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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 41, 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-2017/41\hfil An inverse Sturm-Liouville problem]
{An inverse Sturm-Liouville problem with a generalized symmetric potential}

\author[E. \.I. Esk\.ita\c{s}\c{c}io\u{g}lu, M. A\c{c}\.il \hfil EJDE-2017/41\hfilneg]
{Es\.in \.Inan Esk\.ita\c{s}\c{c}io\u{g}lu, Mehmet A\c{c}\.il}

\address{Esin \.Inan Esk\.ita\c{s}\c{c}io\u{g}lu \newline
Department of Mathematics,
Faculty of Sciences, Y\"uz\"unc\"u Y{\i}l University,
65080, Van, Turkey}
\email{inancinar@mynet.com}

\address{Mehmet A\c{c}\.il \newline
Department of Mathematics,
Faculty of Sciences, Y\"uz\"unc\"u Y{\i}l University,
65080, Van, Turkey}
\email{mehmetacil76@gmail.com}

\dedicatory{Communicated by Mokhtar Kirane}

\thanks{Submitted December 22, 2016. Published February 7, 2017.}
\subjclass[2010]{31A25,  34B24, 34L16}
\keywords{Boundary value and inverse problems; Sturm-Liouville theory;
\hfill\break\indent numerical approximation of eigenvalues}

\begin{abstract}
 We consider the normal form of Sturm-Liouville differential equations with
 separable boundary conditions. For this problem we know that the potential
 function is determined  uniquely by two spectra and that if the potential
 is symmetric, then it is determined uniquely by just one spectrum.
 In this paper firstly we generalize symmetric potential and then investigate
 change of needed data to determine potential function uniquely.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks


\section{Introduction}

The inverse Sturm-Liouville problem was firstly studied by Ambartsumyan in 1929.
 He considered the  boundary value problem
\begin{gather*}
-y''(x)+(\lambda+q(x))y(x)=0, \\
y'(0)=y'(1)=0
\end{gather*}
and showed that if eigenvalues of the problem are $\lambda_{n}=n^2\pi^2$, 
then the potential function $q\equiv0$ \cite{am}. This work naturally 
led to question whether the potential is determined uniquely by one spectrum 
or not. This problem is called the inverse Sturm- Liouville problem. 
Borg showed that it is not true generally and that two spectra are needed 
to determine the potential uniquely by changing one of boundary conditions. 
He also showed that if the potential $q$ is symmetric about midpoint of 
the interval $[0,\pi]$ for his problem, then it is determined uniquely 
by one spectrum in 1949 \cite{bo}. Levinson shortened the proof of Borg 
by using complex analysis techniques\cite{le}. From then on, various versions 
of Borg's work have been considered. Some of them can be seen in \cite{ge, ho, hos}.

The study we conducted here is motivated by \cite{le, lev, mc}. 
Firstly we need to give some theorems and definitions. 
The structure of BVP we considered is as follows:
\begin{gather}
L[y]=\mu y, \label{e1}\\
y'(0)-hy(0)=0, \label{e2}\\
y'(a)+Hy(a)=0, \label{e3}
\end{gather}
where $L[u]=-u''+qu$ and $h, H, a$ are real constants. 
The operator $L$ is a self-adjoint operator defined on $L^2[0,a]$ provided 
that $q$ satisfies suitable regularity conditions. Under these conditions
 $L$ has a discrete spectrum consisting of the simple and real 
eigenvalues $\{\mu_i\}_{i=0}^{\infty}$ \cite{bi}.
Let us consider \eqref{e1} with the  initial conditions
\begin{equation} \label{e4}
y(0)=1,\quad y'(0)=h.
\end{equation}

\begin{theorem} \label{thm1} 
Suppose that $\phi(x,\lambda)$ is the solution of \eqref{e1} satisfying 
the initial conditions \eqref{e4} and that $q(x)$ has m locally integrable 
derivatives. Then there exists function $K(x,t)$ having $m+1$ locally 
integrable derivatives with respect to each of variables such that
\begin{gather} \label{e5}
\phi(x,\lambda)= \cos{\sqrt{\lambda}x}+{\int_{0}^{x}K(x,t)\cos{\sqrt{\lambda}t}dt},\\
\label{e6}
K(x,x)= h+\frac{1}{2}{\int_{0}^{x}q(t)dt}.
\end{gather}
\end{theorem}

The above theorem and other properties of function $K(x,t)$ can be found in \cite{lev}.

\begin{lemma} \label{lem2}
Let $(a,b)$ be a finite interval and $f(x)\in L(a,b)$. Then
\[
\lim_{|\lambda| \mapsto \infty}{\int_{a}^{b}f(x)\cos{\sqrt{\lambda}x}dx}=0.
\]
\end{lemma}

