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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 237, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/237\hfil Inverse scattering problems]
{Inverse scattering problems for energy-dependent Sturm-Liouville equations
with point $\delta $-interaction and eigenparameter-dependent boundary
condition}

\author[M. Dzh. Manafov, A. Kablan \hfil EJDE-2013/237\hfilneg]
{Manaf Dzh. Manafov, Abdullah Kablan}  % in alphabetical order

\dedicatory{In memory of M. G. Gasymov, one of the
pioneers of this subject}\emph{}

\address{Manaf Dzh. Manafov \newline
Faculty of Arts and Sciences, Department of Mathematics,
Adiyaman University, Adiyaman, 02040, Turkey}
\email{mmanafov@adiyaman.edu.tr}

\address{Abdullah Kablan \newline
Faculty of Arts and Sciences, Department of Mathematics, 
Gaziantep University, \newline Gaziantep, 27310, Turkey}
\email{kablan@gantep.edu.tr}

\thanks{Submitted July 29, 2013. Published October 24, 2013.}
\subjclass[2000]{34A55, 34B24, 47E05}
\keywords{Inverse scattering; scattering data;
point delta-interaction; \hfill\break\indent
eigenvalue-dependent boundary condition}

\begin{abstract}
 We consider an inverse problem of the scattering theory for energy-dependent
 Sturm-Liouville equations on the half line $[0,+\infty )$ with point
 $\delta $-interaction and eigenparameter-dependent boundary condition.
 We define the scattering data of the problem first, then consider the
 basic equation and study an algorithm for finding the potentials with the
 given scattering data.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks


\section{Introduction}

We consider inverse scattering problem for the equation
\begin{equation}
-y''+q(x)y=\lambda ^{2}y,\quad  x\in (0,a)\cup (a,+\infty )
\label{a1}
\end{equation}
with the boundary condition
\begin{equation}
U(y):=\lambda ^{2}( y'(0)-hy(0)) -( h_1y'(0)-h_2y(0)) =0  \label{a2}
\end{equation}
and conditions at the point $x=a$,
\begin{equation}
I(y):=\begin{cases}
y(a+0)=y(a-0)=y(a), \\
y'(a+0)-y'(a-0)=2i\alpha \lambda y(a),
\end{cases}  \label{a3}
\end{equation}
where $\lambda $ is a spectral parameter, $q(x)$ is real-valued function
satisfying the condition
\begin{equation*}
\int_0^\infty (1+x)| q(x)|
dx<\infty ,
\end{equation*}
and $\alpha <0$, $h$, $h_1$, $h_2$ are real numbers such that
\begin{equation*}
\delta :=hh_1-h_2>0.
\end{equation*}
Notice that, we can understand problem \eqref{a1} and \eqref{a3} as one of the
study of the equation
\begin{equation}
y''+( \lambda ^{2}-2i\lambda p(x)-q(x)) y=0,\quad  x\in (0,+\infty ),  \label{a1'}
\end{equation}
when $p(x)=\alpha \delta (x-a)$, where $\delta (x)$ is the Dirac function,
(see \cite{Albeverio}).

Sturm-Liouville spectral problems with potentials depending on the spectral
parameter arise in various models of quantum and classical mechanics. For
instance, to this form can be reduced the corresponding evolution equations
(such as the Klein-Gordon equation \cite{Jonas}, \cite{Najman}) that are
used to model interactions between colliding relativistic spinless
particles. Then $\lambda ^{2}$ is related to the energy of the system, this
explaining the term ``energy-dependent'' in \eqref{a1'}.

Problems of the form \eqref{a1'} have also appeared in the physical
literature in the context of scattering of waves and particles. In
particular, in \cite{Jaulent1}-\cite{Jaulent3} the inverse scattering
problems for energy-dependent Schr\"{o}dinger operators on the line are
studied, see also
\cite{Aktosun1,Hryniv,Kamimura,Maksudov,Mee,Nabiyev}.

