\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 67, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/67\hfil Boundedly solvable delay differential operators]
{Boundedly solvable extensions of delay differential operators}

\author[B. \"O. G\"uler, B. Y{\i}lmaz, Z. I. Ismailov \hfil EJDE-2017/67\hfilneg]
{Bahad{\i}r \"O. G\"uler, B{\"u}lent Y{\i}lmaz, Zameddin I. Ismailov}

\address{Bahad{\i}r \"O. G\"uler \newline
Department of Mathematics,
Karadeniz Technical University, Turkey}
\email{boguler@ktu.edu.tr}

\address{B\"ulent Y{\i}lmaz \newline
Department of Mathematics,
Marmara University, Turkey}
\email{bulentyilmaz@marmara.edu.tr}

\address{Zameddin Ismailov \newline
Department of Mathematics,
Karadeniz Technical University, Turkey}
\email{zameddin.ismailov@gmail.com}

\dedicatory{Communicated by Ludmila S. Pulkina}

\thanks{Submitted January 10, 2017. Published March 6, 2017.}
\subjclass[2010]{47A20, 47B38}
\keywords{Delay differential expression; boundedly solvable operator}

\begin{abstract}
 We describe all boundedly solvable extensions of minimal operators 
 generated by first-order linear delay differential operators
 in Hilbert spaces of vector-functions on finite intervals.
 Also, we study the structure of spectrum of these  extensions.
 To do this we use methods from operator theory.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks


\section{Introduction}

It is known that many solvability problems arising in life sciences can be
expressed as boundary value problems for linear functional
(time delay, time proportional, neutral, advanced etc.) equations in
corresponding functional spaces. The general theory of linear functional equations
can be found in \cite{b1,e1,h1}.

The solvability of the considered problems may be seen as boundedly solvability
of  linear differential operators in corresponding functional Banach spaces.
Note that the theory of boundedly solvable extensions of a linear densely
defined closed operator in Hilbert spaces was presented in the important
 works of  Vishik in \cite{v1,v2}.

Let us recall that an operator $S:D(S)\subset H\to H$ on any Hilbert space $H$
is called boundedly solvable, if $S$ is one-to-one and onto, and $S^{-1}\in L(H)$.

The main aim of this work is to describe of all boundedly solvable extensions
of the minimal operator generated by first-order linear delay differential-operator
expression in the Hilbert space of vector-functions
at finite interval in terms of boundary conditions.
Lastly, the structure of spectrum of these extensions will be investigated.

\section{Description of solvable extensions}

In the Hilbert space $L^2(H,(a,b)), a,b\in\mathbb{R}$ of H-valued vector-functions
consider the  linear delay differential-operator expression of first order
in form
\begin{equation} \label{e2.1}
l(u)=(\alpha(t)u(t))'+A(t)u(t-\tau)
\end{equation}
where:
\begin{itemize}
\item[(1)] $H$ is a separable Hilbert space;
\item[(2)]  the function $\alpha:[a,b]\to \mathbb{R}_{+}$ is  Lebesgue measurable;
\item[(3)] there are  positive reel numbers $c$ and $C$ such that for $x\in[a,b]$,
\[
c\leq\alpha(x)\leq C;
\]

\item[(4)] the operator-function $A(\cdot):[a,b]\to L(H)$ is continuous on the
uniform operator topology;

\item[(5)] $\frac{\|A(t)\|}{\alpha(t)}\in L^1(H,(a,b))$;
\item[(6)] $0\leq\tau<b-a$.
\end{itemize}
On the other hand, we shall consider the  differential expression
\begin{equation} \label{e2.2}
m(\cdot)=d/dt
\end{equation}
in the Hilbert space $L^2(H,(a,b))$ corresponding to \eqref{e2.1}.
 Using the standard way, the minimal $M_0$ and the maximal $M$ operators generated
by differential expression \eqref{e2.2} can be defined (see \cite{h2}).

