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数学物理学报, 2020, 40(3): 725-734 doi:

论文

一类p-Laplacian方程解的存在性与对称性研究

钱红丽, 黄小涛,

A Symmetry Result for a Class of p-Laplace Involving Baouendi-Grushin Operators via Constrained Minimization Method

Qian Hongli, Huang Xiaotao,

通讯作者: 黄小涛, E-mail: xiaotao huang2008@hotmail.com

收稿日期: 2019-06-11  

基金资助: 南京航空航天大学青年科技创新基金.  NS2019044
南京航空航天大学研究生创新基地(实验室)开放基金.  KFJJ20180803

Received: 2019-06-11  

Fund supported: the Nanjing University of Aeronautics and Astronautics Youth Technology Innovation Fund.  NS2019044
the Foundation of the Graduate Innovation Center, Nanjing University of Aeronautics and Astronautics.  KFJJ20180803

摘要

该文研究一类含Baouendi-Grushin算子的p-Laplace方程解的存在性与对称性问题.通过重排方程对应的极小化约束泛函,得到此泛函关于某点中心对称的最小极值,从而获得方程正解的一些存在性和对称性结果.此结论推广了经典p-Laplace方程和Baouendi-Grushin型Laplace方程的相关结果.

关键词: p-Laplace方程 ; Baouendi-Grushin算子 ; 存在性和对称性 ; 重排 ; 约束最小化方法

Abstract

The purpose of this paper is to investigate a spacial p-Laplace equation involving Baouendi-Grushin operators. Some existence and symmetry results for positive solutions are obtained by rearrangement of its corresponding constrained minimization. These results are in accordance with those for the classical p-Laplace equations and the Baouendi-Grushin type sub-Laplacian equations.

Keywords: p-Laplace equations ; Baouendi-Grushin operators ; Existence and symmetry ; rearrangement ; Constrained minimization method

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本文引用格式

钱红丽, 黄小涛. 一类p-Laplacian方程解的存在性与对称性研究. 数学物理学报[J], 2020, 40(3): 725-734 doi:

Qian Hongli, Huang Xiaotao. A Symmetry Result for a Class of p-Laplace Involving Baouendi-Grushin Operators via Constrained Minimization Method. Acta Mathematica Scientia[J], 2020, 40(3): 725-734 doi:

1 引言

近年来,诸多学者对p-Laplace方程进行广泛的研究.本文讨论一类推广情形

div(|Au|p2Au)=f(x),
(1.1)

其中1<p<, A={aij(x)}n×n为对称矩阵且满足一致椭圆条件

λ|ξ|2A(x)ξξΛ|ξ|2
(1.2)

对任意ξ,xRn及常数0<λ<Λ<成立.此类偏微分方程一般来自几何、调和映射、流体动力学等方面,其理论已被广泛研究. Ladyzhenskaja和Ural'tzeva在文献[1]中利用DeGiorgi迭代[2]的方法证明了方程(1.1)的弱解是H¨older连续的. Serrin[3]和Trudinger[4]通过Moser迭代[5]证明了方程的非负解满足Harnack原理.文献[6-7]等证明了具有不连续系数的一般p-Laplace方程解的正则性问题.在本文中,我们研究一类含Baouendi-Grushin算子的p-Laplace方程

G(|Gu|p2Gu)=f(z),zRN,

其中z=(x,y)Rn×Rm=RN, 1<p<+. Baouendi-Grushin向量场G定义为(参见文献[8])

G=(x,|x|γy)=(x1,,xn,|x|γy1,,|x|γym),γ0,

则Baouendi-Grushin型sub-Laplace算子定义为

ΔGu=GG(u(x))=Δxu+|x|2γΔyu=(In×n00|x|2γIm×m)(Diju),
(1.3)

其中ΔxΔy分别表示RnRm空间上的一般Laplacian算子.当x0时,算子满足一致椭圆条件;当x=0时,算子是退化的.近年来,关于Baouendi-Grushin算子(1.3)已有大量研究. Jerison和Lee在文献[9]中研究了含Baouendi-Grushin算子的p-Laplace方程

