\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2017 (2017), No. 95, pp. 1--13.\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-2017/95\hfil Dissipative Sturm-Liouville operators] {Dissipative Sturm-Liouville operators with a spectral parameter in the boundary condition on bounded time scales} \author[B. P. Allahverdiev, A. Ery{\i}lmaz, H. Tuna \hfil EJDE-2017/95\hfilneg] {Bilender P. Allahverdiev, Aytekin Ery{\i}lmaz, H\"{u}seyin Tuna} \address{Bilender P. Allahverdiev \newline Department of Mathematics, Faculty of Arts and Science, S\"{u}leyman Demirel University, 32260 Isparta, Turkey} \email{bilenderpasaoglu@sdu.edu.tr} \address{Aytekin Ery{\i}lmaz \newline Department of Mathematics, Nev\c{s}ehir University, Nev\c{s}ehir, Turkey} \email{eryilmazaytekin@gmail.com} \address{H\"{u}seyin Tuna \newline Department of Mathematics, Mehmet Akif Ersoy University, Burdur, Turkey} \email{hustuna@gmail.com} \dedicatory{Communicated by Jerome Goldlstein} \thanks{Submitted February 29, 2016. Published April 5, 2017.} \subjclass[2010]{47A20, 47A40, 47A45, 34B05, 34B10, 39A10} \keywords{Time scale; $\Delta$-differentiable; dilation; dissipative operator; \hfill\break\indent system of eigenvectors; scattering matrix; functional model; characteristic function} \begin{abstract} In this article we consider a second-order Sturm-Liouville operator with a spectral parameter in the boundary condition on bounded time scales. We construct a selfadjoint dilation of the dissipative Sturm-Liouville operators. Using by methods of Pavlov \cite{p1,p2,p3}, we prove the completeness of the system of eigenvectors and associated vectors of the dissipative Sturm-Liouville operators on bounded time scales. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} In the recent years, the study of dynamic equations on time scales have found a noticeable interest and attracted many researches; see for example \cite{a1,a2,b2,t1}. The first fundamental results in this area were obtained by Hilger \cite{h1}. He introduced the idea of time scales as a way to unify continuous and discrete analysis and it allows a simultaneous treatment of differential and difference equations, extending those theories to so called dynamic equations. The study of time scales has led to several important applications, e.g., in the study of neural networks, heat transfer, and insect population models, phytoremediation of metals, wound healing and epidemic models \cite{a2,j1,t1}. For some basic definitions, we refer the reader to consult the reference \cite{a8,b12,b13,d1,g1,l1}. The study of problems involving parameter dependent systems is of great interest to a lot of numerous problems in physics and engineering. A boundary-value problem with a spectral parameter in the boundary condition appears commonly in mathematical models of mechanics. There are many studies about parameter dependent problems \cite{a3,a7,b8,e1,f4,h2,o1,o2,s1,t1}. The spectral analysis of non-selfadjoint (dissipative) operators is based on ideas of the functional model and dilation theory rather than on traditional resolvent analysis and Riesz integrals. Using a functional model of a non-selfadjoint operator as a principal tool, spectral properties of such operators are investigated. The functional model of non-selfadjoint dissipative operators plays an important role within both the abstract operator theory and its more specialized applications in other disciplines. The construction of functional models for dissipative operators, natural analogues of spectral decompositions for selfadjoint operators is based on Sz. Nagy-Foias dilation theory \cite{n1} and Lax-Phillips scattering theory \cite{l2}. Pavlov's approach \cite{p1,p2,p3} to the model construction of dissipative extensions of symmetric operators was followed by Allahverdiev in his works \cite{a3,a4,a5,a6,a7} and some others \cite{e1,o1,o2,t2,t3}. The theory of the dissipative Schr\"{o}dinger operator on a finite interval was applied to the problems arising in the semiconductor physics \cite{b1,b2,b3}. In \cite{b4,b5,b6,b7}, Pavlov's functional model was extended to (general) dissipative operators which are finite dimensional extensions of a symmetric operator, and the corresponding dissipative and Lax-Phillips scattering problems were investigated in some detail. We extend the results \cite{a3,a4,a5,a6,a7,o1,o2,t2,t3} to the more general eigenvalues problem \eqref{e2.2}--\eqref{e2.4} on time scales. While proving our results, we use the machinery and method of \cite{a3,a4,a5,a6,a7}. The organization of this document is as follows: In Section 2, some time scale essentials are included for the convenience of the reader. In Section 3, we construct a selfadjoint dilation of dissipative Sturm-Liouville operator