\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{graphicx} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2017 (2017), No. 264, pp. 1--13.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2017 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2017/264\hfil Sturm Liouville equations with periodic coefficients] {Inequalities among eigenvalues of Sturm Liouville equations with periodic coefficients} \author[Y. Yuan, J. Sun, A. Zettl \hfil EJDE-2017/264\hfilneg] {Yaping Yuan, Jiong Sun, Anton Zettl} \address{Yaping Yuan \newline School of Mathematical Sciences, Inner Mongolia University, Hohhot, China} \email{yaping\_yyp@qq.com} \address{Jiong Sun \newline School of Mathematical Sciences, Inner Mongolia University, Hohhot, China} \email{272454707@qq.com} \address{Anton Zettl \newline Mathematics Deparment, Northern Illinois University, DeKalb, IL, USA} \email{zettl@msn.com} \dedicatory{Communicated by Jerome Goldstein} \thanks{Submitted July 28, 2017. Published October 19, 2017.} \subjclass[2010]{34B20, 34B24, 47B25} \keywords{Periodic coefficients; eigenvalue inequalities and equalities} \begin{abstract} It is well known that for h-periodic coefficients, every periodic eigenvalue on every interval $[a,a+kh]$, $k=2,3,4,\dots$, is also an eigenvalue on the interval $[a,a+h]$ of a periodic, semi-periodic or complex self-adjoint boundary condition. Here we give an explicit 1-1 correspondence between these eigenvalues. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{remark}[theorem]{Remark} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} Consider the equation \begin{equation} -(py')'+qy=\lambda wy,\quad \lambda \in \mathbb{C},\;\text{on }\mathbb{R} \label{1.1} \end{equation} with coefficients satisfying: \begin{equation} \begin{gathered} \frac{1}{p}, q, w \in L_{\rm loc}(\mathbb{R},\mathbb{R}),\quad p>0,\;w>0 \text{ a.e. on }\mathbb{R}, \\ p(t+h) =p(t),\quad q(t+h)=q(t),\quad w(t+h)=w(t),\quad\text{a.e. }t\in \mathbb{R}, \end{gathered}\label{1.2} \end{equation} and for $K=I$, or $K=-I$ and $0\leq \gamma \leq \pi $, the boundary conditions \begin{equation} Y(a+k\,h)=e^{i\gamma }K\,Y(a+(k-1)h),\quad Y=\begin{bmatrix} y \\ (py') \end{bmatrix} \quad k\in \mathbb{N}. \label{1.3} \end{equation} Here $\mathbb{R}$, $\mathbb{C}$ denote the real and complex numbers, respectively, $I$ the identity matrix, $\mathbb{N}=\{1,2,3,\dots\}$, and $L_{\rm loc}(\mathbb{R},\mathbb{R})$ the real valued functions which are Lebesgue integrable on every compact subinterval of $\mathbb{R}$, in particular on the k-intervals $[a+k\,h]$, $k\in \mathbb{N}$. Note that $L_{\rm loc}(\mathbb{R},\mathbb{R})$ contains the piecewise continuous functions on any compact subinterval. Also note that for $\gamma =0$ and $K=I$ the conditions \eqref{1.3} are periodic, for $\gamma =\pi $, $K=I$ as well as for $\gamma =0$ and $K=-I$ the conditions \eqref{1.3} are semi-periodic. For $0<\gamma <\pi $ \eqref{1.3} are complex valued. It is well known that for all these cases the conditions \eqref{1.3} are self-adjoint and for each of these self-adjoint conditions the spectrum is real, discrete, bounded below, not bounded above, has no finite cluster point, and the eigenvalues can be ordered to satisfy \begin{equation} -\infty <\lambda _0\leq \lambda _1\leq \lambda _2\leq \lambda _3\leq \dots \label{1.4} \end{equation} with no consecutive equalities. This ordering determines $\lambda _n$ uniquely. In case of multiplicity $2$ the eigenfunctions are not determined uniquely. Let $\mathbb{N}_0=\{0,1,2,3,\dots \}$ and, for $k\in \mathbb{N}$, $n\in \mathbb{N}_0$, we define \begin{equation} P(k)=\cup _{n=0}^{\infty }\lambda _n^{P}(k),\quad S(k)=\cup _{n=0}^{\infty }\lambda _n^{S}(k), \quad \Gamma (\gamma )=\cup _{n=0}^{\infty }\lambda_n(\gamma ), \label{1.5} \end{equation} where $\lambda _n^{P}(k)$, $\lambda _n^{S}(k)$, denote the periodic and semi-periodic eigenvalues on the k-interval $[a,a+kh]$, respectively, and $ \lambda _n(\gamma )$ denote the eigenvalues on the 1-interval $[a,a+h]$ for $0<\gamma <\pi ;$ we also use the notation $\lambda _n^{P}(1)=\lambda_n^{P}=\lambda _n^{P}(0)$, $\lambda _n^{S}(1)=\lambda _n^{S}=\lambda_n^{S}(\pi )$, $n\in \mathbb{N}_0$, since the periodic eigenvalues correspond to the endpoint $0$ and the semi-periodic eigenvalues to the endpoint $\pi $ of the interval $(0,\pi )$ in a natural sense as we will see below. For reference below we specialize \eqref{1.4} to these three cases \begin{gather} -\infty <\lambda _0^{P}(k)\leq \lambda _1^{P}(k)\leq \lambda _2^{P}(k)\leq \lambda _3^{P}(k)\leq \dots, \label{1.6} \\ -\infty <\lambda _0^{S}(k)\leq \lambda _1^{S}(k)\leq \lambda _2^{S}(k)\leq \lambda _3^{S}(k)\leq \dots , \label{1.7} \\ -\infty <\lambda _0(\gamma )<\lambda _1(\gamma )<\lambda _2(\gamma )<\lambda _3(\gamma )<\dots , \label{1.8} \end{gather} which are of special interest here. Note that in \eqref{1.8} the inequalities are all strict \cite{zett05}. See the book \cite{zett05} for a general discussion of basic results about Sturm-Liouville problems and as a reference for results, definitions, and notation used here. \begin{remark} \label{r1}\rm The eigenvalues \eqref{1.6}, \eqref{1.7}, \eqref{1.8} can be computed with the Bailey- Everitt- Zettl Fortran code SLEIGN2 \cite{baez01}, \cite{bgkz91} which can be downloaded free and comes with a user friendly interface. \end{remark} This paper is a follow up of \cite{yusz16} where we proved, under the general hypothesis \eqref{1.2}, that for every $n\in \mathbb{N}_0$, and every $k\in \mathbb{N}$, every eigenvalue $\lambda _n^{P}(k)$, $\lambda_n^{S}(k)$ on the $k$-interval for $k>1$ is also an eigenvalue on the $k=1$ interval. In this paper we identify which values of $\gamma \in (0,\pi )$ generate periodic and semi-periodic eigenvalues on the intervals $[a+k\,h]$, for $k\in \mathbb{N}$ and construct an explicit 1-1 correspondence between these eigenvalues. Although we are influenced by some of the methods in Eastham's well known book \cite{east73} there are some significant differences in our approach. The boundary conditions \eqref{1.3} are defined in terms of the quasi-derivative $(py')$ rather than the classical derivative $y'$ used in \cite{east73}. This not only allows the use of the much more general hypothesis \eqref{1.2} but has numerous other advantages. Our focus is on the eigenvalues of the boundary conditions \eqref{1.3} and their relationships to each other. Also we use the parameterization $\gamma \in(0,\pi )$, rather $t\in (0,1)$ as in \cite{east73}, directly. This makes our presentation clearer and more transparent. In particular the 1-1 correspondence. The organization of the paper is as follows. This Introduction is followed by general eigenvalue characterizations and inequalities in Section 2, eigenvalue inequalities for different intervals in Section 3, the 1-1 correspondence between these in Section 4. Examples to illustrate the inequalities and the 1-1 correspondence between the eigenvalues for different intervals are given in Section 5. \section{Eigenvalue inequalities and characterizations} Russel Bertrand (1872-1970): A good notation has a subtlety and suggestiveness which at times make it almost seem like a live teacher. In \cite{yusz16} we proved the following two theorems. \begin{theorem} \label{t21} Let \eqref{1.1} to \eqref{1.5} hold. Then, for $k=2s$, $s\geq 1$ and for $k=2s+1$, $s\geq 0$, we have \begin{equation} P(k)=\cup _{l=0}^{s}\Gamma (\frac{2l\pi }{k}). \label{2.1} \end{equation} Furthermore, if $k>2$ then every eigenvalue in $S(k)$ has multiplicity $2$. In particular, for $k=1$ we have $P(1)=\Gamma (0)=\{\lambda _n^{P}(1)=\lambda _n^{P}:n\in \mathbb{N}_0\}$. \end{theorem} For a proof of the above theorem see \cite{yusz16}. The case $k=2$ in Theorem \ref{t21} is `special' in the sense that there is no $\gamma $ in the open $(0,\pi )$ which generates a periodic eigenvalue in the interval $k=2$. For every $k>2$ there is at least one such $\gamma $. It is clear that if $\lambda $ is a periodic eigenvalue for $k=1$ then it is also a periodic eigenvalue for $k=2$. Also if $\lambda $ is a semi-periodic eigenvalue for $k=1$ then $\lambda $ is a periodic eigenvalue for $k=2$. The next corollary shows that the converse is true: If $\lambda $ is a periodic eigenvalue for $k=2$ then it is either a periodic or semi-periodic eigenvalue for $k=1$. \begin{corollary} \label{c21} Let the hypotheses and notation of Theorem \ref{t21} hold. Then \begin{equation*} P(2)=\Gamma (0)\cup \Gamma (\pi )=P(1)\cup S(1). \end{equation*} \end{corollary} The above corollary follows directly from \eqref{2.1}. \begin{theorem}\label{t22} Let \eqref{1.1} to \eqref{1.5} hold. Then, for $k=2s$, $s\geq 1$ and for $k=2s+1$, $s\geq 0$, we have \begin{equation} S(k)=\cup _{l=0}^{s}\Gamma (\frac{(2l+1)\pi }{k}). \label{2.2} \end{equation} Furthermore, if $k>2$, then every eigenvalue in $P(k)$ has multiplicity $2$. In particular, for $k=1$ we have $S(1)=\Gamma (\pi )=\{\lambda _n^{S}(1)=\lambda _n^{S}:n\in \mathbb{N}_0\}$. \end{theorem} For a proof of the above theorem see \cite{yusz16}. The next theorem plays an important role below and is stated here for the benefit of the reader. Fix $a\in \mathbb{R}$, and $\lambda \in \mathbb{C}$ define solutions $u(\cdot ,\lambda )$, $v=v(\cdot ,\lambda )$ of equation \eqref{1.1} with the initial conditions \begin{equation} u(a,\lambda )=1=(pv')(a,\lambda ),\quad v(a,\lambda )=0=(pu')(a,\lambda ). \label{2.3} \end{equation} When $a$ and $\lambda $ are fixed we abbreviate this notation to $u=u(\cdot,\lambda )$, $v=v(\cdot ,\lambda )$ and sometimes to just $u,v$. \begin{theorem}\label{t23} Let \eqref{1.1}--\eqref{1.5} hold. Let $a\in \mathbb{R}$, $k\in\mathbb{N}$, $b=a+k\,h$ and let $K=I$. With $u,v$ determined by \eqref{2.3} define $D(\lambda )$ by \begin{equation} D(\lambda )=\,u(b,\lambda )+\,v^{[1]}(b,\lambda ),\quad \lambda \in \mathbb{R} \label{2.4} \end{equation} Then \begin{enumerate} \item The real number $\lambda =\lambda _n(\gamma )$ for some $n\in\mathbb{N}_0$ and some $\gamma \in (0,\pi )$ if and only if \begin{equation} D(\lambda )=2\cos \gamma ,\\quad -\pi <\gamma <\pi . \label{2.5} \end{equation} In this case \begin{equation} -21$ but, given the ordering \eqref{1.6}, is it the first eigenvalue determined by this ordering?