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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 155, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/155\hfil Inverse problem for the Dirac operator]
{A uniqueness theorem on the inverse problem for the Dirac operator}

\author[Y. P. Wang, M. Sat \hfil EJDE-2016/155\hfilneg]
{Yu Ping Wang, Murat Sat}

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

\address{Murat Sat \newline
Department of Mathematics,
Faculty of Science and Art,
Erzincan University,
Erzincan, 24100, Turkey}
\email{murat\_sat24@hotmail.com}

\thanks{Submitted March 12, 2016. Published June 21, 2016.}
\subjclass[2010]{34A55, 34B24, 34L05, 45C05}
\keywords{ Inverse problem; uniqueness theorem; eigenvalue;
 Dirac operator}

\begin{abstract}
In this article, we consider an inverse problem for the Dirac operator.
 We show that a particular set of eigenvalues is sufficient to
 determine the unknown potential functions.
\end{abstract}

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


\section{Introduction}

The inverse spectral problem for a differential operator consists of recovering
the operator from its spectral data. In 1929, Ambarzumyan \cite{a2}
was the first to discuss the following statement 
\begin{quote}
If $q\in C[ 0,\pi ] $ and $\{ n^2:n=0,1,2,\dots \} $
is the spectral set of the boundary value problem
\begin{equation}
-y''+q(x)y=\lambda y,\quad x\in [ 0,\pi ] , \label{1}
\end{equation}
with Neumann boundary conditions
\begin{equation}
y'(0,\lambda )=y'(\pi ,\lambda )=0,  \label{2}
\end{equation}
then $q(x)\equiv 0$ in $[0,\pi ]$.
\end{quote}
McLaughlin and Rundell \cite{m3} discussed the inverse problem for the
Sturm-Liouville equation \eqref{1} with the boundary conditions 
$y(0,\lambda )=0$ and $y'(\pi ,\lambda )+H_ky(\pi ,\lambda )=0$ and showed
that the spectral data, for a fixed  $n$ $(n=0,1,2,\dots )$, 
$\{\lambda _n(q,H_k)\} _{k=1}^{+\infty }$ is equivalent to two
spectra of boundary value problems with the equation \eqref{1} and one
common boundary condition at $x=0$ and two different boundary conditions at 
$x=\pi$. By using McLaughlin and Rundell's method \cite{m3}, Koyunbakan
\cite{k2} considered a singular Sturm-Liouville problem. Using the
spectral data in \cite{m3} and Hochstadt and Lieberman's method, Wang 
\cite{w1} discussed the inverse problem for indefinite Sturm-Liouville
operators on the finite interval $[ a,b] $. However, we are
motivated by inverse spectral problems for Dirac operators with the above
spectral data which are particular set of eigenvalues. As far as we know,
inverse spectral problems for Dirac operators have not been considered with
the spectral data before.

We consider the system of Dirac operators $L:=L(p,q,H_k) $
\begin{equation}
Ly=By'+Q(x)y=\lambda y,\quad 0\leq x\leq \pi ,  \label{3}
\end{equation}
with the boundary conditions
\begin{gather}
y_1(0,\lambda ) = 0,  \label{4} \\
y_2(\pi ,\lambda )+H_ky_1(\pi ,\lambda ) =0,  \label{5}
\end{gather}
where 
\[
B=\begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix}, \quad 
Q(x)=\begin{pmatrix}
p(x) & q(x) \\
q(x) & -p(x)
\end{pmatrix}, \quad
y(x)=\begin{pmatrix}
y_1(x) \\
y_2(x)
\end{pmatrix}
\]
 and $(p(x),q(x)) $ are potential functions which are
real valued functions in space $L^2[ 0,\pi ] $.

We consider another Dirac operator 
$\widetilde{L}:=\widetilde{L}(
\widetilde{p},\widetilde{q},H_k) $ which is defined as:
\begin{equation}
\widetilde{L}y=B\widetilde{y}'+\widetilde{Q}(x)\widetilde{y}
=\lambda \widetilde{y},\quad 0\leq x\leq \pi ,  \label{6}
\end{equation}
with the boundary conditions
\begin{gather}
\widetilde{y}_1(0,\lambda ) = 0,  \label{7} \\
\widetilde{y}_2(\pi ,\lambda )+H_k\widetilde{y}_1(\pi ,\lambda ) = 0,
\label{8}
\end{gather}
where 
\[
\widetilde{Q}(x)=\begin{pmatrix}
\widetilde{p}(x) & \widetilde{q}(x) \\
\widetilde{q}(x) & -\widetilde{p}(x)
\end{pmatrix},
\]
$H_k\in\mathbb{R}$, $0<H_1<H_2<\dots <H_k<H_{k+1}<\dots $,
the potentials $(\widetilde{p}(x),\widetilde{q}(x))$ are real valued functions, 
$(\widetilde{p}(x),\widetilde{q}(x) ) \in L^2[0,\pi ] $ and 
$\lambda $ is a spectral parameter.

