我们想证明\{u_{\lambda}\}E中有界, V(x)\geq0这个条件在(4.1)式中似乎很重要(E的定义请看第2部分).

2 预备知识

记集合A的Lebesgue测度为|A|. |\cdot| _{p}L^{p}({\Bbb R}^{N} )的范数.首先,我们回顾(-\Delta)^{s}:{\cal S}\rightarrow L^{2}({{\Bbb R}^{N}})定义如下

(-\Delta)^{ s}u(x) = C_{N, s} P.V.\int_{{\Bbb R}^{ N}}\frac{u(x)-u(y)}{|x-y|^{N+2s}}{\rm d}y,

其中{\cal S}是速降C^{ \infty} Schwartz函数空间.

分数阶Sobolev空间H ^{s}({\Bbb R}^{N})定义为

H^{s}({\Bbb R}^{N})= \bigg\{u\in L^{2}({{\Bbb R}^{N}})~:~\frac{|u(x)-u(y)|^{2}}{|x-y|^{N+2s}}\in L^{1}({{\Bbb R}^{N}\times{\Bbb R}^{N}}) \bigg\}.

H ^{s}({\Bbb R}^{N})中定义内积

(u, v)_{H ^{s}}=\int_{{\Bbb R}^{N}}\bigg((-\Delta)^{\frac{s}{2}}u(-\Delta)^{\frac{s}{2}}v+uv\bigg){\rm d}x

和范数

\|u\|_{H ^{s}}^{2}=\int_{{\Bbb R}^{N}}\bigg(|(-\Delta)^{\frac{s}{2}}u|^{2}+u^{2}\bigg){\rm d}x,

其是一个Hilbert空间.

V^{\pm}(x)=\max\{\pm V(x), 0\},则V(x)=V^{+}(x)-V^{-}(x).

E=\bigg\{u\in H ^{s}({\Bbb R}^{N})~:~\int_{{\Bbb R}^{N}}V^{+}(x)u^{2}{\rm d}x<\infty \bigg\},

赋予内积和范数

(u, v)=\int_{{\Bbb R}^{N}}\bigg((-\Delta)^{\frac{s}{2}}u(-\Delta)^{\frac{s}{2}}v+V^{+}(x)uv\bigg){\rm d}x, ~~~~~\|u\|=(u, u)^{\frac{1}{2}}.

对于固定的\lambda,记

(u, v)_{\lambda}=\int_{{\Bbb R}^{N}}\bigg((-\Delta)^{\frac{s}{2}}u(-\Delta)^{\frac{s}{2}}v+\lambda V^{+}(x)uv\bigg){\rm d}x, ~~ \|u\|_{\lambda}=(u, u)_{\lambda}^{\frac{1}{2}}.

E_{\lambda}=(E, \|\cdot\|_{\lambda}).

引理2.1  当p\in[2, 2^{\ast}_{s}]时, E\hookrightarrow H^{s}({\Bbb R}^{N})\hookrightarrow L^{p}({\Bbb R}^{N})连续.此外,当p\in[2, 2^{\ast}_{s})时, E\hookrightarrow L^{p}({\Bbb R}^{N})是局部紧的.

  注意到(V_{2}), V^{+}(x)\not\equiv0.

A(R):=\{x\in{\Bbb R}^{N}~;~|x|>R, V^{+}(x)\geq b\},

B(R):=\{x\in{\Bbb R}^{N}~;~|x|>R, V^{+}(x)< b\}.

因此

\int_{A(R)}u^{2}{\rm d}x\leq\frac{1}{b}\int_{{\Bbb R}^{N}}u^{2}{\rm d}x.

由Hölder不等式和文献[30]中式子(4),结合 |B(R)|<\infty,我们得

\int_{B(R)}u^{2}{\rm d}x\leq C\int_{{\Bbb R}^{N}}|(-\Delta)^{\frac{s}{2}}u|^{2}{\rm d}x.

类似的,有

\int_{|x|\leq R}u^{2}{\rm d}x\leq C\int_{{\Bbb R}^{N}}|(-\Delta)^{\frac{s}{2}}u|^{2}{\rm d}x.

因此E\hookrightarrow H^{s}({\Bbb R}^{N}).众所周知H^{s}({\Bbb R}^{N})\hookrightarrow L^{p}({\Bbb R}^{N})连续.紧嵌入可由文献[5,引理3.2]得到.