\newline The next Corollary answers this question. \begin{corollary} \label{c23} Let the hypotheses and notation of Theorem \ref{t23} hold. Then \begin{equation} \lambda _0^{P}(k)=\lambda _0^{P}(1)=\lambda _0^{P},\quad k\in \mathbb{N} \label{2.11} \end{equation} \end{corollary} \begin{proof} Clearly $\lambda _0^{P}\in P(k)$. By definition $\lambda _0^{P}(k)$ is the lowest eigenvalue determined by the ordering \eqref{1.6}. It follows from \eqref{2.1} and \eqref{2.9} that this is $\lambda _0^{P}$. \end{proof} \section{Inequalities among eigenvalues of different intervals} Note that for both Theorems \ref{t21} and \ref{t22} the eigenvalues on the right side are all from the interval $k=1$ while the eigenvalues on the left are from intervals for $k>1$. Corollary \ref{c23} shows that $\lambda _0^{P}(k)$ stays constant as $k$ changes but how do the other eigenvalues change? More specifically: \begin{itemize} \item Given an eigenvalue $\lambda $ in $P(k)$ for some $k>1$, by Theorem \ref{t21} $\lambda $ is also an eigenvalue for $k=1$, which eigenvalue? \item Given an eigenvalue $\lambda $ in $S(k)$ for some $k>1$, by Theorem \ref{t22} $\lambda $ is also an eigenvalue for $k=1$, which eigenvalue? \end{itemize} These questions are answered in this section. Our proof is based on Theorems \ref{t21}, \ref{t22}, \ref{t23} and develops a method for finding a 1-1 correspondence between these eigenvalues for each fixed $k>1$. This method is used in Section 4 to explicitly construct this 1-1 correspondence. For each $k$ we identify the values of $\gamma $ which generate periodic and semi-periodic eigenvalues on $k$-interval. Note that the set $\cup_{k=1}^{\infty }P(k)$ is a countable union of countable sets and is therefore countable, whereas the set $\Gamma (\gamma )=\{\cup _{n=0}^{\infty}\lambda _n(\gamma ):\gamma \in (0,\pi )\}$ is not countable so there can be no 1-1 correspondence between these two sets. As mentioned above, the inequalities \eqref{1.6}, \eqref{1.7}, \eqref{1.8} determine $\lambda _n^{P}(k)$, $\lambda _n^{S}(k)$ and $\lambda _n(\gamma )$ for each $\gamma \in (0,\pi )$ and each $n\in \mathbb{N}_0$. This is the `natural' ordering which defines $\lambda _n$ for any self-adjoint boundary condition when the eigenvalues are bounded below. In \cite{east73} the assumption that $p$ is positive seems to have been omitted. M\"oller \cite{moll99} has shown that if $p$ is positive and negative each on a set of positive Lebesgue measure then the eigenvalues are unbounded above and below. In this case $\lambda _n$ is not well defined. Using Theorems \ref{t21}, \ref{t22}, and \ref{t23} we will find a different ordering and a 1-1 correspondence between these two orderings. This new correspondence will be illustrated with some examples for both the periodic and the semi-periodic case. We start with a remark. \begin{remark} \label{r31} \rm Although \eqref{1.5} defines $\Gamma (\gamma )$ only for $\gamma $ in the open interval $(0,\pi )$ Theorems \ref{t21} and \ref{t22} show that the `boundary sets' $\Gamma (0)$, $\Gamma (\pi )$ represent the periodic eigenvalues and semi-periodic eigenvalues on the interval $[a,a+h]$, respectively. However, it is important to keep in mind that the eigenvalues when $\gamma \in (0,\pi )$ are all simple but the eigenvalues in $\Gamma(0), $ $\Gamma (\pi )$ may be simple or double, except for $\lambda _0^{P}$ which is always simple. It follows from Theorem \ref{t22} that $\Gamma(0)=\Gamma (2l\pi )$ and $\Gamma (\pi )=\Gamma ((2l+1)\pi )$ for any $l\in\mathbb{Z}=\{\dots -3,-2,-1,0,1,2,3,\dots \}$. \end{remark} In the next two theorems we establish inequalities between the eigenvalues of $P(k)=\cup _{n=0}^{\infty }\lambda _n^{P}(k)$, $S(k)=\cup _{n=0}^{\infty}\lambda _n^{S}(k)$, and $\Gamma (\gamma )=\cup _{n=0}^{\infty }\lambda_n(\gamma )$. \begin{theorem} \label{t31} Let \eqref{1.1}--\eqref{1.5} hold. Fix $k>2$, let $P(k)$, $S(k)$ , $\Gamma (\gamma )$ be defined by \eqref{1.5} and let \begin{equation} \begin{gathered} P(1) =\{\lambda _n^{P}(1):n\in \mathbb{N}_0\}=\Gamma (0)=\{\lambda _n(0):n\in \mathbb{N}_0\}, \\ S(1) =\{\lambda _n^{S}(1):n\in \mathbb{N}_0\}=\Gamma (\pi )=\{\lambda _n(\pi ):n\in \mathbb{N}_0\}. \end{gathered} \label{3.1} \end{equation} (1) If $k=2s$, $s>1$, then \begin{equation} \begin{aligned} &\lambda _0^{P}(0) \\ &= \lambda _0(0)<\lambda _0(2\pi /k)<\lambda _0(4\pi /k)<\dots <\lambda _0(2(s-1)\pi )/k)<\lambda _0(\pi ) \\ &\leq \lambda _1(\pi )<\lambda _1(2(s-1)\pi /k)<\lambda _1(2(s-2)\pi /k)<\dots <\lambda _1(2\pi /k)<\lambda _1(0) \\ &\leq \lambda _2(0)<\lambda _2(2\pi /k)<\lambda _2(4\pi /k)< \dots <\lambda _2(2(s-1)\pi /k)<\lambda _2(\pi ) \\ &\leq \lambda _3(\pi )<\lambda _3(2(s-1)\pi /k)<\lambda _3(2(s-2)\pi /k)\dots <\lambda _3(2\pi /k)<\lambda _3(0) \\ &\leq \lambda _{4}(0)<\lambda _{4}(2\pi /k)<\dots \end{aligned} \label{3.2} \end{equation} Therefore \begin{equation} \begin{gathered} \lambda _0^{P}(k)=\lambda _0^{P}, \\ \lambda_{s}^{P}(k)=\lambda _0(2s\pi /k)=\lambda _0^{S}, \\ \lambda_{s+1}^{P}(k)=\lambda _1(2s\pi /k)=\lambda _1^{S}, \\ \lambda_{s+2}^{P}(k)=\lambda _1((2s-2)\pi /k) \\ \dots \end{gathered} \end{equation} (2) If $k=2s+1$, $s>1$, then \begin{equation} \begin{aligned} \lambda _0^{P} &= \lambda _0(0)<\lambda _0(2\pi /k)<\lambda _0(4\pi /k)<\lambda _0(6\pi /k)\dots <\lambda _0(2s\pi /k) \\ &< \lambda _1(2s\pi /k)<\lambda _1(2(s-1)\pi /k)<\dots <\lambda _1(2\pi /k)<\lambda _1(0) \\ &\leq \lambda _2(0)<\lambda _2((2\pi /k)<\lambda _2(4\pi /k)<\dots <\lambda _2(2s\pi /k) \\ &< \lambda _3(2s\pi /k)<\lambda _3(2(s-1)\pi /k)<\dots <\lambda _3(2\pi /k)<\lambda _3(0) \\ &\leq \lambda _{4}(0)<\lambda _{4}(2\pi /k)\dots \,. \end{aligned} \label{3.3} \end{equation} Therefore \begin{equation} \begin{gathered} \lambda _0^{P}(k)=\lambda _0^{P}, \\ \lambda_{s}^{P}(k)=\lambda _0(2s\pi /k), \\ \lambda _{s+1}^{P}(k)=\lambda_1(2s\pi /k), \\ \lambda _{s+2}^{P}(k)=\lambda _1((2s-2)\pi /k) \\ \dots \end{gathered} \end{equation} \end{theorem} \begin{proof} These inequalities follow from Theorems \ref{t21}, \ref{t22} and \ref{t23}, particularly \eqref{2.8} and \eqref{2.9}. The fact $\lambda _0(\gamma )$ is decreasing, $\lambda _1(\gamma )$ is increasing, $\lambda _2(\gamma )$ decreasing, $\lambda _3(\gamma )$ increasing, $\dots $, for $\gamma \in (0,\pi )$ is reflected in the pattern for the alternating rows in \eqref{3.2}, \eqref{3.3}. This pattern is clearly seen in the examples below. \end{proof} \begin{theorem}\label{t32} Let the hypotheses and notation of Theorem \ref{t31} hold. (1) If $k=2s$, $s>1$, then \begin{equation} \begin{aligned} \lambda _0(\pi /k) &< \lambda _0(3\pi /k)<\dots <\lambda _0((2s-1)\pi /k) \\ &< \lambda _1((2s-1)\pi /k)<\lambda _1((2s-3)\pi /k)<\dots <\lambda _1(\pi /k) \\ &< \lambda _2(\pi /k)<\dots <\lambda _2(3\pi /k)<\dots <\lambda _3((2s-1)\pi /k) \\ &< \lambda _3((2s-1)\pi /k)<\lambda _3((2s-3)\pi /k)<\dots <\lambda _3(\pi /k) \\ &< \lambda _{4}(\pi /k)<\dots <\lambda _{4}(3\pi /k)<\dots <\lambda _{4}((2s-1)\pi /k)\dots \end{aligned} \label{3.4} \end{equation} Therefore \begin{equation} \begin{gathered} \lambda _0^{S}(k) =\lambda _0(\pi /k), \\ \lambda_{s-1}^{S}(k) =\lambda _0((2s-1)\pi/k), \\ \lambda _{s}(k) =\lambda_1((2s-1)\pi/k), \\ \lambda _{s+1}^{S}(k) =\lambda _1((2s-3)\pi/k) \\ \dots \end{gathered} \end{equation} (2) If $k=2s+1$, $s>1$, then \begin{equation} \begin{aligned} \lambda _0(\pi /k) &<\lambda _0(3\pi /k)<\dots <\lambda _0((2s+1)\pi /k)=\lambda _0^{S} \\ &\leq \lambda _1^{S}=\lambda _1(\pi )<\lambda _1((2s-1)\pi /k)<\dots <\lambda _1(\pi /k) \\ &< \lambda _2(\pi /k)<\lambda _2(3\pi /k)<\dots <\lambda _2((2s+1)\pi /k)=\lambda _2^{S} \\ &\leq \lambda _3^{S}=\lambda _3(\pi )<\lambda _3((2s-1)\pi /k)<\dots <\lambda _3(\pi /k) <\dots \end{aligned} \label{3.5} \end{equation} Therefore \begin{equation} \begin{gathered} \lambda _0^{S}(k)=\lambda _0(\pi /k), \\ \lambda_{s}^{S}(k)=\lambda _0^{S}, \\ \lambda _{s+1}^{S}(k)=\lambda _1^{S}, \\ \lambda _{s+2}^{S}(k)=\lambda _1((2s-1)\pi /k), \\ \dots \end{gathered} \end{equation} \end{theorem} \begin{proof} These inequalities follow from Theorems \ref{t21}, \ref{t22} and \ref{t23}. Particularly \eqref{2.8} and \eqref{2.9}. The fact $\lambda _0(\gamma) $ is decreasing, $\lambda _1(\gamma )$ is increasing, $\lambda_2(\gamma )$ decreasing, $\lambda _3(\gamma )$ increasing, $\dots $, for $\gamma \in (0,\pi )$ is reflected in the pattern for the alternating rows in \eqref{3.4}, \eqref{3.5}. This pattern is used in the proofs of Theorems below and illustrated in the examples below. \end{proof} Now we list some examples to illustrate Theorem \ref{t31} and clarify its proof. We start with the periodic case for $k=2$. This case is special and does not illustrate the general pattern because it does not involve $\gamma$. As $k$ gets large the eigenvalues $\lambda _n^{P}(k)$ and $\lambda_n^{S}(k)$ approach $\lambda _0^{P}(1)=\lambda _0^{P}$ from the right. More precisely we have the following