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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 130, pp. 1--19.\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-2017/130\hfil Spectral function ]
{Spectral function for a nonsymmetric differential operator on the half line}

\author[W. Ning \hfil EJDE-2017/130\hfilneg]
{Wuqing Ning}

\address{Wuqing Ning \newline
School of Mathematical Sciences,
University of Science and Technology of China,
 Hefei 230026, China}
\email{wqning@ustc.edu.cn}

\dedicatory{Communicated by Jerome A. Goldstein}

\thanks{Submitted May 2, 2017. Published May 11, 2017.}
\subjclass[2010]{34L10, 47E05}
\keywords{Nonsymmetric first order differential operator; spectral function;
\hfill\break\indent expansion theorem}

\begin{abstract}
 In this article we study the spectral function for a nonsymmetric
 differential operator on the half line. Two cases of the coefficient matrix
 are considered, and for each case we prove by Marchenko's method that,
 to the boundary value problem, there corresponds a spectral function related
 to which a Marchenko-Parseval equality and an expansion formula are established.
 Our results extend the classical spectral theory for self-adjoint Sturm-Liouville
 operators and Dirac operators.
\end{abstract}

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

\section{Introduction}

  As a very essential mathematical problem, the  Weyl-Stone eigenfunction expansion
\cite{stone,weyl} in which the key role is the spectral function for  singular
self-adjoint second order linear differential operators, has been
studied deeply by many renowned mathematicians:  Kodaira \cite{kodaira},
Levinson  \cite{levinson},  Levitan \cite{lev50},   Titchmarsh \cite{titchmarsh},
Yosida \cite{yosida50}  and others.
As for spectral function, one of the well-known classical results is about
the Sturm-Liouville problem on the half line:
  $$
-y''+q(x)y=\lambda y,\quad  x>0;\quad  y(0)=1,\quad  y'(0)=h.
  $$
  Let $y=y(x,\lambda)$ be the solution to the Sturm-Liouville problem.
Then there exists  a spectral function $\rho(\lambda)$ (see e.g. \cite{gelf})
such that, for all real $f\in{L^2(0,\infty)}$, it holds that
$$
\int_0^{\infty} f^2(x){\rm d}x=\int_{-\infty}^{\infty}
\Big[\lim_{n\to \infty}\int_0^n f(x)y(x,\lambda){\rm d}x\Big]^2{\rm d}\rho(\lambda).
$$
 The above equality can be derived as a limiting case of the classical
Sturm-Liouville expansion theorem for the regular operators
(see e.g. \cite{gelf,lev91}), where the Parseval equality for the regular
operators plays a very important role for the proofs. Similar ideas
can also be applied to singular  self-adjoint first order systems,
for example, the Dirac operators \cite{lesch,lev91}.
For general theory of eigenfunction  expansion for self-adjoint and regular
non-self-adjoint operators in Hilbert space,
 we refer to \cite{berez,kotani,me}. For multidimensional cases,
see, e.g., \cite{ike}. Moreover, a two-fold spectral expansion in terms of
principal functions of a Schr\"{o}dinger operator has been derived in \cite{bck}.
Recently,  Kirsten and  Loya \cite{kl}
obtained some interesting results on the spectral zeta function
for a Schr\"{o}dinger operator on the half line.

 However, to the author's knowledge, for singular
 nonsymmetric differential operators  there are only a few results
on eigenfunction expansion. The limiting approach for self-adjoint case
can not be applied even for very simple case of nonsymmetric
differential operators, since in general the corresponding regular spectrum
has irregular behavior on the complex plane.  To
 extend expansion theory to general case,  Marchenko \cite{mar63,mar}
established an excellent method in
dealing with the singular Sturm-Liouville operator with complex-valued
potential. In this paper, inspired by the idea of  Marchenko, we are going
to establish expansion theorem in two cases for a singular nonsymmetric
 differential operator, where the key is to prove the existence of the
corresponding spectral function. Our results can be extended to $2n\times 2n$
systems, and for simplicity we here will
only consider the case of $n=1$. For the regular case of this nonsymmetric
differential operator, recently  we have obtained some results on inverse
spectral problems with applications to inverse problems
for one-dimensional hyperbolic systems, see \cite{n06}--\cite{ny08}.
It is well known that for many differential operators there are intrinsic
relations between their spectral functions and  the corresponding Weyl
functions (often called $m$-functions), and for the recent interesting results on Weyl functions see,
e.g.,  \cite{frei01,gb,gks,ks,sak,yurko92,yurko02}. For the
asymptotic behavior of spectral functions for elliptic operators we
refer to \cite{gording,hormander,miyazaki}.


 In this article we consider boundary value problems generated
by a nonsymmetric differential
 operator on the half line $0\leq x<\infty$:
 $$
(A_P\varphi) (x):=B\frac{{\rm d}\varphi }{{\rm d}
 x}(x)+P(x)\varphi (x)=\lambda\varphi(x),
$$
 where  $B=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ and
$P=\begin{pmatrix} p_{11} & p_{12} \\ p_{21} & p_{22} \end{pmatrix}$
in $(C^1[0,\infty))^4$.
Both the matrix-valued function $P$ and parameter $\lambda$ are complex-valued.
In this article, we  consider only the $C^1$-class case for $P$,
because in this case it is easier to prove the transformation formula
(see Lemma \ref{lem2.1}) while in general case it will be very complicated.
It is directly checked that the adjoint operator of $A_P$
in some suitable Hilbert space is $-B\frac{{\rm d}}{{\rm d} x}+\overline{P^T(x)}$
and consequently  $A_P$ is nonsymmetric. Here and henceforth,
$\overline{c}$ denotes the complex conjugate of $c$ and
 $\cdot^T$ denotes the transpose of a vector or matrix under
 consideration. Here we point out that the spectrum problem for $A_P$ with
compact matrix-valued function $P$ has been studied  in \cite{tr}.

To  describe our results properly, we first give some information on distributions
and  we refer to \cite{mar} for more details.  Let
${\mathbb{K}}^2(0,\infty)$ denote the set of all square integrable
functions in $(0,\infty)$ with compact support. For $\sigma>0$, we set
${\mathbb{K}}^2_\sigma(0,\infty)=\{f\in{\mathbb{K}}^2(0,\infty):
f(x)=0\ \text{for}\ x>\sigma \}$.
The entire function $e(\rho)$ is
 called the \emph{function of exponential type} if
$|e(\rho)|\leq C\exp(\sigma|{\rm Im}\rho|)$ where the positive constants $C$
and $\sigma$ depend on $e(\rho)$. Moreover, the index
$$
\sigma_e=\limsup_{r\to \infty} r^{-1}\ln\big(\max_{|\rho|=r}|e(\rho)|\big)
$$
is called the type of entire function $e(\rho)$. Let linear
topological space $Z$ be the set of all entire exponential type
functions integrable on the real line. The sequence $e_n$
converges to $e$ in $Z$ if
$\lim_{n\to \infty}\int_{-\infty}^\infty|e_n(\rho)-e(\rho)|{\rm
d}\rho=0$
 and the types $\sigma_n$ of the functions $e_n(\rho)$
are bounded: $\sup \sigma_n<\infty$. The set of all linear
continuous functionals defined on the test space $Z$ will be denoted by
$Z'$ whose components are called \emph{distributions} (generalized
functions). The sequence $D_n$ converges to $D$ in $Z'$ if
${\lim_{n\to \infty}<D_n,e(\rho)>=<D,e(\rho)>}$
for all test functions $e\in Z$.

  In this article we consider two cases of the coefficient matrix $P$.
The first case is special and will be described as follows.
Let $P$ be a continuously differentiable    matrix-valued function  satisfying
 $BP=PB$ and $\mu$ be a complex constant. Here it is easy to see that $P$
is of the form  $\begin{pmatrix} a & b \\
b & a \end{pmatrix}$. We consider the boundary value problem
 \begin{equation}\label{eq1}
\begin{gathered}
B{\frac{{\rm d}\varphi}{{\rm d} x}(x)+P(x)\varphi(x)=\lambda\varphi(x), \ 0<x<\infty,}\\
\varphi(0)=\begin{pmatrix} \cosh\mu & \sinh\mu \\ \sinh\mu & \cosh\mu
\end{pmatrix}.
 \end{gathered}
 \end{equation}
 Let $\varphi=\varphi(x,\lambda)$ be the solution to \eqref{eq1} and
$$
\varphi_{[1]}=\begin{pmatrix}
 \varphi_{[1]} ^{(1)}\\
\varphi_{[1]}^{(2)}\end{pmatrix}\quad\text{and}\quad
 \varphi_{[2]}=\begin{pmatrix} \varphi_{[2]}^{(1)}\\
\varphi_{[2]}^{(2)}\end{pmatrix}
$$
be the first and the second column vector of the
matrix $\varphi$, i.e., $\varphi=(\varphi_{[1]}\;\varphi_{[2]})$. Similarly we denote the matrix
 inverse of $\varphi$ by
$\psi=\varphi^{-1}=(\psi_{[1]}\; \psi_{[2]})$. Now for
$$
f=\begin{pmatrix} f^{(1)}
\\ f^{(2)}
\end{pmatrix}\in \left(L^2(0,\infty)\right)^2,\quad
 g =\begin{pmatrix} 
 g^{(1)}\\
 g^{(2)}\end{pmatrix} 
\in \left(L^2(0,\infty)\right)^2
$$
where $\big(L^2(0,\infty)\big)^2$ denotes the product space of $L^2(0,\infty)$,
we set
$$
{\omega_f^{k}(\rho)=\int_0^\infty f^T(x)\psi_{[k]}(x,i\rho){\rm
d}x},\quad
{\eta_{{g}}^k(\rho)=\int_0^\infty
\varphi_{[k]}^T(x,i\rho){g(x)}{\rm d}x}\quad (k=1,2),
$$
where $i=\sqrt{-1}$, $\rho\in\mathbb{R}$. Then we have the first main result
of this article.

 \begin{theorem} \label{thm1}
It holds for the boundary value problem \eqref{eq1} that
 \begin{equation}\label{parseval-1}
\int_0^\infty f^T(x)\overline{g(x)}{\rm d}x
=\frac{1}{2\pi}\sum_{k=1}^{2}\int_{-\infty}^{\infty}\omega_f^{k}(\rho)
\eta_{\overline{g}}^k(\rho){\rm d}\rho.
\end{equation}
Moreover, for $f\in\big({\mathbb{K}}^2(0,\infty)\big)^2$ with
$\omega_f^{k}(\rho),\eta_{{f}}^k(\rho)\in Z\ (k=1,2)$, the following
expansion formula holds:
\begin{equation}\label{expansion-1}
\begin{aligned}
f(x)&=\frac{1}{2\pi}\sum_{k=1}^{2}\int_{-\infty}^{\infty}\omega_f^{k}(\rho)
\varphi_{[k]}(x,i\rho){\rm d}\rho \\
&=\frac{1}{2\pi}\sum_{k=1}^{2}\int_{-\infty}^{\infty}
\eta_f^k(\rho)\psi_{[k]}(x,i\rho){\rm
d}\rho.
\end{aligned}
\end{equation}
\end{theorem}

We often call \eqref{parseval-1} (or \eqref{parseval-2}) the
\emph{Marchenko-Parseval equality} which means that a spectral function exists
in corresponding boundary value problem. Historically,
the concept of spectral function came from the classical theory of Weyl.
Theorem \ref{thm1} implies that $\frac{1}{2\pi}E$ is a spectral function corresponding
to problem \eqref{eq1} with $P$ satisfying $BP=PB$, which is the same as the
case of $P=0$. Here and henceforth $E$ denotes the $2\times 2$ unit matrix.

