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{\bf\small Caspian Journal of Mathematical Sciences (CJMS)}\\
{\bf\small University of Mazandaran, Iran }\\
{\bf\small  \url{http://cjms.journals.umz.ac.ir}}\\
{\bf\small ISSN: 1735-0611}\\
\end{flushright}
{{\small CJMS}. {\bf xx}(x)(201x), xx-xx}\\
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\begin{center}
{\bf\large On the determination of eigenvalues for
differential pencils with the turning point}
 \\[0.5cm]
{A. Neamaty \footnote{ Corresponding author: namaty@umz.ac.ir\\ \qquad Received: xx Month 201x\\ \qquad  Revised: xx Month 201x \\ \qquad Accepted: 16 Apr 2016} and Y. Khalili $^2$\\
$^1$ Department of Mathematics, University of Mazandaran, Babolsar, Iran\\
$^2$ Department of Basic Sciences, Sari
Agricultural Sciences and Natural Resources University, Sari, Iran} \\[2mm]


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\begin{quotation}
\noindent
{\footnotesize {\sc Abstract.}
In this paper, we investigate
the boundary value problem for differential pencils on the half-line
with a turning point. Using a fundamental system of solutions, we
give a asymptotic distribution of eigenvalues.\\

{ Keywords:} Eigenvalues, Differential
pencil, Turning point.\\

\noindent
\textit{2010 Mathematics subject classification: } 34L15,
34K10, 34E20.}
\end{quotation}
\markboth {A. Neamaty and Y. Khalili}{On the determination of eigenvalues}
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\section{Introduction}
\noindent
\hspace*{.5cm}We consider the differential equation
\begin{eqnarray}
y''(x)+(\rho^{2}r(x)+i \rho q_{1}(x)+q_{0}(x))y(x)=0 \ , \ \ \ \
x\geq0,
\end{eqnarray}
on the half-line with a nonlinear dependence on the spectral parameter
$\rho$. Let $ 0<a<1,$ and
\begin{eqnarray}
r(x)=\le\{ \ba{ll} -(x-1)^{2}, &
0 \leq x<a,\\[0.5cm]  1, & x\geq a,  \ea \ri.
\end{eqnarray}
i.e., the sign of the weight function changes in an interior point
$x=a,$ which is called the turning point. The functions $q_{j}(x),\
j=0,1, $ are complex-valued, $ q_{1}(x)$ is absolutely continuous
and $(1+x)q_{j}^{(l)}\in L(0,\infty)$ for $0\leq l\leq j\leq1$.
\\
\hspace*{.5cm}Differential equations with turning points are widely
used to describe many important phenomena and dynamic processes in
physics, geophysics, mechanics (see [5,7] for details). The
classical Sturm-Liouville operators with turning points in the
finite interval have been studied fairly completely in [2].
Indefinite differential pencils with turning points produce
significant qualitative modification in the investigation of the
inverse problem. Some aspects of the inverse problem theory for
differential pencils without turning points have been studied in [3,8].
Here we investigate a boundary value problem for differential
pencils with a turning point. Similar problems for Sturm-Liouville
operators have been studied in [9].  In this work, the weight function is
a polynomial of degree two before the turning point.\\
\hspace*{.5cm}In this paper, we will study the solution for Eq. (1.1). In Section 2, we determine
the asymptotic form of the characteristic function and then give the eigenvalues. \\
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\section{Main result}
\hspace*{.5cm}We consider the boundary value problem $L$ for Eq.
(1.1) with the spectral boundary condition
\begin{eqnarray}
U(y):=y'(0)+(\beta_{1}\rho+\beta_{0})y(0)=0,
\end{eqnarray}
where the coefficients $ \beta_{1}$ and $\beta_{0}$ are complex
numbers and $\beta_{1}\neq \pm 1$.\\
\hspace*{.5cm}Denote $\Pi_{\pm}:=\{\rho:\pm Im\rho>0\}$ and
$\Pi_{0}:=\{\rho:Im\rho=0\}$. By the well-known method (see [4,7]),
we obtain a solution for Eq. (1.1) which is called the
Jost-type solution.\\
\textbf{Theorem 1.} The Eq. (1.1) has a unique solution
$y=e(x,\rho)$, $\rho\in \Pi_{\pm}$, $x\geq a$, with the following
properties :\\
1. For each fixed $x\geq a$, the functions $e^{(\nu)}(x,\rho)$,
$\nu= 0,1$, are holomorphic for $\rho\in{\Pi}_{+}$
 and $\rho\in{\Pi}_{-}$(i.e., they are piecewise holomorphic).\\
2. The functions $e^{(\nu)}(x,\rho)$, $\nu= 0,1, $ are continuous
for $x\geq a $, $\rho\in\overline{\Pi_{+}}$ and
$\rho\in\overline{\Pi_{-}}$. In other words, for real $\rho$, there exist the finite
limits
\begin{eqnarray*}
e^{(\nu)}_{\pm}(x,\rho)=\lim_{z\longrightarrow\rho,\ z\in\Pi_{\pm}}
e^{(\nu)}(x,z).
