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\begin{document}
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{\textbf{\Large Uniqueness Theorem for the Inverse Aftereffect problem and Representation the Nodal Points Form}}\\
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\vspace{10mm}

\begin{center}
\index{A. Neamaty}\index{Sh. Akbarpoor}\index{A. Dabbaghian}
{\text{A. Neamaty, Sh. Akbarpoor and A. Dabbaghian}\\

{\small Department of Mathematics, University of Mazandaran, Babolsar, Iran\\
E-mail: {\tiny Namaty@umz.ac.ir, Sh.akbarpoor@stu.umz.ac.ir,
A.dabbaghian@iauneka.ac.ir } \vspace{0.2cm}} }
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\text{\small \textbf{Abstract}.} {\small In this paper, we consider
a boundary value problem with aftereffect on a finite interval and
study the asymptotic behavior of the solutions, eigenvalues, the
nodal points and the associated nodal length and calculate the
numerical values of the nodal points and the nodal length. Then, we
prove the uniqueness theorem for the inverse aftereffect problem by
applying any dense subset of the nodal points.
\\}

\text{\small \textbf{Keywords}:} {\small Differential equation;
Aftereffect; Uniqueness theorem; Nodal points.}
\\
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\section{\bf{Introduction} }

\noindent

In this work, we consider the equation
\begin{eqnarray}
-y''(x)+q(x)y(x)+\int_{0}^{x}M(x-t)y(t)dt=\lambda y(x),& 0 \leq x
\leq \pi,
\end{eqnarray}
under the separated boundary conditions
\begin{eqnarray}
U(y):=y'(0)-h y(0)=0,
\end{eqnarray}
\begin{eqnarray}
 V(y):=y'(\pi)+H y(\pi)=0,
\end{eqnarray}
where $\lambda=\rho^{2}$ and $ \rho $, $\rho=\sigma+i\tau$, is the
spectral parameter, $ q, M \in W^{2,1}(0,\pi)$ are real functions.
We denote the boundary value problem (1.1)-(1.3) by $ L(q,M,h,H) $.

In fact, in this work, we consider the Sturm-Liouville operator
disorganized by a Volterra integral operator. Uniqueness in the
inverse aftereffect problem is the problem of chek the uniqueness of
the function $ M $. In this work, we suppose the function $ q $ is
the known function and prove the uniqueness theorem for the solution
of the inverse problem i.e $ M $.

Many authors studied the uniqueness of the inverse boundary value
problem for the Sturm-Liouville equations lately (see [1, 3, 7, 9,
16]) but a few of them considered it for the differential equations
with aftereffect (for example see [6]). In this paper, we obtain the
nodal points and the associated nodal length and investigate the
uniqueness of inverse aftereffect problem $L(q,M,h,H)$ with the
separated boundary conditions by using any dense subset of the nodal
points. Proof of the uniqueness and computation of the nodal points
was studied for Sturm-Liouville equations in [2, 4, 8, 10, 11-12,
14-15, 17-18] and other works but it was not considered for the
differential equations with aftereffect.

The form of the differential equation with aftereffect without the
computation of the nodal points were perused in [6]. In [6],
Fereiling and Yurko considered the equation (1.1) under the
Dirichlet boundary conditions, obtained the eigenvalues, proved the
uniqueness theorem by using the transformation operator method. In
[5], we studied the equation (1.1) under the separated boundary
conditions on a finite interval with discontinuity conditions in an
interior point and proved the uniqueness theorem by using the nodal
points but we did not obtain the numerical values of the nodal
points. Whereas in this paper, we consider the differential equation
(1.1) under the boundary conditions (1.2)-(1.3) but without
discontinuity conditions in an interior point and obtain the
numerical values of the nodal points and the nodal length and prove
the uniqueness theorem by applying any dense subset of the nodal
points. In section 2, we obtain the asymptotic solution, the
characteristic function, the eigenvalues, the numerical values of
the nodal points and the nodal length and present a uniqueness
theorem for the solution of inverse aftereffect problem.