on bounded time scales. We present its incoming and outcoming spectral representations which makes it possible to determine the scattering matrix of the dilation according to the Lax and Phillips scheme \cite{l2}. A functional model of this operator is constructed by methods of Pavlov \cite{p1,p2,p3} and define its characteristic functions. Finally, we proved a theorem on completeness of the system of eigenvectors and associated vectors of dissipative operators. \section{Preliminaries} Let us denote a time scale by $\mathbb{T}$. The forward jump operator $\sigma :\mathbb{T}\to \mathbb{T}$ is defined by $\sigma (t) =\inf \{ s\in \mathbb{T}:s>t\}$, $t\in \mathbb{T}$ and the backward jump operator $\rho :\mathbb{T}\to \mathbb{T}$ is defined by $\rho ( t) =\sup \{ s\in \mathbb{T}:s0$ there is a neighborhood $U\subset\mathbb{T}$ of $t$ such that $|f(\sigma (t))-f(s)-f^{\Delta }(t)(\sigma (t)-s)|\leq \varepsilon |\sigma (t)-s|$, $( s\in U)$. Analogously one may define the notion of $\nabla $-differentiability of some function using the backward jump $\rho $. One can show $f^{\Delta }(t)=f^{\nabla }(\sigma (t))$ and $f^{\nabla }(t)=f^{\Delta }(\rho (t))$ for continuously differentiable functions \cite{g1}. Let $f:\mathbb{T}\to\mathbb{R}$ be a function, and $a,b\in \mathbb{T}$. If there exists a function $F:\mathbb{T}\to \mathbb{R}$, such that $F^{\Delta }( t) =f( t) $ for all $t\in \mathbb{T}^{k}$, then $F$ is a $\Delta$-antiderivative of $f$. In this case the integral is given by the formula $\int_a^{b}f( t) \Delta t=F( b)-F( a)$ for $a,b\in \mathbb{T}$. Similarly, one may define the notion of $\nabla $-antiderivative of some function. Let $L_{\Delta }^2(\mathbb{T}^{\ast })$ be the space of all functions defined on $\mathbb{T}^{\ast }$ such that \[ \| f\| :=\Big(\int_a^{b}| f( t) | ^2\Delta t\Big)^{1/2}<\infty . \] The space $L_{\Delta }^2(\mathbb{T}^{\ast })$ is a Hilbert space with the inner product $( f,g):=\int_a^{b}f( t) \overline{g( t) }\Delta t$, $f,g\in L_{\Delta }^2(\mathbb{T}^{\ast })$ (see \cite{r1}). Let $a\leq b$ be fixed points in $\mathbb{T}$ and $a\in \mathbb{T}_k,b\in\mathbb{T}^{k}$. We will consider the Sturm-Liouville equation \begin{equation} l( y) :=-[ p( t) y^{\Delta }( t)] ^{\nabla }+q( t) y( t) ,\quad t\in [ a,b], \label{e2.1} \end{equation} where $q:\mathbb{T}\to \mathbb{C}$ is continuous function, $p:\mathbb{T}\to \mathbb{C}$ is $\nabla $-differentiable on $\mathbb{T}^{k}$, $p(t) \neq 0$ for all $t\in \mathbb{T}$, and $p^{\nabla }:\mathbb{T}_k \to \mathbb{C}$ is continuous. The Wronskian of $y,z$ is defined as $W( y,z) ( t) :=p( t) [ y(t) z^{\Delta }( t) -y^{\Delta }( t) z(t) ] ,\quad t\in \mathbb{T}^{\ast }$ (see \cite{g1}). Let $L_0$ denote the closure of the minimal operator generated by \eqref{e2.1} and by $D_0$ its domain. Moreover, we denote by $D$ the set of all functions $y( t) $ from $L_{\Delta }^2$ $(\mathbb{T}^{\ast })$ such that $l( y) \in L_{\Delta }^2(\mathbb{T}^{\ast })$; $D$ is the domain of the maximal operator $L$. Furthermore $L=L_0^{\ast } $ (see \cite{n2}). Suppose that the operator $L_0$ has defect index $(2,2)$ (see \cite{h3}). For every $y,z\in D$, Lagrange's identity \cite{g1} is defined as $(Ly,z) -( y,Lz) =[y,z]( b) -[y,z]( a)$, where $[y,z]:=p( t) [ y( t) \overline{z^{\Delta }( t) }-y^{\Delta }( t) \overline{z(t) }]$. Consider the boundary-value problem defined by \begin{equation} l(y)=\lambda y,\quad y\in D, \label{e2.2} \end{equation} subject to the boundary conditions \begin{gather} y( b) -hp( b) y^{\Delta }( b) =0,\quad \operatorname{Im}h>0 , \label{e2.3} \\ \alpha _1y( a) -\alpha _2p( a) y^{\Delta }( a) =\lambda (\alpha _1'y( a) -\alpha _2'p( a) y^{\Delta }( a) ) \label{e2.4} \end{gather} where $\lambda $ is spectral parameter and $\alpha _1,\alpha _2,\alpha_1',\alpha _2'\in \mathbb{R}$ and $\alpha $ is defined by \[ \alpha :=\begin{vmatrix} \alpha _1' & \alpha _1 \\ \alpha _2' & \alpha _2 \end{vmatrix} =\alpha _1'\alpha _2-\alpha _1\alpha _2'>0. \] For $\lambda =1$, this problem was investigated by Tuna \cite{t3}. On the other hand, when we restrict the time variable to $t\geq 0$, the usual case of nonnegative continuous time, there is some related literature. For instance, Favini et al.