Cheremnikh E.V. "On normal eigenvalue embedded in continuous spectrum"

On normal eigenvalue embedded in continuous spectrum

The study of spectral singularities of non-selfadjoint Sturm-Liouville's operators was given first in the well known work [1]. Analytic prolongement of the resolvent over the continuous spectrum can have the poles, placed in continuous. In the non-selfadjoint case these poles are not obligatory eigen-value and they complicate the study of the continuous spectrum. In the works [2-3] the authors apply the notion of residue to study the spectral singularities embedded in continuous spectrum. Let $\sigma$ be such singularities of the operator $T:H\to H$ . V.E.Ljance [2] consider the projector $P_{\sigma\pm}$ given by the residue of analytical prolongation of the resolvent

$\begin{displaymath} (P_{\sigma\pm}\varphi,\psi)\equiv-\mathop{\mbox{Res}}\limit... ... (T_\zeta \varphi,\psi)_\pm,\quad T_\zeta\equiv(T-\zeta)^{-1} \end{displaymath}$

(see, for ex. [2]). M. Gasumov and F. Maxudov[3] use the residue of the function ${C^*}^{-1}T_\lambda C^{-1}\varphi$ where the operator C is associated with some equipped space and acts from negative space in the space H. It is well known that if a value $\lambda$ is a pole of the resolvent( embedded in the resolvent set) then the residue of the resolvent generates the projection on first term of the following decomposition

$\begin{displaymath} \hskip2cm H=Z((T-\lambda)^m)\stackrel{\cdot}{+}R((T-\lambda)^m),\quad m\ge1 \hskip3.5cm(0.1) \end{displaymath}$

In the present article the following question is discussed : there exists or not some analog of the decomposition (0.1) in the case of spectral singularity embedded in continuous spectrum. In the section 1 we precise the subject of our study ( some Friedrich's model ). In the section 2 we prove the preliminaries statements which presents some interest itself. In the section 3 the notion of maximal operator is introduced in Friedrichs' model. In the section 4 the main result, i.e. the relation (4.1), is given. In the section 5 using an example we explain the relation (4.1) in terms of solvability certain differential equation. In the case of potential with delay a representation of the solution of Sturm-Liouville' equation is obtained. The section 6 contains the conclusion.

Friedrichs' model

V.E.Ljance [2] considered a Friedrichs' model in the space $L^2(-\infty,\infty)$ . We will consider the model in the space $L^2(0,\infty)$ under the conditions which permit to study Fourier transform of some Sturm-Liouville' operator. Let us consider in the space $H=L_\rho^2(0,\infty)$ the operator

where non-perturbed operator is given by the relation $S\varphi(\tau)\equiv\tau\varphi(\tau)$ , $\tau>0$ (with maximal domain of definition D(S)) and the perturbation admits the factorization V=A^*B where the operators A, B act from H in auxiliary Hilbert space G as follows

$\begin{displaymath} A{\varphi}=\int\limits_0^\infty\varphi(s)\alpha(s)\rho(s)ds... ...\ B{\varphi}=\int\limits_0^\infty\varphi(s)\beta(s)\rho(s)ds \end{displaymath}$

and $\alpha(s),\ \ \beta(s)\in G$ be some vector-functions. Let $\Vert\cdot\Vert$ , $\Vert\cdot\Vert _H$ be norms in G, H. We suppose that the following conditions hold: A₁) $\displaystyle{\int\limits_0^\infty\frac{ds}{\rho(s)(1+s)}<\infty }$ ; A₂) functions $\rho(s)$ , $\alpha(s)$ , $\beta(s)$ have n continuous derivates for some $n\ge1$ ; A₃) $\alpha'(s)$ , $\beta'(s)$ , $\displaystyle{\bigg(\frac{1}{\rho(s)}\bigg)'=o(1) }$ , $\alpha(s)$ , $\displaystyle{\beta(s)=o\bigg(\frac{1}{\ln s}\bigg)}$ , $s\to\infty$ ; A₄) $\sup\limits_{(0,\infty)}\rho(s)\Vert\alpha(s)\Vert<\infty$ , $\sup\limits_{(0,\infty)}\rho(s)\Vert\beta(s)\Vert<\infty$ ; A₅) for all $\varphi\in L_\rho^2$ the integrals

$\begin{displaymath} A\varphi=\int\limits_0^\infty\varphi(s)\alpha(s)\rho(s)ds,\ \ \ B\varphi=\int\limits_0^\infty\varphi(s)\beta(s)\rho(s)ds \end{displaymath}$

converge in the sense of the norm of G and define bounded operators $L_\rho^2\to G$ with the ranges R(A), R(B) dense in G. Note that the idea of factorization of the perturbation is using many years ago. It gives the symmetry to the study, the dimensions of auxiliary Hilbert space and the perturbations are the same as usually.

Preliminaries statements.

We need of properties of operator-function $K(\zeta)=1+BS_\zeta A^*$ , $S_\zeta=(S-\zeta)^{-1}$ . This function is well known in the theory of the operator T=S+A^*B. In view of its bilinear form (see (2.5)) we begin by Cauchy' integral. We denote $\Omega_+(\delta)=\{\zeta:\ \hbox{Im\ }\zeta>0,\Vert\zeta-\sigma\vert<\delta\}$ and $F_+(\sigma)=\lim\limits_{\tau\searrow0}F(\sigma+i\tau)$ for a function F(s), analytical in the domain $\Omega_+(\sigma)$ . Let

$\begin{displaymath} F(\zeta)=\int\limits_a^b\frac{f(s)}{s-\zeta}ds,\ \ \ f\in C^1[a,b],\ \ \zeta\not\in[a,b]. \end{displaymath}$

(2.1)

Lemma 2.1 Let $\sigma\in(a,b)$ be a fixed point. Then for every $\varepsilon>0$ there exists $\delta>0$ such that

$\begin{displaymath} \vert F(\zeta)-F_+(\sigma)\vert<\varepsilon\Vert f\Vert _{C^1[a,b]},\ \ \ \zeta\in\Omega_+(\delta). \end{displaymath}$

(2.2)

Proof. We choose $\varepsilon>0$ such that $\Delta=(\sigma-\varepsilon,\sigma+\varepsilon)\subset[a,b]$ and we decompose integral (2.1):

$\begin{displaymath} F(\zeta)=\int\limits_{\Delta}\frac{f(s)}{s-\zeta}ds+ \int\... ...etilde\Delta}\frac{f(s)}{s-\zeta}ds\equiv G(\zeta)+H(\zeta), \end{displaymath}$

where $\widetilde\Delta=[a,b]\setminus\Delta$ . Suppose that $\delta<\frac{\varepsilon}{2}$ , then the function $H(\zeta)$ is holomorphic in the domain $\vert\zeta-\sigma\vert<\delta$ and

$\begin{displaymath} H(\zeta)-H(\sigma)=(\zeta-\sigma)\int\limits_{\widetilde\Delta}\frac{f(s)} {(s-\zeta)(s-\sigma)}ds. \end{displaymath}$

If $s\in[\sigma+\varepsilon,b]$ then $\vert s-\zeta\vert>s-(\sigma+\delta)$ , $\zeta\in\Omega_+(\delta)$ and

$\begin{displaymath} \left\vert\int\limits_{\sigma+\varepsilon}^b\frac{ds}{(s-\z... ...+\frac{1}{\varepsilon-\delta}< \frac{1}{\varepsilon-\delta}. \end{displaymath}$

$\begin{displaymath} \vert H(\zeta)-H(\sigma)\vert<\frac{2\delta}{\varepsilon-\delta}\Vert f\Vert _{C^1[a,b]}. \end{displaymath}$

According to the principle of maximal module the function $G(\zeta)-G_+(\sigma)$ admits in the domain $\Omega_+(\delta)$ the estimate

$\begin{displaymath} \vert G(\zeta)-G_+(\sigma)\vert<\max\left[\sup\limits_{z\in... ...igma_0\in\Delta_0}\vert G_+(\sigma_0)-G_+(\sigma)\vert\right] \end{displaymath}$

where $\Gamma=\{z:\ \hbox{Im\ }z>0,\ \vert z-\sigma\vert=\delta\}$ , $\Delta_0=[\sigma-\delta,\sigma+\delta]$ . Let $\sigma_1=\hbox{Re\ }z$ , then $G(z)-G_+(\sigma)\le\vert G(z)-G_+(\sigma_1)\vert+ \vert G_+(\sigma_1)-G_+(\sigma)\vert$ and for all $\zeta\in\Omega_+(\delta)$