The Dirac equation is a modern presentation of the relativistic quantum
mechanics of electrons intended new mathematical outcomes accessible to a
wider audience. It treats in some dept the relativistic invariance of a
quantum theory, self-adjointness and spectral theory, qualitative features
of relativistic bound and scattering states and the external field problem
in quantum electrodynamics, without neglecting the interpretational
difficulties and limitations of the theory.

Inverse problems for Dirac system were studied by Moses \cite{m4}, Prats
and Toll \cite{p3}, Verde \cite{v1}, Gasymov and Levitan \cite{g1}
and Panakhov \cite{p1,p2}. It is well known \cite{g2} that two spectra
uniquely determine the matrix-valued potential function. In particular, in
reference \cite{j1}, eigenfunction expansions for one dimensional Dirac
operators describing the motion of a particle in quantum mechanics are
discussed.

Direct or inverse spectral problem for Dirac and Sturm-Liouville operators
were extensively studied in \cite{a1,b1,b2,b3,g3,k1,m1,m4,n1,o1,p4,s1,w1}.
However, the results on direct or inverse spectral problems of the Dirac operator 
are less than classical Sturm-Liouville operator, this leads to additional
difficulties in connection with inverse spectral problem by the spectral
data in \cite{m3}.

In this article, by using the spectral data in \cite{m3} and Hochstadt and
Lieberman's \cite{h1} method a uniqueness theorem for Dirac operator on
the interval $[ 0,\pi ] $ will be established, i.e., for a fixed
index $n$ $(n=0,\pm 1,\pm 2,\dots )$, we show that if the spectral set 
$\{ \lambda _n(p,q,H_k)\} _{k=1}^{+\infty }$ for distinct
$H_k$ can be measured, then the spectral set 
$\{ \lambda_n(p,q,H_k)\} _{k=1}^{+\infty }$ is sufficient to determine the
potential functions $(p(x),q(x)) $.

\begin{lemma}[\cite{l1}] \label{lem1.1}
 Let the function $\varphi
(x,\lambda )=\begin{pmatrix}
\varphi _1(x,\lambda ) \\
\varphi _2(x,\lambda )
\end{pmatrix}$ be the solution of  \eqref{3} satisfying the initial
condition
\begin{equation*}
\varphi (0,\lambda )=\begin{pmatrix}
\varphi _1(0,\lambda ) \\
\varphi _2(0,\lambda )
\end{pmatrix}
=\begin{pmatrix}
0 \\
-1
\end{pmatrix}
=(0,-1) ^{T}\,.
\end{equation*}
Then $\varphi (x,\lambda )$ satisfies the integral equations
\begin{equation}
\varphi (x,\lambda )=\begin{pmatrix}
\sin \lambda x \\
-\cos \lambda x
\end{pmatrix}  +\int_0^x K(x,t) \begin{pmatrix}
\sin \lambda t \\
-\cos \lambda t
\end{pmatrix} dt,  \label{10}
\end{equation}
where kernel $K(x,t) $  is symmetric
matrix-valued functions whose entries are continuously differentiable in
both of its variables.
\end{lemma}


\begin{lemma} \label{lem1.2} 
The eigenvalues $\lambda _n$ $(n\neq 0) $ of the
boundary-value problem \eqref{3}-\eqref{5} for the coefficient $H=H_k$ in 
\eqref{5} are the roots of \eqref{5} and satisfy the asymptotic formulae:
\begin{equation}
\lambda _n=\lambda _n^{0}+\epsilon _n,  \label{9}
\end{equation}
where $\{\epsilon _n\}\in l_2$
$(l_2$ consist of sequences $\{x_n\}$ such that 
$\sum_{n=1}^{\infty }|x_n| ^2<\infty$) 
and $\lambda _n^{0}$ are the zeros
of $\Delta _0(\lambda ):=-\cos \lambda \pi +H\sin \lambda \pi $, i.e.,
\begin{equation*}
\lambda _n^{0}=n+\frac{1}{\pi }\arctan \frac{1}{H}.
\end{equation*}
\end{lemma}

\begin{proof}
 Let  $\varphi (x,\lambda)=
\begin{pmatrix} \varphi _1(x,\lambda ) \\
\varphi _2(x,\lambda )
\end{pmatrix}$ be the solution of \eqref{3} satisfying the initial
condition
\begin{equation*}
\varphi (0,\lambda )=\begin{pmatrix}
\varphi _1(0,\lambda ) \\
\varphi _2(0,\lambda )
\end{pmatrix}
=\begin{pmatrix}
0 \\
-1
\end{pmatrix},
\end{equation*}
and $\varphi _1(0,\lambda )=0,\varphi _2(\pi ,\lambda )+H\varphi
_1(\pi ,\lambda )=0$. The characteristic function $\Delta (\lambda )$ of
the problem $L$ is defined by the following relation:
\begin{equation*}
\Delta (\lambda )=\varphi _2(\pi ,\lambda ) +H\varphi_1(\pi ,\lambda ) ,
\end{equation*}
and the zeros of $\Delta (\lambda )$ coincide with the eigenvalues of the
problem  $L$.

