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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 226, pp. 1--11.\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-2017/226\hfil Uniqueness theorems]
{Uniqueness theorems for Sturm-Liouville operators with interior
twin-dense nodal set}

\author[Y. P. Wang \hfil EJDE-2017/226\hfilneg]
{Yu Ping Wang}

\address{Yu Ping Wang \newline
Department of Applied Mathematics,
Nanjing Forestry University,
Nanjing,  Jiangsu 210037, China}
\email{ypwang@njfu.com.cn}

\thanks{Submitted September 19, 2016. Published September 20, 2017.}
\subjclass[2010]{34A55, 34B24, 47E05}
\keywords{Uniqueness theorem; inverse nodal problem; potential;
\hfill\break\indent Sturm-Liouville operator;
 the interior twin-dense nodal subset}

\begin{abstract}
 We study Inverse  problems for the Sturm-Liouville operator with Robin boundary
 conditions. We establish two uniqueness theorems from the twin-dense nodal
 subset $W_{S}([\frac{1-\varepsilon}{2},\frac{1}{2}])$,
 $ 0<\varepsilon\leq1$, together with parts of either one spectrum, or the
 minimal nodal subset $\{x_n^1\}_{n=1}^\infty$ on the interval $[0,\frac{1}{2}]$.
 In particular, if one spectrum is given a priori, then the potential $q$ on
 the whole interval $[0,1]$ can be uniquely determined by
 $W_{S}([\frac{1-\varepsilon}{2},\frac{1}{2}])$  for any $S$ and arbitrarily
 small $\varepsilon$.
\end{abstract}

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

\section{Introduction}

Consider the Sturm-Liouville operator
$L:=L(q,h,H)$ defined by
\begin{equation}\label{E1.1}
- u'' + q(x){u} = \lambda {u},\quad x\in (0,1)
\end{equation} 
with  boundary conditions
\begin{gather}\label{E1.2}
U_0(u):=u'(0 ,\lambda)-hu(0,\lambda)= 0,\\
U_1(u):=u'(1 ,\lambda )+Hu(1,\lambda) = 0,\label{E1.3}
\end{gather}
where $h,H{\in\mathbb{R}}$, $q(x)$ is a real-valued
function and $q \in L^1[0,1]$.

The inverse nodal problem is to reconstruct this operator from the given nodal 
points(zeros) of its eigenfunctions.  Inverse nodal problems for  differential 
operators  have many applications in many areas, such as mathematics, physics, 
engineering, etc (see \cite{15,8,9,3,10,4,6,16,11,1,7,13,32,5,12,2,14} 
and the references therein). Inverse spectral problems for 
\eqref{E1.1}-\eqref{E1.3} consist in recovering this operator from the given
 data (refer to \cite{22,19,28,24,17,27,18,21,26,25,23,20,30} and other works). 
In particular, McLaughlin \cite{1} discussed  the inverse nodal problem 
for  \eqref{E1.1}-\eqref{E1.3} and showed that a dense subset of nodal points 
of its eigenfunctions is sufficient to determine the potential $q$ up to 
its mean value and coefficients $h,H$ of boundary conditions. 
From the physical point of view this corresponds to finding, e.g., 
the density of a string or a beam from the zero-amplitude positions of 
their eigenvibrations.  Later, X.F. Yang \cite{2} presented an interesting
 theorem for  \eqref{E1.1}-\eqref{E1.3} and showed that the $s$-dense nodal 
subset on the interval $[0,b]$, $\frac{1}{2}<b\leq 1$, is sufficient to 
determine the potential $q$ up to its mean value and coefficients $h,H$ of 
boundary conditions by the Gesztesy-Simon theorem \cite{19}. 
Then  Cheng et al \cite{3} improved the Yang's theorem by the twin-dense 
nodal subset (Similar to definition \ref{D2.1}) instead of the $s$-dense 
nodal subset.  Yang \cite{5} presented a counterexample, which illustrates 
that two operators have the same spectrum and in the subinterval 
$[0,\frac{1-\alpha}{2}]\cup [\frac{1+\alpha}{2},1]$ for any $\alpha,
0<\alpha<1$,  their nodal points are the same, but $q(x)\neq\tilde q(x) $
 on the interval $(\frac{1-\alpha}{2},\frac{1+\alpha}{2})$. 
In \cite{4,31}, Guo and Wei showed that only the twin-dense nodal data on 
a small interval $[a,b]$ containing the midpoint $\frac{1}{2}$ suffices 
to determine the differential operator (potential functions plus boundary 
constants h and H) uniquely. Their method is inspired by analysis of Weyl 
$m$-functions in the work of Gesztesy-Simon\cite{19}. The result of Guo-Wei 
is a big step forward from those in \cite{2,3}, where nodal data on more that
 half of the interval are needed.

In this note, we plan to follow the method of Guo-Wei to show two uniqueness 
results. We shall concentrate on the situation when only information of the 
twin-dense nodal subset $W_S([a,\frac{1}{2}])$ on the left portion 
$[a,\frac{1}{2}]$, still an interior subinterval. As discussed in \cite{4}, 
this is not enough. We add some more information (part of the eigenvalues 
$\lambda_n$, or the sequence of first nodal point $x_{n_k}^1)$. They suffice 
to guarantee the uniqueness of the potential function. There are four types 
of boundary conditions, we shall only concentrate on Case IV: $h,H\in \mathbb R$ 
in \cite{4}. Moreover we shall simplify part of their proof 
(cf. proof of Lemma \ref{L3.1} below).

This article is organized as follows. 
In Section 2, we present preliminaries. We introduce  our main results 
in Section 3, which  will be proved in Section 4.