The non-linear dependence of equation \eqref{a1'} on the spectral parameter
$\lambda $ should be regarded as a spectral problem for a quadratic operator
pencil. The problem with $p(x)\in W_2^{1}(0,1)$ and $q(x)\in L_2(0,1)$
and with Robin boundary conditions was discussed in \cite{Gasymov}. Such
problems for separated and nonseparated boundary conditions were considered
(see \cite{Akhtyamov,Guseinov1,Yang,Yurko} and the references therein).

The inverse scattering problem for equation \eqref{a1}
(in the case $\alpha=0$) with boundary condition $y'(0)-hy(0)=0$
was completely solved in \cite{Marchenko} (in the case $h=\infty $),
in \cite{Levitan} (for arbitrary real number $h$). Similar problems
were dealt with in \cite
{Aktosun2} and references in these works. The inverse scattering problem for
equation \eqref{a1'} (in the case $\alpha =0$) with spectral parameter in
the boundary condition on the half line were examined in \cite{Mamedov1},
\cite{Mamedov2}.

In this paper we consider the inverse problem scattering theory on the half
line $[0,\infty )$ for the \eqref{a1'}, \eqref{a2} boundary-value problem.

\section{Integral representation for the Jost solution and scattering data}

In this section, we will find an integral representation for the Jost
solution and study some properties of scattering data of the problem.

Let us denote by $e_0(x,\lambda)$ the solution of \eqref{a1},
when $q(x)\equiv 0$, satisfying  conditions \eqref{a3} and the condition
at infinity
\begin{equation*}
\lim_{x\to \infty } e_0(x,\lambda )e^{-i\lambda x}=1\,.
\end{equation*}
This function is called the Jost solution.

It is obvious that the function $e_0(x,\lambda )$ can be written as
\begin{equation*}
e_0(x,\lambda )=\begin{cases}
(1-\alpha )e^{i\lambda x}+\alpha e^{i\lambda (2a-x)},&  0<x<a \\
e^{i\lambda x},& x>a
\end{cases}
\end{equation*}
Analogously to \cite{Guseinov2} the following theorem can be proved.

\begin{theorem} \label{thm2.1}
Let $\int_0^\infty (1+x)| q(x)|
dx<\infty $. Then the Jost solution of \eqref{a1} has the form
\begin{equation}
e(x,\lambda )=e_0(x,\lambda )+\int_x^\infty
K(x,t)e^{i\lambda t}dt\quad  (\operatorname{Im}\lambda \geq 0),  \label{b1}
\end{equation}
where for each fixed $x\neq a$ the kernel $K(x,\cdot )$ belong to the
$L_1(x,\infty )$ and
\begin{equation*}
\int_x^\infty | K(x,t)| dt\leq e^{c\sigma _1(x)}-1,
\end{equation*}
where $\sigma _1(x)=\int_0^\infty t| q(t)| dt$, $c=1-\frac{\alpha }{2}$.
\end{theorem}

If the function $q(x)$ is differentiable then the kernel $K(x,t)$ satisfies
the following properties:
\begin{equation} \label{b2}
\begin{gathered}
K_{xx}(x,t)-q(x)K(x,t)=K_{tt}(x,t),\quad  0<x<\infty ,\;  t>x, \\
K(x,x)= \begin{cases}
\frac{1}{2}\int_x^\infty q(s)ds,& x>a \\
\frac{1}{2}(1-\alpha )\int_x^\infty q(s)ds, & x<a
\end{cases}
\\
\begin{aligned}
&K(x,2a-x+0)-K(x,2a-x-0)\\
&=\frac{\alpha }{2}\big\{
 \int_a^\infty q(s)ds-\int_a^x q(s)ds\big\} ,\quad x<a.
\end{aligned}
\end{gathered}
\end{equation}

