考虑带有超线性项和次线性项的Klein-Gordon-Maxwell系统

$ \begin{equation}\label{eq:a1} \left\{\begin{array}{ll} -\triangle u+u-(2\omega +\phi)\phi u=a(x)|u|^{p-2}u+\lambda b(x)|u|^{q-2}u, \quad\ x\in {\Bbb R}^3, \\ -\triangle \phi +\phi u^2=-\omega u^2, \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad x\in {\Bbb R}^3, \end{array}\right. \end{equation} $

(1.1)

这里$4<p<6$, $1<q<2$, $\omega > 0$, $\phi:{\Bbb R}^3\rightarrow {\Bbb R}$, $\lambda>0$是一个参数, $a(x)$、$b(x)$是非负连续函数.该类问题起源于数学物理领域中的某些应用问题. 2001年Benci和Fortunato在文献[1]中为了描述三维空间中非线性Klein-Gordon场与静电场之间相互作用产生的孤立波问题, 提出了半线性Klein-Gordon-Maxwell方程组模型

$ \begin{eqnarray*} \left\{\begin{array}{ll} -\triangle u+[m^2-(\omega+e\phi)^2]u=|u|^{p-2}u, \quad & x\in {\Bbb R}^3, \\ -\triangle \phi +e^2u^2\phi=-e\omega u^2, &x\in {\Bbb R}^3, \end{array}\right. \end{eqnarray*} $

其中$4<p<6$, $m, \omega$均为实数, $m$和$e$分别代表粒子的质量和电量, $\omega$为相位.更多物理方面的详述参见文献[2-6].如下形式的系统

$ \begin{equation}\label{eq:a2} \left\{\begin{array}{ll} -\triangle u+V(x)u-(2\omega+\phi)\phi u=f(x, u), \quad& x\in {\Bbb R}^3, \\ -\triangle \phi =(\omega+\phi)u^2, & x\in {\Bbb R}^3. \end{array}\right. \end{equation} $

(1.2)

贺小明在文献[7]中第一次证明了问题(1.2) 无穷多解的存在性.通过对$V$和$f$加不同的条件, 便可得到众多的解的存在性和多解性的结果.李林等在文献[8]中研究了系统(1.2) 无穷多驻波解的存在性, 其中$f$在无穷远处是超线性的.文献[9]中, 唐春雷等考虑了带有超线性项或次线性项的Schrödinger-Poisson系统解的存在性和多重性.近期, 在文献[10]中, 孙明正、苏加宝和赵雷嘎研究了带有混合项的Schrödinger-Poisson系统解的存在性和多重性.

受文献[7-10]启发, 本文利用对称山路定理研究带有超线性项和次线性项的Klein-Gordon-Maxwell系统无穷多解的存在性.

我们列出以下关于$a(x)$和$b(x)$的假设:

(Ha1) $a(x)\in L^{\frac{6}{6-p}}({\Bbb R}^3)\cap L^{\infty}({\Bbb R}^3)$;

(Ha2) 对任意的$x\in {\Bbb R}^3$, $a(x)> 0$;

(Hb1) $b(x)\in L^{\frac{2}{2-q}}({\Bbb R}^3)\cap L^{\infty}({\Bbb R}^3)$;

(Hb2) 对任意的$x\in {\Bbb R}^3$, $b(x)>0$.

本文的主要结果如下:

定理1.1  假设条件(Ha1)-(Ha2) 以及(Hb1)-(Hb2) 成立, 则存在$\lambda_0>0$, 当$\lambda<\lambda_0$时, 系统(1.1) 有一序列解$\{(u_n, \phi_n)\}\subset H^1({\Bbb R}^3)\times D^{1, 2}({\Bbb R}^3)$, 且对应的能量值趋于无穷大.