易知问题(1.1)的能量泛函为

\begin{eqnarray}\label{eq2.1} \nonumber I_{\lambda}(u)&=&\frac{1}{2}\int_{{\Bbb R}^{N}}\bigg(|(-\Delta)^{\frac{s}{2}}u|^{2}+\lambda V(x)u^{2}+V_{0}(x)u^{2}\bigg){\rm d}x\\ &&-\frac{1}{p}\int_{{\Bbb R}^{N}}P(x)|u|^{p}{\rm d}x-\frac{1}{q}\int_{{\Bbb R}^{N}}Q(x)|u|^{q}{\rm d}x. \end{eqnarray}
(2.1)

显然, I_{\lambda}E上是C^{1}的. u \in H ^{s}({\Bbb R}^{N})是问题(1.1)的弱解当且仅当 I'_{\lambda}(u)=0.由于V变号,二次型

a_{\lambda}(u, u)=\int_{{\Bbb R}^{N}}\bigg(|(-\Delta)^{\frac{s}{2}}u|^{2}+\lambda V(x)u^{2}\bigg){\rm d}x

是不定的.沿用文献[33]的想法,记

D:=\{u\in E~:~{\rm supp}u \subset V^{-1}([0, \infty])\},

\begin{equation}\label{eq2.2} F=\overline{D}^{\|\cdot\|_{\lambda}}=\overline{\{u\in E~:~{\rm supp}u \subset V^{-1}([0, \infty])\}}^{\|\cdot\|_{\lambda}}, \end{equation}
(2.2)

FD在范数\|\cdot\|_{\lambda}的完备化.根据文献[24,定理12.4],存在闭子空间M使得E_{\lambda}=F\oplus M.如果V(x)\geq0,则E=F.我们可以直接跳到第4部分.

考虑双线性型

\begin{eqnarray*} a_{\lambda}(u, v)=\int_{{\Bbb R}^{N}}\bigg((-\Delta)^{\frac{s}{2}}u(-\Delta)^{\frac{s}{2}}v+\lambda V(x)uv\bigg){\rm d}x\end{eqnarray*}

及特征值问题

\begin{equation}\label{eq2.3} (-\Delta)^{s}u+\lambda V^{+}(x)u=\alpha\lambda V^{-}(x)u, ~~~u\in M.\end{equation}
(2.3)

A:={\rm supp}V^{-},由(V_{2}),得|A|<\infty.易证存在C>0,使得对任意的u\in M

\begin{equation}\label{eq2.4} \int_{{\Bbb R}^{N}}\bigg(|(-\Delta)^{\frac{s}{2}}u|^{2}+\lambda V^{+}(x)u^{2}\bigg){\rm d}x\geq C\int_{A}\lambda V^{-}(x)u^{2}{\rm d}x.\end{equation}
(2.4)

\begin{equation}\label{eq2.5} Q(u):=\int_{{\Bbb R}^{N}}\bigg(|(-\Delta)^{\frac{s}{2}}u|^{2}+\lambda V^{+}(x)u^{2}\bigg){\rm d}x. \end{equation}
(2.5)

引理2.2  \alpha_{1}(\lambda)=\inf\limits_{0\neq u\in M}\frac{Q(u)}{\int_{A}\lambda V^{-}(x)u^{2}{\rm d}x}>0可达并且\alpha_{1}(\lambda)是(2.3)式的最小特征值.

  证明类似于文献[22, p74].不同之处在于M不是全空间,但很幸运M是凸闭的.设\{u_{k}\}\subset M是极小化序列.由于M是凸闭的,存在u\in M使得在E_{\lambda}中, u_{k}\rightharpoonup u.结合引理2.1嵌入的紧性,类似于文献[22]的证明易知结论成立.

类似于文献[22, p80], \lambda>0固定,我们定义\alpha_{j}(\lambda) (j=2, 3, \cdots ).我们有如下性质.

性质2.1  每一特征值的特征函数空间是有限维的.

性质2.2  \alpha_{j}(\lambda)\mathop{\longrightarrow}\limits^{j\rightarrow\infty}\infty.

性质2.3  不同特征值对应的特征函数在E_{\lambda}中正交.

