\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 36, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/36\hfil Normal extensions]
{Normal extensions of a singular multipoint differential operator 
 of first order}

\author[Z. I. Ismailov, R. \"Ozt\"urk Mert \hfil EJDE-2012/36\hfilneg]
{Zameddin I. Ismailov, Rukiye \"Ozt\"urk Mert}  % in alphabetical order

\address{Zameddin I. Ismailov \newline
Department of Mathematics, Faculty of Sciences,
Karadeniz Technical University, 61080, Trabzon, Turkey}
\email{zameddin@yahoo.com}

\address{Rukiye \"Ozt\"urk Mert \newline
Department of Mathematics, Art and Science Faculty,
Hitit University, 19030, Corum, Turkey}
\email{rukiyeozturkmert@hitit.edu.tr}

\thanks{Submitted August 10, 2011. Published Match 7, 2012.}
\subjclass[2000]{47A10, 47A20}
\keywords{Multipoint differential operators; selfadjoint and normal extension;
\hfill\break\indent  spectrum}

\begin{abstract}
 In this work, we describe all normal extensions of the minimal
 operator generated by linear singular multipoint formally normal differential
 expression $l=(l_1,l_2,l_3)$, $l_k=\frac{d}{dt}+A_k$ with selfadjoint operator
 coefficients $A_k$ in a Hilbert space.
 This is done as a direct sum of Hilbert spaces of vector-functions
 \[
 L_2(H,(-\infty ,a_1))\oplus L_2(H,(a_2,b_2))
 \oplus L_2(H,(a_3,+\infty))
 \]
 where $-\infty <a_1<a_2<b_2<a_3<+\infty$. 
 Also, we study the structure of the spectrum of these extensions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction}

Many problems arising in modeling processes in multi-particle quantum mechanics,
in quantum field theory, in multipoint boundary value problems for
differential equations, and in the physics of rigid bodies
use normal extensions of formaly normal differential operators
as a direct sum of Hilbert spaces \cite{Alb,Tim,Ze}.
The general theory of these normal extensions of formally normal operators in Hilbert
spaces has been investigated by many mathematicians; see for example
\cite{Bir,Cod,Dav,Kilp,Kilp2,Kilp3}.
Applications of this theory to two-point differential operators in Hilbert space
of functions can be found in \cite{Ism,Ism2,Ism3,Ism4,Ism5,Maks,Maks2,Maks3}.

It is clear that the minimal operators  $L_0(1,0,0)=L_{10}\oplus 0\oplus 0$ and
$L_0(0,0,1)=0\oplus 0\oplus L_{30}$ generated by differential expressions for the
forms $(\frac{d}{dt}+A_1,0,0)$   and ($0,0,\frac{d}{dt}+A_3$) in  Hilbert spaces
of vector-functions ${L_2(1,0,0)=L}_2(H,(-\infty,a_1))\oplus 0\oplus 0$,
$L_2(0,0,1)=0\oplus 0\oplus L_2(H,(a_3$,$\infty ))$ respectively, where
$A_1=A^*_1\le 0$, $A_3=A^*_3\ge 0$, $-\infty <a<b<\infty$,
are maximal formally normal.
Consequently they do not have normal extensions. 
But the minimal operator
$L_0(0,1,0)=0\oplus L_{20}\oplus 0$ generated by differential  expression of
the form $(0,\frac{d}{dt}+A_2,0)$ in the  Hilbert spaces of  vector-functions
$L_2(0,1,0)=0\oplus L_2(H,(a_2,b_2))\oplus 0$ is formally normal and not maximal.

Unfortunately, multipoint situations may occur in different tables in the
following sense. Let $B_1$ , $B_2$ and $B_3$ be minimal operators generated by
the linear differential expression $\frac{d}{dt}$ in the Hilbert space of
functions $L_2(-\infty ,a_1)$ ,$\ L_2(a_2,b_2)$ and $L_2(a_3,+\infty)$ where
 $-\infty <a_1<a_2<b_2<a_3<\infty $ respectively.
Consequently, $B_{{\rm 1}}$ and $B_3$ are maximal formally normal operators,
but are not normal extensions.   However, the direct sum $B_1\oplus B_2\oplus B_3$
of operators $B_1, {B_2\ and\ B}_3$ in a direct sum
$L_2(-\infty ,a_1)\oplus L_2(a_2,b_2)\oplus L_2(a_3,+\infty)$  has a normal extension.
For example, in case $H=\mathbb{C}$ it can be easily shown that an extension of the
minimal operator $B_1\oplus B_2\oplus B_3$ with the boundary conditions
\begin{gather*}
u_3(a_3)=e^{i\varphi }u_1(a_1), \quad \varphi \in [0,2\pi ), \\
u_2(\ b_2)=e^{i\psi }u_2(a_2\ ), \psi \in [0,2\pi ),\\
u=(u_1,u_2,u_3),\quad  u_1\in D(B^*_1),\quad  u_2\in D(B^*_2), u_3\in D(B^*_3)
\end{gather*}
is normal in  $L_2(-\infty ,a_1)\oplus L_2(a_2,b_2)\oplus L_2(a_3,+\infty)$.

