\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 143, pp. 1--8.\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-2017/143\hfil Selfadjoint singular differential operators]
{First-order selfadjoint singular differential operators in a Hilbert
space of vector functions}

\author[P. Ipek,  B. Yilmaz, Z. I. Ismailov \hfil EJDE-2017/143\hfilneg]
{Pembe Ipek,  B\"ulent Yilmaz, Zameddin I. Ismailov}

\address{Pembe Ipek \newline
Karadeniz Technical University,
Institute of Natural Sciences,
61080, Trabzon, Turkey}
\email{ipekpembe@gmail.com}

\address{B\"ulent Yilmaz \newline
Marmara University,
Department of Mathematics,
 Kad{\i}k\"oy, 34722, Istanbul, Turkey}
\email{bulentyilmaz@marmara.edu.tr}

\address{Zameddin I. Ismailov \newline
Karadeniz Technical University,
Department of Mathematics, 61080, Trabzon, Turkey}
\email{zameddin.ismailov@gmail.com}

\dedicatory{Communicated by Ludmila Pulkina }

\thanks{Submitted March 17, 2018. Published June 17, 2017.}
\subjclass[2010]{47A10, 47B25}
\keywords{Multipoint singular differential expression; deficiency indeces;
\hfill\break\indent symmetric and selfadjoint  differential operator;
 spectrum}

\begin{abstract}
 In this article, we give a representation of all selfadjoint extensions of the
 minimal operator generated by first-order linear symmetric multipoint
 singular differential expression, with operator coefficient in
 the direct sum of Hilbert spaces of vector-functions defined at the
 semi-infinite intervals. To this end we use the Calkin-Gorbachuk method.
 Finally, the geometry of spectrum set of such extensions is researched.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}

 In the first years of the previous century,  von Neumann \cite{rf11} and  
Stone \cite{rf10}
investigated the theory of selfadjoint extensions of linear densely defined
closed symmetric operators in a Hilbert spaces. Applications to scalar
linear even order symmetric differential operators and description of
all selfadjoint extensions in terms of boundary conditions due to  Glazman in
his seminal work \cite{rf5} and in the book of  Naimark \cite{rf8}. 
In this sense the famous
Glazman-Krein-Naimark (or Everitt-Krein-Glazman-Naimark) Theorem in the mathematical
literature it is to be noted. In the mathematical literature there is another
method co-called Calkin-Gorbachuk method. (see \cite{rf6,rf9}).

Our motivation for this article originates from the interesting researches
of  Everitt,  Markus,  Zettl,  Sun,  O'Regan,  Agarwal \cite{rf2,rf3,rf4,rf12}
in scalar cases. 
Throughout this paper we consider  Zettl and  Suns's view about these topics
\cite{rf12}.  A selfadjoint ordinary differential operator
in Hilbert space is generated by two things:
\begin{itemize}
\item[(1)] a symmetric (formally selfadjoint) differential expression; 
\item[(2)] a boundary condition which determined selfadjoint differential operators;
\end{itemize}
And also for a given selfadjoint differential operator, a basic question is:
 What is its spectrum?

In this work in Section 3 the representation of all selfadjoint extensions
of a multipoint symmetric quasi-differential operator, generated by first-order
symmetric differential-operator expression (for the definition see \cite{rf4})
in the direct sum of Hilbert spaces of vector-functions defined at the
semi-infinite intervals in terms of boundary conditions are described.
In sec. 4 the structure of spectrum of these selfadjoint extensions is
investigated.

\section{Statement of the problem}

In the direct sum $ \mathcal{H}=L^2(H,(-\infty, a_1))
\oplus L^2(H,(a_2,\infty))$,  $H$ is a separable Hilbert space, and
$ a_1, a_2 \in \mathbb{R} $ will be considered for the  multipoint
differential-operator expression in the form
\begin{gather*}
l(u)=(l_1(u_1),l_2(u_2)), \\
l_{k}(u_{k})=i\rho_{k}u_{k}'+\frac{1}{2}i\rho_{k}'u_{k}+A_{k}u_{k}, \quad k=1,2,
\end{gather*}
where
\begin{itemize}
\item[(1)] $ \rho_1: (-\infty,a_1)\to (0,\infty)$,
$\rho_2: (a_2,\infty)\to (0,\infty)$;