设$E=H^1({\Bbb R}^3)$是通常的Sobolev空间, 其上的范数为$\|u\|^2=\int_{{\Bbb R}^3}|\nabla u|^2+u^2{\rm d}x$. $D^{1, 2}({\Bbb R}^3)$是空间$C_0^{\infty}({\Bbb R}^3)$关于范数$\|u\|^2_{D^{1, 2}}=\int_{{\Bbb R}^3}|\nabla u|^2{\rm d}x$的完备化得到的Hilbert空间.设$1\leq p<\infty$, $L^{p}({\Bbb R}^3)$表示通常的Lebesgue空间, 其上的范数为$\|u\|_p=(\int_{{\Bbb R}^3}|u|^p{\rm d}x)^{1/p}$.显然, 空间嵌入$E\hookrightarrow L^p({\Bbb R}^3)$ $(2\leq p\leq 6)$是连续的. $C$, $C_i$表示不同的正常数.

令$L^p({\Bbb R}^3, a)(2\leq p< 6)$表示加权的Lebesgue空间$\{u:{\Bbb R}^3\rightarrow {\Bbb R}$是可测函数, 且$\int_{{\Bbb R}^3}a(x)|u|^p{\rm d}x<+\infty\}$, 其上的范数定义为

$ \|u\|_{p, {\Bbb R}^3, a}= \bigg(\int_{{\Bbb R}^3}a(x)|u|^p{\rm d}x\bigg)^{\frac{1}{p}}. $

同样, $L^q({\Bbb R}^3, b)(1<q<2)$表示加权的Lebesgue空间$\{u:{\Bbb R}^3\rightarrow {\Bbb R}$是可测函数, 且$\int_{{\Bbb R}^3}b(x)|u|^q{\rm d}x<+\infty\}$, 其上的范数定义为

$ \|u\|_{q, {\Bbb R}^3, b}=\bigg(\int_{R^3}b(x)|u|^q{\rm d}x\bigg)^{\frac{1}{q}}, $

显然$L^p({\Bbb R}^3, a)$, $L^q({\Bbb R}^3, b)$是完备的.

下面证明一个重要的引理.

引理2.1  在(Ha1)-(Ha2) 条件下, 空间嵌入$E\hookrightarrow L^p({\Bbb R}^3, a)\quad (4<p<6)$是紧的.

证  只需证明$\{u_n\}\subset E$, 且在$E$中$u_n\rightharpoonup 0(n\rightarrow \infty)$, 在$L^{p}({\Bbb R}^3, a)$中$u_n\rightarrow 0(n\rightarrow \infty)$.

因为

$ \int_{{\Bbb R}^3}a(x)|u_n|^p{\rm d}x =\int_{B_R}a(x)|u_n|^p{\rm d}x+\int_{B_R^c}a(x)|u_n|^p{\rm d}x, $

其中$B_R=\{x\in {\Bbb R}^3: |x|\leq R\}$.

首先, 由局部紧嵌入定理可知$E\hookrightarrow L_{loc}^p({\Bbb R}^3)(p\in [2, 6))$是紧的.因此, 当$n$充分大时, 有

$ \begin{equation}\label{eq:a3} \int_{B_R}a(x)|u_n|^p{\rm d}x \leq \sup\limits_{B_R} a(x)\int_{B_R}|u_n|^p{\rm d}x\leq \frac{\epsilon}{2}. \end{equation} $

(2.1)

另一方面, 由条件(Ha1), $a(x)\in L^{\frac{6}{6-p}}({\Bbb R}^3)$可得:当$R$充分大时, 有

$ \begin{eqnarray}\label{eq:a4} \int_{B_R^c}a(x)|u_n|^p{\rm d}x &\leq& \bigg (\int_{B_R^c}|a(x)|^{\frac{6}{6-p}}{\rm d}x\bigg)^{\frac{6-p}{6}} \bigg(\int_{{\Bbb R}^3}|u_n|^6{\rm d}x\bigg)^{\frac{p}{6}} \\ &\leq& C\bigg(\int_{B_R^c}|a(x)|^{\frac{6}{6-p}}{\rm d}x\bigg)^{\frac{6-p}{6}}\|u_n\|^p \leq \frac{\epsilon}{2}. \end{eqnarray} $

(2.2)

结合(2.1)–(2.2) 式可得, 存在$R>0$, 当$n$充分大时, 有

$ \begin{eqnarray*} \int_{{\Bbb R}^3}a(x)|u_n|^p{\rm d}x = \int_{B_R}|a(x)||u_n|^p{\rm d}x+\int_{B_R^c}|a(x)||u_n|^p{\rm d}x \leq \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon, \end{eqnarray*} $

即$u_n\rightarrow 0, (n\rightarrow \infty)$在$L^p({\Bbb R}^3, a)$中强收敛.

