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数学物理学报, 2020, 40(2): 340-351 doi:

论文

具有周期系数的左定Sturm-Liouville问题的特征值不等式

孙龙洁,, 郝晓玲,, 牛天,

Inequalities Among Eigenvalues of Left-Definite Sturm-Liouville Problems with Periodic Coefficients

Sun Longjie,, Hao Xiaoling,, Niu Tian,

通讯作者: 郝晓玲, E-mail:xlhao1883@163.com

收稿日期: 2019-06-14  

基金资助: 国家自然科学基金.  11561050
内蒙古自然科学基金.  2018MS01021

Received: 2019-06-14  

Fund supported: the NSFC.  11561050
the NSF of Inner Mongolia.  2018MS01021

作者简介 About authors

孙龙洁,E-mail:ljsun0905@163.com , E-mail:ljsun0905@163.com

牛天,E-mail:396379244@qq.com , E-mail:396379244@qq.com

摘要

该文利用了左定问题与右定问题的联系,得到了具有周期系数的左定Sturm-Liouville问题在区间[aa+kh]上的周期和半周期特征值的描述,阐明了周期特征值之间的不等式关系,并明确给出了区间[aa+kh]上的周期、半周期特征值和区间[aa+h]上特征值的一一对应关系.

关键词: 左定Sturm-Liouville问题 ; 周期系数 ; 特征值不等式 ; 一一对应关系

Abstract

In this paper, by using the relation between left-definite and right-definite problems, the characterization of the periodic and semi-periodic eigenvalues of left-definite Sturm-Liouville problems with periodic coefficients on the interval[a, a+kh] are obtained. For each k >2 we identify explicitly which of the uncountable number of complex conditions generates these periodic eigenvalues. Based on this condition, we give the inequality relation of periodic eigenvalues on the interval[a, a+kh] and 1-1 correspondence between the periodic and semi-periodic eigenvalues on the interval[a, a+kh], k >1 and the corresponding eigenvalues from the interval[a, a+h].

Keywords: Left-definite Sturm-liouville problems ; Periodic coefficients ; Eigenvalues inequalities ; 1-1 correspondence

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本文引用格式

孙龙洁, 郝晓玲, 牛天. 具有周期系数的左定Sturm-Liouville问题的特征值不等式. 数学物理学报[J], 2020, 40(2): 340-351 doi:

Sun Longjie, Hao Xiaoling, Niu Tian. Inequalities Among Eigenvalues of Left-Definite Sturm-Liouville Problems with Periodic Coefficients. Acta Mathematica Scientia[J], 2020, 40(2): 340-351 doi:

1 引言

研究如下二阶左定Sturm-Liouville方程

(py)+qy=λwy,λC,
(1.1)

其中系数函数p, q, w满足条件

1p,q,wLloc(R,R),p>0 a.e. R,wR上改变符号, 
(1.2)

这里R为实数集, Lloc(R,R)是指区间R上的实值Lebesgue可积函数全体构成的空间.方程(1)的系数为周期系数,即hR, 0<h<,系数满足:

p(t+h)=p(t),q(t+h)=q(t),w(t+h)=w(t),tR.
(1.3)

对于γ(0,π), K=I,复共轭边界条件为:

Y(a+h)=eiγKY(a),其中Y=(ypy),0<γ<π.
(1.4)

对任意aR,第k -区间[a,a+kh], kN={1,2,3,},边界条件为

Y(a+kh)=eiγKY(a),
(1.5)

其中当γ=0K=I时,条件(5)为周期边界条件;当γ=0K=Iγ=πK=I时,条件(5)为半周期边界条件.

Yuan等人在文献[1-2]中给出了具有周期边界条件的右定问题在区间[a,a+kh]上的周期和半周期特征值,并得到了特征值不等式,本文根据左定问题与右定问题的关系,给出了左定问题周期特征值的相关结论.

此外, Kong等人在文献[3]中阐明了左定的判定条件,无论边界条件是分离的还是耦合的,方程(1)是左定的当且仅当在同种边界条件下,方程

(py)+qy=ξ|w|y
(1.6)

的最小特征值是正的.为了方便研究左定问题,引入了一个含有两参数的二阶微分方程:

(py)+qyλwy=ξ|w|y,
(1.7)

此方程通过“特征曲线”将左定问题与右定问题联系起来,本文对于具有周期系数的左定问题也能转化成同系数的右定问题来分析,其中右定问题参考文献[4-8],周期问题参考文献[9-12]. Kong等人在文献[3]中指出方程(1)与其相应的边界条件组成的Sturm-Liouville问题是左定问题的充要条件为相同边界条件下方程(7)中的ξ0(0)>0,即当λ=0时,方程(7)的最小特征值是正的.并且若方程(1)是左定的,则存在nN0={0,1,2,3,},使得其特征值是ξn(λ)=0的根.

通过左定问题与右定问题的联系,结合具有周期系数的右定问题的特征值情况,本文首先在第2节中确切地描述了关于具有周期系数的左定问题在区间[a,a+kh]上的周期和半周期特征值.此后在第3节中分别给出了在此区间上的周期和半周期特征值不等式及一一对应关系,最后在第4节中给出几个相关例子加以说明.