\section{ Preliminaries }

Let $S(x,\lambda),\;C(x,\lambda),\; u_{-}(x,\lambda)$ and $u_+ (x,\lambda)$ 
be solutions
of \eqref{E1.1} with the initial conditions:
\begin{gather*}
S(0,\lambda)=0,\quad  S'(0,\lambda)=1,
C(0,\lambda)=1,C'(0,\lambda)=0,\\
u_-(0,\lambda)=1,\quad  u_-'(0,\lambda)=h,\quad  u_+(1,\lambda)=1,\ u_+'
(1,\lambda)=-H.
\end{gather*}
 Clearly,
$U_0(u_-)=U_1(u_+)=0$  and
\begin{gather*}
u_-(x,\lambda)=C(x,\lambda)+hS(x,\lambda),\\
u_+(x,\lambda)=U_1(S)C(x,\lambda)-U_1(C)S(x,\lambda).
\end{gather*}
Denote $\lambda=\rho^2$ and $\tau=|\mathrm{Im} \rho|$. We have the asymptotic 
formulae (see \cite{30}).
\begin{gather}\label{E2.3}
u_-(x,\lambda)=\cos \rho x+\Big(h+\frac{1}{2}\int_0^xq(t)\mathrm{d}t\Big)
\frac{\sin \rho x}{\rho }+o\big(\frac{\mathrm{e}^{\tau x}}{\rho }\big),\quad
 0\leq x\leq1\\
u_-'(x,\lambda)=-\rho\sin\rho x+O(\mathrm{e}^{\tau x}),\quad 0\leq x\leq 1,
\label{E2.4}\\
u_+(x,\lambda)=\cos \rho (1-x)+\Big(H+\frac{1}{2}\int_x^1q(t)\mathrm{d}t\Big)
 \frac{\sin \rho (1-x)}{\rho}+o\big(\frac{\mathrm{e}^{\tau (1-x)}}{\rho }\big),
\nonumber \\
\text{for } 0\leq x\leq1  \nonumber \\
u_+'(x,\lambda)=\rho\sin\rho(1-x)+O(\mathrm{e}^{\tau(1-x)}),\quad 0\leq x\leq1. 
\nonumber
\end{gather}
The following formula is called  the Green's formula
\begin{equation}
\int_0^1(yL(z)-zL(y))=[y,z](1)-[y,z](0),
\end{equation}
 where
$[y,z](x):=y(x)z'(x)-y'(x)z(x)$ is the Wronskian of $y$ and $z$.

Denote
\begin{equation*}
\Delta(\lambda):=[u_+,u_-](x,\lambda).
\end{equation*}
Then $\Delta(\lambda)$ does not depend on $x$ and
\[
\Delta(\lambda)= U_1(u_-)=-U_0(u_+),
\]
 which is called the characteristic function of  $L$. Hence
\begin{equation}
\Delta(\lambda)=-\rho\sin\rho+O(\mathrm{e}^{\tau}).
\end{equation}
Let $\sigma(L):=\{\lambda_n\}_{n=0}^\infty$ be the set of all
eigenvalues of \eqref{E1.1}-\eqref{E1.3}. It is well known that  all zeros
$\lambda_n$ of $\Delta(\lambda)$ are real and simple. For sufficiently
large $n$, we have asymptotic formula for eigenvalues $\lambda_n$
 of \eqref{E1.1}-\eqref{E1.3}
\begin{equation} \label{E2.9}
\sqrt{\lambda_n}=n\pi+\frac{\omega}{n\pi}+o\big(\frac{1}{n}\big),
\end{equation}
where $\omega=h+H+\frac{1}{2}\int_0^1q(t)\mathrm{d}t$.
 Denote $G_\delta:=\{\rho:|\rho-k\pi|>\delta, k\in \mathbb Z\}$.
 For sufficiently small $\delta$, then there exists a constant
$C_\delta$ such that for sufficiently large $|\lambda|$,
\begin{equation}
|\Delta(\lambda)|\geq C_\delta |\rho|\mathrm{e}^{\tau},\quad \forall \rho \in
G_\delta.
\end{equation}
We define the Weyl $m$-function
$m_{\pm}(x,\lambda)$ by
\begin{equation*}
m_{\pm}(x,\lambda)=\pm\frac{u_{\pm}'(x,\lambda )}{u_{\pm}(x,\lambda)}.
\end{equation*}
From  \cite{18,19}, we get the following asymptotic formulae:
\begin{equation}\label{E2.11}
m_{\pm}(x,\lambda)=i\rho+o(1),\quad
\frac{1}{m_{\pm}(x,\lambda)}=-\frac{i}{\rho}+o\big(\frac{1}{\rho^2}\big)
\end{equation}
uniformly in $x\in [0,1-\delta]$ for $m_{+}(x,\lambda)$ (resp., $x\in [\delta,1]$
for $m_{-}(x,\lambda)$), $\delta>0$ as
$|\lambda|\to\infty$ in any sector
$\varepsilon<\arg(\lambda)<\pi-\varepsilon$ for $\varepsilon>0$.

Let $u_-(x,\lambda_n )$ be the
eigenfunction corresponding to the $n$-th eigenvalue $\lambda_n $ of \
eqref{E1.1}-\eqref{E1.3} and $x_n^j$ be the nodal
points of the eigenfunction $u_-(x,\lambda_n )$, i.e.,
$u_-(x_n^j,\lambda_n )=0$, where 
$0<x_n^1<x_n^2<\dots<x_n^j<\dots<x_n^{n}<1, n\geq1$. Denote 
$x_n^0=0$ and $x_n^{n+1}=1$. Additionally, for $j=\overline{0,n}$, let $I_n^j$ be
the nodal interval by $I_n^j=(x_n^j,x_n^{j+1})$ and $l_n^j$ be the
nodal length of the interval $I_n^j$ by $l_n^j=x_n^{j+1}-x_n^j$.
Denote $X:=\{x_n^j\}$ be the set of nodal points of  \eqref{E1.1}-\eqref{E1.3}, 
where $j=j(n),j=\overline{0,n}$.

