研究如下二阶左定Sturm-Liouville方程

(1.1) $ \begin{equation} -(py')'+qy = \lambda wy, \lambda\in{\Bbb C}, \end{equation} $

其中系数函数$ p $, $ q $, $ w $满足条件

(1.2) $ \begin{equation} \frac{1}{p}, q, w\in L_{loc}({{\Bbb R}}, {{\Bbb R}} ), p>0\ \mbox{a.e. ${{\Bbb R}}, w$在${{\Bbb R}} $上改变符号, } \end{equation} $

这里$ {{\Bbb R}} $为实数集, $ L_{loc}({{\Bbb R}}, {{\Bbb R}} ) $是指区间$ {{\Bbb R}} $上的实值Lebesgue可积函数全体构成的空间.方程(1)的系数为周期系数,即$ \exists h\in{{\Bbb R}} $, $ 0<h<\infty $,系数满足:

(1.3) $ \begin{equation} p(t+h) = p(t), q(t+h) = q(t), w(t+h) = w(t), t\in{{\Bbb R}}. \end{equation} $

对于$ \gamma\in(0, \pi) $, $ K = I $,复共轭边界条件为:

(1.4) $ \begin{equation} Y(a+h) = {\rm e}^{{\rm i}\gamma}KY(a), \mbox{其中}\; Y = \left(\begin{array}{cc}y\\py'\end{array}\right)\, , 0<\gamma<\pi. \end{equation} $

对任意$ a\in{{\Bbb R}} $,第$ k $ -区间$ [a, a+kh] $, $ k\in{\Bbb N} = \{1, 2, 3, \cdots \} $,边界条件为

(1.5) $ \begin{equation} Y(a+kh) = {\rm e}^{{\rm i}\gamma}KY(a), \end{equation} $

其中当$ \gamma = 0 $且$ K = I $时,条件(5)为周期边界条件;当$ \gamma = 0 $且$ K = -I $或$ \gamma = \pi $且$ K = 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

(1.6) $ \begin{equation} -(py')'+qy = \xi |w|y \end{equation} $

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:

(1.7) $ \begin{equation} -(py')'+qy-\lambda wy = \xi |w|y, \end{equation} $

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$ \xi_{0}(0)>0 $,即当$ \lambda = 0 $时,方程(7)的最小特征值是正的.并且若方程(1)是左定的,则存在$ n\in{\Bbb N}_{0} = \{0, 1, 2, 3, \cdots \} $,使得其特征值是$ \xi_{n}(\lambda) = 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.

对于左定问题, $ n\in{\Bbb N}_{0} $, $ k\in{\Bbb N} $,设$ \lambda_{\pm n}(\gamma) $是复共轭边界条件下的特征值, $ \lambda^{P}_{\pm n}(k) $和$ \lambda^{S}_{\pm n}(k) $分别为周期和半周期特征值.可以确定$ \lambda_{\pm n}(\gamma) $的重数为$ 1 $,但$ \lambda^{P}_{\pm n}(k) $和$ \lambda^{S}_{\pm n}(k) $重数可能为$ 1 $或$ 2 $.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$ a\in{{\Bbb R}} $, $ \xi\in{\Bbb C} $,定义(7)式的两个线性无关解$ u = u(\cdot, \xi) $, $ v = v(\cdot, \xi) $,并满足初始条件

$ u(a, \xi) = 1 = (pv')(a, \xi), v(a, \xi) = 0 = (pu')(a, \xi), $

并令

$ D(\xi) = u(b, \xi)+(pv')(b, \xi). $

方程(7)与下面的一阶系统等价:

(2.1) $ \begin{equation} Y^{\prime} = PY, P = \left( \begin{array}{cc} 0{\quad}&1/p\\ (q-\lambda\omega-\xi|\omega|){\quad}&0 \end{array} \right), \end{equation} $

这里由(3)式可知系数$ p $, $ q $, $ w $是周期的,故$ |w| $和$ P $也是周期的.我们令$ q_{1} = q-\lambda w $,则

(2.2) $ \begin{equation} P = \left( \begin{array}{cc} 0{\quad}&1/p\\ (q_{1}-\xi|\omega|){\quad}&0 \end{array} \right), \end{equation} $