For general matrix function $P\in{\left(C^1[0,\infty)\right)^4}$ without the
constraint $BP=PB$, we also can  show the  existence of the corresponding  spectral
function. More precisely, let $Q$ be a $2\times 2$ matrix satisfying $QB+BQ=B$ and
 $Q^2=Q$. It is seen by simple computation that there exists matrix $Q$
satisfying the above conditions, and the  simplest one is $Q=$diag$(1,0)$.
It follows easily from $\det B=-1$ that $\det Q=0$. We consider the
boundary value problems
\begin{equation}\label{eq-phi}
\begin{gathered}
B{\frac{{\rm d}\varphi}{{\rm d} x}(x)+P(x)\varphi(x)=\lambda\varphi(x), \quad 0<x<\infty,}\\
\varphi(0)=Q,
\end{gathered}
 \end{equation}
 and
 \begin{equation}\label{eq-phi*}
\begin{gathered}
-{\frac{{\rm d}\widetilde{\varphi}}{{\rm d}
 x}(x)B+\widetilde{\varphi}(x)P(x)=\lambda\widetilde{\varphi}(x), \quad  0<x<\infty,}\\
\widetilde{\varphi}(0)=Q.
\end{gathered}
 \end{equation}
We denote the solutions to problems \eqref{eq-phi} and \eqref{eq-phi*}
by $\varphi(x,\lambda)$ and $\widetilde{\varphi}(x,\lambda)$, respectively.
 For all $2\times 2$ matrices $f,g\in\left(L^2(0,\infty)\right)^4$, we set
 \begin{equation}\label{def1}
\Phi_f(\rho)=\int_0^\infty f(x) \varphi(x,i\rho){\rm d}x,\quad
\widetilde{\Phi}_g(\rho)=\int_0^\infty
\widetilde{\varphi}(x,i\rho)g(x){\rm d}x,
\end{equation}
where $i=\sqrt{-1}$, $\rho\in\mathbb{R}$. Then  we have another main result of
this paper.

 \begin{theorem} \label{thm2}
To the problems \eqref{eq-phi} and \eqref{eq-phi*} there corresponds a
distribution-valued  spectral  function
$D=\left(D_{kl}\right)_{1\leq k,l\leq 2}$ such that
 $D=QDQ$, $D_{kl}\in Z'$ and
\begin{equation}\label{parseval-2}
\int_0^\infty f(x)g(x){\rm d}x=\int_{-\infty}^\infty \Phi_f(\rho)
 D(\rho)
 \widetilde{\Phi}_g(\rho){\rm d}\rho.
\end{equation}
Moreover, for $f\in\left({\mathbb{K}}^2(0,\infty)\right)^4$ with
$\Phi_f(\rho),\widetilde{\Phi}_f(\rho)\in Z^4$, the following expansion
formula holds:
\begin{equation}\label{expansion-2}
f(x)=\int_{-\infty}^\infty \Phi_f(\rho)  D(\rho)
 \widetilde{\varphi}(x,i\rho){\rm d}\rho=\int_{-\infty}^\infty \varphi(x,i\rho)
 D(\rho) \widetilde{\Phi}_f(\rho){\rm d}\rho.
 \end{equation}
\end{theorem}

Although Theorems  \ref{thm1} and  \ref{thm2} have shown the existence of spectral function
for the singular nonsymmetric differential operator in two cases,
we  point out that the uniqueness of spectral function for the operator does
not hold generally, which is the same as
 that for Sturm-Liouville operators (see, e.g., \cite{lev91}).
Moreover, since the spectral function
is distribution-valued, it is not a measure in general, which is different
from the case of self-adjoint Sturm-Liouville operators. Besides, given
singular nonsymmetric differential operators with general $P$, it is still
 an open problem to prove the existence of spectral functions under general
boundary conditions.
 On the other hand, it is interesting to investigate the corresponding inverse
problems, namely, given spectral functions or Weyl functions, find the
differential operators. See \cite{gelf} for the classical inverse problem to
determine the potential of the Sturm-Liouville operator from its spectral
function and \cite{frei12} for determination of singular differential pencils
from the Weyl function. Theorem \ref{thm1} has implied that the uniqueness does not
hold generally for the inverse problems,
and we need impose other assumptions for uniqueness. In a forthcoming
paper we will  study the inverse problems for the singular nonsymmetric
 differential operator.


This article is composed of four sections. In Section 2 we establish
transformation formulae for our boundary value problems.
Sections 3 and 4 are devoted to prove
Theorems \ref{thm1} and  \ref{thm2} by transformation formulae, respectively.

\section{Transformation formulae}

Set
\begin{equation}\label{Omega}
\Omega=\{(x,y)\in\mathbb{R}^2: 0<y<x\}.
\end{equation}
For $P_j=(P_{j,kl})_{1\leq k,l\leq
2}\in\left(C^1[0,\infty)\right)^4$ $(j=1,2)$, we define
\begin{gather}\label{theta1}
\theta_1(x)={\frac{1}{2}\int_0^x\left(P_{2,12}+P_{2,21}-P_{1,12}-P_{1,21}
\right)(s){\rm d}s}, \\
\label{theta2}
\theta_2(x)={\frac{1}{2}\int_0^x\left(P_{2,11}+P_{2,22}-P_{1,11}-P_{1,22}\right)(s)
{\rm d}s}.
\end{gather}
Moreover let us put
\begin{equation}\label{R}
R(P_1,P_2)(x)=\exp\left(-\theta_1(x)\right)
\begin{pmatrix}\cosh\theta_2(x) &
-\sinh\theta_2(x) \\
-\sinh\theta_2(x) & \cosh\theta_2(x) \end{pmatrix}.
\end{equation}
Here we remark that $R(P_1,P_2)(0)=E$,
$R(P_1,P_2)(x)=R^{-1}(P_2,P_1)(x)$ and that
$R\big(-\overline{P_1^T},-\overline{P_2^T}\big)(x)
=\overline{R\big(P_2,P_1\big)(x)}$. Let $M_2(\mathbb{C})$ be the set of
all $2\times 2$ complex-valued matrices. We first prove the following
lemma.

 \begin{lemma} \label{lem2.1}
For any $\lambda\in\mathbb{C}$, $Q\in{ M_2(\mathbb{C})}$ with $\det Q=0$ and $P_j\in
 \left(C^1[0,\infty)\right)^4$ $(j=1,2)$, let $\varphi_j=\varphi_j(x,\lambda)$ satisfy
\begin{equation}\label{eq-Pj}
\begin{gathered}
{B\frac{{\rm d}\varphi_j(x)}{{\rm d}
 x}+P_j(x)\varphi_j(x)=\lambda\varphi_j(x), \quad 0< x<\infty,}\\
\varphi_j(0)=Q.
\end{gathered}
 \end{equation}
Then there exists a unique
$K(P_1,P_2;Q)=\left(K_{kl}(P_1,P_2;Q)\right)_{1\leq k,l\leq
2}\in\left(C^1(\overline{\Omega})\right)^4$ independent of $\lambda$
such that for $0\leq x<\infty$ and all $\lambda\in\mathbb{C}$
\begin{equation}\label{transform}
\varphi_2(x,\lambda)={R(P_1,P_2)(x)\varphi_1(x,\lambda)+\int_0^x
{K(P_1,P_2;Q)(x,y)\varphi_1(y,\lambda)}{\rm d}y}.
\end{equation}
(transformation formula).
Here $R(P_1,P_2)(x)$ is defined by \eqref{R}.

Moreover, the kernel $K(P_1,P_2;Q)$ is
the unique solution to the following problem of first order system
\eqref{eq-K}--\eqref{xx-K}:
\begin{gather}\label{eq-K}
\begin{aligned}
&{B\frac{\partial K(P_1,P_2;Q)}{\partial x}(x,y)
+\frac{\partial K(P_1,P_2;Q)}{\partial y}(x,y)B} \\
&{+P_2(x)K(P_1,P_2;Q)(x,y)-K(P_1,P_2;Q)(x,y)P_1(y)=0, \quad
(x,y)\in \Omega}.
\end{aligned} \\
\label{x0-K}
K(P_1,P_2;Q)(x,0)BQ=0\quad (0\leq x<\infty), \\
\label{xx-K}
\begin{aligned}
& K(P_1,P_2;Q)(x,x)B-BK(P_1,P_2;Q)(x,x)\\
&=B{\frac{{\rm d}R(P_1,P_2)}{{\rm d}x}(x)+P_2(x)R(P_1,P_2)(x)-R(P_1,P_2)(x)P_1(x)}\\
&\quad(0\leq x<\infty).
\end{aligned}
\end{gather}
\end{lemma}

\begin{proof}
 We prove this lemma using  ideas from  \cite{ya88}.
Since $P_j\in\left(C^1[0,\infty)\right)^4$ $(j=1,2)$, it can be
verified directly that, if
$K(P_1,P_2;Q)\in\left(C^1(\overline{\Omega})\right)^4$ is the
unique solution to  problem \eqref{eq-K}--\eqref{xx-K}, then
\eqref{transform} holds. Therefore, it is sufficient to prove the
existence and the uniqueness of the solution to  problem
\eqref{eq-K}--\eqref{xx-K} for each
$P_1,P_2\in{\left(C^1[0,\infty)\right)^4}$.

