\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 159, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/159\hfil Representation of the norming constants]
{Representation of the norming constants by two spectra}

\author[T. N. Harutyunyan\hfil EJDE-2010/159\hfilneg]
{Tigran N. Harutyunyan}

\address{Tigran N. Harutyunyan \newline
Department of Mathematics and Mechanics of
Yerevan State University,\newline
A. Manukian str.1, 0025, Yerevan, Armenia}
\email{hartigr@yahoo.co.uk}


\thanks{Submitted 14 May 2010. Published November 4, 2010.}
\subjclass[2000]{34B24, 34B05, 34B09, 34L15}
\keywords{Sturm-Liouville problem; eigenvalues; norming constants}

\begin{abstract}
 The representation of the norming constants by 2
 spectra  was studied by  Levitan,  Gasymov (and others) for
 the Sturm-Liouville problem with boundary conditions
 $y(0)\cos\alpha +y'(0)\sin\alpha=0$,
 $y(\pi)\cos\beta+y'(\pi)\sin\beta =0$, when
 $\sin \alpha\neq 0$ and $\sin\beta\neq 0$. We investigate the
 representation by 2 spectra without these restrictions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction and statement of the main result}

Let $L(q, \alpha, \beta)$ denote the Sturm-Liouville (S.-L.) problem
\begin{gather}
\ell y\equiv -y''+q(x)y=\mu y,\quad x\in (0, \pi),\; \mu\in \mathbb{C},\label{eq1_1}\\
y(0)\cos\alpha+y'(0)\sin\alpha=0,\quad \alpha\in (0, \pi],\label{eq1_2}\\
y(\pi)\cos\beta +y'(\pi)\sin\beta=0,\quad \beta\in[0, \pi),\label{eq1_3}
\end{gather}
where $q$ is a real-valued, summable on $[0, \pi]$ function
(we write $q\in L^1_{\mathbb{R}}[0, \pi]$). By $L(q, \alpha, \beta)$ we
also denote the self-adjoint operator, generated by the problem
\eqref{eq1_1}-\eqref{eq1_3}, (see \cite{l1, m1}). It is known,
that the spectra of the operator $L(q, \alpha, \beta)$ is discrete and
consists of simple eigenvalues, which we denote by $\mu_n(q, \alpha, \beta)$,
$n=0, 1, 2, \dots$, emphasizing the dependence of $\mu_n$ on $q$, $\alpha$
and $\beta$.

The eigenvalues $\mu_n(q,\alpha,\beta)$ are enumerated in increasing order;
 i.e.,
$$
\mu_0(q,\alpha,\beta)< \mu_1(q,\alpha,\beta) <\dots < \mu_n(q,\alpha,\beta)
<\mu_{n+1}(q,\alpha,\beta)<\dots\,.
$$
In what follows, for brevity, we often use the notation
$\mu_n =\mu_n(\alpha,\beta)=\mu_n(q,\alpha,\beta)$.

By $\varphi(x, \mu, \gamma)$ and $\psi(x, \mu, \delta)$ we denote the solutions
of \eqref{eq1_1}, satisfying the initial conditions
\begin{gather}
\varphi(0, \mu, \gamma)=\sin \gamma,\quad \varphi'(0, \mu, \gamma)=-\cos \gamma, \quad
\gamma\in {\mathbb C},\label{eq1_4}\\
\psi(\pi, \mu, \delta)=\sin\delta,\quad \psi'(\pi, \mu, \delta)=-\cos \delta ,\quad
 \delta\in{\mathbb C},\label{eq1_5}
\end{gather}
correspondingly. The eigenvalues $\mu_n=\mu_n(q, \alpha, \beta)$,
$n=0,1, 2, \dots$, of $L(q, \alpha, \beta)$ are the solutions of the
equation
\begin{equation}\label{eq1_6}
\Phi(\mu)=\Phi(\mu,\alpha,\beta):=\varphi(\pi, \mu, \alpha)\cos \beta+\varphi'(\pi, \mu,\alpha)
\sin\beta=0,
\end{equation}
or the equation
\begin{equation}\label{eq1_7}
\Psi(\mu)=\Psi(\mu,\alpha,\beta):=\psi(0, \mu, \beta)\cos\alpha+\psi'(0, \mu, \beta)\sin \alpha=0.
\end{equation}
According to the  Liouville formula, the
Wronskian $W(x)=W(x,\varphi,\psi)=\varphi\cdot\psi'-\varphi'\psi$ 
of the solutions $\varphi$ and $\psi$ is constant. 
It follows that $W(0)=W(\pi)$ and, consequently,
$$
\Psi(\mu, \alpha, \beta)=-\Phi(\mu, \alpha, \beta).
$$
The functions $\varphi_n(x)=\varphi(x, \mu_n, \alpha)$ and
 $\psi_n(x)=\psi(x,\mu_n, \beta)$, $n=0, 1, 2, \dots$, are 
the eigenfunctions, corresponding to the eigenvalue $\mu_n$. 
The squares of the $L^2$-norms of these eigenfunctions:
\begin{equation}\label{eq1_8}
a_n=a_n(q,\alpha,\beta)=\int_0^{\pi} \vert\varphi_n(x)\vert^2 dx,\quad 
b_n=b_n(q,\alpha,\beta)=\int_0^{\pi}\vert\psi_n(x)\vert^2 dx,
\end{equation}
are called the norming constants.