$\begin{displaymath} \vert G(\zeta)-G_+(\sigma)\vert\le \sup\limits_{z\in\Gamma... ...n\Delta_0}\vert G_+(\sigma_0)-G_+(\sigma)\vert\equiv I_1+I_2. \end{displaymath}$

$\begin{displaymath} G(z)=\int\limits_\Delta\frac{f(s)-f(\sigma_1)}{s-z}ds+f(\sigma_1)\ell(z),\ \ \ \ell(z)=\int\limits_\Delta\frac{ds}{s-z} \end{displaymath}$

$\begin{displaymath} G_+(\sigma_1)=\int\limits_\Delta\frac{f(s)-f(\sigma_1)}{s-\sigma_1}ds+ f(\sigma_1)\ell_+(\sigma_1), \end{displaymath}$

(2.3)

$\begin{displaymath} G(z)-G_+(\sigma_1)=(z-\sigma_1) \int\limits_\Delta\frac{f(... ...gma_1} \frac{ds}{s-z}+f(\sigma_1)[\ell(z)-\ell_+(\sigma_1)]. \end{displaymath}$

As $\vert s-z\vert\ge\tau$ , we have $\displaystyle{\int\limits_\Delta\frac{ds}{\vert s-z\vert}<2\frac{\varepsilon}{\tau} }$ and

$\begin{displaymath} \left\vert(z-\sigma_1)\int\limits_\Delta\frac{f(s)-f(\sigma... ...c{ds}{\vert s-z\vert}\le2\varepsilon\Vert f\Vert _{C^1[a,b]}. \end{displaymath}$

If $\sigma_1\in\Delta_0$ then $\displaystyle{\vert\ell(z)-\ell_+(\sigma_1)\vert< \frac{C\delta}{\varepsilon-\delta} }$ and $I_1\le(2\varepsilon+C\delta/(\varepsilon-\delta))\Vert f\Vert _{C^1[a,b]}$ . It follows from (2.3) that

$\begin{displaymath} \vert G_+(\sigma_0)-G_+(\sigma)\vert\le \left\vert\int\lim... ...nt\limits_\Delta\frac{f(s)-f(\sigma)}{s-\sigma}ds\right\vert+ \end{displaymath}$

$\begin{displaymath} +\vert f(\sigma_0)-f(\sigma)\Vert\ell_+(\sigma_0)\vert+\vert f(\sigma)\Vert\ell_+(\sigma_0)- \ell_+(\sigma)\vert. \end{displaymath}$

Taking into account that $\vert\ell_+(\sigma_1)\vert<C\vert\ln(\varepsilon-\delta)\vert$ we get $I_2\le C(\delta\vert\ln(\varepsilon-\delta)\vert+\delta/(\varepsilon-\delta))\Vert f\Vert _{C^1[a,b]}$ . With $\delta=\varepsilon^2$ we obtain inequality (2.2). Lemma is proved.

Lemma 2.2 Let f(s), f'(s), s>0 be differentiable functions and $F(s)/(1+s)\in L^1(0,\infty)$ , $\displaystyle{f(s)=o\bigg(\frac{1}{\ln s}\bigg)}$ , f'(s)=o(1), $s\to\infty$ . Then the function

$\begin{displaymath} F(\zeta)=\int\limits_0^\infty\frac{f(s)}{s-\zeta}ds \end{displaymath}$

has the limit $\lim\limits_{\vert\zeta\vert\to\infty}F(\zeta)=0$ uniformly with respect to $\hbox{arg\ }\zeta\in[0,2\pi]$ .
Proof. Suppose that $\hbox{Re\ }\zeta\le0$ , then for s₀>0

$\begin{displaymath} \left\vert\int\limits_0^\infty\frac{f(s)}{s-\zeta}ds\right\... ...ert\int\limits_{s_0}^{\infty} \frac{\vert f(s)\vert}{1+s}ds. \end{displaymath}$

$\begin{displaymath} \left\vert\int\limits_0^\infty\frac{f(s)}{s-\zeta}ds\right\... ...{\sqrt N}^\infty\frac{\vert f(s)\vert}{1+s}ds=\varepsilon(N). \end{displaymath}$

We shall denote by $\varepsilon(N)$ every infinitesimal when $N\to\infty$ . Suppose now that $\hbox{Re\ }\zeta\ge0$ , i.e. $\zeta=\sigma+i\tau$ , $\sigma>0$ . We consider the integral on the intervals $\left(0,\frac{\sigma}{2}\right)$ , $\left(\frac{3\sigma}{2},\infty\right)$ and $\left(\frac{\sigma}{2},\frac{3\sigma}{2}\right)$ . If $0<s<\sigma/2$ then $\vert s-\sigma-i\tau\vert>N/2$ . It is sufficiently to consider an interval $(s_1,\sigma/2)$ , $s_1=\hbox{const}$ instead of $(0,\sigma/2)$ :

$\begin{displaymath} \left\vert\int\limits_{s_1}^{\sigma/2}\frac{f(s)}{s-\sigma-... ...{2}{N}\int\limits_{s_1}^{N/2}\frac{ds}{\ln s}=\varepsilon(N). \end{displaymath}$

If $s>3\sigma/2$ then $s-\sigma>s/3$ , $\vert s-\sigma-i\tau\vert>\sqrt{s^2+\tau^2}/3$ and

$\begin{displaymath} \left\vert\int\limits_{3\sigma/2}^{\infty}\frac{f(s)}{s-\si... ...sigma/2}^{\infty}\frac{\vert f(s)\vert}{\sqrt{s^2+\tau^2}}ds. \end{displaymath}$

We have two cases. a) $\frac{3\sigma}{2}>\vert\tau\vert$ , it follows from $\sigma^2+\tau^2=N^2$ that $\frac{3\sigma}{2}>\frac{N}{4}$ and

$\begin{displaymath} \int\limits_{3\sigma/2}^{\infty}\frac{\vert f(s)\vert}{\sqr... ...mits_{N/4}^{\infty}\frac{\vert f(s)\vert}{s}ds=\varepsilon(N) \end{displaymath}$

$\begin{displaymath} \int\limits_{3\sigma/2}^{\infty}\frac{\vert f(s)\vert}{\sqr... ...t\tau\vert}^\infty\frac{\vert f(s)\vert}{s}ds=\varepsilon(N). \end{displaymath}$

Interval $\left(\frac{\sigma}{2},\frac{3\sigma}{2}\right)$ : suppose at first that $\sigma^2>N$ . Let $a=\hbox{const}$ , $a<\sigma/2$ , then

$\begin{displaymath} \int\limits_{\sigma/2}^{3\sigma/2}\frac{f(s)}{s-\zeta}ds= ... ...mits_{\sigma-a}^{\sigma+a} \frac{f(s)-f(\sigma)}{s-\zeta}ds+ \end{displaymath}$

$\begin{displaymath} +f(\sigma)\int\limits_{\sigma-a}^{\sigma+a}\frac{ds}{s-\zeta}\equiv I_1+I_2+I_3. \end{displaymath}$

$\begin{displaymath} \vert I_1\vert\le\max\limits_{[\sigma/2,3\sigma/2]}\vert f(... ...sigma/2]}\vert f(s)\vert \ln\frac{\sigma}{2a}=\varepsilon(N) \end{displaymath}$

$\begin{displaymath} \vert I_2\vert= \left\vert\int\limits_{\sigma-a}^{\sigma+a... ...{[\sigma-a,\sigma+a]}\vert f'(s)\vert\cdot2a =\varepsilon(N) \end{displaymath}$

$\begin{displaymath} \vert I_3\vert=\vert f(\sigma)\vert\cdot2\hbox{arctg}\frac{a}{\vert\tau\vert}=\varepsilon(N). \end{displaymath}$

$\begin{displaymath} \left\vert\int\limits_{\sigma/2}^{3\sigma/2}\frac{f(s)}{s-\... ...sigma}{\vert\tau\vert}< \frac{C}{\sqrt{N-1}}=\varepsilon(N). \end{displaymath}$

$\begin{displaymath} F(\zeta)=\int\limits_0^\infty\frac{f(s)}{s-\zeta}ds=\varepsilon(N),\ \ \ N=\vert\zeta\vert\to\infty \end{displaymath}$

(2.4)

i.e. $F(\zeta)\to0$ uniformly with respect to $\hbox{arg\ }\zeta\in[0,2\pi]$ . Lemma is proved.