For clarity, we reduce the proof to a special case.  By the
condition $\det Q=0$, we may assume that a complex constant $c$
exists such that $q_2=c q_1$ where $q_1,q_2$ are the first column
vector and the second one of $Q$, respectively. Then it is
sufficient to prove the existence and the uniqueness of the solution
to problem \eqref{eq-K}--\eqref{xx-K} in the case
$\varphi_j(0,\lambda)=q_1$, since problem \eqref{eq-Pj} is linear.
Moreover, since a complex constant
$c^*$ exists such that
$q_1=c^*\begin{pmatrix} \cosh\mu\\ 
\sinh\mu \end{pmatrix}$
where $\mu\in\mathbb{C}$, it can be
reduced to the case $\varphi_j(0,\lambda)=\begin{pmatrix} \cosh\mu\\
\sinh\mu \end{pmatrix}$. In this case, we denote the the solution to
problem \eqref{eq-K}--\eqref{xx-K} by $K(P_1,P_2,\mu)(x,y)$, and
\eqref{x0-K} has the form
\begin{equation}\label{x0-K-mu}
\begin{gathered}
 K_{12}(P_1,P_2, \mu)(x,0)=-\tanh \mu\ K_{11}(P_1,P_2,\mu)(x,0),\\
K_{22}(P_1,P_2, \mu)(x,0)=-\tanh \mu\ K_{21}(P_1,P_2,\mu)(x,0).
\end{gathered}
\end{equation}
If we set
\begin{equation}\label{L}
\begin{gathered}
L_1(x,y)=K_{12}(P_1,P_2,\mu)(x,y)- K_{21}(P_1,P_2,\mu)(x,y),\\
L_2(x,y)=K_{11}(P_1,P_2,\mu)(x,y)- K_{22}(P_1,P_2,\mu)(x,y), \\
L_3(x,y)=K_{11}(P_1,P_2,\mu)(x,y)+K_{22}(P_1,P_2,\mu)(x,y),\\
L_4(x,y)=K_{12}(P_1,P_2,\mu)(x,y)+K_{21}(P_1,P_2,\mu)(x,y)
\end{gathered}
\end{equation}
and $L=L(x,y)=\left(L_1(x,y),L_2(x,y),L_3(x,y),L_4(x,y)\right)$, then we can
rewrite \eqref{eq-K}--\eqref{xx-K} as follows:
\begin{gather}\label{L12}
{\frac{\partial L_k(x,y)}{\partial x}-\frac{\partial
L_k(x,y)}{\partial y}=f_k(x,y,L)}\quad   ((x,y)\in{\Omega},\;k=1,2), \\
\label{L34}
{\frac{\partial L_k(x,y)}{\partial x}+\frac{\partial
L_k(x,y)}{\partial y}=f_k(x,y,L)}\quad  ((x,y)\in{\Omega},\; k=3,4), \\
\label{x0-L12} \quad  L_k(x,x)=r_k(x)    \quad  (0\leq x<\infty,\; k=1,2), \\
\label{x0-L34}
\begin{aligned} L_3(x,0)=\sinh(2\mu) L_1(x,0)+\cosh(2\mu) L_2(x,0)
\quad  (0\leq x<\infty),\\
L_4(x,0)= -\cosh(2\mu)L_1(x,0)-\sinh(2\mu) L_2(x,0)
\quad  (0\leq x<\infty),
\end{aligned}
\end{gather}
where $f_k(x,y,L)={\frac{1}{2}\sum_{m=1}^4}\left(a_{km}(y)+b_{km}(x)\right)L_m(x,y)$
$(1\leq k\leq 4)$, here $a_{km}(y)$, $b_{km}(x)$ $(1\leq k,m\leq
4)$ are linear combinations of two elements of the matrix
functions $P_1(y)$ and $P_2(x)$ respectively, and
$r_k\in{C^1[0,\infty)}$ $(k=1,2)$ are dependent only on
$P_1$ and $P_2$.

Integrating \eqref{L12}, \eqref{L34} with \eqref{x0-L12} and
\eqref{x0-L34} along the characteristics $x+y=$const. and
$x-y=$const. respectively, we obtain the following integral
equations:
\begin{equation}\label{int-L12}
\begin{gathered}
L_k(x,y)={\int_y^{\frac{x+y}{2}} f_k(-s+x+y,s,L){\rm
d}s+r_k(\frac{x+y}{2})}\\
 \left((x,y)\in{\overline{\Omega}}, k=1,2\right),
\end{gathered}
\end{equation}
and
\begin{equation}\label{int-L34}
\begin{aligned}
 L_k(x,y)
&={\int_0^y f_k(s+x-y,s,L){\rm d}}s
+\int_0^{\frac{x-y}{2}} \Big\{\alpha_k f_1(-s+x-y,s,L) \\
&\quad +\beta_k f_2(-s+x-y,s,L)\Big\}{\rm d}s
 +\alpha_k r_1(\frac{x-y}{2}) +\beta_k r_2(\frac{x-y}{2}) \\
&\quad \left((x,y)\in{\overline{\Omega}},\; k=3,4\right),
\end{aligned}
\end{equation}
where $\alpha_3=\sinh(2\mu)$, $\beta_3=\cosh(2\mu)$ and
$\alpha_4=-\cosh(2\mu)$, $\beta_4=-\sinh(2\mu)$.

The unique solution $L\in{\left(C^1(\overline{\Omega})\right)^4}$ to
\eqref{int-L12} and \eqref{int-L34} can be obtained by the iteration
method. In fact, setting
\begin{gather*}
L_k^{(0)}(x,y)=0\quad \left((x,y)\in{\overline{\Omega}}, 1\leq k\leq 4\right), \\
 L_k^{(n)}(x,y)={\int_y^{\frac{x+y}{2}}
f_k\left(-s+x+y,s,L^{(n-1)}\right){\rm d}s+r_k(\frac{x+y}{2})}\\
 \left((x,y)\in{\overline{\Omega}},n\geq 1,\; k=1,2\right),
\end{gather*}
and
\begin{align*}
&L_k^{(n)}(x,y)\\
&= {\int_0^y f_k\left(s+x-y,s,L^{(n-1)}\right){\rm d}}s\\
&\quad +{\int_0^{\frac{x-y}{2}} \left\{\alpha_k
f_1\left(-s+x-y,s,L^{(n-1)}\right)+\beta_k
f_2\left(-s+x-y,s,L^{(n-1)}\right)\right\}{\rm
d}s}\\
&\quad {+\alpha_k r_1(\frac{x-y}{2}) +\beta_kr_2(\frac{x-y}{2})}
\quad \left((x,y)\in{\overline{\Omega}}, n\geq 1, k=3,4\right),
\end{align*}
we can obtain by induction the estimates for each $n\geq 1$,
\begin{equation}\label{iter-est}
\big|L_k^{(n)}(x,y)-L_k^{(n-1)}(x,y)\big|
\leq \omega(x)\frac{\zeta^{n-1}(x)}{(n-1)!}\quad
\left((x,y)\in{\overline{\Omega}},\; 1\leq k\leq 4\right),
\end{equation}
where
\begin{gather*}
\omega(x)=\left(|\sinh(2\mu)|+|\cosh(2\mu)|+1\right)
\max_{0\leq s\leq x}\left(|r_1(s)|+|r_2(s)|\right), \\
\zeta(x)=\left(|\sinh(2\mu)|+|\cosh(2\mu)|+1\right)
x \max_{0\leq s\leq x} \frac{1}{2}\sum_{k,l=1}^{2}
\left(|P_{1,kl}(s)|+|P_{2,kl}(s)|\right).
\end{gather*}
Thus
$L_k(x,y)=\lim_{n\to \infty}L_k^{(n)}(x,y)$
$(1\leq k\leq 4)$ exist uniformly for
$(x,y)\in{\overline{\Omega}}$ and we see that $L_k(x,y)$ $(1\leq
k\leq 4)$ satisfy \eqref{int-L12} and \eqref{int-L34} with the
bound $|L_k(x,y)|\leq \omega(x)\exp(\sigma(x))$.

Moreover, differentiating \eqref{int-L12} and \eqref{int-L34} with
respect to $x$ and $y$; similarly we can obtain by induction the following
estimates
\begin{gather}\label{iter-est-px}
\big|{\frac{\partial L_k^{(n)}(x,y)}{\partial
x}-\frac{\partial L_k^{(n-1)}(x,y)}{\partial x}}\big|
\leq \xi(x)\frac{\zeta^{n-1}(x)}{(n-1)!}\quad
\big((x,y)\in{\overline{\Omega}}, 1\leq k\leq 4\big), \\
\label{iter-est-py}
\big|{\frac{\partial L_k^{(n)}(x,y)}{\partial
y}-\frac{\partial L_k^{(n-1)}(x,y)}{\partial y}}\big|
\leq \xi(x)\frac{\zeta^{n-1}(x)}{(n-1)!}\quad
\big((x,y)\in{\overline{\Omega}}, 1\leq k\leq 4\big),
\end{gather}
where
\begin{align*}
\xi(x)
&={\frac{1}{2}\left(|\sinh(2\mu)|+|\cosh(2\mu)|+1\right)}\\
&\quad \times {\Big\{ \max_{0\leq s\leq
x}\left(|r'_1(s)|+|r'_2(s)|\right)
+\frac{1}{2}\omega(x)\exp(\zeta(x))}\\
&\quad  \times{\max_{0\leq s\leq x}
 \sum_{k,l=1}^{2}\left(|P_{1,kl}(s)|+|P_{2,kl}(s)|
 +\left(|P'_{1,kl}(s)|+|P'_{2,kl}(s)|\right)x\right)\Big\}.}
\end{align*}
Therefore,  from \eqref{iter-est-px} and
\eqref{iter-est-py} it follows that
$L\in{(C^1(\overline{\Omega}))^4}$. The uniqueness of the
solution to \eqref{eq-K}--\eqref{xx-K} is shown by
\eqref{iter-est}. 
\end{proof}

\begin{corollary} \label{coro2.2} 
For $j=1,2$, let $\varphi_j$ be the solution to problem \eqref{eq1} with 
$P=P_j\in (C^1[0,\infty))^4$ satisfying
 $P_jB=BP_j$. Then the following
transformation formula holds:
\begin{equation}\label{ecoro2.2}
\varphi_2(x,\lambda)=R(P_1,P_2)(x)\varphi_1(x,\lambda)
\end{equation}
where $R(P_1,P_2)(x)$ is defined by \eqref{R}.
\end{corollary}

The above corollary follows from the fact that
$K(P_1,P_2,\mu)\equiv 0$, which can be derived easily by observing that 
the right hand side of \eqref{xx-K} is $0$
(in this case the condition $\det Q=0$ is not necessary).
Or one may directly verify \eqref{ecoro2.2}. Here we omit the details.


\begin{corollary} \label{coro2.3}
Let $S$ and $\widetilde{S}$ be the solutions corresponding to
$P=0$ in \eqref{eq-phi} and \eqref{eq-phi*}, respectively. 
Then the following transformation formulae hold.