result. \begin{theorem} \label{t33} For any $n\in \mathbb{N}$ we have \begin{equation} \lim_{k\to \infty }\lambda _n^{P}(k)=\lambda_0^{P},\quad \lim_{k\to \infty }\lambda _n^{S}(k)=\lambda_0^{P}. \label{3.6} \end{equation} \end{theorem} \begin{proof} Let $n\in \mathbb{N}$. For $k=2(n+1)=2s$. From \eqref{3.2} we have $\lambda _n^{P}(k)=\lambda _0(2s\pi )/k)$ and therefore \begin{equation} \lim_{k\to \infty }\lambda _n^{P}(k)=\lambda _0^{P}. \label{3.7} \end{equation} For $k=2n+1=2s+1$ from \eqref{3.4} we have $\lambda _n^{P}(k)=\lambda_0(2s\pi /k)$ and \eqref{3.6} follows. By Theorem \ref{t31} $\lambda_n^{P}(k)>\lambda _0^{P}$ for $k$ even or odd; hence the limit in \eqref{3.7} is from the right. The proof of $\lim_{k\to\infty }\lambda _n^{S}(k)=\lambda _0^{P}$ is similar using \eqref{3.3}, \eqref{3.5} and the limit is also from the right. \end{proof} It is well known that equation \eqref{1.1} is oscillatory on $\mathbb{R}$ when $\lambda >\lambda _0^{P}$ and non-oscillatory when $\lambda \leq \lambda _0^{P}$. In the next theorem we give an elementary proof of this using Theorem \ref{t33} valid under our general hypotheses \eqref{1.2}. \begin{theorem} \label{t34} Let the hypotheses and notation of Theorem \ref{t31} hold. Then \eqref{1.1} is oscillatory on $\mathbb{R}$ when $\lambda >\lambda _0^{P}$ and non-oscillatory when $\lambda \leq \lambda _0^{P}$. \end{theorem} \begin{proof} Suppose that $\lambda =\lambda _0^{P}$ and $u$ is an eigenfuntion of $\lambda $. Then by \cite[Theorem 8]{yusz16} $u$ has no zero in the closed interval $[a,a+h]$. Hence the extension of $u$ to $\mathbb{R}$ has no zero on $\mathbb{R}$. By the Sturm Comparison Theorem equation \eqref{1.1} is non-oscillatory for $\lambda \leq \lambda _0^{P}$. Let $\lambda >\lambda_0^{P}$. By Theorems \ref{t21}, \ref{t33}, $\lambda _0^{P}<\lambda_n^{P}(k)<\lambda $ for all sufficiently large $n$ and $k$. Since $\lambda_n^{P}(k)$ has zeros in the interval $[a,kh]$, its extension to $\mathbb{R}$ has infinitely many zeros, i.e. it is oscillatory. \end{proof} \section{Construction of the 1-1 correspondence} The next two theorems give the explicit 1-1 correspondence between the periodic and semi-periodic eigenvalues on the $k$ interval $k>1$ and the corresponding eigenvalues from the interval $k=1$. \begin{theorem}\label{t41} Let the hypotheses and notation of Theorem \ref{t31} hold and let the eigenvalues $\lambda _n^{P}(k)$ be ordered according to \eqref{1.6}. \begin{itemize} \item If $k=2s$, $s\in \mathbb{N}$, then: \begin{enumerate} \item for $m$ even we have \begin{equation} \lambda _{ms+n}^{P}(k)=\lambda (2(n-m)\pi )/k),\quad n=m,m+1,\dots,m+s. \label{4.1} \end{equation} \item for $m$ odd we have \begin{equation} \lambda _{ms+n}^{P}(k)=\lambda (2(m+s-n)\pi )/k),\quad n=m,\,m+1,\dots ,m+s. \label{4.2} \end{equation} \end{enumerate} \item If $k=2s+1$, $s>0$, then: \begin{enumerate} \item for $m$ even and we have \begin{equation} \lambda _{ms+n}^{P}(k)=\lambda (2(n-m)\pi )/k),\quad n=m,m+1,\dots ,m+s. \label{4.3} \end{equation} \item for $m$ odd we have \begin{equation} \lambda _{ms+n}^{P}(k)=\lambda (2(m+s-n)\pi )/k),\quad n=m,m+1,\dots ,m+s. \label{4.4} \end{equation} \end{enumerate} \end{itemize} \end{theorem} \begin{proof} For clarity of presentation we use the notation discussed in Theorem \ref{t31}. Suppose $k=2s$, $s\in \mathbb{N}$. From \eqref{3.2} and the natural ordering \eqref{1.6} it follows that \begin{gather*} \lambda _0^{P}=\lambda^P_0,\quad \lambda _1^{P}(k)=\lambda _0(2\pi/k),\;\dots ,\quad \lambda _{s-1}^{P}(k)=\lambda _0(2(s-1)\pi)/k),\quad \lambda _{s}^{P}(k)=\lambda _0^{S}, \\ \lambda _{s+1}^{P}(k)=\lambda _1^{S},\quad \lambda _{s+2}^{P}(k)=\lambda_1(2(s-1)\pi /k),\;\dots,\\ \lambda _{2s}^{P}(k)=\lambda_1(2\pi /k), \quad \lambda _{2s+1}^{P}(k)=\lambda_1^{P}, \\ \lambda _{2s+2}^{P}(k)=\lambda_2^{P},\quad \lambda _{2s+3}^{P}(k)=\lambda_2(2\pi /k),\;\dots,\\ \lambda _{3s+1}^{P}(k)=\lambda_2(2(s-1)\pi /k),\quad \lambda _{3s+2}^{P}(k)=\lambda _2^{S},\\ \lambda _{3s+3}^{P}(k)=\lambda _3^{S},\quad \lambda _{s+4}^{P}(k)=\lambda_3(2(s-1)\pi /k),\;\dots,\\ \lambda _{4s+2}^{P}(k)=\lambda_3(2\pi /k),\quad \lambda _{4s+3}^{P}(k)=\lambda _3^{P} \end{gather*} and so on. Note that for $\lambda _{ms+n}^{P}(k)$ the values of $\gamma $ increase $0$, $2\pi /k$, $\dots $, $2(s-1)\pi /k$, $2s\pi /k=\pi $ as the index $n$ goes from $m$ to $m+s$ when $m$ is even and decreases $2s\pi/k=\pi $, $2(s-1)\pi /k,\dots ,2\pi /k,0$ when $m$ is odd. This establishes \eqref{4.1} and \eqref{4.2}. Suppose $k=2s+1$, $s>0$. From \eqref{3.3} and the natural ordering \eqref{1.6} it follows that \begin{gather*} \lambda^P_0(k)=\lambda^P_0,\quad \lambda^P_1(k)=\lambda_0(2\pi/k),\;\dots, \\ \lambda^P_{s-1}(k)=\lambda_0(2(s-1)\pi/k),\quad \lambda^P_s(k)=\lambda_0(2s\pi/k),\\ \lambda^P_{s+1}(k)=\lambda_1(2s\pi/k),\quad \lambda^P_{s+2}(k)=\lambda_1(2(s-1)\pi/k),\;\dots,\\ \lambda^P_{2s}(k)=\lambda_1(2\pi/k), \quad \lambda^P_{2s+1}(k)=\lambda^P_1,\\ \lambda^P_{2s+2}(k)=\lambda^P_2, \quad \lambda^P_{2s+3}(k)=\lambda_2(2\pi/k),\;\dots,\\ \lambda^P_{3s+1}(k)=\lambda_2(2(s-1)\pi/k),\quad \lambda^P_{3s+2}(k)=\lambda_2(2s\pi/k), \\ \lambda^P_{3s+3}(k)=\lambda_3(2s\pi/k),\quad \lambda^P_{3s+4}(k)=\lambda_3(2(s-1)\pi/k),\;\dots, \\ \lambda^P_{4s+2}(k)=\lambda_3(2\pi/k),\quad \lambda^P_{4s+3}(k)=\lambda^P_3 \end{gather*} and so on. Note that for $\lambda _{ms+n}^{P}(k)$ the values of $\gamma $ increase $0$, $2\pi /k$, \dots, $2(s-1)\pi /k$, $2s\pi /k=\pi $ as the index $n$ goes from $m$ to $m+s$ when $m$ is even and decreases $2s\pi/k=\pi $, $2(s-1)\pi /k$, \dots , $2\pi /k,0$ when $m$ is odd. This establishes \eqref{4.3} and \eqref{4.4}. \end{proof} \begin{theorem}\label{t42} Let the hypotheses and notation of Theorem \ref{t31} hold and let the eigenvalues $\lambda _n^{S}(k)$ be ordered according to \eqref{1.7}. \begin{itemize} \item If $k=2s$, $s>1$, then: \begin{enumerate} \item for $m$ even we have \begin{equation} \lambda _{ms+n}^{S}(k)=\lambda (2n+1)\pi )/k),\quad n=0,1,\dots ,s-1. \label{4.5} \end{equation} \item for $m$ odd we have \begin{equation} \lambda _{ms+n}^{S}(k)=\lambda (2(s-1-n)\pi )/k),\quad n=0,1,\dots ,s-1. \label{4.6} \end{equation} \end{enumerate} \item If $k=2s+1$, $s>0$, then: \begin{enumerate} \item for $m$ even and $n\in [ m,m+s]$ we have \begin{equation} \lambda _{ms+n}^{S}(k)=\lambda (2(n-m)\pi +1)/k),\quad n=m,\,m+1,\dots ,m+s. \label{4.7} \end{equation} \item for $m$ odd and $n\in [ m,m+s]$ we have \begin{equation} \lambda _{ms+n}^{S}(k)=\lambda (2(m+s-n)+1\pi )/k),\quad n=m,\,m+1,\dots,m+s. \label{4.8} \end{equation} \end{enumerate} \end{itemize} \end{theorem} \begin{proof} For clarity of presentation we use the notation discussed in Theorem \ref{t32}. Suppose $k=2s$, $s\in \mathbb{N}$. From \eqref{3.4} and the natural ordering \eqref{1.7} it follows that \begin{gather*} \lambda^S_0(k)=\lambda_0(\pi/k),\;\dots,\; \lambda^S_{s-2}(k)=\lambda_0((2s-3)\pi/k),\quad \lambda^S_{s-1}(k)=\lambda_0((2s-1)\pi/k), \\ \lambda^S_s(k)=\lambda_1((2s-1)\pi/k),\;\dots,\; \lambda^S_{2s-2}(k)=\lambda_1(3\pi/k),\quad \lambda^S_{2s-1}(k)=\lambda_1(\pi/k), \\ \lambda^S_{2s}(k)=\lambda_2(\pi/k),\;\dots,\; \lambda^S_{3s-2}(k)=\lambda_2((2s-3)\pi/k),\quad \lambda^S_{3s-1}(k)=\lambda_2((2s-1)\pi/k), \\ \lambda^S_{3s}(k)=\lambda_3((2s-1)\pi/k),\;\dots,\; \lambda^S_{4s-2}(k)=\lambda_3(3\pi/k),\quad \lambda^S_{4s-1}(k)=\lambda_3(\pi/k), \end{gather*} and so on. Note that for $\lambda _{ms+n}^{S}(k)$ the values of $\gamma $ increase $\pi/k$, $\dots $, $(2s-1)\pi /k$, as the index $n$ goes from $0$ to $s-1$ when $m$ is even, and decreases $(2s-1)\pi /k$, $\dots$, $\pi/k, $ when $m$ is odd. This establishes \eqref{4.5} and \eqref{4.6}. Suppose $k=2s+1$, $s>0$. From \eqref{3.5} and the natural ordering \eqref{1.7} it follows that \begin{gather*} \lambda^S_0(k)=\lambda_0(\pi/k),\;\dots,\; \lambda^S_{s-1}(k)=\lambda_0((2s-1)\pi/k),\\ \lambda^S_s(k)=\lambda_0((2s+1)\pi/k)=\lambda^S_0, \quad \lambda^S_{s+1}(k)=\lambda^S_1,\\ \lambda^S_{s+2}(k)=\lambda_1((2s-1)\pi/k),\; \dots,\; \lambda^S_{2s}(k)=\lambda_1(3\pi/k), \\ \lambda^S_{2s+1}(k)=\lambda_1(\pi/k), \quad \lambda^S_{2s+2}(k)=\lambda_2(\pi/k),\;\dots,\; \lambda^S_{3s+1}(k)=\lambda_2((2s-1)\pi/k),\\ \lambda^S_{3s+2}(k)=\lambda_2((2s+1)\pi/k)=\lambda^S_2,\quad \lambda^S_{3s+3}(k)=\lambda^S_3,\\ \lambda^S_{3s+4}(k)=\lambda_3((2s-1)\pi/k),\; \dots,\; \lambda^S_{4s+2}(k)=\lambda_3(3\pi/k),\quad \lambda^S_{4s+3}(k)=\lambda_3(\pi/k), \end{gather*} and so on. Note that for $\lambda _{ms+n}^{S}(k)$ the values of $\gamma $ increase $\pi/k$, $\dots $, $(2s-1)\pi /k$, as the index $n$ goes from $m$ to $m+s$ when $m$ is even, and decreases $(2s+1)\pi )/k=\pi$, \dots, $\pi /k$, when $m$ is odd. This establishes \eqref{4.7} and \eqref{4.8}. \end{proof} \section{Examples} In this section we give some examples. First for the cases $k=2,3,4$, then for some higher order cases. There are some key differences between $k$ even and $k$ odd. For the periodic even order case any periodic eigenvalue for $ k=1$ is also a periodic eigenvalue for $k>1$. Also a semi-periodic eigenvalue for $k=1$ is a periodic eigenvalue for even $k$. A more subtle difference is the effect of the inequalities of Theorem \ref{t31} on the 1-1 correspondence. This has to do with the alternating increasing and decreasing values of $\gamma $ for the even and odd order cases. These will be illustrated in the examples below. \begin{example}\label{examp1}\rm $k=2$. As mentioned above the case $k=2$ is special. By Corollary \ref{c21}: $P(2)=P(1)\cup S(1)=\Gamma (0)\cup \Gamma (\pi )$. From this and \ref{2.9}) we get \begin{equation*} \lambda _0^{P}<\lambda _0^{S}\leq \lambda _1^{S}<\lambda _1^{P} \leq \lambda _2^{P}<\lambda _2^{S}\leq \lambda _3^{S}<\lambda _3^{P} \leq \lambda _{4}^{P}<\dots\,. \end{equation*} Hence the 1-1 correspondence is: \begin{gather*} \lambda _0^{P}(2)=\lambda _0^{P}(1)=\lambda _0^{P},\quad \lambda_1^{P}(2)=\lambda _0^{S},\quad \lambda _2^{P}(2)=\lambda _1^{S},\\ \lambda _3^{P}(2)=\lambda _1^{P},\quad \lambda_{4}^{P}(2)=\lambda _2^{P},\,\dots\,. \end{gather*} \end{example} \begin{example}\label{examp2} \rm $k=3$. This case is similar to \ref{examp1}. In this case there is one $\gamma =2\pi /3$ generates the additional eigenvalues rather than the semi-periodic ones which can be identified with $\gamma =\pi $. Thus we have \begin{equation*} \lambda _0^{P}<\lambda _0(2\pi /3)<\lambda _1^{P}\leq \lambda _2^{P}<\lambda _2(2\pi /3)<\lambda _3^{P}\leq \lambda _{4}^{P}<\lambda _{4}(2\pi /3)<\dots\,. \end{equation*} Hence the 1-1 correspondence is: \begin{gather*} \lambda _0^{P}(2)=\lambda _0^{P}(1)=\lambda _0^{P},\quad \lambda_1^{P}(2)=\lambda _0(2\pi /3),\quad \lambda _2^{P}(2)=\lambda_2^{P},\\ \lambda _3^{P}(2)=\lambda _3(2\pi /3),\quad \lambda_{4}^{P}(2)=\lambda _{4}^{P},\,\dots\,. \end{gather*} \end{example} \begin{example} \label{examp3}\rm $k=2s$, $s=4$.This and the next example illustrates the fact that the values of $\gamma $ increase $\pi /k$, $\dots $, $(2s-1)\pi/k$, as the index $n$ goes from $m$ to $m+s$ when $m$ is even and decrease $(2s+1)\pi )/k=\pi ,\dots ,\pi /k$, when $m$ is odd. By Theorem \ref{t32} we have \begin{align*} \lambda _0^{P}(0) &= \lambda _0(0)<\lambda _0(2\pi /8)<\lambda _0(4\pi /8)<\lambda _0(6\pi /8)<\lambda _0(\pi ) \\ &\leq \lambda _1(\pi )<\lambda _1(6\pi /8)<\lambda _1(4\pi /8)<\lambda _1(2\pi /8)<\lambda _1(0) \\ &\leq \lambda _2(0)<\lambda _2(2\pi /8)<\lambda _2(4\pi /8)<\lambda _2(6\pi /8)<\lambda _2(\pi ) \\ &\leq \lambda _3(\pi )<\lambda _3(6\pi /8)<\lambda _3(4\pi /8)<\lambda _3(2\pi /8)<\lambda _3(0) \\ &\leq \lambda _{4}(0)<\lambda _{4}(2\pi /8)<\dots \end{align*} Therefore \begin{enumerate} \item for $m=0$ we have \begin{gather*} \lambda _0^{P}(8)=\lambda _0^{P}, \quad \lambda _1^{P}(8)=\lambda _0(2\pi/8),\quad \lambda _2^{P}(8)=\lambda _0(4\pi /8),\\ \lambda _3^{P}(8)=\lambda_0(6\pi /8),\quad \lambda _{4}^{P}(8)=\lambda _0(8\pi /8)=\lambda _0^{S}; \end{gather*} \item for $m=1$ we have \begin{gather*} \lambda _{5}^{P}(8)=\lambda _1^{S},\quad \lambda _{6}^{P}(8)=\lambda_1(6\pi /8),\quad \lambda _{7}^{P}(8)=\lambda _1(4\pi /k),\\ \lambda_{8}^{P}(8)=\lambda _1(2\pi /8),\quad \lambda _{9}^{P}(8)=\lambda_1(0)=\lambda _1^{P}; \end{gather*} \item for $m=2$ we have \begin{gather*} \lambda _{10}^{P}(8)=\lambda _2^{P},\quad \lambda _{11}^{P}(8)=\lambda _2(2\pi/8),\quad \lambda _{12}^{P}(8)=\lambda _2(4\pi /8), \\ \lambda _{13}^{P}(8)=\lambda_2(6\pi /8),\quad \lambda _{14}^{P}(8)=\lambda_2(\pi )=\lambda _2^{S}; \end{gather*} \item for $m=3$ we have \begin{gather*} \lambda _{15}^{P}(8)=\lambda _3^{S},\quad \lambda _{16}^{P}(8)=\lambda_3(6\pi /8),\quad \lambda _{17}^{P}(8)=\lambda _3(4\pi /8),\\ \lambda_{18}^{P}(8)=\lambda _3(2\pi /8),\quad \lambda _{19}^{P}(8)=\lambda _3^{P}. \end{gather*} \end{enumerate} \end{example} \begin{example} \label{examp4} \rm $k=2s+1$, $s=4$. By Theorem \ref{t32} we have \begin{align*} \lambda _0^{P} &< \lambda _0(2\pi /9)<\lambda _0(4\pi /9)<\lambda _0(6\pi /9)<\lambda _0(8\pi /9)< \\ &< \lambda _1^{S}=\lambda _1(\pi )<\lambda _1((2s-1)\pi /9)<\dots <\lambda _1(\pi /9) \\ &< \lambda _2(\pi /9)<\lambda _2(3\pi /9)<\dots <\lambda _2((2s+1)\pi /9)=\lambda _2^{S} \\ &\leq \lambda _3^{S}=\lambda _3(\pi )<\lambda _3((2s-1)\pi /9)<\dots <\lambda _3(\pi /9) < \dots \end{align*} Therefore \begin{enumerate} \item for $m=0$ we have \begin{gather*} \lambda _0^{P}(9)=\lambda _0^{P},\quad \lambda _1^{P}(9)=\lambda _0(\pi/9),\quad \lambda _2^{P}(9)=\lambda _0(3\pi /9), \\ \lambda _3^{P}(9)=\lambda_0(5\pi /9), \quad \lambda _{4}^{P}(9)=\lambda _0(7\pi /9); \end{gather*} \item for $m=1$ we have \begin{gather*} \lambda _{5}^{P}(9)=\lambda _1(7\pi /9),\quad \lambda _{6}^{P}(9)=\lambda_1(5\pi /9),\\ \lambda _{7}^{P}(9)=\lambda _1(3\pi /9),\quad \lambda_{8}^{P}(9)=\lambda _1(\pi /9)<\lambda _{9}^{P}(9)=\lambda _1^{P}; \end{gather*} \item for $m=2$ we have \begin{gather*} \lambda _{10}^{P}(9)=\lambda _2^{P},\quad \lambda _{11}^{P}(9)=\lambda_2(\pi /9),\quad \lambda _{12}^{P}(9)=\lambda _2(3\pi /9),\\ \lambda_{13}^{P}(9)=\lambda _2(5\pi /9),\quad \lambda _{14}^{P}(9)=\lambda_2(7\pi/9); \end{gather*} \item for $m=3$ we have \begin{gather*} \lambda _{15}^{P}(9)=\lambda _3(7\pi /9),\quad \lambda _{16}^{P}(9)=\lambda_3(5\pi /9),\\ \lambda _{17}^{P}(9)=\lambda _3(3\pi /9),\quad \lambda_{18}^{P}(9)=\lambda _3(\pi /9)<\lambda _{19}^{P}(9)=\lambda _3^{P}. \end{gather*} \end{enumerate} \end{example} The next examples