Let $S_\zeta=(S-\zeta)^{-1}$ , $\zeta\not\in[0,\infty)$ and $K(\zeta)=1+BS_\zeta A^*$ . It is obvious that $A^*c(s)=(c,\alpha(s))$ , where $(\cdot,\cdot)$ is a scalar product in G. Then

$\begin{displaymath} (K(\zeta)c,d)=(c,d)+\int\limits_0^\infty\frac{(c,\alpha(s))... ...a(s),d)} {s-\zeta}\rho(s)ds,\ \ \ \zeta\not\in[0,\infty). \end{displaymath}$

(2.5)

     We pose $c\in D(K_\pm(\sigma))$ if the functional $d\to(K(\sigma)c,d)_\pm$ is bounded in G and we define $K_\pm(\sigma):\ G\to G$ by the relation $(K_\pm(\sigma)c,d)=(K(\sigma)c,d)_\pm$ , $c,d\in G$ .

     Lemma 2.3 1) $\lim\limits_{\zeta\to\sigma\pm i0} \Vert K(\zeta)-K_\pm(\sigma)\Vert=0$ and the operators $K(\zeta)-1$ , $\zeta\not\in[0,\infty)$ and $K_\pm(\sigma)-1$ , $\sigma>0$ , too are completely continuous. 2) $\lim\limits_{\zeta\to\infty}\Vert K(\zeta)-1\Vert=0$ uniformly with respect to $\hbox{arg\ }\zeta\in[0,2\pi]$ .
     Proof. 1) Let $\varepsilon>0$ be fixed.      We choose s₀>0 such that (see A₁), A₄))

$\begin{displaymath} \left\vert\left\vert\int\limits_{s_0}^\infty\frac{c,\alpha(... ...)\vert s-\zeta\vert}\le \frac{\varepsilon}{2}\Vert c\Vert. \end{displaymath}$

(2.6)

$\begin{displaymath}\Vert\alpha(s)-\alpha(s_i)\Vert\le\frac{\varepsilon}{2M},\ \ \ s\in[s_i,s_{i+1}], \end{displaymath}$

(2.7)

where M will be indicated later. Let {e_k} be arbitrary orthonormal basis of G and

$\begin{displaymath} \alpha_n(s)\equiv\alpha(s)-\sum\limits_{k=1}^n(\alpha(s),e_k)e_k. \end{displaymath}$

For every $s\in[0,s_0]$ there exists a number i such that $s\in[s_i,s_{i+1}]$ . Taking into account that for all n $\Vert\alpha_n(s)-\alpha_n(s_i)\Vert\le\Vert\alpha(s)-\alpha(s_i)\Vert$ and in view of (2.7) we get

$\begin{displaymath} \Vert\alpha_n(s)\Vert\le\Vert\alpha_n(s)-\alpha_n(s_i)\Vert+\Vert\alpha_n(s_i)\Vert\le \end{displaymath}$

$\begin{displaymath} \le\Vert\alpha(s)-\alpha(s_i)\Vert+\Vert\alpha_n(s_i)\Vert<\frac{\varepsilon}{2M}+ \Vert\alpha_n(s_i)\Vert. \end{displaymath}$

(2.8)

The set {s_i} is finite, so, there exists N>0 such that for all i $\Vert\alpha_n(s_i)\Vert<\frac{\varepsilon}{2M}$ , n>N. In view of (2.8) $\Vert\alpha_n(s)\Vert<\frac{\varepsilon}{M}$ , $s\in[0,s_0]$ , n>N. We have

$\begin{displaymath} \int\limits_0^{s_0}\frac{(c,\alpha(s))\beta(s)}{s-\zeta}\rh... ...imits_0^{s_0}\frac{(c,\alpha_n(s))}{s-\zeta}\beta(s)\rho(s)ds \end{displaymath}$

$\begin{displaymath} \left\vert\left\vert\int\limits_0^{s_0}\frac{(c,\alpha_n(s)... ...ert s-\zeta\vert}\rho(s)ds<\frac{\varepsilon}{2}\Vert c\Vert. \end{displaymath}$

It follows from (2.5)-(2.6) that the operator $K(\zeta)-1$ is approximated by a finite-dimensional operator to $\varepsilon$ in operator norm. So, the operator $K(\zeta)-1$ is completely continuous. Let us consider the bilinear form

$\begin{displaymath} (J(\zeta)c,d)=\int\limits_a^b\frac{f(s)}{s-\zeta}ds,\ \ \ f(s)=(c,\alpha(s))(\beta(s),d)\rho(s). \end{displaymath}$

(2.9)

According to Lemma 2.2 for every $\varepsilon>0$ there exists $\delta>0$ such that

$\begin{displaymath} \vert(J(\zeta)c,d)-(J(\sigma)c,d)_\pm\vert<\varepsilon\Vert... ...ert _{C^1[a,b]}\le M\varepsilon\Vert c\Vert\cdot\Vert d\Vert \end{displaymath}$

if $\vert\zeta-\sigma\vert<\delta$ , $\hbox{Im\ }\zeta>0$ . It follows from this that the definition of $K_\pm(\sigma)$ is correct and proposition 1) is proved. 2) In view of A₁), A₃) - A₄) the function f(s) (see (2.9)) satisfies the conditions of Lemma 2.2. If follows from the proof of Lemma 2.2 that relation (2.4) becomes

$\begin{displaymath} ((K(\zeta)-1)c,d)=\varepsilon(N)\Vert c\Vert\cdot\Vert d\Vert,\ \ \ \vert\zeta\vert=N\to\infty \end{displaymath}$

Maximal operator and generalized residue of its resolvent on continuous spectrum.

For maximal differential operator corresponding to the Sturm-Liouville's operator with decreasing potential every point outside of half line $[0,\infty)$ is an eigenvalue. However the resolvent of maximal operator on half line $[0,\infty)$ exists as unbounded operator. This resolvent is more useful for us than the resolvent of initial operator as its domain of definition is more larger. We shall introduce an abstract analog of notion of maximal operator.
Definition 3.1. Operator $\widetilde S$ is called the maximal operator, corresponding to the operator S if

$\begin{displaymath} D(\widetilde S)=\bigg\{\varphi\in H:\ \exists c=c(\varphi):... ...u\varphi(\tau)+c(\varphi)\vert^2\rho(\tau)d\tau<\infty\bigg\} \end{displaymath}$

$\begin{displaymath} \widetilde S\varphi(\tau)=\tau\varphi(\tau)+c(\varphi). \end{displaymath}$

Obviously, for every $\varphi\in D(\widetilde S)$ the number $c(\varphi)$ is unique and, in addition, $\varphi\in D(S)\Longleftrightarrow c(\varphi)=0$ . In the case of differential operators the operator $\tilde S$ corresponds to the traditional maximal operator (see (5.2)). Suppose that the function $\psi\in R(\widetilde S-\sigma)$ is continuous, then $(\widetilde S-\sigma)\varphi=\psi$ , or $(s-\sigma)\varphi(s)+c(\varphi)=\psi(s)$ . If $s=\sigma$ then $c(\varphi)=\psi(\sigma)$ , therefore,

$\begin{displaymath} \varphi(s)={\cal R}_\sigma\psi(s)\equiv\frac{\psi(s)-\psi(\sigma)}{s-\sigma}. \end{displaymath}$

The operator $\widetilde S$ is closed and, consequently, the operator $\widetilde S_\sigma\equiv(\widehat S-\sigma)^{-1}$ is the closure of the operator ${\cal R}_\sigma$ . For our purpose it is sufficiently to consider the operator ${\cal R}_\sigma$ itself only. We call

$\begin{displaymath} \widetilde T=\widetilde S+A^*B \end{displaymath}$

the maximal operator, corresponding to T. Let $N(\sigma)=1+B{\cal R}_\sigma A^*$ . Obviously (see (2.5))

$\begin{displaymath} K_\pm(\sigma)c=c+V.p.\int\limits_0^\infty \frac{(c,\alpha(... ...\rho(s)ds+\pi i(c,\alpha(\sigma))\beta(\sigma) \rho(\sigma)= \end{displaymath}$

$\begin{displaymath} =c+\int\limits_0^\infty\bigg(c,\frac{\alpha(s)-\alpha(\sigm... ...^0(\sigma)=N(\sigma)c+ (c,\alpha(\sigma))\beta_\pm^0(\sigma), \end{displaymath}$