2 区间[a,a+kh]上的周期特征值的描述

对于左定问题, nN0, kN,设λ±n(γ)是复共轭边界条件下的特征值, λP±n(k)λS±n(k)分别为周期和半周期特征值.可以确定λ±n(γ)的重数为1,但λP±n(k)λS±n(k)重数可能为12.这里几何重数和代数重数是相等的,因此这里的重数可以是二者中的任意一个.

对于aR, ξC,定义(7)式的两个线性无关解u=u(,ξ), v=v(,ξ),并满足初始条件

u(a,ξ)=1=(pv)(a,ξ),v(a,ξ)=0=(pu)(a,ξ),

并令

D(ξ)=u(b,ξ)+(pv)(b,ξ).

方程(7)与下面的一阶系统等价:

Y=PY,P=(01/p(qλωξ|ω|)0),
(2.1)

这里由(3)式可知系数p, q, w是周期的,故|w|P也是周期的.我们令q1=qλw,则

P=(01/p(q1ξ|ω|)0),
(2.2)

而文献[1]中

P=(01/p(qλω)0),w>0,
(2.3)

(2.2)式中的q1看作(2.3)式的q, ξ看作λ, |w|看作w,这样能得到区间[a,a+kh]上的周期特征值ξPn(k,λ)和半周期特征值ξSn(k,λ),两者都是关于λ的连续函数.下面我们研究对任意nN0, kN, k>1时在区间[a,a+kh]上的每个特征值ξPn, ξSn都是[a,a+h]上复共轭边界条件下的特征值.下面两个定理阐述了对于k>1, γ(0,π)在区间[a,a+h]上复共轭边界条件下哪些γ值能产生区间[a,a+kh]上的周期和半周期特征值.

引理2.1[1]  假设条件(3)成立,令kN且令D(ξ)=u(a+h,ξ)+(pv)(a+h,ξ),则方程(7)在R上有kh -周期解当且仅当

D(ξ)=2cos(2lπ/k),

其中lZ={,3,2,1,0,1,2,3,}.进一步,

(1)若k=2s, s>1, γ=2lπ/2s, l=1,,s1,且对某个nN0,有ξ=ξn(γ),则ξ是区间[a,a+kh]上重数为2的周期特征值,对应的特征函数可以扩展为整个R上的周期解.

(2)若k=2s+1, s1, γ=2lπ/(2s+1), l=1,,s,且对某个nN0,有ξ=ξn(γ),则ξ是区间[a,a+kh]上重数为2的周期特征值,对应的特征函数可以扩展为整个R上的周期解.

引理2.2[1]  假设条件(3)成立,令kN且令D(ξ)=u(a+h,ξ)+(pv)(a+h,ξ),则方程(7)在R上有kh -半周期解当且仅当

D(ξ)=2cos((2l+1)π/k),

其中lZ={,3,2,1,0,1,2,3,}.进一步,有

(1)若k=2s, s1, γ=(2l+1)π/2s, l=1,,s1,且对某个nN0,有ξ=ξn(γ),则ξ是区间[a,a+kh]上重数为2的半周期特征值.

(2)若k=2s+1, s1, γ=(2l+1)π/(2s+1), l=1,,s1,且对某个nN0,有ξ=ξn(γ),则ξ是区间[a,a+kh]上重数为2的半周期特征值.

kN, nN0,令

Pξ(k)=n=0ξPn(k,λ),Sξ(k)=n=0ξSn(k,λ),Γξ(γ)=n=0ξn(γ,λ);

Pλ(k)=n=0λP±n(k),Sλ(k)=n=0λS±n(k),Γλ(γ)=n=0λ±n(γ),

这里ξPn(k,λ)ξSn(k,λ)分别是(7)式在区间[a,a+kh]上的周期和半周期特征值,而ξn(γ,λ)表示(7)式在区间[a,a+kh]上的复共轭边界条件下的特征值.其中ξPn(k,λ), ξSn(k,λ)ξn(γ,λ)都是分别关于λP±n(k), λS±n(k)λ±n(γ)的连续函数. λP±n(k), λS±n(k)λ±n(γ)分别是ξPn(k,λ)=0, ξSn(k,λ)=0ξn(γ,λ)=0的正负根.由文献[3,注3.1]可知,对于任意kN, ξPn(k,λ), ξSn(k,λ)ξn(γ,λ)满足下列不等式:

ξP0(k,λ)ξP1(k,λ)ξP2(k,λ)ξP3(k,λ),ξS0(k,λ)ξS1(k,λ)ξS2(k,λ)ξS3(k,λ),ξ0(γ,λ)ξ1(γ,λ)ξ2(γ,λ)ξ3(γ,λ).

由文献[3,注3.1],方程(1)是左定问题,因此对于任意kN,则有ξPn(k,0)>0, ξSn(k,0)>0ξn(γ,0)>0.

定理2.1  对每一个k=2s, s>1k=2s+1, s>0,则有

Pλ(k)=sl=0Γλ(2lπk),
(2.4)

特别地,对k=1时,有

Pλ(1)=Γλ(0)={λP±n(1)=λP±n:nN0}.

  对于kN,由引理2.1,方程(7)有周期为kh的非平凡解的充要条件为D(ξ)=2cos(2lπ/k),因此存在一个实值γ=2lπ/k(0,π)使得ξn(2lπ/k,λ)是周期特征值.并且对于k=2s, s>1k=2s+1, s>0,有

Pξ(k)=sl=0Γξ(2lπk).