\item[(2)] $ \rho_1\in AC_{loc}(-\infty,a_1)$ and
$ \rho_2\in AC_{loc}(a_2,\infty) $;

\item[(3)] $ \int_{-\infty}^{a_1}\frac{ds}{\rho_1(s)}=\infty$,
$\int_{a_2}^{\infty}\frac{ds}{\rho_2(s)}=\infty $;

\item[(4)] $ A_{k}^{*}=A_{k}:D(A_{k})\subset H\to H$, $k=1,2$.
\end{itemize}

The minimal operators $ L_0^{1} $ and $ L_0^2 $
corresponding to differential-operator expressions 
$ l_1 $ and $ l_2 $  in $ L^2(H,(-\infty, a_1)) $ and
$ L^2(H,(a_2, \infty)) $, respectively,  can be defined
by a standard processes, see\cite{rf7}.
The operators
$ L_1=(L_0^{1})^{*}$ and $L_2=(L_0^2)^{*} $ are maximal operators
corresponding to $ l_1 $ and $ l_2 $ in  $ L^2(H,(-\infty, a_1)) $
and $ L^2(H,(a_2, \infty)) $, respectively. In this case the operators
$$
L_0=L_0^{1}\oplus L_0^2 \quad \text{and} \quad
L=L^{1}\oplus L^2
$$
will be indicating the minimal and maximal operators corresponding to
differential expression on $ \mathcal{H} $, respectively.

It is clear that
\begin{gather*}
D(L^{1})  =  \{ u_1\in L^2(H,(-\infty, a_1)): l_1(u_1)
\in L^2(H,(-\infty, a_1)\},   \\
D(L^{1}_0)  =   \{ u_1\in D(L^{1}):  (\sqrt{\rho_1}u_1)(a_1)=0  \}
\end{gather*}
and
\begin{gather*}
D(L^2)  =  \{ u_2\in L^2(H,(a_2,\infty)):
  l_2(u_2)\in L^2(H,(a_2,\infty)\},  \\
D(L^2_0)  =   \{ u_2\in D(L^2):(\sqrt{\rho_2}u_2)(a_2)=0\} .
\end{gather*}

\section{Description of Selfadjoint Extensions}

In this section using the Calkin-Gorbachuk method will be investigated the
 general representation of selfadjoint extensions of minimal operator $ L_0 $.
 Firstly we prove the  following result.

\begin{lemma} \label{lem3.1}
The deficiency indices of the operators $ L^{1}_0 $ and $ L^2_0 $
are of the form
$$
(m(L^{1}_0),n(L^{1}_0))=(0,\dim H), \quad
(m(L^2_0),n(L^2_0))=(\dim H,0).
$$
\end{lemma}