命题2.1^[7]  对于任何固定的$u\in H^1({\Bbb R}^3)$, 存在唯一的函数$\phi=\phi_u\in D^{1, 2}({\Bbb R}^3)$, 满足方程

$ \begin{equation}\label{eq:a5} -\triangle \phi+u^2\phi=-\omega u^2. \end{equation} $

(2.3)

再者泛函$\Phi:u\in H^1({\Bbb R}^3)\rightarrow \Phi[u]:=\phi_u\in D^{1, 2}({\Bbb R}^3)$连续可微且

(ⅰ) 在集合$\{x| u(x)\neq 0\}$上, $-\omega \leq \phi_u\leq 0$.

(ⅱ) $\|\phi_u\|_{D^{1, 2}}\leq C_1\|u\|^2$, 且$\int_{R^3}|\phi_u|u^2{\rm d}x\leq C_1\|u\|_{12/5}^4\leq C_1\|u\|^4$.

将(2.3) 式乘以$\phi_u$并且利用分部积分可得

$ \int_{{\Bbb R}^3}|\nabla \phi_u|^2{\rm d}x=-\int_{{\Bbb R}^3}\omega \phi_u u^2{\rm d}x-\int_{{\Bbb R}^3}\phi_u^2u^2{\rm d}x. $

因此, 问题(1.1) 所对应的能量泛函$J:E\rightarrow {\Bbb R}$为

$ \begin{eqnarray*} J(u)=\frac{1}{2}\int_{{\Bbb R}^3}(|\nabla u|^2+u^2-\omega \phi_u u^2){\rm d}x -\frac{1}{p}\int_{{\Bbb R}^3}a(x)|u|^p{\rm d}x-\frac{\lambda}{q} \int_{{\Bbb R}^3}b(x)|u|^q{\rm d}x, \end{eqnarray*} $

由引理2.1和命题2.1可知泛函$J$是有意义的.更进一步, $J$是一个$C^1$泛函, 并且对任意$ u, v\in E$, 有

$ \begin{eqnarray*}\langle J'(u), v\rangle &=&\int_{{\Bbb R}^3}\nabla u\cdot \nabla v+uv-(2\omega+\phi_u)\phi_uuv{\rm d}x \\ &&-\int_{{\Bbb R}^3}a(x)|u|^{p-2}uv{\rm d}x -\lambda\int_{{\Bbb R}^3}b(x)|u|^{q-2}uv{\rm d}x. \end{eqnarray*} $

命题2.2^[1]  下面的是两个等价的命题

(ⅰ) $(u, \phi_u)\in H^1({\Bbb R}^3)\times D^{1, 2}({\Bbb R}^3)$是系统(1.1) 的解;

(ⅱ) $u\in H^1({\Bbb R}^3)$是泛函$J$的一个临界点, 并且$\phi=\phi_u$.

命题2.3^[11]  令$E$是无限维的Banach空间且$I\in C^1(E, {\Bbb R})$为偶泛函, 满足(PS)条件且$I(0)=0$.如果$E=Y\oplus Z$, 其中子空间$Y$是有限维的, 并且$I$满足

(ⅰ) 存在常数$\rho, \alpha>0$满足$I_{{\partial B_\rho}\cap Z}\geq \alpha$;

(ⅱ) 对任意的有限维子空间$\widetilde{E}\subset E$, 存在常数$R=R(\widetilde{E})>0$满足:当$u\in \widetilde{E}\backslash {B_R}$时, $I(u)\leq 0$.则泛函$I$有一列无界的临界值.

为证明定理1.1, 我们需要以下引理.

令$\{e_i\}$为空间$E$的一组正交基. $X_i={\Bbb R}e_i$, $Y_k=\bigoplus_{i=1}^{k}X_i$, $Z_k=\bigoplus_{i=k+1}^{\infty}X_i$, $k\in N$.

