\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2012 (2012), No. 36, pp. 1--9.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2012 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2012/36\hfil Normal extensions] {Normal extensions of a singular multipoint differential operator of first order} \author[Z. I. Ismailov, R. \"Ozt\"urk Mert \hfil EJDE-2012/36\hfilneg] {Zameddin I. Ismailov, Rukiye \"Ozt\"urk Mert} % in alphabetical order \address{Zameddin I. Ismailov \newline Department of Mathematics, Faculty of Sciences, Karadeniz Technical University, 61080, Trabzon, Turkey} \email{zameddin@yahoo.com} \address{Rukiye \"Ozt\"urk Mert \newline Department of Mathematics, Art and Science Faculty, Hitit University, 19030, Corum, Turkey} \email{rukiyeozturkmert@hitit.edu.tr} \thanks{Submitted August 10, 2011. Published Match 7, 2012.} \subjclass[2000]{47A10, 47A20} \keywords{Multipoint differential operators; selfadjoint and normal extension; \hfill\break\indent spectrum} \begin{abstract} In this work, we describe all normal extensions of the minimal operator generated by linear singular multipoint formally normal differential expression $l=(l_1,l_2,l_3)$, $l_k=\frac{d}{dt}+A_k$ with selfadjoint operator coefficients $A_k$ in a Hilbert space. This is done as a direct sum of Hilbert spaces of vector-functions $L_2(H,(-\infty ,a_1))\oplus L_2(H,(a_2,b_2)) \oplus L_2(H,(a_3,+\infty))$ where $-\infty a_3$, then it is clear that $(u_1,u_2)\in D(M^*_0)$ and $\gamma_1(u)=f$, $\gamma_2(u)=g$. \end{proof} \begin{lemma}\label{lem2} The triplet $(H,{\Gamma }_1,{\Gamma }_2)$, \begin{gather*} {\Gamma }_1:D(M^{*}_{0}(0,1,0))\to H, \quad \Gamma_1(u)=\frac{1}{i\sqrt{2}}({u_2(b_2)+u}_2(a_2)),\\ {\Gamma }_2:D(M^*_0(0,1,0))\to H, \quad {\Gamma }_2(u)=\frac{1}{\sqrt{2}}(u_2(b_2)-u_2(a_2)),\\ u=(0,u_2,0)\in D(M^*_0(0,1,0)) \end{gather*} is a space of boundary values of the minimal operator $M_0(0,1,0)$ in the direct sum $L_2(0,1,0)$. \end{lemma} \begin{theorem} \label{thm3} If the minimal operators $L_{10}$, $L_{20}$ and $L_{30}$ are formally normal then \begin{gather*} D(L_{10}){\subset W}^1_2(H,(-\infty ,a_1)),\quad A_1D(L_{10})\subset L_2(H,(-\infty ,a_1)),\\\ D(L_{20}){\subset W}^1_2(H,(a_2,b_2)),\quad A_2D(L_{20})\subset L_2(H,(a_2,b_2)),\\ D(L_{30})\subset W^1_2(H,(a_3,\infty )),\quad A_3D(L_{30})\subset L_2(H,(a_3,\infty )). \end{gather*} \end{theorem} \begin{proof} Indeed, in this case for each $u_1\in D(L_{10})\subset D(L^*_{10})$ is true $u'_1+A_1u_1\in L_2(H,(-\infty ,a))\ \ and\ \ u'_1-A_1u_1\in L_2(H,(-\infty ,a)),$ hence $u'_1\in L_2(H,(-\infty ,a))\quad\text{and}\quad A_1u_1\in L_2(H,(-\infty ,a));$ i.e., $D(L_{10}){\subset W}^1_2(H,(-\infty ,a))\ \ and\ \ { A}_1D(L_{10})\subset L_2(H,(-\infty ,a)).