Our aim is to prove the following theorem.

\begin{theorem} \label{thm1}
(1) For arbitrary $\varepsilon\in (0, \pi)$, $\varepsilon\neq \alpha$,
the following representations are true ($n=0,1,2,\dots$):
\begin{equation}
 a_n(q, \alpha, \beta)= \frac{\sin\alpha} {\sin \varepsilon}
\frac{\sin(\alpha-\varepsilon)} {\mu_n(\alpha, \beta)
-\mu_n(\varepsilon,\beta)}
\prod_{k=0,\, k\ne n}^\infty
\frac{\mu_k(\alpha, \beta)-\mu_n(\alpha, \beta)}{\mu_k(\varepsilon,
\beta)-\mu_n(\alpha,\beta)} \label{eq1_9}
\end{equation}
if $ \alpha\in (0,\pi)$, $\beta\in [0,\pi)$.
\begin{equation} \label{eq1_10}
\begin{aligned}
a_n(q, \pi, \beta)&= \frac{4n^2}{\pi\left(n+\frac{1}{2}\right)^2}\cdot
\frac{\mu_0(\pi, \beta)-\mu_n(\pi, \beta)} {\mu_0(\varepsilon,
\beta)-\mu_n(\pi, \beta)} \\
& \quad\cdot \frac{1}{\mu_n(\pi, \beta)-\mu_n(\varepsilon, \beta)}
 \cdot \prod_{\overset{k=1}{k\ne n}}^\infty \frac{k^2}{\left(k+\frac{1}{2}\right)^2}
 \cdot \frac{\mu_k(\pi,\beta)-\mu_n(\pi,\beta)} {\mu_k(\varepsilon,
\beta)-\mu_n(\pi, \beta)}
\end{aligned}
\end{equation}
if $\beta\in (0,\pi)$, $n\neq 0$.
\begin{equation} \label{eq1_11}
a_0(q, \pi, \beta)=\frac{4}{\pi}\frac{1}{\mu_0(\pi, \beta)
-\mu_0(\varepsilon, \beta)} \cdot \prod_{k=1}^\infty
\frac{k^2}{\left(k+\frac{1}{2}\right)^2}\cdot
\frac{\mu_k(\pi,\beta)-\mu_0(\pi,\beta)} {\mu_k(\varepsilon,
\beta)-\mu_0(\pi, \beta)}
\end{equation}
if $ \beta\in (0,\pi)$.
\begin{equation}
\begin{aligned}
&a_n(q, \pi, 0)\\
&=\frac{\left(n+\frac{1}{2}\right)^2}{\left(n+1\right)^2}
\cdot
\frac{\pi}{\mu_n(\pi, 0)-\mu_n(\varepsilon, 0)}\cdot
 \prod_{\overset{k=0}{k\ne n}}^\infty
\frac {\left(k+\frac{1}{2}\right)^2}{\left(k+1\right)^2}\cdot
 \frac{\mu_k(\pi,0)-\mu_n(\pi,0)} {\mu_k(\varepsilon, 0)-\mu_n(\pi, 0)}.
\end{aligned} \label{eq1_12}
\end{equation}

(2) For arbitrary $\eta\in (0, \pi)$, $\eta\neq \beta$,
($n=0,1,2,\dots$):
\begin{equation} \label{eq1_13}
 b_n(q, \alpha, \beta)= \frac{\sin\beta}{\sin \eta}\cdot
\frac{\sin(\beta-\eta)}{\mu_n(\alpha, \beta)-\mu_n(\alpha, \eta)}
\prod_{\overset{k=0}{k\ne n}}^\infty
\frac{\mu_k(\alpha, \beta)-\mu_n(\alpha, \beta)}{\mu_k(\alpha, \eta)
-\mu_n(\alpha,\beta)},
\end{equation}
if $\alpha\in (0,\pi]$, $\beta\in (0,\pi)$.
\begin{equation} \label{eq1_14}
\begin{aligned}
b_n(q, \alpha, 0)
&= \frac{4n^2}{\pi\left(n+\frac{1}{2}\right)^2}\cdot
\frac {\mu_0(\alpha, 0)-\mu_n(\alpha, 0)}{ \mu_n(\alpha, \eta)-\mu_n(\alpha, 0)} \\
&\quad \cdot \frac{1}{\mu_n(\alpha, 0)-\mu_n(\alpha, \eta)} \cdot
\prod_{\overset{k=1}{k\ne n}}^\infty \frac{k^2}{\left(k+\frac{1}{2}\right)^2}
\cdot \frac {\mu_k(\alpha, 0)-\mu_n(\alpha, 0)}
{\mu_k(\alpha, \eta)-\mu_n(\alpha, 0)},
\end{aligned}
\end{equation}
if $\alpha\in (0,\pi)$, $n\neq 0$.
\begin{equation} \label{eq1_15}
 b_0(q, \alpha, 0)=\frac{4}{\pi \left( \mu_0(\alpha, 0)-\mu_0(\alpha, \eta)\right) }
 \cdot \prod_{k=1}^\infty \frac{k^2}{\left(k+\frac{1}{2}\right)^2}\cdot
\frac{\mu_k(\alpha, 0)-\mu_n(\alpha, 0)}{\mu_k(\alpha, \eta)-\mu_n(\alpha, 0)},
\end{equation}
if $\alpha\in (0,\pi)$.
\begin{equation}
\begin{aligned}
&b_n(q, \pi, 0)\\
&=\frac{\left(n+\frac{1}{2}\right)^2}{(n+1)^2}\cdot
\frac{\pi }{\mu_n(\pi, 0)-\mu_n(\pi, \eta)}\cdot
\prod_{\overset{k=0}{k\ne n}}^\infty
\frac{\left(k+\frac{1}{2}\right)^2}{\left(k+1\right)^2}\cdot
\frac{\mu_k(\pi,0)-\mu_n(\pi,0)}{\mu_k(\pi,\eta)-\mu_n(\pi, 0)}.
\end{aligned}\label{eq1_16}
\end{equation}
\end{theorem}