引理3.1  假设条件(Ha1)-(Ha2) 以及(Hb1)-(Hb2) 成立.则任何序列$\{u_n\}\subset E$使得

$ J(u_n)\rightarrow c>0, \quad \langle J'(u_n), u_n\rangle \rightarrow 0 $

有界, 且$\{u_n\}$有一个收敛的子序列.

证  令$\{u_n\}\subset E$且$J(u_n)\rightarrow c>0, \quad \langle J'(u_n), u_n\rangle \rightarrow 0$.我们断言$\{u_n\}$有界.若不然, 则存在一子序列(仍记为$\{u_n\}$)满足$\|u_n\|\rightarrow \infty \ (n\rightarrow \infty)$.

由条件(Hb1) 可得

$ \begin{eqnarray*} \int_{{\Bbb R}^3}b(x)|u_n|^q{\rm d}x\leq \bigg (\int_{{\Bbb R}^3}|b(x)|^{\frac{2}{2-q}}{\rm d}x \bigg)^{\frac{2-q}{2}} \bigg(\int_{{\Bbb R}^3}|u_n|^2{\rm d}x\bigg)^{\frac{q}{2}}\leq C\|u_n\|^q. \end{eqnarray*} $

因此, 当$n$充分大时, 有

$ \begin{eqnarray*} c+1&\geq&J(u_n)-\frac{1}{4}\langle J'(u_n), u_n\rangle\\ &=& \frac{1}{4}\|u_n\|^2+\frac{1}{4}\int_{{\Bbb R}^3}\phi_{u_n}^2u_n^2{\rm d}x+ \Big(\frac{1}{4}-\frac{1}{p}\Big)\int_{{\Bbb R}^3}a(x)|u_n|^p{\rm d}x +\Big(\frac{1}{4}-\frac{1}{q}\Big)\lambda\int_{{\Bbb R}^3}b(x)|u_n|^q{\rm d}x\\ & \geq &\frac{1}{4}\|u_n\|^2+\Big(\frac{1}{4}-\frac{1}{q}\Big)\lambda\int_{{\Bbb R}^3}b(x)|u_n|^q{\rm d}x\\ &\geq &\frac{1}{4}\|u_n\|^2-C\Big(\frac{1}{q}-\frac{1}{4}\Big)\lambda\|u_n\|^q\rightarrow +\infty, \end{eqnarray*} $

这是矛盾的.故序列$\{u_n\}$有界.

由于序列$\{u_n\}$有界, 对任何固定的$R>0$, 令$\xi_R\in C^{\infty}({\Bbb R}^3, {\Bbb R})$满足

$ \begin{eqnarray*} \xi_R(x)=\left\{\begin{array}{ll}0, ~~& \mbox{对}~|x|\leq R/2, \\ 1, & \mbox{对}~|x|>R, \end{array}\right. \end{eqnarray*} $

且$|\nabla \xi_R(x)|\leq \frac{C}{R}$.则对任何$u\in E$, $R\geq 1$, 存在常数$C>0$, 使得$\|\xi_Ru\|\leq C\|u\|$.

因此$\langle J'(u_n), \xi_Ru_n\rangle=o(1)$, 即

$ \begin{eqnarray}\label{eq:a6} o(1)&=&\int_{{\Bbb R}^3}(|\nabla u_n|^2+u_n^2-(2\omega+\phi_{u_n})\phi_{u_n}u_n^2)\xi_R{\rm d}x \\ && -\int_{{\Bbb R}^3}a(x)|u_n|^{p}\xi_R{\rm d}x-\lambda\int_{{\Bbb R}^3}b(x)|u_n|^{q}\xi_R{\rm d}x. \end{eqnarray} $

(3.1)

根据Hölder不等式, 有

$ \begin{equation}\label{eq:a7} \bigg|\int_{{\Bbb R}^3}b(x)|u_n|^q \xi_R{\rm d}x\bigg| \leq \int_{B^c_{R/2}(0)}|b(x)||u_n|^q {\rm d}x\leq C|b|_{L^{\frac{2}{2-q}}(B^c_{R/2}(0))}\|u_n\|^q, \end{equation} $