$ The second and third parts of theorem can be proved in a similar way. \end{proof} The following result can be easily established. \begin{lemma}\label{lem4} Every normal extension of $L_0(1,1,1)$ in $L_2(1,1,1)$ is a direct sum of normal extensions of the minimal operator ${ L}_0(1,0,1)=L_{10}\oplus 0\oplus L_{30}$ in $L_2(1,0,1){=L}_2(H,(-\infty ,a_1))\oplus{0\oplus L}_2(H,(a_3,\infty ))$ and minimal operator $L_0(0,1,0)=0\oplus L_{20}\oplus 0$ in $L_2(0,1,0)={0\oplus L}_2(H,(a_2,b_2))\oplus$0. \end{lemma} Finally, using the method in \cite{Gor,Ism,Ism2,Ism3,Ism4,Ism5,Maks,Maks2,Maks3} and Lemmas \ref{lem1} and \ref{lem2} the following result can be deduced. \begin{theorem} \label{thm5} Let ${(-A_1)}^{1/2}W^1_2(H,(-\infty ,a_1))\subset W^1_2(H,(-\infty ,a_1))$, \begin{gather*} A^{1/2}_2W^1_2(H,(a_2,b_2)) \subset W^1_2(H,(a_2,b_2)),\\ A^{1/2}_3W^1_2(H,(a_3,\infty )) \subset W^1_2(H,(a_3,\infty )). \end{gather*} Each normal extension $\tilde{L}$ of the minimal operator $L_0$ in the Hilbert space $L_2(1,1,1)$ is generated by differential expression \eqref{eq2.1} and boundary conditions \begin{gather} \label{GrindEQ__2_3_} u_3(a_3)=W_1u_1(a_1),\quad u_1(a_1)\in {\ker {(-A_1)}^{1/2}\ }, \quad u_3(a_3)\in \ker A^{1/2}_3, \\ \label{GrindEQ__2_4_} u_2(b_2)=W_2u_1(a_2), \end{gather} where $W_1,W_2:H\to H$ is a unitary operators. Moreover, the unitary operators $W_1,W_2$ in $H$ are determined by the extension $\tilde{L}$; i.e., $\widetilde{\ L}=L_{W_1W_2}$ and vice versa. \end{theorem} \begin{corollary}\label{cor6} If at least one of the operators $A_1$ and $A_3$ is one-to-one mapping in $H$, then minimal operator $L_0(1,1,1)$ is maximally formal normal in $\ L_2(1,1,1)$. \end{corollary} \begin{corollary}\label{cor7} If there exists at least one normal extension of the minimal operator $L_0(1,1,1),$ then $\dim\ker{(-A_1)}^{1/2}=\dim\ker A^{1/2}_3>0.$ \end{corollary} \section{The spectrum of the normal extensions} In this section the structure of the spectrum of the normal extension $L_{W_1W_2}$ in $L_2(1,1,1)$ will be investigated. In this case by the Lemma \ref{lem4} it is clear that $L_{W_1W_2}=L_{W_1}\oplus L_{W_2} ,$ where $L_{W_1}$ and $L_{W_2}$ are normal extensions of the minimal operators $L_0(1,0,1)$ and $L_0(0,1,0)$ in the Hilbert spaces $L_2(1,0,1)$ and $L_2(0,1,0)$ respectively. Later, it will be assumed that $A_1=A^*_1\le 0$, $A_2=A^*_2\ge 0, A_3=A^*_3\ge 0$ and $0\in {\sigma }_p({(-A_1)}^{1/2})\cap {\sigma }_p(A^{1/2}_3)$. First, we have to prove the following result. \begin{theorem} \label{thm3.1} The point spectrum of any normal extension $L_{W_1}$ of the minimal operator $L_0(1,0,1)$ in the Hilbert space $L_2(1,0,1)$ is empty; i.e., $\sigma_p(L_{W_1})=\emptyset$. \end{theorem} \begin{proof} Let us consider the following problem for the spectrum of the normal extension $L_{W_1}$ of the minimal operator $L_0(1,0,1)$ in the Hilbert space $L_2(1,0,1)$, $L_{W_1}u=\lambda u, \quad \lambda ={\lambda }_r+i{\lambda }_i\in {\mathbb C},\quad u=(u_1,0,u_3)\in L_2(1,0,1);$ that is, \begin{gather*} {\tilde{l}}_1(u_1)=u'_1+{\tilde{A}}_1u_1=\lambda u_1,\quad u_1\in L_2(H,(-\infty ,a_1)),\\ {\tilde{l}}_3(u_3)=u'_3+{\tilde{A}}_3u_3=\lambda u_3,\quad u_3\in L_2(H,(a_3,+\infty )), \quad \lambda \in \mathbb{R}, \\ u_3(a_3)=W_1u_1(a_1) ,\quad u_1(a_1)\in {\ker {(-A_1)}^{1/2}},\quad u_3(a_3)\in \ker A^{1/2}_3 . \end{gather*} The general solution of this problem is \begin{gather*} u_1(\lambda ;t)=e^{-({\tilde{A}}_1-\lambda )(t-a_1)}f^*_1,\quad ta_3,\; f^*_3\in H_{{-1}/{2}}(A_3), \\ f^*_3=W_1f^*_1,\quad f^*_1,f^*_3\in H,\quad f^*_1=u_1(\lambda ;a_1),\quad f^*_3=u_3(\lambda ;a_3). \end{gather*} Since $0\in {\sigma }_p({(-A_1)}^{1/2})\cap {\sigma }_p(A^{1/2}_2)$ and ${(-A_1)}^{1/2}f^*_1=0$, $A^{1/2}_2f^*_3=0$, we have \begin{gather*} u_1 ( \lambda ;t)=e^{\lambda (t-a)}f^*_1,\quad tb,\; f^*_3\in H_{-1/2}(A_3),\\ f^*_3=W_1f^*_1,\quad f^*_1=u_1(\lambda ;a_1),\quad f^*_3=u_3(\lambda ;a_3). \end{gather*} In order for $u_1(\lambda ;.)\in L_2(H,(-\infty ,a_1))$ and $u_2(\lambda ;.)\in L_2(H,(a_3,\infty ))$ necessary and sufficient condition is that ${\lambda }_r\ge 0$ and ${\lambda }_r\le 0$ respectively. Hence ${\lambda }_r=0$. Consequently, \begin{gather*} u_1( \lambda ;t)=e^{i{\lambda }_i(t-a_1)}f^*_1,\quad ta_3,\; f^*_3\ ={W_1f}^*_1. \end{gather*} In this case for $u_1(\lambda ;.)\in L_2(H,(-\infty ,a_1))$ and $u_2(\lambda ;.)\in L_2(H,(a_3,\infty ))$, clearly, necessary and sufficient conditions are that $f^*_1=0$, $f^*_3=0$. This implies that $u_1=0$ and $u_2=0$ in $L^2$. Therefore ${\sigma }_p(L_{W_1})=\emptyset$. \end{proof} Since residual spectrum of any normal operators in any Hilbert space is empty, it is sufficient to investigate the continuous spectrum of the normal extensions $L_{W_1}$ of the minimal operator $L_0(1,0,1)$ in the Hilbert space $L_2(1,0,1)$. \begin{theorem} \label{thm3.2} The continuous spectrum of any normal extension $L_{W_1}$ of the minimal operator $L_0(1,0,1)$ in the Hilbert space $L_2(1,0,1)$ is ${\sigma }_c(L_{W_1})=i\mathbb{R}$. \end{theorem} \begin{proof} Assume that $\lambda \in {\sigma }_c(L_{W_1})$. Then by the theorem for the spectrum of normal operators \cite{Dun}, $\sigma (L_{W_1})\subset \sigma (\operatorname{Re}L_{W_1})+i\sigma (\operatorname{Im}L_{W_1}\ ),$ we obtain that ${\lambda }_r\in \sigma (\operatorname{Re}L_{W_1}),\quad {\lambda }_i\in \sigma (\operatorname{Im} L_{W_1}).