\begin{document} 

\title{On the Sturm-Liouville Operator with Summable Potential}
\author{O. A. Veliev \\
%EndAName
{\small Depart. of Math, Fen-Ed. Fak, Marmara University., }\\
{\small \ G\"{o}ztepe Kamp\"{u}s\"{u}, 81040 , Kadik\"{o}y Istanbul , 
Turkey,%
}\\
{\small \ e-mail: oveliev@marmara.edu.tr} \and M.Toppamuk Duman \\
%EndAName
{\small \ Depart. of Math, Fen-Ed. Fak, Dokuz Eyl\"{u}l Univ., }\\
{\small \ Tinaztepe Kam, Kaynaklar, Buca, 35160, \ Izmir, Turkey,}\\
{\small \ e-mail: melda.duman@deu.edu.tr }}
\maketitle 

\begin{abstract}
We investigate the Sturm-Liouville operator
\[
L(q)=-\frac{d^{2}}{dx^{2}}+q(x)
\]
in $L_{2}[0,1]$ with strongly regular boundary conditions and arbitrary
Lebesque integrable Potential $q(x)$. We obtain asymptotic formulas of
arbitrary order for eigenvalues and eigenfunctions of $L(q).$ Besides we
give a simple proof of Riesz basisness of eigenfunctions and associeted
functions of this operator.
\end{abstract} 

\section{ Introduction} 

Consider the boundary value problem
\begin{equation}
 -y^{^{\prime \prime }}(x)+q(x)y(x)=\lambda y(x),\qquad 0\leq x\leq 1
\end{equation} 

\[
U_{1}(y)=a_{1}y^{^{\prime }}(0)+b_{1}y^{^{\prime
}}(1)+a_{0}y(0)+b_{0}y(1)=0,
\] 

\begin{equation}
U_{2}(y)=c_{1}y^{^{\prime }}(0)+d_{1}y^{^{\prime 
}}(1)+c_{0}y(0)+d_{0}y(1)=0,
\end{equation}
where $q(x)$ is a Lebesgue integrable complex-valued function and the
coefficients of the boundary conditions (2)$\ $are complex numbers. We use
the following notations, definitions, and results of [3] ( see p. 56-65 of
[3]). 

The normalized boundary conditons of (2) is
\[
U_{i}(y)=U_{i0}(y)+U_{i1}(y)=0,\text{ }i=1,2,
\]
where $U_{i0}(y)=\alpha _{i}y^{k_{i}}(0)+\alpha _{i,0}y(0)$ , $%
U_{i1}(y)=\beta _{i}y^{k_{i}}(1)+\beta _{i,0}y(1)$ and $k_{i}$ are the
maximal orders such that $0\leq k_{2}\leq k_{1}\leq 1$ and at least one of 
$%
\alpha _{i}$ and $\beta _{i}$ is nonzero for any $i=1,2$. The normalized
boundary conditions is said to be regular if the numbers $\theta _{-1}$ and 
$%
\theta _{1}$ defined by
\begin{equation}
\frac{\theta _{-1}}{s}+\theta _{0}+s\theta _{1}=\left|
\begin{array}{cc}
(\alpha _{1}+s\beta _{1})w_{1}^{k_{1}} & (\alpha _{1}+\frac{1}{s}\beta
_{1})w_{2}^{k_{1}} \\
(\alpha _{2}+s\beta _{2})w_{1}^{k_{2}} & (\alpha _{2}+\frac{1}{s}\beta
_{2})w_{2}^{k_{2}}
\end{array}
\right|  \label{charct}
\end{equation}
are nonzero, where $w_{1}$ and $w_{2}$ are the two distinct roots of -1. 

If the roots $\zeta _{1},\zeta _{2}$ of (3) (or $\theta _{1}\zeta
^{2}+\theta _{0}\zeta +\theta _{-1}=0)$ are simple, that is,
\begin{equation}
\theta _{0}^{2}-4\theta _{-1}\theta _{1}\neq 0,  \label{discrminant}
\end{equation}
then we say that the boundary conditions (2) \ are strongly regular. This
occurs in only following three cases: 

\textbf{1. }$a_{1}d_{1}-b_{1}c_{1}\neq 0.$ Then (2) equivalent to the
conditions\textbf{\ }
\begin{equation}
y^{^{\prime }}(0)+\alpha _{11}y(0)+\alpha _{12}y(1)=0,\text{ }y^{^{\prime
}}(1)+\alpha _{21}y(0)+\alpha _{22}y(1)=0.  \label{bc1}
\end{equation}
Hense $\theta _{0}=0,$ $\theta _{-1}=\theta _{1}=1,$ $\theta
_{0}^{2}-4\theta _{-1}\theta _{1}\neq 0$ 

\textbf{2. }$a_{1}d_{1}-b_{1}c_{1}=0,$ $\left| a_{1}\right| +\left|
b_{1}\right| >0,$ $b_{1}c_{0}+a_{1}d_{0}\neq 0$ , $c_{0}\neq \pm d_{0},$ $%
a_{1}\neq \pm b_{1}$. 

Then (2) equivalent to
\begin{eqnarray}
U_{1}(y) &=&a_{1}y^{^{\prime }}(0)+b_{1}y^{^{\prime
}}(1)+a_{0}y(0)+b_{0}y(1)=0,\text{ }  \nonumber \\
\text{\ }U_{2}(y) &=&c_{0}y(0)+d_{0}y(1)=0,
\end{eqnarray}
and $\theta _{0}=2(a_{1}c_{0}+b_{1}d_{0})w_{1},$ \ $\theta _{-1}=\theta
_{1}=(b_{1}c_{0}+a_{1}d_{0})w_{1}\neq 0.$ Hense
\begin{equation}
\theta _{0}^{2}-4\theta _{-1}\theta
_{1}=4(c_{0}^{2}-d_{0}^{2})(a_{1}^{2}-b_{1}^{2})\neq 0
\end{equation}
if and only if $c_{0}\neq \pm d_{0},$ $a_{1}\neq \pm b_{1}.$ 

\textbf{3. } $a_{1}=b_{1}=c_{1}=d_{1}=0,a_{0}d_{0}+b_{0}c_{0}\neq 0.$ 

Then the \ conditions (2) are equivalent to the Dirichlet conditions and $%
\theta _{0}=0,\theta _{-1}=-\theta _{1}=1.$ 