The representation of the norming constants by two spectra was
investigated in \cite{f1,g1,l2,z1}
with the purpose of using it for (constructive) solution of the
inverse Sturm-Liouville problem by two spectra. They do it by
reducing the inverse problem by 2 spectra to the solution of
the inverse problem by spectral function and solve it
by Gelfand-Levitan method. In these papers the Sturm-Liouville
problem $L(q,\alpha,\beta)$ considered under conditions $\sin\alpha\neq 0$
and $\sin\beta\neq 0$ and as the norming constants the authors
consider (in our notation)
$\tilde{a}_n (q,\alpha,\beta)=\frac{a_n(q,\alpha,\beta)}{\sin^2\alpha}$.

Levitan \cite{l2} studied the representation of
$\tilde{a}_n (q,\alpha,\beta)$ by spectra $\{\mu_k(q,\alpha,\beta)\}^\infty_{k=0}$
and $\{\mu_k (q,\alpha,\beta_1)\}^\infty_{k=0}$, $\beta\neq\beta_1$,
$\alpha,\beta,\beta_1\in (0,\pi)$, when $q\in C_{\mathbb R}[0,\pi]$.

Levitan and Gasymov \cite{g1} study the representation of
$\tilde{a}_n (q,\alpha,\beta)$ by spectra $\{\mu_k(q,\alpha,\beta)\}^\infty_{k=0}$
 and $\{\mu_k (q,\varepsilon,\beta)\}^\infty_{k=0}$, $\alpha\neq\varepsilon$,
$\alpha,\varepsilon,\beta\in (0,\pi)$, when $q\in L^1_{\mathbb R}[0,\pi]$.
The formula, obtained in \cite{g1}, coincides with \eqref{eq1_9}.
The cases, when $\sin\alpha =0$ ($\alpha=\pi$) or $\sin\beta =0$ ($\beta=0$),
in \cite{l2} and \cite{g1} are not considered.
The representation of norming constants $b_n(q,\alpha,\beta)$
(or $ \tilde{b}_n(q,\alpha,\beta)=\frac{b_n(q,\alpha,\beta)}{\sin^2\beta}$)
 are usually not considered (since it is very similar to
investigation of $a_n$, and also it is not required in their
 solution of inverse problem by Gelfand-Levitan method).
One of our aims is the solution of inverse problem by ``the
eigenvalues function (EVF) of the family of Sturm-Liouville
operators" (see \cite{h1}). The statement of the problem and its
solution will be the subject of our future articles. However in
this solution we will need to simultaneously  use the
representations for both $a_n$ and $b_n$, and this is the main
reason for such a detailed formulation of the Theorem of this
article.

Technically the consideration of the cases $\sin\alpha\neq 0$,
$\sin\beta\neq 0$ is distinguished by the fact that in these cases
the (rough) asymptotic behavior of the eigenvalues is
\begin{equation}\label{eq1_17}
\mu_n(q,\alpha,\beta)=n^2+O(1),
\end{equation}
and thus
\begin{gather}
 \mu_n(q,\alpha,\beta)-\mu_n(q,\varepsilon,\beta)=O(1),\label{eq1_18}\\
 \mu_n(q,\alpha,\beta)-\mu_n(q,\alpha,\beta_1)=O(1).\label{eq1_19}
\end{gather}
These relations are important for the convergence in the infinite
product \eqref{eq1_9} and \eqref{eq1_13}. Meanwhile, when
$\sin\alpha=0$ ($\alpha=\pi$) or/and $\sin\beta=0$ ($\beta=0$) the (rough)
asymptotic behavior of the eigenvalues have the form
($n\to\infty$) (in \cite{h1} we have obtained the common formula
for the asymptotic behavior of the eigenvalues $\mu_n(q,\alpha,\beta)$ in
all the cases $\alpha\in (0,\pi]$ and $\beta\in [0,\pi)$, from
which \eqref{eq1_17}, \eqref{eq1_20}-\eqref{eq1_22} follow as
particular cases):
\begin{gather}
 \mu_n(q,\pi,\beta)=\Big(n+\frac{1}{2}\Big)^2+O(1),\quad \mbox{when }
 \sin\beta\neq 0\label{eq1_20}\\
 \mu_n(q,\alpha,0)=\Big(n+\frac{1}{2}\Big)^2+O(1),\quad \mbox{when }
 \sin\alpha\neq 0\label{eq1_21}\\
 \mu_n(q,\pi,0)=\left(n+1\right)^2+O(1).\label{eq1_22}
\end{gather}
The relations \eqref{eq1_18} and \eqref{eq1_19} do not hold in the
general case (for example, $\mu_n(q,\pi,\beta)-\mu_n(q,\alpha,\beta)=O(n)$
when $\alpha\in (0,\pi)$).

Thus to obtain the representation of $a_n$ and $b_n$ for all the
cases by two spectra we need some improvement of the method
in \cite{l2} and \cite{g1}.

