\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 $-\infty 0$. 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 }ta_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 ta_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. $$\label{GrindEQ__3_1_} \begin{gathered} {\tilde{l}}_1(u_1)=iu'_1+{\tilde{A}}_1u_1={\lambda }u_1+f_1, \quad u_1, f_1\in L_2(H,(-\infty ,a_1)), \\ {\tilde{l}}_3(u_3)=iu'_3+{\tilde{A}}_3u_3={\lambda }u_3+f_3, \quad u_3, f_3\in L_2(H,(a_3,+\infty )), \\ \lambda \in \mathbb C, \quad \lambda_i=\operatorname{Im}\lambda >0 \\ u_3(a_3)=W_1u_1(a_1) \end{gathered}$$ 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 ta_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 ta_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_21,\; 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} \begin{thebibliography}{00} \bibitem{Alb} S. 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