Using Lemma \ref{lem1.1}, we obtain
\begin{align*}
\Delta (\lambda ) 
&=-\cos \lambda \pi +H\sin \lambda \pi
+\int_0^{\pi }(K_{21}(\pi ,t) +HK_{11}(\pi,t) ) \sin \lambda tdt \\
&\quad -\int_0^{\pi }(K_{22}(\pi ,t) +HK_{12}(\pi ,t) ) \cos \lambda tdt.
\end{align*}
Since the eigenvalues are zeros of $\Delta (\lambda )$, we can write the
 equation
\begin{align*}
&-\cos \lambda \pi +H\sin \lambda \pi +\int_0^{\pi }
(K_{21}(\pi ,t) +HK_{11}(\pi ,t) ) \sin \lambda t\,dt\\
&-\int_0^{\pi }(K_{22}(\pi ,t) +HK_{12}(\pi ,t) ) \cos \lambda tdt=0.
\end{align*}
Denote
\begin{gather*}
G_n=\{ \lambda \in \mathbb{C}:| \lambda | =| \lambda _n^{0}|
+\beta ,\; n=0,\pm 1,\pm 2,\dots \} , \\
G_{\delta }=\{ \lambda :| \lambda -\lambda _n^{0}|
\geq \delta ,\; n=0,\pm 1,\pm 2,\dots\} ,
\end{gather*}
where $\delta $ is sufficiently small number $(\delta \ll \beta) $.

Since $| \Delta _0(\lambda )| >C_{\delta }\exp (| \tau | \pi ) $ 
for $\lambda \in \overline{G}_{\delta}$, from  \cite{m2},
\begin{align*}
&\lim_{| \lambda | \to \infty }e^{-|
\tau | \pi }(\Delta (\lambda )-\Delta _0(\lambda ))
\\
&=\lim_{| \lambda | \to \infty }
\Big(e^{-| \tau | \pi }\int_0^{\pi }(
K_{21}(\pi ,t) +HK_{11}(\pi ,t) ) \sin \lambda t\,dt \\
&\quad -e^{-| \tau | \pi }\int_0^{\pi }( K_{22}(\pi ,t) 
 +HK_{12}(\pi ,t) ) \cos \lambda t\,dt\Big) 
=0
\end{align*}
and $| \Delta (\lambda )-\Delta _0(\lambda )|
<C_{\delta }\exp (| \tau | \pi ) $ for
sufficiently large $n$ and $\lambda \in G_n$, we have
\begin{equation*}
| \Delta _0(\lambda )| >| \Delta (\lambda)-\Delta _0(\lambda )| ,
\end{equation*}
where $\tau =\operatorname{Im}\lambda $.

Using the Rouch\'{e}'s theorem, we conclude that, for sufficiently large $n$,
the functions $\Delta _0(\lambda )$ and 
$\Delta _0(\lambda )+\{\Delta (\lambda )-\Delta _0(\lambda )\} =\Delta (\lambda )$
have the same number of zeros inside the contour $G_n$, namely $2n+1$ zeros 
$\lambda _{-n},\dots ,\lambda _0,\dots ,\lambda _n$. Thus the eigenvalues 
$\lambda _n$ are of the form $\lambda _n=\lambda _n^{0}+\epsilon _n$,
where
\begin{equation*}
\lim_{n\to \infty }\epsilon _n=0.
\end{equation*}
Substituting $\lambda _n^{0}+\epsilon _n$ for $\lambda _n$ in the last
equality and using the fact that 
$\Delta _0(\lambda _n^{0}+\epsilon_n)=\Delta _0'(\lambda _n^{0})[ 1+o(1)]
\epsilon _n$. We conclude that $\epsilon _n\in l_2$. 
The proof is complete.
\end{proof}

