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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 231, pp. 1--10.\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-2013/231\hfil Selfadjoint extensions]
{Selfadjoint extensions of multipoint singular differential operators}

\author[Z. I. Ismailov \hfil EJDE-2013/231\hfilneg]
{Zameddin I. Ismailov}  % in alphabetical order

\address{Zameddin I. Ismailov \newline
Department of Mathematics, Faculty of Sciences,
Karadeniz Technical University, 61080, Trabzon, Turkey}
\email{zameddin.ismailov@gmail.com}

\thanks{Submitted September 19, 2013. Published October 18, 2013.}
\subjclass[2000]{47A10}
\keywords{Everitt-Zettl and Calkin-Gorbachuk methods; singular multipoint;
 \hfill\break\indent
 differential operators; selfadjoint extension; spectrum}

\begin{abstract}
 This article describes all selfadjoint extensions of the minimal
 operator generated by a linear singular multipoint symmetric
 differential-operator expression for first order in the direct sum
 of Hilbert spaces of vector-functions.
 This description is done in terms of the boundary values, and it uses the
 Everitt-Zettl and the Calkin-Gorbachuk methods.
 Also the structure of the spectrum of these extensions is studied.
\end{abstract}

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


\section{Introduction}

 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,Gor1,Gor2,Nai,Rof,von}. 
Applications of this theory to two point differential operators in Hilbert 
spaces 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 in a Hilbert space, 
a necessary and sufficient condition is an equality of
 deficiency indices \cite{von}. 
However multipoint situations may occur in different tables in
 the following sense: Let $L_1$ and $L_2$ be minimal operators generated 
by the linear differential expression $l(u)=i \frac{d}{dt}$ in the Hilbert 
space of functions $L^2 (-\infty,a_1)$
 and $L^2 (a_2,+\infty)$, $a_1 ,a_2 \in \mathbb{R}$, respectively. 
In this case, it is known that deficiency indices of these
minimal operators are in form $( m(L_1),n(L_1) )=(0,1)$, 
$ (m(L_2),n(L_2) )=(1,0)$.
Consequently, $L_1$ and $L_2$ are maximal symmetric operators, 
but they have no selfadjoint extensions.
However, direct sum $L=L_1 \oplus L_2$ of operators in the direct sum
$L^2 (-\infty,a_1) \oplus L^2 (a_2,+\infty)$ of Hilbert 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 \cite{Bai}
that all selfadjoint extensions of $L$ are in the form 
$$
 u_2 (a_2)=e^{i\varphi} u_1 (a_1),\quad
 \varphi \in[0,2\pi),\quad  u=(u_1,u_2),\quad
 u_1\in D(L_1^*),\quad u_2\in D(L_2^*).
$$

 Note that in the multiinterval linear ordinary differential expression 
case the deficiency indices may be different for each interval, but equal 
in the direct sum Hilbert spaces from the different intervals. 
The selfadjoint extension theory of linear ordinary differential
expression of any order is known from famous work of 
 Everitt and Zettl \cite{Ewe} for any
 number of intervals, finite or infinite, of real-axis. 
This theory is based on the
 Glassman-Krein-Naimark Theorem. Information on the selfadjoint extensions, 
the direct and complete characterizations for the Sturm-Liouville differential 
expression in finite or  infinite interval with interior points or endpoints 
singularities can be found in the  significant monograph of Zettl \cite{Ze}. 
Special cases of problems considered in this paper  has been investigated 
in \cite{Bai,Ism}.

Lastly, note that many problems arising in the modeling processes, 
multi-particle quantum mechanics, quantum field theory, the physics of 
rigid bodies and etc support to study selfadjoint extension of symmetric 
differential operators in direct sum of  Hilbert spaces
(see \cite{Ze} and references in it).

 In section 2 in this work, by the method of Calkin-Gorbachuk Theory,
all selfadjoint extensions of the minimal operator generated by 
linear multipoint singular symmetric differential-operator expression 
of first order in the direct sum of Hilbert spaces
 $L^2 (H_1,(-\infty,a_1))\oplus L^2 (H_2,(a_2 ,+\infty))\oplus 
L^2 (H_3,(a_3,+\infty))$,
 where $H_1$, $H_2$, $H_3$ are a separable Hilbert spaces with condition
 $ 0<\operatorname{dim} H_1=\operatorname{dim}H_2+ \operatorname{dim}H_3 \leq \infty$ and $a_1, a_2,a_3 \in \mathbb{R}$, in terms of boundary
 values are described. In section 3, the spectrum of such extensions is investigated.

In this article, let 
\begin{gather*}
 \Delta _1=(-\infty, a_1), \quad \Delta_2=(a_2, +\infty), \quad
 \Delta_3=(a_3, +\infty) \quad
\text{for } a_k\in \mathbb{R}, \quad k=1,2,3,
\\
\mathfrak{L}(k)=L^{2}(H_k,\Delta_k), \quad k=1,2,3; \quad 
\mathfrak{L}=\mathfrak{L}(1)\oplus \mathfrak{L}(2)\oplus \mathfrak{L}(3).
\end{gather*}

\section{Description of selfadjoint extensions}

 In the Hilbert space $\mathfrak{L}$ of vector-functions let us consider the 
linear  multipoint singular symmetric differential-operator expression
\[
 l(u)=(l_1 (u_1 ), l_2 ( u_2 ), l_3) (u_3)),
\]
where $u=(u_1, u_2, u_3)$, $l_k (u_k)=iu_k' (t)+A_k u_k (t)$,
$t\in \Delta_k$, $k=1,2,3$.

