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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 182, pp. 1--16.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/182\hfil Reconstructing the potential function]
{Reconstructing the potential function for indefinite
 Sturm-Liouville problems using infinite product forms}

\author[M. Dehghan, A. Jodayree \hfil EJDE-2013/182\hfilneg]
{Mohammad Dehghan, Ali Asghar Jodayree }  % in alphabetical order

\address{Mohammad Dehghan \newline
Faculty of Mathematical Sciences, University of Tabriz, Tabriz, Iran}
\email{m-dehghan@tabrizu.ac.ir,  Tel. +989113512619}

\address{Ali Asghar Jodayree \newline
Faculty of Mathematical Sciences, University of Tabriz, Tabriz, Iran}
\email{akbarfam@yahoo.com}

\thanks{Submitted July 15, 2013. Published August 7, 2013.}
\subjclass[2000]{34B24, 34A55, 34E20, 34E05}
\keywords{Indefinite Sturm-Liouville problem; turning point; dual equations;
\hfill\break\indent infinite product form}

\begin{abstract}
 In this article we consider the linear second-order equation of
 Sturm-Liouville type
 $$
 y''+(\lambda\phi^2(t)-q(t))y=0, \quad 0\leq t\leq 1,
 $$
 where $\lambda$ is a real parameter, $q(t)$ is the potential function
 and $\phi^2(t)$ is the weight function. We use the infinite product
 representation of the derivative of the solution to the differential equation
 with Dirichlet-Neumann conditions, and for the system of dual equations
 which is needed for expressing inverse problem and for retrieving potential.
 It must be mentioned that the weight function has a zero whose order
 is an integer called a \emph{turning point}.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}\label{intro}

We consider the indefinite Sturm-Liouville equation
\begin{equation} \label{orig}
ly:=-y''+q(t)y=\lambda \phi^{2}(t)y, \quad  0\leq t\leq x,
\end{equation}
with  Dirichlet conditions
\begin{equation} \label{con1}
   y(0)=y(x)=0,
\end{equation}
and with   Dirichlet-Neumann conditions
\begin{equation} \label{con2}
    y(0)=y'(x)=0,
\end{equation}
where $\lambda=\rho^2$ is the spectral parameter, $x$ is a fixed
point in the interval $(0,1)$ and also the weight function  $\phi^{2}(t)$
and the potential function $q(t)$ satisfies
\begin{itemize}
 \item  $\phi^{2}(t)=(t-t_0)^{l_0}\phi_0(t)$ is real and has one zero,
 $t_0$, so called turning point of odd order $l_0\in \mathbf{N}$ in $[0,1]$
 and also $\phi_0(t)$ is positive and twice continuously differentiable.
 \item  $q(t)$ is bounded and integrable on $[0,1]$.
\end{itemize}

 The asymptotic solutions of  \eqref{orig} depend on a complex parameter
$\rho$ as  $|\rho|\to \infty$.
We assume $t_0$ to be a turning point of type IV; i.e., $l_0$ is
odd. The operator $l$ defined in \eqref{orig} is called the
indefinite Sturm-Liouville operator.
The differential equation \eqref{orig}  with conditions \eqref{con1} and
\eqref{con2}   are denoted by $L_1(\phi^2(t),q(t),x)$ and
$L_2(\phi^2(t),q(t),x)$, respectively.

Inverse spectral theory can be considered as the determination of an operator
(usually differential) from its spectral data. The literature on such
subjects is immense for the definite cases in which the weight function
 $\phi^2(t)$ is positive throughout the interval such as
\cite{levitan,rund,koy,ozk}, while for the indefinite cases is not.
The transformation operator method   and the Gelfand-Levitan integral
equation with respect to the kernel of the transformation operator \cite{kach}
 in this case is not suitable for the solution of the
inverse problems.

The eigenvalue problem for the indefinite Sturm-Liouville problem has
been discussed in \cite{kong}.
The potential $q(x)$ of an indefinite Sturm-Liouville problem has been
determined uniquely by three spectra in {\cite{fu}}.
We present a new approach to reconstruct the operator $l$
(indefinite Sturm-Liouville operator); i.e., retrieving potential
function $q(t)$ in \eqref{orig} from its spectral data.
The question at the core of this paper involves the determination
of the infinite product representation for the derivative of the
solution of the indefinite Sturm-Liouville problem as well as the
reconstruction of the potential function $q(t)$ by means of two spectra,
 while the weight function $\phi^2(t)$ is given.

Differential equations with an indefinite weight function appear in
several mathematical physics problems. For instance, turning points
correspond to the limit of motion of a wave mechanical particle bounded
 by a potential field. Turning points arise also in various fields
such as optics, elasticity, spectroscopy, stratification and radio
engineering problems to design directional couplers for non-uniform
electronic lines (see \cite{daho,lit,mchugh,mesh,sve,wasow}
 for further references).

The presence of turning points yields fundamental qualitative changes
in the study of this kind of differential equation.
In problem $L_1(\phi^2(t),q(t),x)$, for the special case
$\phi^2(t)=t$ in the interval $[-1,1]$, Jodayree et al. obtained the
infinite product representation of the solution in the  closed
form \cite{jodayree1}:
\[
U(x,\lambda)=\begin{cases}
\frac{p(x)}{(-x)^{1/4}}\prod_{k\geq 1}\frac{\lambda-\lambda_k(x)}{z_k^2(x)},
&-1\leq x<0, \\[4pt]
\frac{\pi\sqrt{x}}{6}\prod_{k\geq
1}\frac{(\lambda-r_k(x))p^2(0)}{\tilde{j}^2_k} \prod_{k\geq
1}\frac{f^2(x)(u_k(x)-\lambda)}{\tilde{j}^2_k}, & 0<x\leq 1,
\end{cases}
\]
where $x$ is a fixed point in $(-1,1)$, $U(x,\lambda)$ is the
solution of the differential equation  $L_1(t,q(t),x)$ which satisfies
the initial condition
\[
U(-1,\lambda)=0,\quad
\frac{\partial U}{\partial t}(-1,\lambda)=1.
\]
Here
$p(x)=-(2/3)(-x)^{3/2}+2/3$, $f(x)=(2/3)x^{3/2}$,
$z_k(x)=k\pi/p(x)$, $\tilde{j}_k$ are the positive zeros of
Bessel function $J'_1(z)$, $\{\lambda_k(x)\}$ are the eigenvalues of
problem with the Dirichlet condition on $[-1,x]$ for $x<0$ and
$\{r_k(x)\}$ and $\{u_k(x)\}$ are the negative and positive
eigenvalues of problem, respectively, with the Dirichlet condition on
$[-1,x]$ for $x>0$. Finding the solution in the infinite product
form led to construct the dual equations  which are necessary to
retrieve the potential function $q(t)$ in the inverse problem
\cite{jodayree2}.

Barcilon \cite{barcilon}  introduced
$\{\lambda_n(x)\}$ and $\{\mu_n(x)\}$ as the eigenvalues of
classical Sturm-Liouville equation (vibrating string equation),
\begin{equation} \label{string}
y''+\lambda\phi^2(t)y=0, \quad  x\leq t\leq L,
\end{equation}
with conditions $y(x)=y(L)=0$ and $y'(x)=y(L)=0$, respectively, in
where $x$ is a fixed point in the interval $(0,L)$.
In contrast with problem \eqref{orig}, the function $\phi^2(t)$
is positive throughout the interval $(0,L)$. It has been
shown that if $u(t,\lambda)$ is the solution of equation
\eqref{string} with the initial conditions $u(L,\lambda)=0$ and
$\frac{\partial u}{\partial t}(L,\lambda)=1$, then by using
Hadamard's factorization, for fixed $x$ belonging to $[0,L]$,
it can be written
\begin{gather*}
u(x,\lambda)=-(L-x)\prod_{k=1}^{\infty}(1-\frac{\lambda}{\lambda_n(x)}), \\
u'(x,\lambda)=\prod_{k=1}^{\infty}(1-\frac{\lambda}{\mu_n(x)}),
\end{gather*}
which are the infinite product form of the solution and its derivative for
vibrating string problem \cite{barcilon}. He also derived the dual equations of
problem \eqref{string} in the form
\begin{equation}\label{dual:string}
\begin{gathered}
\frac{d\lambda_n(x)}{dx}=-\frac{\lambda_n(x)}{L-x}
\frac{\prod_{k=1}^{\infty}(1-\frac{\lambda_n(x)}{\mu_k(x)})}
{\prod_{k\neq n}^{\infty}(1-\frac{\lambda_n(x)}{\lambda_k(x)})},  \\
\frac{d\mu_n(x)}{dx}=\mu_n^2(x)\phi^2(x)(L-x)
 \frac{\prod_{k=1}^{\infty}(1-\frac{\mu_n(x)}{\lambda_k(x)})}
{\prod_{k\neq n}^{\infty}(1-\frac{\mu_n(x)}{\mu_k(x)})},
\end{gathered}
\end{equation}
with the initial condition
\[
\lambda_n(0)=\lambda_n,\quad \mu_n(0)=\mu_n.
\]
In fact, the pair of sequences $(\lambda_n(0),\mu_n(0))$
suffices as data to guarantee the existence and uniqueness of
function $\phi^2(t)$ in \eqref{string} {\cite{barcilon}}.
Hence, by using the solution $(\lambda_n(x),\mu_n(x))$
of \eqref{dual:string}, one can construct the original equation
\eqref{string}. For this reason, the equation \eqref{dual:string}
is referred to as dual equation of \eqref{string} in the classical literature.