引理2.3  每一个固定的j, \alpha_{j}(\lambda)\mathop{\longrightarrow}\limits^{\lambda\rightarrow\infty}0.

  不妨设j\geq2.u_{i}\in M\alpha_{i}(\lambda)相应的特征函数(i=1, 2, \cdots , j-1).V_{j-1}={\rm span}\{u_{i}, i=1, 2, \cdots , j-1\}. V_{j-1}^{\perp}V_{j-1}L^{2}的正交补.设u\in C^{\infty}_{0}({\Bbb R}^{N})使得{\rm supp}u\subset {\rm supp}V^{-}, {\rm supp}u\cap {\rm supp}u_{i}=\emptyset, 1\leq i\leq j-1.这意味着

\alpha_{j}(\lambda)\leq\frac{\int_{{\Bbb R}^{N}}|\nabla u|^{2}}{\lambda\int_{{\Bbb R}^{N}} V^{-}(x)u^{2}}\rightarrow0, ~~ \lambda\rightarrow\infty.

引理2.3证毕.

根据引理2.3,存在\Lambda>0使得当\lambda>\Lambda

\hat{E}_{\lambda}:={\rm span}\{e_{j}, \alpha_{j}(\lambda)\leq1\}
(2.6)

非空并且a_{\lambda}(u, u)\hat{E}_{\lambda}上是负半定的,其中e_{j}\alpha_{j}(\lambda)相应的特征函数.记E^{+}_{\lambda}:={\rm span}\{e_{j}, \alpha_{j}(\lambda)>1\}.根据性质2.1和性质2.3, E_{\lambda}=M\oplus F=\hat{E}_{\lambda}\oplus E^{+}_{\lambda}\oplus F并且dim\hat{E}_{\lambda}<\infty.

3 定理1.1的证明

我们将证明I_{\lambda}满足环绕定理[31,定理2.12]中的环绕几何结构和(C)_{c}条件.

引理3.1  对于固定的\lambda,存在\rho_{\lambda}>0\kappa_{\lambda}>0使得当u\in E ^{+}_{\lambda}\oplus F, \|u\|_{\lambda}=\rho_{\lambda}时,有

I _{\lambda} (u)\geq \kappa _{\lambda}.

  由E^{+}_{\lambda}的定义,存在\delta_{\lambda}使得

a_{\lambda}(u, u) \geq \delta_{\lambda} \| u \|^{ 2}_{\lambda}, ~~\forall~~ u \in E^{+}_{\lambda},

a_{\lambda}(u, u)=\| u \|^{ 2}_{\lambda}, ~~\forall~~ u \in F.

因此,当u=v+w\in E^{+}_{\lambda}\oplus F时,利用Sobolev嵌入,我们得

\begin{eqnarray} \nonumber I_{\lambda}(u) &=& \frac{1}{2}a_{\lambda}(v, v)+\frac{1}{2}a_{\lambda}(w, w)+\frac{1}{2}\int_{{\Bbb R}^{N}}V_{0}(x)u^{2}{\rm d}x\\ \nonumber&&-\frac{1}{p}\int_{{\Bbb R}^{N}}P(x)|u|^{p}{\rm d}x-\frac{1}{q}\int_{{\Bbb R}^{N}}Q(x)|u|^{q}{\rm d}x\\ &\geq& \frac{1}{2}\min\{\delta_{\lambda}, 1\}\| u \|^{ 2}_{\lambda}-C_{1} \|u\|_{\lambda}^{p}-C_{2}\|u\|_{\lambda}^{q}.\end{eqnarray}
(3.1)

选取\varepsilon>0, \rho_{\lambda}>0\kappa_{\lambda}>0充分小易知结论成立.

(V_{3}),我们可以取e_{0}\in C_{0}^{\infty}(\Omega),则e_{0}\in F.

引理3.2  对于固定的\lambda,存在R_{\lambda}>0使得

\sup\limits _{u\in \partial Q}I _{\lambda} (u)<\kappa _{\lambda} ,

其中Q=\{u=v+te_{0}~:~v\in\hat{E}_{\lambda}, t\geq0, \|u\|_{\lambda}\leq R_{\lambda}\}.

  根据(P)(Q),存在C_{1}>0, C_{2}>0使得

\frac{1}{p}P(x)|u|^{p}+ \frac{1}{q}Q(x)|u|^{q}\geq C_{1}|u|^{p}-C_{2}|u|^{q}.