Now we learn some properties of scattering data of the problem. Since the
function $q(x)$ is real-valued, it follows that for real $\lambda $ together
with $e(x,\lambda )$ the solution of \eqref{a1} are also $\overline{
e(x,\lambda )}$. Since the Wronskian of the two solutions $y_1(x)$ and
$y_2(x)$ of equation \eqref{a1},
\begin{equation*}
W\{ y_1(x),y_2(x)\} =y_1'(x)y_2(x)-y_1(x)y_2'(x),
\end{equation*}
is independent of $x$, it coincides with its own value as
$x\to \infty $. Therefore
\begin{equation}
W\{ e(x,\lambda ),\overline{e(x,\lambda )}\}
=\lim_{x\to +\infty } [ e'(x,\lambda )\overline{
e(x,\lambda )}-e(x,\lambda )\overline{e'(x,\lambda )} ]
=2i\lambda .  \label{c1}
\end{equation}
Consequently for $0\neq \lambda \in\mathbb{R}$ the pair
$\{e(x,\lambda )$, $\overline{e(x,\lambda )}\}$ is a fundamental
system of solutions of \eqref{a1}.

Let $\varphi (x,\lambda )$ be the solution of \eqref{a1} satisfying the
initial condition
\begin{equation}
\varphi (0,\lambda )=\lambda ^{2}-h_1,\quad
\varphi '(0,\lambda )=\lambda ^{2}h-h_2.  \label{c2}
\end{equation}

\begin{lemma} \label{lem2.2}
The equality
\begin{equation}
\frac{2i\lambda \varphi (x,\lambda )}{(\lambda ^{2}-h_1)e'(0,\lambda )
-(\lambda ^{2}h-h_2)e(0,\lambda )}
=\overline{e(x,\lambda )} -S(\lambda )e(x,\lambda )  \label{c3}
\end{equation}
holds for all real $\lambda \neq 0$, where
\begin{equation}
S(\lambda )=\frac{(\lambda ^{2}-h_1)\overline{e'(0,\lambda )}
-(\lambda ^{2}h-h_2)\overline{e(0,\lambda )}}{(\lambda
^{2}-h_1)e'(0,\lambda )-(\lambda ^{2}h-h_2)e(0,\lambda )}
\label{c4}
\end{equation}
with
\begin{equation*}
S(\lambda )=\overline{S(-\lambda )}=[ S(-\lambda )] ^{-1}
\end{equation*}

\begin{proof}
The functions $e(x,\lambda )$ and $\overline{e(x,\lambda )}$ form a
fundamental system of solutions of \eqref{a1} for all real $\lambda \neq 0$,
then we can write
\begin{equation}
\varphi (x,\lambda )=c_1(\lambda )e(x,\lambda )+c_2(\lambda )\overline{
e(x,\lambda )},  \label{c5}
\end{equation}
where
\begin{gather*}
c_1(\lambda )=-\frac{(\lambda ^{2}-h_1)\overline{e'(0,\lambda )}
-(h\lambda ^{2}-h_2)\overline{e(0,\lambda )}}{2i\lambda },\\
c_2(\lambda )=\frac{(\lambda ^{2}-h_1)e'(0,\lambda )-(h\lambda
^{2}-h_2)e(0,\lambda )}{2i\lambda }.
\end{gather*}
To prove that $\psi (\lambda )\equiv (\lambda ^{2}-h_1)e'(0,\lambda
)-(\lambda ^{2}h-h_2)e(0,\lambda )\neq 0$ for all real $\lambda \neq 0$,
we assume that $\lambda _0\in (-\infty ,+\infty )\backslash \{0\} $
such that
\begin{equation*}
(\lambda _0^{2}-h_1)e'(0,\lambda _0)-(\lambda_0^{2}h-h_2)e(0,\lambda _0)=0
\end{equation*}
or
\begin{equation*}
e'(0,\lambda _0)=\frac{(\lambda _0^{2}h-h_2)}{\lambda
_0^{2}-h_1}e(0,\lambda _0)
\end{equation*}
From  \eqref{c1}, we obtained the equation
\begin{equation*}
| e(0,\lambda _0)| ^{2}\frac{0}{| \lambda
_0^{2}-h_1| ^{2}}=2i\lambda _0;
\end{equation*}
i.e., we have a contradiction. Hence $\psi (\lambda )\neq 0$ for all real
 $ \lambda \neq 0$. Dividing both sides of  \eqref{c5} by
$\psi (\lambda )/2i\lambda $ we obtain \eqref{c3} where
$S(\lambda )$ is defined with \eqref{c4}. Since
\begin{equation*}
\psi (\lambda )=\overline{\psi (-\lambda )}=[ \psi (-\lambda )]
^{-1}
\end{equation*}
we have
\begin{equation*}
S(\lambda )=\overline{S(-\lambda )}=[ S(-\lambda )] ^{-1}.
\end{equation*}
\end{proof}
\end{lemma}