\section{Main results and proofs}

\begin{lemma} \label{lem123} 
Let $\sigma (L_{k_j}) :=\{ \lambda _n(p,q,H_{k_j})\} $
$(j=1,2) $ be the spectrum of the
boundary value problem \eqref{3}-\eqref{5} for the 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{11}
\end{equation}
where $k_j\in \mathbb{N}$, $\emptyset $ denotes an empty set.
\end{lemma}

\begin{lemma}\label{lem234} 
Let $\lambda _n(p,q,H_k)$ be the $n$-th eigenvalue of the
boundary-value problem \eqref{3}-\eqref{5}. Then the spectral set 
$\{\lambda _n(p,q,H_k)\} _{k=1}^{+\infty }$ is a bounded infinite
set.
\end{lemma}

The above lemema plays an important role in the proof of next theorem.

\begin{theorem} \label{thm456} 
Let $\lambda _n(p,q,H_k)$ be the $n$-th eigenvalue of the
boundary-value problem \eqref{3}-\eqref{5} and 
$\lambda _n(\widetilde{p}, \widetilde{q},H_k)$ be the $n$-th eigenvalue 
of the boundary-value problem \eqref{6}-\eqref{8}, for a fixed index 
$n(n\in \mathbb{Z}) $. If
\[
\lambda _n(p,q,H_k)=\lambda _n(\widetilde{p},\widetilde{q},H_k)\quad
\text{for all } k\in \mathbb{N},
\]
then
\[
(p(x),q(x)) =(\widetilde{p}(x),\widetilde{q}(x)) \quad\text{a.e. on } 
[ 0,\pi ].
\]
\end{theorem}


\begin{proof}[Proof of Lemma \ref{lem123}]
Suppose that the conclusion is false.
Denote $\lambda _{n_j}(H_{k_j})=\lambda_{n_j}(p,q,H_{k_j})$, $j=1,2$. 
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}) $ 
$n_j\in \mathbb{Z}$. Let $\varphi _j(x,\lambda _{n_j}(H_{k_j})) $ be the
solution of \eqref{3}-\eqref{5} corresponding to the eigenvalue
$\lambda _{n_j}(H_{k_j})$ and that it satisfies the initial conditions 
$\varphi _{j,1}(0,\lambda _{n_j}(H_{k_j})) =0$ where 
$\varphi _j=(\varphi _{j,1},\varphi _{j,2}) ^{T}$. We get
\begin{equation}
B\varphi _{_1}'(x,\lambda _{n_1}(H_{k_1})
) +Q(x)\varphi _{_1}(x,\lambda _{n_1}(H_{k_1})
) =\lambda _{n_1}(H_{k_1})\varphi _{_1}(x,\lambda
_{n_1}(H_{k_1}) ) ,  \label{12}
\end{equation}
and
\begin{equation}
B\varphi _2'(x,\lambda _{n_2}(H_{k_2})
) +Q(x)\varphi _2(x,\lambda _{n_2}(H_{k_2})
) =\lambda _{n_2}(H_{k_2})\varphi _2(x,\lambda
_{n_2}(H_{k_2}) ) .  \label{13}
\end{equation}
If we multiply \eqref{12} by 
$\varphi _2(x,\lambda _{n_2}( H_{k_2}) ) $, and \eqref{13} by 
$\varphi _{_1}(x,\lambda _{n_1}(H_{k_1}) ) $ respectively 
(in the sense of scalar product i.e. 
\[
\langle (\varphi
_{1,1},\varphi _{1,2}) ^{T},(\varphi
_{2,1},\varphi _{2,2}) ^{T}\rangle =\varphi
_{1,1}\varphi _{2,1}+\varphi _{1,2}\varphi _{2,2}
\]
and subtract from each other and integrate from $0$ to $\pi $, we obtain
\begin{equation}
\varphi _{2,2}(x,\lambda _{n_2}(H_{k_2}) )
\varphi _{1,1}(x,\lambda _{n_1}(H_{k_1}) )
-\varphi _{2,1}(x,\lambda _{n_2}(H_{k_2})
) \varphi _{1,2}(x,\lambda _{n_1}(H_{k_1})
) \mid _{x=0}^{\pi }=0.  \label{14}
\end{equation}
Using the initial conditions, we obtain
\begin{equation}
\varphi _{2,2}(\pi ,\lambda _{n_2}(H_{k_2})
) \varphi _{1,1}(\pi ,\lambda _{n_1}(
H_{k_1}) ) -\varphi _{2,1}(\pi ,\lambda
_{n_2}(H_{k_2}) ) \varphi _{1,2}(\pi
,\lambda _{n_1}(H_{k_1}) ) =0.  \label{15}
\end{equation}
On the other hand, note the equality
\begin{equation}
\begin{aligned}
&\varphi _{2,2}(\pi ,\lambda _{n_2}(H_{k_2})
) \varphi _{1,1}(\pi ,\lambda _{n_1}(
H_{k_1}) ) -\varphi _{2,1}(\pi ,\lambda
_{n_2}(H_{k_2}) ) \varphi _{1,2}(\pi
,\lambda _{n_1}(H_{k_1}) )   \\
&=\varphi _{1,1}(\pi ,\lambda _{n_1}(H_{k_1})
) [ \varphi _{2,2}(\pi ,\lambda _{n_2}(
H_{k_2}) ) +H_{k_2}\varphi _{2,1}(\pi ,\lambda
_{n_2}(H_{k_2}) ) ]   \\
&\quad -\varphi _{2,1}(\pi ,\lambda _{n_2}(H_{k_2})
) [ \varphi _{1,2}(\pi ,\lambda _{n_1}(
H_{k_1}) ) +H_{k_1}\varphi _{1,1}(\pi ,\lambda
_{n_1}(H_{k_1}) ) ]   \\
&\quad +(H_{k_1}-H_{k_2}) \varphi _{1,1}(\pi ,\lambda
_{n_1}(H_{k_1}) ) \varphi _{2,1}(\pi
,\lambda _{n_2}(H_{k_2}) )  \label{16} \\
&=(H_{k_1}-H_{k_2}) \varphi _{1,1}(\pi ,\lambda
_{n_1}(H_{k_1}) ) \varphi _{2,1}(\pi
,\lambda _{n_2}(H_{k_2}) ) .
\end{aligned}
\end{equation}