These improvement are the causes that the formulae \eqref{eq1_9}
and \eqref{eq1_10}-\eqref{eq1_12} (similarly \eqref{eq1_13}
and \eqref{eq1_14}-\eqref{eq1_16}) are different and depend
on the cases:
\begin{itemize}
\item $\alpha,\beta\in(0,\pi)$,

\item $\alpha=\pi$, $\beta\in (0,\pi)$,

\item $\alpha\in (0,\pi)$, $\beta=0$,

\item $\alpha=\pi$, $\beta=0$.
\end{itemize}
Whether it is possible to join all these formulae in one (may be
more complicated, but one), today we do not know.

It was obtained in \cite{h1}, that the lowest eigenvalue
$\mu_0(q,\alpha,\beta)$ has the property: for arbitrary $\beta\in
[0,\pi)$, $\lim_{\alpha\to 0}\mu_0(q,\alpha,\beta)=-\infty$ and for arbitrary
$\alpha\in (0,\pi]$, $\lim_{\beta\to\pi}\mu_0(q,\alpha,\beta)=-\infty$.

In combination with the property
$$
 \frac{1}{a_n(q, \alpha, \beta)}
 =\frac{\partial \mu_n(q,\alpha,\beta)}{\partial\alpha}
$$
(and $ \frac{1}{b_n(q, \alpha, \beta)}
=-\frac{\partial \mu_n(q,\alpha,\beta)}{\partial\beta}$  \cite{h1})
this leads to the fact that the formulae \eqref{eq1_11}
and  \eqref{eq1_15} for $a_0(q,\pi,\beta)$ and $b_0(q,\alpha,0)$
have a special form.


\section{Proof of the main theorem}

The proof of the main theorem is based on the following lemmas.
Here we give the statements of these lemmas and after that we
prove the Theorem. The proofs of the lemmas are presented in \S3.

\begin{lemma}\label{lem1}
 As $t\to\infty$, we have the following:
\begin{gather}
 \Phi(-t^2,\alpha,\beta)=\frac{t e^{\pi t}}{2}
 [ \sin\alpha\cdot \sin\beta+O\big( \frac1{t}\big)],\quad\text{when }
 \alpha, \beta\in (0,\pi), \label{eq2_1}\\
 \Phi(-t^2,\pi,\beta)= \frac{e^{\pi t}}{2}
[ \sin\beta+O\big( \frac1{t}\big)], \quad\text{when }
 \beta\in (0,\pi),\label{eq2_2}\\
 \Phi(-t^2,\alpha,0)= \frac{e^{\pi t}}{2}
[ \sin\alpha+O\big( \frac1{t}\big)], \quad\text{when }
 \alpha\in (0,\pi),\label{eq2_3}\\
 \Phi(-t^2,\pi, 0)= \frac{e^{\pi t}}{2t}
[ 1+O\big( \frac1{t}\big)].  \label{eq2_4}
\end{gather}
\end{lemma}

\begin{lemma}\label{lem2}
 The specification of the spectra $\{\mu_n(q,\alpha,\beta)\}^\infty_{n=0}$
uniquely determines the characteristic functions
$\Phi (\mu,\alpha,\beta)$ by the formulae
\begin{equation}
 \Phi(\mu, \alpha,\beta)=\pi \left( \mu_0
(\alpha,\beta)-\mu\right)\cdot\sin\alpha\cdot\sin\beta\cdot\prod^\infty_{k=1}
\frac{\mu_k(\alpha,\beta)-\mu}{k^2}, \label{eq2_5}
\end{equation}
when $\alpha,\beta\in(0,\pi)$.
\begin{gather}
 \Phi(\mu, \pi,\beta)=\sin\beta\cdot\prod^\infty_{k=0}
 \frac{\mu_k(\pi,\beta)-\mu}{(k+\frac{1}{2})^2},
 \quad\text{when } \beta\in(0,\pi), \label{eq2_6}
\\
 \Phi(\mu, \alpha, 0)=\sin\alpha\cdot\prod^\infty_{k=0} \frac{\mu_k(\alpha,0)
 -\mu}{(k+\frac{1}{2})^2}, \quad\text{when } \alpha\in(0,\pi),
 \label{eq2_7}
\\
 \Phi(\mu, \pi, 0)=\pi \prod^\infty_{k=0}
\frac{\mu_k(\pi,0)-\mu}{( k+1)^2}. \label{eq2_8}
\end{gather}
\end{lemma}

In what follows, by $\dot{f}(x,\mu)$ we denote the derivative
by $\mu$; i.e., $ \dot{f}(x,\mu)=\frac{\partial f(x,\mu)}{\delta \mu}$.

