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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 129, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/129\hfil Selfadjoint extensions]
{Selfadjoint extensions of a singular multipoint differential operator
 of first order}

\author[Z. I. Ismailov, R. \"Ozt\"urk Mert \hfil EJDE-2013/129\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 April 29, 2013. Published May 27, 2013.}
\subjclass[2000]{47A10, 47A20}
\keywords{Quantum field theory; spectrum; multipoint differential operators;
 \hfill\break\indent
 selfadjoint extension}

\begin{abstract}
 In this work, we describe all selfadjoint extensions of the minimal
 operator generated by linear singular multipoint symmetric differential
 expression $l=(l_1,l_2,l_3)$, $l_k=i\frac{d}{dt}+A_k$ with selfadjoint operator
 coefficients $A_k$, $k=1,2,3$ 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{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 selfadjoint
extensions of symmetric differential operators as a direct sum of Hilbert spaces
 \cite{Alb,Tim,Ze1}.

 The general theory of selfadjoint extensions of symmetric operators in Hilbert
spaces and their spectral theory have deeply been investigated by many
mathematicians;  see for example \cite{Dun,Gor,von,Rof}. 
Applications of this theory to two-point  differential operators in Hilbert
space of functions have been even continued up to date.

 It is well-known that for the existence of selfadjoint extension of
any linear closed densely defined symmetric operator $B$ in a Hilbert space $H$,
necessary and sufficient condition is a equality of deficiency indices
$m(B)=n(B)$, where $m(B)=\dim\ker (B^{*}+i)$, $n(B)=\dim\ker (B^{*}-i)$.

 However multipoint situations may occur in different tables in the following
sense. Let $B_1$ and $B_2$ be minimal operators generated by the linear
differential expression $i\frac{d}{dt}$ in the Hilbert space of functions
$L_2(-\infty,a)$ and $L_2(b,+\infty)$, $a<b$, respectively. In this case,
it is known that
\[ 
(m(B_1),n(B_1))=(0,1), \quad (m(B_2),n(B_2))=(1,0).
\]

 Consequently, $B_1$ and $B_2$ are maximal symmetric operators,
but they are not selfadjoint. However, direct sum $B=B_1\oplus B_2$
of operators in the direct sum $\mathfrak{H}=L_2(-\infty,a)\oplus L_2(b,+\infty)$
spaces have equal defect numbers $(1,1)$. Then by the general theory \cite{von}
it has a selfadjoint extension. On the other hand, it can be easily shown that
\[
 u_2(b)=e^{i\varphi}u_1(a), \quad \varphi \in [0,2\pi), \quad
 u=(u_1,u_2), \quad u_1\in D(B_1^{*}), \quad u_2\in D(B_2^{*}). 
\]

 In the singular cases, there has been no investigation so far.
However, in physical and technical processes, many of the problems resulting
from the examination of the solution is of great importance in singular cases.

 The selfadjoint extension theory for ode's is known for any
number of intervals, finite or infinite, and any order expressions,
see \cite{Ze}. This theory is based on the GKN (Glazmann-Krein-Naimark) 
Theory \cite{Nai}.

 In this work in section 2, by the method of  Calkin-Gorbachuk
 (see \cite{Cal,Gor,Rof}), we describe all selfadjoint extensions of the 
minimal operator generated by singular  multipoint symmetric differential 
operator of first order in the direct sum of Hilbert space
 \[ 
L_2(H,(-\infty ,a_1))\oplus L_2(H,(a_2,b_2)) \oplus L_2(H,(a_3,+\infty)), 
\]
 where $-\infty <a_1<a_2<b_2<a_3<+\infty$ in terms of boundary values.In
 section 3, the spectrum of such extensions is studied.

\section{Description of selfadjoint 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
 $L_2(H,(-\infty ,a_1))\oplus L_2(H,(a_2,b_2)) \oplus L_2(H,(a_3,+\infty))$
 of vector-functions let us consider the linear multipoint differential expression
 \[
 l(u)=(l_1(u_1),l_2(u_2),l_3(u_3))=(iu'_1+A_1u_1,iu'_2+A_2u_2,iu'_3+A_3u_3),
 \]
 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 $H$. 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 the Hilbert space $H$.
It is denoted by $H_{k,+}$. Denote $H_{k,-}$ 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,
 ${\tilde{A}}_k:D({\tilde{A}}_k)=H\subset H_{k,-1}\to H_{k,-1}$ 
is a linear selfadjoint operator.

In the direct sum,
 $L_2(H,(-\infty ,a_1))\oplus L_2(H,(a_2,b_2)) \oplus L_2(H,(a_3,+\infty))$ 
define 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 ${\tilde{l}}_1(u_1)=iu'_1+{\tilde{A}}_1u_1$,
${\tilde{l}}_2(u_2)=iu'_2+{\tilde{A}}_2u_2$,
${\tilde{l}}_3(u_3)=iu'_3+{\tilde{A}}_3u_3$.