In the general case of $H$ being the direct sum
 ${L_0(1,1,1)=L}_{10}\oplus L_{20}\oplus L_{30}$ of operators
$L_{10}$, $L_{20}$ and $L_{30}$ is formally normal, is not maximal.
 Moreover it has normal extensions in the direct sum
\[
{L_2(1,1,1)=L}_2(H,(-\infty ,a_1))\ \oplus {L_2(H,(a_2,b_2))\oplus L}_2(H,(a_3,\infty )).
\]
In singular cases, however, there has been no investigation
so far. But the physical and technical process for many of the problems resulting
from the examination of the solution is of great importance for the singular cases.

In this article, it will be considered as a linear multipoint differential-operator
expression
\[
l=(l_1,l_2,l_3), \quad l_k=\frac{d}{dt}+A_k,\quad k=1,2,3
\]
in the direct sum of Hilbert spaces of vector-functions $L_2(1,1,1),$  where
$A_1=A^*_1\le 0$, $A_2=A^*_2\ge 0$, $A_3=A^*_3\ge 0$,
$-\infty <a_1<a_2<b_2<a_3<\infty$.

In the second section, by the method of Calkin-Gorbachuk theory, we describe
all normal extensions of the minimal operator generated by singular multipoint
formally normal differential expression for first order $l(.\ )$
in the direct sum of Hilbert space $L_2(1,1,1)$ in terms of boundary values.
In the third section the spectrum of such extensions is studied.


\section{Description of  normal extensions}

Let $H$ be a separable Hilbert space and $a_1,a_2,b_2,a_3\in\mathbb{R}$,
$a_1<a_2<b_2<a_3$. In the Hilbert space $L_2(1,1,1)$ of vector-functions let us
consider the  linear multipoint differential expression
\begin{gather*}
l(u)=(l_1(u_1),l_2(u_2),l_3(u_3))=(u'_1+A_1u_1,u'_2+A_2u_2,u'_3+A_3u_3),
\end{gather*}
where $u=(u_1,u_2,u_3)$,
 $A_k:D(A_k)\subset H\to H$, $k=1,2,3$ are linear selfadjoint operators
in $H$ and  $A_1=A^*_1\le 0$, $A_2=A^*_2\ge 0$, $A_3=A^*_3\ge 0$.
In the linear manifold $D(A_k)\subset H$ introduce the inner product
\[
{(f,g)}_{k,+}:={(A_kfA_k,g)}_H+{(f,g)}_H,\quad  f,g\in D(A_k),\; k=1,2,3.
\]
For $k=1,2,3$, $D(A_k)$ is a Hilbert space under the positive norm
${\|\cdot \|}_{k,+}$ with  respect to Hilbert space $H$.
It is denoted by $H_{k,+}$. Denote the $H_{k,-}$ a Hilbert space with the negative norm.
It is clear that an operator $A_k$ is continuous from $H_{k,+}$ to $H$ and that
 its adjoint operator ${\tilde{A}}_k:H\to H_{k,-}$ is a extension of the operator $A_k$.
On the other hand, the operator
 ${\tilde{A}}_k:D({\tilde{A}}_k)=H\subset H_{k,-1}\to H_{k,-1}$ is a linear selfadjoint.

In the direct sum, $\ L_2(1,1,1)$ is defined by
\begin{equation} \label{eq2.1}
\tilde{l}(u)=({\tilde{l}}_1(u_1),{\tilde{l}}_2(u_2),{\tilde{l}}_3(u_3)),
\end{equation}
where $u=(u_1,u_2,u_3)$ and ${\tilde{l}}_1(u_1)=u'_1+{\tilde{A}}_1u_1$,
${\tilde{l}}_2(u_2)=u'_2+{\tilde{A}}_2u_2$,
${\tilde{l}}_3(u_3)=u'_3+{\tilde{A}}_3u_3$.

The operators  $L_0(1,1,1)=L_{10}\oplus L_{20}\oplus L_{30}$ and
 $L(1,1,1)=L_1{\oplus L}_2\oplus L_3$ in the space $L_2(1,1,1)$ are called
 minimal (multipoint) and maximal (multipoint) operators generated by the
differential expression \eqref{eq2.1}, respectively.

Here all normal extensions of the minimal operator $L_0(1,1,1)$ in  ${L}_2(1,1,1)$
in terms  of the boundary values are described.