\begin{proof}
 Now for simplicity we assume that $ A_1=A_2=0$.
It is clear that the general solutions of differential equations
\begin{gather*}
i\rho_1(t)u'_{1\pm}(t)+ \frac{1}{2}i\rho'_1(t)u_{1\pm}(t)
\pm iu_{1\pm} (t)=0, \quad t<a_1, \\
i\rho_2(t)u'_{2\pm}(t)+ \frac{1}{2}i\rho'_2(t)u_{2\pm}(t) \pm iu_{2\pm} (t)=0,
\quad  t>a_2
\end{gather*}
in $L^2(H,(-\infty,a_1))$ and $ L^2(H,(a_2,+\infty)) $ are in the form
$$
u_{1\pm}(t)=\exp\Big(  \pm \int_{t}^{c_1}\frac{2\pm \rho_1'(s)}{2\rho_1(s)}
ds\Big)f_1, \quad  f_1\in H, \; t<a_1, \; c_1<a_1
$$
and
$$
u_{2\pm}(t)=\exp\Big(  \mp \int_{c_2}^{t}
\frac{2\pm \rho_2'(s)}{2\rho_2(s)}ds\Big)f_2, \quad f_2\in H, \; t>a_2, \;
 c_2>a_2
$$
respectively.
 From these representations we have
\begin{align*}
&\| u_{1+}\|^2_{L^2(H,(-\infty,a_1))} \\
&=  \int_{-\infty}^{a_1} \| u_{1+}(t)\|^2_{H}dt  \\
&=  \int_{-\infty}^{a_1} \exp\Big(  \int_{t}^{c_1}
 \frac{2+ \rho_1'(s)}{\rho_1(s)}ds\Big)dt \| f_1\|_{H}^2  \\
&=  \int_{-\infty}^{a_1} \frac{\rho_1 (c_1)}{\rho_1 (t)}
 \exp\Big(  \int_{t}^{c_1} \frac{2}{\rho_1 (s)}ds \Big) dt \| f_1\|_{H}^2  \\
&=  \frac{\rho_1 (c_1)}{2}\int_{-\infty}^{a_1} \exp\Big(  \int_{t}^{c_1}
 \frac{2}{\rho_1 (s)}ds \Big) d\Big(  -\int_{t}^{c_1} \frac{2}{\rho_1 (s)}ds
 \Big)  \| f_1\|_{H}^2  \\
&=  -\frac{\rho_1 (c_1)}{2} \Big[ \exp\Big(  \int_{a_1}^{c_1}
\frac{2}{\rho_1 (s)}ds  \Big)- \exp\Big(
\int_{-\infty}^{c_1} \frac{2}{\rho_1 (s)}ds \Big) \Big]
\| f_1\|_{H}^2=\infty.
\end{align*}
Consequently,
$$
\dim\ker(L_0^{1}+iE)=0
$$
 On the other hand it is clear that
\begin{align*}
&\| u_{1-}\|^2_{L^2(H,(-\infty,a_1))}\\
&=  \int_{-\infty}^{a_1} \| u_{1-}(t)\|^2_{H}dt  \\
&=  \int_{-\infty}^{a_1} \exp\Big(  -\int_{t}^{c_1}
 \frac{2+ \rho_1'(s)}{\rho_1(s)}ds\Big)dt \| f_1\|_{H}^2  \\
&=  \int_{-\infty}^{a_1} \frac{\rho_1 (c_1)}{\rho_1 (t)}
 \exp\Big( - \int_{t}^{c_1} \frac{2}{\rho_1 (s)}ds \Big) dt \| f_1\|_{H}^2  \\
&=  \frac{\rho_1 (c_1)}{2}\int_{-\infty}^{a_1}
 \exp\Big( - \int_{t}^{c_1} \frac{2}{\rho_1 (s)}ds \Big)
 d\Big( - \int_{t}^{c_1} \frac{2}{\rho_1 (s)}ds \Big)  \| f_1\|_{H}^2  \\
&=  \frac{\rho_1 (c_1)}{2} \Big[ \exp\Big( - \int_{a_1}^{c_1}
 \frac{2}{\rho_1 (s)}ds  \Big)
 - \exp\Big( - \int_{-\infty}^{c_1} \frac{2}{\rho_1 (s)}ds \Big) \Big]
 \| f_1\|_{H}^2 \\
&=  \frac{\rho_1 (c_1)}{2} \exp\Big( - \int_{a_1}^{c_1} \frac{2}{\rho_1 (s)}ds \Big) 
  \| f_1\|_{H}^2<\infty.
\end{align*}
Therefore,
$$
u_{1-}(t)=\exp\Big( \int_{a_1}^{t}\frac{2 - \rho_1'(s)}{2\rho_1(s)}ds\Big)f_1 
\in L^2(H,(-\infty,a_1)).
$$
Hence
$$
\dim\ker (L_0^{1}-iE)=\dim H
$$
In a similar way it can be shown that
$$
m(L^2_0)=\dim\ker (L_0^2+iE)=\dim H \quad \text{and} \quad
n(L^2_0)=\dim\ker (L_0^2-iE)=0
$$
This completes the proof.
\end{proof}

Consequently, the minimal operator $ L_0 $ has selfadjoint extensions; 
see \cite{rf6}.
To describe these extensions we need to obtain the space of boundary values. 