The function $S(\lambda )$ is called \textit{the scattering function} of the
problem  \eqref{a1}-\eqref{a2}-\eqref{a3}.

\begin{lemma} \label{lem2.3}
The function $\psi (\lambda )$ may have only a finite number of zeros in the
half-plane $\operatorname{Im}\lambda >0$. Moreover, all these zeros are simple
 and lie in the imaginary axis. \label{t1}
\end{lemma}

\begin{proof}
Since for real values of $\lambda \neq 0$ the inequality
$\psi (\lambda )\neq 0$ holds, the only possible real zero of the function
$\psi (\lambda )$ is $\lambda =0$. Since the function $\psi (\lambda )$
 is analytic on the
upper half-plane it follows that its zeros form a (finite or countable) set.

Let us show that this set is bounded. Assume the converse and suppose that
there exist $\lambda _k$ such that $| \lambda _k|
\to \infty $, with $\operatorname{Im}$ $\lambda _k>0$ and
\begin{equation*}
(\lambda _k^{2}-h_1)e'(0,\lambda _k)=(\lambda
_k^{2}h-h_2)e(0,\lambda _k).
\end{equation*}
Then as $| \lambda _k| \to \infty $ yields
$\lim_{k\to \infty } e(0,\lambda _k)=0$. On the other
hand, it follows from  \eqref{c1} that $\lim_{k\to \infty}e(0,\lambda _k)=1$.
 The resulting contradiction shows that
the set $\{ \lambda _k\} $ is bounded. Thus, the zeros of the
function $\psi (\lambda )$ form a bounded finite or countable set whose
unique limiting point can only be zero.

Now let us show that all the zeros of the function $\psi (\lambda )$ lie on
the imaginary axis. Suppose that $\lambda _1$ and $\lambda _2$ are some
zeros of the function $\psi (\lambda )$. Since
\begin{equation*}
-e''(x,\lambda _1)+q(x)e(x,\lambda _1)=\lambda
_1^{2}e(x,\lambda _1),\quad
-\overline{e''(x,\lambda
_2)}+q(x)\overline{e(x,\lambda _2)}=\lambda _2^{2}\overline{
e(x,\lambda _2)}.
\end{equation*}
If the first equation is multiplied by $\overline{e(x,\lambda _2)}$, second
equation is multiplied by $e(x,\lambda _1)$ and subtracting them side by
side and finally integrating over the interval $[0,+\infty )$, the equality
\begin{equation*}
(\lambda _1^{2}-\overline{\lambda }_2^{2})
\int_0^\infty  e(x,\lambda _1)\overline{e(x,\lambda _2)}dx
+W\{e(x,\lambda _1),\overline{e(x,\lambda _2)}\} (|_0^{a}+|_{a}^{+\infty }) =0
\end{equation*}
is attained.

