\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 26, pp. 1--7.\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-2017/26\hfil Inverse problems for Sturm-Liouville operators]
{Inverse problems for Sturm-Liouville operators with boundary
conditions depending on a spectral parameter}

\author[M. Sat \hfil EJDE-2017/26\hfilneg]
{Murat Sat}

\address{Murat Sat \newline
Department of Mathematics,
Faculty of Science and Art,
Erzincan University,
Erzincan, 24100, Turkey}
\email{murat\_sat24@hotmail.com} 

\dedicatory{Communicated by Ira Herbst}

\thanks{Submitted May 5, 2016. Published January 24, 2017.}
\subjclass[2010]{34B05, 34L20, 47E05}
\keywords{Inverse problem; uniqueness theorem; eigenvalue;
 spectral parameter}

\begin{abstract}
 In this article, we study the inverse problem for Sturm-Liouville
 operators with boundary conditions dependent on the spectral
 parameter. We show that the potential $q(x)$ and coefficient
 $\frac{a_1\lambda +b_1}{c_1\lambda +d_1}$ functions can be uniquely
 determined from the particular set of eigenvalues.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

The theory of inverse problem for differential operators takes an important
position in the trend development of the spectral theory of linear
operators. Inverse problems of spectral analysis consist in recovering
operators from their spectral characteristics 
\cite{a1,b5,g2,h1,l1,p1}.
Such problems often come along in mathematical physics, mechanics,
electronics, geophysics and other branches of natural sciences. The inverse
problem  of a regular Sturm-Liouville operator was studied firstly by
Ambarzumyan in 1929 \cite{a2} and secondly by Borg in 1945 \cite{b5}.
From then on, Borg's result has been extended to various versions.

McLaughlin and Rundell in 1986 \cite{m1}, established a new uniqueness
theorem for the inverse Sturm-Liouville problem. They showed that the
measurement of a particular set of eigenvalues was sufficient to define the
obscure potential functions.
They considered the eigenvalue problem
\begin{gather*}
y''+( \lambda -q( x) ) y=\lambda y, \quad 0<x<1, \\
y(0,\lambda )=0,\quad y'(\pi ,\lambda )+H_ky(\pi ,\lambda )=0.
\end{gather*}
They indicated that the spectral knowledge, for a constant index $n$
$(n=0,1,2,\dots )$, $\{ \lambda _n(q,H_k)\} _{k=1}^{+\infty }$
is equivalent to two spectra of boundary value problems with the equation
and the first initial situation (one common boundary situation at $x=0)$ and
the second boundary situation (two different boundary conditions at $x=\pi$).
 In \cite{m1} the spectral data was handled by the Hochstadt and Lieberman
method \cite{h2}. Wang \cite{w1,w2} discussed the inverse
problem for uncertain Sturm-Liouville operators on the finite interval
$[ a,b] $ and diffusion operators. Here, we consider
inverse spectral problems for Sturm Liouville operators with boundary
conditions dependent on the spectral parameter with the above spectral
knowledge. As far as we know, inverse spectral problems for Sturm
Liouville operators with boundary conditions depending on the spectral
parameter have not been studied with the spectral data before.

Eigenvalue dependent boundary conditions have been studied extensively.
References \cite{b1,b2,b3,b4,b6,f1,f3,g1} are
well known examples for problems with boundary conditions depending linearly on
the eigenvalue parameter. Recently inverse problems according to various
spectral knowledge for eigenparameter linearly dependent Sturm-Liouville
operator have been studied  in \cite{b7,f2,g3,o1,w3,w4,w5,y1}.