\\
\section{\small{\bf{ Main results }}}

First, we present some definitions that will be applied in this
paper.
\begin{definition}
$[6]$ The values of the parameter $\lambda$ for which $L$ has
nonzero solutions are called eigenvalues and the corresponding
nontrivial solutions are called eigenfunctions. The set of
eigenvalues is called the spectrum of $L$.
\end{definition}
\begin{definition}
$[13]$ If $f(z)$ and $g(z),$ two functions of a complex number $z,$
which may be a parameter of the problem or an independent variable
defined on some domain $D,$ possess limits as $z\rightarrow z_{0}$
in $D,$ then we say that $f(z)=O(g(z))$ as $z\rightarrow z_{0}$ if
there exist positive constants $K$ and $\delta$ such that $|f|\leq
K|g|$ whenever $0<|z-z_{0}|<\delta$ and if $|f|\leq K|g|$ for all
$z$ in $D,$ we say $f(z)=O(g(z)).$
\end{definition}
\begin{definition}
$[13]$ If $f(z)$ and $g(z)$ are such that, for any $\varepsilon>0,$
$|f|\leq \varepsilon|g|$ whenever $z$ is in a small
$\delta-$neighborhood of $z_{0},$ we say $f(z)=o(g(z))$ as
$z\rightarrow z_{0}.$
\end{definition}
\begin{definition}
$[13]$ A finite or infinite sequence of functions ${\phi_{n}(z)},
n=1,2,...$ is an asymptotic sequence as $z\rightarrow z_{0}$ if, for
all $n,$ $\phi_{n+1}(z)=o(\phi_{n}(z))$ as $z\rightarrow z_{0},$
that is, $\lim_{z\rightarrow z_{0}}\frac{\phi_{n+1}}{\phi_{n}}=0.$
\end{definition}
\begin{definition}
$[13]$ If ${\phi_{n}(z)}, n=1,2,...$ is an asymptotic sequence of
functions as $z\rightarrow z_{0},$ we say that
$\Sigma_{n=1}a_{n}\phi_{n}(z),$ where the $a_{n}$ are constants
(with the upper limit omitted), is an asymptotic expansion or
asymptotic approximation of the function $f(z)$ if for each $N$
$$f(z)=\Sigma_{n=1}^{N}a_{n}\phi_{n}(z)+o(\phi_{n}(z)) \ \ as \ \ z\rightarrow z_{0}.$$
\end{definition}
\subsection{\small{\bf{ The asymptotic form of the solution and the eigenvalues
}}}  Let $ C(x,\rho) $ and $ S(x,\rho) $ be solutions of (1.1) under
the initial conditions $ S(0,\rho)=C'(0,\rho)=0 $ and $
C(0,\rho)=S'(0,\rho)=1 $. In this case, the functions $ C(x,\rho) $
and $ S(x,\rho) $ satisfy  the following integral equations
respectively (see [6]):
\begin{eqnarray*}
C(x,\rho)=\cos\rho x+
\int_{0}^{x}\frac{\sin\rho(x-t)}{\rho}\left(q(t)C(t,\rho)+\int_{0}^{t}M(t-s)C(s,\rho)ds\right)dt,
\end{eqnarray*}
\begin{eqnarray*}
S(x,\rho)=\frac{\sin\rho x}{\rho}+
\int_{0}^{x}\frac{\sin\rho(x-t)}{\rho}\left(q(t)S(t,\rho)+\int_{0}^{t}M(t-s)S(s,\rho)ds\right)dt.
\end{eqnarray*} \ \ \ \
Denote $ \varphi(x,\rho)=C(x,\rho)+h S(x,\rho) $. Thus, the function
$ \varphi(x,\rho) $ is solution of (1.1) under the initial
conditions $ \varphi(0,\rho )=1 $ and $ \varphi'(0,\rho)=h $ and the
solution of the integral equation
\begin{eqnarray*}
\varphi(x,\rho)=\cos\rho x+h\frac{\sin\rho
x}{\rho}+\int_{0}^{x}\frac{\sin\rho(x-t)}{\rho}
\end{eqnarray*}
\begin{eqnarray}
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \times
\left(q(t)\varphi(t,\rho)+\int_{0}^{t}M(t-s)\varphi(s,\rho)ds\right)dt,
\end{eqnarray}
and hence
\begin{eqnarray*}
\varphi'(x,\rho)=-\rho\sin\rho x+h \cos\rho x+
\int_{0}^{x}\cos\rho(x-t)
\end{eqnarray*}
\begin{eqnarray}
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \times
\left(q(t)\varphi(t,\rho)+\int_{0}^{t}M(t-s)\varphi(s,\rho)ds\right)dt.