 $A_k :D(A_k)\subset H_k\to H_k$, $k=1,2,3$ are linear selfadjoint operators
 in $H_k$. In the linear manifold $D(A_k )\subset H_k$ introduce the inner 
product
\[
 (f,g)_{k,+}=(A_k f, A_k g)_{H_k}+(f,g)_{H_k}, \quad f,g\in D(A_k ),\quad
 k=1,2,3.
\]
For $k=1,2,3$, $D(A_k)$ is a Hilbert space under the positive norm 
$\|\cdot \|_{k,+}$
with respect to the Hilbert space $H_k$. It is denoted by $H_{k,+}$, $k=1,2,3$.
Denote $H_{k,-}$, $k=1,2,3$ a Hilbert space with the negative norm
(for information on Hilbert spaces with positive and negative norms, see
for example \cite{Gor2}).
 It is clear that an operator $A_k$ is continuous from $H_{k,+}$ to $H_k$ 
and that its adjoint operator
 ${\tilde{A}}_k: H_k \to H_{k,-}$ is a extension of the operator 
$A_k$, $k=1,2,3$.
 On the other hand, ${\tilde{A}}_k: H_k\subset H_{k,-}\to H_{k,-}$, $k=1,2,3$ is
 a linear selfadjoint operator.

 In the direct sum $\mathfrak{L}$ let us define
\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 ${\tilde{l}}_k (u_k)=iu_k' (t)
+{\tilde{A}}_k u_k (t)$, $t\in \Delta_k$, $k=1,2,3$.

The minimal operator $L_{10}$ ($L_{20}$, $L_{30}$) and maximal operator $L_1$ 
($L_2$, $L_3$) generated by
differential-operator expression ${\tilde{l}}_1 (\cdot)$ (${\tilde{l}}_2 (\cdot)$,
 ${\tilde{l}}_3 (\cdot)$) in $\mathfrak{L}$
 have been investigated in \cite{Gor1} and here established that the minimal 
operator  $L_{10}$ ($L_{20}$, $L_{30}$) is not selfadjoint in $\mathfrak{L}$.
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 $\mathfrak{L}$
are called minimal and maximal (multipoint) operators generated by the 
differential expression \eqref{eq2.1},
respectively. Note that the operator $L_0$ is symmetric in $\mathfrak{L}$.
On the other hand, it is clear that, deficiency indices
$m(L_{10})=\operatorname{dim}H_1$, $n(L_{10})=0$, $m(L_{20})=0$, $n(L_{20})=\operatorname{dim}H_2$.
Consequently, $m(L_{0})=\operatorname{dim}H_1$, $n(L_{0})=\operatorname{dim}H_2+\operatorname{dim}H_3$.
Hence, the minimal operator $L_0$ in the direct sum $\mathfrak{L}$
has a selfadjoint extension \cite{von}.

In this section all selfadjoint extensions of the minimal operator
$L_0$ in $\mathfrak{L}$ in terms of the boundary values are described,
using Calkin-Gorbachuk method.
 Note that in this theory, the space of boundary values is important for the
 description of selfadjoint extensions of linear symmetric differential
operators \cite{Gor1, Gor2, Rof}.
Now give their definition.

\begin{definition}\label{def1} \rm
Let $T:D(T)\subset \mathcal{H}\to \mathcal{H}$ 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 $D(T^*)$ into $\mathfrak{H}$, is called a space of
boundary values for the operator $T$ if for any $f,g\in D(T^*)$
\[
 {(T^*f,g)}_{\mathcal{H}}-{(f,T^*g)}_{\mathcal{H}}=
 {({\gamma }_1(f),{\gamma }_2(g))}_{\mathfrak{H}}-{({\gamma }_2(f),
 {\gamma }_1(g))}_{\mathfrak{H}},
\]
 while for any $F,G\in \mathfrak{H}$, there exists an element $f\in D(T^*)$,
 such that ${\gamma }_1(f)=F\ $and ${\gamma }_2(f)=G$.
\end{definition}

 Note that any symmetric operator with equal deficiency indices has at least one space
 of boundary values \cite{Gor2}.

 Since $H_1$, $H_2$, $H_3$ are separable Hilbert spaces and 
$\operatorname{dim}H_1 =\operatorname{dim}H_2 +\operatorname{dim}H_3$,
 then it is known that there exist an isometric isomorphism 
$V:H_2\oplus H_3\to H_1$ such that $V(H_2\oplus H_3)=H_1$.
In this case the following statement is true.