\begin{definition}[\cite{rf6}] \label{def3.2} \rm 
Let $ H $ be any Hilbert space and $ S:D(S)\subset H\to H $
 be a closed densely defined symmetric operator in the Hilbert space 
$ \mathcal{H} $ having equal finite or infinite deficiency indices. 
A triplet $( \mathcal{B},\gamma_1,\gamma_2)$, where $ \mathcal{B} $ is a
Hilbert space, $ \gamma_1 $ and $ \gamma_2 $ are linear mappings from 
$ D(S^{*}) $ into $ \mathcal{B} $, is called a space of boundary values for 
the operator $ S $ if for any $ f,g\in D(S^{*}) $
$$
(S^{*}f,g)_{H}-(f,S^{*}g)_{H}
=(\gamma_1(f),\gamma_2(g))_{\mathcal{B}}-(\gamma_2(f),\gamma_1(g))_{\mathcal{B}}
$$
while for any $ F_1,F_2\in \mathcal{B} $, there exists an element 
$ f\in D(S^{*}) $ such that $ \gamma_1(f)=F_1 $ and $ \gamma_2(f)=F_2 $.
\end{definition}

\begin{lemma} \label{lem3.3}
Let 
\begin{gather*}
\gamma_1:D(L)\to H, \quad \gamma_1(u)=\frac{1}{i\sqrt{2}}\big( (\sqrt{\rho_1}u_1)
(a_1)+ (\sqrt{\rho_2}u_2)(a_2)\big), \\
\gamma_2:D(L)\to H, \quad \gamma_2(u)= \frac{1}{\sqrt{2}}
\left( (\sqrt{\rho_1}u_1)(a_1) - (\sqrt{\rho_2}u_2)(a_2)\right), 
\end{gather*}
where $u=(u_1,u_2) \in D(L)$. 
Then the triplet $ (H,\gamma_1,\gamma_2)$
is a space of boundary values of the minimal operator $ L_0 $ in $ \mathcal{H}$.
\end{lemma}

\begin{proof} 
For any $ u=(u_1,u_2)$, $v=(v_1,v_2) \in D(L) $
\begin{align*}
& (Lu,v)_{\mathcal{H}} -(u,Lv)_{\mathcal{H}}  \\
& =  (L_1u_1,v_1)_{L^2(H,(-\infty, a_1))} + (L_2u_2,v_2)_{L^2(H,(a_2,\infty))} 
 -(u_1,L_1v_1)_{L^2(H,(-\infty, a_1))} \\
&\quad -(u_2,L_2v_2)_{L^2(H,(a_2,\infty))}   \\
& =  \big[  (i\rho_1 u_1'+\frac{i}{2}\rho_1 'u_1+A_1u_1,v_1)_{L^2(H,(-\infty, a_1))}\\
&\quad -(u_1,i\rho_1 v_1'+\frac{i}{2}\rho_1 'v_1+A_1v_1)_{L^2(H,(-\infty,a_1))} \big]  \\
& \quad + \big[  (i\rho_2 u_2'+\frac{i}{2}\rho_2 'u_2+A_2u_2,v_2)
  _{L^2(H,(a_2,\infty))} \\
&\quad  -(u_2,i\rho_2 v_2'+\frac{i}{2}\rho_2 'v_2+A_2v_2)_{L^2(H,(a_2,\infty))}
 \big]  \\
& =  i \big[ (\rho_1 u_1',v_1)_{L^2(H,(-\infty,a_1))}
 +( u_1,\rho_1v_1')_{L^2(H,(-\infty,a_1))}\big]  \\
&\quad + \frac{i}{2} \big[ (\rho_1'u_1,v_1)_{L^2(H,(-\infty,a_1))}
 + (u_1,\rho_1'v_1)_{L^2(H,(-\infty,a_1))}\big]  \\
&\quad + i \big[ (\rho_2 u_2',v_2)_{L^2(H,(a_2,\infty))}
 +( u_2,\rho_2v_2')_{L^2(H,(a_2,\infty))}\big] \\
&\quad +\frac{i}{2} \big[ (\rho_2'u_2,v_2)_{L^2(H,(a_2,\infty))}
 + (u_2,\rho_2'v_2)_{L^2(H,(a_2,\infty))}\big]   \\
& =  i \big[ (\rho_1 u_1',v_1)_{L^2(H,(-\infty,a_1))}
 +( u_1,\rho_1v_1')_{L^2(H,(-\infty,a_1))}\big]
 + i (\rho_1'u_1,v_1) _{L^2(H,(-\infty,a_1))} \\
& \quad + i \big[ (\rho_2 u_2',v_2)_{L^2(H,(a_2,\infty))}
 +( u_2,\rho_1v_2')_{L^2(H,(a_2,\infty))}\big]
 +i (\rho_2'u_2,v_2)L_{(H,(a_2,\infty))}  \\
& =  i \big[ (\rho_1 u_1'+\rho_1u_1,v_1)_{L^2(H,(-\infty,a_1))}
 +( \rho_1u_1,v_1')_{L^2(H,(-\infty,a_1))}\big]  \\
&\quad + i \big[ (\rho_2 u_2'+\rho_2u_2,v_2)_{L^2(H,(a_2,\infty))}
 +( \rho_2u_2,v_2')_{L^2(H,(a_2,\infty))}\big]  \\
& =  i [  ((\rho_1u_1),v_1)'] _{L^2(H,(-\infty,a_1))}
+i \left[  ((\rho_2u_2),v_2)'\right] _{L^2(H,(a_2,\infty))}  \\
& =  i \big[ \left( (\sqrt{\rho_2}u_2)(a_2),(\sqrt{\rho_2}v_2)(a_2)\right) _{H}
-\left( (\sqrt{\rho_1}u_1)(a_1),(\sqrt{\rho_1}v_1)(a_1)\right) _{H}\big]  \\
& =  (\gamma_1(u),\gamma_2(v))_{H}-(\gamma_2(u),\gamma_1(v))_{H}.
\end{align*}