 (1) For problem \eqref{eq-phi} we have
\begin{equation}\label{f2-f1}
S(x,i\rho)={R(P,0)(x)\varphi(x,i\rho)+\int_0^x
{K(P,0;Q)(x,y)\varphi_(y,i\rho)}{\rm d}y}
\end{equation}
where the kernel
$K(P,0;Q)\in\left(C^1(\overline{\Omega})\right)^4$ satisfies
\begin{equation}\label{K-p1p2}
{B K_x(P,0;Q)(x,y)+K_y(P,0;Q)(x,y)B}-K(P,0;Q)(x,y)P(y)=0,
\end{equation}
for $(x,y)\in \Omega$,
as well as the conditions 
\begin{gather}\label{K-p1p2-0}
 K(P,0;Q)(x,0)Q=K(P,0;Q)(x,0), \\
\label{K-p1p2-xx}
K(P,0;Q)(x,x)B-BK(P,0;Q)(x,x)=B R'(P,0)(x)-R(P,0)(x)P(x)
\end{gather}
for  $0\leq x<\infty$.

(2) For problem \eqref{eq-phi*} we have
\begin{equation}\label{f2-f1*}
\widetilde{S}(x,i\rho)={\widetilde{\varphi}(x,i\rho)R(0,P)(x)
+\int_0^x {\widetilde{\varphi}(y,i\rho)}
\overline{K^T(-\overline{P^T},0;\overline{Q^T})(x,y)}{\rm d}y}
\end{equation}
where the kernel
$\overline{K^T(-\overline{P^T},0;\overline{Q^T})(x,y)}$
satisfies
\begin{equation}\label{K-p1p2*}
Q\overline{K^T(-\overline{P^T},0;\overline{Q^T})(x,0)}
=\overline{K^T(-\overline{P^T},0;\overline{Q^T})(x,0)}.
\end{equation}
\end{corollary}

\begin{proof}  (1) is obvious, since $\det Q=0$ and then Lemma \ref{lem2.1} can be 
applied. Here \eqref{K-p1p2-0} follows
from \eqref{x0-K}, $BQ=B-QB$ and $B^2=E$. Now we prove (2).
Note that by \eqref{eq-phi*} the function $\widetilde{\varphi}(x,i\rho)$ statisfies
\begin{gather*}
B{\frac{{\rm d}\overline{\widetilde{\varphi}^T}}{{\rm d} x}(x)
-\overline{P^T(x)}\overline{\widetilde{\varphi}^T(x)}
=i\rho\overline{\widetilde{\varphi}^T(x)}, \ 0<x<\infty,}\\
\overline{\widetilde{\varphi}^T(0)}=\overline{Q^T}.
\end{gather*}
Then  one  obtains \eqref{f2-f1*} by (1) using that
$R(0,P)(x)=\overline{R(-\overline{P^T},0)(x)}$.
\end{proof}

Since the solutions to the boundary value problems with $P=0$ are 
entire in $\lambda$, by the transformation formulae we obtain easily
the following result.

\begin{corollary} \label{coro2.4} 
For each fixed $x$, all solutions to the boundary value problems under 
consideration are entire in $\lambda$.
\end{corollary}


\section{Proof of Theorem \ref{thm1}}

We divide the proof into four steps.
\smallskip

\noindent\textbf{First step.} We  construct a
regular spectral function. Let $S$ denote the solution of \eqref{eq1} 
corresponding to $P=0$.
Set
$$
\rho=-i\lambda\quad  \text{and}\quad \nu=-i\mu.
$$
 It is easy to see that
\begin{equation}\label{S}
\begin{aligned}
 S=S(x,\lambda)
&=\begin{pmatrix} \cosh(\lambda x+\mu)& \sinh(\lambda x+\mu)\\
 \sinh(\lambda x+\mu) & \cosh(\lambda x+\mu)\end{pmatrix}\\
&=\begin{pmatrix} \cos(\rho
x+\nu) & i\sin(\rho x+\nu)\\  i\sin(\rho x+\nu) & \cos(\rho
x+\nu)\end{pmatrix}
\end{aligned}
\end{equation}
and
\begin{equation}\label{S-1}
S^{-1}=S^{-1}(x,\lambda)=\begin{pmatrix} \cos(\rho
x+\nu) & -i\sin(\rho x+\nu)\\  
-i\sin(\rho x+\nu) & \cos(\rho x+\nu)
\end{pmatrix}.
\end{equation}
 We choose two sufficiently smooth real-valued functions
$\delta_n(x)$ and $\gamma_\sigma(x)$ subject to the following
conditions:
\begin{equation}\label{gamma-sigma}
\begin{gathered}
{\int_0^\infty\delta_n(x){\rm d}x=1,} \\
\delta_n(x)=0\ {\text{for}}\ x=0\ {\text{and}}\
x\geq{\frac{1}{n}},\ \ \delta_n(x)>0\ {\text{for}}\
0<x< {\frac{1}{n}}, \\
\gamma_\sigma(x)=\begin{cases}
1 &\text{for } 0\leq x\leq \sigma,\\
0 &\text{for } x> \sigma+1,
\end{cases}
\end{gathered}
\end{equation}
and it is obvious that $\delta_n(x)$ tends to the Dirac delta function $\delta(x)$
as $n\to \infty$. We set
\begin{equation}\label{Dn}
\begin{aligned}
D_n^\sigma(\rho)
&=\left(D_{n,jm}^\sigma(\rho)\right)_{1\leq j,m\leq 2}\\
&={\frac{1}{2\pi}\int_{0}^{\infty}
\begin{pmatrix} \cos(\rho x+\nu) & -i\sin(\rho x+\nu) \\
-i\sin(\rho x+\nu) & \cos(\rho x+\nu)\end{pmatrix}}\\
&\quad \times R(P,0)(x)\delta_n(x) E \gamma_\sigma(x)
\begin{pmatrix} \cos\nu & i\sin\nu \\
i\sin\nu & \cos\nu
\end{pmatrix}{\rm d}x.
\end{aligned}
\end{equation}
Since the Fourier transform is a one-to-one mapping on the space of 
bounded continuous Lebesgue-integrable functions
and $R(P,0)(x)\delta_n(x) E \gamma_\sigma(x)$ is
a continuously differentiable matrix function with
compact support, it is not hard to see that the matrix
function $D_n^\sigma(\rho)$ is bounded and Lebesgue-integrable on the real line
$-\infty<\rho<\infty$. Hence the integral
$$
{\int_{-\infty}^{\infty} S(x,i\rho)
D_n^\sigma(\rho) \begin{pmatrix} \cos\nu & -i\sin\nu \\
-i\sin\nu & \cos\nu\end{pmatrix}{\rm d}\rho}
$$
converges absolutely. By Corollary \ref{coro2.2} we have
$\varphi(x,i\rho)$$=R(0,P)(x)S(x,i\rho)$$=R^{-1}(P,0)(x)S(x,i\rho)$, which implies by
the Fourier inverse transform that
\begin{equation}\label{inv-f}
{\int_{-\infty}^{\infty}
\varphi(x,i\rho)
D_n^\sigma(\rho) \begin{pmatrix} \cos\nu & -i\sin\nu \\
-i\sin\nu & \cos\nu\end{pmatrix}{\rm d}\rho=\delta_n(x) E\ \ (0\leq
x\leq \sigma).}
\end{equation}
Here and henceforth we repeatedly
make use of the fact that two matrices $P_1$ and $P_2$ in the form of 
$\begin{pmatrix} a & b \\
b & a \end{pmatrix}$ are interchangeable: $P_1P_2=P_2P_1$.
\smallskip

\noindent\textbf{Second step.} Now we  investigate the asymptotic 
behavior of the  matrix function
as $n\to \infty$
\begin{equation}\label{eq-Un-sigma}
\begin{aligned}
 U_n^\sigma(x,y)
=&\left(U_{n,kl}^\sigma(x,y)\right)_{1\leq k,l\leq 2}\\
:=&{\int_{-\infty}^{\infty}
\varphi(x,i\rho) D_n^\sigma(\rho) \varphi^{-1}(y,i\rho){\rm d}\rho\quad
 (0\leq x,y\leq \sigma).}
\end{aligned}
\end{equation}
It is easy to find that
$$
U_n^\sigma(x,0)=\int_{-\infty}^{\infty}
\varphi(x,i\rho) D_n^\sigma(\rho) \begin{pmatrix} \cos\nu & -i\sin\nu \\
-i\sin\nu & \cos\nu\end{pmatrix} {\rm d}\rho=\delta_n(x) E
$$
for $0\leq x\leq \sigma$,
and
$$
U_n^\sigma(0,y)=\int_{-\infty}^{\infty}
 \begin{pmatrix} \cos\nu & i\sin\nu \\
i\sin\nu & \cos\nu\end{pmatrix}D_n^\sigma(\rho)\varphi^{-1}(y,i\rho)  {\rm
d}\rho\quad  (0\leq y\leq \sigma).
$$
Now we show that $U_n^\sigma(0,y)=0$ for all $y\geq 0$. Indeed,
first one can see from \eqref{Dn} that
\begin{align*}
D_n^\sigma(\rho)
&={\frac{1}{2\pi}\int_{0}^{\infty}
\begin{pmatrix} \cos(\rho x) & -i\sin(\rho x) \\
-i\sin(\rho x) & \cos(\rho x)\end{pmatrix}}\\
&\quad \times \left\{
R_{11}(P,0)(x)E+R_{12}(P,0)(x)B\right\}\delta_n(x) \gamma_\sigma(x)
{\rm d}x.
\end{align*}
Moreover, for any continuous
scalar function $u(x)$ with compact support and $u(0)=0$, 
it follows easily from the theory of
the Fourier cosine and sine transforms that
\begin{equation}\label{c-s}
\begin{aligned}
& {\int_{-\infty}^{\infty}\int_0^{\infty}
 \begin{pmatrix} \cos(\rho x) & -i\sin(\rho x) \\
-i\sin(\rho x) & \cos(\rho x)\end{pmatrix}u(x)
\begin{pmatrix} \cos(\rho y) & -i\sin(\rho y) \\
-i\sin(\rho y) & \cos(\rho y)\end{pmatrix} {\rm d}x\, {\rm d}\rho}\\
&=0.
\end{aligned}
\end{equation}
From \eqref{c-s} and 
$\varphi^{-1}(\cdot,i\rho)=S^{-1}(\cdot,i\rho)R(P,0)(\cdot)$ it follows that
 $U_n^\sigma(0,y)R(0,P)(y)=0$ and hence $U_n^\sigma(0,y)=0$, 
since $R(0,P)(y)$ is invertible.