\end{eqnarray}
\begin{lemma}{\label{theor}} For $|\rho|\rightarrow\infty$, the
formula \\
\begin{eqnarray}
\varphi(x,\rho)=\cos\rho
x+O(\frac{1}{|\rho|}e^{|\tau|x})=O(e^{|\tau|x}),
\end{eqnarray}
holds, uniformly with to $x\in[0,\pi]$ as $ \tau=Im\rho $.
\end{lemma}
\begin{proof}
See [6].
\end{proof}
Substituting (2.3) into (2.1) and (2.2), we get
\begin{eqnarray*}
\varphi(x,\rho)=\cos\rho x+q_{1}(x) \frac{\sin\rho
x}{\rho}+\frac{1}{2\rho}\int_{0}^{x}q(t)\sin\rho(x-2t)dt
\end{eqnarray*}
\begin{eqnarray}
+\frac{1}{\rho}\int_{0}^{x}\sin\rho(x-t)\int_{0}^{t}M(t-s)\cos \rho
s dsdt+O(\frac{1}{|\rho|^2}e^{|\tau|x}),
\end{eqnarray}
and also
\begin{eqnarray*}
\varphi'(x,\rho)=-\rho\sin\rho x+q_{1}(x) \cos\rho
x+\frac{1}{2}\int_{0}^{x}q(t)\cos\rho(x-2t)dt
\end{eqnarray*}
\begin{eqnarray}
+\int_{0}^{x}\cos\rho(x-t)\int_{0}^{t}M(t-s)\cos \rho s
dsdt+O(\frac{1}{|\rho|}e^{|\tau|x}),
\end{eqnarray}
where
\begin{eqnarray*}
q_{1}(x)=h+\frac{1}{2}\int_{0}^{x}q(t)dt.
\end{eqnarray*}
 Integration by parts results
\begin{eqnarray*}
\int_{0}^{x}q(t)\sin\rho(x-2t)dt=-\frac{1}{2\rho}q(x)\cos\rho
x+\frac{1}{2\rho}q(0)\cos\rho x
\end{eqnarray*}
\begin{eqnarray}
+\frac{1}{2\rho}\int_{0}^{x}q'(t)\cos\rho(x-2t)dt,
\end{eqnarray}
and
\begin{eqnarray*}
\int_{0}^{x}\sin\rho(x-t)\int_{0}^{t}M(t-s)\cos \rho s
dsdt=\frac{1}{\rho}\int_{0}^{x}M(x-s)\cos\rho s ds
\end{eqnarray*}
\begin{eqnarray}
-\frac{1}{\rho}\int_{0}^{x}\cos\rho(x-t)\left(M(0)\cos\rho
t+\int_{0}^{t}\frac{\partial M}{\partial t}(t-s)\cos\rho s
ds\right)dt.
\end{eqnarray}
We obtain from (2.4)-(2.7) that
\begin{eqnarray}
\varphi(x,\rho)=\cos\rho x+q_{1}(x)\frac{\sin\rho
x}{\rho}+O(\frac{1}{|\rho|^2}e^{|\tau|x}),
\end{eqnarray}
and
\begin{eqnarray}
\varphi'(x,\rho)=-\rho\sin\rho x+q_{1}(x) \cos\rho
x+O(\frac{1}{|\rho|}e^{|\tau|x}).