由文献[3,注3.1],方程(1)有周期为kh的非平凡解的充要条件为存在γ=2lπ/k(0,π)使得λP±n(k)是周期特征值ξPn(k,λ)=0的根且λ±n(γ)ξn(γ,λ)=0的根.因此有

Pλ(k)=sl=0Γλ(2lπk)

特别地,对于k=1,我们有ξPn(1,λ)=ξn(0,λ),则有Pξ(1)=Γξ(0)={ξPn(1)=ξPn:nN0}.由于λP±n(1)λ±n(0)分别是ξPn(1,λ)=0ξn(0,λ)=0的根,因此有

Pλ(1)=Γλ(0)={λP±n(1)=λP±n:nN0}.

定理2.1证毕.

在定理2.1中,当k=2是特殊的,因为对于k=2没有对应γ(0,π)的周期特征值,对于k>2,至少存在一个这样的γ(0,π).显然,若λ是区间[a,a+h]上的周期特征值,则λ也是区间[a,a+2h]上的周期特征值.同样地,若λ[a,a+h]上的半周期特征值,则λ也是区间[a,a+2h]上的周期特征值.下面的推论告诉我们反过来结论依然成立:若λ是区间[a,a+2h]上的周期特征值,则λ是区间[a,a+h]上的周期或半周期特征值.

推论2.1  若定理2.1中的条件成立,则有

Pλ(2)=Pλ(1)Sλ(1).

  由Pλ(k)的表达式可知

Pλ(2)=Γλ(0)Γλ(π)=Pλ(1)Sλ(1),

无论k取何值, λP±0(k)=λ±0.证毕.

推论2.2  若定理2.1中的条件成立,则有

λP±0(k)=λP±0(1)=λP±0,kN.

定理2.2  对每一个k=2s, s>1k=2s+1, s>0,则有

Sλ(k)=sl=0Γλ((2l+1)πk),
(2.5)

特别地,对k=1时,有

Sλ(1)=Γλ(π)={λS±n(1)=λS±n:nN0}.

  对于kN,由引理2.2,方程(7)有半周期为kh的非平凡解的充要条件为D(ξ)=2cos((2l+1)π/k),因此存在一个实值γ=(2l+1)π/k(0,π)使得ξn((2l+1)π/k,λ)是半周期特征值.并且对于k=2s, s>1k=2s+1, s>0,有

Sξ(k)=sl=0Γξ((2l+1)πk).

由文献[3,注3.1],方程(1)有半周期为kh的非平凡解的充要条件为存在γ=(2l+1)π/k(0,π)使得λS±n(k)是半周期特征值ξSn(k,λ)=0的根且λ±n(γ)ξn(γ,λ)=0的根.因此有

Sλ(k)=sl=0Γλ((2l+1)πk),

特别地,对于k=1,我们有ξSn(1,λ)=ξn(π,λ),则有Sξ(1)=Γξ(π)={ξSn(1)=ξSn:nN0}.由于λS±n(1)λ±n(π)分别是ξSn(1,λ)=0ξn(π,λ)=0的根,因此有

Sλ(1)=Γλ(π)={λS±n(1)=λS±n:nN0}.

定理2.2证毕.

3 具有周期系数左定问题的特征值不等式及一一对应关系

下面两个定理给出Pλ(k)=n=0λP±n(k), Sλ(k)=n=0λS±n(k)Γλ(γ)=n=0λ±n(γ)间的不等式关系.对于区间[a,a+kh]上的周期特征值,有如下定理.

定理3.1  对任意的k>2,令

Pλ(1)={λP±n(1):nN0}=Γλ(0)={λ±n(0):nN0};Sλ(1)={λS±n(1):nN0}=Γλ(π)={λ±n(π):nN0}.

(1)若k=2s, s>1,则

λS2<λ2(2(s1)πk)<<λ2(2πk)<λP2λP1<λ1(2πk)<<λ1(2(s1)πk)<λS1λS0<λ0(2(s1)πk)<<λ0(2πk)<λP0(0)<0<λP0(0)<λ0(2πk)<<λ0(2(s1)πk)<λS0λS1<λ1(2(s1)πk)<<λ1(2πk)<λP1λP2<λ2(2πk)<<λ2(2(s1)πk)<λS2,

因此有

λP0(k)=λP0,,λPs=λ0(2sπk)=λS0,λPs+1(k)=λ1(2sπk)=λS1,λPs+2(k)=λ1(2(s1)πk),,λP0(k)=λP0,,λPs=λ0(2sπk)=λS0,λP(s+1)(k)=λ1(2sπk)=λS1,λP(s+2)(k)=λ1(2(s1)πk),.

(2)若k=2s+1, s>1,则

λP1<λ1(2πk)<<λ1(2(s1)πk)<λ1(2sπk)<λ0(2sπk)<λ0(2(s1)πk)<<λ0(2πk)<λP0(0)<0<λP0(0)<λ0(2πk)<<λ0(2(s1)πk)<λ0(2sπk)<λ1(2sπk)<λ1(2(s1)πk)<<λ1(2πk)<λP1,

因此有

λP0(k)=λP0,,λPs(k)=λ0(2sπk)=λS0,λPs+1(k)=λ1(2sπk),λPs+2(k)=λ1(2(s1)πk),,λP0(k)=λ0,,λPs(k)=λ0(2sπk)=λS0,λP(s+1)(k)=λ1(2sπk),λP(s+2)(k)=λ1(2(s1)πk),.