而文献[1]中

(2.3) $ \begin{equation} P = \left( \begin{array}{cc} 0{\quad}&1/p\\ (q-\lambda\omega){\quad}&0 \end{array} \right), w>0, \end{equation} $

把$ (2.2) $式中的$ q_{1} $看作$ (2.3) $式的$ q $, $ \xi $看作$ \lambda $, $ |w| $看作$ w $,这样能得到区间$ [a, a+kh] $上的周期特征值$ \xi_{n}^{P}(k, \lambda) $和半周期特征值$ \xi_{n}^{S}(k, \lambda) $,两者都是关于$ \lambda $的连续函数.下面我们研究对任意$ n\in{\Bbb N}_{0} $, $ k\in{\Bbb N} $, $ k>1 $时在区间$ [a, a+kh] $上的每个特征值$ \xi_{n}^{P} $, $ \xi_{n}^{S} $都是$ [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 $, $ \gamma\in(0, \pi) $在区间$ [a, a+h] $上复共轭边界条件下哪些$ \gamma $值能产生区间$ [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$ k\in{\Bbb N} $且令$ D(\xi) = u(a+h, \xi)+(pv')(a+h, \xi) $,则方程(7)在$ {{\Bbb R}} $上有$ kh $ -周期解当且仅当

$ D(\xi) = 2\cos(2l\pi/k), $

其中$ l\in{\Bbb Z} = \{\cdots , -3, -2, -1, 0, 1, 2, 3, \cdots \} $.进一步,

(1)若$ k = 2s $, $ s>1 $, $ \gamma = 2l\pi/2s $, $ l = 1, \cdots , s-1 $,且对某个$ n\in{\Bbb N}_{0} $,有$ \xi = \xi_{n}(\gamma) $,则$ \xi $是区间$ [a, a+kh] $上重数为$ 2 $的周期特征值,对应的特征函数可以扩展为整个$ {{\Bbb R}} $上的周期解.

(2)若$ k = 2s+1 $, $ s\geq1 $, $ \gamma = 2l\pi/(2s+1) $, $ l = 1, \cdots , s $,且对某个$ n\in{\Bbb N}_{0} $,有$ \xi = \xi_{n}(\gamma) $,则$ \xi $是区间$ [a, a+kh] $上重数为$ 2 $的周期特征值,对应的特征函数可以扩展为整个$ {{\Bbb R}} $上的周期解.

引理2.2$ ^{[1]} $  假设条件(3)成立,令$ k\in{\Bbb N} $且令$ D(\xi) = u(a+h, \xi)+(pv')(a+h, \xi) $,则方程(7)在$ {{\Bbb R}} $上有$ kh $ -半周期解当且仅当

$ D(\xi) = 2\cos((2l+1)\pi/k), $

其中$ l\in{\Bbb Z} = \{\cdots , -3, -2, -1, 0, 1, 2, 3, \cdots \} $.进一步,有

(1)若$ k = 2s $, $ s\geq1 $, $ \gamma = (2l+1)\pi/2s $, $ l = 1, \cdots , s-1 $,且对某个$ n\in{\Bbb N}_{0} $,有$ \xi = \xi_{n}(\gamma) $,则$ \xi $是区间$ [a, a+kh] $上重数为$ 2 $的半周期特征值.

(2)若$ k = 2s+1 $, $ s\geq1 $, $ \gamma = (2l+1)\pi/(2s+1) $, $ l = 1, \cdots , s-1 $,且对某个$ n\in{\Bbb N}_{0} $,有$ \xi = \xi_{n}(\gamma) $,则$ \xi $是区间$ [a, a+kh] $上重数为$ 2 $的半周期特征值.