 Pranger \cite{pranger} studied the recovery of the function
$\phi^2(t)$ from the eigenvalues in equation \eqref{string} with the
Dirichlet boundary condition on the interval $[0,1]$,  replacing
$\{\mu_n\}$ by $\{\lambda'_n\}$ and introducing  the infinite product
form of the solution  to construct the
dual equation
\[
\lambda''_n+\frac{2}{x}\lambda_n+2\lambda_n\lambda'_n\sum_{j\neq
n}(\frac{\lambda'_j}{\lambda^2_j})(1-\frac{\lambda_n}{\lambda_j})^{-1}
-2\frac{(\lambda'_n)^2}{\lambda_n}=0,
\]
where $\{\lambda_n\}$ are eigenvalues of equation \eqref{string} on
the interval $[0,x]$, $0<x\leq 1$. It is well known that if there is
a $c>0$ so that $\phi^2(t)\geq c$ for all $t$ and
$\phi^2(t)\in C^2(0,L)$, then Equation \eqref{string} can be
transformed into the canonical Sturm-Liouville equation \cite{hille}
\[
y''+(\lambda-q)y=0.
\]

In section 2 we introduce some notation which we use throughout this article.
In section 3   we find the infinite product form for the  derivative of
the solution of the indefinite Sturm-Liouville equation \eqref{orig}
 before and after the turning point at the interval $(0,1)$.
 The main results of the paper are expressed by theorems \ref{qphi} and
\ref{theo}. The infinite product representation for the solution of
problem \eqref{orig} in \cite{kheiri} and its derivative given here,
enable us to construct the dual equations of this problem, in section 4,
which this system of equations identifies the two spectra of eigenvalues
 for an arbitrary fixed point in the whole interval.
Using these two spectra, one can retrieve the potential function
 $q(t)$  by the algorithm stated in the end of section 4.

\section{Preliminaries}

Let $\epsilon>0$ be fixed and sufficiently small, and let
$D_\epsilon =[0,t_0-\epsilon]\cup [t_0+\epsilon,1]$. Further, we set
$\mu=\frac{1}{2+l_0}$ ($l_0$ is order of turning point),
$\lambda=\rho^2$ ($\rho$ is a complex parameter) and $\theta=4\mu$.
 We also denote
\begin{gather*}
I_+ =\{t: \phi^{2}(t)>0\},\quad   I_- =\{t: \phi^{2}(t)<0\},\\
\xi(t)=\begin{cases}
0 &  \text{for }   t\in I_+(t), \\
1 &  \text{for }   t\in I_-(t),
\end{cases} \\
\phi^{2}_+(t)=\max (0,\phi^{2}(t)), \quad
\phi^{2}_-(t)=\max (0,-\phi^{2}(t)), \\
K_\pm (t)=\begin{cases}
1 &  \text{for }  t\in I_-(t), \\
\frac{1}{2}\csc(\frac{\pi\mu}{2})\exp(\mp i\frac{\pi}{4}) &
\text{for }   t\in I_+(t),
\end{cases}\\
K_\pm ^*(t)=\begin{cases}
\pm i & \text{for }   t\in I_-(t), \\
2\sin(\frac{\pi\mu}{2})\exp(\pm i\frac{\pi}{4}) & \text{for }
t\in I_+(t),
\end{cases}
\end{gather*}
Let
\[
S_k=\{\rho: \arg \rho \in [\frac{k\pi}{4},\frac{(k+1)\pi}{4}] \},
\quad  k=0,1.
\]
 Here the choice of the root $\phi$ of $\phi^2$ depends on the interval
and the sector under consideration and has to be determined carefully.
Due to the type of turning point $t_0$, we have
\[
\phi(t)=\begin{cases}
|\phi(t)| &  \text{for }   t>t_0,  \\
|\phi(t)| e^{ i\frac{\pi}{2}l_0} & \text{for }   t<t_0.
\end{cases}
\]
In {\cite{eberhard}} it is shown that for each fixed sector
 $S_k$ $(k=0,1)$, there exist  Fundamental Solutions (FS)
of \eqref{orig} $\{z_1(t,\rho), z_2(t,\rho)\}$, $t\in (0,1)$,
$\rho \in S_k $ such that the functions
$(t,\rho)\to z_s ^{(j)}(t,\rho)(s=1,2; j=0,1)$ are
continuous and holomorphic for $t\in (0,1)$, $\rho \in S_k$. Moreover,
for $|\rho|\to \infty$, $\rho \in S_k$, $t\in D_\epsilon$, $j=0,1$
\begin{gather}
\begin{aligned}
z_1^{(j)}(t,\rho)
&=(\pm i\rho)^j |\phi(t)|^{j-\frac{1}{2}}(e^{\mp i\frac{\pi}{2}\xi(t)})^j
 e^{\rho \int_0^t |\phi_-(\tau)|d\tau} \\
&\quad \times e^{\pm i\rho\int_0^t|\phi_+(\tau)|d\tau}K_\pm(t)\kappa(t,\rho),
\end{aligned} \label{E3}
\\
\begin{aligned}
z_2^{(j)}(t,\rho)
&=(\mp i\rho)^j |\phi(t)|^{j-\frac{1}{2}}(e^{\mp i\frac{\pi}{2}\xi(t)})^j
e^{-\rho \int_0^t |\phi_-(\tau)|d\tau}  \\
&\quad \times  e^{\mp i\rho\int_0^t|\phi_+(\tau)|d\tau}K_\pm^*(t)\kappa(t,\rho),
\end{aligned}\label{E4}
\\
\omega(\lambda)=\begin{vmatrix}
   z_1(t,\rho) & z_2(t,\rho) \\
    z_1'(t,\rho) & z_2'(t,\rho)
  \end{vmatrix}
=\mp(2i\rho)[1].\nonumber
\end{gather}
Here and in the following:
\begin{itemize}
  \item[(i)] The upper or lower signs in formulae correspond to the sectors
$S_0$, $S_1$ respectively.
  \item[(ii)] $[1]=1+O(\frac{1}{\rho^\theta})$ uniformly in $t\in D_\epsilon$.
  \item[(iii)] $\kappa(t,\rho)=O(1)$ as $|\rho|\to \infty$, $\rho \in S_k$.
\end{itemize}