\begin{lemma}\label{lem1} 
The triplet $(H_1,\gamma_1,\gamma_2)$, where
\begin{gather*}
 \gamma_1:D(L^*_0)\to H_1,\quad
 \gamma_1(u)=\frac{1}{i\sqrt{2\ }}(u_1(a_1)+V(u_2(a_2),u_3(a_3))),\quad
 u\in D(L^*_0),\\
 \gamma_2 :D(L^*_{0})\to H_1,\quad
 \gamma_2(u)=\frac{1}{\sqrt{2}}(u_1(a_1)-V(u_2(a_2),u_3(a_3))), \quad
 u\in D(L^*_0)
\end{gather*}
 is a space of boundary values of the minimal operator $L_{0}$ 
in direct sum $\mathfrak{L}$.
\end{lemma}

\begin{proof} For arbitrary $u=(u_1,u_2,u_3)$, $v=(v_1,v_2,v_3)\in D(L_0) $
 the validity of the equality
\[
 (Lu,v)_{\mathfrak{L}}-(u,Lv)_{\mathfrak{L}}
={(\gamma_1(u),\gamma_2( v))}_{H_1}-{(\gamma_2(u),\gamma_1(v))}_{H_1}
\]
 can be easily verified. Now for any given elements $F,G\in H_1$,
 we will find the function
 $u=(u_1,u_2,u_3)\in D(L_0)$ such that
\begin{gather*}
 \gamma_1(u)=\frac{1}{i\sqrt{2\ }}(u_1(a_1)+V(u_2(a_2),u_3(a_3)))=F,\\
 \gamma_2(u)=\frac{1}{\sqrt{2}}(u_1(a_1)-V(u_2(a_2),u_3(a_3)))=G.
\end{gather*}
 Indeed, in this case
\[
 V(u_2(a_2),u_3(a_3))={(iF-G)}/{\sqrt{2}},\quad
 u_1(a_1)={(iF+G)}/{\sqrt{2}}
\]
 and from this, since $V$ is the isometric mapping from $H_2\oplus H_3$ 
onto $H_1$, then it implies that there exists unique elements 
$v_1(F,G)\in H_2$ and  $v_2(F,G)\in H_3$ such that
\begin{gather*}
 (u_2(a_2),u_3(a_3))=\frac{1}{\sqrt{2}}V^{-1}(iF-G)=
 (v_1(F,G),v_2(F,G))\in H_2\oplus H_3, \\
 u_1(a_1)=\frac{1}{\sqrt{2}}(iF+G)\in H_1.
\end{gather*}
 If we choose the functions $u(t)=(u_1(t),u_2(t),u_3(t))$ in the form
\begin{gather*}
 u_1(t)=e^{\frac{t-a_1}{2}}\frac{1}{\sqrt{2}}(iF+G), \quad t\in \Delta_1,\\
 u_2(t)=e^{\frac{a_2-t}{2}}v_1(F,G), \quad t\in \Delta_2, \\
 u_3(t)=e^{\frac{a_3-t}{2}}v_2(F,G), \quad t\in \Delta_3,
\end{gather*}
then it is clear that $u(t)=(u_1(t),u_2(t),u_3(t))\in D(L_{0})$ and
$\gamma_1(u)=F$, $\gamma_2(u)=G$.
\end{proof}

Using the method in \cite{Gor2} the following result can be deduced.

\begin{theorem} \label{thm1} 
If ${\tilde{L}}$ is a selfadjoint extension of the minimal
 operator $L_0$ in direct sum $\mathfrak{L}$,
 then it is generated by differential-operator expression \eqref{eq2.1} 
and boundary condition
\[
 u_1(a_1)=WV(u_2(a_2),u_3(a_3)),
\]
 where $W:H_1\to H_1$ is a unitary operator. Moreover, the unitary operator
 $W$ is determined uniquely by extension ${\tilde{L}}$, i.e. ${\tilde{L}}=L_{W}$
 and vice versa.
\end{theorem}


\begin{remark} \label{rmk2.4} \rm
In a similar way, the selfadjoint extensions can be described of minimal operator
generated by multipoint differential-operator expression
\[
 l(u)=(l_1(u_1),l_2(u_2),\dots,l_{n}(u_n);m_1(v_1),m_2(v_2),\dots ,m_k(v_k)),
\]
 where $u=(u_1,u_2,\dots ,u_{n};v_1,v_2,\dots ,v_k)$,
\begin{gather*}
 l_p(u_p)=iu'_p(t)+A_pu_p(t), \quad t\in (-\infty,a_p), \; p=1,2,\dots ,n;\\
 m_{j}(v_j)=iu'_{j}(t)+B_{j}u_{j}(t), \quad t\in (b_{j},+\infty), \;
 j=1,2,\dots ,k,
\end{gather*}
 $A_p:D(A_p)\subset H_p\to H_p$ and $B_{j}:D(B_{j})\subset G_{j}\to G_{j}$
 are linear selfadjoint operators in Hilbert spaces $H_p$, $p=1,2,\dots ,n$ 
and $G_j$, $j=1,2,\dots ,k$  respectively, in direct sum spaces 
$\bigoplus_{p=1}^{n}L^{2}(H_p,(-\infty,a_p))\oplus
 \bigoplus _{j=1}^{k}L^{2}(G_{j},(b_{j},+\infty))$ with condition
$0<\sum_{p=1}^{n}\operatorname{dim}H_p=\sum_{j=1}^{k}
\operatorname{dim}G_{j}\leq\infty$.
\end{remark}

\section{Spectrum of the selfadjoint extensions}

 In this section, we study the structure of the spectrum of the selfadjoint
extension  $L_W$ in a direct sum  $\mathfrak{L}$.
First, we have to prove the following result.