 On the other hand,  by \eqref{eq-Un-sigma} it is easy to see that, for fixed 
$n$ and $\sigma$, $U_n^\sigma(\sigma,\cdot)$ is
  a bounded differentiable function on $[0,\sigma]$ and denoted by 
$\Xi_n(\cdot)$ for simplicity. Therefore, since by \eqref{eq1} we easily show that
$$
{\frac{{\rm d}\varphi^{-1}(x)}{{\rm d}x}B-\varphi^{-1}(x)P(x)=-i\rho\varphi^{-1}(x)},
$$
the above argument implies that the functions
\begin{equation}\label{UnN}
U_{nN}^\sigma(x,y):={ \int_{-N}^{N} \varphi(x,i\rho)
D_n^\sigma(\rho) \varphi^{-1}(y,i\rho){\rm d}\rho}
\end{equation}
are continuously differentiable and satisfy the equation
\begin{equation}\label{eq-Un-N}
 {B\frac{\partial U}{\partial
x}(x,y)+\frac{\partial U}{\partial y}(x,y)B}
{+P(x)U(x,y)-U(x,y)P(y)=0 \ \text{in}\ \Pi_{\sigma}}
\end{equation}
 and the following conditions
\begin{gather}\label{bc-Un-N}
U(x,0)=\delta_{nN}(x) E\quad  (0\leq x\leq\sigma), \\
\label{bc-Un-N-2}
U(0,y)=\Gamma_{nN}(y),\quad U(\sigma,y)=\Xi_{nN}(y) \quad (0\leq y\leq\sigma),
\end{gather}
where $\Pi_{\sigma}=\{(x,y)\in\mathbb{R}^2: 0<x,y<\sigma\}$, the functions
$\delta_{nN},\Gamma_{nN}$ and $\Xi_{nN}$
satisfy the compatibility conditions and
${\lim_{N\to \infty}\delta_{nN}(x)}$$=\delta_n(x)$,
${\lim_{N\to \infty}\Gamma_{nN}(y)}=0$ and 
$\lim_{N\to \infty}\Xi_{nN}(y)=\Xi_n(y)$.
We should note that problem \eqref{eq-Un-N}, \eqref{bc-Un-N} and \eqref{bc-Un-N-2}
can be rewritten as a symmetric hyperbolic
system:
\begin{equation}\label{hyper-V}
\begin{gathered}
{\frac{\partial V}{\partial y}(x,y)+\begin{pmatrix} 0 & E \\
E & 0\end{pmatrix}\frac{\partial V}{\partial x}(x,y)+C(x,y)V(x,y)=0}\quad
 \text{in }\Pi_{\sigma},\\
V(x,0)=\delta_{nN}(x) \overrightarrow{H}\quad  (0\leq x\leq\sigma),\\
V(0,y)=\overrightarrow{\Gamma}_{nN}(y),\quad V(\sigma,y)=\overrightarrow{\Xi}_{nN}(y) 
\quad  (0\leq y\leq\sigma),
\end{gathered}
\end{equation}
where
\begin{gather*}
V(x,y)=\begin{pmatrix} U_{11}(x,y)\\U_{12}(x,y)\\U_{21}(x,y)\\U_{22}(x,y)
\end{pmatrix},\quad
\overrightarrow{H}=\begin{pmatrix} 1\\0\\0\\1 \end{pmatrix},\\
\overrightarrow{\Gamma}_{nN}(y)
=\begin{pmatrix}
\Gamma_{nN,11}(y)\\ \Gamma_{nN,12}(y)\\ \Gamma_{nN,21}(y)\\ \Gamma_{nN,22}(y)
\end{pmatrix},\quad
 \overrightarrow{\Xi}_{nN}(y)
=\begin{pmatrix}
\Xi_{nN,11}(y)\\ 
\Xi_{nN,12}(y)\\ \Xi_{nN,21}(y)\\ \Xi_{nN,22}(y)
\end{pmatrix}
\end{gather*}
and $C(x,y)$ is the  $4\times 4$ matrix-valued function
$$
\begin{pmatrix}
-P_{12}(y) & P_{11}(x)-P_{22}(y) & 0 & P_{12}(x) \\
P_{11}(x)-P_{11}(y) & -P_{21}(y) & P_{12}(x) & 0 \\
0 & P_{21}(x) & -P_{12}(y) & P_{22}(x)-P_{22}(y) \\
P_{21}(x) & 0 & P_{22}(x)-P_{11}(y) & -P_{21}(y)
\end{pmatrix}.
$$
Since $BU_{nN}^\sigma(x,y)=U_{nN}^\sigma(x,y)B$, a direct calculation 
shows that the symmetric hyperbolic system \eqref{hyper-V} is actually 
equivalent to the following normal hyperbolic system
\begin{equation}\label{eq-v}
\begin{gathered}
{\frac{\partial v}{\partial y}(x,y)=\begin{pmatrix} 1 & 0 \\
0 & -1 \end{pmatrix}\frac{\partial v}{\partial x}(x,y)+c(x,y)v(x,y)}\quad
 \text{in}\ \Pi_{\sigma},\\
v(x,0)=\delta_{nN}(x) \overrightarrow{h}\quad  (0\leq x\leq\sigma),\\
v_2(0,y)=2v_1(0,y)-\Gamma_{nN,11}(y)+3\Gamma_{nN,12}(y),\\
v_2(\sigma,y)=2v_1(\sigma,y)-\Xi_{nN,11}(y)+3\Xi_{nN,12}(y) \quad
 (0\leq y\leq\sigma),
\end{gathered}
\end{equation}
where
$$
v(x,y)=\begin{pmatrix} v_1(x,y)\\ v_2(x,y) \end{pmatrix}
=\begin{pmatrix} U_{11}(x,y)-U_{12}(x,y)\\ U_{11}(x,y)+U_{12}(x,y) 
\end{pmatrix},\quad
 \overrightarrow{h}= \begin{pmatrix} 1\\ 1 \end{pmatrix},
$$
and
\begin{align*}
& c(x,y)\\
&=\begin{pmatrix} (P_{11}-P_{12})(x)+(P_{12}-P_{11})(y)  
&  (P_{12}-P_{11})(x)+(P_{22}-P_{21})(y)\\
(P_{11}+P_{12})(y)-(P_{11}+P_{12})(x)  
& (P_{22}+P_{21})(y)-(P_{11}+P_{12})(x)
\end{pmatrix}.
\end{align*}
If we take the variable $y$ as time, then it is not hard to verify that 
the classical Uniform Kreiss Condition holds, and hence from the well-known
results of well-posedness on linear hyperbolic systems (cf. \cite{higdon}
 and references therein) we see that \eqref{eq-v} has a unique solution;
 that is, there exists a
unique solution $U_{nN}^\sigma(x,y)$ to problem \eqref{eq-Un-N},
\eqref{bc-Un-N} and \eqref{bc-Un-N-2} such that
$U_{nN}^\sigma(x,y)\to  U_n^\sigma(x,y)$ as
$N\to \infty$.

 On the other hand, if we set
$W_{nN}^\sigma(x,y)=U_{nN}^\sigma(x,y)-\delta_{nN}(x-y) E$ for
 $0\leq x,y\leq\sigma$ where $\delta_{nN}(x-y)=0$ for 
$0\leq x<y\leq\sigma$, then $W_{nN}^\sigma(x,y)$ satisfies the  equation
\begin{equation}\label{hyper-W}
\begin{aligned}
& { B\frac{\partial W}{\partial x}(x,y)+\frac{\partial W}{\partial y}(x,y)B
+P(x)W(x,y)-W(x,y)P(y)}\\
&=\delta_{nN}(x-y)\left(P(y)-P(x)\right)
\end{aligned}
\end{equation}
and $W(x,0)=0$. From the compatibility conditions it follows  that 
$W_{nN}^\sigma(0,y)\to  0$ as $N,n\to  \infty$. Next we  show that
\begin{equation}\label{w-sigma}
W_{nN}^\sigma(\sigma,y)\to  0 \quad (N,n\to  \infty).
\end{equation}
From \eqref{S}, \eqref{S-1}, \eqref{inv-f} and the transformation formulae
 $\varphi(\cdot,i\rho)$$=R(0,P)(\cdot)S(\cdot,i\rho)$$=R^{-1}(P,0)(\cdot)S(\cdot,i\rho)$, 
$\varphi^{-1}(\cdot,i\rho)=S^{-1}(\cdot,i\rho)R(P,0)(\cdot)$  it follows that
\begin{align*}
\Xi_n(y)
&{=R(0,P)(\sigma)R(P,0)(y)R(P,0)(\sigma-y)}\\
&{\quad \times\int_{-\infty}^{\infty}\varphi(\sigma-y,i\rho)
D_n^\sigma(\rho) \begin{pmatrix} \cos\nu & -i\sin\nu \\
-i\sin\nu & \cos\nu\end{pmatrix}{\rm d}\rho}\\
&=R(0,P)(\sigma)R(P,0)(y)R(P,0)(\sigma-y)\delta_n(\sigma-y) E\quad
 (0\leq x\leq \sigma).
\end{align*}
Therefore,
\begin{equation}\label{key}
\begin{aligned}
&\Xi_n(y)-\delta_n(\sigma-y) E\\
&=\delta_n(\sigma-y)R(0,P)(\sigma)[R(P,0)(y)R(P,0)(\sigma-y)-R(P,0)(\sigma)]
\end{aligned}
\end{equation}
whence \eqref{w-sigma} follows easily. Consequently, by the well-posedness 
of symmetric hyperbolic linear differential equations, we have
$$
W_{nN}^\sigma(x,y)\to  0\quad \text{as } N,n\to  \infty
$$
since $\delta_{nN}(x-y)\to  \delta(x-y)\quad  \text{as }
N,n\to  \infty$ and hence the right hand side of
\eqref{hyper-W} tends to $0$.
Therefore, for $0\leq x,y\leq\sigma$,
\begin{equation}\label{lim-delta}
U_n^\sigma(x,y)\to \delta(x-y) E\quad  (n\to \infty).
\end{equation}