Note that a space of boundary values has an important role in the theory extensions
of the linear symmetric differential operators \cite{Gor} and will be used in the last
investigation.

Let $B:D(B)\subset \mathcal{H}\to \mathcal{H}$ be a closed densely defined symmetric
operator in the Hilbert space $\mathcal{H}$, having equal finite or infinite
deficiency indices. A triplet $(\mathfrak{H},{\gamma }_1,{\gamma }_2)$, where
 $\mathfrak{H}$ is a Hilbert space, ${\gamma }_1$ and ${\gamma }_2$ are linear mappings
of $D(B^*)$ into $\mathfrak{H}$, is called a space of boundary values for the operator
 $B$ if for any $f,g\in D(B^*)$
\[
{(B^*f,g)}_{\mathcal{H}}-{(f,B^*g)}_{\mathcal{H}}
={({\gamma }_1(f),{\gamma }_2(g))}_{\mathfrak{H}}-{({\gamma }_2(f),
{\gamma }_1(g))}_{\mathfrak{H}},
\]
while for any $F,G\in \mathfrak{H}$, there exists an element $f\in D(B^*)$,
such that ${\gamma }_1(f)=F\ $and ${\gamma }_2(f)=G$.

Note that any symmetric operator with equal deficiency indices has at least one
space of boundary values \cite{Gor}.

Now let us construct a space of boundary values for the minimal operators
$M_{0\ }(1,0,1)$ and $M_0(0,1,0)$ generated by linear  singular differential
 expressions of first order in the  form
\begin{gather*}
({m_1(u_1),0,m_3(u}_3)) =(-i\frac{du_1}{dt},0,-i\frac{du_3}{dt} ),\\
(0,m_2(u_2),0)=(0,\ -i\frac{du_2}{dt},0)
\end{gather*}
in the direct sum  $L_2(1,0,1)\ $ and $L_2(0,1,0)$, respectively.
 Note that the minimal operators $M_0 (1,0,1)$ and $M_0(0,1,0)$ are
closed symmetric operators in $L_2(1,0,1)$ and $L_2(0,1,0)$ with deficiency
 indices $(\dim H,\dim H)$ .

\begin{lemma}\label{lem1}
 The triplet   $(H,\gamma_1,\gamma_2)$, where
\begin{gather*}
\gamma_1:D(M^*_0)\to H,\quad \gamma_1(u)=\frac{1}{i\sqrt{2\ }}(u_3(a_3)+u_1(a_1)),\\
\gamma_2 :D(M^*_{0\ })\to H, \quad
\gamma_2(u)=\frac{1}{\sqrt{2}}(u_3(a_3)-u_1(a_1)),\ u=(u_1,0,u_3)\in D(M^*_0)
\end{gather*}
is a space of boundary values of the minimal operator ${M}_0(1,0,1)$  in $L_2(1,0,1)$.
\end{lemma}

\begin{proof}
For arbitrary $u=(u_1,{,0,u}_3)$ and $v=(v_{1,}{0,v}_3) $ from $D(M^*_0(1,0,1))$
the validity of the  equality
\begin{align*}
&{(M^*_0(1,0,1)u,v)}_{L_2(1,0,1)}-({u,M^*_0(1,0,1)v)}_{L_2(1,0,1)}\\
&={(\gamma_1(u),\gamma_2( v))}_H-{(\gamma_2(u),\gamma_1(v))}_H
\end{align*}
can be easily verified. Now for any given elements  $f,g\in H$, we will find the
 function $u=(u_1,{0,u}_3)\in D(M^*_0(1,0,1))\ $ such that
\[
\gamma_1(u)=\frac{1}{i\sqrt{2\ }}(u_3(a_3)+u_1(a_1))=f
\quad\text{and}\quad \gamma_2(u)=\frac{1}{\sqrt{2}}(u_3(a_3)-u_1(a_1))=g;
\]
that is,
\[
u_1(a_1)={(if-g)}/{\sqrt{2}} \quad\text{and}\quad
u_3(a_3)={(if+g)}/{\sqrt{2}}.
\]
If we choose the functions  $u_1(t),u_2(t)$  in the form
$u_1(t)=\int^t_{-\infty }{e^{s-a}}{ds(if-g)}/{\sqrt{2}}$ with
$t<a_1$ and $u_3(t)=\int^{\infty }_t{e^{a_3-t}}ds{(if+g)}/{\sqrt{2}}$ with
$t>a_3$, then it is clear that $(u_1,u_2)\in D(M^*_0)$  and $\gamma_1(u)=f$,
 $\gamma_2(u)=g$.
\end{proof}