Now we define an operator $S_{\tau}:L^2(H,(a,b))\to L^2(H,(a,b))$ by
\[
S_{\tau}u(t):=\begin{cases}
0,&\text{if }  a<t<a+\tau,\\
u(t-\tau), &\text{if }  a+\tau<t<b.
\end{cases}
\]
for $u\in L^2(H,(a,b))$.
It is clear that $S_{\tau}\in L(L^2(H,(a,b)))$ and $\|S_{\tau}\|= 1$.


We also define the minimal $L_0$ and the maximal $L$ operators corresponding
to differential-operator expression
\[
l(u)=(\alpha(t)u(t))'+A(t)S_{\tau}u(t)
\]
in $L^2(H,(a,b))$  (see \cite{h2}).

By $U(t,s)$ with $t,s \in [a,b]$ we denote the  family of evolution operators
corresponding to the homogeneous differential operator equation
\begin{gather*}
\frac{\partial}{\partial t}U(t,s)f+\frac{A(t)S_{\tau}}{\alpha(t)}U(t,s)f=0,\quad
  t,s \in [a,b],\\
U(s,s)f=f, \quad  f\in H.
\end{gather*}
The operator $U(t,s)$  is linear, continuous boundedly invertible and
$$
U^{-1}(t,s)=U(t,s), \quad  t,s \in [a,b]\,.
$$
For a detail analysis see \cite{k1}.

Now we introduce the following operators:
\begin{gather*}
Uz(t):=U(t,0)z(t), \\
Vz(t):=\frac{1}{\alpha(t)}Uz(t), \\
U,V:L^2(H,(a,b))\to L^2(H,(a,b)).
\end{gather*}
In this case it is easy to check that
\begin{align*}
l(Vz)
& = (\alpha Vz)'(t)+A(t)S_{\tau}Vz(t)\\
& = (Uz(t))'+\frac{A(t)S_{\tau}}{\alpha(t)}Uz(t)\\
& = Uz'(t)+(U_t'+\frac{A(t)S_{\tau}}{\alpha(t)}U)z(t) \\
& = Uz'(t)=Um(z)
\end{align*}
Therefore,
\[
U^{-1}l(Vz)=m(z).
\]
Hence it is clear that if $\widetilde{L}$ is some extension of the minimal
operator $L_0$, that is, $L_0\subset\widetilde{L}\subset L$.
Then
\begin{gather*}
U^{-1}L_0V=M_0, \\
M_0\subset U^{-1}\widetilde{L}V=\widetilde{M}\subset M, \\
U^{-1}LV=M.
\end{gather*}
Now we prove the following assertions.

\begin{theorem} \label{thm2.1}
$ \ker L_{0}=[ 0 ] $ and $\overline{\operatorname{Im}(L_{0})}\neq L^2(H,(a,b))$.
\end{theorem}

\begin{proof} If for any $u\in D(L_0)$
\[
L_0u=0,
\]
then from the relation $U^{-1}L_0V=M_0$ it is obtained that $UM_0V^{-1}(u)=0$.
From last equation $M_0V^{-1}(u)=0$.
Since $\ker M_0={0}$, then $V^{-1}u(t)=0$,
Consequently $u=0$. So $\ker L_0={0}$.
To prove the relation $\overline{\operatorname{Im}(L_{0})}\neq L^2(H,(a,b))$,
consider the subspace
$\ker(L_0^\ast)$ in $L^2(H,(a,b))$.

In this case it is clear that the differential equation
\begin{align*}
L_0^\ast u(t)&=(UM_0V^{-1})^\ast u(t)
  =(V^{-1})^\ast M_0^\ast U^\ast u(t) \\
&=-(V^{-1})^\ast(U^\ast z(t))'=0
\end{align*}
has  solution of the form
\[
U^\ast u(t)=g, \quad  g\in H
\]
that is,
\[
u(t)=(U^\ast)^{-1}g, \quad  g\in H, \; t\in (a,b)
\]
This shows that
\[
\ker(L_0^\ast)\neq{0}
\]
From this and the relation
\[
\operatorname{Im}(L_0)\oplus \ker(L_0^\ast)=L^2(H,(a,b)),
\]
we obtain that
\[
\overline{\operatorname{Im}(L_{0})}\neq L^2(H,(a,b)).
\]
\end{proof}