(3.2)

且

$ \begin{equation}\label{eq:a8} \bigg|\int_{{\Bbb R}^3}a(x)|u_n|^p \xi_R{\rm d}x\bigg|\leq \int_{B^c_{R/2}(0)}|a(x)||u_n|^p {\rm d}x\leq C'|a|_{L^{\frac{6}{6-p}}(B^c_{R/2}(0))}\|u_n\|^p. \end{equation} $

(3.3)

结合(3.1)–(3.3) 式可得

$ \begin{eqnarray*} \int_{|x|\geq R}(|\nabla u_n|^2+u_n^2){\rm d}x&\leq &\int_{|x|\geq R}(|\nabla u_n|^2+u_n^2-(2\omega+\phi_{u_n})\phi_{u_n}u_n^2){\rm d}x \nonumber\\ &\leq&C'|a|_{L^{\frac{6}{6-p}}(B^c_{R/2}(0))}\|u_n\|^p+C|b|_{L^{\frac{2}{2-q}}(B^c_{R/2}(0))}\|u_n\|^q+o(1). \end{eqnarray*} $

这意味着, 对任意给定的$\epsilon>0$, 存在常数$R>0$, 使得当$n$充分大时, 有

$ \begin{equation}\label{eq:a9} \int_{|x|\geq R}(|\nabla u_n|^2+u_n^2){\rm d}x\leq \epsilon. \end{equation} $

(3.4)

因为$\{u_n\}$有界, 则存在一子序列(仍记为$\{u_n\}$)满足在$E$中$u_n\rightharpoonup u\ (n\rightarrow \infty)$.根据(3.4) 式和空间局部紧嵌入$H^1({\Bbb R}^3)\hookrightarrow L_{loc}^p({\Bbb R}^3)\quad (p\in [2, 6))$, 有

$ \begin{equation}\label{eq:a10} u_n\rightarrow u \quad (L^p({\Bbb R}^3)), \quad p\in [2, 6). \end{equation} $

(3.5)

要证明在$E$中$u_n\rightarrow u, (n\rightarrow \infty)$, 只需验证$\|u_n\|\rightarrow \|u\|$.因为$\langle J'(u_n), u\rangle=o(1)$, $\langle J'(u_n), u_n\rangle=o(1)$, 我们可以得到

$ \begin{align} &o(1)=\int_{{{\mathbb{R}}^{3}}}{\nabla }{{u}_{n}}\nabla u+{{u}_{n}}u-(2\omega +{{\phi }_{{{u}_{n}}}}){{\phi }_{{{u}_{n}}}}{{u}_{n}}u\text{d}x \\ &\ \ \ \ \ \ \ \ \ \ \ -\int_{{{\mathbb{R}}^{3}}}{a}(x)|{{u}_{n}}{{|}^{p-2}}{{u}_{n}}u\text{d}x-\lambda \int_{{{\mathbb{R}}^{3}}}{b}(x)|{{u}_{n}}{{|}^{q-2}}{{u}_{n}}u\text{d}x, \\ \end{align} $

(3.6)

$ \begin{eqnarray} \label{eq:a12} o(1)&=&\int_{{\Bbb R}^3}|\nabla u_n|^2+u_n^2-(2\omega+\phi_{u_n})\phi_{u_n}u_n^2{\rm d}x \\ && -\int_{{\Bbb R}^3}a(x)|u_n|^{p}{\rm d}x-\lambda\int_{{\Bbb R}^3}b(x)|u_n|^{q}{\rm d}x. \end{eqnarray} $

(3.7)

并且我们有

$ \begin{equation}\label{eq:a13} \int_{{\Bbb R}^3}(2\omega+\phi_{u_n})\phi_{u_n}u_n(u_n-u){\rm d}x=o(1), \end{equation} $

(3.8)

$ \begin{equation}\label{eq:a14} \int_{{\Bbb R}^3}a(x)|u_n|^p-a(x)|u_n|^{p-2}u_nu{\rm d}x=o(1), \end{equation} $