如果u=v+w\in\hat{E}_{\lambda}\oplus Re_{0},注意到

a_{\lambda}(u, u)=a_{\lambda}(v, v)+a_{\lambda}(w, w)\leq \|u\|_{\lambda}^{2},

结合有限维空间中所有范数等价和q<p,我们得

\begin{eqnarray*} I_{\lambda}(u)\leq C\|u\|_{\lambda}^{2}+C_{3}\|u\|_{\lambda}^{q}-C_{4}\|u\|_{\lambda}^{p}\mathop{ \longrightarrow}\limits^{\|u\|_{\lambda}\rightarrow\infty}-\infty. \end{eqnarray*}

因此,存在R_{\lambda}>0使得当u\in\hat{E}_{\lambda}\oplus Re_{0}并且\|u\|_{\lambda}=R_{\lambda}时, I_{\lambda}(u)\leq0,

如果u\in \hat{E}_{\lambda},则I_{\lambda}(u)\leq\frac{1}{2}\int_{{\Bbb R}^{N}}V_{0}(x)u^{2}{\rm d}x+C_{3}\|u\|_{\lambda}^{q}-C_{4}\|u\|_{\lambda}^{p},易知结论成立.

由引理3.1和引理3.2, I_{\lambda}存在(C)_{c}序列\{u_{n}\}\subset E_{\lambda},即

\begin{equation}\label{eq3.2} I_{\lambda}(u_{n})\rightarrow c, ~~~~~~~(1+\|u_{n}\|_{\lambda})I'_{\lambda}(u_{n})\rightarrow 0. \end{equation}
(3.2)

引理3.3  \{u_{n}\}E_{\lambda}中有界(依赖于\lambda).

  对充分大的n,有

\begin{eqnarray}\label{eq3.3} \nonumber I_{\lambda}(u_{n})-\frac{1}{q}\langle I'_{\lambda}(u_{n}), u_{n}\rangle&=&\bigg(\frac{1}{2}-\frac{1}{q}\bigg)\|u_{n}\|_{\lambda}^{2} -\bigg(\frac{1}{2}-\frac{1}{q}\bigg)\int_{{\Bbb R}^{N}}\lambda V^{-}(x)u_{n}^{2}{\rm d}x\\ \nonumber&&+\bigg(\frac{1}{2}-\frac{1}{q}\bigg)\int_{{\Bbb R}^{N}}V_{0}(x)u_{n}^{2}{\rm d}x+\bigg(\frac{1}{q}-\frac{1}{p}\bigg)\int_{{\Bbb R}^{N}}P(x)|u_{n}|^{p}{\rm d}x\\ &\leq&c+1.\end{eqnarray}
(3.3)

结合条件(V_{1})\|u_{n}\|_{\lambda}^{2}\leq C_{1}\int_{{\Bbb R}^{N}}u_{n}^{2}{\rm d}x+ C_{2}.从而只需证明\{u_{n}\}L^{2}({\Bbb R}^{N})中有界即可.事实上,若\int_{{\Bbb R}^{N}}u_{n}^{2}{\rm d}x\mathop{\longrightarrow}\limits^{n\rightarrow\infty}\infty,记v_{n}=\frac{u_{n}}{|u_{n}|_{2}},则|v_{n}|_{2}=1.

一方面,由(3.3)式知

\|v_{n}\|_{\lambda}^{2}+\int_{{\Bbb R}^{N}}V_{0}(x)v_{n}^{2}{\rm d}x-\int_{{\Bbb R}^{N}}\lambda V^{-}(x)v_{n}^{2}{\rm d}x\leq\frac{C}{|u_{n}|^{2}_{2}}.
(3.4)

因此\{\|v_{n}\|_{\lambda}^{2}\}是有界的并且

\int_{{\Bbb R}^{N}}V(x)v_{n}^{2}{\rm d}x\leq\frac{C}{|u_{n}|^{2}_{2}}\mathop{\longrightarrow}\limits^{n\rightarrow\infty}0.
(3.5)

E_{\lambda}自反,在子列意义下可设在E_{\lambda}v_{n}\rightharpoonup v.

