Pure Mathematics
Vol.08 No.03(2018), Article ID:25136,7 pages
10.12677/PM.2018.83038

Many Radial Solutions of Singular k-Hessian Equations

Huayuan Sun, Meiqiang Feng*

School of Applied Science, Beijing Information Science & Technology University, Beijing

Received: May 7th, 2018; accepted: May 21st, 2018; published: May 29th, 2018

ABSTRACT

We prove that many nontrivial radial solutions exist for the singular k-Hessian problem { S k ( D 2 u ) = H ( | x | ) f ( u ) x B u = 0 , x B . Here is the k-Hessian operator, and B is the unit ball in R N ( N 2 ) . The main interest is that the weight function H C ( B ) is unbounded as x B , and many nontrivial radial solutions to the above k-Hessian problem are derived. Our approach to show existence and multiplicity, exploits fixed point index theory.

Keywords:k-Hessian Equation, Singular Weight Function, Many Radial Solutions, Existence, Fixed Point Index Theory

奇异k-Hessian方程的多个径向解

孙华远,冯美强*

北京信息科技大学理学院,北京

收稿日期:2018年5月7日;录用日期:2018年5月21日;发布日期:2018年5月29日

摘 要

本文研究奇异k-Hessian方程多个非平凡径向解的存在性: { S k ( D 2 u ) = H ( | x | ) f ( u ) x B u = 0 , x B . 其中, k { 1 , 2 , , N } , S k ( D 2 u ) 是k-Hessian算子,B表示 R N ( N 2 ) 中的单位球。研究的主要意义在于权函数 H C ( B ) 在B的边界上无界,并且证明上述k-Hessian有多个非平凡解。研究方法主要采用不动点指数定理。

关键词 :k-Hessian方程,奇异权函数,多个径向解,存在性,不动点指数定理

Copyright © 2018 by authors and Hans Publishers Inc.

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

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

1. 引言

K-Hession问题源自几何学,流体力学和其他应用学科。例如,当k = N时,k-Hessian问题可以表示Weingarten曲率或者是反射面形状,参见文献 [1] 。近年来,越来越多作者开始研究k-Hessian问题,并取得了很多优秀的成果,详见文献 [2] - [12] 。

k-Hessian方程的一般形式如下:

{ S k ( D 2 u ) = H ( x ) f ( u ) , x Ω , u = 0 , x Ω .

其中 k = { 1 , 2 , , N }

S k ( D 2 u ) = P k ( Λ ) = 1 i 1 < < i k N λ i 1 λ i k ,

Λ = ( λ 1 , λ 2 , , λ N ) 是Hessian矩阵 D 2 u 的特征值。在文献 [13] [14] 中,Wang证明了 P k ( Λ ) 表示 Λ 中的第k个初等对称多项式。

不难看出,k-Hessian算子是一族算子,包括Laplace算子(当k = 1时)和Monge-Ampère算子(当k = N时),以及其他著名算子。

近几年来,有许多人研究了k-Hessian问题径向解的存在性,包括Clement et al. [15] ,Wang和An [16] ,Wang [17] ,Han,Ma和Dai [18] (当k = N时),Sánchez和Vergara [6] [19] (当 1 k < N / 2 时),Jacobsen [20] (当 1 k < N 时)、Escuder和Torres [21] (当 2 k < N 时)。其中,权函数H和非线性项ƒ都是特殊情形,而没有对它们的一般情形进行讨论。

本文考虑奇异k-Hessian方程

{ S k ( D 2 u ) = H ( | x | ) f ( u ) , x B , u = 0 , x B . (1.1)

多个非平凡径向解的存在性。其中, k { 1 , 2 , , N } , S k ( D 2 u ) 是k-Hessian算子,B表示 R N ( N 2 ) 中的单位球,f是连续函数。

目前对于问题(1.1)的特殊情况,已有不少结果。例如,Zhang和Zhou在文献 [22] 中证明了当权函数H连续且f是连续的增函数时,问题(1.1)有一个径向解。Wei在文献 [23] 中考虑了权函数H恒为1的情况,通过运用Pohozaev型恒等式和单调分离法证明了问题(1.1)满足如下条件时至多有一个径向解:

1) f C 2 ( [ 0 , + ) ) , f ( 0 ) = 0 , f ( s ) > 0 ( 0 s < + )

2) s f ( s ) > k f ( s ) ( s < 0 )

3) s f ( s ) < k f ( s ) ( s < 0 )

基于上述文献的工作,本文主要研究问题(1.1)多个径向解的存在性。我们注意到这可能是第一次讨论k-Hessian方程存在多个径向解的结果,尤其是当权函数H在单位球边界无界时,这方面的结果更少。

2. 几个引理

为了证明我们的主要结论,本节给出几个引理。

首先,考虑k-Hessian算子的径向坐标形式

S k ( D 2 u ) = C N 1 k 1 t 1 N ( t N k k ( u ) k ) , t = | x | , x R N .

进而,我们可以把问题(1.1)写成如下形式:

{ C N 1 k 1 t 1 N ( t N k k ( u ) k ) = λ H ( t ) f ( u ) , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) = 0. (2.1)

令v=-u,则问题(2.1)可化为定义在[0,1]上的如下问题:

{ C N 1 k 1 t 1 N ( t N k k ( v ) k ) = λ H ( t ) f ( v ) , 0 < t < 1 , v ( 0 ) = 0 , v ( 1 ) = 0. (2.2)

这里 k = { 1 , 2 , , N } 。在本文中,我们假定H和f满足下述条件

(H1) f C ( R + , R + ) ,其中 R + = [ 0 , + )

(H2) 非负函数 H C ( [ 0 , 1 ) ) 0 1 H ( t ) d t < + ,且其在[0,1]任意子区间上不恒为0。

由(2.1)和(2.2)式可以得到引理2.1。

引理2.1:假定条件(H1)和(H2)成立,则有

1) 如果v(t)是问题(2.2)的一个解,那么 u ( t ) = v ( t ) 也是问题(2.1)在J上的一个解;

2) 如果u(t)是问题(2.1)的一个解,那么 v ( t ) = u ( t ) 也是问题(2.2)在J上的一个解。

接下来将研究问题(2.2)正解的存在性。

本文是基于 E = C [ 0 , 1 ] 空间进行讨论的。显然,E是实的Banach空间,其范数 定义为

v = max t J | v ( t ) | 。如果 v C 2 ( 0 , 1 ) C 1 [ 0 , 1 ) ,且满足(2.2)式,那么称v是问题(2.2)的解。

通过直接计算可以得到下述结论。

引理2.2:假定条件(H1)和(H2)成立,则v是问题(2.2)的解当且仅当 v E 是下述方程的解

v ( t ) = t 1 ( 0 τ k τ k N s N 1 ( C N 1 k 1 ) 1 H ( s ) f v ( s ) d s ) 1 k d τ (2.3)

并且

min t J θ v ( t ) θ v , (2.4)

其中, θ ( 0 , 1 2 ) J θ = [ θ , 1 θ ]

为了估计问题(2.2)的多个正解的存在性,我们构造E上的锥K如下:

K = { v E : v 0 , min v ( t ) θ v t J θ } .

对正实数 ρ ,我们同样可以定义

Ω ρ = { v K : min t J v ( t ) < γ ρ } = { v E : γ v min v ( t ) < γ ρ t J θ } .

下面的结论的证明请参考文献( [24] ,引理2.5,p. 693)。

引理2.3:( [24] 的引理2.5) Ω ρ 有如下性质:

1) Ω ρ 是K中开集;

2) K γ ρ Ω ρ K ρ

3) v Ω ρ min t J θ v ( t ) = γ ρ

4) 如果 v Ω ρ ,则有

定义算子 T : K E 为:

( T v ) ( t ) = t 1 ( 0 τ k τ k N s N 1 ( C N 1 k 1 ) 1 H ( s ) f ( v ( s ) ) d s ) 1 k d τ (2.5)

由文献 [16] 和简单推导可以得到:

引理2.4:假定条件(H1)和(H2)成立,则算子 T : K K 是全连续的。

引理2.5:( [24] 的引理2.4)设K是实Banach空间X上的锥,D是X的开子集满足 D k = D K D k ¯ K 。假定算子 A : D k ¯ K 是全连续的且 x A x ( x D k ) ,则下列结论成立:

1) 如果对 x D k , A x x ,则 i k ( A , D k ) = 1

2) 如果 使得对 x D k λ > 0 x A x + λ e 成立,则 i k ( A , D k ) = 0

3) 设U是K中的开集且 U ¯ D k 。如果 i k ( A , D k ) = 1 i k ( A , U k ) = 0 ,则A在 D k \ U k ¯ 上有一个不动点。如果条件为 i k ( A , D k ) = 0 i k ( A , U k ) = 1 ,则结论也成立。

3. 主要结论

为了简单起见,定义一些在证明过程中要用到的符号:

d = 0 1 H ( s ) d s , d * = θ 1 θ H ( s ) d s , f γ ρ ρ = min { f ( v ) ρ k : v [ γ ρ , ρ ] } ;

f 0 ρ = max { f ( v ) ρ k : v [ 0 , ρ ] } , f ρ = lim v α f ( v ) v k ( α : = 0 + ) ;

1 l = { d k ( C N 1 k 1 ) 1 } 1 k k 2 k N , 1 L = { d * k ( C N 1 k 1 ) 1 } 1 k k 2 k N [ 1 ( 1 θ ) 2 k N k ] .

定理3.1:假定条件(H1)和(H2), k > N 2 ,并且下列条件之一成立:

(H3)存在 ρ 1 , ρ 2 , ρ 3 ( 0 , + ) ,且 ρ 1 < γ ρ 2 , ρ 2 < ρ 3 使得

f 0 ρ 1 > l k , f γ ρ 2 ρ 2 < L k , f 0 ρ 3 > l k ,

(H4)存在 ρ 1 , ρ 2 , ρ 3 ( 0 , + ) ,且 ρ 1 < ρ 2 < ρ 3 使得

f γ ρ 1 ρ 1 > L k , f 0 ρ 2 < l k , f γ ρ 3 ρ 3 > L k ,

则下述结论成立:

1) 问题(2.2)至少有两个正解 v 1 , v 2 满足 v 1 Ω ρ 2 \ K ρ 1 ¯ , v 2 K ρ 3 \ Ω ρ 2 ¯

2) 问题(1.1)至少有两个非平凡径向解 u 1 , u 2 满足 u 1 = v 1 , u 2 = v 2

证明:这里我们只考虑条件(H3)成立的情况。如果条件(H4)成立的情况,那么证明过程和条件(H3)成立时的证明类似。

首先,我们证明 i k ( T , K ρ 1 ) = 1 。实际上,由(2.5)中的 f 0 ρ 1 < l k 知,对 v K ρ 1 ,有

( T v ) ( t ) = t 1 ( 0 τ k τ k N s N 1 ( C N 1 k 1 ) 1 H ( s ) f ( v ( s ) ) d s ) 1 k d τ 0 1 ( 0 1 k τ k N s N 1 ( C N 1 k 1 ) 1 H ( s ) f ( v ( s ) ) d s ) 1 k d τ < 0 1 ( 0 1 k τ k N s N 1 ( C N 1 k 1 ) 1 H ( s ) l k ρ 1 k d s ) 1 k d τ = ( k ( C N 1 k 1 ) 1 ) 1 k l ρ 1 0 1 τ k N k d τ ( 0 1 s N 1 H ( s ) d s ) 1 k ( k ( C N 1 k 1 ) 1 ) 1 k l ρ 1 0 1 τ k N k d τ ( 0 1 H ( s ) d s ) 1 k = ( d k ( C N 1 k 1 ) 1 ) 1 k l ρ 1 k 2 k N ,

即对 v K ρ 1 T v < v 。由引理2.5的结论(1)可得 i k ( T , K ρ 1 ) = 1

其次证明 i k ( T , Ω ρ 2 ) = 0

e ( t ) 1 ,则 e K 1 。事实上,有

v T v + λ e , v Ω ρ 2 , λ > 0 ,

若不然,则存在 v 0 Ω ρ 2 λ 0 > 0 使得

v 0 = T v 0 + λ 0 e . (3.1)