If the boundary conditions (2) \ are strongly regular then the eigenvalues
of the operator $L(q(x))$ , generated by (1), \ (2) consists of the
sequences $\{\lambda _{n,1}\},\{\lambda _{n,2}\}$ \ satisfying
\begin{equation}
\lambda _{n,1}=(2n\pi -i\ln \zeta _{1})^{2}+O(1),  \label{sq1}
\end{equation}
\begin{equation}
\lambda _{n,2}=(2n\pi -i\ln \zeta _{2})^{2}+O(1),  \label{sq2}
\end{equation}
for $n\geq N,$ $N>>1,$ where $\zeta _{1}$ and $\zeta _{2}$ are the distinct
roots of (3) (see (4) ) . From these formulas, and the same formulas for the
eigenvalues $\{\lambda _{n,1}^{0}\},\{\lambda _{n,2}^{0}\}$ \ of $L(0)$ one
can easily obtain the following inequalities from which we use essentially:
\begin{eqnarray}
\left| \lambda _{n,j}-\lambda _{k,j}^{0}\right| &>&\left| 2(n-k)\pi \right|
\left| 2(n+k)\pi -i2\ln \zeta _{j}\right| -c_{1}>c_{2}n,  \nonumber \\
\text{ }j &=&1,2,\text{\ }\forall k\neq n,\text{ }k=0,1,...,
\end{eqnarray}
\begin{equation}
\left| \lambda _{n,j}-\lambda _{k,s}^{0}\right| \geq \left| J_{1}\right|
\left| J_{2}\right| -c_{3}>c_{4}n,\text{ \ }j\neq s,\forall k=0,1,....
\label{dis2}
\end{equation}
for $n\geq N,$ where $J_{1}=2(n-k)\pi -i(\ln \zeta _{1}-\ln \zeta _{2}),$ 

$J_{2}=2(n+k)\pi -i(\ln \zeta _{1}+\ln \zeta _{2}).$ \ 

Here and in forthcaming relations by $N$ we denote a big positive integer,
that is, $N\gg 1,$ and by $c_{m},$ for $m=1,2,...,$ the positive,
independent on $N,$ constants whose exact value are inessential. Proof of
(10) follows from the relations $n-k\neq 0$, $n+k>n>>1$, $\left| \ln \zeta
_{j}\right| <c_{5}.$ To prove (11) we note that
\[
(2n\pi -i\ln \zeta _{1})^{2}-(2k\pi -i\ln \zeta _{2})^{2}=J_{1}J_{2}.
\]
Here $\left| J_{2}\right| >c_{6}n,$ because $n+k>n>>1$ and $\left| \ln \zeta
_{1}+\ln \zeta _{2}\right| <c_{7}.$ Let $\zeta _{j}=r_{j}e^{it_{j}},$ $0\leq
t_{j}<2\pi $ $(j=1,2).$ Then $\ln \zeta _{j}=\ln r_{j}+it_{j}.$ Since $\zeta
_{1}\neq \zeta _{2},$ there are two possibilities: 

1. $r_{1}\neq r_{2}$ . Then
\[
\left| J_{1}\right| =\left| 2(n-k)\pi +t_{1}-t_{2}-i(\ln r_{1}-\ln
r_{2})\right| \geq \left| \ln r_{1}-\ln r_{2}\right| >c_{8}.
\] 

\ 2. $r_{1}=r_{2}$ . Then $t_{1}-t_{2}\neq 0,$ $\left| t_{1}-t_{2}\right|
<2\pi .$ Therefore 

$\left| J_{1}\right| \geq \left| t_{1}-t_{2}\right| \geq c_{9}$ whether $%
n-k=0$ or not. Hence in all cases we have $\left| J_{1}\right| >c_{10}.$ So 
$%
\left| J_{1}J_{2}\right| >c_{10}c_{6}n.$ 

Moreover from \ (10), (11) it follows that if $k>2n,n\geq N,$ $N>>1$ then
\begin{equation}
\left| \lambda _{n,j}-\lambda _{k,i}^{0}\right| \geq k^{2},\text{ \ }\forall
i,j=1,2,
\end{equation}
since \ $\left| 2(n-k)\pi \right| >2k,\left| 2(n+k)\pi -i2\ln \zeta
_{1}\right| >2k,\left| J_{1}\right| >2k,\left| J_{2}\right| >2k.$ 

The eigenfunctions $F_{n,i}(x)$ belonging to $\lambda _{n},_{j}$ satisfy the
formula
\begin{eqnarray}
F_{n,j}(x) &=&(-i)^{k_{2}}(\alpha _{2}+\frac{1}{\zeta _{j}}\beta _{2}+O(%
\frac{1}{n}))e^{i\rho _{n,j}x}- \\
&&(-i)^{k_{2}}(\alpha _{2}+\zeta _{j}\beta _{2}+O(\frac{1}{n}))e^{-i\rho
_{n,j}x}.  \nonumber
\end{eqnarray}
(see [3, p.77]) , where $(\rho _{n,j})^{2}=\lambda _{n,j}$ , $\rho
_{n,j}=2n\pi -i\ln \zeta _{1}+O(\frac{1}{n}),$ and 

in Case 1 ( see (5)) $k_{2}=1,\alpha _{2}=\beta _{2}=1,\xi _{j}=\pm i.$ 

in Case 2 ( see (6)) $k_{2}=0,\alpha _{2}=c_{0},\beta _{2}=d_{0},\xi
_{j}\neq 0,c_{0}\neq \pm d_{0},$ 

in Case 3 $k_{2}=0,\alpha _{2}=\beta _{2}=1,\xi _{j}=\pm 1$ 

Using these relations one can easily verify that the normalized
eigenfunction $\Psi _{n,j}(x)$ of $L(q)$ corresponding to the eigenvalue $%
\lambda _{n,i}$ satisfy
\begin{equation}
\Psi _{n,j}(x)=A_{j}e^{i(2\pi n+\gamma _{j})x}+B_{j}e^{-i(2\pi n+\gamma
_{j})x}+O(\frac{1}{n}),j=1,2,
\end{equation}
where $A_{j},B_{j}$ are constants and $\gamma _{j}=-i\ln \zeta _{j}.$ 

Using these \i n Section 2 we give a simple proof of Riesz basisness of
eigenfunctions and associeted functions of the operator $L(q)$ with strongly
regular boundary conditions and arbitrary Lebesque integrable Potential $%
q(x) $. For the first time the basisness of eigenfunctions and associeted
functions of \ ordinary differential operators with strongly regular
boundary conditions is investigated in [5,6,7] ( see also [3,4]). 