\section{Infinite product representation}

Let $S(t,\lambda)$ be the solution of equation \eqref{orig} with
initial conditions
 \begin{equation} \label{sinitial}
  S(0,\lambda)=0, \quad S'(0,\lambda)=1.
\end{equation}
Using  $\{z_1(t,\rho), z_2(t,\rho)\}$, we can write
$$
S(t,\lambda)=c_1z_1(t,\rho)+c_2z_2(t,\rho).
$$
By imposing the initial conditions \eqref{sinitial} we have
\begin{gather*}
c_1z_1(0,\rho)+c_2z_2(0,\rho)=0, \\
c_1z'_1(0,\rho)+c_2z'_2(0,\rho)=1.
\end{gather*}
After getting $c_1$ and $c_2$ by using Cramer's rule we obtain
\begin{gather}
S(t,\lambda)=\frac{z_1(0,\rho)z_2(t,\rho)-z_2(0,\rho)
z_1(t,\rho)}{\omega(\lambda)},\label{E6}
\\
S'(t,\lambda)=\frac{z_1(0,\rho)z_2'(t,\rho)
-z_2(0,\rho)z_1'(t,\rho)}{\omega(\lambda)}.\label{E7}
\end{gather}
According to \cite{eberhard}, we can also write the fundamental solutions 
 $\{z_1(t,\rho),z_2(t,\rho)\}$, of \eqref{orig}, in the  asymptotic form
\begin{equation}
 z_1(t,\rho)=\begin{cases}
|\phi(t)|^{-1/2}e^{\rho \int_0^t|\phi(\tau)|d\tau}[1]
& 0\leq t<t_0, \\
\frac{1}{2}\csc(\frac{\pi\mu}{2})|\phi(t)|^{-1/2}e^{\rho
\int_0^{t_0}|\phi(\tau)|d\tau} \\
\times \big\{e^{i\rho \int_{t_0}^t|\phi(\tau)|d\tau-i\frac{\pi}{4}}[1]
+e^{-i\rho\int_{t_0}^t|\phi(\tau)|d\tau+i\frac{\pi}{4}}[1]\big\}
& t_0<t\leq 1,
\end{cases}\label{z1}
\end{equation}
\begin{equation}
z_2(t,\rho)=\begin{cases}
i|\phi(t)|^{-1/2}e^{-\rho \int_0^t|\phi(\tau)|d\tau}[1]
& 0\leq t<t_0, \\
2\sin(\frac{\pi\mu}{2})|\phi(t)|^{-1/2}e^{-\rho
\int_0^{t_0}|\phi(\tau)|d\tau}\\
\times e^{i\rho
\int_{t_0}^t|\phi(\tau)|d\tau-i\frac{\pi}{4}}[1] &
t_0<t\leq 1,
\end{cases}\label{z2}
\end{equation}
\begin{equation}
 z'_1(t,\rho)=\begin{cases}
\rho|\phi(t)|^{1/2}e^{\rho \int_0^t|\phi(\tau)|d\tau}[1]
& 0\leq t<t_0, \\
\frac{i\rho}{2}\csc(\frac{\pi\mu}{2})|\phi(t)|^{1/2}e^{\rho
\int_0^{t_0}|\phi(\tau)|d\tau}\\
\times \big\{e^{i\rho
\int_{t_0}^t|\phi(\tau)|d\tau-i\frac{\pi}{4}}[1]
-e^{-i\rho \int_{t_0}^t|\phi(\tau)|d\tau+i\frac{\pi}{4}}[1]\big\}
& t_0<t\leq 1,
\end{cases} \label{z3}
\end{equation}
\begin{equation}
 z'_2(t,\rho)=\begin{cases}
-i\rho|\phi(t)|^{1/2}e^{-\rho \int_0^t|\phi(\tau)|d\tau}[1]
& 0\leq t<t_0, \\
2i\rho\sin(\frac{\pi\mu}{2})|\phi(t)|^{1/2}\\
\times e^{-\rho\int_0^{t_0}|\phi(\tau)|d\tau}e^{i\rho
\int_{t_0}^t|\phi(\tau)|d\tau-i\frac{\pi}{4}}[1]
& t_0<t\leq 1.\label{z4}
\end{cases}
\end{equation}
Then, by \eqref{E6} and \eqref{E7} and asymptotic
forms of FS in \eqref{z1}-\eqref{z4}, we can write
\begin{equation} \label{asym1}
 S(t,\lambda)=\begin{cases}
\frac{|\phi(0)\phi(t)|^{-1/2}}{\rho}\sinh(\rho
\int_0^t|\phi(\tau)|d\tau)[1]&
0\leq t<t_0,  \\
\frac{|\phi(0)|^{-1/2}|\phi(t)|^{1/2}}{-2i\rho}\big\{D_1(\rho)e^{i\rho
\int_{t_0}^t|\phi(\tau)|d\tau}[1]\\
+D_2(\rho)e^{-i\rho
\int_{t_0}^t|\phi(\tau)|d\tau}[1] \big\}  &t_0<t\leq 1,
\end{cases} %\label{s1}
\end{equation}
\begin{equation} \label{asym2}
 S'(t,\lambda)=\begin{cases}
|\phi(0)|^{-1/2}|\phi(t)|^{1/2}
\cosh\left(\rho\int_0^{t} |\phi(\tau)|d\tau\right)[1] & 0\leq t<t_0,  \\
\frac{|\phi(0)|^{-1/2}|\phi(t)|^{1/2}}{-2}
\big\{D_1(\rho)e^{i\rho \int_{t_0}^t|\phi(\tau)|d\tau}[1] \\
-D_2(\rho)e^{-i\rho \int_{t_0}^t|\phi(\tau)|d\tau}[1] \big\}
 & t_0<t\leq 1,
\end{cases} %\label{s2}
\end{equation}
where
\begin{gather*}
D_1(\rho)=2\sin(\frac{\pi\mu}{2})e^{-\rho
\int_0^{t_0}|\phi(\tau)|d\tau-i\frac{\pi}{4}}
-\frac{1}{2}\csc(\frac{\pi\mu}{2})e^{\rho
\int_0^{t_0}|\phi(\tau)|d\tau-i\frac{\pi}{4}},
\\
D_2(\rho)=-\frac{1}{2}\csc(\frac{\pi\mu}{2})e^{\rho
\int_0^{t_0}|\phi(\tau)|d\tau+i\frac{\pi}{4}}.
\end{gather*}
The functions $S(x,\lambda)$ and $S'(x,\lambda)$ have zero sets for
each fixed point
 $x\in (0,1)$ referred to as $\lambda_n(x)$ and $\mu_n(x)$, respectively;
 i.e., $S(x,\lambda_n(x))=0$ and $S'(x,\mu_n(x))=0$.
These two zero sets correspond
 to the eigenvalues of problems $L_1(\phi^2(t),q(t),x)$ and
$L_2(\phi^2(t),q(t),x)$,  respectively.
 So, for fixed $x$, $x<t_0$, the asymptotic approximation of the
infinite sequence of negative eigenvalues for boundary-value
problem \eqref{orig} associated with boundary conditions
$y(0)=y(x)=0$ can be  obtained from \eqref{asym1}  of the form
\[
 \sqrt{-\lambda_n(x)}=\frac{n\pi}{p(x)}+O(\frac{1}{n}),\quad  \text{as }
n\to \infty,
\]
while for boundary value problem \eqref{orig} associated with boundary conditions
$y(0)=y'(x)=0$,   the asymptotic form of eigenvalues can be derived similarly
from \eqref{asym2}
\[
 \sqrt{-\mu_n(x)}=\frac{n\pi-\frac{\pi}{2}}{p(x)}+O(\frac{1}{n}),\quad
 \text{as } n\to \infty,
\]
where
\begin{equation}\label{p}
p(x)=\int_0^x|\phi(\tau)|d\tau.
\end{equation}
Note that in the case $x<t_0$, the boundary value problem has only
infinitely many negative eigenvalues according to classical results.
For fixed $x$:
$$
\dots<\mu_2(x)<\lambda_2(x)<\mu_1(x)<\lambda_1(x),\quad
\lim_{n\to\infty}\lambda_n(x)=\lim_{n\to\infty}\mu_n(x)=-\infty.
$$
By applying \eqref{E3} and \eqref{E4}, we infer that for
$\rho \in S_k$, $t\in D_\epsilon$, $j=0,1$,
\begin{equation}
\begin{aligned}
 S^{(j)}(t,\lambda)
&=\frac{1}{2}(\pm i\rho)^{j-1}|\phi(0)|^{-1/2}|\phi(t)|^{j-\frac{1}{2}}
(e^{\mp i\frac{\pi}{2}\xi(t)})^j e^{\pm i \frac{\pi}{2}} \\
&\quad \times e^{\rho\int_0^t|\phi_-(\tau)|d\tau} e^{\pm i\rho
\int_0^t|\phi_+(\tau)|d\tau}K_\pm (t)\kappa(t,\rho),
\end{aligned}\label{as}
\end{equation}
and
\begin{equation}
|S^{(j)}(t,\lambda)|\leq C|\rho|^{j-1}|e^{\rho
\int_0^t|\phi_-(\tau)|d\tau}e^{\pm i\rho
\int_0^t|\phi_+(\tau)|d\tau}|.\label{abs}
\end{equation}
It follows from \eqref{abs} that the functions $S^{(j)}(t,.)$ are
entire of order $1/2$. So, by Hadamard's theorem
$S(x,\lambda)$ and $S'(x,\lambda)$ can be represented in the infinite
product form
\begin{gather}\label{ns1}
S(x,\lambda)=C_{1,0}(x)\prod_{n=1}^\infty(1-\frac{\lambda}{\lambda_n(x)}), \\
\label{ns2}
S'(x,\lambda)=C_{2,0}(x)\prod_{n=1}^\infty(1-\frac{\lambda}{\mu_n(x)}),
\end{gather}
where $C_{r,0}$ (r=1,2) are functions of $x$ only. The index
$r$ in  $C_{r,0}$ is denoted to related problem $L_r(\phi^2(t),q(t),x)$
and the index  '0' in $C_{r,0}$ shows that the fixed point $x$
lies before turning point ($x<t_0$).
The function $C_{1,0}(x)$ has been calculated in \cite{kheiri}:
\begin{equation}\label{c10}
C_{1,0}(x)=(\phi(0)\phi(x))^{-1/2}p(x)\prod_n\frac{\lambda_n(x)}{w^2_n(x)},
\end{equation}
where $w_n(x)=n\pi/p(x)$,  and $p(x)$ is defined in \eqref{p}.