\begin{theorem} \label{thm2.2}
For the domains of minimal $L_0$ and the maximal $L$ operators
\begin{align*}
D(L)=\{u\in L^2(H,(a,b)): \alpha u\in {W}_2^{1}(H,(a,b))\}  \ \text{and}\\
D(L_0)=\{u\in D(L): \lim_{t \to a^+}(\alpha u)(t)=\lim_{t \to b^-}(\alpha u)(t)=0\}
\end{align*}
respectively.
\end{theorem}

\begin{proof}
 First of all note that for any $u\in D(L)$, from the relations
$U^{-1}L_0V=M_0$ and $ULV=M$ we obtain
\[
V^{-1}u\in D(M_0), \quad V^{-1}u\in D(M_0)
\]
and vice versa.


From these facts and the relations
\begin{gather*}
D(M_0)=\mathaccent"7017 W_2^{1}(H,(a,b)),\\
D(M)={W}_2^{1}(H,(a,b)),
\end{gather*}
the validity of assertion is obtained.
\end{proof}


\begin{theorem} \label{thm2.3}
Each solvable extension $\widetilde{L}$ of the minimal operator $L_{0}$ in
$L^2(H,(a,b))$ is generated by the differential-operator expression \eqref{e2.1}
and the boundary condition
\begin{equation} \label{e2.3}
 (K+E)(\alpha u)(a)=KU(a,b)(\alpha u)(b),
\end{equation}
where $K\in L(H)$, $E$ is the identity operator in $H$ and
$(\alpha u)(a)=\lim_{t \to a^+}(\alpha u)(t)$,
$(\alpha u)(b)=\lim_{t \to b^-}(\alpha u)(t)$. The operator $K$ is determined
 uniquely by the extension $\widetilde{L}$, i.e $\widetilde{L}=L_{K}$.

  On the contrary, the restriction of the maximal operator $L_{0}$ to the manifold
of vector-functions satisfy the condition \eqref{e2.3} for some bounded
operator $K\in L(H)$ is a boundedly solvable extension of the minimal operator
$L_{0}$ in $L^2(H,(a,b))$.
\end{theorem}

\begin{proof}
Firstly, all boundedly solvable extensions $\widetilde{M}$ of the minimal operator
$M_{0}$ in $L^2(H,(a,b))$ are described in terms of boundary conditions.

Consider the  so-called Cauchy extension $M_{c}$, $M_{c}u=u'(t)$,
\begin{gather*}
M_{c}: D(M_{c}) \to L^{2}(H,(a,b)), \\
D(M_{c})=\{u\in W_2^{1}(H,(a,b)):u(0)=0\}\subset L^{2}(H,(a,b)),
\end{gather*}
of the minimal operator $M_{0}$. It is clear that $M_{c}$ is a solvable extension
of $M_{0}$ and
\begin{gather*}
M_{c}^{-1}f(t)=\int_{a}^{t}f(x)dx, \quad f\in L^{2}(H,(a,b)), \\
M_{c}^{-1}: L^{2}(H,(a,b))\to L^{2}(H,(a,b)).
\end{gather*}
 Now assume that $\widetilde{M}$ is a solvable extension of the minimal
operator $M_{0}$ in $L^{2}(H,(a,b))$. In this case it is known that the domain
of $\widetilde{M}$ can be written as a direct sum
\[
D(\widetilde{M})=D(M_{0}) \oplus (M_{c}^{-1}+K)V,
\]
where $ V=\ker M=H, K\in L(H) $  (see \cite{v2}).
Therefore for each $u(t)\in D(\widetilde{M})$ the following is true
\[
u(t)=u_{0}(t)+M_{c}^{-1}f+Kf, \quad  u_{0}\in D(M_{0}), \quad  f\in H.
\]
That is,
\[
u(t)=u_{0}(t)+tf+Kf, \quad  u_{0}\in D(M_{0}),\quad f\in H.
\]
Hence
\[
u(0)=Kf, \quad  u(1)=f+Kf=(K+E)f
\]
and from these relations it is obtained that
\begin{equation} \label{e2.4}
(K+E)u(a)=Ku(b).
\end{equation}
On the other hand, the uniqueness of the operator $K\in L(H)$ is clear from
the work in \cite{v2}. Therefore $\widetilde{M}=M_{K}$. This completes of necessary
part of this assertion.