\begin{lemma}\label{lem3}
 The following formulae hold
\begin{equation}
\dot{\Phi}(\mu, \alpha,\beta)\big|_{\mu=\mu_0(\alpha,\beta)}=
-\pi \sin\alpha\cdot \sin\beta \prod^\infty_{k=1}
\frac{\mu_k(\alpha,\beta)-\mu}{k^2}, \label{eq2_9}
\end{equation}
when $\alpha,\beta\in(0,\pi)$.
\begin{equation} \label{eq2_10}
\begin{aligned}
\dot{\Phi}(\mu, \alpha,\beta)\big|_{\mu=\mu_n(\alpha,\beta)}
&= -\frac{\pi}{n^2} \left[\mu_0(\alpha,\beta)-\mu_n(\alpha,\beta)\right] \\
& \quad\cdot \sin\alpha\cdot\sin\beta \prod^\infty_{\overset{k=1}{k\ne n}}
\frac{\mu_k(\alpha,\beta)-\mu_n (\alpha,\beta)}{k^2},
\end{aligned}
\end{equation}
when $\alpha,\beta\in(0,\pi)$ and $n\neq 0$.
\begin{equation}
\dot{\Phi}(\mu, \pi, \beta)\big|_{\mu=\mu_n(\pi, \beta)}
=-\frac{\sin\beta}{\left( n+\frac{1}{2}\right)^2}
\prod^\infty_{\overset{k=0}{k\ne n}} \frac{\mu_k(\pi,\beta)
-\mu_n(\pi, \beta)}{(k+\frac{1}{2})^2}, \label{eq2_11}
\end{equation}
when $\beta\in(0,\pi)$.
\begin{equation}
\dot{\Phi}(\mu, \alpha, 0)\big|_{\mu=\mu_n(\alpha, 0)}=
-\frac{\sin\alpha}{\left( n+\frac{1}{2}\right)^2}
\prod^\infty_{\overset{k=0}{k\ne n}} \frac{\mu_k(\alpha,0)
-\mu_n(\alpha, 0)}{(k+\frac{1}{2})^2}, \label{eq2_12}
\end{equation}
when $\alpha\in(0,\pi)$.
\begin{equation}
\dot{\Phi}(\mu, \pi, 0)\big|_{\mu=\mu_n(\pi, 0)}=
-\frac{\pi}{\left( n+1\right)^2} \prod^\infty_{\overset{k=0}{k\ne n}}
\frac{\mu_k(\pi,0)-\mu_n(\pi, 0)}{\left( k+1\right)^2}. \label{eq2_13}
\end{equation}
\end{lemma}

We consider also the meromorphic functions
(see \eqref{eq1_6} and \eqref{eq1_7}):
\begin{gather}
 \label{eq2_14} m_{\alpha,\beta,\varepsilon}(\mu)
 := \frac{\Psi(\mu,\alpha,\beta)}{\Psi(\mu,\varepsilon,\beta)}
 = \frac{\psi(0,\mu,\beta)\cos\alpha+\psi'(0,\mu,\beta)\sin\alpha}{\psi(0,\mu,\beta)
  \cos\varepsilon +\psi'(0,\mu,\beta)\sin\varepsilon}\,,\\
\label{eq2_15} n_{\alpha,\beta,\delta}(\mu)
 := \frac{\Phi(\mu,\alpha,\beta)}{\Phi(\mu,\alpha,\delta)}
 = \frac{\varphi(\pi,\mu,\alpha)\cos\beta+\varphi'(\pi,\mu,\alpha)\sin\beta}{\varphi(\pi,\mu,\alpha)
  \cos\delta +\varphi'(\pi,\mu,\alpha)\sin\delta}\,.
\end{gather}

\begin{lemma}\label{lem4}
 For arbitrary $\varepsilon\in (0,\pi)$, $\varepsilon\neq \alpha$,
\begin{equation}\label{eq2_16}
\frac{\partial m_{\alpha,\beta,\varepsilon}(\mu)}{\partial \mu}\big|_{\mu=\mu_n(\alpha,\beta)}
= \frac{a_n(q,\alpha,\beta)}{\sin(\alpha-\varepsilon)}\,,
\end{equation}
and for arbitrary $\delta\in (0,\pi)$, $\delta\neq\beta$,
\begin{equation}\label{eq2_17}
\frac{\partial n_{\alpha,\beta,\delta}(\mu)}{\partial \mu}\big|_{\mu=\mu_n(\alpha,\beta)}
= \frac{b_n(q,\alpha,\beta)}{\sin(\beta-\delta)}\,.
\end{equation}
\end{lemma}

As it is noted in \S1, $\Psi(\mu,\alpha,\beta)=-\Phi(\mu,\alpha,\beta)$.
So, \eqref{eq2_14} we can be rewritten as
\begin{equation}
m_{\alpha,\beta,\varepsilon}(\mu)=\frac{\Phi(\mu,\alpha,\beta)}{\Phi(\mu,\varepsilon,\beta)}\,.
\label{e2.14'}
\end{equation}
It is easy follows from the \eqref{e2.14'} and \eqref{eq2_15} that
\begin{gather*}
\frac{\partial m_{\alpha,\beta,\varepsilon}(\mu)}{\partial \mu}\big|_{\mu=\mu_n(\alpha,\beta)}
= \frac{\dot{\Phi}(\mu_n(\alpha,\beta),\alpha,\beta)}{\Phi(\mu_n(\alpha,\beta),\varepsilon,\beta)},
\\
\frac{\partial n_{\alpha,\beta,\delta}(\mu)}{\partial \mu}\big|_{\mu=\mu_n(\alpha,\beta)}
= \frac{\dot{\Phi}(\mu_n(\alpha,\beta),\alpha,\beta)}{\Phi(\mu_n(\alpha,\beta),\alpha,\delta)}\,.
\end{gather*}
Then, from \eqref{eq2_16} and \eqref{eq2_17}, we obtain
\begin{gather} \label{eq2_18}
a_n(q,\alpha,\beta)=\sin(\alpha-\varepsilon)\, \frac{\dot{\Phi}(\mu_n(\alpha,\beta),\alpha,\beta)}
{\Phi(\mu_n(\alpha,\beta),\varepsilon,\beta)}\,,\\
\label{eq2_19}
b_n(q,\alpha,\beta)=\sin(\beta-\delta)\, \frac{\dot{\Phi}(\mu_n(\alpha,\beta),\alpha,\beta)}
{\Phi(\mu_n(\alpha,\beta),\alpha,\delta)}\,.
\end{gather}
Substituting corresponding formulae from \eqref{eq2_9}-\eqref{eq2_13}
and \eqref{eq2_5}-\eqref{eq2_8} in \eqref{eq2_18} and \eqref{eq2_19},
we obtain the formulae \eqref{eq1_9}-\eqref{eq1_12} and
\eqref{eq1_13}-\eqref{eq1_16} of Theorem.
Thus, our Theorem is proved if the lemmas \ref{lem1}-\ref{lem4} hold.