\begin{lemma}\label{lem2}
 The triplet $(H,{\Gamma }_1,{\Gamma }_2)$,
\begin{gather*}
{\Gamma }_1:D(M^{*}_{0}(0,1,0))\to H, \quad
\Gamma_1(u)=\frac{1}{i\sqrt{2}}({u_2(b_2)+u}_2(a_2)),\\
{\Gamma }_2:D(M^*_0(0,1,0))\to H, \quad
 {\Gamma }_2(u)=\frac{1}{\sqrt{2}}(u_2(b_2)-u_2(a_2)),\\
  u=(0,u_2,0)\in D(M^*_0(0,1,0))
\end{gather*}
is a space of boundary values of the minimal operator $M_0(0,1,0)$ in the
 direct sum $L_2(0,1,0)$.
\end{lemma}

\begin{theorem} \label{thm3}
If the minimal operators $L_{10}$, $L_{20}$ and $L_{30}$ are formally normal then
\begin{gather*}
D(L_{10}){\subset W}^1_2(H,(-\infty ,a_1)),\quad
A_1D(L_{10})\subset L_2(H,(-\infty ,a_1)),\\\
D(L_{20}){\subset W}^1_2(H,(a_2,b_2)),\quad
A_2D(L_{20})\subset L_2(H,(a_2,b_2)),\\
D(L_{30})\subset W^1_2(H,(a_3,\infty )),\quad
A_3D(L_{30})\subset L_2(H,(a_3,\infty )).
\end{gather*}
\end{theorem}

\begin{proof}
Indeed, in this case for each $u_1\in D(L_{10})\subset D(L^*_{10})$ is true
\[
u'_1+A_1u_1\in L_2(H,(-\infty ,a))\ \ and\ \ u'_1-A_1u_1\in L_2(H,(-\infty ,a)),
\]
hence
\[
u'_1\in L_2(H,(-\infty ,a))\quad\text{and}\quad
A_1u_1\in L_2(H,(-\infty ,a));
\]
i.e.,
\[
D(L_{10}){\subset W}^1_2(H,(-\infty ,a))\ \ and\ \ { A}_1D(L_{10})\subset
L_2(H,(-\infty ,a)).
\]
The second and third parts of theorem can be proved in a similar way.
\end{proof}

The following result can be easily established.

\begin{lemma}\label{lem4}
Every normal extension of $L_0(1,1,1) $ in $L_2(1,1,1) $ is a direct sum of normal
extensions of the minimal operator ${ L}_0(1,0,1)=L_{10}\oplus 0\oplus L_{30}$ in
\[
L_2(1,0,1){=L}_2(H,(-\infty ,a_1))\oplus{0\oplus L}_2(H,(a_3,\infty ))
\]
and minimal operator $L_0(0,1,0)=0\oplus L_{20}\oplus 0$ in
$L_2(0,1,0)={0\oplus L}_2(H,(a_2,b_2))\oplus $0.
\end{lemma}

Finally, using the method in \cite{Gor,Ism,Ism2,Ism3,Ism4,Ism5,Maks,Maks2,Maks3}
and Lemmas \ref{lem1} and \ref{lem2} the following result can be deduced.

\begin{theorem} \label{thm5}
Let ${(-A_1)}^{1/2}W^1_2(H,(-\infty ,a_1))\subset W^1_2(H,(-\infty ,a_1))$,
\begin{gather*}
A^{1/2}_2W^1_2(H,(a_2,b_2)) \subset W^1_2(H,(a_2,b_2)),\\
 A^{1/2}_3W^1_2(H,(a_3,\infty )) \subset W^1_2(H,(a_3,\infty )).
\end{gather*}
Each normal extension $\tilde{L}$ of the minimal operator $L_0$ in the Hilbert
space $L_2(1,1,1)$ is generated by differential expression \eqref{eq2.1} and
boundary conditions
\begin{gather} \label{GrindEQ__2_3_}
u_3(a_3)=W_1u_1(a_1),\quad
u_1(a_1)\in {\ker  {(-A_1)}^{1/2}\ }, \quad
u_3(a_3)\in \ker A^{1/2}_3, \\
\label{GrindEQ__2_4_}
u_2(b_2)=W_2u_1(a_2),
\end{gather}
where $W_1,W_2:H\to H$ is a unitary operators. Moreover, the unitary
operators $W_1,W_2$ in $H$ are determined by the extension $\tilde{L}$;
 i.e., $\widetilde{\ L}=L_{W_1W_2}$ and vice versa.
\end{theorem}

\begin{corollary}\label{cor6}
If at least one of the operators $A_1$ and $A_3$ is one-to-one mapping in $H$,
then minimal operator $L_0(1,1,1)$ is maximally formal normal in $\ L_2(1,1,1)$.
\end{corollary}

\begin{corollary}\label{cor7}
If there exists at least one normal extension of the minimal operator
$L_0(1,1,1),$ then
\[
\dim\ker{(-A_1)}^{1/2}=\dim\ker A^{1/2}_3>0.
\]
\end{corollary}