To estimate $C_{2,0}(x)$ we rewrite the infinite product as
\begin{equation}
S'(x,\lambda)
=C_{2,0}(x) \prod_{n=1}^\infty \frac{\mu_n(x)-\lambda}{\mu_n(x)}
= B_{2,0}(x)\prod_{n=1}^\infty \frac{\lambda-\mu_n(x)}{\tilde{w}_n^2(x)}
\label{prods}
\end{equation}
with
\begin{equation} \label{BC0}
B_{2,0}(x)=C_{2,0}(x) \prod_{n=1}^\infty
\frac{-\tilde{w}_n^2(x)}{\mu_n(x)},
\end{equation}
where
$$
\tilde{w}_n(x)=\frac{n\pi-\frac{\pi}{2}}{p(x)}.
$$
It follows from the asymptotic form of eigenvalues that
$\frac{-\tilde{w}_n^2(x)}{\mu_n(x)}=1+O(\frac{1}{n^2})$, then the
infinite product $\prod_{n=1}^\infty
\frac{-\tilde{w}_n^2(x)}{\mu_n(x)}$ is absolutely convergent on any
compact subinterval of $(0,t_0)$.

\begin{lemma}\label{lem1}
 Let $\tilde{w}_m(x)=\frac{m\pi-\frac{\pi}{2}}{p(x)}$ and $\mu_m(x), 1\leq m$ 
be a sequence of continuous functions such that for each $x$
 $$
\mu_m(x)=-\frac{m^2\pi^2}{p^2(x)}+\frac{m\pi^2}{p^2(x)}+O(1), \quad 0<x<t_0.
$$
Then, the infinite product
$$
\prod_{m=1}^{\infty}(\frac{\lambda-\mu_m(x)}{\tilde{w}_m^2(x)})
$$
is an entire function of $\lambda$ for fixed $x$ in $(0,t_0)$ whose
roots are precisely $\mu_m(x)$, $m\geq 1$. Moreover
$$
\prod_{m=1}^{\infty}(\frac{\lambda-\mu_m(x)}{\tilde{w}_m^2(x)})
=\cosh(\sqrt{\lambda}p(x))(1+O(\frac{\log n}{n})),
$$
 uniformly on the circles
$|\lambda|=\frac{n^2\pi^2}{p^2(x)}$, where $p(x)$ is defined in \eqref{p}.
\end{lemma}

\begin{proof}
 Since $ \mu_m(x)+\tilde{w}_m^2(x)=\frac{\pi^2}{4p^2(x)}+O(1), m\geq 1$ are uniformly bounded, then
 \[
 \sum_{m=1}^{\infty}\Big|\frac{\lambda-\mu_m(x)}{\tilde{w}_m^2(x)}-1\Big|
= \sum_{m=1}^{\infty}\Big|\frac{\lambda-\mu_m(x)
-\tilde{w}_m^2(x)}{\tilde{w}_m^2(x)}\Big|
= \sum_{m=1}^{\infty}\Big|\frac{\lambda+O(1)}{\tilde{w}_m^2(x)}\Big|
 \]
 converges uniformly on bounded subsets of complex plane. 
Therefore, the infinite product converges to an entire function of $\lambda$, 
whose zeroes are precisely $\tilde{w}_m(x), m\geq 1$ (see \cite{eves}).
 By \cite[4.5.69]{abram}, we have
$$
\cosh(p(x)\sqrt{\lambda})=\prod_{m=1}^{\infty}
\big[1+\frac{4p^2(x)\lambda}{(2m-1)^2\pi^2}\big].
$$
On the other hand, since 
$\frac{4p^2(x)}{(2m-1)^2\pi^2}=\frac{1}{\tilde{w}_m^2(x)}$, we obtain
$$
\frac{\prod_{m=1}^{\infty}\frac{\lambda-\mu_m(x)}{\tilde{w}_m^2(x)}}
{\prod_{m=1}^{\infty}[1+\frac{4p^2(x)\lambda}{(2m-1)^2\pi^2}]}=
\prod_{m=1}^{\infty}\frac{-\mu_m(x)+\lambda}{\tilde{w}_m^2(x)+\lambda}.
$$
Furthermore,
$$
\big| \frac{-\mu_m(x)+\lambda}{\tilde{w}_m^2(x)+\lambda}-1\big|
\leq\frac{|O(1)|}{\big||\lambda|-\frac{(2m-1)^2\pi^2}{4p^2(x)}\big|}.
$$
Therefore, on the circles $|\lambda|=\frac{n^2\pi^2}{p^2(x)}$, the
uniform estimates
$$
\frac{-\mu_m(x)+\lambda}{\tilde{w}_m^2(x)+\lambda}
=\begin{cases}
1+O(\frac{1}{n})& \text{if } n=m \\
1+O(\frac{1}{m^2-n^2})& \text{if } n\neq m
\end{cases}
$$
hold. by \cite[page 165]{trub}, we can write
$$
\prod_{1\leq m}\frac{-\mu_m(x)+\lambda}{\lambda+\tilde{w}_m^2(x)}
=1+O\big(\frac{\log n}{n}\big),
$$
uniformly on these circles. Then
$$
\prod_{1\leq m}\frac{\lambda-\mu_m(x)}{\tilde{w}_m^2(x)}=\cosh
(p(x)\sqrt{\lambda})\Big(1+O\big(\frac{\log n}{n}\big)\Big).
$$
\end{proof}

\begin{theorem}\label{teo1}
For $0\leq x<t_0$,
$$
S'(x,\lambda)=|\phi(0)|^{-1/2}|\phi(x)|^{1/2}
\prod_{n=1}^{\infty}-\frac{\mu_n(x)}{\tilde{w}^2_n(x)}
\prod_{n=1}^{\infty}(1-\frac{\lambda}{\mu_n(x)}),
$$
where $\tilde{w}_n(x)=\frac{n\pi-\frac{\pi}{2}}{p(x)}$, $p(x)$ is
defined in \eqref{p},  and $\{\mu_n(x)\}$ is
the sequence of eigenvalues for the Dirichlet-Neumann problem
associated with \eqref{orig} on $[0,x]$.
\end{theorem}

\begin{proof} 
For $0\leq x<t_0$, $\rho\in S_0$ and
$|\rho|\to\infty$, by virtue of \eqref{as} for $j=1$ we
calculate
\begin{equation}
S'(x,\lambda)=\frac{1}{2}|\phi(0)|^{-1/2}|\phi(x)|^{1/2}
e^{\rho\int_0^x|\phi(\tau)|d\tau}\kappa(x,\rho).\label{theos}
\end{equation}
Now from \eqref{prods}, \eqref{theos} and using lemma \ref{lem1}
uniformly on the circles $|\lambda|=\frac{n^2\pi^2}{p^2(x)}$, we
obtain
\[
B_{2,0}(x)=\frac{S'(x,\lambda)}{\prod_{m=1}^\infty
\frac{\lambda-\mu_m(x)}{\tilde{w}_m^2(x)}} =|\phi(0)|^{-1/2}|\phi(x)|^{1/2},
\]
as $|\rho|\to\infty$. So, by \eqref{BC0}, we obtain
\begin{equation} \label{c20}
C_{2,0}(x)=|\phi(0)|^{-1/2}|\phi(x)|^{1/2}
\prod_{n=1}^{\infty}-\frac{\mu_n(x)}{\tilde{w}^2_n(x)}.
\end{equation}
The proof is completed by \eqref{ns2}.
\end{proof}

For $x\in(t_0,1]$, fixed, both of  problems $L_1(\phi^2(t),q(t),x)$
and $L_2(\phi^2(t),q(t),x)$ have an infinite number of positive and
negative eigenvalues, which we denote by $\{\lambda_n^+(x)\}$,
$\{\lambda_n^-(x)\}$ and $\{\mu_n^+(x)\}$, $\{\mu_n^-(x)\}$ respectively.
 In reference to  {\cite{dast,jodayree3}}, we derive
\begin{gather*}
\sqrt{\lambda_n^+(x)}=\frac{n\pi-\frac{\pi}{4}}{f(x)}+O(\frac{1}{n}),\quad
\sqrt{-\lambda_n^-(x)}=\frac{n\pi-\frac{\pi}{4}}{p(t_0)}+O(\frac{1}{n}),
\\
\sqrt{\mu_n^+(x)}=\frac{n\pi-3\frac{\pi}{4}}{f(x)}+O(\frac{1}{n}),\quad 
\sqrt{-\mu_n^-(x)}=\frac{n\pi-\frac{\pi}{4}}{p(t_0)}+O(\frac{1}{n}),
\end{gather*}
where
\begin{equation}
f(x)=\int_{t_0}^x|\phi(\tau)|d\tau \label{f}
\end{equation}
and $p(x)$ is defined in \eqref{p}. By Hadamard's theorem, the
solution of equation \eqref{orig} and its derivative on $[0,x]$ for
$x>t_0$ is of the form
\begin{gather}
S(x,\lambda)=C_{1,1}(x)\prod_{n=1}^\infty(1-\frac{\lambda}{\lambda_n^-(x)})
\prod_{n=1}^\infty(1-\frac{\lambda}{\lambda_n^+(x)}),\label{prods01}
\\
S'(x,\lambda)=C_{2,1}(x)\prod_{n=1}^\infty(1-\frac{\lambda}{\mu_n^-(x)})
\prod_{n=1}^\infty(1-\frac{\lambda}{\mu_n^+(x)}).\label{prods2}
\end{gather}
Index '1' in $C_{r,1} (r=1,2)$ means that the fixed point $x$ lies
after turning point $(x>t_0)$.