In section 3 we obtain the main results of this paper, naimly we obtain
asymptotic formulas of arbitrary order for eigenvalues and eigenfunctions of
$L(q)$ when $q(x)\in L_{1}[0,1],$ that is, when there is not any condition
about smoothness of the\ potential $q(x)$. 

\section{On the Riesz Basisness} 

First we consider the adjoint boundary conditions. Using the formula
\begin{equation}
y^{^{\prime }}(1)\overline{z(1)}-y^{^{\prime }}(0)\overline{z(0)}-y(1)%
\overline{z^{^{\prime }}(1)}+y(0)\overline{z^{^{\prime }}(0)}=0
\end{equation}
one can compute the adjoint boundary conditions of (2). In the case 1, (see
(5)), the adjoint boundary conditions to (2) is
\begin{equation}
y^{^{\prime }}(0)+\overline{\alpha }_{11}y(0)-\overline{\alpha 
}_{21}y(1)=0,%
\text{ }y^{^{\prime }}(1)-\overline{\alpha }_{12}y(0)+\overline{\alpha }%
_{22}y(1)=0.  \label{normbc1}
\end{equation}
Clearly, for this conditions, we have $\theta _{0}=0,$ $\theta _{-1}=\theta
_{1}=1,$ $\theta _{0}^{2}-4\theta _{-1}\theta _{1}\neq 0,$ that is, this is
strongly regular boundary conditions. The case 3, i.e., the Dirichlet
boundary conditions are self adjoint. To find the adjoint boundary
conditions of the case 2 ( see (6)) we supplement the conditions
\[
U_{3}(y)=a_{1}y(0)-b_{1}y(1)=0,U_{4}(y)=\text{ }y^{^{\prime
}}(0)+y^{^{\prime }}(1)=0.
\]
The determinant of the obtained sistem is $%
(a_{1}-b_{1})(b_{1}c_{0}+a_{1}d_{0})\neq 0.$ Finding $y(0),y(1),$ $%
y^{^{\prime }}(0),y^{^{\prime }}(1)$ in term $U_{1},U_{2},U_{3},U_{4}$ and
then putting they in (15) we see that the adjoint conditions of (6) is
\[
\overline{d}_{0}y^{^{\prime }}(0)+\overline{c}_{0}y^{^{\prime }}(1)+\frac{%
\overline{a_{0}d_{0}-b_{0}c_{0}}}{\overline{b_{1}-a_{1}}}(y(1)-y(0))=0,\text 
{
\ }\overline{a}_{1}y(1)+\overline{b}_{1}y(0)=0,
\]
which is strongly regular boundary condition ( see (7)). Hence in all cases
the adjoint boundary conditions of the strongly regular boundary conditions
are also strongly regular boundary condition. Denote by $\Psi _{n,i,0}(x)$ 
$%
\equiv \Psi _{n,i}(x)$ the eigenfunction of $L(q)$ corresponding to the
eigenvalue $\lambda _{n,i}$ and by $\Psi _{n,i,p}(x),$ for $%
p=1,2,...,t(n,i), $ the associated functions of $L(q)$ corresponding to the
eigenfunction $\Psi _{n,i}(x),$ that is,
\begin{equation}
(L(q)-\lambda _{n,i})\Psi _{n,i}=0,
\end{equation}
\begin{equation}
(L(q)-\lambda _{n,i})\Psi _{n,i,p}=\Psi _{n,i,p-1},\text{ \ }%
p=1,2,...,t(n,i).
\end{equation}
The eigenfunctions and the associeted functions of $L(0)$\ are denoted by 

$\Psi _{n,i}^{0}(x)\equiv \Psi _{n,i,0}^{0}(x)$ and $\Psi _{n,i,p}^{0}(x)$
respectively. 

\begin{theorem}
If the conditions (2) are strongly regular then the system of eigenfunctions
and associated functions of \ $L(q)$ form a Riesz basis of $L_{2}[0,1].$
\end{theorem} 

\begin{proof}
Since the strongly regular boundary conditions are nondegenerated boundary
conditions the system of eigenfunctions and associated functions of \ $L(q)$
is complete ( see [2, Chap. 1, Sec.3] ). The system of eigenfunctions and
associated functions of \ $L^{\ast }(q)$ is also complete in $L_{2}[0,1]$
becouse, as we noted above, the \ adjoint boundary conditions are strongly
regular. 

Now we prove that the biortogonal system of the system $\{\Psi 
_{n,i,p}(x)\}$
consist of eigenfunctions and associated functions of \ the operator $%
L^{\ast }(q)$ genereted by the adjoint expression $-y^{^{\prime \prime 
}}(x)+%
\overline{q(x)}y(x)$ \ and the adjoint boundary conditions. Let $\Phi
_{m,i}(x)$ $\equiv \Phi _{m,i,0}(x)$ be the eigenfunction of $L^{\ast }(q)$
corresponding to the eigenvalue $\mu _{m,i}.$ Denote by $\Phi _{m,i,p}(x),$
for $p=1,2,...,$ the associated functions corresponding to the eigenfunction
$\Phi _{m,i}(x),$ that is,
\begin{equation}
(L^{\ast }(q)-\mu _{m,i})\Phi _{m,i}=0,
\end{equation}
\begin{equation}
(L^{\ast }(q)-\mu _{m,i})\Phi _{m,i,p}=\Phi _{m,i,p-1},\text{ \ }p=1,2,...,.
\end{equation}
Multiplying both side of \ (19), (20) by \ $\Psi _{n,i,p}(x),$ for $%
p=0,1,2,...,t(n,i)$ \ and using (17), (18) one can easily verify that if $%
\overline{\mu _{m,i}}\neq \lambda _{n,i}$ then \
\begin{equation}
(\Psi _{n,i,p}(x),\Phi _{m,i,j})=0,\text{ }\forall p,j=0,1,2,...,
\end{equation}
Since the system of eigenfunctions and associated functions of $L^{\ast 
}(q)$%
, is complete in $L_{2}[0,1]$ \ the function $\Psi _{n,i,p}(x)$ can not be
orthogonal to all elements of this system. Therefore no there exists
eigenvalue of $L^{\ast }(q)$ different from all eigenvalues of $L(q)$ and
similarly no there exists eigenvalue of $L(q)$ different from all
eigenvalues of $L^{\ast }(q),$ that is, the eigenvalues of $L^{\ast }(q)$
consists of $\overline{\lambda _{n,i}}.$\ \ Denote by $E$ and $E^{\ast }$
the linear spaces genereted by the eigenfunction and the associated function
of $L(q)$ and $L^{\ast }(q)$ corresponding to $\lambda _{n,i}$ and $%
\overline{\lambda _{n,i}}$ respectively. Consider the operators
\begin{equation}
P=\int_{\gamma }(L(q)-\lambda I)^{-1}d\lambda ,\text{ }T=\int_{\gamma ^{\ast
}}(L^{\ast }(q)-\lambda I)^{-1}d\lambda ,
\end{equation}
where $\gamma $ is the circle belonging to the resolvent set of $L(q)$ and
containing only the eigenvalue $\lambda _{n,i}$ \ and $\gamma ^{\ast
}=\{\lambda \in C:\overline{\lambda }\in \gamma \}.$ Since for $\lambda \in
\gamma $ the operator $(L(q)-\lambda I)^{-1}$ is bounded and continuously
depend on $\lambda $ the operator $P$ is bounded operator. Similarly $T$ \
is bounded. For every $f\in $ $L_{2}[0,1]$ there exists sequence $f_{n}$
consisting of finite lineer combinations of eigenfunctions and associated
functions of $L(q)$ and converging to $f.$ Clearly 