Since $H_{k_1}-H_{k_2}\neq 0$, 
if $\varphi _{_{1,1}}(\pi,\lambda _{n_1}(H_{k_1}) ) \varphi
_{_{2,1}}(\pi ,\lambda _{n_2}(H_{k_2}) )=0$, then
\begin{equation}
\varphi _{1,1}(\pi ,\lambda _{n_1}(H_{k_1})
) =0\text{ or }\varphi _{_{2,1}}(\pi ,\lambda
_{n_2}(H_{k_2}) ) =0.  \label{17}
\end{equation}
This and  \eqref{5} yield
\begin{equation}
\varphi _{1,1}(\pi ,\lambda _{n_1}(H_{k_1})
) =\varphi _{1,2}(\pi ,\lambda _{n_1}(
H_{k_1}) ) =0,  \label{18}
\end{equation}
or
\begin{equation}
\varphi _{_{2,1}}(\pi ,\lambda _{n_2}(H_{k_2})
) =\varphi _{2,2}(\pi ,\lambda _{n_2}(
H_{k_2}) ) =0.  \label{19}
\end{equation}
Then \eqref{18} and \eqref{19} yield
\begin{equation}
\varphi _{_1}(x,\lambda _{n_1}(H_{k_1}) )
\equiv 0\quad \text{or}\quad
\varphi _2(x,\lambda _{n_2}(
H_{k_2}) ) \equiv 0\quad \text{on }[ 0,\pi ] ,
\label{20}
\end{equation}
where $\varphi _{_1}=(\varphi _{1,1},\varphi_{1,2}) ^{T}$ and 
$\varphi _2=(\varphi_{2,1},\varphi _{2,2}) ^{T}$.
This is impossible. Thus, we obtain
\begin{equation}
\varphi _{2,2}(\pi ,\lambda _{n_2}(H_{k_2})
) \varphi _{1,1}(\pi ,\lambda _{n_1}(
H_{k_1}) ) -\varphi _{_{2,1}}(\pi ,\lambda
_{n_2}(H_{k_2}) ) \varphi _{1,2}(\pi
,\lambda _{n_1}(H_{k_1}) ) \neq 0.  \label{21}
\end{equation}
It is obvious that the contradiction between \eqref{15} and \eqref{21}
implies that \eqref{11} holds.  Hence the proof is complete.
\end{proof}

\begin{proof}[Proof of Lemma \ref{lem234}]
We prove the lemma by two steps.
For the problem $L_1:=L_1(q)$, $\mu _n$ is the $n$-th eiegenvalue with
boundary conditions $\varphi _1(0)=\varphi _1(\pi )=0$, for the
problem $L_2:=L_2(q,h)$, $\lambda _n$ is the $n$-th eigenvalue problem
\eqref{3}-\eqref{5} with $H=H_k$.
\smallskip