\section{Proofs of lemmas}

\begin{proof}[Proof of the lemma \ref{lem1}]
Let us denoted by $y_1(x,\lambda)$ and $y_2(x,\lambda)$ the solutions of
the equation $-y''+q(x)y=\lambda^2 y$, satisfying the initial conditions
\begin{gather*}
 y_1(0,\lambda)=1, \quad y'_1(0,\lambda)=0,\\
 y_2(0,\lambda)=0, \quad   y'_2(0,\lambda)=1.
\end{gather*}
For $y_1(x,\lambda)$ and $y_2(x,\lambda)$ it is well known
\cite{f1,l1,m1,p1} (the case $q\in L^1_{\mathbb C}[0,\pi]$ is
considered in detail in \cite{h2})
the asymptotic formulae (when $\vert\lambda\vert\to\infty$)
\begin{equation}\label{eq3_1}
\begin{gathered}
 y_1(\pi,\lambda)=\cos\lambda\pi
+O\Big(\frac{e^{\vert\text{Im}\lambda\vert \pi}}{\vert \lambda\vert} \Big), \quad
 y'_1(\pi,\lambda)=-\lambda\sin\lambda\pi +O\left( e^{\vert\text{Im}\lambda\vert \pi}\right),\\
 y_2(\pi,\lambda)=\frac{\sin\lambda\pi}{\lambda}
+O\Big(\frac{e^{\vert\text{Im}\lambda\vert \pi}}{\vert \lambda\vert^2} \Big), \quad
 y'_2(\pi,\lambda)=\cos\lambda\pi +O\Big(\frac{e^{\vert\text{Im}\lambda\vert \pi}}
{\vert \lambda\vert} \Big).
\end{gathered}
\end{equation}
Since (see \eqref{eq1_4}),
\begin{equation}\label{eq3_2}
\varphi(x,\lambda^2,\alpha)=y_1(x,\lambda)\sin\alpha-y_2(x,\lambda)\cos\alpha
\end{equation}
and $\Phi(\mu,\alpha,\beta)$ is defined by \eqref{eq1_6}, if we
substitute in \eqref{eq3_1}, \eqref{eq3_2} and \eqref{eq1_6}
$\mu=\lambda^2=(it)^2=-t^2$, we obtain for $\Phi(-t^2,\alpha,\beta)$
 the formula
\begin{align*}
 \Phi(-t^2,\alpha,\beta)
&=\Big( \cos(i\pi t)+O\big(\frac{e^{\pi t}}{t}\big)\Big)
\sin\alpha\cos\beta -\Big( \frac{\sin(i\pi t)}{ i t }+O\big(\frac{e^{\pi t}}{t}
\big)\Big) \cos\alpha\cos\beta \\
& + ( -i t\sin(i\pi t)+O(e^{\pi t}))\sin\alpha\cos\beta
-\Big( \cos(i \pi t)+O\big(\frac{e^{\pi t}}{t}\big)\Big) \cos\alpha\sin\beta .
\end{align*}
Taking into account the formulae
$ \cos(i\pi t)=\frac{e^{\pi t}+e^{-\pi t}}{2}$,
$ \sin(i\pi t)=\frac{e^{-\pi t}-e^{\pi t}}{2 i}$ we obtain
the assertions of lemma \ref{lem1}.
\end{proof}


\begin{proof}[Proof of the lemma \ref{lem2}]
The case $\alpha , \beta\in (0,\pi)$ was considered in
\cite{f1,g1,i1}; see \cite[formula (1.1.26)]{f1} for details.
In this case ($\sin\alpha\neq 0$, $\sin\beta\neq 0$)
they  consider as the  characteristic function
$\frac{\Phi(\mu, \alpha,\beta)}{\sin\alpha \cdot \sin\beta}$ (in our notation).

The remaining three results are presented in \cite{f1}, but without
a proof, and in the case $\alpha=\pi$, $\beta=0$ they write
$$
\Phi(\mu,\pi,0) =\pi \prod^\infty_{k=1} \frac{\mu_k(\pi,0)-\mu}{
k^2},
$$
 since they begin the enumeration for $\mu_n(\pi,0)$ from
$n=1$, but not from $n=0$. We use Hadamard's factorization theorem
(it is similar to method of \cite{f1}) to represent the entire
function $\Phi(\mu,\alpha,\beta)$ of order $1/2$ in the form
\begin{equation}\label{eq3_3}
\Phi(\mu,\alpha,\beta)=c\prod^\infty_{k=0} \Big( 1-\frac{\mu}{\mu_k(\alpha,\beta)}
\Big)\,.
\end{equation}
If $\Phi(0,\alpha,\beta)=0$, i.e. for some $k_0$ $\;\mu_{k_0}(\alpha,\beta)=0$, then we must change the $k_0$-th factor by $-\mu$ and it is easy to show that we can take $\mu_{k_0}(\alpha,\beta)=0=\mu_0(\alpha,\beta)$ (change the enumeration for finite number of eigenvalues and correspondingly for finite factors) in
\begin{equation}\label{eq3_4}
\Phi(\mu,\alpha,\beta)
=c_1\mu \prod^\infty_{\overset{k=0}{k\ne k_0}} \Big(
1-\frac{\mu}{\mu_k(\alpha,\beta)} \Big)
= c_2\mu \prod^\infty_{k=1} \Big(
1-\frac{\mu}{\mu_k(\alpha,\beta)} \Big)\,.
\end{equation}
Detailed computations show that results for the cases \eqref{eq3_3}
and \eqref{eq3_4} are the same.