ΔGu=Δxu+|x|2γΔyu=f(x,y)
(1.4)

与Cauchy-Riemann Yamabe问题的关系. Monti和Morbidelli在文献[10]中研究了临界半线性Baouendi-Grushin方程,并通过kelvin变换给出了对称性结果.在文献[11]中, Ferrari和Valdinoci证明了几何Sobolev-Poincaré不等式以及Grushin平面中稳定解的一些对称结果.有关更多详细信息请参见文献[12]和[13].通过使用Carnot-Caratheodory度量, Franchi在文献[14]中证明了加权Sobolev-Poincare不等式、Harnack不等式和弱解的Cα估计.最近,在文献[15]中,王证明了方程(1.4)解的C2,α正则性估计.宋、王等人在文献[16]得到了方程解的最佳梯度估计. p-Laplace型Baouendi-Grushin算子定义为

ΔpG[u]:=G(|Gu|p2Gu).
(1.5)

{(x,y):x0}的区域上,此算子可以看成是经典的p-Laplace算子;而在{(x,y):x=0}区域附近时,算子是退化(p>2)或奇异的(1<p<2).在文献[8]中, Kombe证明了带Baouendi-Grushin算子的非线性抛物p-Laplace方程正解的一些存在性结果.在文献[17]中,黄、马和王研究了p-Laplace型Baouendi-Grushin方程解的梯度估计.

在本文中,我们研究带Baouendi-Grushin算子的Schrödinger型p-Laplace方程

ΔpGu+μu=λ|u|q1u,λ>0,μ>0.
(1.6)

为了利用约束最小化方法研究方程(1.6),我们需要引入一类新的Sobolev空间

S1,pγ(RN):={u:uLp(RN),GuLp(RN)}.

本文的主要存在性和对称性结果如下:

定理1.1  令Q=n+(1+γ)m,1<p<Qp=QpQp1.q(1,p)时,方程(1.6)存在一个对称的正解uS1,pγ(RN).

p=2时,方程(1.6)为Schrödinger型Laplace方程

ΔGu+μu=λ|u|q1u.

此方程的存在性与对称性被广泛研究,具体可参考文献[18].特别的,当p=2,γ=0,方程(1.6)变为Allen-Cahn方程(非线性Schrödinger方程)

Δu+μu=λ|u|q1u.

此方程的解随着半径增加到最大值,并在无穷远处减小且收敛为零.因此,不能利用移动平面法研究此类Schrödinger方程.在文献[19]中, Berestycki和Lions通过约束最小化方法给出了此类方程的一些对称性结果.利用相同的方法, Dipierro、Palatucci和Valdinoci在文献[20]中研究了以下Schrödinger型分数阶Laplace方程

(Δ)su+μu=λ|u|q1u,xRn,s(0,1)

正解的对称性问题.最近,本文的第一作者在文献[21]中给出了分数阶p-Laplace方程

(Δ)spu+μu=λ|u|q1u,xRn,s(0,1)

解的存在性和对称性结果.

在本节的最后,我们介绍与问题(1.6)相关的拉格朗日泛函

L(u)=1pRN|Gu|pdzRNG(u)dz,
(1.7)

其中

G(u)=u0g(t)dt,g(t)=λ|t|qμ|t|.

函数L(u)是与问题(1.6)相关的能量泛函.基于G(u)中存在μ|t|项,导致L(u)在空间S1,pγ(RN)上是无界的,所以无法寻找函数L(u)的临界点.因此本文将问题(1.6)转化为以下约束极小化问题,而不是通过寻找L的临界点来解决问题.

I:=min
(1.8)

其中

M(v) = \int_{{{\Bbb R}}^N}|\nabla_G v|^p{\rm d}z, \; A(v) = \int_{{{\Bbb R}}^N}G(v(z)){\rm d}z, \; G(v) = \frac{\lambda|v|^{q+1}}{q+1}-\frac{\mu|v|^2}{2}.