For sufficiently large $n$, we have asymptotic formulae for zeros $x_n^j$ 
of the eigenfunction $u_-(x,\lambda_n )$ of \eqref{E1.1}-\eqref{E1.3} 
(see \cite{13})
\begin{equation} \label{E2.13}
\begin{aligned}
x_n^j&=\frac{j-\frac{1}{2}}{n}+\frac{1}{2(n\pi)^2}
\Big(2h+\int_0^{x_n^j} {q(t)}\mathrm{d}t\Big) \\
&\quad -\frac{j-\frac{1}{2}}{2n^3\pi^2}
\Big(2\omega-\int_0^1q(t)\cos(2n\pi t)\mathrm{d}t\Big)+o\big(\frac{1}{n^2}\big).
\end{aligned}
\end{equation}

Let $\mathbb N_0=\mathbb N\cup\{0\}$, $\mathbb N_2=\mathbb N\backslash\{1\}$, and  
$S:=\{n_k\in \mathbb N_2:n_k<n_{k+1},k=1,2,\dots,\infty\}$.

\begin{definition}\label{D2.1} \rm
Take $a\in [0,\frac{1}{2})$. We call $W_S([a,\frac{1}{2}])$ a left twin-dense 
nodal subset on the interval $[a,\frac{1}{2}]$ if
\begin{enumerate}
\item $W_S([a,\frac{1}{2}])\subseteq X\cap [a,\frac{1}{2}]$.
\item For all $n_k\in S$, there exists  $j_k$ such that both 
$x_{n_k}^{j_k},x_{n_k}^{j_k+1}\in W_S([a,\frac{1}{2}])$.
\item The set $W_S([a,\frac{1}{2}])$ is dense on $[a,\frac{1}{2}]$, i.e. 
$\overline{W_S([a,\frac{1}{2}])}=[a,\frac{1}{2}]$.
\end{enumerate}
\end{definition}

 In the same way, we define a right twin-dense nodal subset 
$W_S([\frac{1}{2},b])$ on the interval $[\frac{1}{2}, b]$ for some  
$b, \frac{1}{2}<b\leq 1$.

The following two lemmas are  important for proofs of our main results.

\begin{lemma}[\cite{18}]\label{L2.2} 
Let $m_+(\alpha,\lambda)$ $(resp., m_-(1-\alpha,\lambda)),\alpha\in [0,1)$, 
be the Weyl $m$-function of the  problem
\eqref{E1.1}-\eqref{E1.3}.  Then $m_+(\alpha,\lambda)$
 (resp. $m_-(1-\alpha,\lambda)$) uniquely determines
coefficient $H$ (resp. $h$) of the boundary condition as well as $q$
on the interval $[\alpha,1]$ (resp. $[0,1-\alpha]$).
\end{lemma}

\begin{lemma}[{\cite[Proposition B.6]{21}}] \label{L2.3}
Let $f(z)$ be an entire  function such that
\begin{enumerate}
\item $\sup_{|z|=R_k}|f(z)|\leq C_1exp(C_2R_k^\alpha)$ for some
$0<\alpha<1$, some sequence $R_k\to\infty$ as
$k\to \infty$ and $C_1,C_2>0$.
\item
$\lim_{|x|\to\infty}|f(ix)|=0,\ x\in \mathbb R$.
\end{enumerate}
Then $f\equiv0$. 
\end{lemma}

\section{Main results}

 With $L$ we consider here and in the sequel a boundary value problem 
$\widetilde{L}=L(\widetilde{q},\widetilde{h},\widetilde{H})$ 
of the same form but with different coefficients. 
If a certain symbol $\gamma$ denotes an object related to $L$, 
then the corresponding symbol $\widetilde{\gamma}$ with tilde denotes 
the analogous object related to $\widetilde{L}$, and 
$\hat{\gamma}=\gamma-\widetilde{\gamma}$. The so-called 
$W_{S}([a,b])=\widetilde{W}_{\widetilde{S}}([a,b])$
means that for any $x_{n_k}^{j_{k}}\in W_{S}([a,b])$, then at least one 
of  \eqref{E3.1} and \eqref{E3.2} holds. i.e.
\begin{gather}\label{E3.1}
x_{n_k}^{j_{k}}=\widetilde{x}_{\widetilde{n}_k}^{\widetilde{j}_{k}}
 \quad\text{and}\quad x_{n_k}^{j_{k}+1}=\widetilde{x}_{\widetilde{n}_k}^{\widetilde{j}_{k}+1},\quad or\\
x_{n_k}^{j_{k}}=\widetilde{x}_{\widetilde{n}_k}^{\widetilde{j}_{k}}
 \quad\text{and}\quad x_{n_k}^{j_{k}-1}=\widetilde{x}_{\widetilde{n}_k}^{\widetilde{j}_{k}-1},\label{E3.2}
\end{gather}
 where $x_{n_k}^{j_{k}+j}\in W_{S}([a,b])$ and 
$\widetilde{x}_{\widetilde{n}_k}^{\widetilde{j}_{k}+j}
\in \widetilde{W}_{\widetilde{S}}([a,b])$ in this paper. i.e., for each 
fixed $(n_k,j_k)$, there exists $(\widetilde{n}_k,\widetilde{j}_k)$ such that \eqref{E3.1}, 
or \eqref{E3.2}. Next, we present the following Lemma \ref{L3.1} (see \cite{2,3,4}), however we prove it by an improved method.

 \begin{lemma}\label{L3.1} 
If $W_S([\frac{1-\varepsilon}{2},\frac{1}{2}])
=\widetilde W_{\widetilde S}([\frac{1-\varepsilon}{2},\frac{1}{2}])$,  then
\begin{gather}
q(x)-\widetilde q(x)=2\widehat{\omega}\quad \text{a.e.\ on }
 [\frac{1-\varepsilon}{2},\frac{1}{2}],\label{E3.3}\\
\begin{gathered}
\lambda_{n_k}-\widetilde\lambda_{\widetilde n_k}=2\widehat{\omega}
\quad \text{for all } n_k\in S,\\
n_k=\widetilde n_k \text{ except for a finite number of natural numbers } k.
\end{gathered} \label{E3.4}
\end{gather}
\end{lemma}

 Adding the condition \eqref{E3.7}, we establish the following  uniqueness theorem.

\begin{theorem}\label{T3.2} 
Suppose that the following two conditions are satisfied:
\begin{enumerate}
\item $W_{S}([\frac{1-\varepsilon}{2},\frac{1}{2}])
=\widetilde W_{\widetilde S}([\frac{1-\varepsilon}{2},\frac{1}{2}])$,
and
\begin{equation}\label{E3.5}
\sharp\{n_k\in S:n_k\leq n\}\geq(1-\varepsilon)n+\frac{3\varepsilon-1}{2}
\end{equation}
for sufficiently large integer $n>0$.