$Pf_{n}\in E,Pf\in E$ and $P^{2}f=Pf,$ that is $P$ is projector of $%
L_{2}[0,1]$ onto $E.$ Similarly $T$ is projector of $L_{2}[0,1]$ onto $%
E^{\ast }.$ Considering the integrals in (22) as the limit of finite
integral sums one can easily verify that $T=P^{\ast }.$ Therefore ( see [4,
p. 2313]) for the basis $\Psi _{n,i,p}(x),$ $p=0,1,2,...,t(n,i)$ of $E$ one
can find a unique basis $\Phi _{n,i,p}(x),$ $p=0,1,2,...,t(n,i)$ of $E^{\ast
}$ biorthogonal to the above basis of $E$. Thus \ the biortogonal system of
the system $\{\Psi _{n,i,p}(x)\}$ is $\{\Phi _{n,i,p}(x)\}.$\ Now to prove
the theorem, it is enough to show that for every function $f(x)\in
L_{2}[0,1] $ the inequalities
\begin{equation}
\sum\limits_{n=N}^{\infty }\sum\limits_{j=1}^{2}\left| (f,\Psi
_{n,j})\right| ^{2}<\infty \text{ , }\sum\limits_{n=N}^{\infty
}\sum\limits_{j=1}^{2}\left| (f,\Phi _{n,j})\right| ^{2}<\infty
\end{equation}
hold (see [1, p.310, Theorem 2.1]). By the asymptotic formula (14), we have
\begin{eqnarray*}
\sum\limits_{n\geq N}\sum\limits_{j=1,2}\left| (f,\Psi _{n,j})\right| ^{2}
&\leq &c_{11}(\sum\limits_{n\geq N,j=1,2}\text{ }\mid (e^{-i\gamma
_{j}x}f,e^{i2\pi nx})\mid ^{2}+ \\
&\mid &(e^{i\gamma _{j}x}f,e^{-i2\pi nx})\mid ^{2}+\sum\limits_{n\geq
N}\left| (f,O(\frac{1}{n}))\right| ^{2})<c_{12}\parallel f\parallel
^{2}<\infty .
\end{eqnarray*}
To prove the second inequality in (23) we consider the asymptotic formula
for \ $\Phi _{n,j}.$ For the normalized eigenfunction $G_{n,j}(x)$ \ of \ $%
L^{\ast }(q)$ we have
\begin{equation}
G_{n,j}(x)=C_{j}e^{i(2\pi n+\eta _{j})x}+D_{j}e^{-i(2\pi n+\eta _{j})x}+O(%
\frac{1}{n}),j=1,2,
\end{equation}
where $C_{j},D_{j}$ and $\eta _{j}$ are constants ( see (14)). The
normalized eigenfunction $\Psi _{n,j}^{0}(x)$ \ of \ $L(0)$\ and the
normalized eigenfunction $G_{n,j}^{0}(x)$ \ of \ $L^{\ast }(0)$ satisfy
respectively (14) and (24), that is, in (14) and (24) instead of $\Psi
_{n,j}(x)$ \ and $G_{n,j}(x)$ we can write $\Psi _{n,j}^{0}(x)$ \ and $%
G_{n,j}^{0}(x).$ Since the operator $L(0)$ is spectral operator ( see [4,
p.2334] ), that is, \ the projectors $P(\lambda _{n}^{0},_{j})$
corrosponding to $\lambda _{n}^{0},_{j}$ are uniformly bounded it follows
from the well known formula 

$P(\lambda _{n,j}^{0})f=\frac{(f,G_{n,j}^{0})\Psi _{n,j}^{0}}{%
(G_{n,j}^{0},\Psi _{n,j}^{0})}$ ( see [3, p. 39] ) that
\begin{equation}
\mid \frac{1}{(G_{n,j}^{0},\Psi _{n,j}^{0})}\mid <c_{13},\forall n>N,j=1,2,
\end{equation}
Therefore by (14), (24) we have
\begin{equation}
(G_{n,j}^{0},\Psi _{n,j}^{0})=h_{j}+O(\frac{1}{n}),\text{ }(G_{n,j},\Psi
_{n,j})=h_{j}+O(\frac{1}{n}),
\end{equation}
where $h_{j}\neq 0.$ Thus from (24) we get
\begin{equation}
\Phi _{n,j}(x)=\frac{1}{h_{j}}(C_{j}e^{i(2\pi n+\eta _{j})x}+D_{j}e^{-i(2\pi
n+\eta _{j})x})+O(\frac{1}{n}),j=1,2,
\end{equation}
from which follows the second inequality in (23)
\end{proof} 