上式极小化问题的解也是方程(1.6)的解.事实上,我们假定 u 是方程(1.8)的一个解,则存在拉格朗日乘数 \theta ,使得

\begin{eqnarray*} -\Delta_G^pu = \theta g(u), \; \; \; z\in {{\Bbb R}}^N. \end{eqnarray*}

u_\sigma(z) = u(z/\sigma) = u(\xi) ,则有

\begin{eqnarray*} -\Delta_G^pu_\sigma = -\nabla_G(|\nabla_Gu_\sigma|^{p-2}\nabla_Gu_\sigma) & = &-\frac{1}{\sigma^p}\nabla_G(|\nabla_Gu(\xi)|^{p-2}\nabla_Gu(\xi))\\ & = &-\frac{1}{\sigma^p}\Delta_G^pu(\xi), \; \; \; z\in {{\Bbb R}}^N. \end{eqnarray*}

-\Delta_G^pu = \theta g(u), \; \; \; z\in {{\Bbb R}}^N,

由上式可得

-\Delta_G^pu_\sigma = \frac{\theta}{\sigma^p} g(u_\sigma).

\sigma = \theta^{1/p} ,可以得到 u_\sigma 是方程(1.6)的解.因此,本文主要证明约束最小化问题(1.8)存在对称解,从而得到方程(1.6)解的对称性.

2 预备知识

在本节中,我们陈述一些初步结果,这些结果会在下文证明中使用.在全文中,我们规定 Q = n+(1+\gamma)m N = n+m .

2.1 内在度量与Sobolev空间

对任意的 z_1 = (x_1, y_1), z_2 = (x_2, y_2) , Carnot-Caratheodory距离定义为

d_G(z_1, z_2) = |x_1-x_2|+\frac{|y_1-y_2|}{|x_1|^{\gamma}+|x_2|^{\gamma}+|y_1-y_2|^{\frac{\gamma}{1+\gamma}}},

其中 d_G(z) 满足以下放缩性(请参见文献[15])

d_G(rz_1, rz_2) = rd_G(z_1, z_2).

Sobolev空间

S^{1, p}_{{\gamma}}({{\Bbb R}}^N): = \Big\{u:\; u\in L^p({{\Bbb R}}^N), \nabla_G u\in L^p({{\Bbb R}}^N) \Big\}

在范数

||u||_{S^{1, p}_{{\gamma}}({{\Bbb R}}^N)} = \bigg\{ \int_{{{\Bbb R}}^N}|u|^p+\int_{{{\Bbb R}}^N}|\nabla_G u|^p\bigg\}^{\frac{1}{p}}

下是一个Banach空间.若 \Omega 有界且 2<q <\frac{pQ}{Q-p} ,则有以下紧嵌入(参见文献[22])

\begin{equation} S^{1, p}_\gamma(\Omega)\hookrightarrow L^q(\Omega). \end{equation}
(2.1)

2.2 Schwarz对称

再次,我们回顾Schwarz对称的定义和基本性质.

\Omega {{\Bbb R}}^N 中的开集, u:\Omega\rightarrow{{\Bbb R}} 是度量函数,分布函数 \varphi_u(t)

\varphi_u(t) = \left |{x\in{{{\Bbb R}}^N}:|u(x)|>t} \right|,

其中 |\cdot| 表示Lebesgue测度,则方程 \varphi_u 是递减且右连续的. u 的递减重排可以定义为

u^\sharp(s) = \sup\big\{t\geq0:\varphi_u(t)\geq s\big\},

u^*(x) = u^\sharp(\omega_N|x|^N), \; x\in {{\Bbb R}}^N,

其中 u^*(x) 表示 u 的Schwarz对称化,其具有递减性与对称性. Schwarz对称化具有以下基本属性,具体可参考文献[23-24].