We consider the  Sturm-Liouville operator $L:=L( q,H_k) $
defined by
\begin{equation}
Ly\equiv -y''+q( x) y=\lambda y,\quad 0\leq x\leq \pi ,  \label{3}
\end{equation}
with boundary conditions dependent on the spectral parameter
\begin{equation}
( a_1\lambda +b_1) y(0,\lambda )-( c_1\lambda
+d_1) y'(0,\lambda )=0,  \label{4}
\end{equation}
\begin{equation}
y'(\pi ,\lambda )+H_ky(\pi ,\lambda )=0. \label{5}
\end{equation}
Also we consider the Sturm-Liouville operator 
$\widetilde{L}:=\widetilde{L} ( \widetilde{q},H_k) $ defined by
\begin{equation}
\widetilde{L}y\equiv -\widetilde{y}''+\widetilde{q}(
x) \widetilde{y}=\lambda \widetilde{y},\text{ }(0\leq x\leq \pi ),~
\label{6}
\end{equation}
with boundary conditions depending on the spectral parameter
\begin{gather}
( \widetilde{a}_1\lambda +\widetilde{b}_1) \widetilde{y}
(0,\lambda )-( \widetilde{c}_1\lambda +\widetilde{d}_1)
\widetilde{y}'(0,\lambda )=0,  \label{7} \\
\widetilde{y}'(\pi ,\lambda )+H_k\widetilde{y}(\pi ,\lambda )=0,\label{8}
\end{gather}
where $a_1,b_1,c_1,d_1,\widetilde{a}_1,\widetilde{b}_1,
\widetilde{c}_1,\widetilde{d}_1,H_k\in \mathbb{R}$, such that
$\delta _1=a_1d_1-b_1c_1<0$, 
$\widetilde{\delta }_1=\widetilde{a}_1\widetilde{d}_1
-\widetilde{b}_1\widetilde{c}_1<0$, $0<H_1<H_2<\dots <H_k<H_{k+1}<\dots <H_{0}$,
the potentials $q(x)$ and $\widetilde{q}( x)$ are real valued functions, 
$q(x),\widetilde{q}( x) \in L^{1}[0,\pi ] $ and $\lambda $
is a spectral parameter.

For the boundary-value problem \eqref{3}-\eqref{5} with coefficient
 $H=-\frac{( a_2\lambda +b_2) }{( c_2\lambda +d_2) }$ where
 $c_2\neq 0$ and $d_2\neq 0$, instead of $H_k$
describes the actual background of Sturm Liouville operators with boundary
conditions dependent on a spectral parameter; see \cite{f3}.

In this article, we construct a uniqueness theorem for Sturm-Liouville operators with
boundary conditions depending on the spectral parameter on the finite
interval $[0,\pi]$. i.e. for a constant
index $n\in\mathbb{N}$, we demonstrate that if the spectral set
 $\{ \lambda_n(q,H_k)\} _{k=1}^{+\infty }$ for different $H_k$ can be
restrained, then the spectral set 
$\{ \lambda _n(q,H_k)\}_{k=1}^{+\infty }$ is sufficient to define the 
potential $q(x)$ and coefficient $\frac{a_1\lambda +b_1}{c_1\lambda +d_1}$ of the
boundary condition. The techniques used here will be adopted from 
\cite{h2,m1,w5}.

\begin{lemma}[\cite{b2,w5}]\label{lem1.1}
 Eigenvalues $\lambda _n$ $( n\neq 0) $ of the boundary-value problem 
\eqref{3}-\eqref{5} for coefficient 
$H=H_k=-\frac{( a_2\lambda+b_2) }{( c_2\lambda +d_2) }$ in \eqref{5} are roots
of \eqref{5} and satisfy the asymptotic formula
\begin{equation}
\sqrt{\lambda _n}=n+[ 1+O( \frac{1}{n}) ] .\label{9}
\end{equation}
\end{lemma}

\begin{lemma}[\cite{w5}] \label{lem1.2}
The solution to the  \eqref{3} with the initial conditions
$y(0,\lambda )=(c_1\lambda +d_1) $ and 
$y'(0,\lambda )=(a_1\lambda +b_1) $ is
\begin{equation}
\begin{aligned}
y(x,\lambda )
&=( c_1\lambda +d_1) \Big[ \cos \sqrt{\lambda}x
 +\int_{0}^{x}A(x,t)\cos \sqrt{\lambda }t\,dt\Big]   \\
&\quad +( a_1\lambda +b_1) \Big[ \sin \frac{\sqrt{\lambda }x}{
\sqrt{\lambda }}+\frac{1}{\sqrt{\lambda }}\int_{0}^{x}B(x,t)\sin \sqrt{
\lambda }t\,dt\Big]
\end{aligned} \label{y}
\end{equation}
where the kernel $A(x,t)$ satisfies
\begin{equation*}
\frac{\partial ^{2}A(x,t)}{\partial x^{2}}-q( x) A(x,t)=\frac{
\partial ^{2}A(x,t)}{\partial t^{2}},
\end{equation*}
where $q( x) =2\frac{dA( x,x) }{dx}$, $A(
0,0) =h$, $\frac{\partial A(x,t)}{\partial t}\big| _{t=0}=0$;
and the kernel $B(x,t)$ satisfies
\begin{equation*}
\frac{\partial ^{2}B(x,t)}{\partial x^{2}}-q( x) B(x,t)=\frac{
\partial ^{2}B(x,t)}{\partial t^{2}}
\end{equation*}
where $q( x) =2\frac{dB( x,x) }{dx}$, $B(x,0) =0$.
\end{lemma}