另一方面,由(3.4)式和嵌入的紧性知v=0.因此

\begin{eqnarray*} \int_{{\Bbb R}^{N}} V(x)v_{n}^{2}{\rm d}x&=&\int_{\{x ~:~V(x)\geq b\}} V(x)v_{n}^{2}{\rm d}x+\int_{\{x~:~V(x)< b\}}V(x)v_{n}^{2}{\rm d}x\\ &\geq&b\bigg(1-\int_{\{x~:~V(x)< b\}} V(x)v_{n}^{2}{\rm d}x\bigg)+o_{n}(1)\\ &\geq&b+o_{n}(1)>0. \end{eqnarray*}

这与(3.5)式矛盾.

引理3.4  c_{\lambda}关于\lambda一致有界.

  由文献[31,定理2.12], c_{\lambda}\in[\kappa_{\lambda}, \sup\limits _{u\in Q}I _{\lambda} (u)],我们只需证明\sup\limits _{u\in Q}I _{\lambda} (u)有与\lambda无关的正上界.设

J_{\lambda}(u):=\frac{1}{2}a_{\lambda}(u, u)+\frac{1}{2}\int_{{\Bbb R}^{N}}V_{0}(x)u^{2}{\rm d}x-C_{1}\int_{\Omega}|u|^{q}{\rm d}x, ~~u\in E_{\lambda}.

显然, I_{\lambda}(u)\leq J_{\lambda}(u).

任给\eta>0,存在r_{\eta}>0,使得

\begin{eqnarray*}\left\{\begin{array}{ll}C_{ 1}|u|^{q}\geq \frac{1}{2}\eta u^{2}, ~~~&|u|\geq r_{\eta}, \\[3mm]C_{1}|u|^{q}\leq \frac{1}{2}\eta u^{2}, ~~~&|u|\leq r_{\eta}.\end{array}\right.\end{eqnarray*}

u=v+w\in\hat{E}_{\lambda}\oplus Re_{0}.我们得

\begin{eqnarray}\label{eq3.6} \nonumber J_{\lambda}(u)&\leq& \frac{1}{2}\int_{{\Bbb R}^{N}}|(-\Delta)^{\frac{s}{2}}w|^{2}{\rm d}x+\frac{1}{2}\int_{{\Bbb R}^{N}}V_{0}(x)u^{2}{\rm d}x -\frac{1}{2}\eta\int_{\Omega}u^{2}{\rm d}x \\ \nonumber&& +\int_{\{x\in\Omega~:~|u(x)|\leq r_{\eta}\}}\bigg(\frac{1}{2}\eta u^{2}-C_{1}|u|^{q}\bigg){\rm d}x\\ &\leq&\frac{1}{2}\int_{{\Bbb R}^{N}}|(-\Delta)^{\frac{s}{2}}w|^{2}{\rm d}x+C\|V_{0}\|_{L^{\infty}}\|u\|^{2}_{\lambda}-\frac{\eta}{2}\int_{\Omega}u^{2}{\rm d}x+C(\eta). \end{eqnarray}
(3.6)

由于e_{0}\in C_{0}^{\infty}(\Omega),故

\begin{equation}\label{eq3.7} \int_{{\Bbb R}^{N}}|(-\Delta)^{\frac{s}{2}}w|^{2}{\rm d}x =a_{\lambda}(u, w)=\int_{\Omega}u(-\Delta)^{s}w{\rm d}x\leq\|(-\Delta)^{s}w \|_{L^{2}(\Omega)}\|u\|_{L^{2}(\Omega)}.\end{equation}
(3.7)

(\varphi_{k}, \mu_{k})-\Delta\Omega中带有Dirichlet边界条件的特征函数和特征值.根据文献[2,引理3.4,引理3.5],我们得

\|(-\Delta)^{s}w\|_{L^{2}(\Omega)}=(\Sigma_{k=1}^{\infty}a^{2}_{k}\mu^{s}_{k})^{\frac{1}{2}}<\infty,
(3.8)

其中a_{k}=\int_{\Omega}w\varphi_{k}{\rm d}x.0<\mu_{1}\leq\mu_{2}\cdot\cdot\cdot\leq\mu_{k}\cdot\cdot\cdot\mu_{k}\rightarrow\infty,我们有