对$ k\in{\Bbb N} $, $ n\in{\Bbb N}_{0} $,令

$ P_{\xi}(k) = \bigcup\limits_{n = 0}^{\infty}\xi_{n}^{P}(k, \lambda), S_{\xi}(k) = \bigcup\limits_{n = 0}^{\infty}\xi_{n}^{S}(k, \lambda), \Gamma_{\xi}(\gamma) = \bigcup\limits_{n = 0}^{\infty}\xi_{n}(\gamma, \lambda); $

$ P_{\lambda}(k) = \bigcup\limits_{n = 0}^{\infty}\lambda_{\pm n}^{P}(k), S_{\lambda}(k) = \bigcup\limits_{n = 0}^{\infty}\lambda_{\pm n}^{S}(k), \Gamma_{\lambda}(\gamma) = \bigcup\limits_{n = 0}^{\infty}\lambda_{\pm n}(\gamma), $

这里$ \xi_{n}^{P}(k, \lambda) $和$ \xi_{n}^{S}(k, \lambda) $分别是(7)式在区间$ [a, a+kh] $上的周期和半周期特征值,而$ \xi_{n}(\gamma, \lambda) $表示(7)式在区间$ [a, a+kh] $上的复共轭边界条件下的特征值.其中$ \xi_{n}^{P}(k, \lambda) $, $ \xi_{n}^{S}(k, \lambda) $和$ \xi_{n}(\gamma, \lambda) $都是分别关于$ \lambda_{\pm n}^{P}(k) $, $ \lambda_{\pm n}^{S}(k) $和$ \lambda_{\pm n}(\gamma) $的连续函数. $ \lambda_{\pm n}^{P}(k) $, $ \lambda_{\pm n}^{S}(k) $和$ \lambda_{\pm n}(\gamma) $分别是$ \xi_{n}^{P}(k, \lambda) = 0 $, $ \xi_{n}^{S}(k, \lambda) = 0 $和$ \xi_{n}(\gamma, \lambda) = 0 $的正负根.由文献[3,注3.1]可知,对于任意$ k\in{\Bbb N} $, $ \xi_{n}^{P}(k, \lambda) $, $ \xi_{n}^{S}(k, \lambda) $和$ \xi_{n}(\gamma, \lambda) $满足下列不等式:

$ \begin{eqnarray*} &&\xi_{0}^{P}(k, \lambda)\leq\xi_{1}^{P}(k, \lambda)\leq \xi_{2}^{P}(k, \lambda)\leq\xi_{3}^{P}(k, \lambda)\leq\cdots , \\ &&\xi_{0}^{S}(k, \lambda)\leq\xi_{1}^{S}(k, \lambda)\leq \xi_{2}^{S}(k, \lambda)\leq\xi_{3}^{S}(k, \lambda)\leq\cdots , \\ &&\xi_{0}(\gamma, \lambda)\leq\xi_{1}(\gamma, \lambda)\leq \xi_{2}(\gamma, \lambda)\leq\xi_{3}(\gamma, \lambda)\leq\cdots. \end{eqnarray*} $

由文献[3,注3.1],方程(1)是左定问题,因此对于任意$ k\in{\Bbb N} $,则有$ \xi_{n}^{P}(k, 0)>0 $, $ \xi_{n}^{S}(k, 0)>0 $且$ \xi_{n}(\gamma, 0)>0 $.

定理2.1  对每一个$ k = 2s $, $ s>1 $和$ k = 2s+1 $, $ s>0 $,则有

(2.4) $ \begin{equation} P_{\lambda}(k) = \bigcup\limits_{l = 0}^{s}\Gamma_{\lambda}(\frac{2l\pi}{k}), \end{equation} $

特别地,对$ k = 1 $时,有

$ P_{\lambda}(1) = \Gamma_{\lambda}(0) = \{\lambda_{\pm n}^{P}(1) = \lambda_{\pm n}^{P}:n\in{\Bbb N}_{0}\}. $

证  对于$ k\in{\Bbb N} $,由引理$ 2.1 $,方程(7)有周期为$ kh $的非平凡解的充要条件为$ D(\xi) = 2\cos(2l\pi/k) $,因此存在一个实值$ \gamma = 2l\pi/k\in(0, \pi) $使得$ \xi_{n}(2l\pi/k, \lambda) $是周期特征值.并且对于$ k = 2s $, $ s>1 $和$ k = 2s+1 $, $ s>0 $,有

$ P_{\xi}(k) = \bigcup\limits_{l = 0}^{s}\Gamma_{\xi}(\frac{2l\pi}{k}). $

由文献[3,注3.1],方程(1)有周期为$ kh $的非平凡解的充要条件为存在$ \gamma = 2l\pi/k\in(0, \pi) $使得$ \lambda_{\pm n}^{P}(k) $是周期特征值$ \xi_{n}^{P}(k, \lambda) = 0 $的根且$ \lambda_{\pm n}(\gamma) $是$ \xi_{n}(\gamma, \lambda) = 0 $的根.因此有

