ABSTRACT

Under the assumption of the nonlinearity with critical exponential growth, we consider the existence of solutions to a class of quasilinear N-Laplacian equations in ${ℝ}^N$. By symmetric mountain pass lemma and variational argument, the existence of solutions is established.

Keywords:Trudinger-Moser Inequality, Symmetric Mountain Pass Lemma, Critical Exponential Growth

${ℝ}^N$ 上一类拟线性N-拉普拉斯方程的无穷解

杜桂香，李静

临沂大学数学与统计学院，山东 临沂

收稿日期：2020年2月25日；录用日期：2020年3月9日；发布日期：2020年3月16日

摘 要

本文主要研究 ${ℝ}^N$ 上一类拟线性N-拉普拉斯方程，在非线性项为临界指数增长的情况下，借助对称山路引理以及变分法得出多解的存在性。

关键词 :Trudinger-Moser不等式，山路引理，临界指数增长

Copyright © 2020 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言及主要结论

本文主要研究一类拟线性的N-拉普拉斯方程多解的存在性

$-{\Delta }_{N}u+V\left(x\right){\left|u\right|}^{N-2}u-{\Delta }_N\left({\left|u\right|}^{2\alpha }\right){\left|u\right|}^{2\alpha -2}u=g\left(x\right),x\in {ℝ}^N,$ (1.1)

其中 ${\Delta }_{N}u=div\left({\left|\nabla u\right|}^{N-2}\nabla u\right)$ 为N-拉普拉斯算子，参数 $\alpha >\frac{1}{2}$，我们做如下假定

( ${\mathcal{V}}_1$ ) 存在 $V_0>0$ 使得在 ${ℝ}^N$ 中 $V\left(x\right)\ge V_0$。此外 $V\left(x\right)\to \infty$，$\left|x\right|\to \infty$ ；或者更为一般化对每个 $M>0$，$\text{meas}\left(\left\{x\in R^N:V\left(x\right)\le M\right\}\right)<\infty$。“其中meas”为 ${ℝ}^N$ 上Lebesgue测度。

( ${\mathcal{F}}_1$ ) 奇函数 $g\left(s\right)\in C^1\left(ℝ\right)$，$g\left(t\right)>0$，$t>0$ 且存在 $C_0$，${\alpha }_0$，$q>0$ 使得

$\left|g\left(s\right)\right|\le C_0{\left|s\right|}^{q-1}\left[\exp \left({\alpha }_0{\left|s\right|}^{\frac{2\alpha N}{N-1}}\right)-S_{N-2}\left({\alpha }_0,s\right)\right],\forall s\in ℝ,$ (1.2)

其中

$S_{N-2}\left({\alpha }_0,s\right)={\displaystyle {\sum }_{k=0}^{N-2}\frac{{\alpha }_0^k}{k!}{\left|s\right|}^{\frac{2\alpha N}{N-1}}},\forall s\in ℝ.$ (1.3)

( ${\mathcal{F}}_2$ ) 存在 $q>N$ 和 ${\mu }_0>0$ 使得

$0\le qG\left(s\right)\le sg\left(s\right)\le {\mu }_0{\left|s\right|}^q\exp \left({\alpha }_0{\left|s\right|}^{\frac{2\alpha N}{N-1}}\right)$ (1.4)

其中 $G\left(s\right)={\displaystyle {\int }_0^{s}g\left(t\right)\text{d}t}$，在 $ℝ$ 函数上 $sg\left(s\right)$ 递增且有 ${\lim }_{\left|s\right|\to \infty }\frac{G\left(s\right)}{sg\left(s\right)}=0$。

( ${\mathcal{F}}_3$ ) 存在 ${\mu }_1>0$ 和 $q>N$ 使得

$G\left(s\right)\ge {\mu }_1{\left|s\right|}^q,\forall s\in ℝ.$ (1.5)

在过去的几十年间大量的学者都在研究拟线性的Schrödinger方程，详情可参考文献 [1] [2]。在 [3] 中，作者利用约束极小化方法，得出拟线性的Schrödinger方程正基态解的存在性结论。在 [4] 中，运用变量替换把拟线性问题转化为半线问题，在Orlicz空间中借助山路引理得出解的存在性结论。在 [5] 中作者考虑了N = 2的情形，其中非线性项具有临界指数增长，即当 $\left|s\right|\to \infty$ 时形如 $\exp \left(4\pi s^4\right)-1$，运用 ${ℝ}^N$ 上Trudinger-Moser不等式结合山路定理得出解的存在性。特别的在 [4] [6] [7] 中，研究了具有参数 的方程正基态解

$-\text{Δ}u+V\left(x\right)u-\text{Δ}\left({\left|u\right|}^{2\alpha }\right){\left|u\right|}^{2\alpha -2}u=\lambda {\left|u\right|}^{p-1}u,x\in {ℝ}^N$ (1.6)