\|(-\Delta)^{\frac{s}{2}}w\|_{L^{2}(\Omega)}=(\Sigma_{k=1}^{\infty}a^{2}_{k}\mu^{\frac{s}{2}}_{k})^{\frac{1}{2}}<\infty.
(3.9)

因此\|(-\Delta)^{s}w\|_{L^{2}(\Omega)}\leq C_{0}\|(-\Delta)^{\frac{s}{2}}w\|_{L^{2}(\Omega)} (C_{0}只依赖于e_{0}).注意到(3.7)-(3.9)式,结合Young不等式,易证

\int_{{\Bbb R}^{N}}|(-\Delta)^{\frac{s}{2}}w|^{2}{\rm d}x\leq \frac{2}{\eta}C^{2}_{0}\|(-\Delta)^{\frac{s}{2}}w\|^{2}_{L^{2}(\Omega)}+\frac{\eta}{2}\|u\|^{2}_{L^{2}(\Omega)}.
(3.10)

\eta\geq4C_{0}^{2},则\int_{{\Bbb R}^{N}}|(-\Delta)^{\frac{s}{2}}w|^{2}{\rm d}x\leq\eta\|u\|^{2}_{L^{2}(\Omega)}.如果\|V_{0}\|_{L^{\infty}}\leq\frac{2}{CR_{\lambda}^{2}},根据(3.6)式,我们有J_{\lambda}(u)\leq C(\eta)+1.

引理3.5  任给M>0,存在\Lambda=\Lambda(M)>0使得(C)_{c}序列\{u_{n}\}满足,在E_{\lambda}中, u_{n} \rightharpoonup u,其中uI_{\lambda}的非平凡临界点或者I_{\lambda}满足(C)_{c}条件如果c\leq M.

  由引理3.3,在子列意义下,在E_{\lambda}中, u_{n} \rightharpoonup u.由于V变号,我们需要分为两种情形:

情形(ⅰ) I_{\lambda}(u)<0;

情形(ⅱ) I_{\lambda}(u)\geq0.

如果(ⅰ)发生,则u显然是非平凡解.如果(ii)发生,我们将证明在E_{\lambda}中, u_{n} \rightarrow u.事实上,记v_{n}=u_{n}-u.根据(V_{2}),有

\int_{{\Bbb R}^{N}} v^{2}_{n}{\rm d}x=\int_{\{x~:~V(x)\geq b\}}v^{2}_{n}{\rm d}x+\int_{\{x~:~V(x)< b\}} v_{n}{\rm d}x\leq\frac{1}{\lambda b}\|v_{n}\|^{2}_{\lambda}+o_{n}(1).
(3.11)

这样我们利用内插不等式和Sobolev不等式得

|v_{n}|_{p}\leq|v_{n}|^{\sigma}_{2}|v_{n}|^{1-\sigma}_{2^{\ast}_{s}}\leq d|v_{n}|^{\sigma}_{2}|(-\Delta)^{\frac{1}{2}}v_{n}|_{2}^{1-\sigma} \leq d(\lambda b)^{-\frac{\sigma}{2}}\|v_{n}\|_{\lambda}+o_{n}(1),
(3.12)

其中0<\sigma<1并且常数d\lambda无关.根据Brézis-Lieb引理我们得

I_{\lambda}(v_{n})=I_{\lambda}(u_{n})-I_{\lambda}(u)+o_{n}(1), ~~~I_{\lambda}'(v_{n})=I_{\lambda}'(u_{n})+o_{n}(1).
(3.13)

简单计算易知

\begin{eqnarray}\label{eq3.14} \nonumber \bigg(\frac{1}{q}-\frac{1}{p}\bigg)\int_{{\Bbb R}^{N}}P(x)|v_{n}|^{p}{\rm d}x&\leq&I_{\lambda}(v_{n})-\frac{1}{q}\langle I'_{\lambda}(v_{n}), v_{n}\rangle\\ \nonumber &=& c-I_{\lambda}(u)+o_{n}(1)\\ \nonumber &\leq&M+o_{n}(1).\end{eqnarray}

因此

|v_{n}|^{p}_{p}\leq\frac{Mpq}{P_{0}(p-q)}+o_{n}(1).
(3.14)