\end{eqnarray*}
Moreover, the functions $e^{(\nu)}(x,\rho), \nu=0,1,$ are
continuously differentiable with respect to $\rho\in
\overline{\Pi_{+}}\setminus\{0\}$ and
$\rho\in\overline{\Pi_{-}}\setminus\{0\}$.\\
3. For $x\longrightarrow\infty$,
$\rho\in\overline{\Pi_{\pm}}\setminus\{0\}$, $\nu=0,1,$
\begin{eqnarray}
e^{(v)}(x,\rho)=(\pm i\rho)^{v} \exp(\pm(i\rho x-Q(x)))(1+o(1)),
\end{eqnarray}
where \ \ \ $Q(x)=\frac{1}{2}\int_{0}^{x} q_{1}(t) dt.$\\
4. For $|\rho|\longrightarrow\infty,\rho\in\overline{\Pi_{\pm}}$,
$\nu=0,1$, uniformly  in $x\geq a,$
\begin{eqnarray}
e^{(\nu)}(x,\rho)=(\pm i\rho)^{\nu} \exp(\pm(i\rho x-Q(x)))[1],
\end{eqnarray}
where \ \ \ $[1]:=1+O\left(\rho^{-1}\right).$ \\
\hspace*{.5cm}We  extend $e(x,\rho)$ to the segment $[0,a]$ as a
solution of Eq. (1.1) which is smooth for $x\geq 0$, i.e.,
\begin{eqnarray}
e^{(\nu)}(a-0,\rho)=e^{(\nu)}(a+0,\rho), & \nu=0,1.
\end{eqnarray}
Then the properties $1-2$ remain true for $x\geq 0$.\\
\hspace*{.5cm}Let the function $\varphi(x,\rho)$ be the solution of
Eq. (1.1) under the initial conditions $\varphi(0,\rho)=1$ and $
U(\varphi)=0.$ For each fixed $x\geq0,$ the functions
$\varphi^{(\nu)}(x,\rho),\  \nu= 0,1$ are entire in $\rho$.\\
\hspace*{.5cm}Denote
\begin{eqnarray}
\Delta(\rho):=U(e(x,\rho)).
\end{eqnarray}
The function $\Delta(\rho)$ is called the characteristic function
for the boundary value problem $L.$ The function $\Delta(\rho)$ is
holomorphic in $\Pi_{+}$ and  $\Pi_{-}$, and for real $\rho,$ there
exist the finite limits
\begin{eqnarray*}
\Delta_{\pm}(\rho)=\lim_{z\longrightarrow\rho,\ z\in\Pi_{\pm}}
\Delta(z).
\end{eqnarray*}
Moreover, the function $\Delta(\rho)$ is continuously differentiable
for $\rho\in\overline{\Pi_{\pm}}\setminus\{0\}$.\\
\textbf{Definition 1.} The values of the parameter $\rho,$ for which
the Eq. (1.1) has nontrivial solutions satisfying the
conditions $U(y)=0,$ $y(\infty)=0$ (i.e.,
$\lim_{x\longrightarrow\infty}y(x)=0)$ are called eigenvalues of
$L,$ and the corresponding solutions are called eigenfunctions.\\
\textbf{Theorem 2.} For $|\rho|\longrightarrow\infty,
\rho\in\overline{\Pi_{\pm}}$, the following asymptotical formula
holds:
\begin{eqnarray*}
\Delta(\rho)&=&\frac{\rho}{2(a-1)}\exp(\pm(i\rho a-Q(a)))\\
&& \hspace*{.5cm} \times\bigg{(}(a\mp
i)(1+\beta_{1})\exp\Big{(}\frac{\rho}{2}(a^{2}-2a)
-\frac{i Q(a)}{a-1}\Big{)}[1] \\
&& \hspace*{1.25cm}-(a\pm i)(1-\beta_{1})\exp\Big{(}-\frac{\rho}{2}(a^{2}-2a)+\frac{i
Q(a)}{a-1}\Big{)}[1]\bigg{)}.\hspace*{4cm}
\end{eqnarray*}
\textbf{Proof.} Denote ${\Pi}^{1}_{\pm}:=\{\rho:\pm Re\rho>0\}$. It
is known (see [4,7]) that for $x\in [0, a],$ $\nu=0, 1,$
$\rho\in\overline{{\Pi}^{1}_{\pm}},$ $|\rho|\longrightarrow\infty,$
there exists the Birkhoff-type fundamental system  of solutions
$\{y_{k}(x,\rho)\}_{k=1, 2}$ of Eq. (1.1) of the form
\begin{eqnarray}
\nonumber y_{k}^{(\nu)}(x,\rho)=((-1)^{k} \rho(x-1))^{\nu}\exp
\left((-1)^{k}\left(\frac{\rho (x^{2}-2x)}{2}-{\frac{i
Q(x)}{x-1}}\right)\right)[1].\\
\end{eqnarray}
Using these solutions, one has
\begin{eqnarray}
e^{(\nu)}(x,\rho)=h_{1}(\rho)y^{(\nu)}_{1}(x,\rho)+h_{2}(\rho)y^{(\nu)}_{2}(x,\rho),\
\ \ \ x\in[0,a].