\begin{theorem} \label{thm3.1}
 The point spectrum of any selfadjoint extension $L_W$ is empty; i.e.,
\[
 \sigma_p(L_{W})=\emptyset .
\]
\end{theorem}

\begin{proof}
Let us consider the  problem for the spectrum of the selfadjoint extension
 $L_{W}$,
\begin{gather*}
 {\tilde{l}}(u)=\lambda u(t), \quad \lambda \in \mathbb{R},\\
 u_1(a_1)=WV(u_2(a_2),u_3(a_3)),
\end{gather*}
 where $W:H_1\to H_1$ is a unitary operator. Then
\[
 ({\tilde{l}}_1(u_1),{\tilde{l}}_2(u_2),{\tilde{l}}_3(u_3))
=\lambda (u_1,u_2,u_3), \quad 
 u_1(a_1)=WV(u_2(a_2),u_3(a_3))
\]
 and from this we have
\begin{gather*}
 {\tilde{l}}_k(u_k)=iu'_k(t)+{\tilde{A}}_ku_k(t)=\lambda u_k(t),\quad
 t\in \Delta_1,\; k=1,2,3; \\
u_1(a_1)=WV(u_2(a_2),u_3(a_3)), \quad \lambda \in \mathbb{R}.
\end{gather*}
The general solution of the this problem is
\[
 u_k(\lambda;t)=e^{i({\tilde{A}}_k-\lambda)(t-a_k)}f_{k,\lambda}, \quad
 f_{k,\lambda}\in H_k, t\in \Delta_k, \quad k=1,2,3.
\]
The boundary condition is in form 
$f_{1,\lambda}=WV(f_{2,\lambda},f_{3,\lambda})$.
In order for $u_1(\lambda;t)\in \mathfrak{L}(1)$,
 $u_2(\lambda;t)\in \mathfrak{L}(2)$,
 $u_3(\lambda;t)\in \mathfrak{L}(3)$, necessary and sufficient conditions are
 $f_{k,\lambda}=0$, $k=1,2,3$. So for every operator $W$ we have
$ \sigma_p(L_{W})=\emptyset $.
\end{proof}

 Since the residual spectrum of any selfadjoint operator in any Hilbert space
is empty, it is sufficient to investigate the continuous spectrum
of the selfadjoint extensions $L_{W}$ of the minimal operator $L_0$ in the
Hilbert space $\mathfrak{L}$.
 First of all, we prove the following result.

\begin{theorem} \label{thm3.2} 
For the resolvent set $\rho (L_{W})$ it holds
\[
 \rho (L_{W})\supset \{\lambda \in \mathbb{C}:\operatorname{Im}\lambda \neq 0\}.
\]
\end{theorem}