\end{eqnarray}
 Substituting (2.8) into (2.1) and
applying integration by parts, we get
\begin{eqnarray*}
\varphi(x,\rho)=\cos\rho x+q_{1}(x)\frac{\sin\rho
x}{\rho}-\frac{1}{4\rho^{2}}q(x)\cos\rho
x+\frac{1}{4\rho^{2}}q(0)\cos\rho x
\end{eqnarray*}
\begin{eqnarray}
-\frac{\cos\rho
x}{2\rho^{2}}\int_{0}^{x}q(t)q_{1}(t)dt-\frac{M(0)}{2\rho^{2}}x\cos\rho
x+O(\frac{1}{|\rho|^3}e^{|\tau|x}),
\end{eqnarray}
similarly, we get
\begin{eqnarray*}
\varphi'(x,\rho)=-\rho\sin\rho x+q_{1}(x)\cos\rho
x+\frac{3}{4\rho}q(x)\sin\rho x-\frac{1}{4\rho}q(0)\sin\rho x
\end{eqnarray*}
\begin{eqnarray}
+\frac{\sin\rho
x}{2\rho}\int_{0}^{x}q(t)q_{1}(t)dt+\frac{M(0)}{2\rho}x\sin\rho
x+O(\frac{1}{|\rho|^2}e^{|\tau|x}).
\end{eqnarray}
\begin{lemma}{\label{theor}} The following formula holds \\
\begin{eqnarray}
\varphi(x,\rho)=\cos\rho x+\int_{0}^{x}G(x,t)\cos\rho t dt,
\end{eqnarray}
where $G(x,t)$ is a real continuous function with the same
smoothness as $\int_{0}^{x}q(t)dt,$ and
\begin{eqnarray}
G(x,x)=h+\frac{1}{2}\int_{0}^{x}q(t)dt.
\end{eqnarray}
\end{lemma}
\begin{proof}
See [6].
\end{proof}
Let $\varphi(x,\rho)$ and $\psi(x,\rho)$ be solutions of (1.1) under
the initial conditions $\varphi(0,\rho)=1$, $\varphi'(0,\rho)=h $,
$\psi(\pi,\rho)=1$, $\psi'(\pi,\rho)=-H$. Denote
\begin{eqnarray}
\Delta(\rho):=\langle\psi(x,\rho),\varphi(x,\rho)\rangle,
\end{eqnarray}
where
\begin{eqnarray*}
\langle\psi(x,\rho),\varphi(x,\rho)\rangle=\psi(x,\rho)\varphi'(x,\rho)-\psi'(x,\rho)\varphi(x,\rho),
\end{eqnarray*}
is the Wronskian of $\psi(x,\rho)$ and $\varphi(x,\rho)$. Since the
function $\Delta(\rho)$ called the characteristic function for the
boundary value problem $L(q,M,h,H)$ does not depend on $x$, hence,
substituting $x=\pi$ into (2.14), we get
\begin{eqnarray}
\Delta(\rho)=V(\varphi)=\varphi'(\pi,\rho)+H\varphi(\pi,\rho).
\end{eqnarray}
\begin{lemma}{\label{theor}}For $|\rho|\rightarrow\infty$, the representation \\
\begin{eqnarray*}
\Delta(\rho)=-\rho\sin\rho \pi+\omega\cos\rho
\pi+\frac{3}{4\rho}q(\pi)\sin\rho \pi-\frac{1}{4\rho}q(0)\sin\rho
\pi
\end{eqnarray*}
\begin{eqnarray*}
+\frac{\sin\rho
\pi}{2\rho}\int_{0}^{\pi}q(t)q_{1}(t)dt+\frac{M(0)}{2\rho}\pi\sin\rho
\pi
\end{eqnarray*}
\begin{eqnarray}
+H q_{1}(\pi)\frac{\sin\rho
\pi}{\rho}+O(\frac{1}{|\rho|^2}e^{|\tau|\pi}),
\end{eqnarray}
 holds where $\omega=H+q_{1}(\pi)$.
\end{lemma}
\begin{proof} we arrive at (2.16) from (2.10), (2.11) and (2.15) by using
straightforward calculations.