The above lemma and its proof can be found in \cite{gr}.

\begin{lemma} \label{lem3}
If a real sequence $\{\mu_{n}\}$ is the spectrum of BVP \eqref{e1}-\eqref{e3}
 with function $q(x)$ where $q\in L^1(0,a)$, then it satisfies the 
 asymptotic formula
\begin{equation} \label{e7}
\sqrt{\mu_n}=\frac{n\pi}{a}+\frac{aa_0}{n}+o(\frac{1}{n})
\end{equation}
where $\mu_n\neq\mu_m$ for $m\neq n$ and $a_0=(K(a,a)+H(/(a\pi)$.
\end{lemma}

\begin{proof}
By using the boundary conditions directly we derive $\sqrt{\mu_n}$. 
From Theorem \ref{thm1}, we have solution \eqref{e5}. 
By considering \eqref{e3} we obtain
\begin{equation} \label{e8}
\begin{aligned}
&-\sqrt{\mu}\sin{\sqrt{\mu}a}+{\int_{0}^{a}
\frac{\partial K(x,t)}{\partial x}\cos{\sqrt{\mu}t}dt}
+K(a,a)\cos{\sqrt{\mu}a} \\
&+H\big[\cos{\sqrt{\mu}a}+{\int_{0}^{a}K(a,t)\cos{\sqrt{\mu}t}dt}\big]=0.
\end{aligned}
\end{equation}
The numbers $\{\mu_n\}$ are the roots of equation \eqref{e8}. 
Since $\lambda_n\mapsto \infty$ as $n\mapsto \infty$ the first approximation 
for $\mu_n$'s is obtained as follows:
\[
\sin{\sqrt{\mu_n}a}+O\big(\frac{1}{\sqrt{\mu_n}}\big)=0.
\]
Therefore 
\[
\sqrt{\mu_n}=\frac{n\pi}{a}+O\big(\frac{1}{n}\big).
\]
Now let us put
\[
\sqrt{\mu_n}=\frac{n\pi}{a}+\frac{aa_0}{n}+\frac{\gamma_0}{n}.
\]
Then 
\begin{align*}
&(-1)^na\Big[\frac{aa_0+\gamma_n}{n}+O\big(\frac{1}{n^3}\big)\Big]
-\frac{(-1)^n[K(a,a)+H]}{\frac{n\pi}{a}\big(\frac{1+a^2a_0
+\gamma_na}{\pi n^2}\big)}\Big[1+O\big(\frac{1}{n^2}\big)\Big]\\
&-\frac{1}{\sqrt{\mu_n}}{\int_{0}^{a}\big[HK(a,t)
+\frac{\partial}{\partial x}K(x,t)\Big|_{x=a}\big]\cos{\sqrt{\mu_n}t}dt}=0
\end{align*}
from \eqref{e8}. Since 
$[HK(a,t)+\frac{\partial}{\partial x}K(x,t)\Big|_{x=a}\big]\in L^1(0,a)$, 
by using Lemma \ref{lem2} it follows that as $n\to\infty$
\[
\frac{1}{\sqrt{\mu_n}}{\int_{0}^{a}\big[HK(a,t)
+\frac{\partial}{\partial x}K(x,t)\Big|_{x=a}\big]\cos{\sqrt{\mu_n}t}dt}\to 0
\]
Therefore $a_0=\frac{K(a,a)+H}{a\pi}$ and $\gamma_n\to 0$ as $n\to\infty$.
 This completes the proof.
\end{proof}

\begin{remark} \label{rmk4} \rm
There is no $\tilde{a}$ different from $a$ such that asymptotic formula 
\eqref{e7} is admitted by the problem \eqref{e1}-\eqref{e3} with 
$\tilde{a}$ instead of $a$.
\end{remark}

Let us consider a function $ f:[0,1]\mapsto \mathbb{R}$. 
If for $\forall x\in [0,1]$ and $n=1,2,\dots,k$,
\begin{eqnarray*}
f(x)=f\Big(\frac{1}{2^{n-1}}-x\Big),
\end{eqnarray*}
then we will call the function $f$ as $k$th-order symmetric function.