 The minimal $L_{10}$ ($L_{20}$ and $L_{30}$) and maximal $L_1$
($L_2$ and $L_3$) operators generated by differential expression
$\tilde{l}_1$ ($\tilde{l}_2$ and $\tilde{l}_3$) in $L_2(H,(-\infty ,a_1))$
($L_2(H,(a_2 ,b_2))$ and $L_2(H,(b,+\infty ))$) have been investigation in
 \cite{Gor2}.

 The operators $L_0=L_{10}\oplus L_{20}\oplus L_{30}$ and
$L=L_1\oplus L_2\oplus L_3$ in the space
$L_2=L_2(H,(-\infty ,a_1))\oplus L_2(H,(a_2,b_2)) \oplus L_2(H,(a_3,+\infty))$
are called minimal and maximal (multipoint) operators generated by the
differential expression \eqref{eq2.1}, respectively. Note that the operator
$L_0$ is symmetric and $L_{0}^{*}=L$ in $L_2$. On the other hand, it is clear that,
$m(L_{10})=0$, $n(L_{10})=\dim H$, $m(L_{20})=\dim H$, $n(L_{20})=\dim H$, $m(L_{30})=\dim H$,
$n(L_{30})=0$.

 Consequently, $m(L_{0})=n(L_{0})=2\dim H >0$. Hence, the minimal operator $L_0$
has a selfadjoint extension \cite{von}. For example, the differential
expression $\tilde{l}(u)$ with the boundary condition $u(a_1)=u(a_3)$,
$u(a_2)=u(b_2)$ generates a selfadjoint operator in $L_2$.

 All selfadjoint extensions of the minimal operator $L_0$ in $L_2$
in terms of the boundary values are described.

 Note that space of boundary values has an important role in the theory of
selfadjoint extensions of linear symmetric differential operators \cite{Gor,Rof}.

 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}.

 Firstly, note that the following proposition which validity of this claim
can be easily proved.

 \begin{lemma}\label{lem1}
 The triplet $(H,\gamma_1,\gamma_2)$, where
 \begin{gather*}
 \gamma_1:D((L_{10}\oplus0\oplus L_{30})^*)\to H,\quad
 \gamma_1(u)=\frac{1}{i\sqrt{2\ }}(u_1(a_1)+u_3(a_3)),\\
 \gamma_2 :D((L_{10}\oplus0\oplus L_{30})^*)\to H, \quad
 \gamma_2(u)=\frac{1}{\sqrt{2}}(u_1(a_1)-u_3(a_3)),\\
 u=(u_1,u_2,u_3)\in D((L_{10}\oplus0\oplus L_{30})^*)
 \end{gather*}
is a space of boundary values of the minimal operator 
$L_{10}\oplus0\oplus L_{30}$
in the direct sum $L_2(H,(-\infty,a_1))\oplus0\oplus L_2(H,(a_3,+\infty))$.
\end{lemma}

 \begin{proof}
 For arbitrary $u=(u_1,u_2,u_3)$ and $v=(v_1,v_2,v_3) $ from
 $D((L_{10}\oplus0\oplus L_{30})^*)$ the validity of the equality
 \begin{align*}
 &{(Lu,v)}_{L_2(H,(-\infty,a_1))\oplus0\oplus L_2(H,(a_3,+\infty))}
 -({u,Lv)}_{L_2(H,(-\infty,a_1))\oplus0\oplus L_2(H,(a_3,+\infty))}\\
 &={(\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,u_2,u_3)\in D((L_{10}\oplus0\oplus L_{30})^*) $ such that
 \[
 \gamma_1(u)=\frac{1}{i\sqrt{2\ }}(u_1(a_1)+u_3(a_3))=f
 \quad\text{and}\quad \gamma_2(u)=\frac{1}{\sqrt{2}}(u_1(a_1)-u_3(a_3))=g;
 \]
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 $u_1(t),u_3(t)$ in the form
 \begin{gather*}
 u_1(t)=\int^t_{-\infty }{e^{s-a_1}}{ds(if+g)}/{\sqrt{2}} 
\quad\text{with }t<a_1; \\
 u_2(t)=0, \quad\text{with } a_2<t<b_2; \\
 u_3(t)=\int^{\infty }_t{e^{a_3-t}}ds{(if-g)}/{\sqrt{2}} 
\quad\text{with } t>a_3
 \end{gather*}
 then it is clear that $(u_1,u_2,u_3)\in D((L_{10}\oplus0\oplus L_{30})^*)$ 
and $\gamma_1(u)=f$,  $\gamma_2(u)=g$.
 \end{proof}

 Furthermore, using the result which is obtained in \cite{Gor2} the next
 assertion is proved.