\ \cite{f1,f2,f3} obtained some corresponding results in $n$ dimensions. See also \cite{b9,b10,b11}. For the sake of simplicity, we define $R_a(y):=\alpha _1y(a) -\alpha _2p( a) y^{\Delta }( a)$, $R_a'(y):=\alpha _1'y( a) -\alpha _2'p( a) y^{\Delta }( a)$, $N_1^{b}(y):=y( b)$, $N_2^{b}(y):=p( b) y^{\Delta}( b)$, $N_1^{a}(y):=y( a)$, $N_2^{a}(y):=p( a) y^{\Delta}( a)$, $R_{b}(y):=N_2^{b}(y)-hN_1^{b}(y)$. \begin{lemma} \label{lem1} For arbitrary $y,z\in D$ suppose that $R_a(\overline{z})=\overline{R_a(z)}$, $R_a'(\overline{z})=\overline{R_a'(z)}$. Then \begin{equation} [ y,\overline{z}] (a)=\frac{1}{\alpha }[ R_a(y)\overline{ R_a'(z)}-R_a'(y)\overline{R_a(z)}]\,. \label{e2.5} \end{equation} \end{lemma} \begin{proof} Note that \begin{align*} &\frac{1}{\alpha }[ R_a(y)\overline{R_a'(z)} -R_a'(y)\overline{R_a(z)}] \\ &=\frac{1}{\alpha }[( \alpha _1y( a) -\alpha _2p(a) y^{\Delta }( a) ) \overline{(\alpha _1'z( a) -\alpha _2'p( a) z^{\Delta}( a) )}] \\ &\quad -\frac{1}{\alpha }[(\alpha _1'y( a) -\alpha _2'p( a) y^{\Delta }( a) )\overline{ ( \alpha _1z( a) -\alpha _2p( a) z^{\Delta}( a) ) }] \\ &=\frac{1}{\alpha }[ (\alpha _1'\alpha _2-\alpha _1\alpha _2')(y( a) p( a) \overline{ z^{\Delta }( a) }-p( a) y^{\Delta }( a) \overline{z( a) })] \\ &=[ y,z] (a). \end{align*} \end{proof} Let $\theta _1$, $\theta _2$ denote the solutions of \eqref{e2.1} satisfying the conditions $N_1^{a}(\theta _2)=\alpha _2-\alpha_2'\lambda$, $N_2^{a}(\theta _2)=\alpha _1-\alpha_1'\lambda$, $N_1^{b}(\theta _1)=h,\ N_2^{b}(\theta_1)=1$. By equation \eqref{e2.1}, we have \begin{align*} \Delta (\lambda ) &=[\theta _1,\theta _2]( t) =-[\theta_2,\theta _1]( t) =-[\theta _2,\theta _1]( a) \\ &=-\frac{1}{\alpha }[ R_a(\theta _1)\overline{R_a'(\theta _2)} -R_a'(\theta _1)\overline{R_a(\theta _2)}] =R_a(\theta _2)-\lambda R_a'(\theta _2), \end{align*} and $\Delta (\lambda )=[\theta _1,\theta _2]( t) =-[\theta _2,\theta _1]( t) =-[\theta _2,\theta_1]( b) =-( N_2^{a}(\theta _1)-hN_1^{a}(\theta_1)) $. We also let \begin{equation*} G( t,\xi ,\lambda ) =\frac{1}{\Delta (\lambda )} \begin{cases} \theta _2( \xi ,\lambda ) \theta _1( t,\lambda ),& t<\xi \\ \theta _1( t,\lambda ) \theta _2( \xi ,\lambda ),& \xi 0$ and $\widehat{f}= \begin{pmatrix} f^{(1)}( t) \\ f^{(2)} \end{pmatrix}$, $\widehat{g}=\begin{pmatrix} g^{(1)}( t) \\ g^{(2)} \end{pmatrix} \in H$, then the formula \[ ( \widehat{f},\widehat{g}) =\int_a^{b}f^{(1)}(t) \overline{g}^{( 1) }(t) \Delta t +\frac{1}{\alpha }f^{(2)} \overline{g}^{(2)} \] defines an inner product in Hilbert space $H$. In terms of this inner product, linear space $H$ is a Hilbert space. Let us define operator of $A_h:H\to H$ with equalities suitable for boundary-value problem \begin{equation*} D(A_h)=\Big\{ \widehat{f}=\begin{pmatrix} f^{(1)} \\ f^{(2)} \end{pmatrix} \in H:f^{(1)}\in D,R_{b}( f^{(1)}) =0,\; f^{(2)}=R_a'( f^{(1)}) \Big\} \end{equation*} and $A_h\widehat{f}=\widetilde{l}(\widehat{f}):= \begin{pmatrix} l( f^{(1)}) \\ R_a( f^{(1)}) \end{pmatrix}$. Recall that a linear operator $A_h$ with domain $D(A_h) $ in Hilbert space $H$ is called dissipative if $\operatorname{Im}(A_hf,f)\geq 0$ for all $f\in D( A_h) $ and maximal dissipative if it does not have a proper extension. \section{Main Results} \begin{theorem} \label{thm1} The operator $A_h$ is maximal dissipative in the space $H$. \end{theorem} \begin{proof} Let $\widehat{y}\in D( A_h)$. From \eqref{e2.5}, we have \begin{gather*} ( A_h\widehat{y},\widehat{y}) -( \widehat{y},A_h\widehat{ y}) =[y_1,y_1]( b) -[y_1,y_1]( a) \\ +\frac{1}{\alpha }[R_a( y_1) \overline{R_a'( y_1) }-R_a'( y_1) \overline{ R_a( y_1) }] = [y_1,y_1]( b) =2\operatorname{Im} h( p( b) y^{\Delta }( b) ) ^2. \end{gather*} Since $\operatorname{Im}( A_h\widehat{y},\widehat{y}) =\operatorname{Im}h(p( b) y^{\Delta }( b) ) ^2\geq 0, A_h$ is a dissipative operator in $H$. Let us prove that $A_h$ is maximal dissipative operator in the space $H$. To do this, it is enough to control that \begin{equation} ( A_h-\lambda I) D( A_h) =H,\quad \operatorname{Im}\lambda <0\label{e3.1} \end{equation} To prove \eqref{e3.1}, let $F\in H$, $\operatorname{Im}\lambda <0$ and set \[ \Gamma =\begin{pmatrix} ( \widetilde{G}_t,\overline{F}) \\ R_a'[ ( \widetilde{G}_t,\overline{F}) ] \end{pmatrix}, \] where \[ \widetilde{G}_{t,\lambda }= \begin{pmatrix} G( t,\xi ,\lambda ) \\ R_a'[ G( t,\xi ,\lambda ) ] \end{pmatrix}, \quad G( t,\xi ,\lambda ) =\begin{cases} \frac{1}{\Delta (\lambda )}\theta _2( \xi ,\lambda ) \theta_1( t,\lambda ) , & t<\xi \\ \frac{1}{\Delta (\lambda )}\theta _1( t,\lambda ) \theta _2( \xi ,\lambda ) ,& \xi 0$ (see \cite{a3,a4,a5,a6,a7,o1,o2,t2,t3}). \begin{theorem} \label{thm2} The operator $L_h$ is selfadjoint in $H$ and it is a selfadjoint dilation of the operator $A_h$. \end{theorem} \begin{proof} We first prove that $\mathcal{L}_h$ is symmetric in $\mathcal{H}$. Namely $( \mathcal{L}_hf,g) _{\mathcal{H}}-( f,\mathcal{L}_hg) _{_{\mathcal{H}}}=0$. Let $f,g\in D( \mathcal{L}_h) $, $f=\mathcal{\langle }\varphi _{-},\widehat{y},\varphi _{+}\rangle $ and $g=\langle \psi _{-},\widehat{z},\psi _{+}\rangle $. Then we have \begin{align*} &( \mathcal{L}_hf,g) _{\mathcal{H}}-( f,\mathcal{L}_hg) _{_{\mathcal{H}}}\\ &=( \mathcal{L}_h\mathcal{\langle } \varphi _{-},\widehat{y},\varphi _{+}\rangle ,\langle \psi _{-},\widehat{z} ,\psi _{+}\rangle ) -( \mathcal{\langle }\varphi _{-},\widehat{y},\varphi _{+}\rangle , \mathcal{L}_h\langle \psi _{-},\widehat{z},\psi _{+}\rangle ) \\ &=[y_1,z_1]( b) -[y_1,z_1]( a) +\frac{1}{\alpha }[R_a( y_1) \overline{R_a'( z_1) }-R_a'( y_1) \overline{ R_a( z_1) }] \\ &\quad +i\psi _{-}( 0) \overline{\varphi }_{-}( 0) -i\varphi _{+}( 0) \overline{\psi }_{+}(0) \end{align*} Therefore, \begin{equation} ( \mathcal{L}_hf,g) _{\mathcal{H}}-( f,\mathcal{L} _hg) _{_{\mathcal{H}}}=[y_1,z_1]( b) +i\psi _{-}( 0) \overline{\varphi }_{-}( 0) -i\varphi _{+}( 0) \overline{\psi }_{+}( 0) . \label{e3.4} \end{equation} On the other hand, \begin{align*} i\psi _{-}( 0) \overline{\varphi }_{-}( 0) -i\varphi _{+}( 0) \overline{\psi }_{+}( 0) &=\frac{i}{\beta ^2}( y( b) -hp( b) y^{\Delta }( b) ) \overline{( z( b) -hp( b) z^{\Delta }( b) ) } \\ &\quad -\frac{i}{\beta ^2}( y( b) -hp( b) y^{\Delta }( b) ) \overline{( z( b) -\overline{h} p( b) z^{\Delta }( b) ) } \\ &= \frac{i}{\beta ^2}( h-\overline{h}) p( b) [ y( b) \overline{z^{\Delta }( b) }-z( b) \overline{y^{\Delta }( b) }] . \end{align*} By \eqref{e3.4}, we obtain \begin{equation} i\psi _{-}( 0) \overline{\varphi }_{-}( 0) -i\varphi _{+}( 0) \overline{\psi }_{+}( 0) =-[y_1,z_1]( b) , \label{e3.5} \end{equation} and we have $( \mathcal{L}_hf,g) _{\mathcal{H}}-( f, \mathcal{L}_hg) _{_{\mathcal{H}}}=0$. Thus, we prove that $\mathcal{L}_h$ is a symmetric operator. To prove that $\mathcal{L}_h$ is selfadjoint, we need to show that $\mathcal{L}_h^{\ast }\subseteq \mathcal{L}_h$. Now, we consider the bilinear form $( \mathcal{L}_hf,g) _{\mathcal{H}}$ on elements $g=\langle \psi _{-},\widehat{z},\psi _{+}\rangle \in D( \mathcal{L}_h^{\ast }) $, where $f=\mathcal{\langle }\varphi _{-},\widehat{y},\varphi _{+}\rangle \in D( \mathcal{L}_h) $, $\varphi _{\mp }\in W_2^{1}(\mathbb{R} _{\mp }) $, $\varphi _{\mp }( 0) =0$. Integrating by parts, we obtain $\mathcal{L}_h^{\ast }g=\langle i\frac{d\psi _{-}}{d\xi },\widehat{z }^{\ast },i\frac{d\psi _{+}}{d\xi }\rangle $, where $\psi _{\mp }\in W_2^{1}(\mathbb{R}_{\mp }) $, $\widehat{z}^{\ast }\in H$. Similarly, if $f=\langle 0,\widehat{y},0\rangle \in D( \mathcal{L}_h) $, then integrating by parts in $( \mathcal{L}_hf,g) _{\mathcal{H}}$, we obtain \begin{equation} \mathcal{L}_h^{\ast }g=\mathcal{L}_h^{\ast }\langle \psi _{-},\widehat{z} ,\psi _{+}\rangle =\langle i\frac{d\psi _{-}}{d\xi }, \widetilde{l}(\widehat{z}) ,i\frac{d\psi _{+}}{d\xi }\rangle ,\quad z_1\in D,\;z_2=R_a' ( z_1) . \label{e3.6} \end{equation} Consequently, we have $( \mathcal{L}_hf,g) _{\mathcal{H}}=( f,\mathcal{L}_hg) \mathit{\ }_{\mathcal{H}}$, for each $ f\in D( \mathcal{L}_h) $ by \eqref{e3.6}, where the operator $\mathcal{L}_h$ is defined by \eqref{e3.3}. Therefore, the sum of the integrated terms in the bilinear form $( \mathcal{L}_hf,g) _{\mathcal{H}}$ must be equal to zero: \begin{align*} &[ y_1,z_1]( b) -[y_1,z_1]( a) +\frac{1}{ \alpha }[R_a( y_1) \overline{R_a'(z_1) } \\ &-R_a'( y_1) \overline{R_a(z_1) }] +i\varphi _{-}( 0) \overline{\psi }_{-}( 0) -i\varphi _{+}( 0) \overline{\psi }_{+}( 0) =0. \end{align*} Then by \eqref{e3.5}, we obtain \begin{equation} [ y_1,z_1]( b) +i\varphi _{-}( 0) _{-} \overline{\psi }_{-}( 0) -i\varphi _{+}( 0) \overline{ \psi }_{+}( 0) =0. \label{e3.7} \end{equation} From the boundary conditions for $\mathcal{L}_h$, we have \begin{equation*} y( b) =\beta \varphi _{-}( 0) -\frac{h}{i\beta }( \varphi _{-}( 0) -\varphi _{+}( 0) ) ,\ p( b) y^{\Delta }( b) =\frac{i}{\beta }( \varphi _{-}( 0) -\varphi _{+}( 0) ) . \end{equation*} Afterwards, by \eqref{e3.7} we obtain \begin{equation} \begin{aligned} &\beta \varphi _{-}( 0) -\frac{h}{i\beta }( \varphi _{-}( 0) -\varphi _{+}( 0) ) \overline{z( b) }-\frac{i}{\beta }( \varphi _{-}( 0) -\varphi _{+}( 0) ) \overline{p( b) z^{\Delta }(b) }\\ &=i\varphi _{+}( 0) \overline{\psi }_{+}( 0) -i\varphi _{-}( 0) \overline{\psi }_{-}( 0) . \end{aligned}\label{e3.8} \end{equation} We obtain $\frac{i\beta ^2-h}{\beta }z( b) +\frac{1}{\beta } \overline{p( b) z^{\Delta }( b) }=\varphi _{-}(0) $ comparing the coefficients of $\varphi _{-}( 0) $ in \eqref{e3.8} or \begin{equation} z( b) -hp( b) z^{\Delta }( b) =\beta \psi _{-}( 0) .