引理2.1  设 v, w {{\Bbb R}}^N 中的积分函数,并且 g: {{\Bbb R}}^N\rightarrow {{\Bbb R}} 为一个递增的非负函数,则有

\int_{{{\Bbb R}}^N}g(|v(x)|){\rm d}x = \int^{|{{\Bbb R}}^N|}_{0}g(v^*(s)){\rm d}s = \int_{{{\Bbb R}}^N}g(v^\sharp(x)){\rm d}x.

在文献[18]中, Lascialfari和Pardo对Baouendi-Grushin算子进行了研究,并得到了如下重排结果.

引理2.2  假设 u\in S^{1, 2}_{{\gamma}}({{\Bbb R}}^N) ,且 u 是紧支集非负函数,则有

\int_{{{\Bbb R}}^n}|\nabla_G u^*|^2{\rm d}x\leq \int_{{{\Bbb R}}^n}|\nabla_G u|^2{\rm d}x.

Franchi和Lanconelli在文献[22]中证明了如下不等式.

引理2.3  若 u\in S^{1, p}_{{\gamma}}({{\Bbb R}}^N) (p\geq1) ,则 u^* \in S^{1, p}_{{\gamma}}({{\Bbb R}}^N) ,且有

\int_{{{\Bbb R}}^n}|\nabla_G u^*|^p{\rm d}x\leq \int_{{{\Bbb R}}^n}|\nabla_G u|^p{\rm d}x,

此引理也可以由引理2.2推导得到.

3 定理1.1的证明

在本节中,我们给出定理1.1的证明.在引言中提到过定理1.1的证明可以转化为研究以下约束最小化问题

\begin{equation} I: = \min\Big\{M(v):v\in S_\gamma^{1, p}({{\Bbb R}}^N), A(v) = 1\Big\}, \end{equation}
(3.1)

其中

M(v) = \int_{{{\Bbb R}}^N}|\nabla_G v|^p{\rm d}z, \; \; A(v) = \int_{{{\Bbb R}}^N}G(v(z)){\rm d}z, \; \; G(v) = \frac{\lambda|v|^{q+1}}{q+1}-\frac{\mu|v|^2}{2}.

现在我们证明问题(3.1)存在对称正解.本文在引言中介绍了极小化问题(3.1)中解的对称性意味着方程(1.6)解的对称性.

定理1.1的证明分为以下两个步骤:

步骤1  证明非空集合 \big\{v\in S_\gamma^{1, p}({{\Bbb R}}^N), A(v) = 1\big\} 里存在对称的序列 u^*_k .

步骤2  证明此序列子列的极限是问题(3.1)的解.

步骤1  首先证明集合 \big\{v\in S_\gamma^{1, p}({{\Bbb R}}^N), A(v) = 1\big\} 非空.对 \rho>1, d>0 ,定义

\begin{equation} w_\rho(z) = \left\{\begin{array}{ll} d, \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; &|z|\leq \rho, \\ d(\rho+1-|z|), \; \; \; \; &\rho<|z|< \rho+1, \\ 0, \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; &|z|\geq \rho+1. \end{array} \right. \end{equation}
(3.2)

经计算

\int_{{{\Bbb R}}^N}|\nabla_G w_\rho(z)|^p{\rm d}z = C\int^{\rho+1}_\rho\bigg|\frac{{\rm d}w_\rho(r)}{{\rm d}r}\bigg|^pr^{N-1}{\rm d}r = C\int^{\rho+1}_\rho r^{N-1}{\rm d}r<C,

可得到 w_\rho(z)\in S_\gamma^{1, p}({{\Bbb R}}^N) .对任意固定的 \sigma>0 ,定义

w_{\rho, \sigma}(z) = w_\rho\bigg(\frac{z}{\sigma}\bigg),

可得 w_{\rho, \sigma}(z)\in S_\gamma^{1, p}({{\Bbb R}}^N) .由于

\begin{eqnarray*} A(w_\rho(z))& = &\int_{{{\Bbb R}}^N}G(w_\rho(z)){\rm d}z = \int_{B_{\rho+1}}G(w_\rho(z)){\rm d}z\\ & = &\int_{B_\rho}G(w_\rho(z)){\rm d}z+\int_{B_{\rho+1}\backslash B_{\rho}}G(w_\rho(z)){\rm d}z\\ &\geq &C_1\big|B_\rho\big|-C_2\big|B_{\rho+1}\backslash B_\rho\big|\geq C_1\rho^N-C_2\rho^{N-1}, \end{eqnarray*}