(3.1)

where the element $\displaystyle{\beta^0(\zeta)=\int\limits_0^\infty \frac{\beta(s)}{s-\zeta}\rho(s)ds=B\bigg(\frac{1}{s-\zeta}\bigg)}$ exists as value of a bounded operator B on the element $\displaystyle{ \frac{1}{s-\zeta}\in H}$ . In view of Lemma 2.3 it is natural to suppose that the operator $N(\sigma)$ is invertible.
Theorem 3.1 The resolvent of the maximal operator $\widetilde T_\sigma\equiv(\widetilde T-\sigma)^{-1}$ exists iff the operator $N(\sigma)$ is invertible. If the inverse operator $N(\sigma)^{-1}$ exists, then

$\begin{displaymath} \widetilde T_\sigma f=\widetilde S_\sigma f-\widetilde S_\sigma A^*N(\sigma)^{-1}B\widetilde S_\sigma f. \end{displaymath}$

Proof. Let the operator $N(\sigma)$ be invertible. It follows from the equation $(\widetilde T-\sigma)\varphi=f$ or $(\widetilde S-\sigma)\varphi+A^*B\varphi=f$ that $\varphi+\widetilde S_\sigma A^*B\varphi=\widetilde S_\sigma f$ . Now $(1+B{\cal R}_\sigma A^*)B\varphi=B\widetilde S_\sigma f$ , where ${\cal R}_\sigma A^*=\widetilde S_\sigma A^*$ , Let the operator $\widetilde T-\sigma$ , $\sigma\ge0$ , be invertible. If $N(\sigma)c=0$ , $c\ne0$ , i.e. $(1+B\widetilde S_\sigma A^*)c=0$ , then $\varphi=\widetilde S_\sigma A^*c\ne0$ and $(\widetilde T-\sigma)\varphi=(\widetilde S-\sigma)\widetilde S_\sigma A^*c+ A^*B\widetilde S_\sigma A^*c=A^*[c+B\widetilde S_\sigma A^*c]=0$ , what is impossible. Theorem is proved. We call $\sigma_0$ a pole of the function $f(s)/(s-\sigma_0)^m$ if $f\in C^{m+1}$ , $f(\sigma_0)\ne0$ .
Lemma 3.1 Let $\sigma_0$ be a pole of the function $\sigma\to N(\sigma)^{-1}$ , then the following representation takes place

$\begin{displaymath} N(\sigma)^{-1}d=\frac{N_{-m}d}{(\sigma-\sigma_0)^m}+\cdots+... ...a)d,\ \ \ d\in G,\ \ 0<\vert\sigma-\sigma_0\vert<\varepsilon, \end{displaymath}$

(3.2)

where $N_{-j}:\ G\to G$ , $j=1,\ldots,m$ are finite-dimensional bounded operators and $N_0(\sigma)$ is an operator-function continuous in the operator norm at the point $\sigma=\sigma_0$ .
Proof follows the proof of well-known Lemma about holomorphic operator function (see [4], p.37). In view of relation (3.1) and Lemma 2.3 the basis of the subspace $Z(N(\sigma_0))(R(N(\sigma_0))^\perp)$ are finite, we denote their by $e_1,\ldots,e_{n_0}(g_1,\ldots,g_{n_0})$ . If we introduce the operator

$\begin{displaymath} N_1(\sigma)=N(\sigma)+\sum_{k=1}^{n_0}(\cdot,e_k)g_k, \end{displaymath}$

then inverse operator $N_1(\sigma_0)^{-1}$ exists and is bounded. The operator function $\sigma\to N(\sigma)$ is continuons in the operator topology because the operator B is bounded and the operator function $\sigma\to{\cal R}_\sigma A^*c(s)=(c,{\cal R}_\sigma\alpha(s))$ is continuons too. The set of invertible operators is open in the operator topology, so, there exists $\varepsilon>0$ such that the operator $N_1(\sigma)$ is invertible if $\vert\sigma-\sigma_0\vert<\varepsilon$ . The equation $N(\sigma)c=d$ , or

$\begin{displaymath} N_1(\sigma)c-\sum_{k=1}^{n_0}(c,e_k)g_k=d,\ \ \ \vert\sigma-\sigma_0\vert<\varepsilon \end{displaymath}$

$\begin{displaymath} c-\sum_{k=1}^{n_0}(c,e_k)N_1(\sigma)^{-1}g_k= N_1(\sigma)^{-1}d,\ \ \ \vert\sigma-\sigma_0\vert<\varepsilon. \end{displaymath}$

(3.3)

Let $\Delta(\sigma)=\hbox{det\ }(\delta_{ik}-(N_1(\sigma)^{-1}g_k,e_i))$ . Multiplying equation (3.3) by $e_1,\ldots,e_{n_0}$ , we get a system of linear equation for the values $(c,e_1),\ldots,(c,e_{n_0})$ , from which

$\begin{displaymath} (c,e_k)=\frac{1}{\Delta(\sigma)}\sum_{i=1}^{n_0}(N_1(\sigma)^{-1}d,e_i) p_{ik}(\sigma), \end{displaymath}$

where $p_{ik}(\sigma)$ are some functions continuons at the point $\sigma_0$ . Representation (3.2) results now from (3.3). Lemma is proved.

The operator function ${\cal R}_\sigma$ satisfies the Hilbert identity. Therefore,

$\begin{displaymath} \frac{1}{(p-1)!}\bigg\{\bigg(\frac{d}{d\sigma}\bigg)^{p-1}\... ...bigg\}\bigg\vert_{\sigma=\sigma_0}={\cal R}_{\sigma_0}^p\psi, \end{displaymath}$

$\begin{displaymath} \psi(s)=\psi(\sigma_0)+\psi'(\sigma_0)(s-\sigma_0)+\cdots+\... ...s-\sigma_0)^{p-1}+(s-\sigma_0)^p{\cal R}_{\sigma_0}^p\psi. \end{displaymath}$

(3.4)

We consider the functions $\psi$ which admit 2m continuons derivatives at the point $\sigma_0$ . In view of A₂) and the representation

$\begin{displaymath} \widetilde S_{\sigma_0}A^*c(s)=\bigg(c,\frac{\alpha(s)-\alpha(\sigma_0)} {s-\sigma_0}\bigg) \end{displaymath}$

(3.5)

the vector-function $\widetilde T_{\sigma}\varphi$ is 2m times differentiable (see Theorem 3.1). We substitute $\psi(s)=(s-\sigma_0)^m \widetilde T_s\varphi$ , p=2m in relation (3.4). Then

$\begin{displaymath} \begin{array}{l} \displaystyle (s-\sigma_0)^m\widetilde T... ...\sigma_0)^{2m} {\cal R}_{\sigma_0}^{2m}\varphi,\end{array} \end{displaymath}$

(3.6)

$\begin{displaymath} A_k\varphi=\frac{1}{(k+m)!}\bigg\{\bigg(\frac{d}{ds}\bigg)^... ...phi)\bigg\}\bigg\vert_{s=\sigma_0};\ \ k=-m,-m+1,\ldots,m-1. \end{displaymath}$

If $A_{-1}\ne0$ , we call the operator A_-1 a generalized residue of the resolvent $\widetilde T_\sigma$ . The closure of the space of 2m times differentiable functions in the norm

$\begin{displaymath} \Vert\varphi\Vert _{D^m}^2=\Vert\varphi\Vert _H^2+\sum\limits_{k=0}^{m-1}\Vert{\cal R}_{\sigma_0}^k\varphi\Vert _H^2 \end{displaymath}$

(3.7)

will be denoted by D^m. According to Theorem 3.1, Lemma 3.1, and (3.5) the operators $A_{-j}:\ D^m\to D^m$ are defined as bounded finite-dimensional operators. Note that the space D^m is invariant with respect to $\widetilde T$ , i.e. if $\varphi\in D^m\cap D(\widetilde T)$ then $\widetilde T\varphi\in D^m$ . The operator $\widetilde T$ is closed and commutes with $\widetilde T_s$ , so $\widetilde TA_k\varphi=A_k\widetilde T_\varphi$ , $\varphi\in D^m\cap D(\widetilde T)$ .

Eigenvalue of maximal operator as normal eigen value embedded in continuous spectrum.