\text{ } \label{e3.9} \end{equation} Similarly, we obtain \begin{equation} z( b) -\overline{h}p( b) z^{\Delta }( b) =\beta \psi _{+}( 0) \text{ } \label{e3.10} \end{equation} by comparing the coefficients of $\varphi _{+}( 0) $ in \eqref{e3.8}. Consequently, conditions \eqref{e3.9} and \eqref{e3.10} imply $D( \mathcal{L}_h^{\ast }) \subseteq D( \mathcal{L}_h) $, hence $\mathcal{L}_h=\mathcal{L}_h^{\ast }$. The selfadjoint operator $\mathcal{L}_h$ generates on $\mathcal{H}$ a unitary group $U_t=\exp ( i\mathcal{L}_ht) $ ($t\in\mathbb{R}=( -\infty ,\infty )$). Let us denote by $P:\mathcal{H}\to H $ and $P_1:H\to \mathcal{H}$ the mappings defined by $P:\langle \varphi _{-},\widehat{y},\varphi _{+}\rangle \to \widehat{y}$ and $P_1:\widehat{y}\to \langle 0,\widehat{y},0\rangle $. Let $Z_t:=PU_tP_{1}$ $t\geq 0$, by using $U_t$. The family $\{Z_t\} $ $( t\geq 0) $ of operators is a strongly continuous semigroup of completely non-unitary contraction on $H$. Let us denote by $B_h$ the generator of this semigroup: $B_h\widehat{y}= \lim_{t\to +0} ( it) ^{-1}( Z_t\widehat{y}-\widehat{y}) $. The domain of $B_h$ consists of all the vectors for which the limit exists. The operator $B_h$ is dissipative. The operator $\mathcal{L}_h$ is called the selfadjoint dilation of $B_h$ (see \cite{a3,a4,a5,a6,a7,o1,o2,t2,t3}). We show that $B_h$ $=A_h$, hence $\mathcal{L}_h$ is selfadjoint dilation of $B_h$. To show this, it is sufficient to verify the equality \begin{equation} P( \mathcal{L}_h-\lambda I) ^{-1}P_1\widehat{y}=( A_h-\lambda I) ^{-1}\widehat{y},\widehat{y}\in H,\quad \operatorname{Im}h<0. \label{e3.11} \end{equation} To do this, we set $( \mathcal{L}_h-\lambda I) ^{-1}P_1 \widehat{y} =g=\langle \psi _{-},\widehat{z},\psi _{+}\rangle $. Then we have $( \mathcal{L}_h-\lambda I) g=P_1\widehat{y}$, and hence $\widetilde{l}( \widehat{z}) -\lambda \widehat{z}=\widehat{y},\psi _{-}( \xi ) =\psi _{-}( 0) e^{-i\lambda \xi }$ and $\psi _{+}( \xi ) =\psi _{+}( 0) e^{-i\lambda \xi }$. Since $g\in D( \mathcal{L}_h) $, then $\psi _{-}\in L^2( -\infty ,0) $, it follows that $\psi _{-}( 0)=0$, and consequently $\widehat{z}$ satisfies the boundary condition $z( b) -hp( b) z^{\Delta }( b) =0$. Therefore $\widehat{z}\in D( A_h) $, and since point $\lambda $ with $\operatorname{Im}\lambda <0$ cannot be an eigenvalue of dissipative operator, it follows that $\psi _{+}( 0) $ is obtained from the formula $\psi _{+}( 0) =\beta ^{-1}( z( b) -\overline{h}p( b) z^{\Delta }( b) ) $. Thus, we have \begin{equation*} ( \mathcal{L}_h-\lambda I) ^{-1}P_1\widehat{y} =\langle 0,( A_h-\lambda I) ^{-1}\widehat{y},\beta ^{-1}( z(b) -\overline{h}p( b) z^{\Delta }( b) ) \rangle \end{equation*} for $\widehat{y}$ and $\operatorname{Im}\lambda <0$. By applying the mapping $P$, we obtain \eqref{e3.11} and \begin{align*} ( A_h-\lambda I) ^{-1} &= P( \mathcal{L}_h-\lambda I) ^{-1}P_1=-iP\int_0^ \infty U_te^{-i\lambda t}dtP_1 \\ &=-i\int_0^{\infty } Z_te^{-i\lambda t}dt=( B_h-\lambda I) ^{-1},\operatorname{Im}\lambda <0, \end{align*} so this clearly shows that $A_h=B_h$. \end{proof} The unitary group $\{ U_t\} $ has an important property which makes it possible to apply it to the Lax-Phillips \cite{l2}. It can be described as a characteristic function of maximal dissipative operator. The Lax-Phillips scheme has orthogonal incoming and outcoming subspaces $D_{-}=\langle L^2( -\infty ,0) ,0,0\rangle $ and $D_{+}=\langle 0,0,L^2( 0,\infty ) \rangle $ satisfying the following properties \begin{itemize} \item[(1)] $U_tD_{-}\subset D_{-}$, $t\leq 0$ and $ U_tD_{+}\subset D_{+}$, $t\geq 0$; \item[(2)] $\cap_{t\leq 0} U_tD_{-}=\cap_{t\geq 0} U_tD_{+}=\{ 0\}$; \item[(3)] $\overline{\cup_{t\geq 0} U_tD_{-}}= \overline{\cup_{t\leq 0} U_tD_{+}}=\mathcal{H}$; \item[(4)] $D_{-}\perp D_{+}$. \end{itemize} \begin{definition} \label{def2} \rm The linear operator $A$ with domain $D( A) $ acting on the Hilbert space $H$ is called completely nonselfadjoint (or simple) if there is no invariant subspace $M\subseteq D( A) $ $( M\neq \{ 0\} ) $ of the operator $A$ on which the restriction $A$ to $M$ is selfadjoint. \end{definition} \begin{lemma} \label{lem2} The operator $A_h$ is completely nonselfadjoint (simple). \end{lemma} \begin{proof} Let $H'\subset H$ be a nontrivial subspace in which $A_h$ induces a selfadjoint operator $A_h'$ with domain $D( A_h') =H' \cap D( A_h) $. If $\widehat{f}\in D( A_h')$, then $\widehat{f}\in D(A_h'^\ast) $ and \begin{align*} \frac{d}{dt}\Vert e^{iA_h't}\widehat{f}\Vert _{H}^2 &=\frac{d}{dt}( e^{iA_h't}\widehat{f},e^{iA_h't}\widehat{f}) _{H} \\ &=i( A_h'e^{iA_h't}\widehat{f},e^{iA_h't}\widehat{f}) _{H} -i( e^{iA_h't}\widehat{f},A_h'e^{iA_h't}\widehat{f}) _{H}\,. \end{align*} Taking $\widehat{g}=e^{iA_h't}\widehat{f}$, we have \begin{align*} 0 &=i( A_h'\widehat{g},\widehat{g}) _{H}-i( \widehat{g},A_h'\widehat{g}) _{H} \\ &=i[ g_1,g_1] ( b) -i[ g_1,g_1] ( a) +\frac{i}{\alpha }[ R_a( g_1) \overline{R_a'( g_1) } -R_a'(y_1) \overline{R_a( g_1) }] \\ &=-2\operatorname{Im}h(D_{q^{-1}}y_1( a) )^2=-\beta ^2(p( b) y^{\Delta }( b) ) ^2. \end{align*} Since $\widehat{f}\in D( A_h') $, $A_h'$ holds condition above. Moreover, eigenvectors of the operator $A_h'$ should also hold this condition. Therefore, for the eigenvectors $\widehat{y}(\lambda )$ of the operator $A_h$ acting in $H'$ and the eigenvectors