In the case $\alpha=\pi$, $\beta\in (0, \pi)$ ($\sin\beta\neq 0$) we use the formula
\begin{equation}\label{eq3_5}
\cos\lambda\pi = \prod^\infty_{k=0}\Big( 1-\frac{\lambda^2}{(k+\frac{1}{2})^2}\Big)
\end{equation}
and consider the relation
\begin{equation}\label{eq3_6}
\frac{\Phi(\lambda^2, \pi,\beta)}{\cos\lambda\pi}=\frac{ c\prod^\infty_{k=0}
\big( 1-\frac{\lambda^2}{\mu_k(\pi,\beta)}\big)}
{\prod^\infty_{k=0} \frac{(k+\frac{1}{2})^2 -\lambda^2}
 {(k+\frac{1}{2})^2}}
 = c \prod^\infty_{k=0}\frac{(k+\frac{1}{2})^2}{\mu_k(\pi, \beta)}
\cdot \frac{\mu_k(\pi, \beta)-\lambda^2}{(k+\frac{1}{2})^2-\lambda^2}.
\end{equation}
According to  \eqref{eq1_20}, in this case the infinite product
$$
\prod^\infty_{k=0} \frac{\mu_k(\pi,\beta)-\lambda^2}
{(k+\frac{1}{2})^2-\lambda^2}
=\prod^\infty_{k=0} \Big( 1+\frac{\mu_k(\pi,\beta)
- (k+\frac{1}{2})^2}{(k+\frac{1}{2})^2-\lambda^2}\Big)$$
converges uniformly because $\lambda^2=-t^2\in (-\infty,-1])$.
The infinite product
$$
\prod^\infty_{k=0}\frac{(k+\frac{1}{2})^2}{\mu_k(\pi,\beta)}
$$
also converses. So we can write the right hand side of
\eqref{eq3_6} in the form
$$
c \prod^\infty_{k=0}\frac{(k+\frac{1}{2})^2}{\mu_k(\pi, \beta)}
\cdot\prod^\infty_{k=0}\frac{\mu_k(\pi, \beta)-\lambda^2}{(k+\frac{1}{2})^2-\lambda^2}.
$$
Thus, if we pass to the limit in \eqref{eq3_6} when
$\lambda=it$, $t\to\infty$,
then in the left hand side of \eqref{eq3_6}, according to lemma
\ref{lem1}, we obtain
$$
\lim_{t\to\infty} \frac{\Phi(-t^2,\alpha,\beta)}{\cos(i t\pi) }
=\lim_{t\to\infty} \frac{\frac{ e^{t\pi}}{2}
[\sin\beta +O\left( \frac{1}{t}\right)]}{\frac{e^{\pi t}+e^{-\pi t}}{2}}
=\sin\beta ,
$$
and in the right hand side we obtain $ c\prod^\infty_{k=0}
\frac{(k+\frac{1}{2})^2}{\mu_k (\pi,\beta)}$; i.e.,
$ c=\sin\beta \prod^\infty_{k=0} \frac{\mu_k (\pi,\beta)}{(k+\frac{1}{2})^2}$.
Substituting this value of $c$ in \eqref{eq3_3}, we obtain formula
\eqref{eq2_6}.

In the case $\alpha\in (0,\pi)$, $\beta=0$ we follow the same procedure and
obtain formula \eqref{eq2_7}.

In case $\alpha=\pi$, $\beta=0$ we consider the relation of
$$
\Phi(\lambda^2,\pi, 0)= c \prod^\infty_{k=0}\Big( 1-
\frac{\lambda^2}{\mu_k(\pi, 0)}\Big)
$$
 and $ \sin(\lambda\pi)/\lambda$, which we can write in the form
$$
\frac{\sin \lambda\pi}{\lambda}=\pi\prod^\infty_{k=0}
\Big( 1-\frac{\lambda^2}{(k+1)^2} \Big).
$$
According to  \eqref{eq1_22}, the infinite product $
\prod^\infty_{k=0} \frac{(k+1)^2}{\mu_k(\pi, 0)}$ converges and
the product
$$
\prod^{\infty}_{k=0}  \frac{\mu_k(\pi, 0)-\lambda^2}{(k+1)^2-\lambda^2}
= \prod^{\infty}_{k=0}
\Big[ 1+ \frac{\mu_k(\pi, 0)-(k+1)^2}{(k+1)^2-\lambda^2}\Big]\, ,
$$
converges uniformly by $\lambda^2=(it)^2=-t^2\in (-\infty, -1]$.
Thus, we can pass to the limit ($t\to\infty$) in the relation
$$
\frac{\Phi(-t^2,\pi, 0)}{\frac{\sin i \pi t}{it}}
=\frac{c}{\pi} \prod^\infty_{k=0}\frac{(k+1)^2}{\mu_k(\pi, 0)}
\cdot \prod^\infty_{k=0} \frac{\mu_k(\pi, 0)+t^2}{(k+1)^2 +t^2}
$$
and according to \eqref{eq2_4} and uniform convergence of the
right hand side, we obtain $ 1=\frac{c}{\pi}
\prod^\infty_{k=0}\frac{(k+1)^2}{\mu_k(\pi, 0)}$.
Thus, we obtain the formula \eqref{eq2_8}.
Lemma \ref{lem2} is proved.
\end{proof}