其中 $\alpha >\frac{1}{2}$，$\lambda >0$，$2<p+1<\frac{4\alpha N}{N-2}$。

据作者了解，关于方程(1.1)的多解存在性鲜有相关文献及结论，本文主要借助文献 [5] [8] 中的思路。由于拟线性项 ${\Delta }_N\left({\left|u\right|}^{2\alpha }\right){\left|u\right|}^{2\alpha -2}u$ 的存在，使得紧嵌入不再成立，因此我们需要考虑合适的工作空间也就需要更为细致的估测。

本文记 $X=W^{1,n}\left({ℝ}^N\right)$ 并且

$E=\left\{u\in W^{1,n}\left({ℝ}^N\right):{\displaystyle {\int }_{{ℝ}^N}\left({\left|\nabla u\right|}^N+V\left(x\right){\left|u\right|}^N\right)\text{d}x}<\infty \right\}$

具备范数

${\Vert u\Vert }_E={\left({\displaystyle {\int }_{{ℝ}^N}\left({\left|\nabla u\right|}^N+V\left(x\right){\left|u\right|}^N\right)\text{d}x}\right)}^{\frac{1}{N}}.$ (1.7)

主要结论如下

定理1.1：假设( ${\mathcal{V}}_1$ )和( ${\mathcal{F}}_2$ )~( ${\mathcal{F}}_3$ )成立。则方程(1.1)在E中存在无穷多个解。

2. 使用须知

为了方便证明需要引入下列引理

引理2.1：(有界区域上的Trudinger-Moser不等式) [9] [10]。设 $\mu \in W_0^{1,N}\left(\Omega \right),\Omega \subset {ℝ}^N\left(N\ge 2\right)$。为一有界区域，则有

${\displaystyle {\int }_{\Omega }{\text{e}}^{\alpha {\left|u\right|}^{\frac{N}{N-1}}}\text{d}x}<\infty ,\forall \alpha >0$

此外，存在 $C=C\left(N,\left|\Omega \right|\right)$ 使得

$\sup {\Vert u\Vert }_{W_0^{1,N}\text{Ω}}\le 1{\displaystyle {\int }_{\Omega }{\text{e}}^{\alpha {\left|u\right|}^{\frac{N}{N-1}}}\text{d}x}\le C,\forall \alpha \le {\alpha }_N,$

其中 ${\alpha }_N=N{\omega }_{N-1}^{\frac{1}{N-1}}>0$ 且 ${\omega }_{N-1}$ 为 $\left(N-\text{1}\right)$ -球体的 $\left(N-\text{1}\right)$ -维测度。

引理2.2：(无界区域上的Trudinger-Moser不等式) [5] [11]。设 $\mu \in W_0{}^{1,N}\left({\mathcal{R}}^N\right),N\ge 2$，则有

${\displaystyle {\int }_{{ℝ}^N}\left({\text{e}}^{\alpha {\left|u\right|}^{\frac{N}{N-1}}}-S_{N-2}\left(\alpha ,u\right)\right)\text{d}x}<\infty ,\forall \alpha >0$

此外，如果有 ${\left|\nabla u\right|}_N^N\le 1,{\left|u\right|}_N\le M<\infty$ 和 $\alpha \le {\alpha }_N$，则存在正常数 $C=C\left(N,M,\alpha \right)$ 使得

${\displaystyle {\int }_{{ℝ}^N}\left({\text{e}}^{\alpha {\left|u\right|}^{\frac{N}{N-1}}}-S_{N-2}\left(\alpha ,u\right)\right)\text{d}x}\le C,$

其中 $S_{N-2}\left(\alpha ,u\right)={\displaystyle {\sum }_{k=0}^{N-2}\frac{{\alpha }^k}{k!}{\left|u\right|}^{\frac{Nk}{N-1}}}$。

与此同时，联合以上两个引理以及Holder不等式，假如 $\alpha ,q>0$ 那么

${\displaystyle {\int }_{{ℝ}^N}{\left|u\right|}^q\left({\text{e}}^{\alpha {\left|u\right|}^{\frac{N}{N-1}}}-S_{N-2}\left(\alpha ,u\right)\right)\text{d}x}<\infty ,\forall u\in W^{1,N}\left({ℝ}^N\right)$ (2.1)

更确切的，当 $\alpha M^{\frac{N}{N-1}}<{\alpha }_N$ 时，如果有 $\Vert u\Vert W^{1,N}\left({ℝ}^N\right)\le M$ 成立，则存在 $C=C\left(\alpha ,q,N\right)>0$ 使得

${\displaystyle {\int }_{{ℝ}^N}{\left|u\right|}^q\left({\text{e}}^{\alpha {\left|u\right|}^{\frac{N}{N-1}}}-S_{N-2}\left(\alpha ,u\right)\right)\text{d}x}\le C{\Vert u\Vert }_{W^{1,N}}^q\left({ℝ}^N\right),$ (2.2)

其中 $\Vert u\Vert W^{1,N}\left({ℝ}^N\right)={\left({\displaystyle {\int }_{{ℝ}^N}\left({\left|\nabla u\right|}^N+{\left|u\right|}^N\right)\text{d}x}\right)}^{1/N}$ Sobolev空间 $W^{1,N}\left({ℝ}^N\right)$ 的范数。