 On the contrary, if $M_{K}$ is a operator generated by differential expression
\eqref{e2.2} and boundary condition \eqref{e2.4}, then $M_{K}$ is boundedly
invertible and
\begin{gather*}
M_{K}^{-1}:L^{2}(H,(a,b))\to L^{2}(H,(a,b)), \\
M_{K}^{-1}f(t)=\int_{a}^{t}f(x)dx
+K\int_{a}^{b}f(x)dx, \quad  f\in L^{2}(H,(a,b)).
\end{gather*}

Consequently, all solvable extensions of the minimal operator $M_{0}$
in $L^{2}(H,(a,b))$ is generated by the differential expression \eqref{e2.2}
and the boundary condition \eqref{e2.4} for any linear bounded operator $K$.


The extension $\widetilde{L}$ of the minimal operator $L_{0}$ is solvable in
 $L^{2}(H,(a,b))$ if and only if the operator $\widetilde{M}=U^{-1}\widetilde{L}V$
is an extension of the minimal operator $M_{0}$ in $L^{2}(H,(a,b))$.
Then $u\in D(\widetilde{L}$) if and only if
\[
V^{-1}u\in D(\widetilde{M}),
\]
Hence there exists $K\in L(H)$ such that
\[
(K+E)V^{-1}u(a)=KV^{-1}u(b).
\]
Consequently,
\[
(K+E)U^{-1}(a,a)(\alpha u)(a)=KU^{-1}(b,a)(\alpha u)(b).
\]
From the above equality,
\[
(K+E)(\alpha u)(a)=KU(a,b)(\alpha u)(b).
\]
This proves the validity of the claims in the theorem.
\end{proof}


\begin{remark} \rm
Now consider in $L^{2}(H,(a,b))$ the  differential expression
\[
l(u)=(\alpha(x)u(x))'+A(t)u(t),
\]
where
$\alpha(x)=0$, $x\in (c,d)$ and $a<c<d<b$ with corresponding conditions.
Assume that for any $t\in [c,d]$,  $A(t)$ is boundedly invertible in $H$
and $\|A^{-1}(t)\|\in L^{2}(c,d)$.
In this case, all boundedly solvable extensions of the minimal operator in
$L^{2}(H,(a,b))=L^{2}(H,(a,c))\oplus L^{2}(H,(c,d))\oplus L^{2}(H,(d,b))$
are generated by the differential-operator expression $l(\cdot)$ and the
 boundary conditions
\begin{gather*}
(K_1+E)(\alpha u)(a)=K_1U_1(a,c)(\alpha u)(c),\\
(K_2+E)(\alpha u)(d)=K_2U_2(d,b)(\alpha u)(b),
\end{gather*}
 where $K_1,K_2\in L(H)$; $E$ is an identity operator in $H$ and $U_1,U_2$
constitute a family of evolution operators generated by corresponding
differential equations in $L^{2}(H,(a,c))$ and $L^{2}(H,(d,b))$ respectively.
\end{remark}

\section{Structure of spectrum of boundedly solvable extensions}

In this section we investigated the geometric form in complex plane of
boundedly solvable extensions of the minimal operators $L_0$ in $L^2(H,(a,b))$.
First let us prove the following assertion.