We give the translation of the proof of well known theorem on residue of the resolvent (see[5], §VIII, 8) adapted for the case of generalized pole. We note that every value $\zeta\not\in[0,\infty)$ is eigenvalue of the maximal operator $\widetilde T$ . So, we call $\sigma_0$ an isolated eigenvalue of $\widetilde T$ if $\sigma_0$ is isolated from other eigenvalues on half line $[0,\infty)$ (i.e. the continuous spectrum of T).
Theorem 4.1 If $\sigma_0\ge 0$ is an isolated eigenvalue and a pole of the resolvent of the maximal operator $\widetilde T$ then for some $m\ge1$ the space D^m is a direct sum of the following closed spaces:

$\begin{displaymath} D^m=Z((\widetilde T-\sigma_0)^m)\dot+R((\widetilde T-\sigma_0)^m). \end{displaymath}$

(4.1)

Proof. At the beginning let $\sigma_0\ge 0$ be an arbitrary number. We consider the relation $\widetilde S_{\sigma_0}(\widetilde T-\sigma_0)\varphi_n=\psi_n$ where the sequence $\{\psi_n\}$ converges in H. It follows from $\widetilde S_{\sigma_0}(\widetilde S-\sigma_0+V)\varphi_n=\psi_n$ that

$\begin{displaymath} \widetilde S_{\sigma_0}(\widetilde T-\sigma_0)\varphi_n\equiv\varphi_n+\widetilde S_{\sigma_0}A^*B\varphi_n=\psi_n \end{displaymath}$

(4.2)

and, multiplying by B, $N(\sigma_0)B\varphi_n=B\psi_n$ . In view of Lemma 3.1 we get

$\begin{displaymath} B\varphi_n+\sum\limits_{k=1}^{n_0}(B\varphi_n,e_k)N_1(\sigma_0)^{-1}g_k= N_1(\sigma_0)^{-1}B\psi_n. \end{displaymath}$

We choose a subsequence {n_i} such that the numerical sequences $(B\varphi_{n_i},e_k)$ converge for $k=1,2,\ldots,n_0$ . Then $\{B\varphi_{n_i}\}$ and, as result, $\{\varphi_{n_i}\}$ converge too. We will use below this property: existence of convergent subsequence $\{\varphi_{n_i}\}$ for relation (4.2). A relation more general is

$\begin{displaymath}\widetilde S_{\sigma_0}^k(\widetilde T-\sigma_0)^k\varphi=\v... ...detilde S_{\sigma_0}^jV(\widetilde T-\sigma_0)^{j-1}\varphi \end{displaymath}$

(4.3)

(proof by induction: multiply by $\widetilde S_{\sigma_0}$ on the left and by $\widetilde T-\sigma_0$ on the right). We get at once from (4.3) and the form V=A^*B that $R((\widetilde T-\sigma_0)^m)\subset D^m$ . We will prove that the subspace $R((\widetilde T-\sigma_0)^m)$ is closed in D^m. Let $\{\psi_n\}\subset R((\widetilde T-\sigma_0)^m)$ be some fundamental sequence in D^m. Then the sequence $\{\psi_n\}$ is fundamental in the space H too and it is sufficiently to verify that its limit in H appertains to the space $R((\widetilde T-\sigma_0)^m)$ . Denote (formally)

$\begin{displaymath} \psi_n\equiv(\widetilde T-\sigma_0)^m\varphi_n,\ \ \ \psi_... ...^i\varphi_n,\ \ \ \psi^{k,i}\equiv\lim\limits_n\psi_n^{k,i}. \end{displaymath}$

The convergence of $\{\psi_n\}\subset D^m$ signifies the convergence of $\{\psi_n^{\ell,m}\}\subset H$ for $\ell=0,1,\ldots,m$ if $n\to\infty$ . Due to (4.3) we have for $2\le\ell\le m$ :

$\begin{displaymath} \psi_n^{\ell,m}=\widetilde S_{\sigma_0}^\ell(\widetilde T-... ...e T-\sigma_0)^\ell(\widetilde T-\sigma_0)^{m-\ell}\varphi_n= \end{displaymath}$

$\begin{displaymath} =(\widetilde T-\sigma_0)^{m-\ell}\varphi_n+\sum\limits_{j=1... ...T-\sigma_0)^{j-1} (\widetilde T-\sigma_0)^{m-\ell}\varphi_n= \end{displaymath}$

$\begin{displaymath} =(1+\widetilde S_{\sigma_0}V)\psi_n^{0,m-\ell}+ \sum\limits_{j=2}^\ell\widetilde S_{\sigma_0}^jV\psi_n^{0,m-\ell+(j-1)}. \end{displaymath}$

If $\ell=1$ then $\psi_n^{1,m}=(1+\widetilde S_{\sigma_0}V)\psi_n^{0,m-1}$ , i.e. the relation under form (4.2). As $\{\psi_n^{1,m}\}$ converges if $n\to\infty$ then there exists a convergent subsequence $\{\psi_{n_i}^{0,m-1}\}$ . Let $\{\psi_{n}^{0,m-1}\}$ be convergent itself. In view of the relation $(\widetilde T-\sigma_0)\psi_n^{0,m-1}=\psi_n^{0,m}$ we get $(\widetilde T-\sigma_0)\psi^{0,m-1}=\psi^{0,m}$ (the operator $\widetilde T$ is closed). If $\ell=2$ then $\psi_n^{2,m}=(1+\widetilde S_{\sigma_0}V)\psi_n^{0,m-2}+\widetilde S_{\sigma_0}^2V\psi_n^{0,m-1}$ . The term $\widetilde S_{\sigma_0}^2V\psi_n^{0,m-1}$ converges because of (3.5) and due to the form V=A^*B, so, the term $(1+\widetilde S_{\sigma_0}V)\psi_n^{0,m-2}$ converges too. As above we find a convergent subsequence, let $\{\psi_n^{0,m-2}\}$ be convergent itself, then the relation $(\widetilde T-\sigma_0)\psi_n^{0,m-2}=\psi_n^{0,m-1}$ becomes $(\widetilde T-\sigma_0)\psi^{0,m-2}=\psi^{0,m-1}$ . By analogy we get a chain $\psi^{0,m},\ldots,\psi^{0,0}$ such that $(\widetilde T-\sigma_0)\psi^{0,k}=\psi^{0,k+1}$ . Now $\psi^{0,m}=(\widetilde T-\sigma_0)^m\psi^{0,0}$ and as result $\lim\limits_n\psi_n=\lim\limits_n\psi_n^{0,m}=\psi^{0,m}\in R((\widetilde T-\sigma_0)^m)$ . Therefore, the space $R((\widetilde T-\sigma_0)^m)$ is closed in D^m. Now let $\sigma_0\ge 0$ be a pole of the resolvent $\widetilde T_\sigma$ . Myltiplying $\widetilde T_\sigma$ by B and A^* and using the definition $N(\sigma)=1+B\widetilde S_\sigma A^*$ , we obtain the relation (see theorem 3.1) $B\widetilde T_\sigma A^* d=N(\sigma)d-d- (N(\sigma)-1)N(\sigma)^{-1}(N(\sigma)-1)d$ from which

$\begin{displaymath}N(\sigma)^{-1}d=d-B\widetilde T_\sigma A^* d,\quad d\in G. \end{displaymath}$

So, $\sigma_0$ is an isolated pole of $N(\sigma)^{-1}$ , i.e. the assumption of Lemma 3.1 holds. Let $N_{-j}=\sum\limits_\ell(\cdot,p_{j\ell})q_{j\ell}$ (see(3.2)), it follows from (3.6) that

$\begin{displaymath} A_{-j}\varphi=\sum C_{kir\ell}^j((B\widetilde S_\sigma\var... ...idetilde S_\sigma A^*q_{r\ell})^{(i)}\vert_{\sigma=\sigma_0}. \end{displaymath}$

So, $A_{-j}:\ D^m\to D^m$ is a finite-dimentional bounded operator. Due to the property $(\widetilde S_\sigma)'=\widetilde S_\sigma^2$ we have $R(A_{-j})\subset R(\widetilde S_{\sigma_0})\subset D(\widetilde S)=D(\widetilde T)$ . If we substitite (3.6) into the Hilbert identity $\widetilde T_\sigma\varphi-\widetilde T_\mu\varphi=(\sigma-\mu)\widetilde T_\sigma\widetilde T_\mu\varphi$ , we get $A_{-1}^2\varphi=-A_{-1}\varphi$ (as coefficient at $(\sigma-\sigma_0)^{-1}(\mu-\sigma_0)^{-1}$ ), i.e. $P\equiv-A_{-1}:\ D^m\to D^m$ is a finite-dimensional bounded projection. We will prove that (4.1) coincides with the following decomposition