引理2.3 [12]：设 $u\in W_0^{1,N}\left(\Omega \right)\cap L^r\left(\Omega \right)$，其中 $r\ge 1,\Omega \subseteq {ℝ}^N$ 为任意区域，那么对 $q\ge r$ 会有

${\Vert u\Vert }_q\le c\left(N,r\right)q^{1-1/N}{\Vert \nabla u\Vert }_N^{1-\frac{r}{q}}{\Vert u\Vert }_r^{\frac{r}{q}}$. (2.3)

$q^{1-1/N}$ 为最佳指数，特别的有，

$c\left(N,N\right)=\frac{1}{\sqrt{\pi }}{\left(\frac{\Gamma \left(\frac{N}{2}\right)\Gamma \left(2N\right)}{2\Gamma {\left(N\right)}^2}\right)}^{1/N}\equiv d_N.$ (2.4)

注2.1：由上面的引理可知，当 $q\ge N$ 时 $X\to L^q\left({ℝ}^N\right)$ 为连续嵌入且有

${\Vert u\Vert }_q\le d_N{q}^{1-1/N}{\Vert u\Vert }_Y.$

考察方程(1.1)所对应的能量泛函 $I(u)$

$I\left(u\right)=\frac{1}{N}{\displaystyle {\int }_{{ℝ}^N}\left(1+{\left(2\alpha \right)}^{N-1}{\left|u\right|}^{\left(2\alpha -1\right)N}\right){\left|\nabla u\right|}^N\text{d}x}+\frac{1}{N}{\displaystyle {\int }_{{ℝ}^N}V\left(x\right){\left|u\right|}^N\text{d}x}-{\displaystyle {\int }_{{ℝ}^N}G\left(u\right)\text{d}x}.$ (2.5)

需要指出的是 $I\left(u\right)$ 在X中无法定义，为了克服这一困难需引入变量替换 [2] $u=f\left(v\right)$ 或者 $v=f^{-1}\left(u\right)$，其中f为

$f^{\prime }\left(t\right)={\left(1+{\left(2\alpha \right)}^{N-1}{\left|f\left(t\right)\right|}^{\left(2\alpha -1\right)N}\right)}^{-1/N},t\ge 0,f\left(0\right)=0$ (2.6)

$f\left(t\right)=-f\left(-t\right),t\in \left(-\infty ,0\right].$

引理2.4：函数 $f\left(t\right)$ 满足

( $f_1$ ) $f\in C^2$ 唯一确定且在 上可逆，

( $f_2$ ) $0<f^{\prime }\left(t\right)\le 1$，$\forall t\in {ℝ}^N$，

( $f_3$ ) $\left|f\left(t\right)\right|\le \left|t\right|$，$\forall t\in {ℝ}^N$，

( $f_4$ ) 当 $t\to 0$ 时， $\frac{f\left(t\right)}{t}\to 1$，

( $f_5$ ) $\left|f\left(t\right)\right|\le {\left(2\alpha \right)}^{1/2\alpha N}{\left|t\right|}^{1/2\alpha },\forall t\in {ℝ}^N$，

( $f_6$ ) $\frac{1}{2}f\left(t\right)\le \alpha t{f}^{\prime }\left(t\right)\le \alpha f\left(t\right),\forall t\in \left[0,\infty \right)$ 以及 $\alpha f\left(t\right)\le \alpha t{f}^{\prime }\left(t\right)\le \frac{1}{2}f\left(t\right),\forall t\in \left(-\infty ,0\right]$，

( $f_7$ ) $\exists a\in \left(0,{\left(2\alpha \right)}^{1/2\alpha N}\right)$ 使得 $\frac{f\left(t\right)}{t^{1/2\alpha }}\to a,t\to +\infty$，

( $f_8$ ) $\exists b_0>0$ 使得

$\left|f\left(t\right)\right|\ge \{\begin{array}{l}{b}_0\left|t\right|\left|t\right|\le 1,\\ b_0{\left|t\right|}^{1/2\alpha }\left|t\right|\ge 1.\end{array}$

通过变量替换后 $I\left(u\right)$ 变为

$J\left(v\right)\equiv I\left(f\left(v\right)\right)=\frac{1}{N}{\displaystyle {\int }_{{ℝ}^N}{\left|\nabla v\right|}^N\text{d}x}+\frac{1}{N}{\displaystyle {\int }_{{ℝ}^N}V\left(x\right){\left|f\left(v\right)\right|}^N\text{d}x}-{\displaystyle {\int }_{{ℝ}^N}G\left(f\left(v\right)\right)\text{d}x}$ (2.7)

易见在假设( ${\mathcal{F}}_1$ )~( ${\mathcal{F}}_2$ )的条件下， $J\left(v\right)$ 在X上有定义。

若 $v\in W^{1,N}\left({ℝ}^N\right)\cap L_{loc}^{\infty }\left({ℝ}^N\right)\left(p>1\right)$ 为泛函J的临界点，即对所有的 $\phi \in W^{1,N}\left({ℝ}^N\right)$ 有 $J^{\prime }\left(v\right)\phi =0$ 其中