\section{Main results and their proofs}

\begin{theorem} \label{thm2.1}
Let $\sigma ( L_{k_{j}}) :=\{ \lambda_n(q,H_{k_{j}})\} $ $( j=1,2) $ 
be the spectrum of the boundary value problem \eqref{3}-\eqref{5} 
with coefficient $H_{k_{j}}$. If $H_{k_1}\neq H_{k_2}$, then
\begin{equation}
\sigma ( L_{k_1}) \cap \sigma ( L_{k_2}) =\emptyset  \label{x}
\end{equation}
where $k_{j}\in \mathbb{N}$, and $\emptyset $ denotes an empty set.
\end{theorem}

\begin{lemma} \label{lem2.1}
Let $\lambda _n(q,H_k)$ be the $n$-th eigenvalue of the boundary-value
problem \eqref{3}-\eqref{5}. Then the spectral set 
$\{ \lambda_n(q,H_k)\} _{k=1}^{+\infty }$ is a bounded infinite set, where 
$0<H_1<H_2<\dots <H_k<H_{k+1}<\dots <H_{0}$.
\end{lemma}

The above Lemma carries a significant part in the proof of the next theorem.

\begin{theorem} \label{thm2.2}
Let $\lambda _n(q,H_k)$ be the $n$-th eigenvalue of the boundary-value
problem \eqref{3}-\eqref{5} and $\lambda _n(\widetilde{q},H_k)$ be the 
$n$-th eigenvalue of the boundary-value problem \eqref{6}-\eqref{8}, for a
constant index $n( n\in \mathbb{N}) $. 
If $\lambda _n(q,H_k)=\lambda _n(\widetilde{q},H_k)$ for all $k\in\mathbb{N}$,
then 
\begin{gather*}
q(x)=\widetilde{q}(x) \text{a.e. on } [ 0,\pi ], \\
\frac{\widetilde{a}_1\lambda +\widetilde{b}_1}{\widetilde{c}_1\lambda +
\widetilde{d}_1}=\frac{a_1\lambda +b_1}{c_1\lambda +d_1},\quad
\forall \lambda \in C.
\end{gather*}
\end{theorem}