\section{The Asymptotic Formulas for Eigenfunctions and Eigenvalues} 

Let $\Phi _{k,j,p}^{0}(x),$ for $p=1,2,...,t(k,j),$ be the associated
function of $L^{\ast }(0)$ corresponding to the eigenfunction $\Phi
_{k,j}^{0}(x)=\Phi _{k,j,0}^{0}(x).$ Since only finite number of eigenvalues
of $L(0)$ and $L(q)$ are multiple there is $n_{0}$ such that $\lambda
_{n,1}^{0}$ and $\lambda _{n,1}$ are simple eigenvalues for $n\geq n_{0}.$
Multiplying the equation
\begin{equation}
\lambda _{n,i}\Psi _{n,i}=(L(0)+q(x))\Psi _{n,i}
\end{equation}
(see (17)) by $\Phi _{k,j}^{0}(x)$ we get 

\begin{equation}
(\lambda _{n},_{i}-\lambda _{k}^{0},_{j})(\Psi _{n,i}(x),\Phi
_{k,j}^{0}(x))=(q(x)\Psi _{n,i}(x),\Phi _{k,j}^{0}(x)),  \label{main1}
\end{equation}
\begin{equation}
(\Psi _{n,i}(x),\Phi _{k,j}^{0}(x))=\frac{(q(x)\Psi _{n,i}(x),\Phi
_{k,j}^{0}(x))}{\lambda _{n},_{i}-\lambda _{k}^{0},_{j}},
\end{equation}
for\ $i,j=1,2.$ Now multiplying (28) by $\Phi _{k,j,1}^{0}(x)$ and using
(19), (20) we get
\[
(\lambda _{n},_{i}-\lambda _{k}^{0},_{j})(\Psi _{n,i},\Phi
_{k,j,1}^{0})=(q(x)\Psi _{n,i},\Phi _{k,j,1}^{0})+\frac{(q(x)\Psi
_{n,i}(x),\Phi _{k,j}^{0}(x))}{\lambda _{n},_{i}-\lambda _{k}^{0},_{j}},
\]
\[
(\Psi _{n,i}(x),\Phi _{k,j,1}^{0}(x))=\frac{(q(x)\Psi _{n,i}(x),\Phi
_{k,j,1}^{0}(x))}{\lambda _{n},_{i}-\lambda _{k}^{0},_{j}}+\frac{(q(x)\Psi
_{n,i}(x),\Phi _{k,j}^{0}(x))}{(\lambda _{n},_{i}-\lambda 
_{k}^{0},_{j})^{2}}%
.
\]
In this way one can redues the formulas
\begin{equation}
(\Psi _{n,i}(x),\Phi _{k,j,m}^{0}(x))=\sum_{p=0}^{m}\frac{(q(x)\Psi
_{n,i}(x),\Phi _{k,j,p}^{0}(x))}{(\lambda _{n},_{i}-\lambda
_{k}^{0},_{j})^{m+1-p}}
\end{equation}
for $m=0,1,2,...,t(k,j).$ Clearly
\begin{equation}
n_{0}<c_{14},\sum_{j=1}^{2}\sum_{k=0}^{n_{0}-1}t(k,j)<c_{14}.
\end{equation}
\ To obtain the asymptotic formulas we use (29), (30), (31), and the
following 

\begin{lemma}
\bigskip Let $n>N,$ where $N\gg 1.$ For $i=1,2$ the equality
\[
(q(x)\Psi _{n,i},\Phi
_{k,j,m}^{0})=\sum\limits_{j_{1}=1}^{2}\sum\limits_{k_{1}=0}^{n_{0}-1}%
\sum_{m_{1}=1}^{t(k_{1},j_{1})}c(k,j,m,k_{1},j_{1},m_{1})(\Psi _{n,i},\Phi
_{k_{1},j_{1},m_{1}}^{0})
\]
\begin{equation}
+\sum\limits_{j_{1}=1}^{2}\sum\limits_{k_{1}=n_{0}}^{\infty
}c(k,j,m,k_{1},j_{1},0)(\Psi _{n,i}(x),\Phi _{k_{1},j_{1},0}^{0}(x))
\end{equation}
holds, where $\overline{c(k,j,m,k_{1},j_{1},m_{1})}=(\overline{q}(x)\Phi
_{k,j,m}^{0}(x),\Psi _{k_{1},j_{1},m_{1}}^{0}(x)).$ Moreover
\begin{equation}
\mid c(k,j,m,k_{1},j_{1},m_{1})\mid <c_{15},\forall k,j,m,k_{1},j_{1},m_{1},
\end{equation}
\begin{equation}
\left| (\Psi _{n,i}(x),\Phi _{k,j,m}^{0}(x))\right| <c_{15},\forall k,j,m,
\end{equation}
\begin{equation}
\left| (q(x)\Psi _{n,i}(x),\Phi _{k,j,m}^{0}(x)\right| <c_{15},\forall 
k,j,m,
\label{boundary}
\end{equation}
$.$
\end{lemma} 

\begin{proof}
The inequality (34) follows from
\begin{equation}
\sup_{x\in \lbrack 0,1]}\mid \Phi _{k,j,m}^{0}(x)\mid <c_{16},\forall k,j,m,
\end{equation}
\begin{equation}
\sup_{x\in \lbrack 0,1]}\mid \Psi _{k_{1},j_{1},m_{1}}^{0}(x)\mid
<c_{17},\forall k_{1},j_{1},m_{1},
\end{equation}
which are simple results of (27), (14). Since \ $\parallel \Psi
_{n,i}(x)\parallel =1$ the inequality (35) follows from Schwarz inequality
and (37). 

Let $n$ be fixed and $n>N\gg 1$. It is well known that (see [2, Chap.1,
Sec.3]), $q(x)\Psi _{n,i}(x)\in L_{1}[0,1]$ and $(q(x)\Psi _{n,i}(x),\Phi
_{k,j}^{0}(x))\rightarrow 0$ as $k\rightarrow \infty .$ Therefore there is a
constant $C(n,i)$ and $k_{0},j_{0},m_{0}$ such that
\begin{equation}
\stackunder{k,j,m}{\max }\left| (q(x)\Psi _{n,i}(x),\Phi
_{k,j,m}^{0}(x))\right| =\left| (q(x)\Psi _{n,i}(x),\Phi
_{k_{0},j_{0},m_{0}}^{0}(x))\right| =C(n,i).  \label{max1}
\end{equation}
Then by (30), (34), and (10), (11) we have
\[
\left| (\Psi _{n,i}(x),\Phi _{k,j}^{0}(x))\right| <\frac{C(n,i)}{\left|
\lambda _{n,i}-\lambda _{k,j}^{0}(x)\right| },\forall k\geq n_{0},(n,i)\neq
(k,j).
\]
Therefore using (12) we get
\[
\sum\limits_{j=1}^{2}\sum\limits_{k>2n_{1}}\left| (\Psi _{n,i}(x),\Phi
_{k,j}^{0}(x))\right| <\sum\limits_{k>2n_{1}}\frac{2C(n,i)}{k^{2}}<\frac{%
c_{18}C(n,i)}{n_{1}},
\]
where $n_{1}>n,$ hence $k>2n_{1}>2n.$ This with (38) imply that the
decomposition of $\Psi _{n,i}(x)$ by the basis $\{\Psi _{k,j,m}^{0}(x)\}$ is
of the form
\begin{equation}
\Psi _{n,i}(x)=\sum\limits_{j_{1}=1}^{2}\sum\limits_{k_{1}\leq
2n_{1}}\sum_{m_{1}=0}^{t(k_{1}j_{1})}(\Psi _{n,i}(x),\Phi
_{k_{1},j_{1},m_{1}}^{0}(x))\Psi _{k_{1},j_{1},m_{1}}^{0}(x)+g_{i}(x),
\label{decomp}
\end{equation}
where$\stackunder{x\in \lbrack 0,1]}{\sup }\left| g_{i}(x)\right| <\frac{%
c_{17}c_{18}C(n,i)}{n_{1}}.$ Using this in $(q(x)\Psi _{n,i}(x),\Phi
_{k,j,m}^{0}(x))$ and tending $n_{1}\rightarrow \infty $, we obtain (33). 