 The proof of the lemma \ref{lem3} coincides with simple
differentiation with respect to $\mu$ at the point
$\mu=\mu_n(\alpha,\beta)$ in
\eqref{eq2_5}-\eqref{eq2_8}. With this aim we represent the
infinite product $ \prod^\infty_{k=0}a_k$ in the form
$$
\prod^\infty_{k=0} a_k
=\prod^n_{k=0} a_k\cdot\prod^\infty_{k=n+1}a_k
= a_n \prod^{n-1}_{k=0}a_k \prod^\infty_{k=n+1} a_k
=a_n \prod^\infty_{\overset{k=0}{k\ne n}} a_k .
$$
For example,
$$
\Phi(\mu,\alpha,0)=\frac{\mu_n(\pi,0)-\mu}{\left(
n+\frac{1}{2}\right)^2}\,\cdot\sin\alpha\,\cdot \prod_{k\neq n}
\frac{\mu_k(\pi,0)-\mu}{(k+\frac{1}{2})^2}.
$$
 Now differentiating with respect to $\mu$ and taking the result
at $\mu=\mu_n(\alpha, 0)$ we obtain
\begin{align*}
&\dot{\Phi}(\mu_n(\alpha,0),\alpha,0) \\
&=\frac{-1}{\left( n+\frac{1}{2}\right)^2}\,\sin\alpha\,\cdot
 \prod^\infty_{\overset{k=0}{k\ne n}} \frac{\mu_k(\alpha,0)
 -\mu}{(k+\frac{1}{2})^2}\big|_{\mu=\mu_n(\alpha,0)}
+\frac{\mu_n(\alpha,0)-\mu}{\left( n+\frac{1}{2}\right)^2}\cdot
 \mathcal{P}(\mu)\big|_{\mu=\mu(\alpha,0)}\\
&= -\frac{\sin\alpha}{\left( n+\frac{1}{2}\right)^2} 
\prod^\infty_{\overset{k\ne n}{k=0}} \frac{\mu_k(\alpha,0)-\mu_n(\alpha,0)}{(k+\frac{1}{2})^2}\,,
\end{align*}
where
$$
\mathcal{P}(\mu)=\frac{\partial}{\partial\mu}\Big(\sin\alpha
\prod^\infty_{\overset{k=0}{k\ne n}} \frac{\mu_k(\alpha,0)-\mu}
{(k+\frac{1}{2})^2}\Big)\,.
$$
The process is analogous in other cases. Lemma \ref{lem3} is proved.


\begin{proof}[Proof of lemma \ref{lem4}]

Since all the eigenvalues of $L(q,\alpha,\beta)$ are simple,
there exist the constants $c_n=c_n(q,\alpha,\beta)$, $n=0,1,2,\dots$,
such that
\begin{equation}\label{eq3_9}
\varphi_n(x)=c_n\cdot \psi_n(x)\,.
\end{equation}
By direct computation, from \eqref{eq2_14} we obtain
\begin{equation}\label{eq3_7}
\frac{\partial m_{\alpha,\beta,\varepsilon}(\mu)}{\partial \mu}
=\frac{[\dot{\psi}'(0,\mu,\beta)\cdot\psi(0,\mu,\beta)
 -\dot{\psi}(0,\mu,\beta)\cdot \psi'(0,\mu,\beta) ] \sin(\alpha-\varepsilon)}
{[\psi(0,\mu,\beta)\cos\varepsilon +\psi'(0,\mu,\beta)\sin\varepsilon]^2}\,.
\end{equation}
On the other hand, by standard methods, we obtain
\begin{equation}\label{eq3_8}
\int^\pi_0\psi^2(x,\mu,\beta)\, dx
=\dot{\psi}'(0,\mu,\beta)\cdot\psi(0,\mu,\beta)-\dot{\psi}(0,\mu,\beta)\cdot
\psi'(0,\mu,\beta)\,.
\end{equation}
Since for $\mu=\mu_n(q,\alpha,\beta)$ we have \eqref{eq3_9}, using
\eqref{eq1_4}, from \eqref{eq3_7} and \eqref{eq3_8} we obtain
\begin{align*}
\frac{\partial m_{\alpha,\beta,\varepsilon}(\mu)}{\partial \mu}
\big|_{\mu=\mu_n(\alpha,\beta)}
&= \frac{\int^\infty_0 \psi^2_n(x) dx\cdot \sin(\alpha-\varepsilon)}
{[\psi_n(0)\cos\varepsilon+\psi'_n(0)\sin\varepsilon]^2}\\
&= \frac{c^2_n\int^\infty_0
\varphi^2_n(x) dx\cdot \sin(\alpha-\varepsilon)}{c^2_n[
\varphi_n(0)\cos\varepsilon+\varphi'_n(0)\sin\varepsilon]^2}\\
&= \frac{a_n(q,\alpha,\beta)}{\sin(\alpha-\varepsilon)}\,.
\end{align*}
We prove \eqref{eq2_17} in
a similar way. Lemma \ref{lem4} is proved.
\end{proof}

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\end{document}