\begin{proof}[Proof of  Theorem \ref{thm2.1}]
 Suppose the argument of Theorem \ref{thm2.1} is false. 
Then there exists $\lambda _{n_1}(H_{k_1})=\lambda _{n_2}(H_{k_2})\in\mathbb{R}$, 
where $\lambda _{n_{j}}(H_{k_{j}})\in \sigma ( L_{k_{j}}) $  for
$j=1,2$ and  $n_{j}\in\mathbb{N}$.
 Let $y_{k_{j}}( x,\lambda _{n_{j}}(H_{k_{j}})) $ be the solution of 
 \eqref{3}-\eqref{5} with the eigenvalue $\lambda _{n_{j}}(H_{k_{j}})$ and 
satisfy the initial conditions 
$y_{k_{j}}(0,\lambda _{n_{j}}(H_{k_{j}}))=(c_1\lambda +d_1) $ and 
$y_{k_{j}}'(0,\lambda_{n_{j}}(H_{k_{j}}))=( a_1\lambda +b_1) $. For a fixed index 
$n$, we have
\begin{equation}
-y_{k_1}''( x,\lambda _{n_1}( H_{k_1})
) +q(x)y_{k_1}( x,\lambda _{n_1}( H_{k_1})) 
=\lambda _{n_1}(H_{k_1})y_{k_1}( x,\lambda _{n_1}(H_{k_1}) )   \label{40}
\end{equation}
and
\begin{equation}
-y_{k_2}''( x,\lambda _{n_2}( H_{k_2})
) +q(x)y_{k_2}( x,\lambda _{n_2}( H_{k_2})
) =\lambda _{n_2}(H_{k_2})y_{k_2}( x,\lambda _{n_2}(
H_{k_2}) ) .  \label{41}
\end{equation}
By multiplying \eqref{40} by 
$y_{k_2}( x,\lambda _{n_2}(H_{k_2}) ) $ and \eqref{41} by $y_{k_1}( x,\lambda
_{n_1}( H_{k_1}) ) $ respectively and subtracting and
integrating from $0$ to $\pi $, we obtain
\begin{equation}
( y_{k_2}y_{k_1}'-y_{k_1}y_{k_2}') \big| _{0}^{\pi } =0.  \label{42}
\end{equation}
Using the initial conditions, we obtain
\begin{equation}
y_{k_2}( \pi ,\lambda _{n_2}( H_{k_2}) )
y_{k_1}'( \pi ,\lambda _{n_1}( H_{k_1})
) -y_{k_1}( \pi ,\lambda _{n_1}( H_{k_1})
) y_{k_2}'( \pi ,\lambda _{n_2}(
H_{k_2}) ) =0.  \label{43}
\end{equation}
On the other hand, we have the equality
\begin{equation} \label{44}
\begin{aligned}
&y_{k_2}( \pi ,\lambda _{n_2}( H_{k_2}) )
y_{k_1}'( \pi ,\lambda _{n_1}( H_{k_1})
) -y_{k_1}( \pi ,\lambda _{n_1}( H_{k_1})
) y_{k_2}'( \pi ,\lambda _{n_2}(
H_{k_2}) )  \\
&= y_{k_2}( \pi ,\lambda _{n_2}( H_{k_2}) )
[ y_{k_1}'( \pi ,\lambda _{n_1}(
H_{k_1}) ) +H_{k_1}y_{k_1}( \pi ,\lambda
_{n_1}( H_{k_1}) ) ]   \\
&\quad -y_{k_1}( \pi ,\lambda _{n_1}( H_{k_1}) )
[ y_{k_2}'( \pi ,\lambda _{n_2}(
H_{k_2}) ) +H_{k_2}y_{k_2}( \pi ,\lambda
_{n_2}( H_{k_2}) ) ]   \\
&\quad +( H_{k_2}-H_{k_1}) y_{k_1}( \pi ,\lambda
_{n_1}( H_{k_1}) ) y_{k_2}( \pi ,\lambda
_{n_2}( H_{k_2}) )   \\
&=( H_{k_2}-H_{k_1}) y_{k_1}( \pi ,\lambda
_{n_1}( H_{k_1}) ) y_{k_2}( \pi ,\lambda
_{n_2}( H_{k_2}) ) .
\end{aligned}
\end{equation}

Since $H_{k_2}-H_{k_1}\neq 0$, if $y_{k_1}( \pi ,\lambda
_{n_1}( H_{k_1}) ) y_{k_2}( \pi ,\lambda
_{n_2}( H_{k_2}) ) =0$, it follows that
\begin{equation}
y_{k_1}( \pi ,\lambda _{n_1}( H_{k_1}) ) =0\text{
or }y_{k_2}( \pi ,\lambda _{n_2}( H_{k_2}) ) .
\label{45}
\end{equation}
By virtue of \eqref{45} together with \eqref{5}, this yields
\begin{equation}
y_{k_1}( \pi ,\lambda _{n_1}( H_{k_1}) )
= y_{k_1}'( \pi ,\lambda _{n_1}( H_{k_1})) =0  \label{46}
\end{equation}
or
\begin{equation}
y_{k_2}( \pi ,\lambda _{n_2}( H_{k_2}) )
=y_{k_2}'( \pi ,\lambda _{n_2}( H_{k_2})
) =0.  \label{47}
\end{equation}