We remark that there is another and simpler way to prove \eqref{lim-delta} 
in which it is not needed to consider \eqref{eq-Un-N}.
The key idea is based on considering \eqref{eq-Un-sigma}, \eqref{c-s} 
and \eqref{key} with replacing $\sigma$ by $x$. We leave
the details to the reader.
\smallskip

\noindent\textbf{Third step.} 
We prove the Marchenko-Parseval equality \eqref{parseval-1}. Assuming that
$f,g\in{\left({\mathbb{K}}^2_\sigma(0,\infty)\right)^2}$
 have compact support,  by changing the
order of integration we have 
\begin{align*}
&{\int_0^\infty\int_0^\infty\sum_{k,l=1}^{2}U_{n,kl}^\sigma(x,y)
 \overline{g^{(k)}(x)}f^{(l)}(y){\rm d}x{\rm d}y}\\
&={\int_0^\infty\int_0^\infty\sum_{k,l=1}^{2}
 \Big(\int_{-\infty}^{\infty}\sum_{j,m=1}^{2}
D_{n,jm}^\sigma(\rho) \varphi_{[j]}^{(k)}(x,i\rho)
\psi_{[m]}^{(l)}(y,i\rho){\rm d}\rho\Big)}
\overline{g^{(k)}(x)}f^{(l)}(y){\rm d}x{\rm d}y\\
&={\sum_{j,m=1}^{2}\int_{-\infty}^{\infty}{\rm d}\rho D_{n,jm}^\sigma(\rho)}
{\Big(\int_0^\infty\int_0^\infty\sum_{k,l=1}^{2}
f^{(l)}(y)
\psi_{[m]}^{(l)}(y,i\rho)\varphi_{[j]}^{(k)}(x,i\rho)\overline{g^{(k)}(x)}{\rm
d}x{\rm d}y\Big) }\\
&={\sum_{j,m=1}^{2}\int_{-\infty}^{\infty}D_{n,jm}^\sigma(\rho)\omega_f^{m}(\rho)
\eta_{\overline{g}}^j(\rho){\rm d}\rho}.
\end{align*}
Therefore, in view of \eqref{lim-delta}, 
by letting $n\to \infty$ we obtain that for any 
$f,g$ in $\left({\mathbb{K}}^2_\sigma(0,\infty)\right)^2$,
\begin{equation}\label{Par-1}
{\int_0^\infty f^T(x)\overline{g(x)}{\rm d}x
=\lim_{n\to \infty}\sum_{j,m=1}^{2}\int_{-\infty}^{\infty}D_{n,jm}^\sigma
(\rho)\omega_f^{m}(\rho)
\eta_{\overline{g}}^j(\rho){\rm d}\rho}.
\end{equation}
By the definition of $D_n^\sigma(\rho)$ (see \eqref{Dn}), we easily see  that
\begin{equation}\label{lim-1}
\lim_{n\to \infty}D_n^\sigma(\rho)=\frac{1}{2\pi}R(P,0)(0)=\frac{1}{2\pi}E.
\end{equation}
On the other hand, since the Fourier transform is a continuous mapping of 
$L^2(\mathbb{R})$ into $L^2(\mathbb{R})$, it follows easily from Corollary \ref{coro2.2}
and the zero extensions of $f$ and $g$ on $\mathbb{R}$ that both
$\omega_f^{m}(\rho)$ and $\eta_{\overline{g}}^j(\rho)$ belong to $L^2(\mathbb{R})$.
Therefore, combining \eqref{Par-1} and \eqref{lim-1} and letting 
$\sigma\to  \infty$,
 we can assert \eqref{parseval-1} by the boundedness  of $D_n^\sigma(\cdot)$, 
the dominated convergence theorem and the fact that 
$\left({\mathbb{K}}^2(0,\infty)\right)^2$ is dense in 
$\left(L^2(0,\infty)\right)^2$.
\smallskip

\noindent\textbf{Forth step.} 
We prove the expansion \eqref{expansion-1}.
First we assume that $f\in{\left(C_0[0,\infty)\right)^2}$, where 
$\left(C_0[0,\infty)\right)^2$ denotes the product space of the set of 
all continuous functions with compact support. For any fixed real number 
$x\geq 0$ and $\delta>0$, set
\begin{equation}\label{varsigma}
\varsigma(t)=\begin{cases}
1/\delta  &\text{for } t\in (x,x+\delta),\\
0 &\text{otherwise}.
\end{cases}
\end{equation}
In \eqref{parseval-1}  first letting
$g^{(1)}(t)=\varsigma(t),g^{(2)}(t)=0$ and then letting
$g^{(1)}(t)=0,g^{(2)}(t)=\varsigma(t)$, we have
$$
{\frac{1}{\delta}\int_x^{x+\delta}f(t){\rm d}t=
\frac{1}{2\pi}\sum_{k=1}^{2}\int_{-\infty}^{\infty}\omega_f^{k}(\rho)
\frac{1}{\delta} \int_x^{x+\delta} \varphi_{[k]}(t,i\rho){\rm d}t\,{\rm d}\rho}.
$$
Since
$$
{\lim_{\delta\to  0}\frac{1}{\delta}\int_x^{x+\delta}f(t){\rm d}t=f(x)}
$$
and  in $Z$,
$$
{\lim_{\delta\to
0}\omega_f^{k}(\rho)\frac{1}{\delta} \int_x^{x+\delta}
\varphi_{[k]}(t,i\rho){\rm d}t=\omega_f^{k}(\rho)\varphi_{[k]}(x,i\rho)},
$$
we prove the first part of \eqref{expansion-1} by the dominated convergence 
theorem if
$f\in{\left(C_0[0,\infty)\right)^2}$. For the case of 
$f\in\left({\mathbb{K}}^2(0,\infty)\right)^2$ we can
approximate $f$ by the functions in
$\left(C_0[0,\infty)\right)^2$. The second part of
\eqref{expansion-1} can be proved similarly.

\section{Proof of Theorem \ref{thm2}}

First we prove the theorem for a special case. 
Recall that $S$ and $\widetilde{S}$ are the solutions corresponding to
$P=0$ in \eqref{eq-phi} and \eqref{eq-phi*}, respectively.

\begin{lemma} \label{lem4.1} 
For $f,g\in{\left(L^2(0,\infty)\right)^4}$,
it holds 
$$
\int_0^\infty f(x)g(x){\rm d}x
=\frac{1}{\pi}\int_{-\infty}^\infty
\Theta_f(\rho)\widetilde{\Theta}_g(\rho){\rm d}\rho
=\frac{1}{\pi}\int_{-\infty}^\infty
\Theta_f(\rho)Q\widetilde{\Theta}_g(\rho){\rm d}\rho
$$ 
and for $x>0$,
$$
f(x)=\frac{1}{\pi}\int_{-\infty}^\infty
\Theta_f(\rho)\widetilde{S}(x,i\rho){\rm d}\rho=
\frac{1}{\pi}\int_{-\infty}^\infty
S(x,i\rho)\widetilde{\Theta}_f(\rho){\rm d}\rho,
$$
where $\Theta_f(\rho)$ and $\widetilde{\Theta}_g(\rho)$ are defined by
\begin{equation}\label{Theta}
\Theta_f(\rho)=\int_0^\infty f(x)S(x,i\rho){\rm d}x,\quad
 \widetilde{\Theta}_g(\rho)=\int_0^\infty \widetilde{S}(x,i\rho) g(x){\rm d}x.
\end{equation}
\end{lemma}

 \begin{proof}  
It is easy to find that
\begin{equation}\label{SS}
\begin{gathered}
S(x,i\rho)=Q\cosh(i\rho x)+BQ\sinh(i\rho x),\\
\widetilde{S}(x,i\rho) =Q\cosh(i\rho x)-QB\sinh(i\rho x),
\end{gathered}
\end{equation}
then we have
\begin{gather*}
\Theta_f(\rho)=\int_0^\infty f(x)S(x,i\rho){\rm
d}x=\frac{1}{2}\widehat{f}(\rho)(Q-BQ)+\frac{1}{2}\widehat{f}(-\rho)(Q+BQ), \\
\widetilde{\Theta}_g(\rho)=\int_0^\infty \widetilde{S}(x,i\rho) g(x){\rm
d}x=\frac{1}{2}(Q+QB)\widehat{g}(\rho)+\frac{1}{2}(Q-QB)\widehat{g}(-\rho),
\end{gather*}
where $\widehat{f}(\rho)=\int_0^\infty f(x)\exp(-i\rho x){\rm d}x$ 
denotes the Fourier transform of $f(x)$.
Therefore, by the well-known Parseval equality
$$
{\int_0^\infty f(x)g(x){\rm
d}x=\frac{1}{2\pi} \int_{-\infty}^\infty
\widehat{f}(\rho)\widehat{g}(-\rho){\rm d}\rho}
$$
 and the identity for $u,v\in{L^2(0,\infty)}$
 $$
\int_{-\infty}^\infty
\widehat{u}(\rho)\widehat{v}(\rho){\rm d}\rho=0,
 $$
 we obtain easily  (note that $Q^2=Q$)
$$
\frac{1}{\pi}\int_{-\infty}^\infty
\Theta_f(\rho)\widetilde{\Theta}_g(\rho){\rm
d}\rho=\frac{1}{\pi}\int_{-\infty}^\infty
\Theta_f(\rho)Q\widetilde{\Theta}_g(\rho){\rm d}\rho=\int_0^\infty
f(x)g(x){\rm d}x.
$$
On the other hand, since for all $u\in{L^2(0,\infty)}$ and $x>0$ it
holds that
$$
\int_{-\infty}^\infty\widehat{u}(\rho)\exp(-i\rho x){\rm d}\rho
=\int_{-\infty}^\infty\widehat{u}(-\rho)\exp(i\rho x){\rm d}\rho=0,
$$
we have
\begin{align*}
&{\frac{1}{\pi}\int_{-\infty}^\infty
\Theta_f(\rho)\widetilde{S}(x,i\rho){\rm d}\rho}\\
\\
&= {\frac{1}{2}f(x)(Q-BQ)(Q-QB)+\frac{1}{2}f(x)(Q+BQ)(Q+QB)=f(x).}
\end{align*}
 Similarly, we can show that
${\frac{1}{\pi}\int_{-\infty}^\infty
S(x,i\rho)\widetilde{\Theta}_f(\rho){\rm d}\rho=f(x)}$. 
\end{proof}