\noindent\textbf{Step 1.}
We  show that (see \cite{f1})
\begin{equation}
\mu _n<\lambda _n\leq \mu _{n+1}.  \label{az}
\end{equation}
From the Green identity, we have
\begin{align*}
&[ \varphi _2(x,\lambda ) \varphi _1(x,\mu
) -\varphi _1(x,\lambda ) \varphi _2(x,\mu
) ] \big| _{x=0}^{x=\pi }  \\
&= (\mu -\lambda ) \int_0^{\pi }[ \varphi
_1(x,\mu ) \varphi _1(x,\lambda ) +\varphi
_2(x,\mu ) \varphi _2(x,\lambda ) ] dx.
\end{align*}
Hence, we have 
\begin{align*}
&(\mu -\lambda ) \int_0^{\pi }[ \varphi
_1(x,\mu ) \varphi _1(x,\lambda ) +\varphi
_2(x,\mu ) \varphi _2(x,\lambda ) ] dx \\
&=[ \varphi _2(\pi ,\lambda ) \varphi _1(\pi
,\mu ) -\varphi _1(\pi ,\lambda ) \varphi _2(\pi
,\mu ) ] =d(\mu ) \Delta (\lambda )-d(\lambda ) \Delta (\mu ),
\end{align*}
where $d(\mu ) =\varphi _1(\pi ,\mu ) $, 
$\Delta (\lambda )=\varphi _2(\pi ,\lambda ) +H\varphi_1(\pi ,\lambda ) $.

When $\lambda \to \mu$, we obtain
\begin{equation*}
\int_0^{\pi }[ \varphi _1^2(x,\mu ) +\varphi
_2^2(x,\mu ) ] dx=d^{\cdot }(\mu )\Delta (\mu
)-d(\mu )\Delta ^{\cdot }(\mu ),
\end{equation*}
with $\Delta ^{\cdot }(\mu )=\frac{d}{d\mu }\Delta (\mu )$ and $
d^{\cdot }(\mu ) =\frac{d}{d\mu }d(\mu )$.
In particular, this yields
\begin{gather*}
\alpha _n=-\Delta ^{\cdot }(\mu _n)d(\mu _n), \\
\frac{1}{d^2(\mu )}\int_0^{\pi }[ \varphi _1^2(
x,\mu ) +\varphi _2^2(x,\mu ) ] dx=-\frac{d}{
d\mu }(\frac{\Delta (\mu )}{d(\mu )}) ,
\end{gather*}
for $-\infty <\mu <\infty$ and $d(\mu ) \neq 0$,
where $\alpha _n$ are norming constants.

Thus the function $\frac{\Delta (\mu )}{d(\mu )}$ is monotonically
decreasing on $R-\{ \lambda _n:n\in Z\} $ with
\begin{equation*}
\lim_{\mu \to \lambda _n}\frac{\Delta (\mu )}{d(\mu )}=\pm
\infty .
\end{equation*}
Consequently from the asymptotic behavior  of $\lambda _n$ and $\mu _n$, we
prove \eqref{az}.
\smallskip

\noindent\textbf{Step 2.}
We  show that the following formula holds, 
\begin{equation}
\lambda _n(H_0)<\dots <\lambda _n(H_{k+1})<\lambda _n(H_k)<\dots
.  \label{a}
\end{equation}
Let $\varphi (x,\lambda _n(H)) $ be the solution of the
boundary value problem \eqref{3}-\eqref{5} corresponding to
 the eigenvalue 
$\lambda_n(H) $ and that it satisfies the initial conditions 
$\varphi _1(0,\lambda_n(H)) =0$, $\varphi _2(0,\lambda _n(H)) =-1$ and 
$\varphi _1(0,\lambda _n(H+\Delta H)) =0$, 
$\varphi_2(0,\lambda _n(H+\Delta H)) =-1$.
We have
\begin{gather}
B\varphi '(x,\lambda _n(H) ) +Q(x)\varphi (x,\lambda _n(H) ) 
=\lambda _n(H) \varphi (x,\lambda _n(H) ) ,  \label{b} \\
\begin{aligned}
&B\varphi '(x,\lambda _n(H+\Delta H) )
+Q(x)\varphi (x,\lambda _n(H+\Delta H) ) \\ 
&=\lambda_n(H+\Delta H) \varphi (x,\lambda _n(H+\Delta H) ) , 
\end{aligned} \label{c}
\end{gather}
where $\Delta H$ is the increment of $H$. Multiplying \eqref{b} by 
$\varphi(x,\lambda _n(H+\Delta H) ) $, and multiplying 
\eqref{c} by $\varphi (x,\lambda _n(H) ) $ and
subtracting from each other and integrating from $0$ to $\pi $, we obtain,
 from the initial conditions at zero,
\begin{equation} \label{d2}
\begin{aligned}
&\Delta \lambda _n(H) \int_0^{\pi }\Big[ \varphi
_1(x,\lambda _n(H) ) \varphi _1(x,\lambda _n(H+\Delta H) )  \\
& +\varphi _2(x,\lambda _n(H) ) \varphi _2(x,\lambda _n(H+\Delta H) ) \Big] dx
   \\
&=\Delta H\varphi _1(\pi ,\lambda _n(H) )
\varphi _1(\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 known that $\varphi (x,\lambda _n(H) ) $
and $\lambda _n(H) $ are real and continuous with respect to $H$. 
Letting $\Delta H\to 0$, we have
\begin{equation}
\frac{\partial \lambda _n(H) }{\partial H}
=\frac{\varphi_1^2(\pi ,\lambda _n(H) ) }{\int_0^{\pi }
[ \varphi _1^2(x,\lambda _n(H) ) +\varphi _2^2(x,\lambda _n(H)) ] dx}>0.  \label{e}
\end{equation}
This implies that \eqref{a} holds. Therefore, from Step 1 and Step 2 we have that
the spectral set $\{ \lambda _n(p,q,H_k) \}_{k=1}^{\infty }$ is a bounded 
infinite set. The proof is complete
\end{proof}