\item For the infinite set $\mathbb N_0\backslash S$,
 \begin{equation}\label{E3.7}
\lambda_n=\widetilde \lambda_n,\quad n\in \mathbb N_0\backslash S.
\end{equation}
\end{enumerate}
 Then
\begin{equation*}
 q(x)=\widetilde q(x)\quad \text{a.e.  on }[0,1],\quad 
h=\widetilde h\quad \text{and} \quad H=\widetilde H.
 \end{equation*}
\end{theorem}

\begin{remark} \rm
(1) For either case $(h,H)=(\infty,H)$, or $(h,\infty)$, or $(\infty,\infty)$, 
if we  modify the condition \eqref{E3.5} suitably, 
then one obtains a similar results.

(2) We obtain an analogous results with the right twin-dense nodal subset 
on the interval $[\frac{1}{2},\frac{1+\varepsilon}{2}]$ instead of the left 
twin-dense nodal subset in Theorem \ref{T3.2}.
\end{remark}

We have the following corollary from Theorem \ref{T3.2}, i.e. if 
one spectrum is given a priori, then potential $q$ on the whole interval 
$[0,1]$ can be uniquely determined by $W_{S}([\frac{1-\varepsilon}{2},\frac{1}{2}])$ 
 for any $S$ and any arbitrarily
small $\varepsilon$.

\begin{corollary}\label{C3.3} 
If one spectrum $\sigma(L)$ is given a priori, then the potential $q$ and 
coefficients $h,H$ can be uniquely determined by the left twin-dense nodal subset 
$W_{S}([\frac{1-\varepsilon}{2},\frac{1}{2}])$ for any $S$ and arbitrarily 
small $\varepsilon$.
\end{corollary}

For any $n\in\mathbb N$, let $x_n^{1}$ and $x_n^{n}$ be the minimal and 
 maximal nodal point of the corresponding eigenvalue $\lambda_n$, respectively.
 From the Sturm's oscillation theorem (see  \cite[Lemma 1.1.4, pp. 18]{2}),
 we see that if $0<x_1^{1}\leq\frac{1}{2}$, then $0<x_n^{1}\leq\frac{1}{2}$ 
for all $n>1$ and if  $\frac{1}{2}\leq x_1^{1}<1$, then $\frac{1}{2}\leq x_n^{n}<1$ 
for all $n>1$. Adding the condition $0<x_1^{1}\leq\frac{1}{2}$ and \eqref{E3.6}, 
we obtain the following uniqueness theorem.

\begin{theorem}\label{T3.5} 
If the following three conditions are satisfied:
\begin{enumerate}
\item $H=\widetilde H$ and $0<x_{1}^1\leq\frac{1}{2}$,
 \item $W_S([\frac{1-\varepsilon}{2},\frac{1}{2}])
=\widetilde W_{\widetilde S}([\frac{1-\varepsilon}{2},\frac{1}{2}])$ and \eqref{E3.5} holds.
\item For all $n\in \mathbb N\backslash S$,
 \begin{equation}\label{E3.6}
x_{n}^{1}=\widetilde x_{\widetilde n}^{1},
\end{equation}
\end{enumerate}
 then
\begin{equation*}
 q(x)-\int_0^1q(t)\mathrm{d}t=\widetilde q(x)
-\int_0^1\widetilde q(t)\mathrm{d}t\quad \text{a.e.  on }
 [0,1], \text{ and } h=\widetilde h.
 \end{equation*}
\end{theorem}

\section{Proofs of main results}

In this section, we present proofs of  our main results. 
Firstly, we  prove Lemma \ref{L3.1} by the improved method.