\section{The spectrum of the normal extensions}

In this section the structure of the spectrum  of the normal extension $L_{W_1W_2}$
in $L_2(1,1,1)$  will be investigated. In this case by the Lemma \ref{lem4} it is
clear that
\[
L_{W_1W_2}=L_{W_1}\oplus L_{W_2}  ,
\]
where $L_{W_1}$  and $L_{W_2}$ are normal extensions of the minimal operators
 $L_0(1,0,1)$ and $L_0(0,1,0)$  in the Hilbert spaces $L_2(1,0,1)$ and
$L_2(0,1,0)$ respectively.
Later, it will be assumed that  $A_1=A^*_1\le 0$, $A_2=A^*_2\ge 0, A_3=A^*_3\ge 0$
and  $0\in {\sigma }_p({(-A_1)}^{1/2})\cap {\sigma }_p(A^{1/2}_3)$.
 First, we have to prove the following result.

\begin{theorem} \label{thm3.1}
The point spectrum of any normal extension $L_{W_1}$ of the minimal operator
$L_0(1,0,1)$ in the Hilbert space $L_2(1,0,1)$  is empty; i.e.,
$\sigma_p(L_{W_1})=\emptyset $.
\end{theorem}

\begin{proof}
Let us consider the following problem for the spectrum of the normal extension
 $L_{W_1}$ of the minimal operator  $L_0(1,0,1)$ in the Hilbert space $L_2(1,0,1)$,
\[
L_{W_1}u=\lambda u, \quad \lambda ={\lambda }_r+i{\lambda }_i\in {\mathbb C},\quad
 u=(u_1,0,u_3)\in L_2(1,0,1);
\]
that is,
\begin{gather*}
{\tilde{l}}_1(u_1)=u'_1+{\tilde{A}}_1u_1=\lambda u_1,\quad
 u_1\in L_2(H,(-\infty ,a_1)),\\
{\tilde{l}}_3(u_3)=u'_3+{\tilde{A}}_3u_3=\lambda u_3,\quad u_3\in  L_2(H,(a_3,+\infty )),
\quad \lambda \in \mathbb{R}, \\
u_3(a_3)=W_1u_1(a_1) ,\quad u_1(a_1)\in {\ker  {(-A_1)}^{1/2}},\quad
u_3(a_3)\in \ker A^{1/2}_3 .
\end{gather*}
The general solution of this problem is
\begin{gather*}
u_1(\lambda ;t)=e^{-({\tilde{A}}_1-\lambda )(t-a_1)}f^*_1,\quad t<a_1,
\; f^*_1\in H_{{-1}/{2}}(-A_1), \\
u_3(\lambda ;t)=e^{-({\tilde{A}}_3-\lambda )(t-a_3)}f^*_3,\quad t>a_3,\;
 f^*_3\in H_{{-1}/{2}}(A_3), \\
f^*_3=W_1f^*_1,\quad f^*_1,f^*_3\in H,\quad f^*_1=u_1(\lambda ;a_1),\quad
 f^*_3=u_3(\lambda ;a_3).
\end{gather*}
  Since $0\in {\sigma }_p({(-A_1)}^{1/2})\cap {\sigma }_p(A^{1/2}_2)$ and 
${(-A_1)}^{1/2}f^*_1=0$,  $A^{1/2}_2f^*_3=0$, we have
\begin{gather*}
u_1 ( \lambda ;t)=e^{\lambda (t-a)}f^*_1,\quad t<a,\; f^*_1\in H_{-1/2}({(-A}_1)),\\
u_2 ( \lambda ;t)=e^{\lambda (t-b)}f^*_3,\quad t>b,\; f^*_3\in H_{-1/2}(A_3),\\
f^*_3=W_1f^*_1,\quad f^*_1=u_1(\lambda ;a_1),\quad f^*_3=u_3(\lambda ;a_3).
\end{gather*}
 In order for $ u_1(\lambda ;.)\in  L_2(H,(-\infty ,a_1))$ and 
$u_2(\lambda ;.)\in  L_2(H,(a_3,\infty ))$  necessary and sufficient condition 
is  that ${\lambda }_r\ge 0$ and ${\lambda }_r\le 0$ respectively.
Hence ${\lambda }_r=0$.
Consequently,
\begin{gather*}
 u_1( \lambda ;t)=e^{i{\lambda }_i(t-a_1)}f^*_1,\quad t<a_1,\\
u_2( \lambda ;t)=e^{i{\lambda }_i(t-a_3)}f^*_3,\quad t>a_3,\; f^*_3\ ={W_1f}^*_1.
\end{gather*}
In this case for  $u_1(\lambda ;.)\in L_2(H,(-\infty ,a_1))$
 and $u_2(\lambda ;.)\in L_2(H,(a_3,\infty ))$, clearly, necessary and sufficient 
conditions are that $f^*_1=0$, $f^*_3=0$. 
This implies that  $u_1=0$ and $u_2=0$ in $L^2$.
 Therefore ${\sigma }_p(L_{W_1})=\emptyset $.
\end{proof}