If conditions \eqref{a3} and
\begin{equation*}
W\{ e(x,\lambda _1),\overline{e(x,\lambda _2)}\}
_{x=a-0}=W\{ e(x,\lambda _1),\overline{e(x,\lambda _2)}\}
_{x=a+0}
\end{equation*}
are considered, then
\begin{equation}
(\lambda _1^{2}-\overline{\lambda }_2^{2})
\int_0^\infty e(x,\lambda _1)\overline{e(x,\lambda _2)}dx
+W\{e(x,\lambda _1),\overline{e(x,\lambda _2)}\} _{x=0}=0  \label{c6}
\end{equation}
is obtained. On the other hand, according to the definition of the function
$\psi (\lambda )$, the following relation holds
\begin{equation*}
\psi (\lambda _{j})=(\lambda _{j}^{2}-h_1)e'(0,\lambda
_{j})-(\lambda _{j}^{2}h-h_0)e(0,\lambda _{j})=0,\quad  j=1,2.
\end{equation*}
Also, by  \eqref{a2} we can write
\begin{equation}
W\{ e(x,\lambda _1),\overline{e(x,\lambda _2)}\} _{x=0}
=\frac{[e'(0,\lambda _1)-he(0,\lambda _1)]
[\overline{e'(0,\lambda _2)}-h\overline{e(0,\lambda _2)}]}{\delta }(\lambda
_2^{-2}-\lambda _1^{2}).  \label{c7}
\end{equation}
Therefore, using \eqref{c6} and \eqref{c7} we have
\begin{equation}
(\lambda _1^{2}-\lambda _2^{-2})\{ \int_0^\infty
e(x,\lambda _1)\overline{e(x,\lambda _2)}dx+\frac{1}{\delta }
[e'(0,\lambda _1)-he(0,\lambda _1)][\overline{e'(0,\lambda _2)}
-h\overline{e(0,\lambda _2)}]\} =0.  \label{c8}
\end{equation}
In particular, the choice $\lambda _2=\lambda _1$ at \eqref{c8} implies
that $\lambda _1=ik_1$, where $k_1\geq 0$. Therefore, zeros of the
function $\psi (\lambda )$ can lie only on the imaginary axis. Now, let us
prove that function $\psi (\lambda )$ has zeros in finite numbers. Since we
can give an estimate for the distance between the neighboring zeros of the
function $\psi (\lambda )$, it follows that the number of zeros is finite,
(see \cite[p. 186]{Marchenko}).
\end{proof}


Let
\begin{equation}
m_{p}^{-2}\equiv \int_0^\infty |
e(x,ik_{p})| ^{2}dx+\frac{1}{\delta }| e'(0,ik_{p})-he(0,ik_{p})| ^{2},\quad
p=1,2,\ldots ,n.
\label{c9}
\end{equation}
These numbers are called the \textit{norming constants} for the boundary
problem \eqref{a1'}, \eqref{a2}.

The collections $\{ S(\lambda ), (-\infty <\lambda <+\infty
); k_{p}, m_{p} (p=1,2,\ldots ,n)\} $ are called the
\textit{scattering data} of the boundary value problem
\eqref{a1}-\eqref{a2}-\eqref{a3} or \eqref{a1'}-\eqref{a2}.