By putting
\begin{equation}\label{fF}
{f(x)=F(x)R(P,0)(x)+\int_x^\infty F(t)K(P,0;Q)(t,x){\rm d}t}
\end{equation}
 and
\begin{equation}\label{gG}
{g(x)=R(0,P)(x)G(x)+\int_x^\infty
\overline{K^T\big(-\overline{P^T},0;\overline{Q^T}\big)(t,x)}G(t){\rm
d}t},
\end{equation}
where $F$ and $G$ can be obtained by solving the above Volterra equations 
of the second kind,
it follows from changing the order of integration and the transformation formulae
\eqref{f2-f1} and \eqref{f2-f1*} that
\begin{equation}\label{fFgG}\Phi_f(\rho)=\Theta_F(\rho),\
\widetilde{\Phi}_g(\rho)=\widetilde{\Theta}_G(\rho).
\end{equation}
Furthermore, we have the following lemma.

\begin{lemma} \label{lem4.2} 
For $f,g\in{\left(L^2(0,\infty)\right)^4}$, it holds that
\begin{equation}\label{fg}
\int_0^\infty f(x)g(x){\rm d}x=\int_0^\infty F(x)G(x){\rm d}x
+\int_0^\infty\int_0^\infty F(y) {\mathfrak{F}}(x,y)G(x){\rm d}x{\rm d}y,
\end{equation}
where 
\begin{equation}\label{Fxy}
\mathfrak{F}(x,y)
=\begin{cases}
R(P,0)(y)\overline{K^T\big(-\overline{P^T},0;\overline{Q^T}\big)(x,y)}\\
+\int_0^y K(P,0;Q)(y,t)\overline{K^T
\big(-\overline{P^T},0;\overline{Q^T}\big)(x,t)}{\rm d}t,& 0\leq y\leq x,\\[4pt]
 K(P,0;Q)(y,x)R(0,P)(x) \\
+\int_0^x
K(P,0;Q)(y,t)\overline{K^T\big(-\overline{P^T},0;\overline{Q^T}\big)(x,t)}
{\rm d}t, & 0\leq x\leq y. 
\end{cases}
\end{equation}
\end{lemma}

\begin{proof}  On the one hand, since
$R(P,0)(\cdot)=R^{-1}(0,P)(\cdot)$, we have by changing of the order
of integration
\begin{align*}
&{\int_0^\infty f(x)g(x){\rm d}x}\\
&= {\int_0^\infty \Big\{ F(x)R(P,0)(x)+\int_x^\infty
F(t)K(P,0;Q)(t,x){\rm d}t\Big\}}\\
&\quad \times \Big\{
R(0,P)(x)G(x)+\int_x^\infty
\overline{K^T\big(-\overline{P^T},0;\overline{Q^T}\big)(t,x)}G(t){\rm
d}t\Big\}{\rm d}x\\
\\
&={\int_0^\infty F(x)G(x){\rm d}x
 +\int_0^\infty\int_x^\infty F(x)R(P,0)(x)\overline{K^T
 \big(-\overline{P^T},0;\overline{Q^T}\big)(t,x)}G(t)
{\rm d}t {\rm d}x }\\
&\quad  +{\int_0^\infty\int_0^t
F(t)K(P,0;Q)(t,x)R(0,P)(x)G(x){\rm d}x {\rm d}t}\\
&\quad +{\int_0^\infty\int_x^\infty\int_x^\infty
F(t)K(P,0;Q)(t,x)\overline{K^T\big(-\overline{P^T},0;\overline{Q^T}\big)(s,x)}G(s)
{\rm d}t{\rm d}s{\rm d}x}.
\end{align*}
 On the other hand, by \eqref{Fxy},
\begin{align*}
& {\int_0^\infty\int_0^\infty
F(y) {\mathfrak{F}}(x,y)G(x){\rm d}x{\rm d}y}\\
&={\int_0^\infty\int_0^y
F(y) \Big\{ K(P,0;Q)(y,x)R(0,P)(x)}\\
&\quad  +{\int_0^x
K(P,0;Q)(y,t)\overline{K^T\big(-\overline{P^T},0;\overline{Q^T}\big)(x,t)}
{\rm d}t \Big\} G(x){\rm d}x{\rm d}y}\\
&\quad +{\int_0^\infty\int_y^\infty
F(y) \Big\{ R(P,0)(y)\overline{K^T
\big(-\overline{P^T},0;\overline{Q^T}\big)(x,y)}}\\
&\quad +{\int_0^y K(P,0;Q)(y,t)\overline{K^T
\big(-\overline{P^T},0;\overline{Q^T}\big)(x,t)}{\rm
d}t  \Big\}G(x){\rm d}x{\rm d}y}. 
\end{align*}
Therefore,  proving \eqref{fg}, it is equivalent to showing that
\begin{align*}
& {\int_0^\infty\int_x^\infty\int_x^\infty
F(t)K(P,0;Q)(t,x)\overline{K^T\big(-\overline{P^T},0;\overline{Q^T}\big)(s,x)}G(s)
{\rm d}t{\rm d}s{\rm d}x}\\
&={\int_0^\infty\int_0^y\int_0^x F(y)
K(P,0;Q)(y,t)\overline{K^T\big(-\overline{P^T},0;\overline{Q^T}\big)(x,t)}
 G(x){\rm d}t {\rm d}x{\rm d}y}\\
&\quad +{\int_0^\infty\int_y^\infty  \int_0^y F(y)
K(P,0;Q)(y,t)\overline{K^T\big(-\overline{P^T},0;\overline{Q^T}\big)(x,t)}
G(x){\rm d}t {\rm d}x{\rm d}y },
\end{align*}
which can be easily proved by changing of the order of integration.
\end{proof}

From \eqref{R} and Lemma \ref{lem2.1} it follows that 
${\mathfrak{F}}(\cdot,\cdot)\in\left(C^1(\overline{\Omega})\right)^4$ and 
${\mathfrak{F}}(\cdot,\cdot)\in
\big(C^1(\overline{{\mathbb{R}^2_{+}}\setminus\Omega})\big)^4$.

\begin{lemma} \label{lem4.3} 
For ${\mathfrak{F}}(x,y)$ defined by \eqref{Fxy},
it holds that
\begin{gather}\label{eq-Fxy}
\frac{\partial {\mathfrak{F}}}{\partial x }(x,y)B
+B\frac{\partial {\mathfrak{F}}}{\partial y }(x,y)=0, \\
\label{cond-Fxy}
{\mathfrak{F}}(x,0)={\mathcal{J}}(x),\ \
{\mathfrak{F}}(0,y)={\mathcal{L}}(y),
\end{gather}
where
\begin{equation}\label{L-L}
{\mathcal{J}}(x)=\overline{K^T\big(-\overline{P^T},0;\overline{Q^T}\big)(x,0)},\
\ {\mathcal{L}}(y)=K(P,0;Q)(y,0).
\end{equation}
Moreover, 
\begin{equation}\label{r-L-L}
{\mathcal{J}}(x)-B{\mathcal{J}}(x)B={\mathcal{L}}(x)-B{\mathcal{L}}(x)B.
\end{equation}
\end{lemma}

\begin{proof}  
For $y\leq x$, in view of \eqref{eq-K}--\eqref{xx-K} in Lemma \ref{lem2.1}, we
see by integration by parts that
\begin{align*}
& \frac{\partial {\mathfrak{F}}}{\partial x }(x,y)B
+B\frac{\partial {\mathfrak{F}}}{\partial y }(x,y) \\
&=\Big\{ R(P,0)(y)\overline{
K^T_x\big(-\overline{P^T},0;\overline{Q^T}\big)(x,y)}\\
&\quad +\int_0^y K(P,0;Q)(y,s)\overline{
K^T_x\big(-\overline{P^T},0;\overline{Q^T}\big)(x,s)}{\rm
d}s\Big\}B \\
&\quad  +B \Big\{ {R'(P,0)(y)\overline{ K^T\big(-\overline{P^T},0;
 \overline{Q^T}\big)(x,y)}}\\
&\quad +K(P,0;Q)(y,y)\overline{ K^T\big(-\overline{P^T},0;
 \overline{Q^T}\big)(x,y)} \Big\}\\
&\quad +B \Big\{ {  R(P,0)(y)\overline{
K^T_y\big(-\overline{P^T},0;\overline{Q^T}\big)(x,y)}}\\
&\quad +\int_0^y K_y(P,0;Q)(y,s)\overline{K^T\big(-\overline{P^T},0;
\overline{Q^T}\big)(x,s)}{\rm d}s \Big\}\\
&=\big\{ BR'(P,0)(y)-R(P,0)(y)P(y)+BK(P,0;Q)(y,y)-K(P,0;Q)(y,y)B\big\}\\
&\quad \times \overline{
K^T\big(-\overline{P^T},0;\overline{Q^T}\big)(x,y)}+K(P,0;Q)(y,0)B\overline{
K^T\big(-\overline{P^T},0;\overline{Q^T}\big)(x,0)}\\
&\quad +{\int_0^y
\big\{BK_y(P,0;Q)(y,s)+K_s(P,0;Q)(y,s)B-K(P,0;Q)(y,s)P(s)\big\}}\\
& \quad \times \overline{
K^T\big(-\overline{P^T},0;\overline{Q^T}\big)(x,s)}{\rm d}s 
=0,
\end{align*}
where we have  used  the relation: $B=QB+BQ$. For the case $x\leq y$, 
the proof of \eqref{eq-Fxy} is similar. On the other hand,
\eqref{cond-Fxy} is obvious by \eqref{Fxy}.