其中 C_1>0 , C_2>0 . \rho_1>0 足够大,对任意的 \rho>\rho_1 ,有 \int_{{{\Bbb R}}^N}A(w_\rho(z)){\rm d}z>0 .取合适的常数 \sigma>0 ,可以得到

\begin{eqnarray*} \int_{{{\Bbb R}}^N}G(w_{\rho, \sigma}(z)){\rm d}z = \sigma^N\int_{{{\Bbb R}}^N}G(w_\rho(z)){\rm d}z = 1, \end{eqnarray*}

所以 w_{\rho, \sigma}(z) 是集合 \big\{v\in S_\gamma^{1, p}({{\Bbb R}}^N), A(v) = 1\big\} 中的元素.

其次证明此集合中存在对称的序列 u_k 为问题(3.1)的解,并对 {u_k} 作先验估计.由于集合 \{v\in S_\gamma^{1, p}({{\Bbb R}}^N), A(v) = 1\} 非空,我们选择 \{u_k \}\subset S_\gamma^{1, p}({{\Bbb R}}^N) 使得 \int_{{{\Bbb R}}^N}G(u_k (z)){\rm d}z = 1 且满足

\begin{eqnarray*} \lim\limits_{k\rightarrow \infty}[u_k ]^p_{S_\gamma^{1, p}({{\Bbb R}}^N)} = \inf\bigg\{[v]^p_{S_\gamma^{1, p}({{\Bbb R}}^N)}: v\in S_\gamma^{1, p}({{\Bbb R}}^N), \int_{{{\Bbb R}}^N}G(v) = 1 \bigg\} = I>0. \end{eqnarray*}

u^*_k 表示 u_k 的Schwarz重排.由Schwarz对称性可知 u^*_k 具有对称性与递减性,并且有

\begin{eqnarray*} \int_{{{\Bbb R}}^N}|\nabla_G u^*_k |^p\leq \int_{{{\Bbb R}}^N}|\nabla_G u_k |^p<C, \; \; \int_{{{\Bbb R}}^N}G(u^*_k ){\rm d}z = \int_{{{\Bbb R}}^N}G(u_k ){\rm d}z, \end{eqnarray*}

结合上式,可得到

u^*_k \in {S_\gamma^{1, p}({{\Bbb R}}^N)}, \; \; \int_{{{\Bbb R}}^N}G(u^*_k ){\rm d}z = 1,

这意味着 u^* _k 也是方程(3.1)的解.在不失一般性的情况下,我们可以选择一个正序列 \{u^*_k\} ,使其满足对称性,此外有

\lim\limits_{k\rightarrow \infty}[u^*_k ]^p_{S_\gamma^{1, p}({{\Bbb R}}^N)} = \inf\bigg\{[v]^p_{S_\gamma^{1, p}({{\Bbb R}}^N)}: v\in S_\gamma^{1, p}({{\Bbb R}}^N), \int_{{{\Bbb R}}^N}G(v) = 1 \bigg\}.

对任意 q\in[2, \frac{Qp}{Q-p}] ,现在估计 ||u^*_k ||_{L^q({{\Bbb R}}^N)} ||u^*_k ||_{S_\gamma^{1, p}({{\Bbb R}}^N)} 的有界性.由 u^*_k 的选择可得到

[u^*_k ]_{S_\gamma^{1, p}({{\Bbb R}}^N)}\leq C.