$\begin{displaymath}D^m=R(P)\dot+Z(P). \end{displaymath}$

(4.4)

$\begin{displaymath} \varphi=[(\widetilde T-\sigma_0)-(s-\sigma_0)]\bigg[\frac{... ...)^{m-1}+(s-\sigma_0)^m{\cal R}_{\sigma_0}^{2m}\varphi\bigg]. \end{displaymath}$

$\begin{displaymath}\cases{ (\widetilde T-\sigma_0)A_{-m}\varphi=0\cr (\widetild... ...(\widetilde T-\sigma_0)A_{m-1}\varphi-A_{m-2}\varphi=0\cr}. \end{displaymath}$

(4.5)

These equations and further transformations follow [5, $\S VIII$ . 8]. In view of (4.5)

$\begin{displaymath} \varphi-P\varphi=\varphi+A_{-1}\varphi= (\widetilde T-\sigma_0)A_{0}\varphi=(\widetilde T-\sigma_0)A_{1}\varphi=\cdots \end{displaymath}$

$\begin{displaymath}=(\widetilde T-\sigma_0)^mA_{m-1}\varphi=A_{m-1}(\widetilde T-\sigma_0)^m\varphi. \end{displaymath}$

(4.6)

We have these relations for all elements $z\in Z((\widetilde T-\sigma_0)^m)$ . Indeed, if $(\widetilde T-\sigma_0)e_{n+1}=(\widetilde S-\sigma_0)e_{n+1}+Ve_{n+1}=e_n$ , $n=0,1,\ldots,e_0=0$ and $R(V)\subset C^{2m}$ then $z=e_n\in C^m$ . So, $z-Pz=A_{m-1}(\widetilde T-\sigma_0)^mz=0$ , from which $z=Pz\in R(P)$ i.e. $Z((\widetilde T-\sigma_0)^m)\subset R(P)$ . Inversely, if $\varphi\in R(P)$ then $\varphi=-A_{-1}\varphi$ and according to (4.5) $R(A_{-1})\subset D(\widetilde T^m)$ , $(\widetilde T-\sigma_0)^m\varphi=-(\widetilde T-\sigma_0)^mA_{-1}\varphi=\cdots=- (\widetilde T-\sigma_0)A_{-m}\varphi=0$ , i.e. $R(P)\subset Z((\widetilde T-\sigma_0)^m)$ . Finally,

$\begin{displaymath}R(P)=Z((\widetilde T-\sigma_0)^m). \end{displaymath}$

(4.7)

Let $\varphi\in R((\widetilde T-\sigma_0)^m)$ , we consider the decomposition $\varphi=\varphi_1+\varphi_2$ where $\varphi_1\in Z(P)$ , $\varphi_2\in R(P)$ (we remind that $R((\widetilde T-\sigma_0)^m)\subset D^m$ ). If $\psi\in C^{2m}\cap H$ then according to (4.6)

$\begin{displaymath} \psi-P\psi=(\widetilde T-\sigma_0)^mA_{m-1}\psi\in R((\widetilde T-\sigma_0)^m). \end{displaymath}$

We can choose $\psi_k\in C^{2m}\cap H$ such that $D^m-\lim\limits_{k}\psi_k=\varphi_1$ . The space $R((\widetilde T-\sigma_0)^m)$ is closed in D^m, therefore,

$\begin{displaymath} \varphi_1=\varphi_1-P\varphi_1=D^m-\lim\limits_{k}(\psi_k-P\psi_k) \in R((\widetilde T-\sigma_0)^m). \end{displaymath}$

Due to (4.7) $\varphi_2\in R(P)=Z((\widetilde T-\sigma_0)^m)$ , i.e. $(\widetilde T-\sigma_0)^m\varphi_2=0$ . More again we have $\varphi_2=\varphi-\varphi_1\in R((\widetilde T-\sigma_0)^m)$ or $\varphi_2=(\widetilde T-\sigma_0)^m\psi$ for some element $\psi$ . So, $(\widetilde T-\sigma_0)^{2m}\psi=0$ . Taking into account the relation

$\begin{displaymath}Z((\widetilde T-\sigma_0)^{m+k})=Z((\widetilde T-\sigma_0)^m),\ \ \ k=1,2,\ldots \end{displaymath}$

(4.8)

we get $(\widetilde T-\sigma_0)^m\psi=0$ , or $\varphi_2=0$ , $\varphi=\varphi_1\in Z(P)$ . Finally $R((\widetilde T-\sigma_0)^m)\subset Z(P)$ . The inverse inclusion follows from the equality $(\widetilde T-\sigma_0)^mA_{m-1}=1+A_{-1}=1-P$ as $Z(P)=R(1-P)\subset R((\widetilde T-\sigma_0)^m)$ . So, $R((\widetilde T-\sigma_0)^m)=Z(P)$ and (4.1) follows from (4.4), (4.7). It remains to prove (4.8), or the inclusion

$\begin{displaymath} Z((\widetilde T-\sigma_0)^{m+k})\subset Z((\widetilde T-\sigma_0)^m), \end{displaymath}$

(4.9)

as the inverse inclusion is trivial. Suppose that $(\widetilde T-\sigma_0)^{m+k}\varphi=0$ but $(\widetilde T-\sigma_0)^m\varphi=y\ne0$ . Let $\ell\ge0$ be the maximal number such that $(\widetilde T-\sigma_0)^\ell y=z\ne0$ . It follows from $(\widetilde T-\sigma_0)z=0$ that $\widetilde T_\sigma z=\frac{z}{\sigma_0-\sigma}$ . Further, it follows from $(\widetilde T-\sigma_0)\varphi_1=z$ that $(\widetilde T-\sigma)\varphi_1+(\sigma-\sigma_0)\varphi_1=z$ , or

$\begin{displaymath} \widetilde T_\sigma\varphi_1=\frac{\varphi_1}{\sigma_0-\sig... ... \frac{\varphi_1}{\sigma_0-z}-\frac{z}{(\sigma_0-\sigma)^2}. \end{displaymath}$

We choose $\varphi_2$ such that $(\widetilde T-\sigma_0)\varphi_2=\varphi_1$ etc. In view of $(\widetilde T-\sigma_0)^{m+\ell}\varphi=z$ we obtain finally the function $\sigma\to\widetilde T_\sigma\varphi_{m+\ell}$ such that the order of the pole is not less than $m+\ell+1$ , what is impossible. Therefore y=0, $(\widetilde T-\sigma_0)^m\varphi=0$ i.e. $\varphi\in Z((\widetilde T-\sigma_0)^m)$ , what proves (4.9). Theorem is proved.

Example of normal eigenvalue embedded in continuous spectrum.

$\begin{displaymath} -y''+q(x)y(x-\Delta)-\sigma y(x)=z(x),\quad x>0,\quad \Delta\geq0 \end{displaymath}$

(5.1)

in the space $L^2(0,\infty)$ if $z\in L^2(0,\infty)$ . Suppose that $y(x-\Delta)\equiv0$ , $0<x<\Delta$ and the function q(x) decreases exponentialy in $(0,\infty)$ . With a view to construct an auxiliary model T=S+V we introduce some boundary condition, for example y(0)=0. The operator L generated by the expression $\ell y=-y''$ , y(0)=0 is equivalent to the operator $S:\ L_\rho^2(0,\infty)\to L_\rho^2(0,\infty)$ , $\rho(s)=\frac{1}{\pi}\sqrt s$ namely $S={\cal F}L{\cal F}^{-1}$ where

$\begin{displaymath} {\cal F}z(s)=\int\limits_0^\infty z(x)\frac{\sin x\sqrt s}{\sqrt s}dx. \end{displaymath}$

It is easy to verity that $\widetilde S={\cal F}L_{max}{\cal F}^{-1}$ , i.e. the equation

$\begin{displaymath} -y''-\sigma y=z,\ \ \ y\in D(L_{max}) \end{displaymath}$

(5.2)

is equivalent to the equation $(\widetilde S-\sigma)\psi=\varphi$ , $\psi\in D(\widetilde S)$ . The maximal operator is closed, so, if the function ${\cal F}z(s)$ is continuous in $\sigma$ and if

$\begin{displaymath}({\cal F}z(s)-{\cal F}z(\sigma))/(s-\sigma)\in L_{\rho}^2(0,\infty), \end{displaymath}$