 \begin{lemma}\label{lem2}
The triplet $(H,{\Gamma }_1,{\Gamma }_2)$,
 \begin{gather*}
 {\Gamma }_1:D((0\oplus L_{20}\oplus0)^{*})\to H, \quad
 \Gamma_1(u)=\frac{1}{i\sqrt{2}}({u_2(a_2)+u}_2(b_2)),\\
 {\Gamma }_2:D((0\oplus L_{20}\oplus0)^{*})\to H, \quad
 {\Gamma }_2(u)=\frac{1}{\sqrt{2}}(u_2(a_2)-u_2(b_2)),\\
 u=(u_1,u_2,u_3)\in D((0\oplus L_{20}\oplus0)^{*})
 \end{gather*}
 is a space of boundary values of the minimal operator $0\oplus L_{0}\oplus0$ 
in the  direct sum $0\oplus L_2(H,(a_2,b_2))\oplus 0$.
 \end{lemma}

 The following result can be easily established.

 \begin{lemma}\label{lem3}
Every selfadjoint extension of $L_0$ in
$$
L_2=L_2(H,(-\infty ,a_1))\oplus L_2(H,(a_2,b_2)) \oplus L_2(H,(a_3,+\infty))
$$
is a direct sum of selfadjoint extensions of the minimal operator
$L_{10}\oplus 0\oplus L_{30}$ in
$L_2(H,(-\infty ,a_1))\oplus 0 \oplus L_2(H,(a_3,+\infty))$
and minimal operator $0\oplus L_{0}\oplus 0$ in
$0\oplus L_2(H,(a_2,b_2))\oplus 0$.
\end{lemma}

 Finally, using the method in \cite{Gor}
the following result can be deduced.

 \begin{theorem} \label{thm4}
If $\tilde{L}$ is a selfadjoint extension of the minimal operator $L_0$ in $L_2$,
then it generates by differential expression \eqref{eq2.1} and boundary conditions
 \begin{gather*}
 u_3(a_3)=W_1u_1(a_1),\\
 u_2(b_2)=W_2u_2(a_2),
 \end{gather*}
where $W_1,W_2:H\to H$ are a unitary operators. Moreover, the unitary
operators $W_1,W_2$ in $H$ are determined uniquely by the extension $\tilde{L}$;
i.e. $\tilde{L}=L_{W_1W_2}$ and vice versa.
 \end{theorem}



\section{The spectrum of the selfadjoint extensions}

 In this section the structure of the spectrum of the selfadjoint extension
$L_{W_1W_2}$ in $L_2$ will be investigated. In this case by the Lemma \ref{lem3}
 it is clear that
 \[
 L_{W_1W_2}=L_{W_1}\oplus L_{W_2} ,
 \]
 where $L_{W_1}$ and $L_{W_2}$ are selfadjoint extensions of the minimal 
operators
 $L_0(1,0,1)=L_{10}\oplus0 \oplus L_{30}$ and $L_0(0,1,0)=0 \oplus L_{0}\oplus 0$
 in the Hilbert spaces 
$L_2(1,0,1)=L_2(H,(-\infty,a_1))\oplus 0 \oplus L_2(H,(a_3,+\infty))$
 and $L_2(0,1,0)=0\oplus L_2(H,(a_2,b_2))\oplus 0$, respectively.

 First, we have to prove the following result.

 \begin{theorem} \label{thm3.1}
 The point spectrum of any selfadjoint extension $L_{W_1}$ 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 selfadjoint 
extension  $L_{W_1}$ of the minimal operator $L_0(1,0,1)$ in the Hilbert 
space $L_2(1,0,1)$,
 \[
 L_{W_1}u=\lambda u, \quad  u=(u_1,0,u_3)\in L_2(1,0,1);
 \]
 that is,
 \begin{gather*}
 {\tilde{l}}_1(u_1)=iu'_1+{\tilde{A}}_1u_1=\lambda u_1,\quad
 u_1\in L_2(H,(-\infty ,a_1)),\\
 {\tilde{l}}_3(u_3)=iu'_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).
 \end{gather*}
The general solution of this problem is
 \begin{gather*}
 u_1(\lambda ;t)=e^{i({\tilde{A}}_1-\lambda )(t-a_1)}f^*_1,\quad t<a_1, \\
 u_3(\lambda ;t)=e^{i({\tilde{A}}_3-\lambda )(t-a_3)}f^*_3,\quad t>a_3, \\
 f^*_3=W_1f^*_1,\quad f^*_1,f^*_3\in H.
 \end{gather*}
It is clear that for the $f^{*}_1\neq 0$, $f^{*}_3\neq 0$ the functions
$ u_1(\lambda ;.) \notin L_2(H,(-\infty ,a_1))$, $ u_2(\lambda ;.) \notin L_2(H,(a_3, +\infty))$.
So for every unitary operator $W_1$ we have $\sigma_p(L_{W_1})=\emptyset$.
 \end{proof}