Finally, using Lemma \ref{lem234}, the properties of entire functions and the
result of \cite{y1}, we have shown that Theorem \ref{thm456} holds.

\begin{proof}[Proof of Theorem \ref{thm456}]
By Lemma \ref{lem1.1} the solutions to Equation \eqref{3} satisfying 
$\varphi (0,\lambda )=(0,-1) ^{T}$, and solutions of  \eqref{6}
satisfying $\widetilde{\varphi }(0,\lambda )=(0,-1) ^{T}$ can be
respectively expressed in the integral forms:
\begin{gather}
\varphi (x,\lambda )=\begin{pmatrix}
\sin \lambda x \\
-\cos \lambda x
\end{pmatrix}
+\int_0^x K(x,t) \begin{pmatrix}
\sin \lambda t \\
-\cos \lambda t
\end{pmatrix} dt,  \label{1.9} 
\\
\widetilde{\varphi }(x,\lambda )
=\begin{pmatrix}
\sin \lambda x \\
-\cos \lambda x
\end{pmatrix}
 +\int_0^x \widetilde{K}(x,t) 
\begin{pmatrix}
\sin \lambda t \\
-\cos \lambda t
\end{pmatrix} dt,  \label{22}
\end{gather}
where kernels $K(x,t) $ and $\widetilde{K}(x,t) $
are symmetric matrix-valued functions whose entries are continuously
differentiable in both of its variables.

If we multiply \eqref{3} by $\widetilde{\varphi }(x,\lambda ) $
and \eqref{6} by $\varphi (x,\lambda ) $ respectively (in the
sense of scalar product in $\mathbb{R}^2$) and subtract from each other, 
then we obtain
\begin{equation}
\frac{d}{dx}\{ \varphi _1(x,\lambda )\widetilde{\varphi }
_2(x,\lambda )-\widetilde{\varphi }_1(x,\lambda )\varphi _2(x,\lambda
)\} 
=\langle [ Q(x)-\widetilde{Q}(x)] \varphi
(x,\lambda ),\widetilde{\varphi }(x,\lambda )\rangle .  \label{24}
\end{equation}
Integrating the last equality from $0$ to $\pi $ with respect to the
variable $x$, we give
\begin{equation}
\{ \varphi _1(x,\lambda )\widetilde{\varphi }_2(x,\lambda )-
\widetilde{\varphi }_1(x,\lambda )\varphi _2(x,\lambda )\} \big|
_{x=0}^{\pi }
=\int_0^{\pi }\langle [ Q(x)-\widetilde{Q}
(x)] \varphi (x,\lambda ),\widetilde{\varphi }(x,\lambda
)\rangle dx.  \label{25}
\end{equation}

Because $\varphi (x,\lambda )$ and $\widetilde{\varphi }(x,\lambda )$
satisfy the same initial conditions, it follows that
\begin{equation}
\varphi _1(0,\lambda )\widetilde{\varphi }_2(0,\lambda )-\widetilde{
\varphi }_1(0,\lambda )\varphi _2(0,\lambda )=0.  \label{26}
\end{equation}
Define
\begin{equation}
P(x)=Q(x)-\widetilde{Q}(x),\quad
p_1(x)=p(x)-\widetilde{p}(x),\quad
q_1(x)=q(x)- \widetilde{q}(x),  \label{27}
\end{equation}
and
\begin{equation}
H(\lambda ):=\int_0^{\pi }\langle P(x)\varphi (x,\lambda ),
\widetilde{\varphi }(x,\lambda )\rangle dx.  \label{28}
\end{equation}

Considering the properties of $\varphi (x,\lambda )$ and 
$\widetilde{\varphi}(x,\lambda )$, we conclude that $H(\lambda )$ 
is an entire function in $\lambda $. Because the first term of \eqref{25} 
for $\lambda =\lambda _n(p,q,H_k)$ and $x=\pi $ is zero, then
\begin{equation*}
H(\lambda _n(p,q,H_k))=0.
\end{equation*}