通过Sobolev嵌入定理,我们有

||u^*_k ||_{L^{p^*}({{\Bbb R}}^N)}\leq C [u^*_k ]_{S_\gamma^{1, p}({{\Bbb R}}^N)}<C,

其中 p^* = \frac{Qp}{Q-p} .

现在我们证明 ||u^*_k ||_{L^2({{\Bbb R}}^N)} 是有界的.定义

g_1(t) = \lambda|t|^{q-1}t, g_2(t) = \mu t, G_1(t) = \frac{\lambda|t|^{q+1}}{q+1}, \; G_2(t) = \frac{\mu|t|^2}{2}.

经计算有

g_2(t) = g_1(t)-g(t), \; G_2(u) = G_1(u)-G(u).

对任意的 q\in(1, \frac{Qp}{Q-p}-1) 及固定的 \epsilon>0 ,存在常数 C(\epsilon)>0 ,有

\lambda|t|^{q}\leq \mu\epsilon |t|+{\frac{ C(\epsilon)Qp}{Q-p}}|t|^{\frac{Qp}{Q-p}-1}.

对上不等式从 0 u 进行积分,则有

G_1(u)\leq \epsilon G_2(u)+C(\epsilon)|u|^{\frac{Qp}{Q-p}}.

结合条件 \int_{{{\Bbb R}}^N}G(u^*_k ){\rm d}z = 1 ,计算可得

\begin{eqnarray*} \int_{{{\Bbb R}}^N}G_2(u^*_k ){\rm d}z& = &\int_{{{\Bbb R}}^N}G_1(u^*_k ){\rm d}z-\int_{{{\Bbb R}}^N}G(u^*_k ){\rm d}z = \int_{{{\Bbb R}}^N}G_1(u^*_k){\rm d}z-1\\ &\leq&\int_{{{\Bbb R}}^N}G_1(u^*_k){\rm d}z\leq \epsilon\int_{{{\Bbb R}}^N} G_2(u^*_k){\rm d}z+C(\epsilon)\int_{{{\Bbb R}}^N}|u^*_k |^{\frac{Qp}{Q-p}}{\rm d}z. \end{eqnarray*}

\epsilon = \frac{1}{2} ,则 ||u^*_k ||_{L^2({{\Bbb R}}^N)} 有界,即

\int_{{{\Bbb R}}^N}|u^*_k |^2{\rm d}z = \frac{2}{\mu}\int_{{{\Bbb R}}^N}G_2(u^*_k ){\rm d}z\leq C\int_{{{\Bbb R}}^N}|u^*_k |^{\frac{Qp}{Q-p}}{\rm d}z<C.

最后由 ||u^*_k ||_{L^2({{\Bbb R}}^N)} ||u^*_k ||_{L^{p^*}({{\Bbb R}}^N)} 的有界性,可得对任意的 s\in[2, p^*] ,有

||u^*_k ||_{L^s({{\Bbb R}}^N)}<C.

步骤2  证明步骤1中序列的极限是问题(3.1)的解.在此步骤中,证明存在正函数 \bar u ,其满足对称性且 \bar u 是最小化问题(3.1)的一个解.则意味等式(1.6)具有正解 \bar u_\sigma 且满足对称性.

由步骤1中选取的 \{u^*_k \} ,可知它是对称的且满足

\lim\limits_{k\rightarrow \infty}[u^*_k ]^p_{S_\gamma^{1, p}({{\Bbb R}}^N)} = I,

易知 \{u^*_k\} S_\gamma^{1, p}({{\Bbb R}}^N) 中是有界的,且存在序列 u^*_k (仍然用 u^*_k 表示)和函数 \bar u\in S_\gamma^{1, p}({{\Bbb R}}^N)

\begin{eqnarray*} &&u^*_k \rightharpoonup \bar u, \; \; \; \; {\rm in}\; \; S_\gamma^{1, p}({{\Bbb R}}^N), \\ &&u^*_k \rightarrow \bar u, \; \; \; \; {\rm a.e. \; \; in}\; \; {{\Bbb R}}^N. \end{eqnarray*}