 Since residual spectrum of any selfadjoint operator in any Hilbert space is
empty,  it is sufficient to investigate the continuous spectrum of the 
selfadjoint 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 selfadjoint extension $L_{W_1}$ of the minimal
operator $L_0(1,0,1)$ in the Hilbert space $L_2(1,0,1)$ is 
${\sigma }_c(L_{W_1})=\mathbb{R}$.
 \end{theorem}

 \begin{proof}
Firstly, we search for the resolvent operator of the
extension $L_{W_1}$ generated by the differential expression 
$({\tilde{l}}_1,0,{\tilde{l}}_3)$ and the boundary condition
 \[
 u_3(a_3)=W_1u_1(a_1)
 \]
in the Hilbert space $L_2(1,0,1)$; i.e.
 \begin{equation} \label{GrindEQ__3_1_}
\begin{gathered}
 {\tilde{l}}_1(u_1)=iu'_1+{\tilde{A}}_1u_1={\lambda }u_1+f_1,
 \quad u_1, f_1\in L_2(H,(-\infty ,a_1)),  \\
 {\tilde{l}}_3(u_3)=iu'_3+{\tilde{A}}_3u_3={\lambda }u_3+f_3,
 \quad u_3, f_3\in L_2(H,(a_3,+\infty )), \\
 \lambda \in \mathbb C, \quad \lambda_i=\operatorname{Im}\lambda >0  \\
 u_3(a_3)=W_1u_1(a_1)
\end{gathered}
\end{equation}

 Now, we will show that the  function
 \[
 u(\lambda ;t)=(u_1(\lambda ;t),0,u_3(\lambda ;t)),
 \]
where
 \begin{gather*}
 u_1({\lambda };t)=e^{i({\tilde{A}}_1-{\lambda })(t-a_1)}f^{*}_1+
 i\int^{a_1}_t{e^{i({\tilde{A}}_1-{\lambda })(t-s)}f_1(s)ds},\quad t<a_1,\\
 u_3({\lambda };t)=i\int^{\infty}_t{{\rm e}}^{i({\tilde{A}}_3
 -{\lambda })(t- s)}f_3(s) ds,\quad t>a_3,\\
 f^{*}_1=W^{*} \Big(i\int^{\infty}_{a_3}{{\rm e}}^{i({\tilde{A}}_3
 -{\lambda })(t- s)(b-s)}f_3(s) ds\Big)
 \end{gather*}
is a solution of the boundary value problem \eqref{GrindEQ__3_1_} in
the Hilbert space $L_2(1,0,1)$. It is sufficient to show that
 \begin{gather*}
 u_1(\lambda ;t) \in L_2(H,(-\infty ,a_1)),\\
 u_3(\lambda ;t) \in L_2(H,(a_3, +\infty))
 \end{gather*}
 for $\lambda_i>0$. Indeed, in this case
 \begin{align*}
 \|f^{*}_1\|^2_{H}
&=\Big\|\int^{\infty}_{a_3}{{\rm e}}^{i({\tilde{A}}_3
 -{\lambda })(a_3- s)}f(s) ds\Big\|^2_{H}
\leq \Big(\int^{\infty}_{a_3} e^{\lambda_i(a_3-s)}\|f(s)\|_{H}ds\Big)^2 \\
&\leq \Big( \int^{\infty}_{a_3}e^{2\lambda_i(a_3-s)} ds \Big)
 \Big(\int^{\infty}_{a_3}\|f(s)\|^2_{H}ds\Big)
 =\frac{1}{2\lambda_i}\|f\|^2_{L_2(H,(a_3,+\infty))}<\infty,
\end{align*}