\end{proof}

Since the eigenvalues $ \{\rho_{n}\}_{n\geq1} $ of the boundary
value problem coincide with the zeros of the function $ \Delta(\rho)
$, using some straightforward calculations (see [6]), we obtain
\begin{eqnarray}
\rho_{n}=n+\frac{\omega}{n\pi}+O(\frac{1}{n^{2}}), \ \
n\rightarrow\infty.
\end{eqnarray}
\subsection{\small{\bf{The asymptotic form of the nodal points and the associated nodal length }}}
Denote
\begin{eqnarray}
\varphi_{n}(x)=\varphi(x,\rho_{n}),
\end{eqnarray}
where $\varphi_{n}$ is the eigenfunction corresponding to the
eigenvalue $\lambda_{n}.$ Let
$\lambda_{0}<\lambda_{1}<\ldots\rightarrow\infty$ be the eigenvalues
of the aftereffect problem (1.1)-(1.3) and also let the nodal points
of the $n$th eigenfunction $\varphi_{n}$ be shown with
$0<x_{n}^1<x_{n}^2<\cdots<x_{n}^j<\pi,$ $j=1,2,...,n-1.$ The set of
all nodal points $\{x_{n}^{j}\}_{n>1,j=\overline{1,n-1}}$ is dense
in $[0,\pi]$ (see [10]).
\begin{theorem}{\label{theor}}The nodal points of the aftereffect
problem (1.1)-(1.3) are
\begin{eqnarray*}
x_{n}^{j}=(j-\frac{1}{2})\frac{\pi}{n}+\frac{1}{n^{2}}q_{1}(x_{n}^{j})+\frac{1}{2n^{2}}\int_{0}^{x_{n}^{j}}q(t)\cos
2nt dt
\end{eqnarray*}
\begin{eqnarray}
\ \ \ \ \ \ \ \
+\frac{1}{n^{2}}\int_{0}^{x_{n}^{j}}\int_{0}^{t}M(t-s)\cos nt \cos
ns ds dt+O(\frac{1}{n^{3}}),
\end{eqnarray}
and the nodal length is
\begin{eqnarray*}
l_{n}^{j}=\frac{\pi}{n}+\frac{1}{2
n^{2}}\int_{x_{n}^{j}}^{x_{n}^{j+1}}q(t)
dt+\frac{1}{2n^{2}}\int_{x_{n}^{j}}^{x_{n}^{j+1}}q(t)\cos 2nt dt
\end{eqnarray*}
\begin{eqnarray}
\ \ \ \ \ \ \ \
+\frac{1}{n^{2}}\int_{x_{n}^{j}}^{x_{n}^{j+1}}\int_{0}^{t}M(t-s)\cos
nt \cos ns ds dt+O(\frac{1}{n^{3}}).
\end{eqnarray}
\end{theorem}
\begin{proof}
Since the nodal points are the zeroes of the eigenfunctions, then we
get from (2.1) that
\begin{eqnarray*}
\varphi_{n}(x)=\cos\rho_{n} x+h\frac{\sin\rho_{n}
x}{\rho_{n}}+\int_{0}^{x}\frac{\sin\rho_{n}(x-t)}{\rho_{n}}
\end{eqnarray*}
\begin{eqnarray*}
\ \ \ \ \ \ \ \ \ \times
\left(q(t)\varphi_{n}(t)+\int_{0}^{t}M(t-s)\varphi_{n}(s)ds\right)dt
\end{eqnarray*}
\begin{eqnarray*}
\ \ \ \ \ \ \ \ \ \ \ \ \ =\cos\rho_{n} x+h\frac{\sin\rho_{n}
x}{\rho_{n}}+\frac{\sin\rho_{n}(x)}{\rho_{n}}\int_{0}^{x}\cos
\rho_{n}(t)
\end{eqnarray*}
\begin{eqnarray*}
\ \ \ \ \ \ \ \ \ \times
\left(q(t)\varphi_{n}(t)+\int_{0}^{t}M(t-s)\varphi_{n}(s)ds\right)dt
\end{eqnarray*}
\begin{eqnarray*}
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
-\frac{\cos\rho_{n}(x)}{\rho_{n}}\int_{0}^{x}\sin\rho_{n}(t)\left(q(t)\varphi_{n}(t)+\int_{0}^{t}M(t-s)\varphi_{n}(s)ds\right)dt.