易知 \bar u 具有对称性与递减性.接下来证明 \bar u 是问题(3.1)的解,即

\begin{equation} [\bar u ]^p_{S_\gamma^{1, p}({{\Bbb R}}^N)} = \inf\bigg\{[v]^p_{S_\gamma^{1, p}({{\Bbb R}}^N)}: v\in S_\gamma^{1, p}({{\Bbb R}}^N), \int_{{{\Bbb R}}^N}G(v) = 1 \bigg\}. \end{equation}
(3.3)

其次本文需要证明

\int_{{{\Bbb R}}^N}|u^*_k |^{q+1}{\rm d}z\rightarrow \int_{{{\Bbb R}}^N}|\bar u|^{q+1}{\rm d}z, \; \; \; \; k\rightarrow +\infty.

q\in (1, \frac{Qp}{Q-p}-1) ,可得

\frac{|u^*_k |^{q+1}}{|u^*_k |^2+|u^*_k |^{\frac{Qp}{Q-p}}}\rightarrow 0, \; \; \; u^*_k \rightarrow 0 \; \mbox{或} \; u^*_k \rightarrow \infty.

对任意的 x\in {{\Bbb R}}^N ,经计算可以得到

|u^*_k (z)|^{q+1}\leq C_1|u^*_k (z)|^2+C_2|u^*_k (z)|^{\frac{Qp}{Q-p}}.

对任意的 q\in (1, \frac{Qp}{Q-p}-1) ,利用Fatou's引理,计算可得

\int_{{{\Bbb R}}^N}|u^*_k (z)|^{q+1}{\rm d}z<C\int_{{{\Bbb R}}^N}|u^*_k |^{2}{\rm d}z+C\int_{{{\Bbb R}}^N}|u^*_k |^{\frac{Qp}{Q-p}}{\rm d}z<C,

\int_{{{\Bbb R}}^N}|\bar u(z)|^{q+1}{\rm d}z<C\int_{{{\Bbb R}}^N}|u^*_k |^{2}{\rm d}z+C\int_{{{\Bbb R}}^N}|u^*_k |^{\frac{Qp}{Q-p}}{\rm d}z<C,

对任意的 \epsilon>0 及固定的 R_1>0 ,存在 N>0 ,对所有 n>N ,有

\begin{equation} \int_{|z|\leq R_1}|(u^*_k) ^{q+1}-{\bar u}^{q+1}|{\rm d}z\leq C\epsilon. \end{equation}
(3.4)

对任意的 s\in [2, p^*] ,我们由步骤1可知 u^*_k \in L^{s}({{\Bbb R}}^N) 是递减函数,则对任意 r = |z|>0 ,有

\|u^*_k\|^{s}_{L^{s}({{\Bbb R}}^N)} = \int_{{{\Bbb R}}^N}|u^*_k|^{s}(z){\rm d}z\geq \alpha(N)\int^{R}_{0}|u^*_k|^{s}(r)r^{N-1}{\rm d}r\geq \alpha(N)|u^*_k|^{s}(R)\frac{R^N}{N},

简单计算有

|u^*_k|^{s}(R)\leq \frac{N}{\alpha(N)R^N}\|u^*_k\|^{s}_{L^{s}({{\Bbb R}}^N)}.

整理得

|u^*_k(z)|\leq C|z|^{-\frac{N}{s}}||u^*_k||_{L^{s}({{\Bbb R}}^N)}<C|z|^{-\frac{N}{s}},

u^*_k (z)\rightarrow0\; \; as \; \; |z|\rightarrow+\infty.