The function $C_{1,1}(x)$ has been estimated in \cite{kheiri}:
\begin{equation} \label{c11}
\begin{aligned}
C_{1,1}(x)&=\frac{1}{16}\pi|\phi(0)\phi(x)|^{-1/2}\csc(\frac{\pi
\mu}{2})p(t_0)^{1/2}f(x)^{1/2} \\
&\quad \times\prod_{n=1}^{\infty}-\frac{\lambda_n^-(x)p^2(t_0)}{\tilde{j}_n^2}
\prod_{n=1}^{\infty}\frac{\lambda_n^+(x)f^2(x)}{\tilde{j}_n^2},
\end{aligned}
\end{equation}
where $p(x)$ and $f(x)$ are defined in \eqref{p} and \eqref{f}
respectively and $\tilde{j}_n(n=1,2,\dots)$ are the positive zeros
of the derivative of the Bessel function of first kind $(J_1'(z))$.

Let $J_\nu(z)$ and $J'_\nu(z)$ be the Bessel function of order
$\nu$ and its derivative, respectively. From \cite{abram} we have
\[
J_\nu(z)=\frac{(z/2)^\nu}{\Gamma(\nu+1)}\prod_{m=1}^\infty
(1-\frac{z^2}{j_{\nu,m}^2}),
\]
where
\begin{gather*}
j_{\nu,m}\sim
\beta-\frac{\alpha-1}{8\beta}-\frac{4(\alpha-1)(7\alpha-31)}{3(8\beta)^3}
-\dots, \\
\beta=(m+\frac{\nu}{2}-\frac{1}{4})\pi,\quad  \alpha=4\nu^2.
\end{gather*}
By inserting $\nu=0$, we can write
\[
J_0(z)=\prod_{m=1}^\infty(1-\frac{z^2}{j_{0,m}^2}),
\]
where
\[
j_{0,m}^2=m^2\pi^2-\frac{m\pi^2}{2}+O(1),\quad  m=1,2,\dots,
\]
are the positive zeros of $J_0(z)$. Also, from  \cite{abram}, We have
\[
J'_\nu(z)=\frac{(z/2)^{\nu-1}}{2\Gamma(\nu)}
\prod_{m=1}^\infty(1-\frac{z^2}{\tilde{j}_{\nu,m}^2}),\quad \nu>0,
\]
where
\begin{gather*}
\tilde{j}_{\nu,m}\sim\beta'-\frac{\alpha+3}{8\beta'}
-\frac{4(7\alpha^2+82\alpha-9)}{3(8\beta')^3}-\dots, \\
\beta'=(m+\frac{\nu}{2}-\frac{3}{4})\pi,\quad \alpha=4\nu^2.
\end{gather*}
In reference to  \cite{abram}, as a result of   $J'_0(z)=-J_1(z)$ 
the  zeros of $J_1(z)$ and $J'_0(z)$ are the same, namely
$\tilde{j}_{0,m}=j_{1,m}$ for $m=1,2,\dots$. Therefore, we can write
\[
J'_0(z)=-J_1(z)=-\frac{z}{2}\prod_{m=1}^\infty(1-\frac{z^2}{j_{1,m}^2}),
\]
where
\[
j_{1,m}=(m+\frac{1}{4})\pi+\dots,\quad  m=1,2,\dots.
\]
Replacing $m$ by $m-1$ in the previous relation we obtain
\begin{gather*}
j_{1,m-1}=(m-\frac{3}{4})\pi+\dots,\quad  m=2,3,\dots, \\
j_{1,m-1}^2=m^2\pi^2-\frac{3}{2}m\pi^2+O(1),\quad  m=2,3,\dots.
\end{gather*}
Consequently,
\[
\frac{-j_{0,n}^2}{p^2(t_0)\mu_n^-(x)}=1+O(\frac{1}{n^2}), \quad
\frac{j_{1,n-1}^2}{f^2(x)\mu_n^+(x)}=1+O(\frac{1}{n^2}).
\]
Therefore, the infinite products
$\prod_{n=1}^\infty\frac{-j_{0,n}^2}{p^2(t_0)\mu_n^-(x)}$ and
$\prod_{n=2}^\infty\frac{j_{1,n-1}^2}{f^2(x)\mu_n^+(x)}$ are
absolutely convergent for each $x>t_0$. Then,  from
\eqref{prods2}, we may write
\begin{equation}
  S'(x,\lambda)=B_{2,1}(x)(1-\frac{\lambda}{\mu_1^+})\prod_{n=1}^\infty
\frac{(\lambda-\mu_n^-(x))p^2(t_0)}{j_{0,n}^2}
\prod_{n=2}^\infty\frac{(\mu_n^+(x)-\lambda)f^2(x)}{j_{1,n-1}^2},\label{prods1}
\end{equation}
where
\begin{equation} \label{bc1}
B_{2,1}(x)=C_{2,1}(x)\prod_{n=1}^\infty
\frac{-j_{0,n}^2}{p^2(t_0)\mu_n^-(x)}\prod_{n=2}^\infty
\frac{j_{1,n-1}^2}{f^2(x)\mu_n^+(x)}.
\end{equation}

\begin{lemma}\label{lem2}
Let $j_{0,m}$ be the positive zeros of $J_0(z)$  and for fixed $x$
in $(t_0,1)$
\[
\mu_m^-(x)=-\frac{m^2\pi^2}{p^2(t_0)}+\frac{3}{2}\frac{m\pi^2}{p^2(t_0)}+O(1),
m\geq1,
\]
be a negative sequence of continuous functions. The infinite product
\[
\prod_{m=1}^\infty\frac{(\lambda-\mu_m^-(x))p^2(t_0)}{j_{0,m}^2}
\]
is an entire function of $\lambda$ for fixed $x$, whose roots are
precisely $\mu_m^-(x)$, $m\geq1$. Moreover,
\[
\prod_{m=1}^\infty\frac{(\lambda-\mu_m^-(x))p^2(t_0)}{j_{0,m}^2}
=J_0(i\sqrt{\lambda}p(t_0))(1+O(\frac{\log n}{n})),
\]
uniformly on the circles $|\lambda|=\frac{n^2\pi^2}{p^2(t_0)}$.
\end{lemma}

\begin{proof} 
This follows from using the method of the proof of
lemma \ref{lem1}. For more details, see \cite{jodayree1}.
\end{proof}

\begin{lemma}\label{lem3}
Let $j_{1,m}$ be the positive zeros of $J_1(z)$ and for fixed $x$ in
$(t_0,1)$
\[
\mu_m^+(x)=\frac{m^2\pi^2}{f^2(x)}-\frac{m\pi^2}{2f^2(x)}+O(1),\quad m\geq1,
\]
be a positive sequence of continuous functions. Then, the infinite
product
\[
\prod_{m=2}^\infty\frac{(\mu_m^+(x)-\lambda)f^2(x)}{j_{1,m-1}^2}
\]
is an entire function of $\lambda$ for fixed $x$, whose roots are
precisely $\mu_m^+(x)$, $m\geq1$. Moreover,
\[
\prod_{m=2}^\infty\frac{(\mu_m^+(x)-\lambda)f^2(x)}{j_{1,m-1}^2}
=-\frac{2}{\sqrt{\lambda}f(x)}J'_0(\sqrt{\lambda}f(x))
(1+O(\frac{\log n}{n})),
\]
uniformly on the circles $|\lambda|=\frac{n^2\pi^2}{p^2(t_0)}$.
\end{lemma}