By \eqref{46} and \eqref{47}, this yields
\begin{equation}
y_{k_1}( x,\lambda _{n_1}( H_{k_1}) ) =0\text{
or }y_{k_2}( x,\lambda _{n_2}( H_{k_2}) ) =0
\quad \text{on }[ 0,\pi ] ,  \label{48}
\end{equation}
This is impossible. Thus, we obtain
\begin{equation}
y_{k_2}( \pi ,\lambda _{n_2}( H_{k_2}) )
y_{k_1}'( \pi ,\lambda _{n_1}( H_{k_1})
) -y_{k_1}( \pi ,\lambda _{n_1}( H_{k_1})
) y_{k_2}'( \pi ,\lambda _{n_2}(
H_{k_2}) ) \neq 0.  \label{49}
\end{equation}
Clearly, this  contradicts  \eqref{43}; therefore \eqref{x}) holds. 
The proof is complete. 
\end{proof}


\begin{proof}[Proof of Lemma \ref{lem2.1}]
 We will show that the following formula holds
\begin{equation}
\lambda _n(H_{0})<\dots <\lambda _n(H_{k+1})<\lambda _n(H_k)<\dots
<\lambda _n(H_1).  \label{50}
\end{equation}
Let $y( x,\lambda _n(H)) $ be the solution of the boundary
value problem \eqref{3}-\eqref{5} of the eigenvalue $\lambda _n(H)$ and
satisfies the initial conditions $y(0,\lambda _n(H))=( c_1\lambda
+d_1) $ and $y'(0,\lambda _n(H))=( a_1\lambda
+b_1) $. We have
\begin{gather}
-y''( x,\lambda _n( H) ) +q(x)y(
x,\lambda _n( H) ) =\lambda _n( H) y(
x,\lambda _n( H) ) ,  \label{51} \\
\begin{aligned}
&-y''( x,\lambda _n( H+\Delta H) )
+q(x)y( x,\lambda _n( H+\Delta H) ) \\
&=\lambda_n( H+\Delta H) y( x,\lambda _n( H+\Delta H))  
\end{aligned} \label{52}
\end{gather}
where $\Delta H$ is the enhancement of $H$.
 Multiplying \eqref{51} by $y( x,\lambda _n( H+\Delta H) ) $ and
 multiplying \eqref{52} by $y( x,\lambda _n( H) ) $ and
subtracting from each other and integrating from $0$ to $\pi $, we obtain
\begin{equation} \label{53}
\begin{aligned}
&\Delta \lambda _n(H)\int_{0}^{^{\pi }}y(x,\lambda _n(H))y(x,\lambda
_n(H+\Delta H))dx   \\
&=\Delta Hy(\pi ,\lambda _n(H)y(\pi ,\lambda _n(H+\Delta H)),
\end{aligned}
\end{equation}
where $\Delta \lambda _n( H) =\lambda _n( H+\Delta H) -\lambda _n( H)$.

It is well understood that $y( x,\lambda _n( H) ) $
and $\lambda _n( H) $ are real and continuous with respect to 
$H$. Dividing \eqref{53} by $\Delta H$, and letting $\Delta H\to 0$
in \eqref{53}, we have
\begin{equation}
\frac{\partial \lambda _n(H)}{\partial H}\int_{0}^{^{\pi }}y^{2}(x,\lambda
_n(H))dx=y^{2}(\pi ,\lambda _n(H)).  \label{54}
\end{equation}
If $y(\pi ,\lambda _n(H))=0$, then $y'(\pi ,\lambda _n(H))=0$.
By the uniqueness theorem, this yields
\[
y(x,\lambda _n(H))\equiv 0.
\]
This contradicts  the eigenfunction $y(x,\lambda _n(H)\neq 0$
corresponding to eigenvalue $\lambda _n(H)$. 
Hence $y^{2}(\pi ,\lambda_n(H))>0$ and 
$\int_{0}^{^{\pi }}y^{2}(x,\lambda _n(H))dx>0$. From \eqref{54}, we have
\begin{equation*}
\frac{\partial \lambda _n(H)}{\partial H}>0.
\end{equation*}
This implies that \eqref{50} holds. Therefore the spectral set 
$\{ \lambda_n(q,H_k)\} _{k=1}^{+\infty }$ is a bounded infinite set. 
The proof is complete.
\end{proof}