\begin{align*}
 \|e^{i({\tilde{A}}_1-{\lambda })(t-a_1)}f^{*}_1\|^2_{L_2(H,(-\infty,a_1))}
 &= \|e^{-i\lambda (t-a_1)}f^{*}_1\|^2_{L_2(H,(-\infty,a_1))}\\
 &=\int^{a_1}_{-\infty}\|e^{-i\lambda (t-a_1)}f^{*}_1\|^2_{H}dt \\
 &= \int^{a_1}_{-\infty}e^{2\lambda_i (t-a_1)}dt\|f^{*}_1\|^2_H\\
 &=\frac{1}{2\lambda_i}\|f^{*}_1\|^2_{H}<\infty
 \end{align*}
and
\begin{align*}
& \Big\|i\int^{a_1}_te^{i({\tilde{A}}_1-{\lambda })(t-s)}f_1(s)ds
\Big\|^2_{L_2(H,(-\infty,a_1))}\\
&\leq \int^{a_1}_{-\infty}
 \Big(\int^{a_1}_te^{\lambda_i(t-s)}\|f_1(s)\|_{H}ds\Big)^2dt \\
&\leq \int^{a_1}_{-\infty}\Big(\int^{a_1}_te^{\lambda_i(t-s)}ds \Big)
 \Big(\int^{a_1}_te^{\lambda_i(t-s)}\|f_1(s)\|^2ds \Big)dt \\
&=\frac{1}{\lambda_i}\int^{a_1}_{-\infty}\int^{a_1}_t
e^{\lambda_i(t-s)}\|f_1(s)\|^2ds dt
 =\frac{1}{\lambda_i}\int^{a_1}_{-\infty}
\Big(\int^{s}_{-\infty}e^{\lambda_i(t-s)}\|f_1(s)\|^2dt\Big)ds \\
&=\frac{1}{\lambda_i}\int^{a_1}_{-\infty}
 \Big(\int^{s}_{-\infty}e^{\lambda_i(t-s)}dt\Big) \|f_1(s)\|^2ds\\
&=\frac{1}{\lambda_i^2}\int^{a_1}_{-\infty}\|f_1(s)\|^2ds \\
& =\frac{1}{\lambda_i^2}\|f_1\|^2_{L_2(H,(-\infty,a_1))}<\infty.
 \end{align*}
Furthermore,
\begin{align*}
&\Big\|i\int^{\infty}_te^{i({\tilde{A}}_3-{\lambda })(t-s)}f_3(s)ds
 \Big\|^2_{L_2(H,(a_3,+\infty))} \\
&\leq \int^{\infty}_{a_3}
 \Big(\int^{\infty}_te^{\lambda_i(t-s)}\|f_3(s)\|_{H}ds\Big)^2dt \\
&\leq \int^{\infty}_{a_3}\Big(\int^{\infty}_te^{\lambda_i(t-s)}ds \Big)
 \Big(\int^{\infty}_te^{\lambda_i(t-s)}\|f_3(s)\|^2ds \Big)dt \\
&=\frac{1}{\lambda_i}\int^{\infty}_{a_3}
 \Big(\int^{\infty}_te^{\lambda_i(t-s)}\|f_3(s)\|^2ds \Big)dt\\
&=\frac{1}{\lambda_i}\int^{\infty}_{a_3}\Big(\int^{s}_{a_3}e^{\lambda_i(t-s)}
 \|f_3(s)\|^2dt\Big)ds \\
&=\frac{1}{\lambda_i}\int^{\infty}_{a_3}
 \Big(\int^{s}_{a_3}e^{\lambda_i(t-s)}dt\Big) \|f_3(s)\|^2ds\\
&=\frac{1}{\lambda_i^2}\int^{\infty}_{a_3}(1-e^{\lambda_i(a_3-s)})\|f_3(s)\|^2ds \\
&\leq \frac{1}{\lambda_i^2}\|f_3\|^2_{L_2(H,(a_3,+\infty))}<\infty.
 \end{align*}
The above calculations imply that $u_1(\lambda ;t) \in L_2(H,(-\infty ,a_1))$,
and that $u_3(\lambda ;t) \in L_2(H,(a_3, +\infty))$ for $\lambda \in \mathbb{C}$,
 $\lambda_i=\operatorname{Im}\lambda> 0$.  On the other hand, one can easily verify that
$u(\lambda;t)=(u_1(\lambda;t),0,u_3(\lambda;t))$
is a solution of boundary-value problem \eqref{GrindEQ__3_1_}.

 When $\lambda \in \mathbb{C}$, $\lambda_i=\operatorname{Im}\lambda< 0$ is true solution
of the boundary-value problem
\begin{gather*}
 L_{W_1}u=\lambda u +f, \quad u=(u_1,0,u_3), \quad f=(f_1,0,f_3)\in L_2(1,0,1) \\
 u_3(a_3)=W_1u_1(a_1),
 \end{gather*}
where $W_1$ is a unitary operator in $H$, is in the form 
$u(\lambda;t)=(u_1(\lambda;t),0,u_3(\lambda;t))$,
\begin{gather*}
 u_1(\lambda;t)=-i\int^{t}_{-\infty}e^{i({\tilde{A}}_1-\lambda)(t-s)}f_1(s)ds, 
\quad t<a_1\\
 u_3(\lambda;t)=e^{i({\tilde{A}}_3-\lambda)(t-a_3)}f^{*}_3
 -i\int^{t}_{a_3}e^{i({\tilde{A}}_3-\lambda)(t-s)}f_3(s)ds, \quad t>a_3,
 \end{gather*}
where
\[
 f^{*}_3=W\Big(-i\int^{a_1}_{-\infty}e^{i({\tilde{A}}_1-\lambda)(a_1-s)}f_1(s)ds 
\Big)\,.
 \]