\begin{proof}[Proof of Lemma \ref{L3.1}]
For each fixed $x\in [\frac{1-\varepsilon}{2},\frac{1}{2}]$, we choose 
$x_{n_k}^{j_k}\in  W_{S}([\frac{1-\varepsilon}{2},\frac{1}{2}])$ such that 
$\lim_{k\to\infty}{x_{n_k}^{j_k}}=x$. From \eqref{E2.13}, we have
$$
\lim_{k\to\infty}{\frac{j_k-\frac{1}{2}}{n_k}}=x.
$$
By using the Riemann-Lebesgue lemma together with \eqref{E2.13}, we get
\begin{equation} \label{E4.1}
\begin{aligned}
f(x):=&\lim_{k\to\infty}{
\big[2(n_k\pi )^2x_{n_k}^{j_k}-2n_k\pi^2\big(j_k-\frac{1}{2}\big)\big]} \\
=&\lim_{k\to\infty}\Big[2h+\int_0^{x_{n_k}^{j_k}} q(t)\mathrm{d}t
-\frac{j_k-\frac{1}{2}}{n_k}\Big(2\omega-\int_0^1 q(t)\cos(2n_k\pi t)
 \mathrm{d}t\Big)\\
& +o(1)\Big] \\
=&\int_0^x q(t)\mathrm{d}t+2h-2\omega x,\quad
x\in [\frac{1-\varepsilon}{2},\frac{1}{2}].
\end{aligned}
\end{equation}
Since $\int_0^{x} q(t)\mathrm{d}t+2h-x\int_0^1{q(t)}\mathrm{d}t$
(a.e. on $x\in [\frac{1-\varepsilon}{2},\frac{1}{2}]$) with respect to $x$
is differentiable, $f(x)$ with respect to $x$ is also differentiable.
By taking derivatives for \eqref{E4.1}, we obtain
\begin{align*}
f'(x)=q(x)-2\omega\quad \text{a.e. on }  [\frac{1-\varepsilon}{2},\frac{1}{2}].
\end{align*}
 Since
$$
W_{S}\big([\frac{1-\varepsilon}{2},\frac{1}{2}]\big)
=\widetilde W_{\widetilde S}\big([\frac{1-\varepsilon}{2},\frac{1}{2}]\big),
$$
it follows that
$f(x)=\widetilde f(x)$ for $x\in [\frac{1-\varepsilon}{2},\frac{1}{2}]$.
Therefore
\begin{align*}
f'(x)=\widetilde f'(x)\quad \text{a.e. on }\quad
[\frac{1-\varepsilon}{2},\frac{1}{2}].
\end{align*}
This implies
\begin{align}\label{E4.2}
q(x)-\widetilde q(x)=2\widehat{\omega}\quad \text{a.e. on }
 [\frac{1-\varepsilon}{2},\frac{1}{2}].
\end{align}
Consider two Dirichlet boundary value problems defined on the interval
$[x_{n_k}^{j_k}, x_{n_k}^{j_k+1}]\subseteq[\frac{1-\varepsilon}{2},\frac{1}{2}]$,
\begin{gather}
-u''_-(x,\lambda_{n_k})+q(x)u_-(x,\lambda_{n_k})
=\lambda_{n_k} u_-(x,\lambda_{n_k}),\label{E4.3}\\
u_-(x_{n_k}^{j_k},\lambda_{n_k})=u_-(x_{n_k}^{j_k+1},\lambda_{n_k})=0,\label{E4.4}
\end{gather}
and
\begin{gather}
-\widetilde u''_-(x,\widetilde \lambda_{\widetilde n_k})
 +\widetilde q(x)\widetilde u_-(x,\widetilde \lambda_{\widetilde n_k})
 =\widetilde \lambda_{\widetilde n_k} \widetilde u_
 -(x,\widetilde \lambda_{\widetilde n_k}),\label{E4.5}\\
\widetilde u_-(x_{n_k}^{j_k},\widetilde \lambda_{\widetilde n_k})
 =\widetilde u_-(x_{n_k}^{j_k+1},\widetilde \lambda_{\widetilde n_k})=0.\label{E4.6}
\end{gather}

Multiplying  \eqref{E4.3} by $\widetilde u_-(x,\widetilde \lambda_{\widetilde n_k})$ 
and  \eqref{E4.5} by $u_-(x,\lambda_{n_k})$, subtracting and integrating it 
from $x_{n_k}^{j_k}$ to $x_{n_k}^{j_k+1}$ together \eqref{E4.4} and \eqref{E4.6},
 we have
\begin{equation} \label{E4.7}
\int_{x_{n_k}^{j_k}}^{x_{n_k}^{j_k+1}}[(q(x)-\widetilde q(x))
-(\lambda_{n_k}-\widetilde \lambda_{\widetilde n_k})]u_-(x,\lambda_{n_k})
\widetilde u_-(x,\widetilde \lambda_{\widetilde n_k})dx=0.
\end{equation}
By  \eqref{E4.7} and $q(x)-\widetilde q(x)=2\widehat{\omega}$ a.e. on
$[\frac{1-\varepsilon}{2},\frac{1}{2}]$, this yields
\begin{equation}
[2\widehat{\omega}-(\lambda_{n_k}-\widetilde \lambda_{\widetilde n_k})]
\int_{x_{n_k}^{j_k}}^{x_{n_k}^{j_k+1}}u_-(x,\lambda_{n_k})\widetilde u_
-(x,\widetilde \lambda_{\widetilde n_k})dx=0.
\end{equation}
Since both $u_-(x,\lambda_{n_k})$ and
$\widetilde u_-(x,\widetilde \lambda_{\widetilde n_k})$ have no zero in
the interval $(x_{n_k}^{j_k},x_{n_k}^{j_k+1})$, we get
\begin{align*}
u_-(x,\lambda_{n_k})\widetilde u_-(x,\widetilde \lambda_{\widetilde n_k})>0\quad
 \text{or}\quad  u_-(x,\lambda_{n_k})\widetilde u_-(x,\widetilde
 \lambda_{\widetilde n_k})<0\quad  \text{for }
 x\in (x_{n_k}^{j_k},x_{n_k}^{j_k+1}).
\end{align*}
This implies
\begin{equation}
\int_{x_{n_k}^{j_k}}^{x_{n_k}^{j_k+1}}u_-(x,\lambda_{n_k})
\widetilde u_-(x,\widetilde \lambda_{\widetilde n_k})dx\neq0.
\end{equation}
Therefore,
\begin{equation} \label{E4.10}
\lambda_{n_k}=\widetilde\lambda_{\widetilde n_k}+2\widehat{\omega},\quad \forall
 n_k\in S.
\end{equation}
By  \eqref{E4.10} and \eqref{E2.9}, for sufficiently large $k$, this yields
$n_k=\widetilde n_k$.
Thus, the proof of Lemma \ref{L3.1} is complete.
\end{proof}

Next we show that  Theorem \ref{T3.2} holds.