对任意 \epsilon>0 ,存在 R_1>0 足够大使得任意 |z|>R_1 ,有

\begin{equation} \int_{|z|>R_1}| u^*_k |^{q+1}{\rm d}z\leq C \int_{|z|>R_1}| u^*_k |^{2}{\rm d}z\leq C\epsilon. \end{equation}
(3.5)

通过Fatou's引理,计算得

\begin{equation} \int_{|z|>R_1}|\bar u|^{q+1}{\rm d}z\leq C\epsilon. \end{equation}
(3.6)

整理(3.4)–(3.6)式,计算可知对任意得 \epsilon>0 ,有

\begin{eqnarray*} \int_{{{\Bbb R}}^N}\big||u^*_k |^{q+1}-|\bar u|^{q+1} \big|{\rm d}z\leq C\epsilon, \end{eqnarray*}

k\rightarrow +\infty 时,

\int_{{{\Bbb R}}^N}|u^*_k |^{q+1}{\rm d}z\rightarrow \int_{{{\Bbb R}}^N}|\bar u|^{q+1}{\rm d}z.

由前面条件

G_1(u) = \frac{\lambda|u|^{q}u}{q+1}, G(u^*_k ) = G_1(u^*_k )-G_2(u^*_k ), \int_{{{\Bbb R}}^N}G(u^*_k ){\rm d}z = 1,

通过Fatou's引理,计算得

\int_{{{\Bbb R}}^N}\frac{\lambda|{\bar u}|^{q+1}}{q+1}{\rm d}z = \int_{{{\Bbb R}}^N}G_1(\bar u){\rm d}z\geq \int_{{{\Bbb R}}^N}G_2(\bar u){\rm d}z+1,

\int_{{{\Bbb R}}^N}|\nabla_G \bar u|^p{\rm d}z\leq \lim\limits_{n\rightarrow \infty}\int_{{{\Bbb R}}^N}|\nabla_G u^*_k |^p{\rm d}z = I.

最后利用反证法证明

\int_{{{\Bbb R}}^N}G(\bar u){\rm d}z = 1.

假设 \int_{{{\Bbb R}}^N}G(\bar u){\rm d}z>1 . \bar u_{\sigma}(z) = \bar u(z/\sigma) .则存在 \sigma\in(0, 1) 使得

\begin{equation} \int_{{{\Bbb R}}^N}G(\bar u_\sigma(z)){\rm d}z = \sigma^N\int_{{{\Bbb R}}^N}G(\bar u(z/\sigma)){\rm d}(z/\sigma) = 1, \end{equation}
(3.7)

且有

\int_{{{\Bbb R}}^N}|\nabla_G \bar u_\sigma|^p{\rm d}z = \sigma^{N-p}\int_{{{\Bbb R}}^N}|\nabla_G \bar u(z)|^p{\rm d}z\leq \sigma^{N-p} I.

通过 I 的定义及不等式(3.7),可得

\begin{eqnarray*} \int_{{{\Bbb R}}^N}|\nabla_G \bar u_\sigma|^p{\rm d}z\geq I = \inf\bigg\{\int_{{{\Bbb R}}^N}|\nabla_G v|^p{\rm d}z : \int_{{{\Bbb R}}^N}G(v){\rm d}z = 1\bigg\}. \end{eqnarray*}

合并上式,对给定的 \sigma\in(0, 1) ,有

\begin{eqnarray*} \int_{{{\Bbb R}}^N}|\nabla_G \bar u_\sigma|^p{\rm d}z = \sigma^{N-p}\int_{{{\Bbb R}}^N}|\nabla_G \bar u(z)|^p{\rm d}z\leq \sigma^{N-p} \int_{{{\Bbb R}}^N}|\nabla_G \bar u_\sigma|^p{\rm d}z. \end{eqnarray*}

整理得

\int_{{{\Bbb R}}^N}|\nabla_G \bar u_\sigma|^p{\rm d}z = 0,

因此产生矛盾,所有

\int_{{{\Bbb R}}^N}G(\bar u){\rm d}z = 1,

进一步可以得到

\int_{{{\Bbb R}}^N}|\nabla_G \bar u|^p{\rm d}z = \lim\limits_{n\rightarrow \infty}\int_{{{\Bbb R}}^N}|\nabla_G u^*_k |^p{\rm d}z = I,

故式(3.3)得证,即完成定理1.1的证明.

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