Now we prove (36). Using (39), (33), (34), (35) and (32) we get 

\[
C(n,i)\leq \sum\limits_{j_{1}=1}^{2}\sum\limits_{k_{1}=n_{0}}^{\infty }\mid
c(k_{0},j_{0},m_{0},k_{1},j_{1},0)(\Psi _{n,i}(x),\Phi
_{k_{1},j_{1},0}^{0}(x))\mid +c_{19}
\]
Isolating the term with multiplicant $(\Psi _{n,i}(x),\Phi _{n,i}^{0}(x))$
and using (30) for $(\Psi _{n,i}(x),\Phi _{k_{1},j_{1}}^{0}(x))$ with $%
(k_{1},j_{1})\neq (n,i)$ we obtain \
\[
C(n,i)\leq c_{20}+\sum_{k_{1},j_{1}:(k_{1},j_{1})\neq (n,i)}\frac{%
c_{21}C(n,i)}{\mid \lambda _{n,i}-\lambda _{k_{1},j_{1}}^{0}\mid }.
\]
Using (10), (11) it is not hard to verify that
\begin{equation}
\sum_{k,j,(k,j)\neq (n,i)}\frac{1}{\left| \lambda _{n,j}-\lambda
_{k,j}^{0}\right| }=O(\frac{\ln n}{n}).
\end{equation}
Since $n\gg 1,$ the last two relations give 
$C(n,1)<c_{20}+\frac{C(n,i)}{2}.$
This inequality and (39) imply (36). The lemma is proved
\end{proof} 