  已知方程(1)是左定的,由文献[3,注3.1]可知,对于任意的kN, ξP0(k,0)>0,因此我们有ξP0(0,0)>0.k=2s,由文献[1]有

0<ξP0(0,λ)<ξ0(2πk,λ)<<ξ0(2(s1)πk,λ)<ξS0ξS1<ξ1(2(s1)πk,λ)<<ξ1(2πk,λ)<ξP1ξP2<ξ2(2πk,λ)<<ξ2(2(s1)πk,λ)<ξS2.

由文献[4]知对任意nN0,有

lim

\xi_{n}(\gamma, \lambda) 的连续性,则有对于每一个 n\in{\Bbb N}_{0} ,存在 \lambda_{\pm n}(\gamma) 满足 \lambda_{-n}(\gamma)<0<\lambda_{n}(\gamma) \xi_{n}(\gamma, \lambda_{\pm n}) = 0 .由文献[3,注3.1]可知 \lambda_{\pm n}(\gamma) 是左定方程(1)的周期特征值.同理可得, \lambda_{\pm n}^{P}(k) \lambda_{\pm n}^{S}(k) 也分别为左定方程(1)的周期特征值.因此不等式的排序由以上不等式可得.

对于 \gamma\neq0 \gamma\neq\pi ,我们下面证明 \lambda_{\pm n}(\gamma) 是唯一的,即 \lambda_{\pm n}(\gamma) 是单重的.若 \lambda_{\pm n}(\gamma) 不是唯一的,不失一般性,假设 0<\lambda_{*}<\lambda_{**} 是两个相邻顺序的特征值,并且有 \xi_{n}(\gamma, \lambda_{*}) = \xi_{n}(\gamma, \lambda_{**}) = 0 .由于 \xi_{n}(\gamma, 0)>0 . \xi_{n}(\gamma, \lambda_{*})<\xi_{0}(\gamma, 0) ,由文献[3,推论 3.1 ]知,对于 \lambda\geq\lambda_{*}>0 , \xi_{n}(\gamma, \lambda) 是严格减的,与 \xi_{n}(\gamma, \lambda_{*}) = \xi_{n}(\gamma, \lambda_{**}) = 0 矛盾.

对于 k = 2s+1 ,同理可证得不等式成立.证毕.

对于区间 [a, a+kh] 上的半周期特征值,也有类似的定理.

定理3.2  设定理 3.1 上的条件成立,则有

(1)若 k = 2s , s>1 ,则有

\begin{eqnarray*} \cdots &<&\lambda_{-1}(\frac{\pi}{k})<\lambda_{-1}(\frac{3\pi}{k})< \cdots <\lambda_{-1}(\frac{(2s-3)\pi}{k})<\lambda_{-1}(\frac{(2s-1)\pi}{k}) \\ &<&\lambda_{-0}(\frac{(2s-1)\pi}{k})<\lambda_{-0}(\frac{(2s-3)\pi}{k})< \cdots <\lambda_{-0}(\frac{3\pi}{k})<\lambda_{-0}(\frac{\pi}{k})\\ &<&0<\lambda_{0}(\frac{\pi}{k})<\lambda_{0}(\frac{3\pi}{k})<\cdots < \lambda_{0}(\frac{(2s-3)\pi}{k})<\lambda_{0}(\frac{(2s-1)\pi}{k})\\ &<&\lambda_{1}(\frac{(2s-1)\pi}{k})<\lambda_{1}(\frac{(2s-3)\pi}{k})<\cdots <\lambda_{1}(\frac{3\pi}{k})<\lambda_{1}(\frac{\pi}{k})<\cdots , \end{eqnarray*}

因此有

\begin{eqnarray*} &&\lambda_{0}^{S}(k) = \lambda_{0}(\frac{\pi}{k}), \cdots , \lambda_{s-1}^{S}(k) = \lambda_{0}(\frac{(2s-1)\pi}{k}), \\ &&\lambda_{s}^{S}(k) = \lambda_{1}(\frac{(2s-1)\pi}{k}) = \lambda_{1}^{S}, \lambda_{s+1}^{S}(k) = \lambda_{0}(\frac{(2s-3)\pi}{k}), \cdots.\\ &&\lambda_{-0}^{S}(k) = \lambda_{-0}(\frac{\pi}{k}), \cdots , \lambda_{-(s-1)}^{S}(k) = \lambda_{-0}(\frac{(2s-1)\pi}{k}), \\ &&\lambda_{-s}^{S}(k) = \lambda_{-1}(\frac{(2s-1)\pi}{k}) = \lambda_{1}^{S}, \lambda_{-(s+1)}^{S}(k) = \lambda_{-1}(\frac{(2s-3)\pi}{k}), \cdots. \end{eqnarray*}