\end{eqnarray*}
Now, we set $\varphi_{n}(x)=0.$ Thus, we obtain
\begin{eqnarray*}
\cot\rho_{n} x+\frac{h}{\rho_{n}}+\frac{1}{\rho_{n}}\int_{0}^{x}\cos
\rho_{n}(t)\left(q(t)\varphi_{n}(t)+\int_{0}^{t}M(t-s)\varphi_{n}(s)ds\right)dt
\end{eqnarray*}
\begin{eqnarray*}
-\frac{\cot\rho_{n}(x)}{\rho_{n}}\int_{0}^{x}\sin\rho_{n}(t)\left(q(t)\varphi_{n}(t)+\int_{0}^{t}M(t-s)\varphi_{n}(s)ds\right)dt=0,
\end{eqnarray*}
then, for $n\rightarrow\infty$, we get
\begin{eqnarray*}
x_{n}^{j}=(j-\frac{1}{2})\frac{\pi}{\rho_{n}}+\frac{h}{\rho_{n}^{2}}+\frac{1}{\rho_{n}^{2}}\int_{0}^{x_{n}^{j}}\cos
\rho_{n}(t)\left(q(t)\varphi_{n}(t)+\int_{0}^{t}M(t-s)\varphi_{n}(s)ds\right)dt.
\end{eqnarray*}
From (2.3) and (2.17), we obtain
\begin{eqnarray*}
x_{n}^{j}=(j-\frac{1}{2})\frac{\pi}{n}+\frac{1}{n^{2}}q_{1}(x_{n}^{j})+\frac{1}{2n^{2}}\int_{0}^{x_{n}^{j}}q(t)\cos
2nt dt
\end{eqnarray*}
\begin{eqnarray*}
\ \ \ \ \ \ \ \
+\frac{1}{n^{2}}\int_{0}^{x_{n}^{j}}\int_{0}^{t}M(t-s)\cos nt \cos
ns ds dt+O(\frac{1}{n^{3}}).
\end{eqnarray*}
Also, the nodal length is
\begin{eqnarray*}
l_{n}^{j}=x_{n}^{j+1}-x_{n}^{j},
\end{eqnarray*}
therefore
\begin{eqnarray*}
l_{n}^{j}=\frac{\pi}{n}+\frac{1}{2n^{2}}\int_{x_{n}^{j}}^{x_{n}^{j+1}}q(t)dt+\frac{1}{2n^{2}}\int_{x_{n}^{j}}^{x_{n}^{j+1}}q(t)\cos
2nt dt
\end{eqnarray*}
\begin{eqnarray*}
\ \ \ \ \ \ \ \
+\frac{1}{n^{2}}\int_{x_{n}^{j}}^{x_{n}^{j+1}}\int_{0}^{t}M(t-s)\cos
nt \cos n s ds dt+O(\frac{1}{n^{3}}),
\end{eqnarray*}
and consequently, the theorem is proved.