\begin{theorem}  \label{thm3.1}
If $\widetilde{L}$ is a boundedly solvable extension of the minimal operator
$L_{0}$ and $\widetilde{M}=U^{-1}\widetilde{L}V$ is the corresponding boundedly
solvable extension of the minimal operator $M_{0}$, then
in order for $\lambda\in\sigma(\widetilde{L})$ the necessary and sufficient
 condition is
$0\in\sigma(\widetilde{M}-\lambda T_\alpha)$, where an operator
$T_\alpha:L^2(H,(a,b))\to L^2(H,(a,b))$ is a multiplication operator to
 $1/\alpha(t)$.
\end{theorem}

\begin{proof}
If $\widetilde{L}=L_{K}$ is a boundedly solvable extension of the minimal operator
 $L_{0}$ and $\lambda\in\mathbb{C}$, then it is clear that
\begin{align*}
L_{K}-\lambda E& = UM_KV^{-1}-\lambda E\\
& = U(M_K-\lambda U^{-1}V)V^{-1}\\
& =  U(M_K-\lambda\frac{1}{\alpha(t)}E)V^{-1}
\end{align*}
The last relation explains the validity of the theorem.
\end{proof}

Now prove the main theorem on the spectrum structure of extensions of 
the minimal operator $L_0$.


\begin{theorem} \label{thm3.2}
The spectrum of the boundedly solvable extension $L_{K}$ of the minimal 
operator $L_{0}$ in  $L^{2}(H,(a,b))$ has the form
\begin{align*}
\sigma(L_{K})=\Big\{&\Big(\int_a^b\frac{ds}{\alpha(s)}\Big)^{-1}
\big[\ln|{\frac{\mu+1}{\mu}}|+ i \arg({\frac{\mu+1}{\mu}})+2n\pi i\big] \\
&: \mu\in \sigma(K)\backslash\{0,-1\},~ n\in \mathbb{Z}\Big\}
\end{align*}
\end{theorem}

\begin{proof}
By  Theorem \ref{thm3.1} for the description the spectrum of boundedly solvable extension 
$L_K$ in $L^{2}(H,(a,b))$ it is sufficient to investigate of boundedly solvability 
of the operator $M_K-\lambda T_\alpha$ in $L^{2}(H,(a,b))$ for $\lambda\in\mathbb{C}$.
Now consider the spectral problem
\[
M_K u=\lambda T_\alpha u+f, \lambda\in\mathbb{C}, f\in L^{2}(H,(a,b)).
\]
From this, it is clear that
\begin{gather*}
u'(t)=\lambda\frac{1}{\alpha(t)}u(t)+f(t), \\
(K+E)u(a)=Ku(b), \\
\lambda\in\mathbb{C}, f\in L^{2}(H,(a,b)), K\in L(H).
\end{gather*}
In this case it is evident that a general solution of above differential equation 
in $L^{2}(H,(a,b))$ has a form
\[
u_{\lambda}(t)=e^{\lambda\int_a^t\frac{ds}{\alpha(s)}}f_0
+\int_a^te^{\lambda\int_\tau^t\frac{ds}{\alpha(s)}}f(\tau)d\tau, f_0\in H.
\]
Therefore from the boundary condition, we obtain the expression
\[
\Big(E+K\Big(1-e^{\lambda\int_a^t\frac{ds}{\alpha(s)}}\Big)\Big)f_0
= K\int_a^be^{\lambda\int_\tau^t\frac{ds}{\alpha(s)}}f(\tau)d\tau
\]

For $\lambda_m=2m{\pi}i\big(\int_a^b\frac{ds}{\alpha(s)}\big)^{-1}$ with
 $m\in\mathbb{Z}$, from the above  relation,  it follows that
\[
f_0^{(m)}=K\int_a^b e^{\lambda_m\int_{\tau}^b\frac{ds}{\alpha(s)}}f(\tau)d\tau,
\quad  m\in\mathbb{Z}.
\]
Consequently, in this case the inverse operator $(M_K-\lambda T_\alpha)^{-1}$ 
is of the form
\[
(M_K-\lambda_m T_\alpha)^{-1}f(t)=Ke^{\lambda_m\int_a^t\frac{ds}{\alpha(s)}}
\Big(\int_a^b e^{\lambda_m\int_{\tau}^b\frac{ds}{\alpha(s)}}f(\tau)d\tau\Big)
+\int_a^t e^{\lambda_m\int_{\tau}^t\frac{ds}{\alpha(s)}}f(\tau)d\tau,
\]
$\lambda_m\in\mathbb{Z}$, $f\in L^{2}(H,(a,b))$,
and it is clear that for this $\lambda_m$,  $m\in\mathbb{Z}$,
\[
(M_K-\lambda_m T_\alpha)^{-1}\in L(L^{2}(H,(a,b))).
\]