 Now for any element $ f_1,  f_2 \in H $ let us find the function 
$ u=(u_1,u_2)\in D(L) $ such that
\begin{gather*}
\gamma_1(u)=\frac{1}{i\sqrt{2}}\left( (\sqrt{\rho_1}u_1)(a_1)
 + (\sqrt{\rho_2}u_2)(a_2)\right) =f_1 , \\
\gamma_2(u)=\frac{1}{\sqrt{2}}\left( (\sqrt{\rho_1}u_1)(a_1)
- (\sqrt{\rho_2}u_2)(a_2)\right)=f_2
\end{gather*}
From here the following two expressions are obtained
$$
(\sqrt{\rho_1}u_1)(a_1)=(i f_1+f_2)/\sqrt{2}, \quad 
(\sqrt{\rho_2}u_2)(a_2)=(i f_1-f_2)/\sqrt{2}.
$$
If we choose the functions $ u_1(\cdot), \ u_2(\cdot) $ in the following forms
\begin{gather*}
u_1(t)=\frac{1}{\sqrt{\rho_1(t)}}e^{t-a_1}(if_1+f_2)/\sqrt{2}, \quad t<a_1, \\
u_2(t)=\frac{1}{\sqrt{\rho_2(t)}}e^{a_2-t}(if_1-f_2)/\sqrt{2}, \quad t>a_2,
\end{gather*}
then it is clear that $ u=(\sqrt{\rho_1}u_1,\sqrt{\rho_2}u_2)\in D(L) $ and
$ \gamma_1(u)=f_1$, $\gamma_2(u)=f_2$.
\end{proof}

\begin{theorem} \label{thm3.4}
If $ \widetilde{L} $ is a selfadjoint extension of the minimal operator 
$ L_0 $ in $ \mathcal{H} $ , then it is generated by the differential-operator
 expression $ l( \cdot ) $ and the following boundary condition
$$
(\sqrt{\rho_2}u_2)(a_2)=W(\sqrt{\rho_1}u_1)(a_1),
$$
where $ W:H\to H $ is a unitary operator. Moreover, the unitary operator
 $ W $ in $ H $ is determined uniquely by the extension $ \widetilde{L} $,
 i.e. $ \widetilde{L}=L_{W}$ and vice versa.
\end{theorem}