\begin{proof}
 This follows from using the method of the proof of
lemma $\ref{lem1}$. For more details, see \cite{jodayree1}.
\end{proof}

\begin{theorem}\label{teo2}
Let $S'(t,\lambda)$ be the derivative of the solution of problem \eqref{orig}
in association with initial condition \eqref{sinitial}. Then, for each fixed
$x>t_0$,
\begin{align*}
S'(x,\lambda)
&=-\frac{1}{4}|\phi(0)|^{-1/2}|\phi(x)|^{1/2}i^{1/2}\pi
\mu_1^+(x) e^{i\frac{\pi}{4}}\csc(\frac{\pi\mu}{2})f^{3/2}(x)p^{1/2}(t_0)\\
&\quad \times\prod_{n=1}^\infty
-\frac{\mu_n^-(x)p^2(t_0)}{j^2_{0,n}}\prod_{n=2}^\infty
\frac{\mu^+_n(x)f^2(x)}{j^2_{1,n-1}}
\prod_{n=1}^\infty (1-\frac{\lambda}{\mu_n^-(x)})\prod_{n=2}^\infty
(1-\frac{\lambda}{\mu_n^+(x)}),
\end{align*}
where $p(x)$ and $f(x)$ is defined in \eqref{p} and \eqref{f}.
Sequences $\{\mu_n^+(x)\}$ and $\{\mu_n^-(x)\}$ represent the
positive and negative eigenvalues of
$L_2(\phi^2(t),q(t),x)$, respectively and $j_{\nu,n} (\nu=0,1)$ are
the positive zeros of $J_{\nu}(z)$.
\end{theorem}

\begin{proof}
Using \eqref{asym2} for $t_0<x<1$ it is obtained that
\begin{align*}
S'(x,\lambda)&=\frac{|\phi(0)|^{1/2}|\phi(x)|^{1/2}}{-2}
\big\{(2\sin(\frac{\pi\mu}{2})
e^{-\sqrt{\lambda}p(t_0) -i\frac{\pi}{4}}\\
&\quad -\frac{1}{2}\csc(\frac{\pi\mu}{2})e^{\sqrt{\lambda}p(t_0)
-i\frac{\pi}{4}}) e^{i\sqrt{\lambda}f(x)}[1] \\
&\quad + \frac{1}{2}\csc(\frac{\pi\mu}{2})e^{\sqrt{\lambda}p(t_0)
+i\frac{\pi}{4}}e^{-i\sqrt{\lambda}f(x)}[1]\}.
\end{align*}
As $|\lambda|\to\infty$, the first expression in the accolade
tends to zero, resulting in
\begin{equation}
S'(x,\lambda)=\frac{|\phi(0)|^{-1/2}|\phi(x)|^{1/2}i}{2}
\csc(\frac{\pi\mu}{2})e^{\sqrt{\lambda}p(t_0)}
\sin(\sqrt{\lambda}f(x)-\frac{\pi}{4})[1].\label{E22}
\end{equation}
On the other hand, by use of \eqref{prods1} and lemma \ref{lem2},
lemma \ref{lem3}, on the circles
$|\lambda|=\min\{\frac{n^2\pi^2}{p^2(t_0)},\frac{n^2\pi^2}{f^2(x)}\}$,
we obtain
\[
S'(x,\lambda)=-\frac{2}{\sqrt{\lambda}f(x)}B_{2,1}(x)
(1-\frac{\lambda}{\mu_1^+})J_0(i\sqrt{\lambda}p(t_0))J'_0(\sqrt{\lambda}f(x))
1+O(\frac{\log n}{n})).
\]
Using the asymptotic form of the Bessel function and its derivative
in the  previous relation, we have
\begin{align*}
 S'(x,\lambda)&=\frac{2}{\sqrt{\lambda}f(x)}B_{2,1}(x)(1-\frac{\lambda}{\mu_1^+})(\frac{2}{i\sqrt{\lambda}p(t_0)\pi})^{1/2}
\cos(i\sqrt{\lambda}p(t_0)-\frac{\pi}{4})\\
&\quad \times(\frac{2}{\sqrt{\lambda}f(x)\pi})^{1/2}
\sin(\sqrt{\lambda}f(x)-\frac{\pi}{4})[1](1+O(\frac{\log n}{n})).
\end{align*}
After some  calculations, we  obtain
\begin{equation}
\begin{aligned}
S'(x,\lambda)
&=\frac{2}{\lambda\pi
i^{1/2}f^{3/2}(x)p^{1/2}(t_0)}B_{2,1}(x)(1-\frac{\lambda}{\mu_1^+})
(e^{\sqrt{\lambda}p(t_0)+i\frac{\pi}{4}}+
e^{-\sqrt{\lambda}p(t_0)-i\frac{\pi}{4}})\\
&\quad \times\sin(\sqrt{\lambda}f(x)-\frac{\pi}{4}) [1](1+O(\frac{\log
n}{n})).
\end{aligned}\label{E23}
\end{equation}
We know that in the above relation, the  expression
$\exp\big(-\sqrt{\lambda}p(t_0)-i\frac{\pi}{4}\big)$  vanishes as
$|\lambda|\to\infty$. Comparing \eqref{E22} and
\eqref{E23} and  considering $|\lambda|\to\infty$ we obtain
\[
B_{2,1}(x)=-\frac{1}{4}|\phi(0)|^{-1/2}|\phi(x)|^{1/2}i^{1/2}\pi
\mu_1^+ e^{i\frac{\pi}{4}}\csc(\frac{\pi\mu}{2})f^{3/2}(x)p^{1/2}(t_0).
\]
So, by \eqref{bc1}, we obtain
\begin{equation} \label{c21}
\begin{aligned}
C_{2,1}(x)
&=-\frac{1}{4}|\phi(0)|^{-1/2}|\phi(x)|^{1/2}i^{1/2}\pi
\mu_1^+
e^{i\frac{\pi}{4}}\csc(\frac{\pi\mu}{2})\\
&\quad\times f^{3/2}(x)p^{1/2}(t_0) \prod_{n=1}^\infty
-\frac{\mu_n^-p^2(t_0)}{j^2_{0,n}}\prod_{n=2}^\infty
\frac{\mu^+_nf^2(x)}{j^2_{1,n-1}},
\end{aligned}
\end{equation}
 which  depends only on $x$. By\eqref{prods2}
 the proof is complete.
\end{proof}

\section{Dual equations}

In this section, we derive the dual equations associated with
problem \eqref{orig} by using the infinite product representation.
First we prove some lemmas which are necessary to present the
main theorem.

\begin{lemma}\label{cs}
$C_{2,j}(x)=C'_{1,j}(x)(j=0,1)$, where $C_{1,0}(x)$, $C_{2,0}(x)$,
$C_{1,1}(x)$ and $C_{2,1}(x)$ are defined in \eqref{c10},
\eqref{c20}, \eqref{c11} and \eqref{c21}, respectively.
\end{lemma}

\begin{proof}  
If one inserts $\lambda=0$ in \eqref{ns1}, \eqref{ns2}, \eqref{prods01}
 and \eqref{prods2}, the proof becomes trivial.
\end{proof}

\begin{theorem}\label{qphi}
The functions $q(x)$ and $\phi^2(x)$ in problem \eqref{orig}
 satisfy the following relations:
\begin{gather}
 q(x)=\begin{cases}
\frac{C'_{2,0}(x)}{C_{1,0}(x)}=\frac{C''_{1,0}(x)}{C_{1,0}(x)}&
0\leq x<t_0,\\ 
\frac{C'_{2,1}(x)}{C_{1,1}(x)}=\frac{C''_{1,1}(x)}{C_{1,1}(x)}
& t_0<x\leq 1,
\end{cases} \label{q}
\\
\phi^2(x)=\begin{cases}
\frac{C'_{2,0}(x)}{C_{1,0}(x)}\sum_i(\frac{1}{\mu_i(x)}
-\frac{1}{\lambda_i(x)})-\frac{C_{2,0}(x)}{C_{1,0}(x)}
\sum_i\frac{\mu'_i(x)}{\mu^2_i(x)}&
0\leq x<t_0,  \\ 
 \frac{C'_{2,1}(x)}{C_{1,1}(x)}\sum_i(\frac{1}{\mu^-_i(x)}
+\frac{1}{\mu^+_i(x)}-\frac{1}{\lambda^-_i(x)}-\frac{1}{\lambda^+_i(x)})\\
-\frac{C_{2,1}(x)}{C_{1,1}(x)}
\sum_i(\frac{\mu^{-'}_i(x)}{{\mu_i^-}^2(x)}
+\frac{\mu^{+'}_i(x)}{{\mu_i^+}^2(x)})
& t_0<x\leq 1,
\end{cases} \label{phi}
\end{gather}
where $C_{1,0}(x)$, $C_{2,0}(x)$, $C_{1,1}(x)$ and $C_{2,1}(x)$ are defined in
\eqref{c10}, \eqref{c20}, \eqref{c11} and \eqref{c21}.
\end{theorem}