On the other hand, if $\lambda\neq 2m{\pi}i$, $m\in\mathbb{Z}$, 
$\lambda\in\mathbb{C}$, then from 
\[
\Big(E+K\Big(1-e^{\lambda\int_a^b\frac{ds}{\alpha(s)}}\Big)\Big)f_0
=K\int_a^be^{\lambda\int_\tau^b\frac{ds}{\alpha(s)}}f(\tau)d\tau,
\]
we have
\[
\Big(K-\frac{1}{e^{\lambda\int_a^b\frac{ds}{\alpha(s)}}-1}\Big)f_0
=\Big(\frac{1}{1-e^{\lambda\int_a^b\frac{ds}{\alpha(s)}}}\Big)
K\int_a^be^{\lambda\int_\tau^b\frac{ds}{\alpha(s)}}f(\tau)d\tau, \quad f_0\in H.
\]
This implies:  $0\in\sigma(M_K-\lambda_m T_\alpha)$
if and only if $\mu=\frac{1}{e^{\lambda\int_a^b\frac{ds}{\alpha(s)}}-1}\in\sigma(K)$.
Hence in this case we have
\[
\lambda_n=\Big(\int_a^b\frac{ds}{\alpha(s)}\Big)^{-1}
\big[\ln|{\frac{\mu+1}{\mu}}|+ i \arg({\frac{\mu+1}{\mu}})+2n\pi i\big],
\]
where $\mu\in\sigma(K)$, $n\in\mathbb{Z}$.
From this and Theorem \ref{thm3.1} the validity of the claim is evident.
\end{proof}

\begin{corollary} \label{coro3.1}
(1) If $\sigma(K)\subset\{0,-1\}$, then for the spectrum corresponding boundedly 
solvable extension $L_K$ is true $\sigma(L_K)=\emptyset$.
 (2) If $\sigma(K)\neq\{0,-1\}\neq\emptyset$, then $\sigma(L_K)$ is infinite.
\end{corollary}

\begin{example} \label{examp3.1} \rm
All boundedly solvable extensions of the minimal operator $L_0$ in $L^{2}(0,1)$ 
generated by  differential expression
\begin{align*}
l(u)=\big((|x-\tfrac{1}{2}|+|x-\tfrac{1}{3}|)u(x)\big)'
+\int_0^xa(t)u(t-\tau)d\tau,  a\in C[0,1]
\end{align*}
are generated by the integro-differential expression $l(.)$ and the 
boundary condition
\[
(k+1)\left((|x-\tfrac{1}{2}|+|x-\tfrac{1}{3}|)u(x)\right)(0)
=kU(0,1)\left((|x-\tfrac{1}{2}|+|x-\tfrac{1}{3}|)u(x)\right)(1), \quad k\in\mathbb{C}
\]
and $U(\cdot,\cdot)$ are the corresponding evolution operators in the Hilbert 
space $L^{2}(0,1)$. In this case, the spectrum $\sigma(L_k)$ of the extension 
$L_k$ when $k\neq0,-1$  is of the form
\[
\sigma(L_k)=\Big\{\frac{1}{\ln(e\sqrt{35})}
\big[\ln|{\frac{k+1}{k}}|+ i \arg({\frac{k+1}{k}})+2n\pi i\big] :
n\in \mathbb{Z}\Big\}\,.
\]
When $k=0$ or $k=-1$, the spectrum of this extension is empty by 
Corollary \ref{coro3.1}.
\end{example}

When $\alpha(t)=1$ for  $t\in(a,b)$, Theorems  \ref{thm2.3} and  \ref{thm3.2}
 have been  proven in \cite{i1}.

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\end{document}