注意到条件(Q)\int_{{\Bbb R}^{N}}\lambda V^{-}(x)v^{2}_{n}{\rm d}x=o_{n}(1),结合(3.11)-(3.14)式,我们得

\begin{eqnarray}\label{eq3.16} \nonumber o_{n}(1)&=&\langle I_{\lambda}'(v_{n}), v_{n}\rangle\geq \|v_{n}\|^{2}_{\lambda}-C_{1}|v_{n}|^{p}_{p}+o_{n}(1)\\ &\geq& \bigg(1-C_{1}\bigg(\frac{Mpq}{P_{0}(p-q)}\bigg)^{\frac{p-2}{p}}\frac{d^{2}}{(\lambda b)^{\sigma}}\bigg)\|v_{n}\|^{2}_{\lambda}+o_{n}(1).\end{eqnarray}
(3.15)

选择\lambda充分大,我们得在E_{\lambda}v_{n}\rightarrow0.

4 定理1.2的证明

根据定理1.2的假设,我们知I_{\lambda}具有山路几何结构.借用文献[25,定理2.2], I_{\lambda}存在一个(C)_{c}序列.注意到\hat{E_{\lambda}}=\{0\},其不影响引理3.3至引理3.5的结果.类似于引理3.3至引理3.5,当\lambda充分大时,我们获得I_{\lambda}有一个非平凡临界点u_{\lambda}并且I_{\lambda}(u_{\lambda})\in[\kappa_{\lambda}, C_{0}],其中\kappa_{\lambda}\geq0依赖于\lambdaC_{0}\lambda无关.记u_{n}:=u_{\lambda_{n}}.

步骤1  我们首先证明\{u_{n}\}E中有界.直接计算我们得

c_{\lambda_{n}}=I_{\lambda_{n}}(u_{n})-\frac{1}{q}\langle I'_{\lambda_{n}}(u_{n}), u_{n}\rangle \geq \bigg(\frac{1}{2}-\frac{1}{q}\bigg)\|u_{n}\|_{\lambda_{n}}^{2}.
(4.1)

由(4.1)式知\{u_{n}\}E中有界.在子列意义下,我们设在Eu_{n}\rightharpoonup \bar{u}.

步骤2  \bar{u}是方程(1.4)的弱解.根据Fatou's引理,我们有

\int_{{\Bbb R}^{N}}V(x)\bar{u}^{2}{\rm d}x\leq\liminf\limits_{n\rightarrow\infty}\int_{{\Bbb R}^{N}}V(x)\bar{u}_{n}^{2}{\rm d}x \leq\liminf\limits_{n\rightarrow\infty}\frac{\|u_{n}\|^{2}_{\lambda_{n}}}{\lambda_{n}}=0.
(4.2)

因此,注意到(V_{3}), \bar{u}=0 a.e. {\Bbb R}^{N}\setminus V^{-1}(0), \bar{u}\in H_{0}^{s}(\Omega).任给\varphi\in C_{0}^{\infty}(\Omega), \langle I'_{\lambda_{n}}(u_{n}), \varphi\rangle =0,易知\bar{u}是方程(1.4)的弱解.

步骤3  在Eu_{n}\rightarrow \bar{u}并且\bar{u}\neq0.

类似于文献[21, 33]或[34].首先,易证在L^{t}({\Bbb R}^{N})u_{n}\rightarrow \bar{u} (2<t<2^{\ast}_{s}).其次,结合

\langle I'_{\lambda_{n}}(u_{n}), u_{n}\rangle=0, ~~\langle I'_{\lambda_{n}}(u_{n}), \bar{u}\rangle=0,

我们得

\lim\limits_{n\rightarrow\infty}\|u_{n}\|^{2}_{\lambda_{n}}= \lim\limits_{n\rightarrow\infty}(u_{n}, \bar{u})_{\lambda_{n}}=\|\bar{u}\|^{2}.

众所周知

\|\bar{u}\|^{2}\leq\liminf\limits_{n\rightarrow\infty}\|u_{n}\|^{2}\leq \lim\limits_{n\rightarrow\infty}\|u_{n}\|^{2}_{\lambda_{n}}.

因此,在Eu_{n}\rightarrow \bar{u}.最后,当n充分大时,有

\|u_{n}\|^{2}\leq \|u_{n}\|^{2}_{\lambda_{n}}\leq C' \|u_{n}\|^{p}+C'' \|u_{n}\|^{q},

这蕴含了\bar{u}\neq0.

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