\begin{proof}
For this, we research the existence of the resolvent operator
of $L_W$ generated by the differential-operator expression 
${\tilde{l}}(\cdot)$ and boundary condition $u_1(a_1)=W V(u_2 (a_2),u_3 (a_3))$ 
in $\mathfrak{L}$, in case when $\lambda \in \mathbb{C}$, 
$\operatorname{Im}\lambda \neq 0$. Firstly, consider the spectral 
problem in form
 \begin{equation} \label{GrindEQ__3_1_}
\begin{gathered}
 {\tilde{l}}(u)={\lambda }u(t)+f(t), \quad f=(f_1 ,f_2 ,f_3) \quad
 \lambda \in \mathbb{C},\quad \lambda_{i}=\operatorname{Im}\lambda >0 \\
 u_1(a_1)=WV(u_2(a_2),u_3(a_3))
\end{gathered}
\end{equation}
 Now, we will show that the function
\[
 u(\lambda ;t)=(u_1(\lambda ;t),u_2(\lambda ;t),u_3(\lambda ;t)),
\]
where
\begin{gather*}
 u_1({\lambda };t)=e^{i({\tilde{A}}_1-{\lambda })(t-a_1)}f_{\lambda}+
 i\int^{a_1}_t{e^{i({\tilde{A}}_1-{\lambda })(t-s)}f_1(s)ds},\quad t\in \Delta_1,\\
 u_2({\lambda };t)=i\int^{\infty}_t{{\rm e}}^{i({\tilde{A}}_2
 -{\lambda })(t- s)}f_2(s) ds,\quad t\in \Delta_2,\\
 u_3({\lambda };t)=i\int^{\infty}_t{{\rm e}}^{i({\tilde{A}}_3
 -{\lambda })(t- s)}f_3(s) ds,\quad t\in \Delta_3,\\
 f_{\lambda}=WV \Big(i\int^{\infty}_{a_2}{{\rm e}}^{i({\tilde{A}}_2
 -{\lambda })(a_2- s)}f_2(s)ds , i\int^{\infty}_{a_3}{{\rm e}}^{i({\tilde{A}}_3
 -{\lambda })(a_3- s)}f_3(s)ds \Big),
\end{gather*}
is a solution of boundary value problem \eqref{GrindEQ__3_1_} in $\mathfrak{L}$.
 It is sufficient to show that $u_1(\lambda ;t)\in \mathfrak{L}(1)$,
 $u_2(\lambda ;t)\in \mathfrak{L}(2)$, $u_3(\lambda ;t)\in \mathfrak{L}(3)$,
 for $\lambda_{i}>0$. Indeed, in this case
 \begin{align*}
 \|f_{\lambda}\|^2_{H_1}
 &=\big\|\int^{\infty}_{a_2}{{\rm e}}^{i({\tilde{A}}_2
 -{\lambda })(a_2- s)}f_2(s) ds\big\|^2_{H_2}+
 \big\|\int^{\infty}_{a_3}{{\rm e}}^{i({\tilde{A}}_3
  -{\lambda })(a_3- s)}f_3(s) ds\big\|^2_{H_3} \\
 &\leq \Big(\int^{\infty}_{a_2} e^{\lambda_i(a_2-s)}\|f_2(s)\|_{H_2}ds\Big)^2+
 \Big(\int^{\infty}_{a_3} e^{\lambda_i(a_3-s)}\|f_3(s)\|_{H_3}ds\Big)^2 \\
 &\leq \Big( \int^{\infty}_{a_2}e^{2\lambda_i(a_2-s)} ds \Big)
 \Big(\int^{\infty}_{a_2}\|f_2(s)\|^2_{H_2}ds\Big)\\
&\quad + \Big( \int^{\infty}_{a_3}e^{2\lambda_i(a_3-s)} ds \Big) 
\Big(\int^{\infty}_{a_3}\|f_3(s)\|^2_{H_3}ds\Big) \\
 &=\frac{1}{2\lambda_i}(\|f_2\|^2_{\mathfrak{L}(2)}
+\|f_3\|^2_{\mathfrak{L}(3)})<\infty,
\end{align*}
\begin{align*}
 \|e^{i({\tilde{A}}_1-{\lambda })(t-a_1)}f_{\lambda}\|^2_{\mathfrak{L}(1)}
 &= \|e^{-i\lambda (t-a_1)}f_{\lambda}\|^2_{\mathfrak{L}(1)}
 =\int^{a_1}_{-\infty}\|e^{-i\lambda (t-a_1)}f_{\lambda}\|^2_{H_1}dt \\
 &= \int^{a_1}_{-\infty}e^{2\lambda_i (t-a_1)}dt\|f_{\lambda}\|^2_{H_1}
 =\frac{1}{2\lambda_i}\|f_{\lambda}\|^2_{H_1}<\infty
 \end{align*}
 and
\begin{align*}
 &\big\|i\int^{a_1}_t e^{i({\tilde{A}}_1-{\lambda })(t-s)}f_1(s)ds
 \big\|^2_{\mathfrak{L}(1)}\\
 &\leq \int^{a_1}_{-\infty}
 \Big(\int^{a_1}_t e^{\lambda_i(t-s)}\|f_1(s)\|_{H_1}ds\Big)^2dt \\
 &\leq \int^{a_1}_{-\infty}\Big(\int^{a_1}_t e^{\lambda_i(t-s)}ds \Big)
 \Big(\int^{a_1}_t e^{\lambda_i(t-s)}\|f_1(s)\|^2_{H_1} ds \Big)dt \\
 &=\frac{1}{\lambda_i}\int^{a_1}_{-\infty}\int^{a_1}_t
 e^{\lambda_i(t-s)}\|f_1(s)\|^2_{H_1}ds dt
 =\frac{1}{\lambda_i}\int^{a_1}_{-\infty}
 \Big(\int^{s}_{-\infty}e^{\lambda_i(t-s)}\|f_1(s)\|^2_{H_1}dt\Big)ds \\
 &=\frac{1}{\lambda_i}\int^{a_1}_{-\infty}
 \Big(\int^{s}_{-\infty}e^{\lambda_i(t-s)}dt\Big) \|f_1(s)\|^2_{H_1}ds\\
 &=\frac{1}{\lambda_i^2}\int^{a_1}_{-\infty}\|f_1(s)\|^2_{H_1}ds