那么,根据引理2.3和式3.1有

v 0 = T v 0 + λ 0 e γ T v 0 + λ 0 e γ 1 θ 1 ( θ 1 θ k τ k N s N 1 ( C N 1 k 1 ) 1 H ( s ) f ( v 0 ( s ) ) d s ) 1 k d τ + λ 0 > γ L ρ 2 1 θ 1 ( θ 1 θ k τ k N s N 1 ( C N 1 k 1 ) 1 H ( s ) d s ) 1 k d τ + λ 0 = γ L ρ 2 ( k ( C N 1 k 1 ) 1 ) 1 k 1 θ 1 τ k N k d τ ( θ 1 θ s N 1 H ( s ) d s ) 1 k + λ 0

γ L ρ 2 ( k ( C N 1 k 1 ) 1 ) 1 k 1 θ 1 τ k N k d τ ( θ 1 θ H ( s ) d s ) 1 k + λ 0 = γ L ρ 2 ( k ( C N 1 k 1 ) 1 ) 1 k k 2 k N [ 1 ( 1 θ ) 2 k N k ] + λ 0 = γ ρ 2 + λ 0 ,

这说明 γ ρ 2 > γ ρ 2 + λ 0 ,显然矛盾。所以由引理2.5的结论(2)可以得出 i k ( T , Ω ρ 2 ) = 0

最后,我们类似地可以证明 i k ( T , K ρ 3 ) = 1 。由于 ρ 1 < γ ρ 2 ,可以推出 K ρ 1 ¯ K γ ρ 2 Ω ρ 2 。因此,利用引理2.5可以得到问题(2.2)含有至少有2个正解 v 1 , v 2 ,且有 v 1 Ω ρ 2 \ K ρ 1 ¯ , v 2 Ω ρ 3 \ K ρ 2 ¯ 。再由 v = u 得到问题(1.1)至少存在两个非负径向解,满足 u 1 = v 1 , u 2 = v 2

若(H4)成立时,我们同样可以证明定理3.1成立。证毕。

注释3.1:从定理3.1的证明中可以看出,问题(2.2)在 K ρ 1 上有第三个非负解 v 3 ,进而得到问题(1.1)存在第三个径向解 u 3 。由于 v 3 K ρ 1 ,所以 u 3 可能是平凡的径向解。

定理3.1:的可以推广到多解的情况。

定理3.2:假定条件(H1)和(H2)成立,且 k > N 2 ,则有如下结论:

1) 如果存在 { ρ i } i = 1 2 m 0 ( 0 , + ) , 满足 ρ 1 < γ ρ 2 < ρ 2 < ρ 3 < γ ρ 4 < < ρ 2 m 0 ,使得

f 0 ρ 2 m 1 < l k , f γ ρ 2 m ρ 2 m > L k , m = 1 , 2 , , m 0 ,

则有

i) 问题(2.2)在K中至少有2m0个正解,

ii) 问题(1.1)至少有2m0个非负径向解。

2) 如果存在 { ρ i } i = 1 2 m 0 ( 0 , ) , 满足 ρ 1 < ρ 2 ρ 2 < γ ρ 3 < ρ 3 < ρ 4 < γ ρ 5 < < ρ 2 m 0 + 2 ,使得

f γ ρ 2 m 1 ρ 2 m 1 < L k , f 0 ρ m k > l k , m = 1 , 2 , , m 0 ,

则有

i) 问题(2.2)在K中至少有2m0-1个正解,

ii) 问题(1.1)至少有2m0-1个非负径向解。

基金项目

本文由国家自然科学基金(11301178),北京市自然科学基金(1163007)和北京市教育委员会科技面上基金(KM201611232017)资助。

文章引用

孙华远,冯美强. 奇异k-Hessian方程的多个径向解
Many Radial Solutions of Singular k-Hessian Equations[J]. 理论数学, 2018, 08(03): 289-295. https://doi.org/10.12677/PM.2018.83038

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