 Since residual spectrum of any normal operators in any Hilbert space is empty,
 it is sufficient to  investigate the continuous spectrum of the normal extensions 
$L_{W_1}$ of the minimal operator $L_0(1,0,1)$  in the Hilbert space $L_2(1,0,1)$.

\begin{theorem} \label{thm3.2} 
The continuous spectrum of any normal extension $L_{W_1}$ of the minimal operator
 $L_0(1,0,1)$  in the Hilbert space $L_2(1,0,1)$ is  ${\sigma }_c(L_{W_1})=i\mathbb{R}$.
\end{theorem}

\begin{proof} 
Assume that $\lambda \in {\sigma }_c(L_{W_1})$. 
Then by the  theorem for the spectrum of normal operators \cite{Dun},
\[
\sigma (L_{W_1})\subset \sigma (\operatorname{Re}L_{W_1})+i\sigma (\operatorname{Im}L_{W_1}\ ),
\]
we obtain that
\[
{\lambda }_r\in \sigma (\operatorname{Re}L_{W_1}),\quad {\lambda }_i\in 
\sigma (\operatorname{Im} L_{W_1}).
\]
This implies that  ${\lambda }_r\in \sigma (A_1)$ and ${\lambda }_r\in \sigma (A_3)$,
hence by the conditions to the operators $A_1$ and $A_3$ we have ${\lambda }_r=0$. 
On the other hand from the proof of previous theorem we see that 
${\ker  (\ L_{W_1}-\lambda )\ }=\{0\}$ for any $\lambda\in {\mathbb C}$.
 Consequently, ${\sigma }_c(\ L_{W_1})\subset i\mathbb {R}$. 
Furthermore, it is clear that for the $\lambda =i{\lambda }_i\in {\mathbb C}$ 
the general solution of the boundary value problem
\begin{gather*}
u'_1+A_1u_1=i{\lambda }_iu_1+f_1, \quad u_1, f_1\in L_2(H,(-\infty ,a_1)),\\
u'_3+A_3u_3=i{\lambda }_iu_3+f_3,\quad u_3, f_3\in L_2(H,(a_3,\infty )),\;
 {\lambda }_i\in \mathbb R,
\\
u_3(a_3)=W_1u_1(a_1),\ u_1(a_1)\in {\ker  {(-A_1)}^{1/2}\ },u_3(a_3)\in \ker A^{1/2}_3
\end{gather*}
will be of the  form
\begin{gather*}
u_1(i{\lambda }_i;t)=e^{-(A_1-{i\lambda }_i)(t-a_1)}f_{{i\lambda }_i}
-\int^{a_1}_t{e^{-(A_1-{i\lambda }_i)(t-s)}f_1(s)ds},\quad t<a_1,\\
u_3(i{\lambda }_i;t)={{\rm e}}^{-({{\rm A}}_{3}- i {\lambda }_{ i })(t-{a}_{3})}
g_{{ i \lambda }_{ i }}{\rm +}\int^{t}_{a_3}{{{\rm e}}^{-({{\rm A}}_{3}
- i {\lambda }_{ i })(t-{\rm s})}f_3(s){\rm ds,\quad t}>a_3,}\\
g_{{ i \lambda }_{ i }}=W_1f_{{i\lambda }_i}.
\end{gather*}
In this case,
\[
e^{-(A_1-{i\lambda }_i)(t-a_1)}f_{{i\lambda }_i}\in L_2(H,(-\infty ,a_1)),\quad
{{\rm e}}^{-({{\rm A}}_{{\rm 2}}- i {\lambda }_{ i })(t-a_3)}g_{{ i \lambda }_{ i }}
\in L_2(H,(a_3,\infty ))
\]
for any $g_{{ i \lambda }_{ i }}$,$f_{{i\lambda }_i}\in H$.
If choose   $f_1(t)=e^{i{\lambda }_it}e^{-(t-a_1)}f^*$,
 $f^*\in {\ker  {(-A_1)}^{1/2}\ }$, $t<a_1$, then
\begin{align*}
\int^{a_1}_t{e^{-(A_1-{i\lambda }_i)(t-s)}f_1(s)ds}
&=e^{-i{\lambda }_it}\int^{a_1}_t e^{-(s-a_1)}f^*ds\\
&=e^{-i{\lambda }_it}(e^{-(t-a_1)}-1)f^*, \quad t<a_1.
\end{align*}
Therefore,
\begin{align*}
\int^{a_1}_{-\infty }{{\|e^{-i{\lambda }_it}(e^{-(t-a_1)}-1)f^*\|}^2}dt
&=\int^{a_1}_{-\infty }{{\|e^{-i{\lambda }_it}(e^{-(t-a_1)}-1)f^*\|}^2}dt \\
&=\ \int^{a_1}_{-\infty } (e^{-2(t-a_1)}-2e^{-(t-a_1)}+1)dt{\|f^*\|}^2=\infty 
\end{align*}
Consequently, we have   $f_1(t)\in L_2(H,(-\infty ,a_1))$,
 $u_1(i{\lambda }_i;t)\notin L_2(H,(-\infty ,a_1))$. This implies that for 
any $\lambda\in {\mathbb C}$, an operator $L_{W_1}-\lambda $ is one-to-one in
${L}_{2}(1,0,1)$, but it is not an onto transformation. On the other hand, 
since the residual spectrum ${\sigma }_r(L_{W_1})$ is empty, we have 
 $\sigma (\ L_{W_1})={\sigma }_c(L_{W_1}\ )=i{\mathbb R}$.
\end{proof}