 =\frac{1}{\lambda_i^2}\|f_1\|^2_{\mathfrak{L}(1)}<\infty.
\end{align*}
 Hence, $\|u_1(\lambda ;t)\|_{\mathfrak{L}(1)}<\infty$.
Furthermore,
\begin{align*}
 \|u_2(\lambda;t)\|^{2}_{\mathfrak{L}(2)}
&= \big\|i\int^{\infty}_t e^{i({\tilde{A}}_2-{\lambda })(t-s)}f_2(s)ds
 \big\|^2_{\mathfrak{L}(2)} \\
&\leq \int^{\infty}_{a_2}
 \Big(\int^{\infty}_t e^{\lambda_i(t-s)}\|f_2(s)\|_{H_2}ds\Big)^2dt \\
&\leq \int^{\infty}_{a_2}\Big(\int^{\infty}_t e^{\lambda_i(t-s)}ds \Big)
 \Big(\int^{\infty}_t e^{\lambda_i(t-s)}\|f_2(s)\|_{H_2}^2ds \Big)dt \\
&=\frac{1}{\lambda_i}\int^{\infty}_{a_2}
 \Big(\int^{\infty}_t e^{\lambda_i(t-s)}\|f_2(s)\|^2_{H_2}ds \Big)dt\\
&=\frac{1}{\lambda_i}\int^{\infty}_{a_2}\Big(\int^{s}_{a_2}e^{\lambda_i(t-s)}
 \|f_2(s)\|^2_{H_2}dt\Big)ds \\
&=\frac{1}{\lambda_i}\int^{\infty}_{a_2}
 \Big(\int^{s}_{a_2}e^{\lambda_i(t-s)}dt\Big) \|f_2(s)\|^2_{H_2}ds\\
&=\frac{1}{\lambda_i^2}\int^{\infty}_{a_2}(1-e^{\lambda_i(a_2-s)})\|f_2(s)\|_{H_2}^2ds \\
&\leq \frac{1}{\lambda_i^2}\|f_2\|^2_{\mathfrak{L}(2)}<\infty
\end{align*}
 and
\begin{align*}
 \|u_3(\lambda;t)\|^{2}_{\mathfrak{L}(3)}
&= \big\|i\int^{\infty}_t e^{i({\tilde{A}}_3-{\lambda })(t-s)}f_3(s)ds
 \big\|^2_{\mathfrak{L}(3)} \\
&\leq \int^{\infty}_{a_3}
 \Big(\int^{\infty}_t e^{\lambda_i(t-s)}\|f_3(s)\|_{H_3}ds\Big)^2dt \\
&\leq \int^{\infty}_{a_3}\Big(\int^{\infty}_t e^{\lambda_i(t-s)}ds \Big)
 \Big(\int^{\infty}_t e^{\lambda_i(t-s)}\|f_3(s)\|^2_{H_3}ds \Big)dt \\
&=\frac{1}{\lambda_i}\int^{\infty}_{a_3}
 \Big(\int^{\infty}_t e^{\lambda_i(t-s)}\|f_3(s)\|^2_{H_3}ds \Big)dt\\
&=\frac{1}{\lambda_i}\int^{\infty}_{a_3}\Big(\int^{s}_{a_3}e^{\lambda_i(t-s)}
 \|f_3(s)\|^2_{H_3}dt\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)\|^2_{H_3}ds\\
&=\frac{1}{\lambda_i^2}\int^{\infty}_{a_3}(1-e^{\lambda_i(a_3-s)})
 \|f_3(s)\|^2_{H_3}ds \\
&\leq \frac{1}{\lambda_i^2}\|f_3\|^2_{\mathfrak{L}(3)}<\infty.
\end{align*}
 The above calculations imply that $u_1(\lambda ;t) \in \mathfrak{L}(1)$,
 $u_2(\lambda ;t) \in \mathfrak{L}(2)$ and 
 $u_3(\lambda ;t) \in \mathfrak{L}(1)$ for $\lambda \in \mathbb{C}$,
 $\lambda_i=\operatorname{Im}\lambda> 0$. On the other hand, one can easily 
verify that
 $u(\lambda;t)=(u_1(\lambda;t),u_2(\lambda;t),u_3(\lambda;t))$
 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 \eqref{GrindEQ__3_1_} 
is in the form
 $u(\lambda;t)=(u_1(\lambda;t),u_2(\lambda;t),u_3(\lambda;t))$,
\begin{gather*}
u_1(\lambda;t)=-i\int^{t}_{-\infty}e^{i({\tilde{A}}_1-\lambda)(t-s)}f_1(s)ds,
 \quad t\in \Delta_1\\
u_2(\lambda;t)=e^{i({\tilde{A}}_2-\lambda)(t-a_2)}g_{\lambda} 
 -i\int^{t}_{a_2}e^{i({\tilde{A}}_2-\lambda)(t-s)}f_2(s)ds, \quad t\in \Delta_2, \\
u_3(\lambda;t)=e^{i({\tilde{A}}_3-\lambda)(t-a_3)}h_{\lambda}
 -i\int^{t}_{a_3}e^{i({\tilde{A}}_3-\lambda)(t-s)}f_3(s)ds, \quad t\in \Delta_3,
\end{gather*}
 where $-i\int^{a_1}_{-\infty}e^{i({\tilde{A}}_1-\lambda)(a_1-s)}f_1(s)ds
=WV(g_{\lambda},h_{\lambda})$
 and since
\begin{align*} 
\big\|-i\int^{a_1}_{-\infty} e^{i({\tilde{A}}_1-\lambda)(a_1-s)}f_1(s)ds
 \big\|^{2}_{H_1}
&\leq \Big(\int^{a_1}_{-\infty}e^{\lambda_i(a_1-s)}\|f_1(s)\|_{H_1}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_1}ds\Big) \\
&\leq \frac{1}{2|\lambda_i|}\|f_1\|^2_{\mathfrak{L}(1)}<\infty,
\end{align*}
we have 
\[
(g_{\lambda},h_{\lambda})=V^{-1}W^{*}\Big(-i\int^{a_1}_{-\infty}
e^{i({\tilde{A}}_1-\lambda)(a_1-s)}f_1(s)ds\Big)
 \in H_2\oplus H_3.
\]