(5.3)

then equation (5.2) has a solution $y(x)={\cal F}^{-1} (({\cal F}z(s)-{\cal F}z(\sigma))/(s-\sigma))(x)$ . Let $Qy(x)=\chi_{[\Delta,\infty]}(x)q(x)y(x-\Delta)$ , we will choose the operators A, B such that $A^*B={\cal F}Q{\cal F}^{-1}$ . Let $G=L^2([0,\infty)\cap\hbox{supp\ }q)$ , $q(x)=\overline{q_1(x)}q_2(x)$ , |q₁(x)|=|q₂(x)| and $(\alpha(s))(x)=\chi_{[\Delta,\infty)}(x)q_1(x)\times$ $\times (\sin x\sqrt s)/\sqrt s$ , $(\beta(s))(x)=\chi_{[\Delta,\infty)}(x)q_2(x) (\sin (x-\Delta)\sqrt s)/\sqrt s$ . Then the conditions A₁)-A₄) are trivial, the condition A₅) results from the properties of the operator ${\cal F}$ . As $A^*c(s)=(c,\alpha(s))$ we have

$\begin{displaymath} (N(\sigma)c)(x)=c(x)+(B\widetilde S_\sigma A^*c)(x)=c(x)+ \end{displaymath}$

$\begin{displaymath} +\frac{1}{\pi}q_2(x)\int\limits_\Delta^\infty c(y)\overline... ...\sigma}}{\sqrt{\sigma}}\bigg)\sin(x-\Delta)\sqrt sds\bigg)dy. \end{displaymath}$

Let M=L+Q and $\widetilde T={\cal F}M_{max}{\cal F}^{-1}$ . As q(x) decreases exponentially, i.e. $\vert q(x)\vert<C e^{-\varepsilon x}$ , x>0, $\varepsilon>0$ , the operator function $N(\sigma)$ admits analytical extension in some neighbourhood of half line $[0,\infty)$ . In view of Lemma about holomorphic operator function (mentioned above) and Lemma 2.3 we obtain from Theorem 4.1 the following
Corollary 5.1. If the potential q(x) satisfies the condition $\vert q(x)\vert<C e^{-\varepsilon x}$ , x>0, $\varepsilon>0$ , then the set of eigenvalues $\sigma$ , i.e. such $\sigma$ that the homogenous equation

$\begin{displaymath} -y^{\prime\prime}(x)+q(x)y(x-\Delta)-\sigma y(x)=0,\quad y\in D(M_{max}), \quad \Delta\geq0 \end{displaymath}$

(5.4)

has the nontrivial solution $y\in L^2(0,\infty)$ is finite. a) If $\sigma$ is an eigenvalue then non homogenous equation (5.1) has a solution $y\in L^2(0,\infty)$ if 1) ${\cal R}_\sigma^j{\cal F}z\in H=L^2_\rho(0,\infty)$ , $j=0,1,\ldots,m$ , for some $m\ge1$ and 2) for a finite set of some elements $f_{jk}\in H$ the conditions $({\cal R}_\sigma^j{\cal F}z,f_{jk})_h=0$ hold. b) If $\sigma$ is not an eigenvalue then non homogeneous equation (5.1) has a solution $y\in L^2(0,\infty)$ if the condition (5.3) holds.
Proof. Proposition b) follows from the relation $D(\widetilde T_\sigma)=D(\widetilde S_\sigma)$ (see Theorem 3.1) and the equality $\widetilde S_\sigma\varphi={\cal R}_\sigma\varphi$ for smooth elements $\varphi$ .

Example. Let us give a simple example when equation (5.4) has a non trivial solution. Let $\Delta>0$ and

$\begin{displaymath} y(x)=\cases{ C_1\cos\sqrt\sigma x+C_2\sin\sqrt\sigma x, & $0<x<\Delta$\cr e^{-x^2}, & $x>\Delta$\cr}, \end{displaymath}$

where C_1,2 are such that y(x), y'(x) are continuous at the point $\Delta$ . We choose arbitrary $\sigma>0$ . The function q(x) is defined in the intervals $(\Delta,2\Delta)$ , $(2\Delta,\infty)$ directly from (5.4), in particular,

$\begin{displaymath} q(x)=(\sigma+2-4x^2)e^{-2\Delta x+\Delta^2},\ \ \ x>2\Delta, \end{displaymath}$

therefore q(x) decreases exponentialy if $x\to\infty$ . One can consider y(x)=e^-kx², k>0, too. So, y(x) is an eigen function. Note that for the search of eigenvalues of the maximal operator we can use the equation $N(\sigma)c=0$ as well directly equation (5.1). Practically, to use the Corollary 5.1 one must solve integral equation $N(\sigma)c=d$ and calculate generalized residue of $\widetilde T_\sigma \varphi$ . We won't make it for any particular potential. However, we continue our consideration to obtain (in addition to Theorem 4.1) the equation for eigenvalues too, taking into account that these eigenvalues are embedded in the continuous spectrum. Let $\Delta>0$ . We use the method of successive integration (see [6,p.17]). Denote $I_k=((k-1)\Delta,k\Delta)\equiv(x_{k-1},x_k)$ . For $x\in I_k$ we use the substitution $x=x_{k-1}+\tau$ , $0<\tau<\Delta$ and the notation $y_k(\tau)=y(x_{k-1}+\tau)$ , $k=1,2,\ldots$ We obtain the equations (see (5.4))

$\begin{displaymath} y_k''(\tau)+\sigma y_k(\tau)=f_k(\tau),\ \ \ 0<\tau<\Delta \end{displaymath}$

(5.5)

where $f_k(\tau)=q(x_{k-1}+\tau)y_{k-1}(\tau)$ , $k=1,2,\ldots$ , $y_0(\tau)\equiv0$ . Using the coincidence of the following values

$\begin{displaymath} \cases{ y_{k-1}(\Delta)=y_k(0)\cr y_{k-1}'(\Delta)=y_k'(0)\cr} \end{displaymath}$

and the notation $a_k=y_k(\Delta)$ , $b_k=\frac{1}{\sqrt\sigma}y_k'(\Delta)$ , we obtain from (5.5) for $k=2,3,\ldots$

$\begin{displaymath}\cases{ \displaystyle{y_k(\tau)=-\frac{1}{\sqrt\sigma}\int\l... ...-a_{k-1}\sin\sqrt\sigma\tau+ b_{k-1}\cos\sqrt\sigma\tau}.\cr} \end{displaymath}$

(5.6)

Note that a₁, b₁ are arbitrary constants for the solution of (5.5). Let us introduce the space ${\cal H}_0=C(\Delta,{\mbox{\bf C}}^2)$ containing the elements

$\begin{displaymath} u_k(\tau)=\pmatrix{ y_k(\tau)\cr \frac{1}{\sqrt\sigma}y_k'(\tau)\cr},\ \ \ 0<\tau<\Delta \end{displaymath}$

with the norm $\Vert u_k(\cdot)\Vert _0=\sup\limits_\tau\Big[\vert y_k(\tau)\vert^2+ \frac{1}{\sigma}\vert y_k'(\tau)\vert^2\Big]^{1/2}$ and the space ${\cal H}={\cal H}_0\stackrel{\cdot}{+}{\mbox{\bf C}}^2$ containing the elements

$\begin{displaymath} u_k=\pmatrix{ u_k(\tau)\cr u_k(\Delta)\cr} \end{displaymath}$

with the norm $\Vert u_k\Vert=\max\Big[\Vert u_k(\cdot)\Vert, (\vert a_k\vert^2+\vert b_k\vert^2)^{1/2}\Big]$ . We introduce in the space ${\cal H}_0$ the operators

$\begin{displaymath} G_0(\tau)=-\frac{1}{\sqrt\sigma}\pmatrix{ \sin\sqrt\sigma\... ...sigma\tau\cr -\sin\sqrt\sigma\tau & \cos\sqrt\sigma\tau\cr}. \end{displaymath}$

$\begin{displaymath}\cases{ \displaystyle{u_k(\tau)=\int\limits_0^\tau q(x_{k-1}... ...}+t)G_0(\Delta-t) u_{k-1}(t)dt+F(\Delta)u_{k-1}(\Delta)}.\cr} \end{displaymath}$

(5.7)

$\begin{displaymath} F=\pmatrix{ 0 & F(\tau) \cr 0 & F(\Delta)\cr}:\ {\cal H}\to{\cal H} \end{displaymath}$

and by $G_k:\ {\cal H}\to{\cal H}$ the transformation given by the integrals in (5.7), then

Obviously, $\Vert F\Vert\le1$ . We shall construct the equation for eigenvalues if the value $\Delta$ is sufficiently small only. More exactly we suppose (due to the properties of q(x)) that $\Vert G_k\Vert\le\theta^k$ , $k=1,2,\ldots,0<\theta<1$ and

$\begin{displaymath} g\equiv\sum\limits_{k=2}^\infty\Vert G_k\Vert\le\theta^2/(1-\theta)<1. \end{displaymath}$

$\begin{displaymath}v_k=u_k-Fu_{k-1},\ \ \ k=2,3,\ldots,\ \ \ v_1=u_1 \end{displaymath}$

(5.8)

v_k=G_ku_k-1.