\section{Spectrum of the Selfadjoint Extensions}

 In this section the structure of the spectrum of the selfadjoint extensions 
$ L_{W} $ of the minimal operator $ L_0 $ in $ \mathcal{H} $ will be investigated.
 First let us prove the following results.

\begin{theorem} \label{thm4.1}
The point spectrum of the selfadjoint extension $ L_{W} $ is empty, i.e.
$\sigma_{p}(L_{W})=\emptyset$.
\end{theorem}

\begin{proof} 
Consider the eigenvalue problem
$$
l(u)=\lambda u, \quad u=(u_1,u_2)\in \mathcal{H}, \quad \lambda\in \mathbb{R}
$$
with the boundary condition
$$
(\sqrt{\rho_2}u_2)(a_2)=W(\sqrt{\rho_1}u_1)(a_1).
$$
From here the following expressions are obtained 
\begin{gather*}
i\rho_1(t)u_1'(t)+\frac{1}{2}i\rho_1'(t)u_1(t)+A_1u_1(t)=\lambda u_1(t), \quad t<a_1,\\
i\rho_2(t)u_2'(t)+\frac{1}{2}i\rho_2'(t)u_2(t)+A_2u_2(t)=\lambda u_2(t), \quad t>a_2,\\
(\sqrt{\rho_2}u_2)(a_2)=W(\sqrt{\rho_1}u_1)(a_1).
\end{gather*}
The general solutions of these equations are in the form
\begin{gather*}
u_1(t;\lambda)=\sqrt{\frac{\rho_1(c)}{\rho_1(t)}}
\exp\Big(-i(A_1-\lambda)\int_{t}^{c}\frac{ds}{\rho_1(s)} \Big) 
f^{1}_{\lambda}, \quad f^{1}_{\lambda} \in H, \quad t<a_1, \; c<a_1, \\
u_2(t;\lambda)=\sqrt{\frac{\rho_2(c)}{\rho_2(t)}}
\exp\Big(i(A_2-\lambda)\int_{c}^{t}\frac{ds}{\rho_2(s)} \Big) f^2_{\lambda}, \quad
 f^2_{\lambda}\in H, \quad t>a_2, \; c>a_2, \\
(\sqrt{\rho_2}u_2)(a_2)=W(\sqrt{\rho_1}u_1)(a_1).
\end{gather*}
It is clear that for the $ f^{1}_{\lambda} \neq 0 $ and $ f^2_{\lambda} \neq 0 $ 
the solutions are $ u_1(\cdot ;\lambda)\notin L^2(H,(-\infty,a_1))$ and
 $ u_2(\cdot ;\lambda)\notin L^2(H,(a_2,\infty))$.
Therefore for every unitary operator $ W $ we have 
$ \sigma_{p}(L_{W})=\emptyset $.
\end{proof}

Since the residual spectrum for any selfadjoint operator in any Hilbert space 
is empty, then we have to investigate the continuous spectrum of 
selfadjoint extensions $ L_{W} $ of the minimal operator $ L_0 $ is 
investigated. On the other hand from the general theory of linear 
selfadjoint operators in Hilbert spaces for the resolvent set
 $ \rho(L_{W})$ of any selfadjoint extension $ L_{W} $ is true
$$
\rho (L_{W})\supset \{ \lambda\in \mathbb{C}: \operatorname{Im} \lambda \neq 0 \}.
$$
For the continuous spectrum of selfadjoint extensions we have the following 
statement. 


\begin{theorem} \label{thm4.2}
The continuous spectrum of any selfadjoint extension $ L_{W} $ in of the form
$$
\sigma_{c}(L_{W})=\mathbb{R}.
$$
\end{theorem}