\begin{proof} 
We prove it for the case $0\leq x<t_0$. There is a similar
proof for $t_0<x\leq 1$. We know that $S(x,\lambda)$ satisfies the
original problem \eqref{orig}, so
\[ % \label{satisfy}
\frac{\partial}{\partial
x}S'(x,\lambda)+(\lambda\phi^2(x)-q(x))S(x,\lambda)=0.
\]
Using \eqref{ns1} and \eqref{ns2} in the previous relation we obtain
\begin{equation} \label{original}
\begin{aligned}
&C'_{2,0}(x)\prod_{k=1}(1-\frac{\lambda}{\mu_k(x)})
+C_{2,0}(x)\sum_i\frac{\mu'_i(x)}{\mu^2_i(x)}\cdot\lambda
\prod_{k\neq i,1\leq k}(1-\frac{\lambda}{\mu_k(x)})\\
&+[\lambda\phi^2(x)-q(x)]C_{1,0}(x)\prod_{k=1}(1-\frac{\lambda}{\lambda_k(x)})=0.
\end{aligned}
\end{equation}
By putting the coefficients of different powers of $\lambda$ equal
to zero in the previous relation, we obtain

Coefficient of $\lambda^0$:
$C'_{2,0}(x)-q(x).C_{1,0}(x)=0$ which implies
\[
q(x)=\frac{C'_{2,0}(x)}{C_{1,0}(x)},
\]
and from Lemma \ref{cs},
\[
q(x)=\frac{C''_{1,0}(x)}{C_{1,0}(x)}.
\]
From the coefficient of $\lambda^1$,
\[
-C'_{2,0}(x)\sum_i\frac{1}{\mu_i(x)}+C_{2,0}(x)\sum_i
\frac{\mu'_i(x)}{\mu^2_i(x)}+\phi^2(x)C_{1,0}(x)+q(x)C_{1,0}(x)
\sum_i\frac{1}{\lambda_i(x)}=0,
\]
 we obtain
\[
\phi^2(x)=\frac{C'_{2,0}(x)}{C_{1,0}(x)}\sum_i(\frac{1}{\mu_i(x)}-\frac{1}{\lambda_i(x)})-\frac{C_{2,0}(x)}{C_{1,0}(x)}
\sum_i\frac{\mu'_i(x)}{\mu^2_i(x)}.
\]
The proof is complete.
\end{proof}


\begin{theorem}\label{theo}
\quad (a) For fixed value  $x$ in $[0,t_0)$, the sequences
$\{\lambda_n(x)\}_{n=1}^\infty$ and
 $\{\mu_n(x)\}_{n=1}^\infty$ which are the negative eigenvalues of
 problems $L_1(\phi^2(t),q(t),x)$ and $L_2(\phi^2(t),q(t),x)$,
 respectively, satisfy the system of equations:
\begin{equation}\label{dual0}
\begin{gathered}
\frac{d\lambda_n(x)}{dx}=\frac{C_{2,0}(x)}{C_{1,0}(x)}\lambda_n(x)
\frac{\prod_{k=1}^\infty(1-\frac{\lambda_n(x)}{\mu_k(x)})}
{\prod_{k\neq n}^\infty(1-\frac{\lambda_n(x)}{\lambda_k(x)})},  \\ 
\frac{d\mu_n(x)}{dx}=-\frac{C_{1,0}(x)}{C_{2,0}(x)}[\mu^2_n(x)\phi^2(x)
-\mu_n(x)\frac{C'_{2,0}(x)}{C_{1,0}(x)}]
\frac{\prod_{k=1}^\infty(1-\frac{\mu_n(x)}{\lambda_k(x)})}
{\prod_{k\neq n}^\infty(1-\frac{\mu_n(x)}{\mu_k(x)})}.
\end{gathered}
\end{equation}

(b) For  fixed value  $x$ in $(t_0,1]$, the sequences
$\{\lambda^-_n(x)\}_{n=1}^\infty$, $\{\lambda^+_n(x)\}_{n=1}^\infty$
and $\{\mu^-_n(x)\}_{n=1}^\infty$, $\{\mu^+_n(x)\}_{n=1}^\infty$
 which are the negative, positive eigenvalues of problems
 $L_1(\phi^2(t),q(t),x)$ and $L_2(\phi^2(t),q(t),x)$,
 respectively, satisfy the system of equations:
\begin{equation} \label{dual1}
\begin{gathered}
\frac{d\lambda^-_n(x)}{dx}=\frac{C_{2,1}(x)}{C_{1,1}(x)}
\lambda^-_n(x)\frac{\prod_{k=1}^\infty(1-\frac{\lambda^-_n(x)}{\mu^-_k(x)})
\prod_{k=1}^\infty(1-\frac{\lambda^-_n(x)}{\mu^+_k(x)})}
{\prod_{k\neq n}^\infty(1-\frac{\lambda^-_n(x)}{\lambda^-_k(x)})
\prod_{k=1}^\infty(1-\frac{\lambda^-_n(x)}{\lambda^+_k(x)})},
\\
\frac{d\lambda^+_n(x)}{dx}=\frac{C_{2,1}(x)}{C_{1,1}(x)}\lambda^+_n(x)
\frac{\prod_{k=1}^\infty(1-\frac{\lambda^+_n(x)}{\mu^-_k(x)})
\prod_{k=1}^\infty(1-\frac{\lambda^+_n(x)}{\mu^+_k(x)})}
{\prod_{k=1}^\infty(1-\frac{\lambda^+_n(x)}{\lambda^-_k(x)})
\prod_{k\neq n}^\infty(1-\frac{\lambda^+_n(x)}{\lambda^+_k(x)})},
 \\
\begin{aligned}
\frac{d\mu^-_n(x)}{dx}
&=-\frac{C_{1,1}(x)}{C_{2,1}(x)}[{\mu^-_n}^2(x)\phi^2(x)
 -\mu^-_n(x)\frac{C'_{2,1}}{C_{1,1}}]\\
&\quad\times \frac{\prod_{k=1}^\infty(1-\frac{\mu^-_n(x)}{\lambda^-_k(x)})
\prod_{k=1}^\infty(1-\frac{\mu^-_n(x)}{\lambda^+_k(x)})}
{\prod_{k\neq n}^\infty(1-\frac{\mu^-_n(x)}{\mu^-_k(x)})
\prod_{k=1}^\infty(1-\frac{\mu^-_n(x)}{\mu^+_k(x)})},
\end{aligned}
  \\
\begin{aligned}
\frac{d\mu^+_n(x)}{dx}
&=-\frac{C_{1,1}(x)}{C_{2,1}(x)}[{\mu^+_n}^2(x)\phi^2(x)
-\mu^+_n(x)\frac{C'_{2,1}(x)}{C_{1,1}(x)}]\\
&\quad\times \frac{\prod_{k=1}^\infty(1-\frac{\mu^+_n(x)}{\lambda^-_k(x)})
\prod_{k=1}^\infty(1-\frac{\mu^+_n(x)}{\lambda^+_k(x)})}
{\prod_{k=1}^\infty(1-\frac{\mu^+_n(x)}{\mu^-_k(x)}) \prod_{k\neq
n}^\infty(1-\frac{\mu^+_n(x)}{\mu^+_k(x)})}.
\end{aligned}
\end{gathered}
\end{equation}
\end{theorem}