\begin{example} \label{examp1} \rm
 Define the function
\begin{displaymath}
f(x)=  \begin{cases}
x^2+x+0.01 &  0<x<0.25,\\
x^2-2x+0.76 &  0.25<x<0.5, \\
x^2-0.24 &  0.5<x<0.75, \\
x^2-3x+1.01 &  0.75<x<1, 
\end{cases} 
\end{displaymath}
Then  $f$ is a second-order symmetric function. It can be seen in Figure 
ref{fig1}.
\end{example}

\begin{figure}[ht]
\begin{center}
 \includegraphics[width=0.5\textwidth]{fig1} % sekil.eps}
\end{center}
\caption{A second-order symmetric function.}
\label{fig1}
\end{figure}

\section{Main results}

In this part we consider the BVP
\begin{gather}
y''(x)+(\lambda-q(x))y(x)=0,  \label{e9}\\
y'(0)-hy(0)=0, \quad y'(1)+Hy(1)=0. \label{e10}
\end{gather}

\begin{theorem} \label{thm5}
Let be $q\in L^1[0,1]$, $q$ is a second-order symmetric function and $h=H$. 
Then for  \eqref{e9}-\eqref{e10} the potential $q$ is determined uniquely 
by its half  spectrum except finite numbers of its eigenvalues.
\end{theorem}

\begin{proof}
To prove this theorem let us consider the BVP
\begin{gather} \label{e11}
-y''(x)+q(x)y(x)=\tilde{\lambda} y(x), \\
\label{e12}
y'(0)-\frac{h}{2}y(0)=0,\quad y'(1/2)+\frac{H}{2}y(1/2)=0.
\end{gather}
From the theory about symmetric potential we know that for all 
$x\in [0,1]$ when $q$ is symmetric about midpoint, one spectrum is sufficient 
to determine potential $q$ uniquely. Furthermore since $q(x)=q(\frac{1}{2}-x)$ 
for all $x\in [0,1/2]$, the same situation holds for (11)-(12).
We intend to show that the spectrum of (11)-(12) represents half spectrum 
of the problem \eqref{e9}-\eqref{e10} except for a finite numbers of its eigenvalues. 
From Lemma \ref{lem2} we see that spectrum of (11)-(12) has the asymptotic form
\[
\sqrt{\tilde{\lambda}_k}=2k\pi+\frac{\tilde{a}_0}{2k}+o\Big(\frac{1}{k}\Big)
\]
and that the spectrum of \eqref{e9}-\eqref{e10} has also asymptotic form
\[
\sqrt{\lambda_k}=n\pi+\frac{a_0}{n}+o\Big(\frac{1}{n}\Big).
\]
Also we have
\begin{align*}
\tilde{a}_0
&=\frac{\tilde{K}(1/2,1/2)+H/2}{\pi/2}
=\frac{1}{\pi}\big[2\tilde{K}(1/2,1/2)+H\big]\\
&=\frac{1}{\pi}\Big[2\Big(\frac{h}{2}
+\frac{1}{2}{\int_0^{1/2}q(x)dx}\Big)+H\Big] \\
&=\frac{1}{\pi}\Big[H+\Big(h+\frac{1}{2}{\int_0^{1}q(x)dx}\Big)\Big]\\
&=\frac{1}{\pi}\big[H+K(1,1)\big]=a_0
\end{align*}
where $K$ and $\tilde{K}$ are kernels of solutions which represent 
\eqref{e9}-\eqref{e10} and (11)-(12) respectively. 
Having the above results we obtain
\[
\sqrt{\tilde{\lambda}_n}-\sqrt{\lambda_{2n}}=o\Big(\frac{1}{n}\Big)\to 0
\]
for large $n$. This completes the proof.
\end{proof}

\begin{corollary} \label{cor6}
Let be $q\in L^1[0,1]$, $q$ is a $k$th-order symmetric function and $h=H$. 
Then for  \eqref{e9}-\eqref{e10} the potential $q$ is determined uniquely 
by its $\frac{1}{2^{k-1}}$ spectrum except for a finite numbers of eigenvalues 
of spectrum.
\end{corollary}

The above corollary can be shown easily by using induction with the above process.
This result leads to a question: What is the situation when $q$ is an 
infinite-order symmetric function? Our related theorem is as follows.