\begin{proof} 
For  $ \lambda\in \mathbb{C}$,  $\lambda_{i}=\operatorname{Im}\lambda>0 $ and 
$ f=(f_1,f_2)\in \mathcal{H} $ the norm of function $ R_{\lambda} (L_{W})f(t) $ 
in $ \mathcal{H} $  we have
\begin{align*}
&\| R_{\lambda} (L_{W}))f(t) \|_{\mathcal{H}}^2 \\
& =  \| \frac{1}{\rho_1(t)}\exp \Big( i(\lambda-A_1)
 \int_{t}^{a_1}\frac{ds}{\rho_1(s)}\Big) f^{1}_{\lambda} \\
&\quad + \frac{i}{\sqrt{\rho_1(t)}}\int_{t}^{a_1}
\exp \Big( i(A_1-\lambda)\int_{s}^{t}\frac{d\tau}{\rho_1(\tau)}\Big) 
\frac{f_1(s)}{\sqrt{\rho_1(s)}}ds\|^2_{L^2(H,(-\infty,a_1))} \\
&\quad + \| \frac{i}{\sqrt{\rho_2(t)}} \int_{t}^{\infty}
\exp \Big( i(A_2-\lambda)\int_{s}^{t}\frac{d\tau}{\rho_2(\tau)}\Big) 
\frac{f_2(s)}{\sqrt{\rho_2(s)}}ds\|^2_{L^2(H,(a_2,\infty))} \\
&\geq  \| \frac{i}{\sqrt{\rho_2(t)}} \int_{t}^{\infty}
 \exp \Big( i(A_2-\lambda)\int_{s}^{t}\frac{d\tau}{\rho_2(\tau)}\Big) 
\frac{f_2(s)}{\sqrt{\rho_2(s)}}ds\|^2_{L^2(H,(a_2,\infty))}.
\end{align*}
The vector functions $ f^{*}(t; \lambda )$ in the form 
\[
 f^{*}(t; \lambda )=\Big( 0, \frac{1}{\sqrt{\rho_2(t)}}
\exp\Big(-i(\overline{\lambda}-A_2)\int_{a_2}^{t}\frac{ds}{\rho_2(s)} 
\Big)f \Big),
\]
 with $\lambda\in \mathbb{C}$,
 $ \lambda_{i}=\operatorname{Im}\lambda>0$, $f\in H $ belong to $ \mathcal{H}$.
 Indeed,
\begin{align*}
 \| f^{*}(t; \lambda )\|^2_{\mathcal{H}} 
& =  \int_{a_2}^{\infty} \frac{1}{\rho_2(t)}\| 
 \exp\Big(-i(\overline{\lambda}-A_2)\int_{a_2}^{t}\frac{ds}{\rho_2(s)} \Big)
 f \|^2_{H}dt  \\
& =   \int_{a_2}^{\infty} \frac{1}{\rho_2(t)} 
 \exp\Big(-2\lambda_{i}\int_{a_2}^{t}\frac{ds}{\rho_2(s)} \Big)dt\| f\|^2_{H}  \\
& =  \frac{1}{2\lambda_{i}}\| f\|^2_{H}<\infty .
\end{align*}
For such functions $ f^{*}( \cdot; \lambda )$ we have
\begin{align*}
&\| R_{\lambda} (L_{W})f^{*}(\lambda;  \cdot ) \|^2_{\mathcal{H}} \\
&\geq   \| \frac{i}{\sqrt{\rho_2(t)}}\int_{t}^{\infty}\frac{1}{\rho_2(s)}
 \exp\Big( i(A_2-\lambda)\int_{s}^{t}\frac{d\tau}{\rho_2(\tau)} \\
&\quad -i(\overline{\lambda}-A_2)\int_{a_2}^{s}\frac{d\tau}{\rho_2(\tau)}\Big) 
 fds\|^2_{L^2(H,(a_2,\infty))} \\
& =   \| \frac{1}{\sqrt{\rho_2(t)}}\exp\Big( -i\lambda \int_{a_2}^{t}
 \frac{d\tau}{\rho_2(\tau)}+iA_2\int_{a_2}^{t}\frac{d\tau}{\rho_2(\tau)}\big)  \\
& \quad \times \int_{t}^{\infty}\frac{1}{\rho_2(s)}
 \exp \Big( -2\lambda_{i}\int_{a_2}^{s}\frac{d\tau}{\rho_2(\tau)}\Big) 
 fds\|^2_{L^2(H,(a_2,\infty))}\\
& =   \| \frac{1}{\sqrt{\rho_2(t)}}\exp\Big( \lambda_{i} \int_{a_2}^{t}
 \frac{d\tau}{\rho_2(\tau)}\Big) \int_{t}^{\infty}\frac{1}{\rho_2(s)}
 \exp \Big( -2\lambda_{i}\int_{a_2}^{s}\frac{d\tau}{\rho_2(\tau)}\Big) 
 ds\|^2_{L^2(H,(a_2,\infty))} \\
&\quad\times \| f\|_{H}^2 \\
& =  \| \frac{1}{2\lambda_{i}}\exp
\Big( -\lambda_{i} \int_{a_2}^{t}\frac{d\tau}{\rho_2(\tau)}\Big)
 \|^2_{L^2(H,(a_2,\infty))}\| f\|_{H}^2 \\
& =   \frac{1}{4\lambda_{i}^2}\int_{a_2}^{\infty}\frac{1}{\rho_2(t)}
 \exp\Big( -2\lambda_{i} \int_{a_2}^{t}\frac{d\tau}{\rho_2(\tau)}\Big)dt 
 \| f\|_{H}^2 \\
& =   \frac{1}{8\lambda_{i}^{3}}\| f\|_{H}^2 .
\end{align*}
From this we have
$$
\| R_{\lambda} (L_{W})f^{*}( \cdot; \lambda )\|_{\mathcal{H}}   
\geq \frac{\| f\|_{H}^2}{2\sqrt{2}\lambda_{i}\sqrt{\lambda_{i}}} 
=\frac{1}{2\lambda_{i}} \| f^{*}(t;\lambda )\|_{\mathcal{H}}.
$$
Then for $ \lambda_{i}=\operatorname{Im}\lambda>0 $ and $ f\neq 0 $ 
the following inequality is valid
$$
\frac{\| R_{\lambda} (L_{W})f^{*}( \cdot, \lambda )\|_{\mathcal{H}} }
 { \| f^{*}(\lambda; t )\|_{\mathcal{H}}}\geq\frac{1}{2\lambda_{i}}.
$$
On the other hand it is clear that
$$
\| R_{\lambda} (L_{W})\|\geq\frac{\| R_{\lambda} 
(L_{W})f^{*}(\cdot; \lambda )\|_{\mathcal{H}} }
{ \| f^{*}(\cdot; \lambda ) \|_{\mathcal{H}}}, \quad f\neq 0.
$$
Consequently,  for $ \lambda\in \mathbb{C}$ and
 $\lambda_{i}=\operatorname{Im}\lambda>0 $ we have
$$
\| R_{\lambda} (L_{W})\|\geq\frac{1}{2\lambda_{i}}.
$$
\end{proof}