 First, we prove that $u(\lambda;t)\in \mathfrak{L}$. In this case,
 \begin{align*}
 \|u_1(\lambda ;t)\|^2_{\mathfrak{L}(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_1}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_1}ds\Big)dt \\
 &= \frac{1}{|\lambda_i|}\int^{a_1}_{-\infty}\int^{t}_{-\infty}e^{\lambda_i(t-s)}
  \|f_1(s)\|^2_{H_1}\,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_1}dt\Big)ds\\
 &= \frac{1}{|\lambda_i|}\int^{a_1}_{-\infty} \Big( \int^{a_1}_{s} e^{\lambda_i(t-s)}dt \Big)
  \|f_1(s)\|^2_{H_1}ds \\
 &= \frac{1}{|\lambda_i|^2}\int^{a_1}_{-\infty}
  (1-e^{\lambda_i(a_1-s)})\|f_1(s)\|^2_{H_1}ds\\
 &\leq \frac{1}{|\lambda_i|^2}\|f_1\|^2_{\mathfrak{L}(1)}< \infty,
\end{align*}
\[
 \|e^{i({\tilde{A}}_2-\lambda)(t-a_2)}g_{\lambda}\|^2_{\mathfrak{L}(2)}
 \leq \int^{\infty}_{a_2}e^{2\lambda_i(t-a_2)}dt\|g_{\lambda}\|^2_{H_2}\\
 =\frac{1}{2|\lambda_i|}\|g_{\lambda}\|^2_{H_2}<\infty
\]
 and
\begin{align*}
&\big\|-i\int^{t}_{a_2}e^{i({\tilde{A}}_2-\lambda)(t-s)}f_2(s)ds
 \big\|^2_{\mathfrak{L}(2)}\\
&\leq \int^{\infty}_{a_2}
 \Big(\int^{t}_{a_2}e^{\lambda_i(t-s)}\|f_2(s)\|_{H_2}ds \Big)^2dt \\
&\leq \int^{\infty}_{a_2}\Big(\int^{t}_{a_2}e^{\lambda_i(t-s)}ds\Big)
 \Big(\int^{t}_{a_2}e^{\lambda_i(t-s)}\|f_2(s)\|^2_{H_2}ds\Big)dt \\
&=\int^{\infty}_{a_2}\Big(\frac{1}{\lambda_i}(1-e^{\lambda_i(t-a_2)}) \Big)
\Big(\int^{t}_{a_2}e^{\lambda_i(t-s)}\|f_2(s)\|^2_{H_2}ds\Big)dt \\
&\leq \frac{1}{|\lambda_i|}\int^{\infty}_{a_2}
 \Big(\int^{t}_{a_2}e^{\lambda_i(t-a_2)}\|f_2(s)\|^2_{H_2}ds\Big)dt \\
&=\frac{1}{|\lambda_i|}\int^{\infty}_{a_2}
 \Big(\int^{\infty}_{s}e^{\lambda_i(t-s)}\|f_2(s)\|^2_{H_2}dt\Big)ds\\
&=\frac{1}{|\lambda_i|}\int^{\infty}_{a_2}
 \Big(\int^{a_2}_{s}e^{\lambda_i(t-s)}dt\Big)\|f_2(s)\|^2_{H_2}ds \\
&=\frac{1}{|\lambda_i|^2}\|f_2\|^2_{\mathfrak{L}(2)}<\infty.
 \end{align*}
 In a similar way, it can be shown that 
\[
\|e^{i({\tilde{A}}_3-\lambda)(t-a_3)}
 h_{\lambda}\|_{\mathfrak{L}(3)}<\infty, \quad 
\|-i\int^{t}_{a_3}e^{i({\tilde{A}}_3-\lambda)(t-s)}f_3(s)ds\|_{\mathfrak{L}(3)}
<\infty.
\]
The above calculations show that 
$u_1(\lambda;\cdot)\in \mathfrak{L}(1)$,
 $u_2(\lambda;\cdot)\in \mathfrak{L}(2)$ and that 
$u_3(\lambda;\cdot)\in \mathfrak{L}(3)$; i.e.,
 $u(\lambda;\cdot)=(u_1(\lambda;\cdot),
u_2(\lambda;\cdot),u_3(\lambda,\cdot))\in \mathfrak{L}$
 for $\lambda \in \mathbb{C}$, $\lambda_i=\operatorname{Im}\lambda< 0$.
 On the other hand it can be verified that the function $u(\lambda;\cdot)$
 satisfies the equation and boundary condition in \eqref{GrindEQ__3_1_}.
\end{proof}