\begin{theorem} \label{thm7}
Let  $q\in C[0,1]$, $H=-h$ and $q$ be an infinite-order symmetric function.
 Then for problem \eqref{e9}-\eqref{e10} the potential function $q$ is 
determined uniquely by just its one eigenvalue.
\end{theorem}

\begin{proof} 
Since $q\in C[0,1]$ from Weierstrass approximation theorem (see \cite{je}) 
there exists polynomial $P$ with degree $m$ on $[a,b]$ which represents 
potential $q$. Then for $\forall k$ it follows that 
$P(x)=P\big(\frac{1}{2^{k-1}}-x\big)$. By rewriting this for $k=1,2 ,\dots,n$ 
and then adding those for $P(x)=c_0+c_1x+\dots+c_mx^m$ we obtain
\begin{align*}
P(x)&=\frac{1}{n}\Big\{nc_0+c_1\big[(1-x)+(\frac{1}{2}-x)+\dots
+(\frac{1}{2^{n-1}})\big]+\dots\\
&\quad +c_m\big[(1-x)^m+\dots+(\frac{1}{2^{n-1}})^m\big]\Big\}\\
&=c_0+\frac{1}{n}\sum_{k=1}^{m}c_k
\big[(1-x)^k+\dots+\big(\frac{1}{2^{n-1}}-x\big)^k\big].
\end{align*}
By taking limits for the general term of the sum as $n\mapsto \infty$ 
it follows that
\begin{align*}
&\lim_{n \mapsto \infty}\frac{(1-x)^k+\dots+\big(\frac{1}{2^{n-1}}-x\big)^k}{n} \\
&=\lim_{n \mapsto \infty}\frac{1}{n}\Big\{\big(1+\frac{1}{2^k}+\dots
 +\frac{1}{(2^k)^{n-1}}\big) \\
&\quad -x\big(1+\frac{1}{2^{k-1}}+\dots+\frac{1}{(2^{k-1})^{n-1}}\big)+
\dots+(-1)^{k}x^{k}n\Big\} \\
&=(-1)^{k}x^{k}
\end{align*}
by using Stolz-Ces\'aro theorem for each term in parenthesis.
 We can assume that $m=2k$ without loss of generality. Thus we see that 
$P(x)=c_0+c_2x^2+\dots+c_{2k}x^{2k}$. Besides we have 
$P(0)=P\big(\frac{1}{2^{n-1}}\big)$ for $x=0$. 
So for $n=2,3,\dots,k+1$ we obtain
\[
\begin{pmatrix}
2^{-2}  & 2^{-4}  & \dots & 2^{-2k} \\
2^{-4}  & 2^{-8}  & \dots & 2^{-4k} \\
\vdots  & \vdots  & \vdots & \vdots \\
2^{-2k} & 2^{-4k} & \dots & 2^{-2k^2}
\end{pmatrix} 
\begin{pmatrix}
c_2 \\
c_4 \\
\vdots  \\
c_{2k}
\end{pmatrix} 
=\begin{pmatrix}
0 \\
0 \\
\vdots  \\
0
\end{pmatrix}.
\]
By using the LU decomposition \cite{va} for coefficient matrix, say $A$, 
it follows that
\begin{align*}
\det(A)
&=  \begin{vmatrix}
2^{-2k}  & 2^{-4k} & 2^{-6k} & \dots & 2^{-2k^2} \\
0 & 3\cdot2^{-4k+2} & 2^{-6k+2}  & \dots & \cdot \\
0 & 0 & 45\cdot2^{-6k+6}  & \dots & \cdot \\
\vdots  & \vdots  & \vdots & \vdots & \vdots\\
0 & 0 & 0 &\dots & M\cdot 2^{-k^2-k}
\end{vmatrix}  \\
&=(2^2-1)^{k-1}(2^4-1)^{k-2}\dots(2^{2(k-1)}-1)2^{-\frac{k(k+1)(2k+1)}{3}}\neq 0
\end{align*}
where $M=(2^2-1)(2^4-1)\dots (2^{2(k-1)}-1)$. So
\[
c_2=c_4=\dots=c_{2k}=0.
\]
Consequently $q(x)=c_0$. Finally by using \eqref{e9}-\eqref{e10} 
with the fact $h=-H$ we see that potential $q$ can be determined by 
just its one eigenvalue.
\end{proof}

\subsection*{Conclusion}
In this study, we considered the variation of needed data to be the spectrum 
of the operator to determine uniqueness of the potential function 
that has property $q(x)=q(\frac{1}{2^{n-1}}-x)$ by changing natural numbers $n$. 
In the recent times the studies which include reconstruction of the potential 
have been conducted. Some of them can be seen in \cite{ef,fr,ro}. 
For an application of this study, Theorem \ref{thm5}, Corollary \ref{cor6}, 
and Theorem \ref{thm7} can be considered. 
We think that one can obtain an approximation to the potential by using the 
sets $\{\lambda_1\}$, $\{\lambda_1, \lambda_2\}$, 
$\{\lambda_1, \lambda_2, \lambda_3, \lambda_4\}, \dots$. 
It is clear that obtaining a good approximation is difficult. 
However it can be considered to be an initial potential in numerical algorithms 
like the algorithm described in \cite{ro}.

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\end{document}