$J^{\prime }\left(v\right)\phi ={\displaystyle {\int }_{R^N}{\left|\nabla v\right|}^{N-2}\nabla v\nabla \phi \text{d}x}+{\displaystyle {\int }_{R^N}V\left(x\right){\left|f\left(v\right)\right|}^{N-2}f\left(v\right)f^{\prime }\left(v\right)\phi \text{d}x}-{\displaystyle {\int }_{R^N}g\left(f\left(v\right)\right)f^{\prime }\left(v\right)\phi \text{d}x}$

那么v方程

$-{\Delta }_{N}v=h\left(v\right),x\in {ℝ}^N,$ (2.8)

的解，其中 $h\left(s\right)=V\left(x\right){\left|f\left(s\right)\right|}^{N-2}f\left(s\right)f^{\prime }\left(s\right)+g\left(f\left(s\right)\right)f^{\prime }\left(s\right)$，$s\in ℝ$。

由此可知 $u=f\left(v\right)$ 为方程(1.1)的弱解。

引理2.5：假设( ${\mathcal{F}}_1$ )和( ${\mathcal{F}}_2$ )成立如果 $\left\{v_n\right\}\subset E$ 为一PS序列，那么 $\left\{v_n\right\}$ 在E中有界。

证：不失一般性，对所有的 $n\in \mathbb{N}$ 假设 $v_n\ne 0$。令 ${\phi }_n\left(x\right)=\frac{f\left(v_n\left(x\right)\right)}{f^{\prime }\left(v_n\left(x\right)\right)}$，则由引理2.4可得

$\left|{\phi }_n\left(x\right)\right|\le 2\alpha \left|v_n\left(x\right)\right|,\left|\nabla {\phi }_n\left(x\right)\right|\le 2\left|\nabla v_n\left(x\right)\right|,{\Vert {\phi }_n\left(x\right)\Vert }_E\le 2\alpha {\Vert v_n\left(x\right)\Vert }_E,\forall n\in \mathbb{N}$

由于 $\left\{v_n\right\}$ 为一PS序列，则 $\exists c>0$ 使得

$\begin{array}{c}c\ge J\left(v_n\right)-\frac{1}{q}J\left(v_n\right)v_n\\ \ge \left(\frac{1}{N}-\frac{2\alpha }{q}\right){\displaystyle {\int }_{{ℝ}^N}{\left|\nabla v_n\right|}^N\text{d}x}+\left(\frac{1}{N}-\frac{1}{q}\right){\displaystyle {\int }_{{ℝ}^N}V\left(x\right){\left|f\left(v\right)\right|}^N\text{d}x}\\ +\frac{1}{q}{\displaystyle {\int }_{{ℝ}^N}\left[g\left(f\left(v_n\right)\right)f\left(v_n\right)-qG\left(f\left(v_n\right)\right)\right]\text{d}x}\\ \ge \left(\frac{1}{N}-\frac{2\alpha }{q}\right){\Vert \nabla v_n\Vert }_N^N+\left(\frac{1}{N}-\frac{1}{q}\right){\displaystyle {\int }_{{ℝ}^N}V\left(x\right){\left|f\left(v\right)\right|}^N\text{d}x}\end{array}$ (2.9)

由 $q>2\alpha N>N$ 可推出序列 $\left\{{\Vert \nabla v_n\Vert }_N\right\}$ 以及 $\left\{{\displaystyle {\int }_{{ℝ}^N}V\left(x\right){\left|f\left(v_n\right)\right|}^N}\right\}$ 有界，亦如此，其中

$A_n^N={\displaystyle {\int }_{{ℝ}^N}\left({\left|\nabla v_n\right|}^N+V\left(x\right){\left|f\left(v_n\right)\right|}^N\right)\text{d}x}$ (2.10)

可得 $\left\{v_n\right\}$ 在 $W^{1,N}\left({ℝ}^N\right)$ 中有界，接下来证明 $\left\{v_n\right\}$ 在 $L^N\left({ℝ}^N\right)$ 中有界。因为

${\displaystyle {\int }_{{ℝ}^N}{\left|v_n\right|}^N\text{d}x}={\displaystyle {\int }_{\Vert v_n\Vert \le 1}{\left|v_n\right|}^N\text{d}x}+{\displaystyle {\int }_{\Vert v_n\Vert >1}{\left|v_n\right|}^N\text{d}x}$ (2.11)

注意由条件( ${\mathcal{F}}_2$ )，可知 $\exists c>0$ 使得 $G\left(s\right)\ge C{s}^N$，$\left|s\right|\ge 1$，因此有

${\displaystyle {\int }_{\Vert v_n\Vert >1}{\left|v_n\right|}^N\text{d}x}\le \frac{1}{C}{\displaystyle {\int }_{\Vert v_n\Vert >1}G\left(f\left(v_n\right)\right)\text{d}x}\le \frac{1}{C}{\displaystyle {\int }_{{ℝ}^N}G\left(f\left(v_n\right)\right)\text{d}x}$