\begin{proof}[Proof of Theorem \ref{T3.2}]
Denote $\Lambda=\{\lambda_n:n\in S,\lambda_n\in \sigma(L)\}$ and 
$N_\Lambda(t)=\sharp\{\lambda_n:\lambda_n\in\Lambda, 
\lambda_n\leq t,\lambda_n\in \sigma(L)\}$ for all sufficiently large
 $t\in\mathbb R$. By calculating $N_\Lambda(t)$, we have
\begin{align}\label{E4.11}
N_\Lambda(t)\geq (1-\varepsilon)N_{\sigma(L)}(t)-\frac{1-\varepsilon}{2},
\end{align}
 By the assumption in Theorem \ref{T3.2}, Lemma \ref{L3.1} yields
\begin{gather}
q(x)-\widetilde q(x)=2\widehat{\omega}\quad \text{a.e. on }
 [\frac{1-\varepsilon}{2},\frac{1}{2}],\label{E4.12}\\
\lambda_{n_k}-\widetilde\lambda_{\widetilde n_k}=2\widehat{\omega},\quad 
\forall n_k\in S.\label{E4.13}
\end{gather}
Since the set $\mathbb N_0\backslash S$ is an infinite set , from
\eqref{E4.13}, \eqref{E3.7} and \eqref{E2.9}, we get
\begin{align}\label{E4.14}
\widehat{\omega}=0.
\end{align}
By \eqref{E4.12}-\eqref{E4.14}, we have
\begin{align}\label{E4.15}
q(x)-\widetilde q(x)=0\ \text{a.e. on }
 [\frac{1-\varepsilon}{2},\frac{1}{2}] \text{ and }
\lambda_{n}-{\lambda}_{\widetilde n}=0,\quad \forall\ n\in \mathbb N,
\end{align}
 Denote
\[
F(u_-,\widetilde u_-,x,\lambda)=[u_-,\widetilde u_-](x,\lambda).
\]
Let us prove
\begin{equation*}
F\big(u_-,\widetilde u_-,\frac{1-\varepsilon}{2},\lambda_{n_k}\big)=0,\quad 
\forall n_k\in S.
\end{equation*}
Indeed, since $ W_{S}([\frac{1-\varepsilon}{2},\frac{1}{2}])
=\widetilde W_{\widetilde S}([\frac{1-\varepsilon}{2},\frac{1}{2}])$
 is a left twin-dense nodal subset, we choose 
$x_{n_k}^{j_{n_k}}\in  W_{S}([\frac{1-\varepsilon}{2},\frac{1}{2}])$. 
By the Green's formula, we obtain
\begin{align}\label{E4.16}
F\big(u_-,\widetilde u_-,\frac{1-\varepsilon}{2},\lambda_{n_k}\big)
=-\int_{\frac{1-\varepsilon}{2}}^{x_{n_k}^{j_{n_k}}}{\widehat{q}(x)u_
-(x,\lambda_{{n_k}} )\widetilde u_-(x, \lambda_{n_k})\mathrm{d}x}.
\end{align}
 By \eqref{E4.16} and $\widehat{q}(x)=0$  a.e. on  
$[\frac{1-\varepsilon}{2},\frac{1}{2}]$, we get
\begin{align}\label{E4.17}
F\big(u_-,\widetilde u_-,\frac{1-\varepsilon}{2},\lambda_{n_k}\big)=0,\quad 
\forall n_k\in S.
\end{align}
Next we prove $q(x)-\widetilde q(x)=0$ a.e. on  $[0,1]$, 
$h=\widetilde h$ and $H=\widetilde H$.

Without loss of generality, we  assume that $\lambda_{n}\neq 0$ for all 
$n\in \sigma(L)$. Define the functions $G_S(\lambda)$ and $K_1(\lambda )$ by
\begin{gather}\label{E4.18}
G_S(\lambda)=\prod_{n_k\in
S}{\big(1-\frac{\lambda}{\lambda_{n_k}}\big)}, \\
\label{E4.19}
K_1(\lambda ) = \frac{F\big(u_-,\widetilde u_-,\frac{1-\varepsilon}{2},
\lambda\big)}{G_S(\lambda)}.
\end{gather}
Hence \eqref{E4.17}, \eqref{E4.18} and \eqref{E4.19} imply that $K_1(\lambda )$ 
is an entire function in $\lambda$. Note that
\begin{equation} \label{E4.20}
\begin{aligned}
&F\big(u_-,\widetilde u_-,\frac{1-\varepsilon}{2},\lambda\big) \\
&= u_-\big(\frac{1-\varepsilon}{2},\lambda\big)\widetilde u_-'
 \big(\frac{1-\varepsilon}{2},\lambda\big)-u_-'
 \big(\frac{1-\varepsilon}{2},\lambda\big)\widetilde u_-\big(\frac{1-\varepsilon}{2},\lambda\big) \\
&= u_-'\big(\frac{1-\varepsilon}{2},\lambda\big)\widetilde u_-'
 \big(\frac{1-\varepsilon}{2},\lambda\big)\Big(m_-^{-1}
\big(\frac{1-\varepsilon}{2},\lambda\big)-\widetilde m_-^{-1}
 \big(\frac{1-\varepsilon}{2},\lambda\big)\Big).
\end{aligned}
\end{equation}
From \eqref{E2.4}, \eqref{E2.11} and \eqref{E4.20}, we have
\begin{equation*}
\big|{F\big(u_-,\widetilde u_-,\frac{1-\varepsilon}{2},\lambda\big)}\big|
=o\big(\mathrm{e}^{(1-\varepsilon)\tau}\big)
\end{equation*}
 as $|\lambda|\to\infty$ in any sector
$\varepsilon<\arg(\lambda)<\pi-\varepsilon$. This implies
\begin{equation}\label{E4.21}
\big|F\big(u_-,\widetilde u_-,\frac{1-\varepsilon}{2},iy\big)\big|
=o\big(\mathrm{e}^{(1-\varepsilon)\mathrm{Im}{\sqrt{i}}|y|^{1/2}}\big)
\end{equation}
 for sufficiently large $y\in\mathbb R$.   We analogously calculate
$G_S(iy)$ from \eqref{E4.11} and  get the following formula (see \cite{19})
\begin{equation*}
|G_S(iy)|\geq c|y|^{1/2}\mathrm{e}^{(1-\varepsilon)\mathrm{Im}{\sqrt{i}}|y|^{1/2}},
\end{equation*}
where $c$ is a constant. Therefore
\begin{equation}\label{E4.22}
 |K_1(iy)| = o\big(|y|^{-1/2}\big).
\end{equation}
It is easy to prove the following formula (see \cite{19}):
\begin{equation}\label{E4.23}
\sup_{|z|=R_k}|K_1(z)|\leq C_1\exp(C_2R_k^\alpha)
\end{equation}
for some $0<\alpha<1$, some sequence $R_k\to\infty$ as
$k\to \infty$ and $C_1,C_2>0$.