$ \begin{eqnarray*} P_{\lambda}(k) = \bigcup\limits_{l = 0}^{s}\Gamma_{\lambda}(\frac{2l\pi}{k}) \end{eqnarray*} $

特别地，对于$ k = 1 $,我们有$ \xi_{n}^{P}(1, \lambda) = \xi_{n}(0, \lambda) $,则有$ P_{\xi}(1) = \Gamma_{\xi}(0) = \{\xi_{n}^{P}(1) = \xi_{n}^{P}:n\in{\Bbb N}_{0}\} $.由于$ \lambda_{\pm n}^{P}(1) $和$ \lambda_{\pm n}(0) $分别是$ \xi_{n}^{P}(1, \lambda) = 0 $和$ \xi_{n}(0, \lambda) = 0 $的根,因此有

$ \begin{eqnarray*} P_{\lambda}(1) = \Gamma_{\lambda}(0) = \{\lambda_{\pm n}^{P}(1) = \lambda_{\pm n}^{P}:n\in{\Bbb N}_{0}\}. \end{eqnarray*} $

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 $没有对应$ \gamma\in(0, \pi) $的周期特征值,对于$ k>2 $,至少存在一个这样的$ \gamma\in(0, \pi) $.显然,若$ \lambda $是区间$ [a, a+h] $上的周期特征值,则$ \lambda $也是区间$ [a, a+2h] $上的周期特征值.同样地,若$ \lambda $是$ [a, a+h] $上的半周期特征值,则$ \lambda $也是区间$ [a, a+2h] $上的周期特征值.下面的推论告诉我们反过来结论依然成立:若$ \lambda $是区间$ [a, a+2h] $上的周期特征值,则$ \lambda $是区间$ [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

$ \begin{eqnarray*} P_{\lambda}(2) = P_{\lambda}(1)\bigcup S_{\lambda}(1). \end{eqnarray*} $

证  由$ P_{\lambda}(k) $的表达式可知

$ \begin{eqnarray*} P_{\lambda}(2) = \Gamma_{\lambda}(0)\bigcup\Gamma_{\lambda}(\pi) = P_{\lambda}(1)\bigcup S_{\lambda}(1), \end{eqnarray*} $

无论$ k $取何值, $ {\lambda}_{\pm0}^{P}(k) = \lambda_{\pm0} $.证毕.

推论2.2  若定理$ 2.1 $中的条件成立,则有

$ \begin{eqnarray*} \lambda_{\pm0}^{P}(k) = \lambda_{\pm0}^{P}(1) = \lambda_{\pm0}^{P}, k\in{\Bbb N}. \end{eqnarray*} $

定理2.2  对每一个$ k = 2s $, $ s>1 $和$ k = 2s+1 $, $ s>0 $,则有

(2.5) $ \begin{equation} S_{\lambda}(k) = \bigcup\limits_{l = 0}^{s}\Gamma_{\lambda}(\frac{(2l+1)\pi}{k}), \end{equation} $

特别地,对$ k = 1 $时,有

$ \begin{eqnarray*} S_{\lambda}(1) = \Gamma_{\lambda}(\pi) = \{\lambda_{\pm n}^{S}(1) = \lambda_{\pm n}^{S}:n\in{\Bbb N}_{0}\}. \end{eqnarray*} $

证  对于$ k\in{\Bbb N} $,由引理$ 2.2 $,方程(7)有半周期为$ kh $的非平凡解的充要条件为$ D(\xi) = 2\cos((2l+1)\pi/k) $,因此存在一个实值$ \gamma = (2l+1)\pi/k\in(0, \pi) $使得$ \xi_{n}((2l+1)\pi/k, \lambda) $是半周期特征值.并且对于$ k = 2s $, $ s>1 $和$ k = 2s+1 $, $ s>0 $,有