应用条件( $f_7$ ) $\left|f\left(s\right)\right|\ge C\left|s\right|$，对某些 $C>0,s<1$，有

${\displaystyle {\int }_{\Vert v_n\Vert \le 1}{\left|v_n\right|}^N\text{d}x}\le \frac{1}{C}{\displaystyle {\int }_{\Vert v_n\Vert \le 1}{\left|f\left(v_n\right)\right|}^N\text{d}x}\le \frac{1}{C}{\displaystyle {\int }_{{ℝ}^N}{\displaystyle {\int }_{\Vert v_n\Vert \le 1}{\left|f\left(v_n\right)\right|}^N\text{d}x}}$

由此可得出 $\left\{v_n\right\}$ 在 $L^N\left({ℝ}^N\right)$ 中有界，因此在E中有界。利用文献 [13] 类似的方法，可知 $\exists C_0>0$ 使得

${\displaystyle {\int }_{{ℝ}^N}{\left|\nabla v_n\right|}^N+V\left(x\right){\left|f\left(v_n\right)\right|}^N\text{d}x}\ge C_0{\Vert v_n\Vert }_E^N$ (2.12)

于是 $\left\{v_n\right\}$ 在E中有界，同样适用于 $\left\{v_n\right\}$ 的子序列仍记为 $\left\{v_n\right\}$，使得 ${\Vert v_n\Vert }_E，{\Vert v\Vert }_E\le M$ 以及

$v_n\rightharpoonup v\text{in}E,v_n\rightharpoonup v\text{in}{L}_{loc}^q\left({ℝ}^N\right),\forall q\in \left[N,+\infty \right),v_n\left(x\right)\rightharpoonup v\left(x\right)a.e.\text{in}{ℝ}^N$ (2.13)

引理2.6：假设( $V_1$ )和( $F_1$ )~( $F_3$ )成立。如果序列 $\left\{v_n\right\}$ 满足(2.13)则

$\underset{n\to \infty }{\lim }{\displaystyle {\int }_{R^N}g\left(f\left(v_n\right)\right)f^{\prime }\left(v_n\right)v_n\text{d}x}={\displaystyle {\int }_{R^N}g\left(f\left(v\right)\right)f^{\prime }\left(v\right)v\text{d}x}$ (2.14)

证：由条件( $F_1$ )~( $F_3$ )对任意小的 $\epsilon >0$，$\exists S_0>s_0>0$ 使得 $\left|G\left(s\right)\right|\le \epsilon {\left|s\right|}^N,\left|s\right|\le s_0$ 以及 $\left|G\left(s\right)\right|\le \epsilon sg\left(s\right),\left|s\right|\ge S_0$ 说明

$\left|G\left(s\right)\right|\le \epsilon \left({\left|s\right|}^N+sg\left(s\right)\right)+{\chi }_{\left[s_0,S_0\right]}\left(\left|s\right|\right)\left|sg\left(s\right)\right|,\forall s\in ℝ$ (2.15)

其中 ${\chi }_A$ 记为可测子集 $A\subset R$ 的特征函数，这样对 $n\in \mathbb{N}$ 可以得到

$\left|g\left(f\left(v_n\right)\right)f^{\prime }\left(v_n\right)v_n\right|\le \left|g\left(f\left(v_n\right)\right)f\left(v_n\right)\right|\le \epsilon \left(V{\left|f\left(v_n\right)\right|}^N+f\left(v_n\right)g\left(f\left(v_n\right)\right)\right)+{\chi }_{\left[s_0,S_0\right]}\left(\left|f\left(v_n\right)\right|\right)\left|f\left(v_n\right)g\left(f\left(v_n\right)\right)\right|$,

以及

${\displaystyle {\int }_{B_r^c}\left|g\left(f\left(v_n\right)\right)f^{\prime }\left(v_n\right)v_n\right|}\le \epsilon \left|Q\left(v_n\right)\right|+S_{0}g\left(S_0\right){\displaystyle {\int }_{A_n\cap B_r^c}{\chi }_{\left[s_0,S_0\right]}\left(\left|f\left(v_n\right)\right|\right)\text{d}x}$ (2.16)

其中 $A_n=\left\{x\in R^N:s_0\le \left|f\left(v_n\right)\right|\le S_0\right\}$ 还有

$Q\left(v_n\right)={\displaystyle {\int }_{{ℝ}^N}\left(V\left(x\right){\left|f\left(v_n\right)\right|}^N+f\left(v_n\right)g\left(f\left(v_n\right)\right)\right)\text{d}x}$

此外，

$s_{0}g\left(s_0\right)\left|A_n\right|\le {\displaystyle {\int }_{{ℝ}^N}f\left(v_n\right)g\left(f\left(v_n\right)\right)\text{d}x}\le M_1$