\end{eqnarray}
Taking Cramer’s rule, we calculate
\begin{eqnarray*}
h_{1}(\rho)= \frac{a\pm i}{2(a-1)}\exp(\pm(i\rho
a-Q(a)))\exp\left(\frac{-\rho}{2}(a^{2}-2a)+\frac{i Q(a)}{a-1}\right)[1],\hspace*{2cm}\\
h_{2}(\rho)= \frac{a\mp i}{2(a-1)}\exp(\pm(i\rho
a-Q(a)))\exp\left(\frac{\rho}{2}(a^{2}-2a)-\frac{i
Q(a)}{a-1}\right)[1].\hspace*{2.4cm}
\end{eqnarray*}
Now, substituting (2.6) and coefficients $h_{j}(\rho),\ j=1, 2$ in
(2.7), we have
\begin{eqnarray*}
e^{(\nu)}(x,\rho)&=&\frac{(\rho(x-1))^{\nu}}{2(a-1)}\exp(\pm(i\rho
a-Q(a)))\\
&&\times((-1)^{\nu}(a\mp i)\exp(k(x,\rho))
[1]+(a\pm
i)\exp(-k(x,\rho))[1]),
\end{eqnarray*}
where
\begin{eqnarray*}
k(x,\rho)=\frac{-\rho}{2}\left((x^{2}-2x)-(a^{2}-2a)\right)
+i \left(\frac{Q(x)}{x-1}-\frac{Q(a)}{a-1}\right).
\end{eqnarray*}
Together with (2.1) and (2.5), this yields the characteristic function. Theorem 2 is proved. $\diamondsuit$\\
\hspace*{.5cm}Now we obtain the eigenvalues for the boundary value problem $L.$\\
 \textbf{Theorem 3.} 1) For sufficiently large $k$,
the function $\Delta(\rho)$ has simple zeros of the form
\begin{eqnarray}
\rho_{k}=\frac{1}{a^{2}-2a}\left(2k\pi i+ 2\frac{i
Q(a)}{a-1}+\kappa_{1}\pm\kappa_{2}\right)+O\left(k^{-1}\right),
\end{eqnarray}
where
\begin{eqnarray*}
\kappa_{1}=ln\frac{1-\beta_{1}}{1+\beta_{1}}, \qquad
\kappa_{2}=ln\frac{a+i}{a-i}.
\end{eqnarray*}
2) For real $\rho\neq0,$ $L$ has no eigenvalues.\\
3) Let $\ \Lambda^{'}=\Lambda^{'}_{+}\cup\Lambda^{'}_{-},$ where
$\Lambda^{'}_{\pm}=\{\rho\in\Pi_{\pm};\Delta(\rho)=0\}.$ The set
$\Lambda^{'}$ coincides with the set of all non-zero eigenvalues of
$L.$ For $\rho_{k}\in\Lambda^{'},$ the functions $e(x,\rho_{k})$ and
$ \varphi(x,\rho_{k})$ are eigenfunctions and
\begin{eqnarray}
e(x,\rho_{k})= \gamma_{k}\varphi(x,\rho_{k}),\ \ \ \gamma_{k}\neq0.
\end{eqnarray}
\textbf{Proof.} Using  characteristic function  and Rouche's theorem
[1], we obtain a countable set of the zeros for the function
$\bigtriangleup(\rho)$ of the form (2.8). Let $\rho_{0}\neq0$ be
real. Then the function
$y(x,\rho_{0})=c_{1}e_{+}(x,\rho_{0})+c_{2}e_{-}(x,\rho_{0})$
vanishes at infinity only if $c_{1}=c_{2}=0$. Thus for real
$\rho\neq0,$ BVP($L$) has no eigenvalues. For prove part 3, let
$\rho_{k}\in\Lambda^{'}$. Therefore
$\Delta(\rho_{k})=U(e(x,\rho_{k}))=0.$ Also
$\lim_{x\longrightarrow\infty}e(x,\rho_{k})=0$. Thus $\rho_{k}$ is
an eigenvalue. Since the Wronskian of the functions
$\varphi(x,\rho)$ and $e(x,\rho)$, i.e.,
$<\varphi(x,\rho),e(x,\rho)>=\bigtriangleup(\rho)$\ (see [6]), we
arrive at (2.9). Conversely, let $\rho_{k}$ (complex value) be an
eigenvalue and $y(x,\rho_{k})$ be a corresponding eigenfunction.
Since $U(y(x,\rho_{k}))=0$ and
$\lim_{x\longrightarrow\infty}y(x,\rho_{k})=0,$ one gets
$y(x,\rho_{k})=c_{k0}\varphi(x,\rho_{k})$ and
$y(x,\rho_{k})=c_{k1}e(x,\rho_{k})$ for $c_{ks}\neq0,\ s=0,1,$
respectively. These yield (2.9) and $\Delta(\rho_{k})=U(e(x,\rho_{k}))=0.$ The proof of Theorem 3 is completed. $\diamondsuit$\\

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\end{document}