\begin{proof} 
We prove it for the case $0\leq x<t_0$. There is a  similar
proof for $t_0<x\leq 1$. Since $\{\lambda_n(x)\}$ is the eigenvalues
of problem \eqref{orig} associated with the boundary condition
\eqref{con1}, we have
\[
S(x,\lambda_n(x))=0.
\]
By differentiating,
\[
\frac{\partial}{\partial x}S(x,\lambda)+\frac{\partial}{\partial
\lambda}S(x,\lambda)\lambda'_n(x)=0.
\]
Therefore, at the points $(x,\lambda_n(x))$, we obtain
\begin{equation} \label{lam}
\lambda'_n(x)=-\frac{\frac{\partial}{\partial
x}S(x,\lambda_n(x))}{\frac{\partial}{\partial
\lambda}S(x,\lambda_n(x))}
=-\frac{S'(x,\lambda_n(x)}{\frac{\partial}{\partial
\lambda}S(x,\lambda_n(x))}.
\end{equation}
We calculate $\frac{\partial S}{\partial \lambda}$ at the points
$(x,\lambda_n(x))$. Using \eqref{ns1}, we reach
\[
\frac{\partial S}{\partial
\lambda}=C_{1,0}(x)\sum_{i=1}^\infty\frac{-1}{\lambda_i(x)}\prod_{k\neq
i,1\leq k}(1-\frac{\lambda}{\lambda_k(x)}).
\]
So, we have
\begin{equation}\label{slam}
\frac{\partial}{\partial
\lambda}S(x,\lambda_n(x))=-\frac{C_{1,0}(x)}{\lambda_n(x)}\prod_{k\neq
n,1\leq k}(1-\frac{\lambda_n(x)}{\lambda_k(x)}).
\end{equation}
Therefore, substituting \eqref{slam} and \eqref{ns2} in
\eqref{lam} we obtain
\[
\lambda'_n(x)=\frac{C_{2,0}(x)}{C_{1,0}(x)}\lambda_n(x)\frac{\prod_{k=1}^\infty(1-\frac{\lambda_n(x)}{\mu_k(x)})}
{\prod_{k\neq n}^\infty(1-\frac{\lambda_n(x)}{\lambda_k(x)})}.
\]
On the other hand,  replacing $\lambda$ by $\mu_n(x)$ in
\eqref{original}, the first statement in the relation vanishes and we
have
\begin{align*}
&C_{2,0}(x)\frac{\mu'_n(x)}{\mu^2_n(x)}\cdot\mu_n(x)\prod_{k\neq n,1\leq
k}(1-\frac{\mu_n(x)}{\mu_k(x)})\\
&+[\mu_n(x)\phi^2(x)-q(x)]C_{1,0}(x)\prod_{k\geq
1}(1-\frac{\mu_n(x)}{\lambda_k(x)})=0,
\end{align*}
so, we obtain
\[
\mu'_n(x)=-\frac{C_{1,0}(x)}{C_{2,0}(x)}[\mu^2_n(x)\phi^2(x)-\mu_n(x)q(x)]
\frac{\prod_{k=1}^\infty(1-\frac{\mu_n(x)}{\lambda_k(x)})}
{\prod_{k\neq n}^\infty(1-\frac{\mu_n(x)}{\mu_k(x)})}.
\]
By inserting $q(x)=\frac{C'_{2,0}(x)}{C_{1,0}(x)}$ from theorem \ref{qphi}, 
the proof is complete.
\end{proof}

 Theorem \ref{theo} which is the main result of this article
gives us an algorithm for the solution of the inverse problem, i.e.,
retrieving  $q(x)$ in $(0,1)$.

\subsection*{Algorithm}
Suppose that $\phi^2(t)=(t-t_0)^{l_0}\phi_0(t)$ is given where 
$l_0$ is odd and $\phi^2(t)(t-t_0)^{-l_0}>0$ in $[0,t_0)\cup(t_0,1]$;
 i.e., $t_0$ is a turning point of type IV and the sequences 
$\{\lambda_n^-\}$, $\{\lambda_n^+\}$, $\{\mu_n^-\}$ and $\{\mu_n^+\}$ 
satisfy the following relations:
\begin{gather*}
\sqrt{\lambda_n^+}=\frac{n\pi-\frac{\pi}{4}}{\int_{t_0}^1|\phi(\tau)|d\tau}
+O(\frac{1}{n}),\quad 
\sqrt{-\lambda_n^-}=\frac{n\pi-\frac{\pi}{4}}{\int^{t_0}_0|\phi(\tau)|d\tau}
+O(\frac{1}{n}),
\\
\sqrt{\mu_n^+}=\frac{n\pi-3\frac{\pi}{4}}{\int_{t_0}^1|\phi(\tau)|d\tau}
+O(\frac{1}{n}),\quad
\sqrt{-\mu_n^-}=\frac{n\pi-\frac{\pi}{4}}{\int^{t_0}_0|\phi(\tau)|d\tau)}
+O(\frac{1}{n}).
\end{gather*}
\begin{enumerate}
\item By solving the dual equation \eqref{dual1} with initial conditions
\begin{equation} \label{initial1}
\lambda^-_n(1)=\lambda^-_n,\quad
\lambda^+_n(1)=\lambda^+_n,\quad
\mu^-_n(1)=\mu^-_n,\quad
\mu^+_n(1)=\mu^+_n,
\end{equation}
we find $\lambda_n^-(x)$, $\lambda_n^+(x)$, $\mu_n^-(x)$ and
$\mu_n^+(x)$ for $x\in (t_0,1)$.

\item Calculate $q(x)=\frac{C'_{2,1}(x)}{C_{1,1}(x)}$ where 
$C_{1,1}(x)$ and $C_{2,1}(x)$ are defined in \eqref{c11} and \eqref{c21}, 
respectively.

\item By solving the dual equation \eqref{dual0} with initial conditions
\begin{equation} \label{initial2}
\lambda_n(t_0)=\lim_{x\to t^+_0}\lambda_n^-(x),\quad
\mu_n(t_0)=\lim_{x\to t^+_0}\mu_n^-(x),
\end{equation}
we  find $\lambda_n(x)$ and $\mu_n(x)$ for $x\in (0,t_0)$.

\item Calculate $q(x)=\frac{C'_{2,0}(x)}{C_{1,0}(x)}$ where 
$C_{1,0}(x)$ and $C_{2,0}(x)$ are defined in \eqref{c10} and \eqref{c20}, 
respectively.
\end{enumerate}

\begin{remark}\label{rmk1} \rm
 It is obvious that the  system of equations
\eqref{dual0} are dual equations for indefinite
Sturm-Liouville equation \eqref{orig}, corresponds to the system of
equations \eqref{dual:string} in the classic Sturm-Liouville case
(vibrating string). It means that the classical result is a particular 
case of our result; i.e., by inserting $q(x)\equiv 0$, $C_{1,0}(x)=-(L-x)$ 
and $C_{2,0}(x)=1$ in \eqref{dual0}, one can obtain \eqref{dual:string}.
 We can use the method stated in \cite{barcilon} to show that the 
systems of equations \eqref{dual0} and \eqref{dual1} with initial conditions 
\eqref{initial2} and \eqref{initial1}, respectively, satisfy the 
Lipschitz condition  which guarantees the existence of a unique 
solution to the initial value problem.
\end{remark}

\begin{proposition} \label{prop1}
Putting \eqref{q} in \eqref{phi} for $0<x<t_0$,
we obtain
\[
\phi^2(x)=q(x)\sum_i(\frac{1}{\mu_i(x)}-\frac{1}{\lambda_i(x)})
-\frac{C_{2,0}(x)}{C_{1,0}(x)}
\sum_i\frac{\mu'_i(x)}{\mu^2_i(x)},
\]
which shows the relationship between weight function $\phi^2(x)$
and potential function $q(x)$ by means of eigenvalues
$\{\lambda_n(x)\}$ and $\{\mu_n(x)\}$. The same relation can be
written for $t_0<x<1$.
\end{proposition}

\begin{proposition} \label{prop2}
 By differentiating relation \eqref{ns1} with respect to $x$, $0<x<t_0$, 
and then replacing $\lambda$ by $\mu_n(x)$
for each $n\in \mathbf{N}$, we can write
\[
S'(x,\mu_n(x))=C'_{1,0}(x)\prod_{k=1}^\infty(1-\frac{\mu_n(x)}{\lambda_k(x)})
+\mu_n(x)\cdot C_{1,0}(x)\sum_i \frac{\lambda'_i(x)}{\lambda^2_i(x)}\prod_{k\neq
i,1\leq k}(1-\frac{\mu_n(x)}{\lambda_k(x)}).
\]
On the other hand, $S'(x,\mu_n(x))=0$, so
\[
\prod_k(1-\frac{\mu_n(x)}{\lambda_k(x)})\{C'_{1,0}(x)+\mu_n(x)\cdot
C_{1,0}(x)\sum_i\frac{\lambda'_i(x)}{\lambda_i(x)}
\cdot\frac{1}{\lambda_i(x)-\mu_n(x)}\}=0.
\]
From lemma \ref{cs} this implies
\[
C_{2,0}(x)+\mu_n(x)\cdot
C_{1,0}(x)\sum_i\frac{\lambda'_i(x)}{\lambda_i(x)}
\frac{1}{\lambda_i(x)-\mu_n(x)}=0,\quad \forall n\in
\mathbf{N},
\]
which represents the relationship between eigenvalues and
coefficients $C_{1,0}(x)$ and $C_{2,0}(x)$. The same relation can be
written for $t_0<x<1$.
\end{proposition}


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\end{document}