(2)若 k = 2s+1 , s>1 ,则有

\begin{eqnarray*} \cdots &<&\lambda_{-1}(\frac{\pi}{k})<\lambda_{-1}(\frac{3\pi}{k})< \cdots <\lambda_{-1}(\frac{(2s-1)\pi}{k})<\lambda_{1}^{S}\\ &\leq&\lambda_{-0}^{S}<\lambda_{-0}(\frac{(2s-1)\pi}{k})<\cdots < \lambda_{-0}(\frac{3\pi}{k})<\lambda_{-0}(\frac{\pi}{k})\\ &<&0<\lambda_{0}(\frac{\pi}{k})<\lambda_{0}(\frac{3\pi}{k})<\cdots < \lambda_{0}(\frac{(2s-1)\pi}{k})<\lambda_{0}^{S}\\ &\leq&\lambda_{1}^{S}<\lambda_{1}(\frac{(2s-1)\pi}{k})<\cdots < \lambda_{1}(\frac{3\pi}{k})<\lambda_{1}(\frac{\pi}{k})<\cdots, \end{eqnarray*}

因此有

\begin{eqnarray*} &&\lambda_{0}^{S}(k) = \lambda_{0}(\frac{\pi}{k}), \cdots, \lambda_{s}^{S}(k) = \lambda_{0}^{S}, \\ &&\lambda_{s+1}^{S}(k) = \lambda_{1}^{S}, \lambda_{s+2}^{S}(k) = \lambda_{1}(\frac{(2s-1)\pi}{k}), \cdots. \\ &&\lambda_{0-}^{S}(k) = \lambda_{-0}(\frac{\pi}{k}), \cdots , \lambda_{-s}^{S}(k) = \lambda_{-0}^{S}, \\ &&\lambda_{-(s+1)}^{S}(k) = \lambda_{-1}^{S}, \lambda_{-(s+2)}^{S}(k) = \lambda_{-1}(\frac{(2s-1)\pi}{k}), \cdots. \end{eqnarray*}

下面的定理给出了区间 [a, a+kh] 上周期、半周期特征值和区间 [a, a+h] 上特征值的一一对应关系.

定理3.3  设定理 3.1 中的条件成立,则有

(1)若 k = 2s , s\in{\Bbb N} ,则

(i)当 m 是偶数时,有

{ } \lambda_{ms+n}^{P}(k) = \lambda(\frac{2(n-m)\pi}{k}), \lambda_{-(ms+n)}^{P}(k) = -\lambda(\frac{2(n-m)\pi}{k}), n = m, m+1, \cdots , m+s.

(ii)当 m 是奇数时,有

\begin{eqnarray*} &{ } \lambda_{ms+n}^{P}(k) = \lambda(\frac{2(m+s-n)\pi}{k}), \lambda_{-(ms+n)}^{P}(k) = -\lambda(\frac{2(m+s-n)\pi}{k}), \\ & n = m, m+1, \cdots , m+s. \end{eqnarray*}

(2)若 k = 2s+1 , s\in{\Bbb N} ,则

(i)当 m 是偶数时,有

\begin{eqnarray*} \lambda_{ms+n}^{P}(k) = \lambda(\frac{2(n-m)\pi}{k}), \lambda_{-(ms+n)}^{P}(k) = -\lambda(\frac{2(n-m)\pi}{k}), n = m, m+1, \cdots , m+s. \end{eqnarray*}

(ii)当 m 是奇数时,有

\begin{eqnarray*} &{ } \lambda_{ms+n}^{P}(k) = \lambda(\frac{2(m+s-n)\pi}{k}), \lambda_{-(ms+n)}^{P}(k) = -\lambda(\frac{2(m+s-n)\pi}{k}), \\ &n = m, m+1, \cdots , m+s. \end{eqnarray*}

定理3.4  设定理 3.2 中的条件成立,则有

(1)若 k = 2s , s\in{\Bbb N} ,则

(i)当 m 是偶数时,有

\begin{eqnarray*} \lambda_{ms+n}^{S}(k) = \lambda(\frac{(2n+1)\pi}{k}), \lambda_{-(ms+n)}^{S}(k) = -\lambda(\frac{(2n+1)\pi}{k}), n = 0, 1, \cdots , s-1. \end{eqnarray*}

(ii)当 m 是奇数时,有

\begin{eqnarray*} \lambda_{ms+n}^{S}(k) = \lambda(\frac{(2(s-n)-1)\pi}{k}), \lambda_{-(ms+n)}^{S}(k) = -\lambda(\frac{(2(s-n)-1)\pi}{k}), n = 0, 1, \cdots , s-1. \end{eqnarray*}

(2)若 k = 2s+1 , s\in{\Bbb N} ,则

(i)当 m 是偶数时,有

\begin{eqnarray*} &{ } \lambda_{ms+n}^{S}(k) = \lambda(\frac{(2(n-m)+1)\pi}{k}), \lambda_{-(ms+n)}^{S}(k) = -\lambda(\frac{(2(n-m)+1)\pi}{k}), \\ & n = m, m+1, \cdots , m+s. \end{eqnarray*}

(ii)当 m 是奇数时,有

\begin{eqnarray*} &{ } \lambda_{ms+n}^{S}(k) = \lambda(\frac{(2(m+s-n)+1)\pi}{k}), \lambda_{-(ms+n)}^{S}(k) = -\lambda(\frac{(2(m+s-n)+1)\pi}{k}), \\ &n = m, \cdots , m+s. \end{eqnarray*}