First, we prove that $u(\lambda;t)\in L_2(1,0,1)$. In this case,
\begin{align*}
 \|u_1(\lambda ;t)\|^2_{L_2(H,(-\infty ,a_1))}
&=\int^{a_1}_{-\infty}
 \Big\|-i\int^{t}_{-\infty}e^{i({\tilde{A}}_1-\lambda)(t-s)}f_1(s)ds\Big\|^2_{H}dt
 \\
&\leq \int^{a_1}_{-\infty}\Big(\int^{t}_{-\infty}e^{\lambda_i(t-s)}ds\Big)
 \Big(\int^{t}_{-\infty}e^{\lambda_i(t-s)}\|f_1(s)\|^2_{H}ds\Big)dt \\
&= \frac{1}{|\lambda_i|}\int^{a_1}_{-\infty}\int^{t}_{-\infty}e^{\lambda_i(t-s)}
 \|f_1(s)\|^2_{H}\,ds\,dt \\
&= \frac{1}{|\lambda_i|}\int^{a_1}_{-\infty}
 \Big(\int^{a_1}_{s}e^{\lambda_i(t-s)}\|f_1(s)\|^2_{H}dt\Big)ds\\
&= \frac{1}{|\lambda_i|}\int^{a_1}_{-\infty}\big(e^{\lambda_i(t-s)}\big)dt
 \|f_1(s)\|^2_{H}ds \\
&= \frac{1}{|\lambda_i|^2}\int^{a_1}_{-\infty}
(1-e^{\lambda_i(a_1-s)})\|f_1(s)\|^2_{H}ds\\
&\leq \frac{1}{|\lambda_i|^2}\|f_1\|^2_{L_2(H,(-\infty,a_1))}< \infty,
 \end{align*}
\begin{align*}
 \|f^{*}_3\|^2_{H}
&=\Big\|\int^{a_1}_{-\infty}e^{i({\tilde{A}}_1-\lambda)(a_1-s)}f_1(s)ds
 \Big\|^2_{H}\\
&\leq \Big(\int^{a_1}_{-\infty}e^{\lambda_i(a_1-s)}\|f_1(s)\|_{H}ds \Big)^2 \\
&\leq \Big(\int^{a_1}_{-\infty}e^{2\lambda_i(a_1-s)}ds\Big)
 \Big(\int^{a_1}_{-\infty} \|f_1(s)\|^2_{H}ds\Big) \\
&= \frac{1}{2|\lambda_i|}\|f_1\|^2_{L_2(H,(-\infty,a_1))}<\infty ,
 \end{align*}
\begin{align*}
\|e^{i({\tilde{A}}_3-\lambda)(t-a_3)}f^{*}_3\|^2_{L_2(H,(a_3,+\infty))}
&\leq \int^{\infty}_{a_3}e^{2\lambda_i(t-a_3)}dt\|f^{*}_3\|^2_{H}\\
&=\frac{1}{2|\lambda_i|}\|f^{*}_3\|^2_{H} \\
&\leq \frac{1}{4|\lambda_i|^2}\|f\|^2_{L_2(H,(a_3,+\infty))}<\infty
\end{align*}
and
\begin{align*}
&\Big\|\int^{t}_{a_3}e^{i({\tilde{A}}_3-\lambda)(t-s)}f_3(s)ds
 \Big\|^2_{L_2(H,(a_3,+\infty))}\\
&\leq \int^{\infty}_{a_3}
 \Big(\int^{t}_{a_3}e^{\lambda_i(t-s)}\|f_3(s)\|_{H}ds \Big)^2dt \\
&\leq \int^{\infty}_{a_3}\Big(\int^{t}_{a_3}e^{\lambda_i(t-s)}ds\Big)
 \Big(\int^{t}_{a_3}e^{\lambda_i(t-s)}\|f_3(s)\|^2_{H}ds\Big)dt \\
&=\int^{\infty}_{a_3}\Big(\frac{1}{\lambda_i}(1-e^{\lambda_i(t-a_3)}) \Big)
\Big(\int^{t}_{a_3}e^{\lambda_i(t-s)}\|f_3(s)\|^2_{H}ds\Big)dt \\
&\leq \frac{1}{|\lambda_i|}\int^{\infty}_{a_3}
 \Big(\int^{t}_{a_3}e^{\lambda_i(t-a_3)}\|f_3(s)\|^2_{H}ds\Big)dt \\
&=\frac{1}{|\lambda_i|}\int^{\infty}_{a_3}
 \Big(\int^{\infty}_{s}e^{\lambda_i(t-s)}\|f_3(s)\|^2_{H}dt\Big)ds\\
&=\frac{1}{|\lambda_i|}\int^{\infty}_{a_3}
 \Big(\int^{a_3}_{s}e^{\lambda_i(t-s)}dt\Big)\|f_3(s)\|^2_{H}ds \\
&=\frac{1}{|\lambda_i|^2}\|f_3\|^2_{L_2(H,(a_3,+\infty))}<\infty.
 \end{align*}
The above calculations show that $u_1(\lambda;\cdot)\in L_2(H,(-\infty,a_1))$,
and that $u_3(\lambda;\cdot)\in L_2(H,(a_3,+\infty))$; i.e.,
$u(\lambda;\cdot)=(u_1(\lambda;\cdot),0,u_3(\lambda,\cdot))\in L_2(1,0,1)$ 
in case  $\lambda \in \mathbb{C}$, $\lambda_i=\operatorname{Im}\lambda< 0$.
 On the other hand it can be verified  that the function $u(\lambda;\cdot)$ 
satisfies the equation $L_{W_1}u=\lambda u(\lambda;\cdot)+f$
and $u_3(a_3)=W_1u_1(a_1)$.