说明 ${\sup }_{n\in N}\left|A_n\right|\le \frac{M_1}{s_{0}g\left(s_0\right)}=\mu <\infty ,\left|A_n\right|=\text{meas}\left(A_n\right)$。由此可以断定对所有的 $n\in N$，具有一致性。首先我们来证明

$\underset{r\to \infty }{\lim }\left|A_n\cap B_r^c\right|=0$ (2.17)

事实上，如果不成立，则 $\exists n_0\ge 1,\delta >0,r_j\uparrow \infty$ 使得

$\left|A_{n_0}\cap B_{r_j}^c\right|\ge \delta ,\forall j\in N$

不失一般性令 $n_0=1$ 显然有 $\left|A_{n_0}\cap B_{r_j}^c\right|\le \left|A_1\right|\le \mu ,\forall j\in N$。

记 ${\Omega }_j=B_{r_j}^c\backslash \overline{B_{r_{j+1}}^c}$，易见

$B_{r_j}^c={\displaystyle \underset{l=j}{\overset{\infty }{\cup }}{\Omega }_l},\forall j\in N,{\Omega }_l\cap {\Omega }_k=\varnothing ,l\ne k$

那么会有

$\left|A_1\cap B_{r_j}^c\right|={\displaystyle {\sum }_{l=j}^{\infty }\left|A_1\cap {\Omega }_l\right|}\ge \delta ,\forall j\in N$ (2.18)

以及序列 ${\displaystyle {\sum }_{l=j}^{\infty }\left|A_1\cap {\Omega }_l\right|}=\infty$，产生矛盾。于是极限(2.17)证毕。

其次，说明极限(2.17)的一致性。实际上由(2.3)可知 $v_n,v\in L^N\left(R^N\right)$ 在 $R^N$ 上 $v_n\left(x\right)\to v\left(x\right)$ 几乎处处成立，因此对于任意小的 $ϵ>0,\exists r_0\ge 1$ 使得对 $r\ge r_0$ 有

${\displaystyle {\int }_{B_r^c}{\left|v\right|}^N\text{d}x}\le ϵ$ (2.19)

对该 $ϵ>0$ 选择 $t_1=r_0,t_j\uparrow \infty$ 使得 $D_j=B_{t_j}^c\backslash \overline{B_{t_{j+1}}^c},B_{r_0}^c={\displaystyle {\cup }_{j=1}^{\infty }{D}_j}$ 和

${\displaystyle {\int }_{D_j}{\left|v\right|}^N\text{d}x}\le \frac{ϵ}{2^j},\forall j\in N$ (2.20)

由Fatou引理，可得对任意 $j\in N$，会有

$\underset{n\to \infty }{\lim }\sup {\displaystyle {\int }_{D_j\cap A_n}{\left|f\left(v_n\right)\right|}^N\text{d}x}\le {\displaystyle {\int }_{D_j}\underset{n\to \infty }{\lim }\sup {\left|f\left(v_n\right)\right|}^N\text{d}x}\le {\displaystyle {\int }_{D_j}{\left|f\left(v\right)\right|}^N\text{d}x}\le \frac{ϵ}{2^j}$ (2.21)

于是

$\begin{array}{c}{s}_0^N\underset{n\to \infty }{\lim }\sup \left|A_n\cap B_{r_0}^c\right|\le \underset{n\to \infty }{\lim }\underset{n\to \infty }{\lim }\sup {\displaystyle {\int }_{B_{r_0}^c\cap A_n}{\left|f\left(v_n\right)\right|}^N\text{d}x}\\ =\underset{n\to \infty }{\lim }\sup {\displaystyle {\sum }_{j=1}^{\infty }{\displaystyle {\int }_{D_j\cap A_n}{\left|f\left(v_n\right)\right|}^N\text{d}x}}\\ \le {\displaystyle {\sum }_{j=1}^{\infty }\underset{n\to \infty }{\lim }\sup {\displaystyle {\int }_{D_j\cap A_n}{\left|f\left(v_n\right)\right|}^N\text{d}x}}\\ \le {\displaystyle {\sum }_{j=1}^{\infty }{\displaystyle {\int }_{D_j}{\left|f\left(v_n\right)\right|}^N\text{d}x}}\le {\displaystyle {\sum }_{j=1}^{\infty }\frac{ϵ}{2^j}}=ϵ\end{array}$ (2.22)

注意对 $r\ge r_0,n\in \mathbb{N}$，可得 $\left(A_n\cap B_r^c\right)\subset \left(A_n\cap B_{r_0}^c\right)$，即得 ${\lim }_{r\to \infty }\left|A_n\cap B_r^c\right|=0$ 对 $n\in N$ 具有一致性。

此外，由积分的决定连续性可知对 $\forall ϵ>0$，存在常数 $0<\delta <ϵ\backslash S_{0}g\left(S_0\right)ϵ>0$ 使得对 $r\ge r_0$ 有 $\text{meas}\left(A_n\cap B_r^c\right)<\delta$ 以及

${\displaystyle {\int }_{A_n\cap B_r^c}{\chi }_{\left[s_0,S_0\right]}\left(\left|f\left(v_n\right)\right|\right)\text{d}x}<\frac{ϵ}{S_{0}g\left(S_0\right)}$ (2.23)