(3.9)

$ \begin{equation}\label{eq:a15} \int_{{\Bbb R}^3}b(x)|u_n|^q-b(x)|u_n|^{q-2}u_nu{\rm d}x=o(1). \end{equation} $

(3.10)

事实上, 由Hölder不等式, 命题2.1和Sobolev不等式, 有

$ \int_{{\Bbb R}^3}\phi_{u_n}u_n(u_n-u){\rm d}x \leq \|\phi_{u_n}(u_n-u)\|_2\|u_n\|_2 \leq \|\phi_{u_n}\|_6\|u_n-u\|_3\|u_n\|_2, $

$ \int_{{\Bbb R}^3}\phi_{u_n}^2u_n(u_n-u){\rm d}x \leq \|\phi_{u_n}^2u_n\|_{3/2}(u_n-u)\|_3 \leq \|\phi_{u_n}\|_6^2\|u_n\|_3\|u_n-u\|_3. $

根据(3.5) 式, $u_n\rightarrow u$在$L^p({\Bbb R}^3)$对所有的$2\leq p<6$, 我们有

$ \int_{{\Bbb R}^3}\phi_{u_n}u_n(u_n-u){\rm d}x \rightarrow 0, \quad n\rightarrow \infty $

和

$ \int_{{\Bbb R}^3}\phi_{u_n}^2u_n(u_n-u){\rm d}x \rightarrow 0, \quad n\rightarrow \infty. $

根据条件(Ha1)-(Ha2), (Hb1)-(Hb2) 可得

$ \begin{eqnarray*} &&\bigg|\int_{{\Bbb R}^3}(a(x)|u_n|^p-a(x)|u_n|^{p-2}u_nu){\rm d}x\bigg|\\ &\leq&\int_{{\Bbb R}^3}|a(x)||u_n|^{p-1}|u_n-u|{\rm d}x\\ &\leq &\bigg(\int_{{\Bbb R}^3}|a(x)||u_n|^{p}\bigg)^{\frac{p-1}{p}} \bigg(\int_{{\Bbb R}^3}|a(x)||u_n-u|^{p}\bigg)^{\frac{1}{p}}, \end{eqnarray*} $

因为$u_n\rightharpoonup u(E)$, 根据引理2.1, $E\hookrightarrow L^p ({\Bbb R}^2, a)$是紧嵌入的, 所以

$ \begin{eqnarray*} \int_{{\Bbb R}^3}(a(x)|u_n|^p-a(x)|u_n|^{p-2}u_nu){\rm d}x\rightarrow 0, \quad n\rightarrow \infty. \end{eqnarray*} $

又根据(3.5) 式和Hölder不等式, 有

$ \begin{eqnarray*} &&\bigg|\int_{{\Bbb R}^3}(b(x)|u_n|^q-b(x)|u_n|^{q-2}u_nu){\rm d}x\bigg|\\ &\leq&\int_{{\Bbb R}^3}|b(x)||u_n|^{q-1}|u_n-u|{\rm d}x\\ &\leq &|b(x)|_{2/2-q}|u_n|^{q-1}_2|u_n-u|_2\rightarrow 0,\quad n\rightarrow \infty. \end{eqnarray*} $

结合(3.6)–(3.10) 式可推出$\|u_n\|\rightarrow \|u\|$.因此, $u_n\rightarrow u$在$E$中强收敛.

引理3.2  假设条件(Ha1)-(Ha2) 和(Hb1)-(Hb2) 成立, 则对任意有限维子空间$\widetilde{E}\subset E$, 存在$R=R(\widetilde{E})>0$使得

$ J(u)\leq 0, \quad \forall u\in \widetilde{E}, \quad \|u\|_E\geq R. $

证  因为有限维空间$\widetilde{E}\subset E$上所有的范数等价, 所以存在常数$C_p>0$, $C_q>0$使得$\|u\|_{p, {\Bbb R}^3, a}\geq C_p\|u\|_E$, $\|u\|_{q, {\Bbb R}^3, b}\geq C_q\|u\|_E$.