Now, we investigate the  spectrum of normal extensions $L_{W_2}$ of the minimal operator 
$L_0(0,1,0)$ in ${L}_2(1,0,1)$.


\begin{theorem} \label{thm3.3} 
The spectrum of the normal extension ${\ L}_{W_{2\ }}$ of the minimal operator 
$L_0(0,1,0)$ in the Hilbert space $L_2(0,1,0)$ is of the form
\begin{align*}
\sigma (L_{W_{2\ }})
=\big\{&\lambda \in{\mathbb C}{\rm :}{\rm \ }\lambda 
 =\frac{1}{a_2-b_2}({\ln |\mu|+i\ }\arg\mu+2n\pi i), n\in {\mathbb Z}, \\
 &\mu \in \sigma (W^*_2e^{-{\tilde{A}}_2(b_2-a_2)}), \;
 0\leq \arg \mu<2\pi \big\} 
\end{align*}
\end{theorem}

\begin{proof}
 The general solution of the problem spectrum  of the normal extension
$ L_{W_{2}}$,
\begin{gather*}
{\tilde{l}}_2(u_2)=u'_{2}+{\tilde{A}}_2u_2=\lambda u_2+f_2, \quad
{u}_2, f_2\in L_2(H,(a_2,b_2))\\
u_2(b_2)=W_2u_2(a_2), \quad \lambda \in \mathbb {C}
\end{gather*}
is of the form
\begin{gather*}
u_2(t)=e^{-({\tilde{A}}_2-\lambda )(t-a_2)}f^*_2
+\int^t_{a_2}{e^{-({\tilde{A}}_2-\lambda )(t-s)}}\ f_2(s)ds,\\
 a_2<t<b_2, \; f^*_2\in H_{{-1}/{2}}(A_2)\\
(e^{-\lambda (b_2-a_2)}-W^*_2e^{-{\tilde{A}}_2(b_2-a_2)})f^*_2
=W^*_2e^{-\lambda (b_2-a_2)}\int^{b_2}_{a_2}{e^{-({\tilde{A}}_2-\lambda )(b_2-s)}f_2(s)ds}
\end{gather*}
This implies that  $\lambda \in \sigma (L_{W_2})$  if and only if  
$\lambda $ is a solution of  the equation  $e^{-\lambda (b_2-a_2)}=\mu$,
 where $ \mu \in \sigma (W^*_2e^{-{\tilde{A}}_2(b_2-a_2)})$. 
 We obtain that
\[
\lambda  =\frac{1}{{a_2-b}_2} ({\ln |\mu|+i}\arg \mu+2n\pi i),\quad
 n\in \mathbb Z,\; \mu \in \sigma (W^*_2e^{-{\tilde{A}}_2(b_2-a_2)}).
\]
\end{proof}

\begin{theorem} \label{thm3.4} 
For the spectrum $\sigma(L_{W_1W_2})$ of any normal extension
$L_{W_1W_2}=L_{W_1}\oplus L_{W_2}$, it is true that
\[
{\sigma }_p(L_{W_1W_2})=\ {\sigma }_p(L_{W_2}) ,\quad
\ {\sigma }_c(L_{W_1W_2})=\{{[{\sigma }_p(L_{W_2})]}^c\cap [i{\mathbb R}]\}
\cup {\sigma }_c(L_{W_2})
\]
\end{theorem}