Using (29), (33) we get \
\[
(\lambda _{n},_{i}-\lambda _{n}^{0},_{i})(\Psi _{n,i}(x),\Phi
_{n,i}^{0}(x))=(q(x)\Psi _{n,i}(x),\Phi _{n,i}^{0}(x))=
\]
\begin{equation}
\sum\limits_{j_{1}=1}^{2}\sum\limits_{k_{1}=0}^{\infty
}\sum_{m_{1}=1}^{t(k_{1},j_{1})}c(n,i,0,k_{1},j_{1},m_{1})(\Psi
_{n,i}(x),\Phi _{k_{1},j_{1},m_{1}}^{0}(x)).
\end{equation}
Isolating the term with multiplicant $(\Psi _{n,i}(x),\Phi _{n,i}^{0}(x)),$
then using (30) and (31) for $(\Psi _{n,i}(x),\Phi
_{k_{1},j_{1},m_{1}}^{0}(x))$ with $m_{1}=0,(k_{1},j_{1})\neq (n,i)$ and $%
m_{1}\neq 0$ respectively we obtain
\begin{equation}
(\lambda _{n},_{i}-\lambda _{n}^{0},_{i})(\Psi _{n,i}(x),\Phi
_{n,i}^{0}(x))=c(n,i)(\Psi _{n,i},\Phi _{n,i}^{0})+C_{1}(\lambda _{n},_{i}),
\end{equation}
where $c(n,i)=c(n,i,0,n,i,0),$
\[
C_{1}(\lambda
_{n},_{i})=\sum\limits_{j_{1}=1}^{2}(\sum\limits_{(k_{1},j_{1})\neq
(n,i),k_{1}\geq n_{0}}c(n,i,0,k_{1},j_{1},0)\frac{(q(x)\Psi _{n,i}(x),\Phi
_{k_{1},j_{1}}^{0}(x))}{\lambda _{n},_{i}-\lambda _{k_{1},j_{1}}^{0}})+
\]
\[
+\sum\limits_{j_{1}=1}^{2}\sum\limits_{k_{1}=0}^{n_{0}-1}%
\sum_{m_{1}=0}^{t(k_{1},j_{1})}\sum%
\limits_{p_{1}=0}^{m_{1}}c(n,i,0,k_{1},j_{1},p_{1})\frac{(q(x)\Psi
_{n,i}(x),\Phi _{k_{1},j_{1},p_{1}}^{0}(x))}{(\lambda _{n},_{i}-\lambda
_{k_{1},j_{1}}^{0})^{m_{1}+1-p_{1}}}
\]
Instead of $(q(x)\Psi _{n,i}(x),\Phi _{k_{1},j_{1}}^{0}(x))$ , $(q(x)\Psi
_{n,i}(x),\Phi _{k_{1},j_{1},p_{1}}^{0}(x))$\ in $C_{1}(\lambda _{n},_{i})$
writing there expression from (33) and again isolating the term with
multiplicant $(\Psi _{n,i}(x),\Phi _{n,i}^{0}(x)),$ then using (30) and (31)
we get
\begin{equation}
(\lambda _{n},_{i}-\lambda _{n}^{0},_{i})(\Psi _{n,i},\Phi
_{n,i}^{0})=(c(n,i)+S_{1}(\lambda _{n},_{i}))(\Psi _{n,i},\Phi
_{n,i}^{0})+C_{2}(\lambda _{n},_{i})
\end{equation}
where $S_{1}(\lambda _{n},_{i})=S_{1,1}(\lambda _{n},_{i})+S_{1,2}(\lambda
_{n},_{i}),$%
\[
S_{1,1}(\lambda
_{n},_{i})=\sum\limits_{j_{1}=1}^{2}(\sum\limits_{(k_{1},j_{1})\neq
(n,i),k_{1}\geq n_{0}}\frac{c(n,i,0,k_{1},j_{1},0)c(k_{1},j_{1},0,n,i,0)}{%
\lambda _{n},_{i}-\lambda _{k_{1},j_{1}}^{0}}),
\]
\[
S_{1,2}(\lambda
_{n},_{i})=\sum\limits_{j_{1}=1}^{2}\sum\limits_{k_{1}=0}^{n_{0}-1}%
\sum_{m_{1}=0}^{t(k_{1},j_{1})}\sum\limits_{p_{1}=0}^{m_{1}}\frac{%
c(n,i,0,k_{1},j_{1},p_{1})c(k_{1},j_{1},p_{1},n,i,0)}{(\lambda
_{n},_{i}-\lambda _{k_{1},j_{1}}^{0})^{m_{1}+1-p_{1}}},
\]
\[
C_{2}(\lambda
_{n},_{i})=\sum\limits_{j_{1},j_{2}=1}^{2}(\sum\limits_{k_{1},k_{2}\geq
n_{0}}\frac{c(n,i,0,k_{1},j_{1},0)c(k_{1},j_{1},0,k_{2},j_{2},0)(q(x)\Psi
_{n,i},\Phi _{k_{2},j_{2}}^{0})}{(\lambda _{n},_{i}-\lambda
_{k_{1},j_{1}}^{0})(\lambda _{n},_{i}-\lambda _{k_{2},j_{2}}^{0})})+
\]
\begin{eqnarray*}
&&+\sum\limits_{j_{1},j_{2}=1}^{2}\sum\limits_{k_{1}\geq
n_{0}}\sum\limits_{k_{2}=1}^{n_{0}-1}\sum_{m_{2}=0}^{t(k_{2},j_{2})}\sum%
\limits_{p_{2}=0}^{m_{2}}\frac{%
c(n,i,0,k_{1},j_{1},0)c(k_{1},j_{1},0,k_{2},j_{2},p_{2})}{\lambda
_{n},_{i}-\lambda _{k_{1},j_{1}}^{0}}\times \\
&&\frac{(q(x)\Psi _{n,i}(x),\Phi _{k_{2},j_{2},p_{2}}^{0}(x))}{(\lambda
_{n},_{i}-\lambda _{k_{2},j_{2}}^{0})^{m_{2}+1-p_{2}}}+
\end{eqnarray*}
\begin{eqnarray*}
&&+\sum\limits_{j_{1},j_{2}=1}^{2}\sum\limits_{k_{1}=0}^{n_{0}-1}\sum%
\limits_{k_{2}\geq
n_{0}}\sum_{m_{1}=1}^{t(k_{1},j_{1})}\sum\limits_{p_{1}=0}^{m_{1}}\frac{%
c(n,i,0,k_{1},j_{1},p_{1})c(k_{1},j_{1},p_{1},k_{2},j_{2},0)}{(\lambda
_{n},_{i}-\lambda _{k_{1},j_{1}}^{0})^{m_{1}+1-p_{1}}}\times \\
&&\frac{(q(x)\Psi _{n,i}(x),\Phi _{k_{2},j_{2},0}^{0}(x))}{\lambda
_{n},_{i}-\lambda _{k_{2},j_{2}}^{0}}+
\end{eqnarray*}
\begin{eqnarray*}
&&+\sum\limits_{j_{1},j_{2}=1}^{2}\sum\limits_{k_{1},k_{2}=0}^{n_{0}-1}%
\sum_{m_{1}=0}^{t(k_{1},j_{1})}\sum\limits_{p_{1}=0}^{m_{1}}%
\sum_{m_{2}=0}^{t(k_{2},j_{2})}\sum\limits_{p_{2}=0}^{m_{2}}\frac{%
c(n,i,0,k_{1},j_{1},p_{1})c(k_{1},j_{1},p_{1},k_{2},j_{2},p_{2})}{(\lambda
_{n},_{i}-\lambda _{k_{1},j_{1}}^{0})^{m_{1}+1-p_{1}}}\times \\
&&\frac{(q(x)\Psi _{n,i}(x),\Phi _{k_{2},j_{2},p_{2}}^{0}(x))}{(\lambda
_{n},_{i}-\lambda _{k_{2},j_{2}}^{0})^{m_{2}+1-p_{2}}}.
\end{eqnarray*}
To write briefly the expressions $S_{1}(\lambda _{n},_{i}),$ $C_{2}(\lambda
_{n},_{i})$ we introduse the notations
\begin{equation}
a(k,j,m,k_{1},j_{1},p_{1})=\sum_{m_{1}=p_{1}}^{t(k_{1},j_{1})}\frac{%
c(k,j,m,k_{1},j_{1},p_{1})}{(\lambda _{n},_{i}-\lambda
_{k_{1},j_{1}}^{0})^{m_{1}+1-p_{1}}},
\end{equation}
\begin{equation}
b(k,j,m,k_{1},j_{1},p_{1})=\sum_{m_{1}=p_{1}}^{t(k_{1},j_{1})}\frac{%
c(k,j,m,k_{1},j_{1},p_{1})(q(x)\Psi _{n,i}(x),\Phi
_{k_{1},j_{1},p_{1}}^{0}(x))}{(\lambda _{n},_{i}-\lambda
_{k_{1},j_{1}}^{0})^{m_{1}+1-p_{1}}}.
\end{equation}
Note that if $k_{1}\geq n_{0}$\ then $t(k_{1},j_{1})=0,p_{1}=0$ and
\begin{equation}
a(k,j,m,k_{1},j_{1},p_{1})=\frac{c(k,j,m,k_{1},j_{1},p_{1})}{\lambda
_{n},_{i}-\lambda _{k_{1},j_{1}}^{0}},
\end{equation}
\begin{equation}
b(k,j,m,k_{1},j_{1},p_{1})=\frac{c(k,j,m,k_{1},j_{1},p_{1})(q(x)\Psi
_{n,i}(x),\Phi _{k_{1},j_{1},p_{1}}^{0}(x))}{\lambda _{n},_{i}-\lambda
_{k_{1},j_{1}}^{0}}.
\end{equation}
In this notation we have
\begin{equation}
S_{1}(\lambda
_{n},_{i})=\sum\limits_{j_{1}=1}^{2}\sum\limits_{k_{1}=0}^{\infty
}\sum%
\limits_{p_{1}=0}^{t(k_{1},j_{1})}a(n,i,0,k_{1},j_{1},p_{1})c(k_{1},j_{1},p_ 
{1},n,i,0),
\end{equation} 

\begin{equation}
C_{2}(\lambda
_{n},_{i})=\sum\limits_{j_{1},j_{2}=1}^{2}\sum\limits_{k_{1},k_{2}=0}^{%
\infty
}\sum\limits_{p_{1}=0}^{t(k_{1},j_{1})}\sum%
\limits_{p_{2}=0}^{t(k_{2},j_{2})}a(n,i,0,k_{1},j_{1},p_{1})b(k_{1},j_{1},p_ 
{1},k_{2},j_{2},p_{2})
\end{equation} 