By Lemma \ref{L2.3}, \eqref{E4.22} and \eqref{E4.23}, we have
$ K_1(\lambda)=0$ for all $\lambda\in \mathbb C$.
Therefore,
\begin{equation}
 F\big(u_-,\widetilde u_-,\frac{1-\varepsilon}{2},\lambda\big)=0,\quad\forall 
\lambda\in \mathbb C.
\end{equation}
This implies
\begin{equation}\label{E4.25}
 m_-\big(\frac{1-\varepsilon}{2},\lambda\big)
=\widetilde m_-\big(\frac{1-\varepsilon}{2},\lambda\big),\quad\forall 
\lambda\in \mathbb C.
\end{equation}
From Lemma \ref{L2.2} and \eqref{E4.25}, we obtain
\begin{equation*}
q(x)-\widetilde q(x)=0\quad  \text{a.e. on }[0,\frac{1-\varepsilon}{2}]
\quad\text{and }  h=\widetilde h.
\end{equation*}
Therefore,
\begin{equation} \label{E4.26}
q(x)-\widetilde q(x)=0\quad \text{a.e. on } [0,\frac{1}{2}],\quad
h=\widetilde h,\quad  \lambda_n=\lambda_n,\quad  n\in \mathbb N_0.
\end{equation}
By the  Hochstadt-Lieberman theorem \cite{17} and  \eqref{E4.26}, we get
\begin{equation*}
q(x)-\widetilde q(x)=0\quad \text{a.e. on } [0,1],\quad \text{and}\quad
H=\widetilde H.
\end{equation*}
Thus the proof of Theorem \ref{T3.2} is  complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{T3.5}]
 From Lemma \ref{L3.1}, we have
\begin{gather}
q(x)-\widetilde q(x)=2\widehat{\omega}\quad \text{a.e. on }
 [\frac{1-\varepsilon}{2},\frac{1}{2}] \\
\lambda_{n_k}-\widetilde\lambda_{\widetilde n_k}=2\widehat{\omega},\quad
 \forall n_k\in S.
\end{gather}
Define the potential $\widetilde q_1(x)$ by
$\widetilde q_1(x)=\widetilde q(x)+2\widehat{\omega}$.
This implies
\begin{equation} \label{E4.27}
q(x)-\widetilde q_1(x)=0\quad \text{a.e. on } [\frac{1-\varepsilon}{2},\frac{1}{2}]
\quad \text{and}\quad
\lambda_{n_k}-\widetilde{\lambda}_{1,\widetilde n_k}=0,\quad \forall n_k\in S,
\end{equation}
where $\widetilde{\lambda}_{1,\widetilde n_k}
=\widetilde\lambda_{\widetilde n_k}+2\widehat{\omega}$,
which is the eigenvalue of equation \eqref{E1.1} corresponding to
$\widetilde q_1$ with boundary conditions \eqref{E1.2} and \eqref{E1.3}.
Analogous to the proof in Theorem \ref{T3.2}, we have
\begin{equation}\label{E4.28}
q(x)-\widetilde q_1(x)=0\quad  \text{a.e. on } [0,\frac{1}{2}],\quad \text{and}\quad
 h=\widetilde h.
\end{equation}
Next, we  prove $\lambda_n=\widetilde{\lambda}_{1,\widetilde n_k}$,
$n\geq 1$. From the assumption
of Theorem \ref{T3.5}, there exists the nodal point $x_{n_k}^{1}$ of the
corresponding  eigenvalue $\lambda_{n_k}$ such that
\begin{equation*}
x_{n_k}^{1}=\widetilde x_{\widetilde n_k}^{1},\quad
\forall n_k\in \mathbb N\backslash S,\quad 0<x_{n_k}^1\leq\frac{1}{2}.
\end{equation*}
Let us consider two  boundary value problems defined on the interval
$[0,x_{n_k}^{1}]$,
\begin{gather}
-u''_-(x,\lambda_{n_k})+q(x)u_-(x,\lambda_{n_k})=\lambda_{n_k}
u_-(x,\lambda_{n_k}),\quad x\in (0,x_{n_k}^{1})\label{E4.29}\\
u'_-(0,\lambda_{n_k})-hu_-(0,\lambda_{n_k})=u_-(x_{n_k}^{1},\lambda_{n_k})=0,
\label{E4.30}
\end{gather}
and
\begin{gather}
-\widetilde u''_-(x,\widetilde{\lambda}_{1,\widetilde n_k})
+\widetilde q_1(x)\widetilde u_-(x,\widetilde{\lambda}_{1,\widetilde n_k})
=\overline\lambda_{\widetilde n_k} \widetilde u_-
 (x,\widetilde{\lambda}_{1,\widetilde n_k}),\quad x\in (0,x_{n_k}^{1})\label{E4.31}\\
\widetilde u'_-(0,\widetilde{\lambda}_{1,\widetilde n_k})-hu_-
 (0,\overline\lambda_{\widetilde n_k})=\widetilde u_-
 (x_{n_k}^{1},\widetilde{\lambda}_{1,\widetilde n_k})=0.\label{E4.32}
\end{gather}
Multiplying equation \eqref{E4.29} by
$\widetilde u_-(x,\overline\lambda_{\widetilde n_k})$ and equation
\eqref{E4.31} by $u_-(x,\lambda_{n_k})$, subtracting and integrating
it from $0$ to $x_{n_k}^{1}$ together with \eqref{E4.30} and \eqref{E4.32}, we have
\begin{align}\label{E4.33}
\int_0^{x_{n_k}^{1}}[(q(x)-\widetilde q_1(x))
-(\lambda_{n_k}-\widetilde{\lambda}_{1,\widetilde n_k})]u_-(x,\lambda_{n_k})
\widetilde u_-(x,\widetilde{\lambda}_{1,\widetilde n_k})dx=0.
\end{align}
By \eqref{E4.33} and $q(x)-\widetilde q_1(x)=0$ a.e. on the interval
$[0,\frac{1}{2}]$, this yields
\begin{align}\label{E4.34}
(\lambda_{n_k}-\widetilde{\lambda}_{1,\widetilde n_k})\int_0^{x_{n_k}^{1}}
u_-(x,\lambda_{n_k})\widetilde u_-(x,\widetilde{\lambda}_{1,\widetilde n_k})dx=0.
\end{align}
Since both $u_-(x,\lambda_{n_k})$ and
$\widetilde u_-(x,\widetilde{\lambda}_{1,\widetilde n_k})$ have no zero in the
interval $(0,x_{n_k}^{1})$, we get
\begin{align*}
u_-(x,\lambda_{n_k})\widetilde u_-(x,\widetilde{\lambda}_{1,\widetilde n_k})>0\quad
 \text{for } x\in (0,x_{n_k}^{1}).
\end{align*}
This implies
\begin{equation} \label{E4.35}
\int_0^{x_{n_k}^{1}}u_-(x,\lambda_{n_k})\widetilde
 u_-(x,\widetilde{\lambda}_{1,\widetilde n_k})dx>0.
\end{equation}
By \eqref{E4.34} and \eqref{E4.35}, this yields
$\lambda_{n_k}=\widetilde{\lambda}_{1,\widetilde n_k}$ for all
$n_k\in \mathbb N\backslash S$.
Thus we obtain
\begin{align}\label{E4.36}
\lambda_n=\widetilde{\lambda}_{1,\widetilde n_k},\quad n=1,2,\dots.
\end{align}
By \cite[Theorem 2.1]{20}, or the related Theorem in \cite[Section 4]{21}
together with \eqref{E4.28}, \eqref{E4.36}, and given coefficients $H=\widetilde H$,
we have
\begin{equation*}
q(x)-\widetilde q_1(x)=0\quad  \text{a.e. on } [\frac{1}{2},1].
\end{equation*}
Therefore,
\begin{equation*}
q(x)-\widetilde q_1(x)=0\quad \text{a.e. on } [0,1],\quad\text{and}\quad
h=\widetilde h.
\end{equation*}
This completes the proof of Theorem \ref{T3.5}.
\end{proof}