 Therefore, the following result has been proved that for the resolvent 
set $\rho(L_{W_1})$
 \[
 \rho(L_{W_1})\supset \{\lambda\in \mathbb{C}:\operatorname{Im}\lambda \neq 0\}.
 \]
Now, we will study continuous spectrum $\sigma_{c}(L_{W_1})$ of the
extension $L_{W_1}$. For $\lambda \in \mathbb{C}$, $\lambda_i=\operatorname{Im}\lambda> 0$,
norm of the resolvent operator $R_{\lambda}(L_{W_1})$ of the 
$L_{W_1}$ is of the form
\begin{align*}
\|R_{\lambda}(L_{W_1})f(t)\|^2_{L_2}
&=\Big\|e^{i({\tilde{A}}_1-\lambda)(t-a_1)}f^{*}_1
 +i\int^{a_1}_te^{i({\tilde{A}}_1-\lambda)(t-s)}f_1(s)ds
\Big\|^2_{L_2(H,(-\infty,a_1))} \\
&\quad +\Big\|i\int^{\infty}_te^{i({\tilde{A}}_3-\lambda)(t-s)}f_3(s)ds
\Big\|^2_{L_2(H,(a_3,+\infty))},
\end{align*}
where $f=(f_1,0,f_3)\in L_2(1,0,1)$.
Then, it is clear that for any $f=(f_1,0,f_3)$ in 
$L_2(1,0,1)$ the following inequality is true.
 \[
 \|R_{\lambda}(L_{W_1})f(t)\|^2_{L_2}\geq \Big\|i\int^{\infty}_t
e^{i({\tilde{A}}_3-\lambda)(t-s)}f_3(s)ds \Big\|^2_{L_2(H,(a_3,+\infty))}.
 \]
The vector functions $f^{*}(\lambda;t)$ which is of the form 
$f^{*}(\lambda;t)=(0,0,e^{i({\tilde{A}}_3-\bar{\lambda})t}f_3)$,
 $\lambda \in \mathbb{C}$, $\lambda_i=\operatorname{Im}\lambda> 0$, $f_3\in H$
belong to $L_2(1,0,1)$. Indeed,
\begin{align*}
 \|f^{*}(\lambda;t)\|^2_{L_2}
&=\int^{\infty}_{a_3}\|e^{i({\tilde{A}}_3-\bar{\lambda})t}f_3\|^2_{H}dt
 =\int^{\infty}_{a_3}e^{-2\lambda_it}dt\|f_3\|^2_{H}\\
&=\frac{1}{2\lambda_i}e^{-2\lambda_ia_3}\|f_3\|^2_{H} < \infty.
\end{align*}
For such functions $f^{*}(\lambda;\cdot)$, we have
\begin{align*}
&\|R_{\lambda}(L_{W_1})f^{*}(\lambda;t)\|^2_{L_2(H,(a_3+\infty))}\\
&\geq \Big\|i\int^{\infty}_te^{i({\tilde{A}}_3-\lambda)(t-s)}
 e^{i(\tilde{A}_3-\bar{\lambda})s}f_3ds \Big\|^2_{L_2(H,(a_3,+\infty))} \\
&=\Big\|\int^{\infty}_te^{-i\lambda t}e^{-2\lambda_is}e^{i\tilde{A}_3t}f_3ds 
 \Big\|^2_{L_2(H,(a_3,+\infty))}\\
&=\Big\|e^{-i\lambda t}e^{i\tilde{A}_3t}\int^{\infty}_te^{-2\lambda_is}
 f_3ds \Big\|^2_{L_2(H,(a_3,+\infty))} \\
&=\Big\|e^{-i\lambda t}\int^{\infty}_te^{-2\lambda_is}ds
 \Big\|^2_{L_2(H,(a_3,+\infty))}\|f_3\|^2_{H}\\
&=\frac{1}{4\lambda^2_i}\int^{\infty}_{a_3}e^{-2\lambda_it}dt\|f_3\|^2_{H}\\
&=\frac{1}{8\lambda^{3}_i}e^{-2\lambda_ia_3}\|f_3\|^2_{H}.
\end{align*}
From this we obtain
 \[
 \|R_{\lambda}(L_{W_1})f^{*}(\lambda;\cdot)\|_{L_2}
 \geq \frac{e^{-\lambda_ia_3}}{2\sqrt{2}\lambda_i\sqrt{\lambda_i}}\|f\|_{H}
 =\frac{1}{2\lambda_i}\|f^{*}(\lambda;\cdot)\|_{L_2};
 \]
 i.e., for $\lambda_i=\operatorname{Im}\lambda >0$ and $f\neq 0$,
 \[
 \frac{\|R_{\lambda}(L_{W_1})f^{*}(\lambda;\cdot)\|_{L_2}}{\|f^{*}(\lambda;\cdot)\|_{L_2}}
 \geq\frac{1}{2\lambda_i}.
 \]
 is valid.
 On the other hand, it is clear that
 \[
 \|R_{\lambda}(L_{W_1})\|\geq \frac{\|R_{\lambda}(L_{W_1})
f^{*}(\lambda;\cdot)\|_{L_2}}{\|f^{*}(\lambda;\cdot)\|_{L_2}},
 \quad f_3\neq 0.
 \]
 Consequently, 
 \[
 \|R_{\lambda}(L_{W_1})\|\geq \frac{1}{2\lambda_i} \quad\text{for }
 \lambda\in \mathbb{C}, \quad  \lambda_i=\operatorname{Im}\lambda >0.
 \]
\end{proof}
 The spectrum of selfadjoint extensions of the minimal operator
 $L_{0}(0,1,0)$ will be investigated next.