Finally, using Theorem \ref{thm2.1}, Lemma \ref{lem2.1} and the properties of entire
functions, we  show that Theorem \ref{thm2.2} holds.

\begin{proof}[Proof of Theorem \ref{thm2.2}]

According to Lemma \ref{lem1.2}, solutions to equation \eqref{3} with boundary
condition \eqref{4} and the equation \eqref{6} with boundary condition 
\eqref{7} can be stated in the integral forms:
\begin{equation}
\begin{aligned}
y(x,\lambda ) &=( c_1\lambda +d_1) \Big[ \cos \sqrt{\lambda
}x+\int_{0}^{x}A(x,t)\cos \sqrt{\lambda }t\,dt\Big]    \\
&\quad +( a_1\lambda +b_1) \Big[ \sin \frac{\sqrt{\lambda }x}{
\sqrt{\lambda }}+\frac{1}{\sqrt{\lambda }}\int_{0}^{x}B(x,t)\sin \sqrt{
\lambda }t\,dt\Big]
\end{aligned}  \label{55}
\end{equation}
and
\begin{equation}
\begin{aligned}
\widetilde{y}(x,\lambda )
&=( \widetilde{c}_1\lambda +\widetilde{d}_1)
\Big[ \cos \sqrt{\lambda }x+\int_{0}^{x}\widetilde{A}(x,t)\cos
\sqrt{\lambda }t\,dt\Big]   \\
&\quad +( \widetilde{a}_1\lambda +\widetilde{b}_1) \Big[ \sin
\frac{\sqrt{\lambda }x}{\sqrt{\lambda }}+\frac{1}{\sqrt{\lambda }}
\int_{0}^{x}\widetilde{B}(x,t)\sin \sqrt{\lambda }t\,dt\Big]
\end{aligned} \label{56}
\end{equation}
respectively. Let $\lambda =s^{2}$. From \eqref{55}, \eqref{56} and
\cite[proof of Theorem 2.1]{w5} we obtain
\begin{equation}
\begin{aligned}
y\widetilde{y}
&=\frac{( c_1s^{2}+d_1) ( \widetilde{c}
_1s^{2}+\widetilde{d}_1) }{2}
\Big[ 1+\cos 2sx+\int_{0}^{x}k(x,\tau )\cos 2s\tau d\tau \Big]   \\
&\quad +\frac{( a_1s^{2}+b_1) ( \widetilde{a}_1s^{2}+
\widetilde{b}_1) }{2s^{2}}\Big[ 1-\cos 2sx+\int_{0}^{x}h(x,\tau
)\cos 2s\tau d\tau \Big]   \\
&\quad +\frac{1}{2s}( c_1s^{2}+d_1) ( \widetilde{a}_1s^{2}+
\widetilde{b}_1) \Big[ \sin 2sx+\int_{0}^{x}l(x,\tau )\sin 2s\tau
d\tau \Big]   \\
&\quad +\frac{1}{2s}( \widetilde{c}_1s^{2}+\widetilde{d}_1) (
a_1s^{2}+b_1) \Big[ \sin 2sx+\int_{0}^{x}m(x,\tau )\sin 2s\tau
d\tau \Big] ,
\end{aligned}  \label{62}
\end{equation}
where the functions $k(x,\tau )$, $h(x,\tau )$, $l(x,\tau )$ and $m(x,\tau )$
are continuous functions.