In the remainder of this section, we present an example for reconstructing 
the potential $q$ from the  twin-dense nodal subset. 
Let $\varepsilon=1/4$ and
\begin{equation} \label{E4.42}
S_0:=\{2n: n\geq 10,n\in\mathbb N\}\cup\{2k_i-1:2k_i-1>10,k_i\in\mathbb N\}_{i=1}^{10}.
\end{equation}

\begin{example} \rm
Let  $W_{S_0}([\frac{1}{4},\frac{1}{2}])
=\widetilde{W}_{\widetilde{S}_0}([\frac{1}{4},\frac{1}{2}])\subseteq X
=\{x_n^j\}$, $n\in \mathbb N$, $j=1,2,\dots,n$, be the left twin-dense nodal 
subset of the operator $L(q,h,1)$, where
\begin{equation}\label{E4.43}
x_n^j=\frac{j-\frac{1}{2}}{n}+\frac{1}{2n^2\pi^2}
\Big(2+\big(\frac{j-\frac{1}{2}}{n}\big)^2\Big)
-\frac{5(j-\frac{1}{2})}{2n^3\pi^2}+o\big(\frac{1}{n^2}\big),\quad \forall n\in S_0
\end{equation}
and
\begin{equation} \label{E4.44}
x_n^1=\frac{1}{2n}+\frac{1}{2n^2\pi^2}
\big(2+\frac{1}{4n^2}\big)-\frac{5}{4n^3\pi^2}+o\big(\frac{1}{n^2}\big)<\frac{1}{2}
\end{equation}
for all $n\in \mathbb N\backslash S_0$. By  \eqref{E4.1} together with
\eqref{E4.43}, we have
\begin{equation} \label{E4.45}
\begin{aligned}
f_1(x):=&\lim_{k\to\infty}{\big[2(n_k\pi )^2x_{n_k}^{j_k}-2n_k\pi^2
\big(j_k-\frac{1}{2}\big)\big]} \\
=&x^2+2-5x,\quad x\in \big[\frac{1}{4},\frac{1}{2}\big].
\end{aligned}
\end{equation}
By \eqref{E4.1} and \eqref{E4.45} again, this yields
\begin{align}\label{E4.46}
h=1\quad \text{and}\quad\omega=\frac{5}{2}.
\end{align}
By the given condition $H=1$ and \eqref{E4.46}, we get
\begin{align}\label{E4.47}
\int_0^1q(t)\mathrm{d}t=1.
\end{align}
By taking derivatives for \eqref{E4.45} together with \eqref{E4.47}, we obtain
\begin{align}\label{E4.48}
q(x)=2x\quad  \text{a.e. on } [\frac{1}{4},\frac{1}{2}].
\end{align}
 By \eqref{E4.42}-\eqref{E4.44}, \eqref{E4.48} and
 $W_{S_0}([\frac{1}{4},\frac{1}{2}])
=\widetilde{W}_{\widetilde{S}_0}([\frac{1}{4},\frac{1}{2}])$,
we see that all assumptions in Theorem \ref{T3.5} hold. Thus we have
\begin{equation*}
 q(x)=2x\quad \text{a.e. on }  [0,1],\quad \text{and}\quad h=1.
 \end{equation*}
\end{example}

\subsection*{Acknowledgements} 
The author would like to thank the anonymous referees  for valuable suggestions
which help to improve
the readability and quality of this article.

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\end{document}