 \begin{theorem} \label{thm3.3}
 The spectrum of the selfadjoint 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{R}}{\rm :}{\rm \ }\lambda
 =\frac{1}{b_2-a_2}\arg\mu+\frac{2n\pi}{b_2-a_2},\ \ n\in {\mathbb Z}, \\
 &\mu \in \sigma (W^*_2e^{i{\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 following problem to spectrum of the
 selfadjoint extension  $ L_{W_2}$,
 \begin{gather*}
 {\tilde{l}}_2(u_2)=iu'_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 {R}
 \end{gather*}
 is of the form
 \begin{gather*}
 u_2(t)=e^{i({\tilde{A}}_2-\lambda )(t-a_2)}f^*_2
 +\int^t_{a_2}{e^{i({\tilde{A}}_2-\lambda )(t-s)}}\ f_2(s)ds,\\
 a_2<t<b_2, \\
 (e^{i\lambda (b_2-a_2)}-W^*_2e^{i{\tilde{A}}_2(b_2-a_2)})f^*_2
 =W^*_2e^{i\lambda (b_2-a_2)}\int^{b_2}_{a_2}{e^{i({\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 $e^{i\lambda (b_2-a_2)}=\mu$,
 where $ \mu \in \sigma (W^*_2e^{i{\tilde{A}}_2(b_2-a_2)})$.
 We obtain that
 \[
 \lambda =\frac{1}{b_2-a_2} \arg \mu+\frac{2n\pi}{b_2-a_2} ,\quad
 n\in \mathbb Z,\; \mu \in \sigma (W^*_2e^{i{\tilde{A}}_2(b_2-a_2)}).
 \]
 \end{proof}

 \begin{theorem} \label{thm3.4}
 Spectrum $\sigma(L_{W_1W_2})$ of any selfadjoint extension
 $L_{W_1W_2}=L_{W_1}\oplus L_{W_2}$ coincides with $\mathbb{R}$.
 \end{theorem}

 \begin{proof}
 Validity of this assertion is a simple result of the following claim that 
a proof of  which it is clear. 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
 By the last theorem the spectrum of following boundary-value problem
 \begin{gather*}
 i\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],\\
 i\frac{\partial u(t,x)}{\partial t}-\frac{{\partial }^2u(t,x)}{\partial x^2}=f(t,x),
 \quad |t|<1/2,\; x\in [0,1],\\
 u(1/2,x)=e^{i\psi }u(-1/2,x),\quad \psi \in [0,2\pi ),\\
 u(1,x)=e^{i\varphi }u(-1,x),\quad \varphi \in [0,2\pi ),\\
 u_x(t,0)=u_x(t,1)=0,\quad |t|>1,\; |t|<1/2
 \end{gather*}
 in the space $L_2((-\infty,-1)\times (0,1))\oplus {L_2((-1/2,1/2)
 \times (0,1))\oplus L}_2((1,\infty)\times (0,1))$ coincides with $\mathbb{R}$.
 \end{example}

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\end{document}