因此我们有

$ \begin{eqnarray*} J(u)&=&\frac{1}{2}\int_{{\Bbb R}^3}(|\nabla u|^2+u^2-\omega \phi_u u^2){\rm d}x-\frac{1}{p}\int_{{\Bbb R}^3}a(x)|u|^p{\rm d}x-\frac{\lambda}{q}\int_{{\Bbb R}^3}b(x)|u|^q{\rm d}x\\ &\leq&\frac{1}{2}\|u\|^2+C_1\omega\|u\|^4-C_p^p\|u\|^p-\lambda C_q^q\|u\|^q\\ &\rightarrow&-\infty, \quad \mbox{当}~ \|u\|\rightarrow \infty. \end{eqnarray*} $

证毕.

引理3.3  在条件(Ha1)-(Ha2), (Hb1)-(Hb2) 下, 存在常数$\rho, \alpha>0$使得$J|_{\partial B_{\rho}\cap Z_k}\geq \alpha$.

证  根据引理2.1和命题2.1, 有

$ \begin{eqnarray*} J(u)&=&\frac{1}{2}\int_{{\Bbb R}^3}(|\nabla u|^2+u^2-\omega \phi_u u^2){\rm d}x-\frac{1}{p}\int_{{\Bbb R}^3}a(x)|u|^p{\rm d}x-\frac{\lambda}{q}\int_{{\Bbb R}^3}b(x)|u|^q{\rm d}x\\ &\geq&\frac{1}{2}\|u\|^2-\frac{1}{p}\int_{{\Bbb R}^3}a(x)|u|^p{\rm d}x-\frac{\lambda}{q}\int_{{\Bbb R}^3}b(x)|u|^q{\rm d}x\\ &\geq& \frac{1}{2}\|u\|^2-C_2|a|_{6/6-p}\|u\|^p-C_3\lambda|b|_{2/2-q}\|u\|^q\\ &\geq&\Big (\frac{1}{2}-C_2|a|_{6/6-p}\|u\|^{p-2}-C_3\lambda|b|_{2/2-q}\|u\|^{q-2} \Big)\|u\|^2. \end{eqnarray*} $

令

$ g(t)=\frac{1}{2}-C_2|a|_{6/6-p}t^{p-2}-C_3\lambda|b|_{2/2-q}t^{q-2}, \quad t>0. $

因为$1<q<2<p$, 所以函数$g(t)$在$t_0\in (0, +\infty)$取到最大值.而且当$\lambda<\lambda_0:=\frac{\frac{1}{2}-C_2|a|_{6/6-p}t_0^{p-2}}{C_3|b|_{2/2-q}t_0^{q-2}}$时, 最大值

$ g(t_0)=\max\limits_{t\in (0, +\infty)}g(t)>0. $

令$\rho=t_0$, $\alpha=g(\rho)\rho^2$可得$J(u)\geq \alpha, \quad \forall u\in Z_k, \quad \|u\|=\rho$.

定理1.1的证明  根据引理2.1和命题2.1可知泛函$J\in C^1(E, {\Bbb R})$为偶泛函.其次, 引理3.1说明$J$满足(PS)条件.最后, 由引理3.2和3.3可知泛函$J$满足命题2.3的几何条件, 即在$E$的有限维子空间$\widetilde{E}$上, 当$\|u\|_E$足够大时, $J(u)\leq 0$; 令$\lambda_0:=\frac{\frac{1}{2}-C_2|a|_{6/6-p}t_0^{p-2}}{C_3|b|_{2/2-q}t_0^{q-2}}$, $t_0\in (0, +\infty)$, 当$\lambda<\lambda_0$时, 存在常数$\rho=t_0$, $\alpha=g(\rho)\rho^2$, 使得当$u\in Z_k$且$\|u\|=\rho$时, $J(u)\geq \alpha$.

至此, 泛函$J$满足命题2.3的所有假设, 所以问题(1.1) 有穷多解$\{(u_n, \phi_n)\}\subset H^1({\Bbb R}^3)\times D^{1, 2}({\Bbb R}^3)$, 且对应的能量值趋于无穷大.