\begin{proof} 
The validity of this assertion is a simple result of the following claim. 
If $S_1$ and $S_2$ are linear closed operators in any Hilbert spaces $H_1$ and $H_2$ 
respectively, then we have
\begin{gather*}
\sigma_p(S_1 \oplus S_2)={\sigma }_p(S_1)\cup {\sigma }_p(S_2),\\
{\sigma }_c(S_1\oplus S_2)={({\sigma }_p(S_1)\cup {\sigma }_p(S_2))}^c
\cap {({\sigma }_r(S_1)\cup {\sigma }_r(S_2))}^c\cap ({\sigma }_c(S_1)\cup {\sigma }_c(S_2)).
\end{gather*}
\end{proof}

 Note that for the singular differential operators for $n$-th order in scalar case 
in the finite interval has been studied in \cite{Sho}.


\begin{example} \label{exa3.5} \rm
 Consider the  boundary-value problem for the differential operator $L_{\varphi \psi }$,
\begin{gather*}
L_{\varphi \psi } :     \frac{\partial u(t,x)}{\partial t}
+\operatorname{sgn}t\frac{{\partial }^2u(t,x)}{\partial x^2}=f(t,x), 
\quad  |t|>1,\;  x\in [0,1],\\
\frac{\partial u(t,x)}{\partial t}-\frac{{\partial }^2u(t,x)}{\partial x^2}+u(t,x)=f(t,x),
 \quad |t|<1/2,\;  x\in [0,1],\\
u(1,x)=e^{i\varphi }u(-1,x),\quad \varphi \in [0,2\pi ),\\
u(1/2,x)=e^{i\psi }u(-1/2,x),\quad  \psi \in [0,2\pi ),\\
u_x(t,0)=u_x(t,1)=0,\quad |t|>1,\; |t|<1/2
\end{gather*}
in the space $L_2((-\infty,-1)\times (0,1))\oplus {L_2((-1/2,1/2)
\times (0,1))\oplus L}_2((1,\infty)\times (0,1))$.
In this case it is clear that in the space $L_2(0,1)$, for the operators
\begin{gather*}
A_1= \frac{{\partial }^2u(.,x)}{\partial x^2} ,\quad  x\in [0,1] ,\; u_x(.,0)=u_x(.,1)=0,\\
A_2=- \frac{{\partial }^2u(.,x)}{\partial x^2}  +u(.,x),\quad x\in [0,1] ,\;
  u_x(.,0)=u_x(.,1)=0,\\
A_3=- \frac{{\partial }^2u(.,x)}{\partial x^2},\quad x\in [0,1] ,\; u_x(.,0)=u_x(.,1)=0
\end{gather*}
we have
\begin{gather*}
A_1=A^*_1\le 0,\quad A_2=A^*_2\ge 1,\quad
A_3=A^*_3\ge 0, \quad {\ker  {(-A_1)}^{1/2}\ }\ne \{0\}, \\
\ker A^{1/2}_3\ne \{0\},\quad
0\in {\sigma }_p({(-A_1)}^{1/2})\cap  {\sigma }_p(A^{1/2}_3).
\end{gather*}
On the other hand, since  
$A^{-1}_2\in {\sigma }_{\infty }(L_2(0,1))$,
$\sigma (L_{\psi })={\sigma }_p(L_{\psi })$, 
 ${\sigma }_c(L_{\psi })=\emptyset $ and
\begin{align*}
\sigma (L_{\psi })
&=\Big\{\lambda \in{\mathbb C}:\lambda = {\ln  |\mu|+i}
\arg\mu+2n\pi i, n\in \mathbb{Z},\,
\mu \in \sigma (e^{i\psi }e^{-{\tilde{A}}_2(b_2-a_2)}),\\
&\qquad 0\leq \arg \mu<2\pi  \Big\}\\
&\subset \big\{\lambda \in \mathbb {C}: \operatorname{Re}\lambda \geq 1\big\},
\end{align*}
then
${[{\sigma }_p(L_{\psi })]}^c\cap [i{\mathbb R}]=i{\mathbb R}$.
Therefore, by the Theorem \ref{thm3.4}, we obtain
\[
{\sigma }_p(L_{\varphi \psi })=\ {\sigma }_p(L_{\psi }) ,\quad
 {\sigma }_c(L_{\varphi \psi })=\ i{\mathbb R}.
\]
\end{example}


\subsection*{Acknowledgments}
The authors want to thank Professors M. L. Gorbachuk 
(Institute of Mathematics NASU, Kiev, Ukraine)
and B. O. Guler (KTU, Trabzon, Turkey) 
for their advice and enthusiastic support.

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\end{document}