From Lemma \ref{lem234}, we see that the spectral set 
$\{ \lambda_n(p,q,H_k)\} _{k=1}^{+\infty }$ is a bounded infinite set. Hence,
there exists $\lambda _{n0}(p,q)\in \mathbb{R}$, such that
 $\lambda _{n0}(p,q)$ is a finite accumulation point of the
spectrum set $\{ \lambda _n(p,q,H_k)\} _{k=1}^{+\infty }$. It
is well known that the set of zeros of every entire function which is not
identically zero hasn't any finite accumulation point.
Therefore
\begin{equation*}
H(\lambda )=0,\quad \forall \lambda \in\mathbb{C}.
\end{equation*}
We can show from \eqref{28} that
\begin{equation}
\begin{aligned}
H(\lambda )
&=\int_0^{\pi }p_1(x)\Big\{ -\cos 2\lambda x
 +\int_0^x R_1(x,t)\exp (2i\lambda t)dt\\
&\quad -\int_0^x R_2(x,t)\exp (-2i\lambda t)dt\Big\} dx
+\int_0^{\pi }q_1(x)\Big\{ -\sin 2\lambda x\\
&\quad +\int_0^x R_{3}(x,t)\exp (2i\lambda t)dt
+\int_0^x R_{4}(x,t)\exp (-2i\lambda t)dt\Big\} dx
=0
\end{aligned}\label{29}
\end{equation}
where $R_{l}(x,t)$, $l=\overline{1,4}$ are piecewise-continuously
differentiable on $0\leq t\leq x\leq \pi $.
Moreover, by using  Euler's formula, \eqref{29} can be written as
\begin{equation} \label{30}
\begin{aligned}
&\int_0^{\pi }f_1(x)\Big\{ \exp (2i\lambda x)
 +\int_0^x S_{11}(x,t)\exp (2i\lambda t)dt \\
&+\int_0^x S_{12}(x,t)\exp (-2i\lambda t)dt\Big\} dx
+\int_0^{\pi }f_2(x)\{ \exp (-2i\lambda x) \\
&+\int_0^x S_{21}(x,t)\exp (2i\lambda t)dt
+\int_0^x S_{22}(x,t)\exp (-2i\lambda t)dt\} dx =0,
\end{aligned}
\end{equation}
where
\begin{equation}
f_1(x)=-\frac{1}{2i}(q_1(x)+ip_1(x)) ,\quad
f_2(x)=\frac{1}{2i}(q_1(x)-ip_1(x)) ,\quad
i=\sqrt{-1},
\label{31}
\end{equation}
and the matrix $S(x,t) =(S_{ij}(x,t) )$,
$i,j=1,2$ with entries being piecewise-con\-tinuously differentiable on
$0\leq t\leq x\leq \pi $. By changing the order of integration, \eqref{30} can be
written as
\begin{align*}
&\int_0^{\pi }\exp (2i\lambda t)
\Big[ f_1(t)+\int_{t}^{\pi }(f_1(x)S_{11}(x,t)+f_2(x)S_{21}(x,t)) dx] dt \\
&+\int_0^{\pi }\exp (-2i\lambda t)\Big[ f_2(t)+\int
_{t}^{\pi }(f_1(x)S_{12}(x,t)+f_2(x)S_{22}(x,t)) dx \Big] dt
=0,
\end{align*}
or
\begin{equation}
\int_0^{\pi }\big\langle e_0(\lambda t),\text{ }
f(t)+\int_{t}^{\pi }S(x,t)f(x)dx\big\rangle dt=0.  \label{32}
\end{equation}
Here $e_0(\lambda t)=(\exp (2i\lambda t),\exp (-2i\lambda t)) $
$^{T}$ and $f(x)=(f_1(x),\text{ }f_2(x)) $ $^{T}$. Thus from
the completeness of the functions $e_0(\lambda t)$, it follows that
\begin{equation*}
f(t)+\int_{t}^{\pi }S(x,t)f(x)dx=0,\quad \text{for }0<t<\pi .
\end{equation*}
But this equation is a homogeneous Volterra integral equation and has only
the zero solution. Thus we have
 $f(x)=(f_1(x)$, $f_2(x))^{T}=0$ on the interval $[ 0,\pi ] $.
From \eqref{31}), it holds
\begin{equation*}
q_1(x)+ip_1(x)=0=q_1(x)-ip_1(x),
\end{equation*}
i.e. $q_1(x)=0$  and $p_1(x)=0$.
From \eqref{27} we obtain
\begin{equation*}
(p(x),q(x)) =(\widetilde{p}(x),\widetilde{q}(x)) ,
\end{equation*}
a.e. on $[ 0,\pi ] $. This result completes the proof.
\end{proof}

\subsection*{Acknowledgements}
The authors would like to thank the anonymous referees for their careful 
reading and valuable comments in improving the original manuscript.

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\end{document}