\section{Basic equation of the inverse scattering problem}

To derive the basic equation for the kernel $K(x,t)$ of the solution
\eqref{b1}, we use equality \eqref{c3}. Substituting expression \eqref{b1} for
$e(x,\lambda )$ into this equality, we get
\begin{equation} \label{d1}
\begin{aligned}
&\frac{2i\lambda \varphi (x,\lambda )}{\psi (\lambda )}-\overline{
e_0(x,\lambda )}+S_0(\lambda )e_0(x,\lambda ) \\
&= \int_x^\infty K(x,t)e^{-i\lambda t}dt+[
S_0(\lambda )-S(\lambda )] e_0(x,\lambda )-S_0(\lambda )
\int_x^\infty  K(x,t)e^{i\lambda t}dt   \\
&\quad +( S_0(\lambda )-S(\lambda ))
\int_x^\infty K(x,t)e^{i\lambda t}dt.
\end{aligned}
\end{equation}
Multiplying this equality by $(1/2\pi )e^{i\lambda r}$ and integrating over $
\lambda $ from $-\infty $ to $+\infty $, for $r>x$, at the right-hand side we
obtain
\begin{equation}  \label{d2}
\begin{aligned}
&K(x,r)+\frac{1}{2\pi }\int_{-\infty}^{\infty}
[S_0(\lambda )-S(\lambda )] e_0(x,r)e^{i\lambda r}d\lambda \\
&-\int_x^{\infty} K(x,t)\Big\{ \frac{1}{2\pi }
\int_{-\infty}^{\infty} S_0(\lambda )e^{i\lambda (t+r)}d\lambda \Big\} dt   \\
&+\int_x^\infty K(x,t)\Big\{ \frac{1}{2\pi }
\int_{-\infty}^{\infty} [ S_0(\lambda
)-S(\lambda )] e^{i\lambda (t+r)}d\lambda \Big\} dt.
\end{aligned}
\end{equation}
Now we will compute the integral $(1/2\pi )\int_{-\infty}^{\infty}
S_0(\lambda )e^{i\lambda (t+r)}d\lambda $. By elementary
transforms we obtain
\begin{equation*}
S_0(\lambda )=\frac{\tau ^{2}-1}{1+\tau e^{2i\lambda a}}+\tau
e^{-2i\lambda a}=(\tau ^{2}-1)\sum_{k=0}^\infty
(-1)^{k}\tau ^{k}e^{2i\lambda ak}+\tau e^{-2i\lambda a}
\end{equation*}
where $\tau =\alpha/(\alpha -1)$. Thus we have
\begin{equation*}
\frac{1}{2\pi }\int_{-\infty}^{\infty} S_0(\lambda
)e^{i\lambda (t+r)}d\lambda =(\tau ^{2}-1)\underset{k=0}{\overset{\infty }{
\sum }}(-1)^{k}\tau ^{k}\delta (t+r+2ak)+\tau \delta (t+r-2a).
\end{equation*}
Consequently \eqref{d2} can be written as
\begin{align*}
&K(x,r)+F_{S}(x,r)+\int_x^\infty
K(x,t)F_{0S}(t+r)dt-\tau K(x,2a-r) \\
&-(\tau ^{2}-1)\sum_{k=0}^\infty (-1)^{k}\tau ^{k}K(x,-2ak-r),
\end{align*}
where
\begin{gather*}
F_{0S}(x)\equiv \frac{1}{2\pi }\int_{-\infty}^\infty
[ S_0(\lambda )-S(\lambda )] e^{i\lambda x}d\lambda ,
\\
F_{S}(x,r)\equiv (1-\alpha )F_{0S}(x+r)+\alpha F_{0S}(2a-x+r).
\end{gather*}
We note that $K(x,r)=0$ for $r<x$ (see \cite{Guseinov2}). Therefore, for
$r>x $, \eqref{c2} takes the form
\begin{equation}
K(x,r)+F_{S}(x,r)+\int_x^\infty
K(x,r)F_{0S}(t+r)dt-\tau K(x,2a-r).  \label{d3}
\end{equation}