4 例子

例4.1   k = 2 ,由推论 2.1 知, k = 2 是一种特殊情况,有 P_{\lambda}(2) = P_{\lambda}(1) \bigcup S_{\lambda}(1) ,因此不等式为

\begin{eqnarray*} \cdots \leq\lambda_{-2}^{S}<\lambda_{-2}^{P}\leq\lambda_{-1}^{P}< \lambda_{-1}^{S}\leq\lambda_{-0}^{S}<\lambda_{-0}^{P}<0<\lambda_{0}^{P}< \lambda_{0}^{S}\leq\lambda_{1}^{S}<\lambda_{1}^{P}\leq\lambda_{2}^{P}< \lambda_{2}^{S}\leq\cdots. \end{eqnarray*}

一一对应关系为:

\begin{eqnarray*} \lambda_{\pm 0}^{P}(2) = \lambda_{\pm 0}^{P}(1) = \lambda_{\pm 0}^{P}, \lambda_{\pm 1}^{P}(2) = \lambda_{\pm 0}^{S}, \lambda_{\pm 2}^{P}(2) = \lambda_{\pm 1}^{S}, \lambda_{\pm 3}^{P}(2) = \lambda_{\pm 1}^{S}, \lambda_{\pm 4}^{P}(2) = \lambda_{\pm 2}^{P}, \cdots. \end{eqnarray*}

例4.2   k = 2s , s = 4 .由定理 3.1

\begin{eqnarray*} \cdots &\leq&\lambda_{-2}(\pi)<\lambda_{-2}(\frac{6\pi}{8})< \lambda_{-2}(\frac{4\pi}{8})<\lambda_{-2}(\frac{2\pi}{8})<\lambda_{-2}(0)\\ &\leq&\lambda_{-1}(0)<\lambda_{-1}(\frac{2\pi}{8})<\lambda_{-1}(\frac{4\pi}{8})< \lambda_{-1}(\frac{6\pi}{8})<\lambda_{-1}(\pi)\\ &\leq&\lambda_{-0}(\pi)<\lambda_{-0}(\frac{6 \pi}{8})<\lambda_{-0} (\frac{4 \pi}{8})<\lambda_{-0}(\frac{2\pi}{8})<\lambda_{-0}(0)\\ &<&0<\lambda_{0}(0)<\lambda_{0}(\frac{2\pi}{8})<\lambda_{0}(\frac{4\pi}{8})< \lambda_{0}(\frac{6\pi}{8})<\lambda_{0}(\pi)\\ &\leq&\lambda_{1}(\pi)<\lambda_{1}(\frac{6\pi}{8})<\lambda_{1}(\frac{4\pi}{8})< \lambda_{1}(\frac{2\pi}{8})<\lambda_{1}(0)\\ &\leq&\lambda_{2}(0)<\lambda_{2}(\frac{2\pi}{8})<\lambda_{2}(\frac{4\pi}{8})< \lambda_{2}(\frac{6\pi}{8})<\lambda_{2}(\pi)\leq\cdots . \end{eqnarray*}

一一对应关系为:

(1)当 m = 0 ,有

\begin{eqnarray*} &&\lambda_{\pm 0}^{P}(8) = \lambda_{\pm 0}^{P}, \lambda_{\pm 1}^{P}(8) = \lambda_{\pm 0}(\frac{2\pi}{8}), \lambda_{\pm 2}^{P}(8) = \lambda_{\pm 0}(\frac{4\pi}{8}), \\ &&\lambda_{\pm 3}^{P}(8) = \lambda_{\pm 0}(\frac{6\pi}{8}) , \lambda_{\pm 4}^{P}(8) = \lambda_{\pm 0}(\pi) = \lambda_{\pm 0}^{S}; \end{eqnarray*}

(2)当 m = 1 ,有

\begin{eqnarray*} &&\lambda_{\pm 5}^{P}(8) = \lambda_{\pm 1}^{S}, \lambda_{\pm 6}^{P}(8) = \lambda_{\pm 1}(\frac{6\pi}{8}), \lambda_{\pm 7}^{P}(8) = \lambda_{\pm 1}(\frac{4\pi}{8}), \\ &&\lambda_{\pm 8}^{P}(8) = \lambda_{\pm 1}(\frac{2\pi}{8}) , \lambda_{\pm 9}^{P}(8) = \lambda_{\pm 1}(0) = \lambda_{\pm 1}^{P}; \end{eqnarray*}

依次下去,当 m 为奇数时, \gamma 值次序为 \pi , \frac{6\pi}{8} , \frac{4\pi}{8} , \frac{2\pi}{8} , 0 ;当 m 为偶数时, \gamma 值次序为 0 , \frac{2\pi}{8} , \frac{4\pi}{8} , \frac{6\pi}{8} , \pi .