$ S_{\xi}(k) = \bigcup\limits_{l = 0}^{s}\Gamma_{\xi}(\frac{(2l+1)\pi}{k}). $

由文献[3,注3.1],方程(1)有半周期为$ kh $的非平凡解的充要条件为存在$ \gamma = (2l+1)\pi/k\in(0, \pi) $使得$ \lambda_{\pm n}^{S}(k) $是半周期特征值$ \xi_{n}^{S}(k, \lambda) = 0 $的根且$ \lambda_{\pm n}(\gamma) $是$ \xi_{n}(\gamma, \lambda) = 0 $的根.因此有

$ S_{\lambda}(k) = \bigcup\limits_{l = 0}^{s}\Gamma_{\lambda}(\frac{(2l+1)\pi}{k}), $

特别地,对于$ k = 1 $,我们有$ \xi_{n}^{S}(1, \lambda) = \xi_{n}(\pi, \lambda) $,则有$ S_{\xi}(1) = \Gamma_{\xi}(\pi) = \{\xi_{n}^{S}(1) = \xi_{n}^{S}:n\in{\Bbb N}_{0}\} $.由于$ \lambda_{\pm n}^{S}(1) $和$ \lambda_{\pm n}(\pi) $分别是$ \xi_{n}^{S}(1, \lambda) = 0 $和$ \xi_{n}(\pi, \lambda) = 0 $的根,因此有

$ S_{\lambda}(1) = \Gamma_{\lambda}(\pi) = \{\lambda_{\pm n}^{S}(1) = \lambda_{\pm n}^{S}:n\in{\Bbb N}_{0}\}. $

定理2.2证毕.

下面两个定理给出$ P_{\lambda}(k) = \bigcup\limits_{n = 0}^{\infty}\lambda_{\pm n}^{P}(k) $, $ S_{\lambda}(k) = \bigcup\limits_{n = 0}^{\infty}\lambda_{\pm n}^{S}(k) $和$ \Gamma_{\lambda}(\gamma) = \bigcup\limits_{n = 0}^{\infty}\lambda_{\pm n}(\gamma) $间的不等式关系.对于区间$ [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

$ \begin{eqnarray*} &&P_{\lambda}(1) = \{\lambda_{\pm n}^{P}(1):n\in{\Bbb N}_{0}\} = \Gamma_{\lambda}(0) = \{\lambda_{\pm n}(0):n\in{\Bbb N}_{0}\};\\ &&S_{\lambda}(1) = \{\lambda_{\pm n}^{S}(1):n\in{\Bbb N}_{0}\} = \Gamma_{\lambda}(\pi) = \{\lambda_{\pm n}(\pi):n\in{\Bbb N}_{0}\}. \end{eqnarray*} $

(1)若$ k = 2s $, $ s>1 $,则

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

因此有

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

(2)若$ k = 2s+1 $, $ s>1 $,则

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

因此有

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

证  已知方程(1)是左定的,由文献[3,注3.1]可知,对于任意的$ k\in{\Bbb N} $, $ \xi_{0}^{P}(k, 0)>0 $,因此我们有$ \xi_{0}^{P}(0, 0)>0 $.若$ k = 2s $,由文献[1]有

$ \begin{eqnarray*} 0&<&\xi_{0}^{P}(0, \lambda)<\xi_{0}(\frac{2\pi}{k}, \lambda)<\cdots < \xi_{0}(\frac{2(s-1)\pi}{k}, \lambda)<\xi_{0}^{S}\\ &\leq&\xi_{1}^{S}<\xi_{1}(\frac{2(s-1)\pi}{k}, \lambda)<\cdots < \xi_{1}(\frac{2\pi}{k}, \lambda)<\xi_{1}^{P} \\ &\leq&\xi_{2}^{P}<\xi_{2}(\frac{2\pi}{k}, \lambda)<\cdots < \xi_{2}(\frac{2(s-1)\pi}{k}, \lambda)<\xi_{2}^{S}\leq\cdots. \end{eqnarray*} $

由文献[4]知对任意$ n\in{\Bbb N}_{0} $,有

$ {\lim\limits_{\lambda\to\pm\infty}\xi_{n}(k, \lambda) = -\infty}. $

由$ \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) $且$ \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) $和$ \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 $且$ \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 $.有$ \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*} $

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 $有

$ \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 $有

$ \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 $有

$ \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 $有

$ \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*} $