那么由(2.16)和(2.23)可推出

${\displaystyle {\int }_{B_r^c}\left|g\left(f\left(v_n\right)\right)f^{\prime }\left(v_n\right)v_n\right|\text{d}x}\le {\displaystyle {\int }_{B_r^c}\left|g\left(f\left(v_n\right)\right)f\left(v_n\right)\right|\text{d}x}\le ϵ\left(M_2+1\right),r\ge r_0$ (2.24)

和

${\displaystyle {\int }_{B_r^c}\left|g\left(f\left(v\right)\right)f^{\prime }\left(v\right)v\right|\text{d}x}\le {\displaystyle {\int }_{B_r^c}\left|g\left(f\left(v\right)\right)f\left(v\right)\right|\text{d}x}\le {\displaystyle {\int }_{B_r^c}\left|g\left(f\left(v_n\right)\right)f\left(v_n\right)\right|\text{d}x}\le ϵ\left(M_2+1\right),r\ge r_0$ (2.25)

另一方面，由(2.17)对所有的 $r>0$，

$\underset{n\to \infty }{\lim }{\displaystyle {\int }_{B_r}g\left(f\left(v_n\right)\right)f^{\prime }\left(v_n\right)v_n\text{d}x}={\displaystyle {\int }_{B_r}g\left(f\left(v\right)\right)f^{\prime }\left(v\right)v\text{d}x}$ (2.26)

那么从(2.24)~(2.26)，可推出(2.14)。

引理2.7：假设( $V_1$ )和( $F_1$ )~( $F_3$ )成立，令 $\left\{v_n\right\}$ 为一PS序列，则有

$\underset{n\to \infty }{\lim }\sup {\displaystyle {\int }_{{ℝ}^N}V\left(x\right){\left|f\left(v_n\right)\right|}^N\text{d}x}={\displaystyle {\int }_{{ℝ}^N}V\left(x\right){\left|f\left(v\right)\right|}^N\text{d}x}$ (2.27)

和

$\underset{n\to \infty }{\lim }{\displaystyle {\int }_{{ℝ}^N}V\left(x\right){\left|f\left(v_n\right)\right|}^{N-2}f\left(v_n\right)f^{\prime }\left(v_n\right)v_n\text{d}x}={\displaystyle {\int }_{{ℝ}^N}V\left(x\right){\left|f\left(v\right)\right|}^{N}f\left(v\right)f^{\prime }\left(v\right)v\text{d}x}$ (2.28)

证：对 $r>1$，可选择函数 ${\eta }_r={\eta }_r\left(\left|x\right|\right)\in C^1\left({ℝ}^N\right)$ 使得

${\eta }_r\left(\left|x\right|\right)\equiv 1,x\in B_{2r}^c;{\eta }_r\left(\left|x\right|\right)=0,x\in B_r;0\le {\eta }_r\le 1,\left|\nabla {\eta }_r\right|\le \frac{2}{r},\text{in}{ℝ}^N$ (2.29)

由于序列 $\left\{v_n\right\}$ 在中E有界，则序列 $\left\{{\eta }_r{\phi }_n\right\}$，${\phi }_n=\frac{f\left(v_n\right)}{f^{\prime }\left(v_n\right)}$，也在E中有界，那么会有 $J\left(v_n\right)\left({\eta }_r{\phi }_n\right)=o_n\left(1\right)$，即

$\begin{array}{l}{\displaystyle {\int }_{{ℝ}^N}{\left|\nabla v_n\right|}^{N-2}\nabla v_n\nabla {\phi }_n{\eta }_r\text{d}x}+{\displaystyle {\int }_{{ℝ}^N}V\left(x\right){\left|f\left(v_n\right)\right|}^N{\eta }_r\text{d}x}\\ =-{\displaystyle {\int }_{{ℝ}^N}{\left|\nabla v_n\right|}^{N-2}\nabla v_n\nabla {\eta }_r{\phi }_n\text{d}x}+{\displaystyle {\int }_{{ℝ}^N}g\left(f\left(v_n\right)\right)f\left(v_n\right){\eta }_r\text{d}x}+o_n\left(1\right)\end{array}$ (2.30)

其中 $\nabla {\phi }_n=\left(1+\frac{\left(2\alpha -1\right)\left(2{\alpha }^{N-1}\right){\left|f\left(v_n\right)\right|}^{\left(2\alpha -1\right)N}}{1+2{\alpha }^{N-1}{\left|f\left(v_n\right)\right|}^{\left(2\alpha -1\right)N}}\right)\nabla v_n$ 以及 $\omega \left(r\right)={\displaystyle {\int }_{{ℝ}^N}g\left(f\left(v_n\right)\right)f\left(v_n\right){\eta }_r\text{d}x}\to 0,n\to \infty$ 那么(2.30)的极限说明