例4.3   k = 2s+1 , s = 4 .由定理 3.1

\begin{eqnarray*} \cdots &<&\lambda_{-2}(\frac{8\pi}{9})<\lambda_{-2}(\frac{6\pi}{9})< \lambda_{-2}(\frac{4\pi}{9})<\lambda_{-2}(\frac{2\pi}{9})<\lambda_{-2}^{P}\\ &\leq&\lambda_{-1}^{P}<\lambda_{-1}(\frac{2\pi}{9})<\lambda_{-1}(\frac{4\pi}{9})< \lambda_{-1}(\frac{6\pi}{9})<\lambda_{-1}(\frac{8\pi}{9})\\ &<&\lambda_{-0}(\frac{8\pi}{9})<\lambda_{-0}(\frac{6 \pi}{9})<\lambda_{-0} (\frac{4 \pi}{9})<\lambda_{-0}(\frac{2\pi}{9})<\lambda_{-0}^{P}\\ &<&0<\lambda_{0}^{P}<\lambda_{0}(\frac{2\pi}{9})<\lambda_{0}(\frac{4\pi}{9})< \lambda_{0}(\frac{6\pi}{9})<\lambda_{0}(\frac{8\pi}{9}) \\ &\leq&\lambda_{1}(\frac{8\pi}{9})<\lambda_{1}(\frac{6\pi}{9})<\lambda_{1}(\frac{4\pi}{9})< \lambda_{1}(\frac{2\pi}{9})<\lambda_{1}^{P}\\ &\leq&\lambda_{2}^{P}<\lambda_{2}(\frac{2\pi}{9})<\lambda_{2}(\frac{4\pi}{9})< \lambda_{2}(\frac{6\pi}{9})<\lambda_{2}(\frac{8\pi}{9})<\cdots . \end{eqnarray*}

一一对应关系为:

(1)当 m = 0 ,有

\begin{eqnarray*} &&\lambda_{\pm 0}^{P}(9) = \lambda_{\pm 0}^{P}, \lambda_{\pm 1}^{P}(9) = \lambda_{\pm 0}(\frac{2\pi}{9}), \lambda_{\pm 2}^{P}(9) = \lambda_{\pm 0}(\frac{4\pi}{9}), \\ &&\lambda_{\pm 3}^{P}(9) = \lambda_{\pm 0}(\frac{6\pi}{9}) , \lambda_{\pm 4}^{P}(9) = \lambda_{\pm 0}(\frac{8\pi}{9}); \end{eqnarray*}

(2)当 m = 1 ,有

\begin{eqnarray*} &&\lambda_{\pm 5}^{P}(9) = \lambda_{\pm 1}(\frac{8\pi}{9}), \lambda_{\pm 6}^{P}(9) = \lambda_{\pm 1}(\frac{6\pi}{9}), \lambda_{\pm 7}^{P}(9) = \lambda_{\pm 1}(\frac{4\pi}{9}), \\ &&\lambda_{\pm 8}^{P}(9) = \lambda_{\pm 1}(\frac{2\pi}{9}) , \lambda_{\pm 9}^{P}(9) = \lambda_{\pm 1}^{P}; \end{eqnarray*}

(3)当 m = 2 ,有

\begin{eqnarray*} &&\lambda_{\pm 10}^{P}(9) = \lambda_{\pm 2}^{P}, \lambda_{\pm 11}^{P}(9) = \lambda_{\pm 2}(\frac{2\pi}{9}), \lambda_{\pm 12}^{P}(9) = \lambda_{\pm 2}(\frac{4\pi}{9}), \\ &&\lambda_{\pm 13}^{P}(9) = \lambda_{\pm 2}(\frac{6\pi}{9}) , \lambda_{\pm 14}^{P}(9) = \lambda_{\pm 2}(\frac{8\pi}{9}). \end{eqnarray*}

例4.4   k = 2s , s = 4 .由定理 3.2

\begin{eqnarray*} S_{\lambda}(8) = \Gamma_{\lambda}(\frac{\pi}{8}) \bigcup\Gamma_{\lambda}(\frac{3\pi}{8})\bigcup\Gamma_{\lambda}(\frac{5\pi}{8}) \bigcup\Gamma_{\lambda}(\frac{7\pi}{8}), \end{eqnarray*}

特征值不等式为

\begin{eqnarray*} \cdots &\leq&\lambda_{-1}(\frac{\pi}{8})<\lambda_{-1}(\frac{3\pi}{8}) <\lambda_{-1}(\frac{5\pi}{8})< \lambda_{-1}(\frac{7\pi}{8})<\lambda_{-1}(\pi)\\ &\leq&\lambda_{-0}(\frac{7\pi}{8})<\lambda_{-0}(\frac{5 \pi}{8})<\lambda_{-0} (\frac{3\pi}{8})<\lambda_{-0}(\frac{\pi}{8}) \\ &<&0<\lambda_{0}(\frac{\pi}{8})<\lambda_{0}(\frac{3\pi}{8})<\lambda_{0}(\frac{5\pi}{8})< \lambda_{0}(\frac{7\pi}{8}) \\ &\leq&\lambda_{1}(\frac{7\pi}{8})<\lambda_{1}(\frac{5\pi}{8})<\lambda_{1}(\frac{3\pi}{8})< \lambda_{1}(\frac{\pi}{8})<\cdots . \end{eqnarray*}

一一对应关系为:

(1)当 m = 0 ,有

\begin{eqnarray*} \lambda_{\pm 0}^{S}(8) = \lambda_{\pm 0}(\frac{\pi}{8}), \lambda_{\pm 1}^{S}(8) = \lambda_{\pm 0}(\frac{3\pi}{8}), \lambda_{\pm 2}^{S}(8) = \lambda_{\pm 0}(\frac{5\pi}{8}), \lambda_{\pm 3}^{S}(8) = \lambda_{\pm 0}(\frac{7\pi}{8}); \end{eqnarray*}

(2)当 m = 1 ,有

\begin{eqnarray*} \lambda_{\pm 4}^{S}(8) = \lambda_{\pm 1}(\frac{7\pi}{8}), \lambda_{\pm 5}^{S}(8) = \lambda_{\pm 1}(\frac{5\pi}{8}), \lambda_{\pm 6}^{S}(8) = \lambda_{\pm 1}(\frac{3\pi}{8}), \lambda_{\pm 7}^{S}(8) = \lambda_{\pm 1}(\frac{\pi}{8}); \end{eqnarray*}

依次下去,当 m 为奇数时, \gamma 值次序为 \frac{7\pi}{8} , \frac{5\pi}{8} , \frac{3\pi}{8} , \frac{\pi}{8} ;当 m 为偶数时, \gamma 值次序为 \frac{\pi}{8} , \frac{3\pi}{8} , \frac{5\pi}{8} , \frac{7\pi}{8} .

例4.5   k = 2s+1 , s = 4 .由定理 3.2

\begin{eqnarray*} S_{\lambda}(9) = \Gamma_{\lambda}(\pi)\bigcup\Gamma_{\lambda}(\frac{\pi}{9}) \bigcup\Gamma_{\lambda}(\frac{3\pi}{9})\bigcup\Gamma_{\lambda}(\frac{5\pi}{9}) \bigcup\Gamma_{\lambda}(\frac{7\pi}{9}), \end{eqnarray*}

特征值不等式为

\begin{eqnarray*} \cdots &\leq&\lambda_{-1}(\frac{\pi}{9})<\lambda_{-1}(\frac{3\pi}{9})<\lambda_{-1}(\frac{5\pi}{9})< \lambda_{-1}(\frac{7\pi}{9})<\lambda_{-1}^{S}\\ &\leq&\lambda_{-0}^{S}<\lambda_{-0}(\frac{7\pi}{9})<\lambda_{-0} (\frac{5\pi}{9})<\lambda_{-0}(\frac{3\pi}{9})<\lambda_{-0}(\frac{\pi}{9})\\ &<&0<\lambda_{0}(\frac{\pi}{9})<\lambda_{0}(\frac{3\pi}{9})<\lambda_{0}(\frac{5\pi}{9})< \lambda_{0}(\frac{7\pi}{9})<\lambda_{0}^{S}\\ &\leq&\lambda_{1}^{S}<\lambda_{1}(\frac{7\pi}{9})<\lambda_{1}(\frac{5\pi}{9})<\lambda_{1}(\frac{3\pi}{9})< \lambda_{1}(\frac{\pi}{9})\\ &\leq&\lambda_{2}(\frac{\pi}{9})<\lambda_{2}(\frac{3\pi}{9})<\lambda_{2}(\frac{5\pi}{9})< \lambda_{2}(\frac{7\pi}{9})<\lambda_{2}^{S}\leq\cdots . \end{eqnarray*}

一一对应关系为:

(1)当 m = 0 ,有

\begin{eqnarray*} &&\lambda_{\pm 0}^{S}(9) = \lambda_{\pm 0}(\frac{\pi}{9}), \lambda_{\pm 1}^{S}(9) = \lambda_{\pm 0}(\frac{3\pi}{9}), \lambda_{\pm 2}^{S}(9) = \lambda_{\pm 0}(\frac{5\pi}{9}), \\ &&\lambda_{\pm 3}^{S}(9) = \lambda_{\pm 0}(\frac{7\pi}{9}) , \lambda_{\pm 4}^{S}(9) = \lambda_{\pm 0}^{S}; \end{eqnarray*}

(2)当 m = 1 ,有

\begin{eqnarray*} &&\lambda_{\pm 5}^{S}(9) = \lambda_{\pm 1}^{S}, \lambda_{\pm 6}^{S}(9) = \lambda_{\pm 1}(\frac{7\pi}{9}), \lambda_{\pm 7}^{S}(9) = \lambda_{\pm 1}(\frac{5\pi}{9}), \\ &&\lambda_{\pm 8}^{S}(9) = \lambda_{\pm 1}(\frac{3\pi}{9}) , \lambda_{\pm 9}^{S}(9) = \lambda_{\pm 1}(\frac{\pi}{9}); \end{eqnarray*}

(3)当 m = 2 ,有

\begin{eqnarray*} &&\lambda_{\pm 10}^{S}(9) = \lambda_{\pm 2}(\frac{\pi}{9}), \lambda_{\pm 11}^{S}(9) = \lambda_{\pm 2}(\frac{3\pi}{9}), \lambda_{\pm 12}^{S}(9) = \lambda_{\pm 2}(\frac{5\pi}{9}), \\ &&\lambda_{\pm 13}^{S}(9) = \lambda_{\pm 2}(\frac{7\pi}{9}) , \lambda_{\pm 14}^{S}(9) = \lambda_{\pm 2}^{S}. \end{eqnarray*}

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