We define the function 
\begin{equation*}
w( \lambda ) =( a_2\lambda +b_2) y(\pi ,\lambda
)-( c_2\lambda +d_2) y'(\pi ,\lambda ).
\end{equation*}
From \eqref{55}, we obtain the  asymptotic forms
\begin{gather*}
y(\pi ,\lambda )=( c_1\lambda +d_1) \cos \sqrt{\lambda }\pi
+O( \sqrt{\lambda }e^{| \operatorname{Im}\sqrt{\lambda }|\pi }), \\
y'(\pi ,\lambda )=-( c_1\lambda +d_1) \sqrt{\lambda }
\sin \sqrt{\lambda }\pi +O( \sqrt{\lambda }e^{| \operatorname{Im}\sqrt{
\lambda }| \pi }) .
\end{gather*}
Hence
\begin{equation}
w( \lambda ) =( c_1\lambda +d_1) (
c_2\lambda +d_2) \sqrt{\lambda }\sin \sqrt{\lambda }\pi +O(
| \lambda | ^{2}e^{| \operatorname{Im}\sqrt{\lambda }
| \pi }) .  \label{63}
\end{equation}
Zeros of $w( \lambda ) $ are the eigenvalues of the
Sturm-Liouville problem \eqref{3}-\eqref{5} where 
$H_k=H=-\frac{(a_2\lambda +b_2) }{( c_2\lambda +d_2) }$. 
$w(\lambda ) $ is an entire function of order $\frac{1}{2}$ of $\lambda $.
Multiplying \eqref{6} by $y$, \eqref{3} by $\widetilde{y}$ and subtracting
and integrating from $0$ to $\pi $, we take
\[
( \widetilde{y}y'-y\widetilde{y}') \big|_{0}^{\pi } 
+\int_{0}^{\pi }( \widetilde{q}-q) y\widetilde{y}dx=0.
\]
Using $y(0,\lambda )=( c_1\lambda +d_1) $, 
$\widetilde{y}(0,\lambda )=( \widetilde{c}_1\lambda +\widetilde{d}_1) $, 
$y'(0,\lambda )=( a_1\lambda +b_1) $ and 
$\widetilde{y}'(0,\lambda )=( \widetilde{a}_1\lambda +\widetilde{b}_1) $, 
this yields
\begin{equation} \label{64}
\begin{aligned}
0 &=[ \widetilde{y}(\pi ,\lambda )y'(\pi ,\lambda )-y(\pi
,\lambda )\widetilde{y}'(\pi ,\lambda )] +( a_1\lambda
+b_1) ( \widetilde{c}_1\lambda +\widetilde{d}_1)\\
&\quad -( c_1\lambda +d_1) ( \widetilde{a}_1\lambda +
\widetilde{b}_1) +\int_{0}^{\pi }( \widetilde{q}(x)-q(x))
y\widetilde{y}dx.
\end{aligned}
\end{equation}
Let $Q(x)=( \widetilde{q}(x)-q(x))$ and
\begin{equation}
\begin{aligned}
K(\lambda )
&= ( a_1\widetilde{c}_1-\widetilde{a}_1c_1)
\lambda ^{2}+( a_1\widetilde{d}_1+b_1\widetilde{c}_1-\widetilde{
a}_1d_1-\widetilde{b}_1c_1) \lambda    \\
&\quad +( b_1\widetilde{d}_1-\widetilde{b}_1d_1)
+\int_{0}^{\pi }Q(x)y\widetilde{y}dx.  \label{65}
\end{aligned}
\end{equation}
Clearly, the function $K(\lambda )$ is an entire function. Because the first
term of equation \eqref{64} for $\lambda =\lambda _n( q,H_k) $ is zero, then
\begin{equation*}
K(\lambda _n( q,H_k) )=0.
\end{equation*}

From Lemmas \ref{lem1.1} and \ref{lem2.1}, we see that the spectral set 
$\{ \lambda _n(q,H_k)\} _{k=1}^{+\infty }$ is a bounded infinite set.
Therefore, if consists of $\lambda _{n0}(q)\in\mathbb{R}$, such that 
$\lambda _{n0}(q)$ is a finite accumulation dot of the spectrum
set $\{ \lambda _n(q,H_k)\} _{k=1}^{+\infty }$. It is well
understood that the set of zeros of every entire function which is not
identically zero hasn't any finite accumulation dot.

Hence
\begin{equation}
K(\lambda )=0,~\forall \lambda \in \mathbb{C}. \label{66}
\end{equation}
From \eqref{65}, \eqref{66} and \cite[proof of Theorem 2.1]{w5}, we have
\begin{gather*}
Q(x)=\widetilde{q}(x)-q(x)=0,\quad\text{a.e. on }[ 0,\pi ] , \\
\frac{\widetilde{a}_1\lambda +\widetilde{b}_1}{\widetilde{c}_1\lambda +
\widetilde{d}_1}=\frac{a_1\lambda +b_1}{c_1\lambda +d_1},\quad \forall
\lambda \in C.
\end{gather*}
The proof is complete.
\end{proof}

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\end{document}