Furthermore, it can be directly verified  that the unique solution of 
problem \eqref{eq-Fxy} and \eqref{cond-Fxy} is
$$
{\mathfrak{F}}(x,y)
=\begin{cases} \frac{1}{2}\{{\mathcal{J}}(x+y)+
{\mathcal{J}}(x-y)\}-\frac{1}{2}B\{{\mathcal{J}}(x+y)-
{\mathcal{J}}(x-y) \}B,& y\leq x, \\[4pt]
\frac{1}{2}\{{\mathcal{L}}(x+y)+{\mathcal{L}}(y-x)\}
-\frac{1}{2}B\{{\mathcal{L}}(x+y)
-{\mathcal{L}}(y-x)\}B,& x\leq y.
\end{cases}
$$
Consequently, \eqref{r-L-L} follows from the continuity of
${\mathfrak{F}}(x,y)$ at $x=y$.
\end{proof}

Now we apply Lemma \ref{lem4.3} to show the following lemma.

\begin{lemma} \label{lem4.4} 
It holds that
$$
\Theta_{{\mathcal{J}}}(\rho)Q=\Theta_{{\mathcal{J}}}(\rho)
=\widetilde{\Theta}_{\mathcal{L}}(\rho)=Q\widetilde{\Theta}_{\mathcal{L}}(\rho).
$$
\end{lemma}

\begin{proof}   
By \eqref{Theta}, we have
$$
\Theta_{{\mathcal{J}}}(\rho)
=\int_0^\infty {{\mathcal{J}}}(x) S(x,i\rho){\rm d}x,\quad
\widetilde{\Theta}_{\mathcal{L}}(\rho)=\int_0^\infty
\widetilde{S}(x,i\rho){\mathcal{L}}(x){\rm d}x,
$$
where $S(x,i\rho)$ and $\widetilde{S}(x,i\rho)$ are given by \eqref{SS}.
 Since $Q^2=Q$, it is sufficient to prove that for all $x\geq 0$,
 $$
{{\mathcal{J}}}(x)Q=Q{\mathcal{L}}(x),\quad
 {{\mathcal{J}}}(x)BQ=-QB{\mathcal{L}}(x).
 $$
First, multiplying right \eqref{r-L-L} by $Q$, we obtain by
$QB+BQ=B$ and ${\mathcal{L}}(x)Q={\mathcal{L}}(x)$ which follows from 
\eqref{K-p1p2-0} and \eqref{L-L} that
\begin{equation}\label{L=}
\{{{\mathcal{J}}}(x)-B{{\mathcal{J}}}(x)B\}Q=
{\mathcal{L}}(x)Q-B{\mathcal{L}}(x)(B-QB)={\mathcal{L}}(x).
\end{equation}
Second, since it follows from \eqref{K-p1p2*} and \eqref{L-L} that 
$Q{{\mathcal{J}}}(x)={{\mathcal{J}}}(x)$, we have
$QB{{\mathcal{J}}}(x)=(B-BQ){{\mathcal{J}}}(x)=0$.
Consequently, it follows from \eqref{L=} that
$$
Q{\mathcal{L}}(x)=Q\{{{\mathcal{J}}}(x)-B{{\mathcal{J}}}(x)B\}Q
={{\mathcal{J}}}(x)Q-QB{{\mathcal{J}}}(x)BQ={{\mathcal{J}}}(x)Q.
$$
On the other hand, multiplying left \eqref{L=} by $B$,  by
$B^2=E$ we have
$$
B{{\mathcal{J}}}(x)Q-{{\mathcal{J}}}(x)BQ
=B{\mathcal{L}}(x)=(QB+BQ){\mathcal{L}}(x);
$$
that is,
$$
{{\mathcal{J}}}(x)BQ+QB{\mathcal{L}}(x)=B\{{{\mathcal{J}}}(x)Q-Q{\mathcal{L}}(x)\}=0.
$$
Thus the proof is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2}]
 Let
${\mathcal{L}}_\sigma(x)=\gamma_\sigma(x){\mathcal{L}}(x)$,
${{\mathcal{J}}}_\sigma(x)=\gamma_\sigma(x){{\mathcal{J}}}(x)$, 
where the scalar function $\gamma_\sigma(x)$ is defined by \eqref{gamma-sigma}. 
It is obvious that both ${\mathcal{L}}_\sigma$ and ${{\mathcal{J}}}_\sigma$ 
are continuously differentiable matrix-valued functions with compact support.
Then it follows easily from Lemma \ref{lem4.4} that 
$\Theta_{{\mathcal{J}}_\sigma}(\rho)=\widetilde{\Theta}_{{\mathcal{L}}_\sigma}(\rho)$.
Hence, combining \eqref{eq-phi}, \eqref{eq-phi*}, \eqref{SS}, 
$Q{{\mathcal{J}}}(\cdot)={{\mathcal{J}}}(\cdot)$,
 ${\mathcal{L}}(\cdot)Q={\mathcal{L}}(\cdot)$ and Lemma \ref{lem4.1}, we conclude easily  
that the  following matrix-valued function
$$
{\mathfrak{F}}_\sigma(x,y):=\frac{1}{\pi}\int_{-\infty}^\infty
S(y,i\rho)\Theta_{{\mathcal{J}}_\sigma}(\rho)\widetilde{S}(x,i\rho){\rm
d}\rho=\frac{1}{\pi}\int_{-\infty}^\infty
S(y,i\rho)\widetilde{\Theta}_{{\mathcal{L}}_\sigma}(\rho)\widetilde{S}(x,i\rho){\rm
d}\rho
$$
satisfies the equation
$$
U_xB+BU_y=0,
$$
and the conditions
$$
U(x,0)={{\mathcal{J}}}_\sigma(x),\quad U(0,y)={\mathcal{L}}_\sigma(y)
$$
for all $x,y>0$. Therefore, if we define
${\mathfrak{F}}_\sigma(0,0)={\mathcal{L}}(0)={{\mathcal{J}}}(0)$,
then  ${\mathfrak{F}}_\sigma(x,y)={\mathfrak{F}}(x,y)$ in the
domain $0\leq x,y\leq\sigma$, since $\gamma_\sigma(x)\equiv 1$ on 
$[0,\sigma]$ and the two matrix-valued functions
satisfy the same boundary problem as that in Lemma \ref{lem4.3}. Moreover, 
if $f,g\in{\left({\mathbb{K}}^2_\sigma(0,\infty)\right)^4}$, 
then it follows from \eqref{fF} and \eqref{gG} that $F(x)=G(x)=0$ for
$x>\sigma$. Consequently, it follows from \eqref{Theta}, \eqref{fFgG}, 
Lemmas \ref{lem4.1} and \ref{lem4.2} that
\begin{equation}\label{expan-1}
\begin{aligned}
\int_0^\infty f(x)g(x){\rm d}x
&={\int_0^\infty F(x)G(x){\rm d}x+\int_0^\infty\int_0^\infty
F(y) {\mathfrak{F}}(x,y)G(x){\rm d}x{\rm d}y}\\
&={\int_0^\infty F(x)G(x){\rm d}x+\int_0^\infty\int_0^\infty
F(y) {\mathfrak{F}}_\sigma(x,y)G(x){\rm d}x{\rm d}y}\\
&={\frac{1}{\pi}\int_{-\infty}^\infty
\Theta_F(\rho)\{Q+\Theta_{{\mathcal{J}}_\sigma}(\rho)\}
 \widetilde{\Theta}_G(\rho){\rm d}\rho}\\
&={\frac{1}{\pi}\int_{-\infty}^\infty
\Phi_f(\rho)\{Q+\widetilde{\Theta}_{\mathcal{L}_\sigma}(\rho)\}
 \widetilde{\Phi}_g(\rho){\rm d}\rho}.
\end{aligned}
\end{equation}
Now we define
\begin{equation}\label{D}
D(\rho)=\frac{1}{\pi}\lim_{\sigma\to  \infty}
\{Q+\Theta_{{\mathcal{J}}_\sigma}(\rho)\}
=\frac{1}{\pi}\lim_{\sigma\to  \infty}\{Q+\widetilde{\Theta}_{\mathcal{L}_\sigma}
(\rho)\}
\end{equation}
where the limits exist in the sense of convergence of distributions. 
 Indeed, by \eqref{Theta} and \eqref{SS} we see that both 
$\Theta_{{\mathcal{J}}_\sigma}(\rho)$ and 
$\widetilde{\Theta}_{\mathcal{L}_\sigma}(\rho)$
  are linear combination of the Fourier cosine and  sine transform of some 
matrix-valued function with compact support. Then it follows from the property 
of the Fourier transform (see e.g. Page 105 in \cite{mar}) that 
$\Theta_{{\mathcal{J}}_\sigma}(\rho)\to \Theta_{\mathcal{J}}(\rho)$ and 
$\widetilde{\Theta}_{\mathcal{L}_\sigma}(\rho)
\to \widetilde{\Theta}_{\mathcal{L}}(\rho)$ as 
$ \sigma\to \infty$ in the sense of distributions, whence $D(\rho)\in{(Z')^4}$. 
Therefore, by \eqref{D} we see
\begin{equation}\label{D1}
D(\rho)={\frac{1}{\pi}\{Q+\Theta_{{\mathcal{J}}}(\rho)\}=\frac{1}{\pi}
\{Q+\widetilde{\Theta}_{\mathcal{L}}(\rho)\}}.
\end{equation}
Therefore, by \eqref{expan-1}, \eqref{D} and \eqref{D1} we can prove the 
Marchenko-Parseval equality \eqref{parseval-2} similarly to \eqref{parseval-1}.
 Moreover, if we let $g(t)=\varsigma(t)E$ or $f(t)=\varsigma(t)E$ where
$\varsigma(t)$ is defined by \eqref{varsigma}, then we can prove
\eqref{expansion-2} similarly to \eqref{expansion-1}.
\end{proof}

\subsection*{Acknowledgements}
The author would like to thank heartily Professors Vladimir Alexandrovich Marchenko 
and  Masahiro Yamamoto for their comments and suggestions. 
This work was partially supported by the National Natural Science Foundation 
of China (Grant No. 11101390), the Fundamental Research Funds for the 
Central Universities and JSPS Fellowship P05297.

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\end{thebibliography}

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