$\begin{array}{l}{\displaystyle {\int }_{B_r^c}\left({\left|\nabla v_n\right|}^N+V\left(x\right){\left|f\left(v_n\right)\right|}^N\right){\eta }_r\text{d}x}\\ \le {\displaystyle {\int }_{B_r^c}{\left|\nabla v_n\right|}^{N-2}\left|v_n\right|\left|\nabla {\eta }_r\right|\text{d}x}+o_n\left(1\right)+\omega \left(r\right)\\ \le \frac{4\alpha }{r}{\displaystyle {\int }_{B_r^c}{\left|\nabla v_n\right|}^{N-2}\left|v_n\right|\text{d}x}+o_n\left(1\right)+\omega \left(r\right)\\ \le \frac{4\alpha }{r}{\Vert \nabla v_n\Vert }_{L^N\left({\Omega }_r\right)}^{N-1}{\Vert v_n\Vert }_{L^N\left({\Omega }_r\right)}+o_n\left(1\right)+\omega \left(r\right)\\ \le \frac{4\alpha }{r}{\Vert v_n\Vert }_E^N+o_n\left(1\right)+\omega \left(r\right)\\ \le \frac{4\alpha }{r}{M}^p+o_n\left(1\right)+\omega \left(r\right),n\to \infty ,r\to \infty \end{array}$ (2.31)

其中 ${\Omega }_r=B_r^c\backslash \overline{B_{2r}^c}$，可推出 ${\lim }_{n\to \infty }\sup {\displaystyle {\int }_{B_{2r}^c}\left({\left|\nabla v_n\right|}^N+V\left(x\right){\left|f\left(v_n\right)\right|}^N\right)\text{d}x}<ϵ$。

此外，该极限还可得

${\lim }_{n\to \infty }\sup {\displaystyle {\int }_{B_{2r}^c}\left(V\left(x\right){\left|f\left(v_n\right)\right|}^N\right)\text{d}x}\le ϵ$ (2.32)

便有，

${\displaystyle {\int }_{B_{2r}^c}\left(V\left(x\right){\left|f\left(v\right)\right|}^N\right)\text{d}x}\le ϵ$ (2.33)

由于 $L^N\left(B_{2r}\right)$ 中， $v_n\to v$，则有

${\lim }_{n\to \infty }{\displaystyle {\int }_{B_{2r}}V\left(x\right){\left|f\left(v_n\right)\right|}^N\text{d}x}={\displaystyle {\int }_{B_{2r}}V\left(x\right){\left|f\left(v\right)\right|}^N\text{d}x}$ (2.34)

对 $\forall ϵ>0$，极限(2.32)~(2.34)产生

${\lim }_{n\to \infty }\sup {\displaystyle {\int }_{R^N}\left|V\left(x\right)\left({\left|f\left(v_n\right)\right|}^N-{\left|f\left(v\right)\right|}^N\right)\text{d}x\right|}\le 3ϵ$. (2.35)

接下来证明(2.28)，由 $f\left(6\right)$ 和 $\omega \left(r\right)\to 0$ 可得

${\displaystyle {\int }_{{ℝ}^N}g\left(f\left(v_n\right)\right)f^{\prime }\left(v_n\right)v_n{\eta }_r\text{d}x}\le {\displaystyle {\int }_{{ℝ}^N}g\left(f\left(v_n\right)\right)f\left(v_n\right){\eta }_r\text{d}x}=\omega \left(r\right)\to 0$ (2.36)

有由 $J\left(v_n\right)\left({\eta }_r{v}_n\right)=o_n\left(1\right)$ 可得

$\begin{array}{l}{\displaystyle {\int }_{{ℝ}^N}\left[{\left|\nabla v_n\right|}^N+V\left(x\right){\left|f\left(v_n\right)\right|}^{N-2}f\left(v_n\right)f^{\prime }\left(v_n\right)\right]{\eta }_r\text{d}x}\\ =-{\displaystyle {\int }_{{ℝ}^N}{\left|\nabla v_n\right|}^{N-2}\nabla v_n\nabla {\eta }_r{v}_n\text{d}x}+{\displaystyle {\int }_{{ℝ}^N}g\left(f\left(v_n\right)\right)f^{\prime }\left(v_n\right){\eta }_r\text{d}x}+o_n\left(1\right)\\ \le {\displaystyle {\int }_{{ℝ}^N}{\left|\nabla v_n\right|}^{N-1}\left|\nabla v_n\right|\left|\nabla {\eta }_r\right|\text{d}x}+o_n\left(1\right)+\omega \left(r\right)\\ \le \frac{4\alpha }{r}{M}^p+o_n\left(1\right)+\omega (r)\end{array}$

那么对 $\forall ϵ>0,\exists r_0\ge 1$ 使得 $r>r_0$，

${\lim }_{n\to \infty }\sup {\displaystyle {\int }_{B_{2r}^c}\left[{\left|\nabla v_n\right|}^N{\eta }_r+V\left(x\right){\left|f\left(v_n\right)\right|}^{N-2}f\left(v_n\right)f^{\prime }\left(v_n\right)v_n\right]\text{d}x}\le ϵ$ (2.37)