\begin{remark} \rm
In the special case  $ \rho_{k}=1$, $k=1,2 $, similar results have been obtained 
in \cite{rf1}.
\end{remark}

As an example all selfadjoint extensions $ L_{\varphi} $ of the minimal operator 
$ L_0 $, generated by the multipoint differential expression
\begin{align*}
l(u)& = (l_1(u_1),l_2(u_2))  \\
& =  \Big( itu_1'(t,x)+\frac{1}{2}iu_1(t,x)-\frac{\partial^2 u_1}
{\partial x^2}(t,x), i\sqrt{t}u_2'(t,x) \\
&\quad +\frac{1}{4\sqrt{t}}iu_2(t,x)-\frac{\partial^2 u_2}{\partial x^2}(t,x) \Big),
\end{align*}
with boundary conditions
\begin{gather*}
u_1(t,0)  =  u_1(t,1), \quad u_1'(t,0)=u_1'(t,1), \quad  t<-1,  \\
u_2(t,0)  =  u_2(t,1), \quad  u_2'(t,0)=u_2'(t,1), \quad  t>1
\end{gather*}
in the direct sum $ L^2((-\infty,-1)\times (0,1))\oplus 
L^2((1,\infty)\times (0,1)) $ in terms of boundary conditions are described 
the  boundary condition
$$
(t^{1/4}u_2(t))(1,x)=e^{i\varphi}(\sqrt{t}u_1(t))(-1,x), \quad
 \varphi\in [0,2\pi), \; x\in (0,1).
$$
Moreover, the spectrum of such extension is
$$
\sigma (L_{\varphi})=\sigma_{c} (L_{\varphi})= \mathbb{R}.
$$

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\end{document}