\end{proof}
\begin{example}
Let $h=1$ and $q(x)=M(x)=x.$ Using (2.19) and (2.20), we obtain \\
\begin{center}
\begin{tabular}{c c c c c c c c c c}
  \hline
  % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
  \multicolumn{6}{c}{The numerical values of the nodal points} \\
  \cline{1-7}
  $x_{n}^{j}$ & j=1 & j=2 & j=3 & j=4 & j=5 & j=6 & j=7 & j=8 & j=9 \\ \hline
  n=2 & 1.0581 &  &  &  &  &  &  &  &  \\
  n=3 & 0.6391 & 0.7473 &  &  &  &  &  &  &  \\
  n=4 & 0.4566 & 1.2613 & 2.0852 &  &  &  &  &  &  \\
  n=5 & 0.3547 & 0.9909 & 1.6351 & 2.2870 &  &  &  &  &  \\
  n=6 & 0.2899 & 0.8166 & 1.3483 & 1.8821 & 2.4220 &  &  &  &  \\
  n=7 & 0.2450 & 0.6955 & 1.1486 & 1.6030 & 2.0606 & 2.5188 &  &  &  \\
  n=8 & 0.2121 & 0.6058 & 1.0011 & 1.3970 & 1.7949 & 2.1931 & 2.5935 &  &  \\
  n=9 & 0.1869 & 0.5366 & 0.8873 & 1.2384 & 1.5907 & 1.9432 & 2.2971 & 2.6510 &  \\
  n=10 & 0.1671 & 0.4817 & 0.7969 & 1.1124 & 1.4287 & 1.7451 & 2.0624 & 2.3798 & 2.6982 \\
  \hline
\end{tabular}
\end{center}

and also \\
\begin{center}
    \begin{tabular}{c c c c c c c c c}
  \hline
  % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
  \multicolumn{6}{c}{The numerical values of the nodal length} \\
  \cline{1-7}
  $l_{n}^{j}$ & j=1 & j=2 & j=3 & j=4 & j=5 & j=6 & j=7 & j=8 \\ \hline
  n=3 & 1.0895 &  &  &  &  &  &  &  \\
  n=4 & 0.8000 & 0.8268 &  &  &  &  &  &  \\
  n=5 & 0.6345 & 0.6452 & 0.6493 &  &  &  &  &  \\
  n=6 & 0.5267 & 0.5317 & 0.5338 & 0.5398 &  &  &  &  \\
  n=7 & 0.4505 & 0.4532 & 0.4544 & 0.4576 & 0.4582 &  &  &  \\
  n=8 & 0.3937 & 0.3953 & 0.3960 & 0.3978 & 0.3983 & 0.4004 &  &  \\
  n=9 & 0.3497 & 0.3507 & 0.3511 & 0.3523 & 0.3525 & 0.3539 & 0.3539 &  \\
  n=10 & 0.3146 & 0.3152 & 0.3155 & 0.3162 & 0.3164 & 0.3173 & 0.3173 & 0.3183  \\
  \hline
\end{tabular}
\end{center}
\end{example}
\subsection{\small{\bf{The uniqueness theorem for the inverse aftereffect problem }}}
In this section, we consider two boundary value problems
$L(q,M,h,H)$ and $ \tilde{L}(q,\tilde{M},\tilde{h},\tilde{H}) $
where $ \tilde{M} $ has same properties of $ M $ and prove the
uniqueness theorem by using any dense subset of the nodal points.
\begin{theorem}{\label{theor}}
Let $q(x)=\tilde{q}(x)$ on $[0,\pi]$. Then the function $M$ and the
numbers $h,$ $H$ are uniquely determined by any dense subset of the
nodes in $[0,\pi]$.
\end{theorem}
\begin{proof}
Consider the problems
\begin{eqnarray}
-y''(x)+q(x)y(x)+\int_{0}^{x}M(x-t)y(t)dt=\rho^{2} y(x),& 0\leq
x\leq \pi,
\end{eqnarray}
\begin{eqnarray}
 y'(0)-h y(0)=0,\ \ \ \ \ \ \ \ y'(\pi)+H y(\pi)=0,
\end{eqnarray}
and
\begin{eqnarray}
-y''(x)+q(x)y(x)+\int_{0}^{x}\tilde{M}(x-t)y(t)dt=\tilde{\rho}^{2}
y(x),& 0\leq x\leq \pi,
\end{eqnarray}
\begin{eqnarray}
y'(0)-\tilde{h} y(0)=0,\ \ \ \ \ \ \ \ y'(\pi)+\tilde{H} y(\pi)=0.