$ This implies that ${\lambda }_r\in \sigma (A_1)$ and ${\lambda }_r\in \sigma (A_3)$, hence by the conditions to the operators $A_1$ and $A_3$ we have ${\lambda }_r=0$. On the other hand from the proof of previous theorem we see that ${\ker (\ L_{W_1}-\lambda )\ }=\{0\}$ for any $\lambda\in {\mathbb C}$. Consequently, ${\sigma }_c(\ L_{W_1})\subset i\mathbb {R}$. Furthermore, it is clear that for the $\lambda =i{\lambda }_i\in {\mathbb C}$ the general solution of the boundary value problem \begin{gather*} u'_1+A_1u_1=i{\lambda }_iu_1+f_1, \quad u_1, f_1\in L_2(H,(-\infty ,a_1)),\\ u'_3+A_3u_3=i{\lambda }_iu_3+f_3,\quad u_3, f_3\in L_2(H,(a_3,\infty )),\; {\lambda }_i\in \mathbb R, \\ u_3(a_3)=W_1u_1(a_1),\ u_1(a_1)\in {\ker {(-A_1)}^{1/2}\ },u_3(a_3)\in \ker A^{1/2}_3 \end{gather*} will be of the form \begin{gather*} u_1(i{\lambda }_i;t)=e^{-(A_1-{i\lambda }_i)(t-a_1)}f_{{i\lambda }_i} -\int^{a_1}_t{e^{-(A_1-{i\lambda }_i)(t-s)}f_1(s)ds},\quad ta_3,}\\ g_{{ i \lambda }_{ i }}=W_1f_{{i\lambda }_i}. \end{gather*} In this case, $e^{-(A_1-{i\lambda }_i)(t-a_1)}f_{{i\lambda }_i}\in L_2(H,(-\infty ,a_1)),\quad {{\rm e}}^{-({{\rm A}}_{{\rm 2}}- i {\lambda }_{ i })(t-a_3)}g_{{ i \lambda }_{ i }} \in L_2(H,(a_3,\infty ))$ for any $g_{{ i \lambda }_{ i }}$,$f_{{i\lambda }_i}\in H$. If choose $f_1(t)=e^{i{\lambda }_it}e^{-(t-a_1)}f^*$, $f^*\in {\ker {(-A_1)}^{1/2}\ }$, $t1,\; x\in [0,1],\\ \frac{\partial u(t,x)}{\partial t}-\frac{{\partial }^2u(t,x)}{\partial x^2}+u(t,x)=f(t,x), \quad |t|<1/2,\; x\in [0,1],\\ u(1,x)=e^{i\varphi }u(-1,x),\quad \varphi \in [0,2\pi ),\\ u(1/2,x)=e^{i\psi }u(-1/2,x),\quad \psi \in [0,2\pi ),\\ u_x(t,0)=u_x(t,1)=0,\quad |t|>1,\; |t|<1/2 \end{gather*} in the space$L_2((-\infty,-1)\times (0,1))\oplus {L_2((-1/2,1/2) \times (0,1))\oplus L}_2((1,\infty)\times (0,1))$. In this case it is clear that in the space$L_2(0,1)$, for the operators \begin{gather*} A_1= \frac{{\partial }^2u(.,x)}{\partial x^2} ,\quad x\in [0,1] ,\; u_x(.,0)=u_x(.,1)=0,\\ A_2=- \frac{{\partial }^2u(.,x)}{\partial x^2} +u(.,x),\quad x\in [0,1] ,\; u_x(.,0)=u_x(.,1)=0,\\ A_3=- \frac{{\partial }^2u(.,x)}{\partial x^2},\quad x\in [0,1] ,\; u_x(.,0)=u_x(.,1)=0 \end{gather*} we have \begin{gather*} A_1=A^*_1\le 0,\quad A_2=A^*_2\ge 1,\quad A_3=A^*_3\ge 0, \quad {\ker {(-A_1)}^{1/2}\ }\ne \{0\}, \\ \ker A^{1/2}_3\ne \{0\},\quad 0\in {\sigma }_p({(-A_1)}^{1/2})\cap {\sigma }_p(A^{1/2}_3). \end{gather*} On the other hand, since$A^{-1}_2\in {\sigma }_{\infty }(L_2(0,1))$,$\sigma (L_{\psi })={\sigma }_p(L_{\psi })$,${\sigma }_c(L_{\psi })=\emptyset and \begin{align*} \sigma (L_{\psi }) &=\Big\{\lambda \in{\mathbb C}:\lambda = {\ln |\mu|+i} \arg\mu+2n\pi i, n\in \mathbb{Z},\, \mu \in \sigma (e^{i\psi }e^{-{\tilde{A}}_2(b_2-a_2)}),\\ &\qquad 0\leq \arg \mu<2\pi \Big\}\\ &\subset \big\{\lambda \in \mathbb {C}: \operatorname{Re}\lambda \geq 1\big\}, \end{align*} then{[{\sigma }_p(L_{\psi })]}^c\cap [i{\mathbb R}]=i{\mathbb R}\$. 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