of the operator $A_h'$, we have $p( b) y^{\Delta }( b) =0$. From the boundary conditions, we obtain $y( b) =0$ and $\widehat{y}(t,\lambda ) =0$. Consequently, by the theorem on expansion in the eigenvectors of the selfadjoint operator $A_h'$, we obtain $H'=\{ 0\} $. Hence the operator $A_h$ is simple. The proof is complete. \end{proof} Let us define \[ H_{-}=\overline{\cup_{t\geq 0} U_tD_{-}}\,, \quad H_{+}=\overline{\cup_{t\leq 0} U_tD_{+}} \] where $D_{-}=\langle L^2(-\infty ,0),0,0\rangle $ and $D_{+}=\langle 0,0,L^2(0,\infty )\rangle $. By using Lemma \ref{lem2}, one can obtain $H_{-}+H_{+}= \mathcal{H}$. Assume that $\varphi ( x,\lambda ) $ and $\psi ( x,\lambda) $ are solutions of $l( y) =\lambda y$, satisfying the conditions \begin{gather*} \varphi ( a,\lambda ) =\frac{\alpha _2'}{\alpha },\quad p( a) \varphi ^{\Delta }( a,\lambda ) =\frac{\alpha_1'}{\alpha },\\ \psi ( a,\lambda ) =\alpha _2-\alpha_2'\lambda ,\quad p( a) \psi ^{\Delta }( a,\lambda) =\alpha _1-\alpha _1'\lambda , \end{gather*} Let us adopt the notation \begin{equation*} \widehat{\psi }( x,\lambda ) :=\begin{pmatrix} \psi ( x,\lambda ) \\ \alpha \end{pmatrix} ,\quad n_{b}(\lambda )=\frac{p( b) \varphi ^{\Delta }(b,\lambda ) }{\varphi ( b,\lambda ) }, \quad m_{b}(\lambda )= \frac{\psi ( b,\lambda ) }{p( b) \psi ^{\Delta }(b,\lambda ) }. \end{equation*} The functions $m_{b}(\lambda )$ is a meromorphic function on the complex plane $\mathbb{C}$ with a countable number of poles on the real axis. Further, it is possible to show that the function $m_{b}(\lambda )$ possesses the following properties: $\operatorname{Im}$ $m_{b}(\lambda )\geq 0$ for all $\operatorname{Im}\lambda >0$, and $\overline{m_{b}(\lambda )}=m_{b}( \overline{\lambda }) $ for all $\lambda \in \mathbb{C}$, except the real poles $m_{b}(\lambda )$. We set \begin{gather} S_h(\lambda ):=\frac{m_{b}(\lambda )-h}{m_{b}(\lambda )-\overline{h}}. \label{e3.12} \\ U_{\lambda }^{-}( t,\xi ,\zeta ) =\langle e^{-i\lambda \xi },\alpha n_{b}(\lambda )\{( m_{b}(\lambda )-h) p( b) \varphi ^{\Delta }( b,\lambda ) \}^{-1}\widehat{\psi }( t,\lambda ) \overline{,S_h}(\lambda )e^{-i\lambda \zeta }\rangle . \nonumber \end{gather} We note that the vectors $U_{\lambda }^{-}( t,\xi ,\zeta )$ for real $\lambda $ do not belong to the space $\mathcal{H}$. However, $U_{\lambda }^{-}( t,\xi ,\zeta ) $ satisfies the equation $\mathcal{L}_hU=\lambda U$ and the corresponding boundary conditions for the operator $\mathcal{L}_{H}$. By means of vector $U_{\lambda }^{-}(t,\xi ,\zeta ) $, we define the transformation $F_{-}:f\to \widetilde{f_{-}}(\lambda )$ by \begin{equation*} ( F_{-}f) (\lambda ):=\widetilde{f_{-}}(\lambda ):= \frac{1}{\sqrt{2\pi }}( f,U\overline{_{\lambda }}) _{\mathcal{H}} \end{equation*} on the vectors $f=\mathcal{\langle }\varphi _{-},\widehat{y},\varphi _{+}\rangle $ in which $\varphi _{-}( \xi )$, $\varphi _{+}(\zeta )$, $y( x) $ are smooth, compactly supported functions \begin{lemma} \label{lem3} \rm The transformation $F_{-}$ isometrically maps $H_{-}$ onto $L^2(\mathbb{R})$. For all vectors $f,g\in H_{-}$ the Parseval equality and the inversion formulae hold: \begin{equation*} ( f,g) _{\mathcal{H}} =(\widetilde{f_{-}},\widetilde{g}_{-})_{L^2} =\inf_{-\infty}^{\infty } \widetilde{f_{-}}(\lambda )\overline{\widetilde{g_{-}} (\lambda )}d\lambda ,\quad f=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{\infty } \widetilde{f_{-}}(\lambda )U_{\overline{\lambda}} d\lambda , \end{equation*} where $\widetilde{f_{-}}(\lambda )=( F_{-}f)(\lambda )$ and $\widetilde{g_{-}}(\lambda)=( F_{-}g) (\lambda )$. \end{lemma} \begin{proof} For $f,g\in D_{-}$, $f=\mathcal{\langle }\varphi _{-},0,0\rangle $, $g=\mathcal{\langle }\psi _{+},0,0\rangle $, with Paley-Wiener theorem, we have \begin{equation*} \widetilde{f_{-}}(\lambda ) =\frac{1}{\sqrt{2\pi }}( f,U \overline{_{\lambda }}) _{\mathcal{H}} =\frac{1}{\sqrt{2\pi }}\int_{-\infty }^0 \varphi _{-}( \xi ) e^{-i\lambda \xi}d\xi \in H_{-}^2, \end{equation*} and by using the usual Parseval equality for Fourier integrals, we have \begin{equation*} ( f,g) _{\mathcal{H}} =\int_{-\infty }^{\infty } \varphi _{-}( \xi ) \overline{\psi _{-}( \xi ) }d\xi =\int_{-\infty }^{\infty } \widetilde{f_{-}} (\lambda )\overline{\widetilde{g_{-}}(\lambda )}d\lambda =( F_{-}f,F_{-}g) _{L^2}, \end{equation*} Here, $H_{\pm }^2$ denotes the Hardy classes in $L^2(\mathbb{R}) $ consisting of the functions analytically extendible to the upper and lower half-planes, respectively. We now extend to the Parseval equality to the whole of $H_{-}$. We consider in $H_{-}$ the dense set of $H_{-}'$ of the vectors obtained as follows from the smooth, compactly supported functions in $D_{-}:f\in H_{-}'$ if $f=U_{T}$ $f_0$, $f_0=\mathcal{\langle }\varphi _{-},0,0\rangle $, $\varphi _{-}\in C_0^{\infty }( -\infty ,0) $, where $T=T_{f}$ is a nonnegative number depending on $f$. If $f,g\in H_{-}'$ , then for $T>T_{f}$ and $T>T_{g}$ we have $U_{-T}f,U_{-T}g\in D_{-}$, moreover, the first components of these vectors belong to $C_0^{\infty}( -\infty ,0)$. Therefore, since the operators $U_t$ $(t\in\mathbb{R}) $ are unitary, by the equality $F_{-}U_tf=( U_tf,U \overline{_{\lambda }}) _{\mathcal{H}} =e^{i\lambda t}(f,U_{\lambda }^{-}) _{\mathcal{H}}=e^{i\lambda t}F_{-}f$, we have \begin{gather} ( f,g) _{\mathcal{H}}=( U_{-T} f,U_{-T} g) _{\mathcal{H}} =( F_{-}U_{-T} f,F_{-}U_{-T} g) _{L^2} , \nonumber \\ (e^{i\lambda T}F_{-}f,e^{i\lambda T}F_{-}g)_{L^2} =(\widetilde{f}, \widetilde{g})_{L^2}. \label{e3.13} \end{gather} By taking the closure \eqref{e3.13}, we obtain the Parseval equality for the space $H_{-}$. The inversion formula is obtained from the Parseval equality if all integrals in it are considered as limits in the of integrals over finite intervals. Finally $F_{-}H_{-}=\overline{\cup_{t\geq 0}F_{-}U_tD_{-}} =\overline{\cup_{t\geq 0} e^{i\lambda t}H_{-}^2} =L^2(\mathbb{R}) $, that is $F_{-}$ maps $H_{-}$ onto the whole of $L^2(\mathbb{R}) $. The proof is complete. \end{proof} We set \begin{equation*} U_{\lambda }^{+}( t,\xi ,\zeta ) =\langle S_h(\lambda )e^{-i\lambda \xi },\alpha n_{b}(\lambda )( m_{b}(\lambda )-h) p( b) \varphi ^{\Delta }( b,\lambda ) \widehat{\psi } ( t,\lambda ) ,e^{-i\lambda \zeta }\rangle . \end{equation*} We note that the vectors $U_{\lambda }^{+}( t,\xi ,\zeta ) $ for real $\lambda $ do not belong to the space $\mathcal{H}$. However, $U_{\lambda }^{+}( t,\xi ,\zeta ) $ satisfies the equation $\mathcal{L}_hU=\lambda U$ and the corresponding boundary conditions for the operator $\mathcal{L}_{H}$. With the help of vector $U_{\lambda}^{+}( t,\xi ,\zeta ) $, we define the transformation $F_{+}:f\to \widetilde{f}_{+}(\lambda )$ by $(F_{+}f) (\lambda ):=\widetilde{f}_{+}(\lambda ):=\frac{1}{ \sqrt{2\pi }}( f,U_{\lambda }^{+}) _{\mathcal{H}}$ on the vectors $f=\mathcal{\langle }\varphi _{-},\widehat{y},\varphi _{+}\rangle $ in which $\varphi _{-}( \xi ) ,\varphi _{+}( \zeta ) $ and $y( x) $ are smooth, compactly supported functions. \begin{lemma} \label{lem4} The transformation $F_{+}$ isometrically maps $H_{+}$ onto $L^2(\mathbb{R})$. For all vectors $f,g\in H_{+}$ the Parseval equality and the inversion formula hold: \begin{equation*} ( f,g) _{\mathcal{H}}=(\widetilde{f}_{+},\widetilde{g}_{+})_{L^2} =\int_{-\infty }^{\infty } \widetilde{f}_{+}(\lambda )\overline{\widetilde{g_{+}} (\lambda )}d\lambda ,\quad f=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{\infty } \widetilde{f}_{+}(\lambda )U_{\lambda}^{+}d\lambda , \end{equation*} where $\widetilde{f}_{+}(\lambda )=( F_{+}f)(\lambda )$ and $\widetilde{g_{+}}(\lambda )=(F_{+}g) (\lambda )$. \end{lemma} The proof of the above lemma is analogous to the Lemma \ref{lem3}, and it is omitted. It is obvious that $| S_h(\lambda )| =1$ for $\lambda \in\mathbb{R}$. Therefore, it explicitly follows from the formulae for the vectors $U_{\lambda }^{-}$ and $U_{\lambda }^{+}$ that \begin{equation} U_{\lambda }^{+}=\overline{S_h}(\lambda )U_{\lambda }^{-}. \label{e3.14} \end{equation} It follows from Lemmas \ref{lem3} and \ref{lem4} that $H_{-}=H_{+}$. Together with Lemma \ref{lem2}, it can be concluded that $H_{-}=H_{+}=\mathcal{H}$. Thus, the transformation $F_{-}$ isometrically maps $H_{-}$ onto $L^2(\mathbb{R}) $ with the subspace $D_{-}$ mapped onto $H_{-}^2$ and the operators $U_t$ are transformed into the operators of multiplication by $e^{i\lambda t}$. This means that $F_{-}$ is the incoming spectral representation for the group $\{ U_t\} $. Similarly, $F_{+}$ is the outgoing spectral representation for the group $\{ U_t\} $. It follows from \eqref{e3.14} that the passage from the $F_{-}$ representation of an element $f\in \mathcal{H}$ to its $F_{+}\ $representation is accomplished as $\widetilde{f}_{+}(\lambda )=\overline{S_h}(\lambda ) \widetilde{f}_{-}(\lambda )$. Consequently, according to \cite{l2}, we have proved the following Theorem. \begin{theorem} \label{thm3} The function $\overline{S_h}(\lambda )$ is the scattering matrix of the group $\{U_t\} $ (of the selfadjoint operator $\mathcal{L}_h$). \end{theorem} Let $S(\lambda )$ be an arbitrary non-constant inner function (see \cite{n1}) on the upper half-plane (the analytic function $S(\lambda )$ on the upper half-plane $\mathbb{C}_{+}$ is called \textit{inner function} on $\mathbb{C}_{+}$ if $| S(\lambda )| \leq 1$ for all $\lambda \in \mathbb{C}_{+}$ and $| S(\lambda )| =1$ for almost all $\lambda \in \mathbb{R})$. Define $K=H_{+}^2 \ominus SH_{+}^2$. Then $K\neq \{0\} $ is a subspace of the Hilbert space $H_{+}^2$. We consider the semigroup of operators $Z_t$ $( t\geq 0) $ acting in $K$ according to the formula $Z_t\varphi =P[ e^{i\lambda t}\varphi ]$, $\varphi =\varphi (\lambda )\in K$, where $P$ is the orthogonal projection from $H_{+}^2$ onto $K$. The generator of the semigroup $\{ Z_t\} $ is denoted by $T\varphi =\lim_{t\to +0}( it) ^{-1}( Z_t\varphi -\varphi )$, which $T $ is a maximal dissipative operator acting in $K$ and with the domain $D(T)$ consisting of all functions $\varphi \in K$, such that the limit exists. The operator $T$ is called \textit{a model dissipative operator} (we remark that this model dissipative operator, which is associated with the names of Lax-Phillips \cite{l2}, is a special case of a more general model dissipative operator constructed by Nagy and Foia\c{s} \cite{n1}). The basic assertion is that $S(\lambda )$ is the \textit{characteristic function }of the operator $T$. Let $K=\mathcal{\langle }0,H,0\rangle $, so that $\mathcal{H=}D_{-}\oplus K\oplus D_{+}$. It follows from the explicit form of the unitary transformation $F_{-}$ under the mapping $F_{-}$ that \begin{equation} \begin{gathered} \mathcal{H} \to L^2(\mathbb{R}),\quad f\to \widetilde{f_{-}}(\lambda )=(F_{-}f) (\lambda ),\quad D_{-}\to H_{-}^2,\quad D_{+}\to S_hH_{+}^2, \\ K \to H_{+}^2\ominus S_hH_{+}^2,\quad U_t\to (F_{-}U_tF_{-}^{-1}\widetilde{f_{-}})(\lambda )=e^{i\lambda t} \widetilde{f_{-}}(\lambda ). \end{gathered} \label{e3.15} \end{equation} The formulas \eqref{e3.15} show that operator $A_h$ is a unitarily equivalent to the model dissipative operator with the characteristic function $S_h(\lambda )$. Since the characteristic functions of unitary equivalent dissipative operator coincide (see \cite{n1}), we have thus proved following theorem. \begin{theorem} \label{thm4} The characteristic function of the maximal dissipative operator $A_h$ coincides with the function $S_h(\lambda )$ defined in \eqref{e3.12}. \end{theorem} Using the characteristic function, we investigate the spectral properties of the maximal dissipative operator $A_h$. We know that the characteristic function of a maximal dissipative operator carries information about the spectral properties of this operator. To prove completeness of the system of eigenvectors\ and associated vectors of the operator $A_h$ in the space $H$, we must show that there exists no singular factor $s(\lambda )$ of the characteristic function $S_h(\lambda )$ in the factorization $S_h(\lambda )=s(\lambda )B(\lambda )$ ($B(\lambda )$ is a Blaschke product) (see \cite{a3,a4,a5,a6,a7,k1,n1}). The characteristic function $S_h(\lambda )$ of the maximal dissipative operator $A_h$ has the form \begin{equation*} S_h(\lambda ):=\frac{m_{b}(\lambda )-h}{m_{b}(\lambda )-\overline{h}}, \end{equation*} where $\operatorname{Im}h>0$. From \eqref{e3.12}, it is clear that $S_h(\lambda )$ is an inner function in the upper half-plane, and it is meromorphic in the whole complex $\lambda $-plane. \begin{theorem} \label{thm5} For all the values of $h$ with $\operatorname{Im}h>0$, except possibly for a single value $h=h_0$, the characteristic function $S_h(\lambda )$ of the maximal dissipative operator $A_h$ is a Blaschke product. The spectrum of $A_h$is purely discrete and belongs to the open upper half-plane. The operator $A_h$ has a countable number of isolated eigenvalues with finite multiplicity and limit points at infinity. The system of all eigenvectors and associated vectors of the operator $A_h$ is complete in the space $H$ (see \cite{a3,a4,a5,a6,a7,e1,o1,o2,t2,t3}). \end{theorem} \begin{proof} Since $S_h(\lambda )$ is an inner function, it can be factored in the form \begin{equation} S_h(\lambda )=e^{i\lambda c}B_h(\lambda ),\quad c=c( h) \geq 0, \label{e3.16} \end{equation} where $B_h(\lambda )$ is a Blaschke product. It follows from \eqref{e3.16} that \begin{equation} | S_h(\lambda )| =| e^{i\lambda c}|\,| B_h(\lambda )| \leq e^{-c( h) \operatorname{Im}\lambda },\quad \operatorname{Im}\lambda \geq 0. \label{e3.17} \end{equation} Further, for $m_{b}(\lambda )$ in terms of $S_h(\lambda )$, we find from \eqref{e3.12} that \begin{equation} m_{b}(\lambda )=\frac{h-\overline{h}S_h(\lambda )}{S_h(\lambda )-1}. \label{e3.18} \end{equation} If $c( h) >0$ for a given value $h$ ($\operatorname{Im}h>0$), then \eqref{e3.17} implies that $\lim_{t\to +\infty } S_h(it) =0$, and then \eqref{e3.18} gives us that $\lim_{t\to +\infty} m_{b}( it) =-h$. Since $m_{b}(\lambda )$ does not depend on $h$, this implies that $c( h) $ can be nonzero at not more than a single point $h=h_0$ (and further $h_0=-\lim_{t\to +\infty } m_{b}( it) $). The proof is complete. \end{proof} \begin{thebibliography}{99} \bibitem{a1} R. P. Agarwal, M. 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