\end{eqnarray}
Let $\varphi_{n}(x)$ and $\tilde{\varphi}_{n}(x)$ be the solutions
of (2.21) and (2.23) under the initial conditions $
\varphi_{n}(0,\rho)=1,$ $\varphi_{n}'(0,\rho)=h$ and
$\tilde{\varphi}_{n}(0,\rho)=1,$
$\tilde{\varphi}'_{n}(0,\rho)=\tilde{h}$, respectively and also let
$ x_{n}^{j}=\tilde{x}_{n}^{j}, $ for $n>1$ and $j=1,2,...,n-1,$ be a
dense set in $[0,\pi]$ (see [10]). Then from (2.21) and (2.23), we
obtain
\begin{eqnarray*}
[\tilde{\varphi}_{n}'(x)\varphi_{n}(x)-\tilde{\varphi}_{n}(x)\varphi'_{n}(x)]'=
\int_{0}^{x}[\tilde{M}(x-t)\varphi_{n}(x)\tilde{\varphi}_{n}(t)
\end{eqnarray*}
\begin{eqnarray}
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
-M(x-t)\varphi_{n}(t)\tilde{\varphi}_{n}(x)]
dt+(\rho_{n}^{2}-\tilde{\rho}_{n}^{2})\varphi_{n}\tilde{\varphi}_{n},
\end{eqnarray}
integrating (2.25) from $ 0 $ to $ x_{n}^{j} $ and using (2.22) and
(2.24), we get
\begin{eqnarray*}
(h-\tilde{h})\varphi_{n}(0)\tilde{\varphi}_{n}(0)=\int_{0}^{x_{n}^{j}}
\int_{0}^{x}[\tilde{M}(x-t)\varphi_{n}(x)\tilde{\varphi}_{n}(t)
\end{eqnarray*}
\begin{eqnarray}
\ \ \ \ \ \ \ \ \ \ \ \ \ \
-M(x-t)\varphi_{n}(t)\tilde{\varphi}_{n}(x)] dt
dx+(\rho_{n}^{2}-\tilde{\rho}_{n}^{2})\int_{0}^{x_{n}^{j}}\varphi_{n}(x)\tilde{\varphi}_{n}(x)dx.
\end{eqnarray}
Now, we select a subsequence of the nodal points from the dense set
that tends to $0$, then from (2.26), we get $h=\tilde{h}.$
Integrating both sides of (2.25) from $ x_{n}^{j} $ to $ \pi $, we
obtain
\begin{eqnarray*}
(H-\tilde{H})\varphi_{n}(\pi)\tilde{\varphi}_{n}(\pi)=\int_{x_{n}^{j}}^{\pi}
\int_{0}^{x}[\tilde{M}(x-t)\varphi_{n}(x)\tilde{\varphi}_{n}(t)
\end{eqnarray*}
\begin{eqnarray}
\ \ \ \ \ \ \ \ \ \ \ \ \ \
-M(x-t)\varphi_{n}(t)\tilde{\varphi}_{n}(x)] dt
dx+(\rho_{n}^{2}-\tilde{\rho}_{n}^{2})\int_{x_{n}^{j}}^{\pi}\varphi_{n}(x)\tilde{\varphi}_{n}(x)dx.
\end{eqnarray}
We select a subsequence of the nodal points that tends to $\pi$.
Thus, we get $H=\tilde{H}.$ Consequently
$\rho_{n}=\tilde{\rho_{n}}.$ Now, Integrating (2.25) from $ 0 $ to $
x_{n}^{j} $, we obtain
\begin{eqnarray}
\int_{0}^{x_{n}^{j}}\int_{0}^{x}[\tilde{M}(x-t)\varphi_{n}(x)\tilde{\varphi}_{n}(t)-M(x-t)\varphi_{n}(t)\tilde{\varphi}_{n}(x)]dtdx=0.
\end{eqnarray}
We take a sequence $ \{x_{n}^{j}\}_{n>1,j=\overline{1,n-1}} $
accumulating at an arbitrary $ b\in[0,\pi] $. Then, using (2.3) for
$n\rightarrow\infty$, we get
\begin{eqnarray}
\int_{0}^{b}\int_{0}^{x}[\tilde{M}(x-t)-M(x-t)]\cos n x \cos n t
dtdx=0.
\end{eqnarray}
Consequently, from the completeness of the function $ \cos, $ we can
conclude that $ M $ is uniquely determined on $ [0,\pi] $.
\end{proof}
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\end{document}