(5.9)

$\begin{displaymath} v_k+\sum\limits_{p=2}^{k-1}F^{k-p}v_p=u_k-Fu_{k-1}+ \sum\l... ...}u_p- \sum\limits_{p=2}^{k-1}F^{k-p+1}u_{p-1}=u_k-F^{k-1}u_1 \end{displaymath}$

$\begin{displaymath}u_k=v_k+\sum\limits_{p=1}^{k-1}F^{k-p}v_p= \sum\limits_{p=1}^{k}F^{k-p}v_p. \end{displaymath}$

(5.10)

$\begin{displaymath} v_k=G_k\bigg(\sum\limits_{p=1}^{k-1}F^{k-p-1}v_p\bigg), \end{displaymath}$

$\begin{displaymath} \Vert v_k\Vert\le\Vert G_k\Vert\sum\limits_{p=1}^{k-1}\Vert v_p\Vert \end{displaymath}$

(5.11)

$\begin{displaymath} \sum\limits_{k=2}^N\Vert v_k\Vert\le \sum\limits_{k=2}^N\V... ...=1}^{N-1} \Vert v_p\Vert\sum\limits_{k=p+1}^N\Vert G_k\Vert. \end{displaymath}$

$\begin{displaymath} (1-g)\sum\limits_{k=2}^N\Vert v_k\Vert\le g\Vert v_1\Vert \end{displaymath}$

and $\sum\limits_{k=2}^\infty\Vert v_k\Vert<\infty$ . Now we obtain from (5.11) that $\Vert v_k\Vert\le C\theta^k$ , $k=1,2,\ldots$ As

$\begin{displaymath}F^n=\pmatrix{ 0 & F(\tau)F(\Delta)^{n-1} \cr 0 & F(\Delta)^n... ... = \pmatrix{ 0 & F(\tau)F((n-1)\Delta) \cr 0 & F(n\Delta)\cr},\end{displaymath}$

$\begin{displaymath} u_k(\tau)=v_k(\tau)+F(\tau)\sum\limits_{p=1}^{k-1}F((k-p)\Delta)F(\Delta)v_p(\Delta) \end{displaymath}$

$\begin{displaymath} u_k(\tau)=\widetilde v_0(\tau)+F(k\Delta+\tau)\sum\limits_{... ...infty F(-p\Delta)F(\Delta)v_p(\Delta),\,\,\,\,x=k\Delta+\tau \end{displaymath}$

we can pose $\Delta\to0$ , then $y(x)=\widetilde y(x)+a\cos x\sqrt{\sigma}+ b\sin x\sqrt{\sigma}$ where $\widetilde y\in L^2(0,\infty)$ ). Let E and $\overline{E}$ be respectively a matrix and its complex conjugate such that

$\begin{displaymath} F(\tau)=e^{i\tau\sqrt{\sigma}}E+e^{-i\tau\sqrt{\sigma}}\overline{E}. \end{displaymath}$

Denote $w_p(\sigma)\equiv F(\Delta)v_p(\Delta)$ (values (5.8) depend on $\sigma$ ), then

$\begin{displaymath} u_k(\tau)=v_k(\tau)+F(\tau)\Big[e^{ik\Delta\sqrt\sigma} \s... ...}^{k-1} e^{ip\Delta\sqrt\sigma}\overline{E}w_p(\sigma)\Big]. \end{displaymath}$

$\begin{displaymath}\pmatrix{P_1(\sigma)\cr P_2(\sigma)} =\sum\limits_{p=1}^{\in... ...igma) &Q_1(\sigma)\cr P_2(\sigma) &Q_2(\sigma)}\right\vert, \end{displaymath}$

$\begin{displaymath} u_k(\tau)=\widetilde u_k(\tau)+F(\tau)\Big[e^{ik\Delta\sqrt... ...Delta\sqrt\sigma} \pmatrix{Q_1(\sigma)\cr Q_2(\sigma)}\Big], \end{displaymath}$

where $\sum\limits_{k=1}^\infty\Vert\widetilde u_k(\cdot)\Vert^2_0<\infty$ . So, the following theorem is proved. Theorem 5.1 Suppose q(x) is expotentially decreasing function than the homogen equation (5.4) where $\sigma>0$ has a solution with following asymtotic behaviour

$\displaystyle \left(\begin{array}{cc}y\big((k-1)\Delta+\tau\big)\\ \frac1{\sqr... ...eft(\begin{array}{cc}Q_1(\sigma)\\ Q_2(\sigma)\end{array}\right)\right] +o(1),$

$\displaystyle \qquad\qquad\qquad 0<\tau<\Delta,\quad k\to\infty\qquad$

(5.12)

uniformly for $\tau\in[0,\Delta]$ , $0<\Delta<\Delta_0$ for some $\Delta_0>0$ . The coefficients $P_{1,2}(\sigma)$ , $Q_{1,2}(\sigma)$ are continuous in $(0,\infty)$ . The determinant $D(\sigma)$ depends on a₁, b₁ (see (5.5)-(5.6)). If we introduce some operator $M_\theta\subset M_{\mbox{\rm max}}$ with the condition $y^\prime(0)-\theta y(0)=0$ (see (5.4)), then $D(\sigma)$ is determined up to a constant factor. If $D(\sigma)\ne0$ , considering that the matrix $F(\tau)$ is unitary, we have

$\begin{displaymath} \Vert u_k(\cdot)-\widetilde u_k(\cdot)\Vert _0=\left\Vert e... ... \equiv \eta(\alpha)\ne0, \quad \alpha=k\Delta\sqrt{\sigma}. \end{displaymath}$

The function $\eta(\alpha)$ is continuous and periodic, therefore there exists a number $\eta_0>0$ such that $\eta(\alpha)\ge\eta_0$ , $\alpha\in[0,\infty)$ . Then $\sum\limits_{k=1}^\infty\Vert u_k(\cdot)-\widetilde u_k(\cdot)\Vert _0^2=\infty$ and $y\not\in L^2(0,\infty)$ . Finally, we can state the following corollary:
Corollary 5.2 a) if $\sigma$ is an eigenvalue of the operator $M_\theta$ , then $D(\sigma)=0$ ; b) if $P_1(\sigma)=P_2(\sigma)=Q_1(\sigma)=Q_2(\sigma)=0$ , then $\sigma$ is an eigenvalue of the operator $M_\theta$ . So, the corollary 5.1 shows that the analogy of (0.1) for the spectral singularities i.e. (4.1) may be used for analysis of differential equations with deviation.

Conclusion.

The answer for the posed question is: in the case of spectral singularity within Friedrichs' model we have the analog (see (4.1)) of the decomposition (0.1), but here we must extend the domain of definition of the operator by one element and change the norm of the space. In the case of Sturm-Liouville's operator the relation (4.1) gives sufficient and necessaire conditions of solvability of the equation $-y''(x)+q(x)y(x-\Delta)-\sigma y=f$ , $\sigma>0$ , $\Delta\geq0$ in the space $L^2(0,\infty)$ under the condition that homogene equation has a non-trivial solution. The element $f\in L^2(0,\infty)$ is supposed to have a smooth Fourier transformation. Additionally, for homogen equation $-y''(x)+q(x)y(x-\Delta)-\sigma y=0$ , $\sigma>0$ , $\Delta>0$ the asymptotic behaviour (5.12) of the solution is obtained. The function of the operators (defined in connection with (5.4)) may be described and estimated by the method of [7] where the model T=S+V is considered.

Bibliography

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The translation was initiated by root on 2001-11-13

$\displaystyle \left(\begin{array}{cc}y\big((k-1)\Delta+\tau\big)\\ \frac1{\sqr... ...eft(\begin{array}{cc}Q_1(\sigma)\\ Q_2(\sigma)\end{array}\right)\right] +o(1),$
$\displaystyle \qquad\qquad\qquad 0<\tau<\Delta,\quad k\to\infty\qquad$			(5.12)