On the left-hand side of \eqref{d1} with help of Jordan's lemma and the
residue theorem and by taking Lemma \ref{t1} into account for $r>x$, we
obtain
\begin{equation}
-\sum_{p=1}^n \frac{2ik_{p}\varphi (x,ik_{p})}{\psi
'(ik_{p})}e^{-k_{p}r}.  \label{d4}
\end{equation}
From the definition of norming constants $m_{p}$ $(p=1,2,\ldots ,n)$
in \eqref{c9} we have
\begin{equation}  \label{d5}
\begin{aligned}
&-\sum_{p=1}^n \frac{2ik_{p}\varphi (x,ik_{p})}{\psi
'(ik_{p})}e^{-k_{p}r}\\
&=-\sum_{p=1}^n \frac{ 2ik_{p}e^{-k_{p}r}e(x,ik_{p})}
 {[ e'(0,ik_{p})-he(0,ik_{p})] \psi '(ik_{p})}
\\
&=-\sum_{p=1}^n m_{p}^{2}e(x,ik_{p})e^{-k_{p}r}\\
&=- \sum_{p=1}^n m_{p}^{2}\{
e(x,ik_{p})e^{-k_{p}(x+r)}+\int_x^\infty
K(x,t)e^{-k_{p}(x+r)}dt\} .
\end{aligned}
\end{equation}
Thus, for $x<\min (r,2a-r)$, by taking \eqref{d3} and \eqref{d5} into
account, from \eqref{d2} we derive the relation
\begin{equation}
K(x,r)+F(x,r)+\int_x^\infty K(x,t)F_0(t+r)dt+
\frac{\alpha }{1-\alpha }K(x,2a-r)=0  \label{d6}
\end{equation}
where
\begin{equation}
\begin{gathered}
F_0(x)=F_{0S}(x)+\sum_{p=1}^n m_{p}^{2}e^{-k_{p}r}, \\
F(x,r)=F_{S}(x,r)+\sum_{p=1}^n
m_{p}^{2}e_0(x,ik_{p})e^{-k_{p}(x+r)}.
\end{gathered} \label{d7}
\end{equation}
Equation \eqref{d6} is called the \textit{basic equation} of the inverse
problem of the scattering theory for the boundary problem
\eqref{a1}-\eqref{a2}-\eqref{a3} or \eqref{a1'}-\eqref{a2}.
The basic equation is different
from the classical equation of Marchenko (see \cite{Marchenko}).
Thus, we have proved  the following theorem.

\begin{theorem} \label{thm3.1}
For each $x\geq 0$, the kernel $K(x,r)$ of the solution \eqref{b1} satisfies
the basic equation \eqref{d6}.
\end{theorem}

Obviously, to form the basic equation \eqref{d6}, it suffices to know the
functions $F_0(x)$ and $F(x,r)$. In turn, in order to find the functions
$F_0(x)$, $F(x,r)$, it suffices to know only the scattering data
$\{S(\lambda )\, (-\infty <\lambda <\infty ); k_{p}, m_{p}\, (p=1,2,\ldots
,n)\}$. Given the scattering data, we can use formulas \eqref{d7} to
construct the functions $F_0(x)$, $F(x,r)$ and write down the basic
equation \eqref{d6} for the unknown solution $K(x,r)$, the function $q(x)$
may be found from formula \eqref{b2}.

\begin{theorem} \label{thm3.2}
Equation \eqref{d6} has a unique solution $K(x,\cdot )\in L_1(x,+\infty )$
for each fixed $x\geq 0$. \label{t2}
\end{theorem}

\begin{proof}
It is easy to show that for each fixed $x\geq 0$ the operator
\begin{equation*}
(M_{x}f)(r)=\begin{cases}
f(r), & x>a \\
f(r)+\frac{\alpha }{1-\alpha }f(2a-r), &  x<a,
\end{cases}
\end{equation*}
acting in the space $L_1(x,+\infty )$ (and also in $L_2(x,+\infty )$) is
invertible. Therefore, basic equation \eqref{d6} is equivalent to
\begin{equation*}
K(x,r)+(M_{x})^{-1}F(x,r)+(M_{x})^{-1}(F_0K(x,\cdot ))(r)=0;
\end{equation*}
i.e. to the equation with a compact operator $(M_{x})^{-1}F$
 (for the compactness of $F$, see \cite[Lemma 3.3.1]{Marchenko}). To prove the
theorem, it is sufficient to show that the homogeneous equation
\begin{equation}
f_{x}(r)+\frac{\alpha }{1-\alpha }f_{x}(2a-r)
+\int_x^\infty f_{x}(t)F_0(t+r)dt=0,\quad  x<\min (r,2a-r),  \label{d8}
\end{equation}
has only the trivial solution $f_{x}(r)\in L_1(x,+\infty )$. We can show
that (see \cite{Marchenko}) the function $F_0(r)$ belongs to the space
$L_2[0,+\infty )$, is absolutely continuous on all the intervals not
containing the point $2a$; and for all $\beta \geq 0$
\begin{equation*}
\int_\beta^\infty  | F_0(r)| dr<+\infty ,
\quad \int_\beta^\infty (1+r)|F'(r)| dr<+\infty .
\end{equation*}
Therefore, the function $F(r)$ and the solution $f_{x}(r)$ are together
bounded on the semi-axis $x\leq y<+\infty $. Consequently,
$f_{x}(r)\in L_2(x,+\infty )$.