illustrate the semi-periodic case. For $S(2)=\Gamma (\frac{\pi }{2})$ the 1-1 correspondence is just the identity so we start with $S(3)$. \begin{example} \label{examp5} \rm $k=3$. For $S(3)=S(1)\cup \Gamma (\frac{\pi }{3})=\Gamma (\pi )\cup \Gamma (\frac{\pi }{3})$ and from Theorem \ref{t23} we get the inequalities: \begin{align*} \lambda _0(\pi /3) &<\lambda _0(\pi )=\lambda _0^{S}\leq \lambda _1^{S}=\lambda _1(\pi )<\lambda _1(\pi /3)<\lambda _2(\pi /3)<\lambda _2(\pi )=\lambda _2^{S} \\ &\leq \lambda _3^{S}=\lambda _3(\pi )<\lambda _3(\pi /3)<\lambda _{4}(\pi /3)<\lambda _{4}(\pi )=\lambda _{4}^{S}\leq \lambda _{5}^{S}<\dots \end{align*} Hence $\lambda _0^{S}(3)=\lambda _0(\pi /3)$, $\lambda_1^{S}(3)=\lambda _0^{S}$, $\lambda _2^{S}(3)=\lambda _1^{S}$, $\lambda _{4}^{S}(3)=\lambda _2(\pi /3)$, \dots\,. \end{example} \begin{example} \label{examp6} \rm $k=2s$, $s=4$. By Theorem {\ref{t22}} we have \begin{equation*} S(8)=\Gamma (\pi /8)\cup \Gamma (3\pi /8)\cup \Gamma (5\pi /8)\cup \Gamma (7\pi /8). \end{equation*} By Theorem {\ref{t23}} we have the inequalities: \begin{align*} \lambda _0(\pi /8) &< \lambda _0(3\pi /8)<\lambda _0(5\pi /8)<\lambda _0(7\pi /8) \\ &< \lambda _1(7\pi /8)<\lambda _1(5\pi /8)<\lambda _1(3\pi /8)<\lambda _1(\pi /8) \\ &< \lambda _2(\pi /8)<\lambda _2(3\pi /8)<\lambda _2(5\pi /8)<\lambda _2(7\pi /8) \\ &< \lambda _3(7\pi /8)<\lambda _3(5\pi /8)<\lambda _3(3\pi /8)<\lambda _3(\pi /8) \\ &< \lambda _{4}(\pi /8)<\lambda _{4}(3\pi /8)<\lambda _{4}(5\pi /8)<\lambda _{4}(7\pi /8)<\dots \end{align*} From these inequalities and Theorem \ref{t42}: \begin{enumerate} \item for $m=0$ we have \begin{equation*} \lambda _0^{S}(k)=\lambda _0(\pi /k),\,\lambda _1^{S}(k)=\lambda _0(3\pi /k),\,\lambda _2^{S}(k)=\lambda _0(5\pi /k),\lambda _3^{S}(k)=\lambda _0(7\pi /k); \end{equation*} \item for $m=1$ we have \begin{equation*} \lambda _{4}^{S}(k)=\lambda _1(7\pi /k),\,\;\lambda _{5}^{S}(k)=\lambda _1(5\pi /k),\,\;\lambda _{6}^{S}(k)=\lambda _1(3\pi /k),\,\;\lambda _{7}^{S}(k)=\lambda _0(\pi /k); \end{equation*} \item for $m=2$ we have \begin{equation*} \lambda _{8}^{S}(k)=\lambda _2(\pi /k),\,\lambda _{9}^{S}(k)=\lambda _2(3\pi /k),\,\lambda _{10}^{S}(k)=\lambda _2(5\pi /k),\lambda _{11}^{S}(k)=\lambda _2(7\pi /k); \end{equation*} \item for $m=3$ we have \begin{equation*} \lambda _{12}^{S}(k)=\lambda _3(7\pi /k),\,\;\lambda _{13}^{S}(k)=\lambda _3(5\pi /k),\,\;\lambda _{14}^{S}(k)=\lambda _3(3\pi /k),\,\;\lambda _{15}^{S}(k)=\lambda _3(\pi /k). \end{equation*} \end{enumerate} \end{example} \begin{example}\label{examp7} \rm $k=2s+1$, $s=4$. From Theorem {\ref{t22}} we have: \begin{align*} S(9)=&S(1)\cup \Gamma (\frac{\pi }{9})\cup \Gamma (\frac{3\pi }{9})\cup \Gamma (\frac{5\pi }{9})\cup \Gamma (\frac{7\pi }{9}) \\ &=\Gamma (\pi )\cup \Gamma (\frac{\pi }{9})\cup \Gamma (\frac{3\pi }{9})\cup \Gamma (\frac{5\pi }{9})\cup \Gamma (\frac{7\pi }{9}) \end{align*} This and Theorem {\ref{t23}} yields the inequalities: \begin{align*} \lambda _0(\pi /9) &<\lambda _0(3\pi /9)<\lambda _0(5\pi /9)<\lambda _0(7\pi /9)<\lambda _0(9\pi /9)=\lambda _0(\pi )\leq \lambda _1(\pi ) \\ &<\lambda _1(7\pi /9)<\lambda _1(5\pi /9)<\lambda _1(3\pi /9)<\lambda _1(1\pi /9) \\ &<\lambda _2(1\pi /9)<\lambda _2(3\pi /9)<\lambda _2(5\pi /9)<\lambda _2(7\pi /9)<\lambda _2(\pi )\leq \lambda _3(\pi ) \\ &<\lambda _3(7\pi /9)<\lambda _3(5\pi /9)<\lambda _3(3\pi /9)<\lambda _3(1\pi /9) \\ &<\lambda _{4}(1\pi /9)<\lambda _{4}(3\pi /9)<\lambda _{4}(5\pi /9)<\lambda _{4}(7\pi /9)<\lambda _{4}(\pi )\leq \lambda _{5}(\pi )<\dots \end{align*} From these inequalities and Theorem \ref{t42}: \begin{enumerate} \item for $m=0$ we have \begin{gather*} \lambda _0^{S}(9)=\lambda _0(\pi /9),\quad \lambda _1^{S}(9)=\lambda_0(3\pi /9),\quad \lambda _2^{S}(9)=\lambda _0(5\pi /9),\\ \lambda_3^{S}(9)=\lambda _0(7\pi /9), \quad \lambda _{4}^{S}(9)=\lambda _0^{S}; \end{gather*} \item $m=1:$ \begin{gather*} \lambda _{5}^{S}(9)=\lambda _1^{S},\quad \lambda _{6}^{S}(9)=\lambda_1(7\pi /9),\quad \lambda _{7}^{S}(9)=\lambda _1(5\pi /9),\\ \lambda_{8}^{S}(9)=\lambda _1(3\pi /9),\quad \lambda _{9}^{S}(9)=\lambda _1(\pi/9); \end{gather*} \item for $m=2$ we have \begin{gather*} \lambda _{10}^{S}(9)=\lambda _2(\pi /9),\quad \lambda _{11}^{S}(9)=\lambda_2(3\pi /9),\quad \lambda _{12}^{S}(9)=\lambda _2(5\pi /9),\\ \lambda_{13}^{S}(9)=\lambda _2(7\pi /9),\quad \lambda _{14}^{S}(9)=\lambda _2^{S}; \end{gather*} \item for $m=3$ we have \begin{gather*} \lambda _{15}^{S}(9)=\lambda _3^{S},\quad \lambda _{16}^{S}(9)=\lambda_3(7\pi /9),\quad \lambda _{17}^{S}(9)=\lambda _3(5\pi /9),\\ \lambda_{18}^{S}(9)=\lambda _3(3\pi /9),\quad \lambda _{19}^{S}(9)=\lambda _3(\pi/9); \end{gather*} \end{enumerate} \end{example} \subsection*{Acknowledgements} Y. 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