 Now, we will study continuous spectrum $\sigma_{c}(L_W)$ of the
 selfadjoint extension $L_W$.

\begin{theorem} \label{thm3.3} 
The continuous spectrum of any selfadjoint extension $L_W$ is
\[
 \sigma_{c}(L_W)=\mathbb{R}.
\]
\end{theorem}
\begin{proof}
 For $\lambda \in \mathbb{C}$, $\lambda_i=\operatorname{Im}\lambda> 0$,
 norm of the resolvent operator $R_{\lambda}(L_W)$ of the $L_W$ is of the form
\begin{align*}
 \|R_{\lambda}(L_{W})f(t)\|^2_{\mathfrak{L}}
 &=\big\|e^{i({\tilde{A}}_1-\lambda)(t-a_1)}f_{\lambda}
 +i\int^{a_1}_t e^{i({\tilde{A}}_1-\lambda)(t-s)}f_1(s)ds
 \big\|^2_{\mathfrak{L}(1)} \\
 &\quad +\big\|i\int^{\infty}_t e^{i({\tilde{A}}_2-\lambda)(t-s)}f_2(s)ds
 \big\|^2_{\mathfrak{L}(2)}\\
&\quad +\big\|i\int^{\infty}_t e^{i({\tilde{A}}_3-\lambda)(t-s)}f_3(s)ds \big\|^2_{\mathfrak{L}(3)},
\end{align*}
 where $f=(f_1,f_2,f_3)\in \mathfrak{L}$, 
\[
f_{\lambda}=WV\Big(i\int^{\infty}_{a_2}e^{i({\tilde{A}}_2-\lambda)(a_2-s)}
f_2(s)ds, \,
 i\int^{\infty}_{a_3}e^{i({\tilde{A}}_3-\lambda)(a_3-s)}f_3(s)ds \Big).
\]
 Then, it is clear that for any $f=(f_1,f_2,f_3)\in \mathfrak{L}$,
\[
 \|R_{\lambda}(L_{W})f(t)\|^2_{\mathfrak{L}}\geq \big\|i\int^{\infty}_t
 e^{i({\tilde{A}}_2-\lambda)(t-s)}f_2(s)ds \big\|^2_{\mathfrak{L}(2)}.
\]
 The vector functions $f^{*}(\lambda;t)$ which is of the form
 $f^{*}(\lambda;t)=(0,e^{-i(\bar{\lambda}-{\tilde{A}}_2)t}f,0)$,
 $\lambda \in \mathbb{C}$, $\lambda_i=\operatorname{Im}\lambda> 0$, $f\in H_2$
 belong to $\mathfrak{L}$. Indeed,
\begin{align*}
 \|f^{*}(\lambda;t)\|^2_{\mathfrak{L}}
 &=\int^{\infty}_{a_2}\|e^{-i(\bar{\lambda}-{\tilde{A}}_2)t}f\|^2_{H_2}dt
 =\int^{\infty}_{a_2}e^{-2\lambda_it}dt\|f\|^2_{H_2}\\
 &=\frac{1}{2\lambda_i}e^{-2\lambda_ia_2}\|f\|^2_{H_2} < \infty.
\end{align*}
 For such functions $f^{*}(\lambda;\cdot)$, we have
\begin{align*}
 \|R_{\lambda}(L_{W})f^{*}(\lambda;t)\|^2_{\mathfrak{L}(2)}
 &\geq \big\|i\int^{\infty}_t e^{-i(\lambda-{\tilde{A}}_2)(t-s)}
 e^{-i(\bar{\lambda}-\tilde{A}_2)s}fds \big\|^2_{\mathfrak{L}} \\
 &=\big\|\int^{\infty}_te^{-i\lambda t}e^{-2\lambda_is}e^{i\tilde{A}_2t}fds
 \big\|^2_{\mathfrak{L}(2)}\\
 &=\big\|e^{-i\lambda t}e^{i\tilde{A}_2t}\int^{\infty}_te^{-2\lambda_is}
 fds \big\|^2_{\mathfrak{L}(2)} \\
 &=\big\|e^{-i\lambda t}\int^{\infty}_te^{-2\lambda_is}ds
 \big\|^2_{\mathfrak{L}(2)}\|f\|^2_{H_2}\\
 &=\frac{1}{4\lambda^2_i}\int^{\infty}_{a_2}e^{-2\lambda_it}dt\|f\|^2_{H_2}\\
 &=\frac{1}{8\lambda^{3}_i}e^{-2\lambda_ia_2}\|f\|^2_{H_2}.
\end{align*}
 From this we obtain
\[
 \|R_{\lambda}(L_{W})f^{*}(\lambda;\cdot)\|_{\mathfrak{L}}
 \geq \frac{e^{-\lambda_ia_2}}{2\sqrt{2}\lambda_i\sqrt{\lambda_i}}\|f\|_{H_2}
 =\frac{1}{2\lambda_i}\|f^{*}(\lambda;\cdot)\|_{\mathfrak{L}};
\]
 i.e., for $\lambda_i=\operatorname{Im}\lambda >0$ and $f\neq 0$,
\[
 \frac{\|R_{\lambda}(L_{W})f^{*}(\lambda;\cdot)\|_{\mathfrak{L}}}{\|f^{*}(\lambda;\cdot)\|_{\mathfrak{L}}}
 \geq\frac{1}{2\lambda_i}.
\]
 is valid.
 On the other hand, it is clear that
\[
 \|R_{\lambda}(L_{W})\|\geq \frac{\|R_{\lambda}(L_{W})
 f^{*}(\lambda;\cdot)\|_{\mathfrak{L}}}{\|f^{*}(\lambda;\cdot)\|_{\mathfrak{L}}},
 \quad f\neq 0.
\]
 Consequently, we have
 \[
 \|R_{\lambda}(L_{W})\|\geq \frac{1}{2\lambda_i} \quad\text{for }
 \lambda\in \mathbb{C}, \quad \lambda_i=\operatorname{Im}\lambda >0.
 \]
 The last relation implies the validity of assertion.
\end{proof}

\begin{example} \label{examp3.4} \rm 
By the previous theorem, the spectrum of the  boundary-value problem
\begin{gather*}
 i\frac{\partial u(t,x)}{\partial t}-\frac{{\partial }^2u(t,x)}{\partial x^2}=f(t,x),
 \quad |t|>1,\; x\in [0,1],\\
 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,\;
\end{gather*}
 in the space $L_2((-\infty,-1)\times (0,1))\oplus {L_2((1,+\infty)\times(0,1))}$
 is continuous and coincides with $\mathbb{R}$. This corresponds to the case
 when $H_1=H_2=\mathbb{C}$ and $H_3=\{0\}$.
\end{example}


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\end{document}