Now let us multiply  \eqref{d8} by $\overline{f_{x}(r)}$ and
integrate with respect to $y$ over the interval $(x,+\infty )$.
Using \eqref{d6}, \eqref{d7} and Parseval's identity
\begin{gather*}
\int_x^\infty | f_{x}(r)|
^{2}dr = \frac{1}{2\pi }\int_{-\infty}^{\infty}
| \widetilde{f}_{x}(\lambda )| ^{2}d\lambda , \\
\frac{\alpha }{1-\alpha }\int_x^\infty f_{x}(2a-r)
\overline{f_{x}(r)}dr = \frac{1}{2\pi }\int_{-\infty}^{\infty} S_0(\lambda )\overline{\widetilde{f}_{x}(\lambda )}\widetilde{f}
_{x}(-\lambda )d\lambda ,
\end{gather*}
where
\begin{equation*}
\widetilde{f}_{x}(\lambda )=\int_x^\infty
f_{x}(t)e^{-i\lambda t}dt,
\end{equation*}
we obtain
\begin{equation*}
\frac{1}{2\pi }\int_{-\infty}^{\infty} |
\widetilde{f}_{x}(\lambda )| ^{2}d\lambda
+\sum_{p=1}^n m_{p}| \widetilde{f}_{x}(-ik_{p})| ^{2}+\frac{1
}{2\pi }\int_{-\infty}^{\infty} S(\lambda )\widetilde{
f}_{x}(-\lambda )\overline{\widetilde{f}_{x}(\lambda )}d\lambda =0.
\end{equation*}
Since $| S(\lambda )| =| S(-\lambda )|$, we obtain the estimate
\begin{align*}
\frac{1}{2\pi }\int_{-\infty}^{\infty} |
\widetilde{f}_{x}(\lambda )| ^{2}d\lambda
 &\leq \frac{1}{2\pi } \int_{-\infty}^{\infty} | S(\lambda
)| | \overline{\widetilde{f}_{x}(-\lambda )}|
| \widetilde{f}_{x}(\lambda )| d\lambda \\
&\leq \frac{1}{2\pi }\int_{-\infty}^{\infty}
| S(\lambda )| \frac{| \widetilde{f}
_{x}(-\lambda )| ^{2}+| \widetilde{f}_{x}(\lambda
)| ^{2}}{2}d\lambda \\
&= \frac{1}{2\pi }\int_{-\infty}^{\infty} |
S(\lambda )| | \widetilde{f}_{x}(\lambda )|
^{2}d\lambda
\end{align*}
or
\begin{equation*}
\frac{1}{2\pi }\int_{-\infty}^{\infty} \{1-| S(\lambda )| \} | \widetilde{f}
_{x}(\lambda )| ^{2}d\lambda \leq 0.
\end{equation*}
It follows from the above that $\widetilde{f}_{x}(\lambda )\equiv 0$ since
$1-| S(\lambda )| >0$ for all $\lambda \neq 0$. Thus, the
basic equation \eqref{d6} is uniquely solvable.
\end{proof}

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\end{document}