Repeating this process ( that is, instead of $(q(x)\Psi _{n,i}(x),\Phi
_{k_{2},j_{2},p_{2}}^{0}(x))$\ in (44), more pricisely in $C_{2}(\lambda
_{n},_{i}),$ \ (see (50), (48)) writing there expression from (33) and again
isolating the term with multiplicant $(\Psi _{n,i}(x),\Phi _{n,i}^{0}(x)),$
then using (30) and (31) ) $m$ times, we obtain
\begin{equation}
(\lambda _{n},_{i}-\lambda _{n}^{0},_{i})(\Psi _{n,i}(x),\Phi
_{n,i}^{0}(x))=(c(n,i)+A_{m}(\lambda _{n}{}_{,i}))(\Psi _{n,i},\Phi
_{n,i}^{0})+C_{m},
\end{equation}
where \ $A_{m}(\lambda _{n}{}_{,i})=\sum_{l=1}^{m-1}S_{l}(\lambda 
_{n},_{i})$
\ for $m>1$, $S_{1}(\lambda _{n},_{i})$ is defined in (49) and
\[
S_{l}(\lambda
_{n}{}_{,i})=\sum\limits_{j_{1},j_{2},...,j_{l}}\sum%
\limits_{k_{1},k_{2},...,k_{l}}\sum%
\limits_{p_{1},p_{2},...,p_{l}}a(n,i,0,k_{1},j_{1},p_{1})a(k_{1},j_{1},p_{1} 
,k_{2},j_{2},p_{2})\times
\]
\begin{equation}
a(k_{l-1},j_{l-1},p_{l-1},k_{l},j_{l},p_{l})c(k_{l},j_{l},p_{l},n,i,0)
\end{equation}
\[
C_{l}=\sum\limits_{j_{1},j_{2},...,j_{l}}\sum\limits_{k_{1},k_{2},...,k_{l}} 
%
\sum%
\limits_{p_{1},p_{2},...,p_{l}}a(n,i,0,k_{1},j_{1},p_{1})a(k_{1},j_{1},p_{1} 
,k_{2},j_{2},p_{2})\times
\]
\begin{equation}
a(k_{l-2},j_{l-2},p_{l-2},k_{l-1},j_{l-1},p_{l-1})b(_{l-1},j_{l-1},p_{l-1},k 
_{l},j_{l},p_{l})
\end{equation}
for $l>1.$ Here the sums are taken under the conditions $%
j_{1},j_{2},...,j_{l}=1,2;$ $k_{1},k_{2},...,k_{l}=1,2,...,\infty ;$ and $%
p_{s}=0,1,...,t(k_{s},j_{s})$ for $s=1,2,....$ Moreover $(k_{s},j_{s})$ $%
\neq (n,i).$ From the formula (51) we obtain the following 

\begin{theorem}
The eigenvalue $\lambda _{n,i}$ satisfies
\begin{equation}
\lambda _{n,i}=\lambda _{n,i}^{0}+c(n,i)+F_{m}+O\left( (\frac{\ln n}{n}%
)^{m}\right) ,\ \forall m=1,2,....
\end{equation}
where $F_{1}=0,$ $F_{k}=A_{k}(\lambda _{n,i}^{0}+c(n,i)+F_{k-1}),$ \ $%
\forall k=2,3,....$ Moreover for $m=2$ the formula (54)\ can be written in
the form
\[
\lambda _{n,i}=\lambda _{n,i}^{0}+c(n,i)+\sum\limits\Sb (k,j)\neq (n,i),  \\ 
%
k\geq n_{0}  \endSb \frac{c(n,i,0,k,j,0)c(k,j,0,n,i,0)}{\lambda
_{n},_{i}-\lambda _{k,,j}^{0}}+O\left( (\frac{\ln n}{n})^{2}\right)
\]
\end{theorem} 

\begin{proof}
\bigskip It follows from (45)- (48), (32), (34), and (36) that
\begin{eqnarray*}
&\mid &a(k,j,m,k_{1},j_{1},p_{1})\mid \leq \mid \frac{c_{22}}{\mid \lambda
_{n}{}_{,i}-\lambda _{k_{1},j_{1}}^{0}\mid }, \\
&\mid &b(k,j,m,k_{1},j_{1},p_{1})\mid \leq \frac{c_{22}}{\mid \lambda
_{n}{}_{,i}-\lambda _{k_{1},j_{1}}^{0}\mid },
\end{eqnarray*}
for all $k,j,m,k_{1},j_{1},p_{1},$ $(k_{1},j_{1})\neq (n,i).$ Therefore
using (41) we get \
\begin{equation}
S_{m}=O\left( (\frac{\ln n}{n})^{m}\right) ,C_{m}=O\left( (\frac{\ln n}{n}%
)^{m}\right) ,\forall m=1,2,...
\end{equation}
Since both $\Psi _{n,i}(x)$ and $\Psi _{n,i}^{0}(x)$ satisfy the formula
(14) we have
\begin{equation}
\Psi _{n,i}(x)=\Psi _{n,i}^{0}(x)+O(\frac{1}{n}),(\Psi _{n,i},\Phi
_{n,i}^{0})=1+O(\frac{1}{n})
\end{equation}
Hence dividing both part of (43) by $(\Psi _{n,i},\Phi _{n,i}^{0})$ and
using (55) we get (54) for $m=1.$ Similarly dividing both part of (51) by $%
(\Psi _{n,i},\Phi _{n,i}^{0})$ \ we obtain
\begin{equation}
(\lambda _{n}{}_{,i}-\lambda _{n}^{0}{}_{,i})=c(n,i)+A_{m}(\lambda
_{n}{}_{,i})+O\left( (\frac{\ln n}{n})^{m}\right) .
\end{equation}
First we consider \ this formula for $m=2.$ Clierly for $k_{1}<n_{0}$ we
have $\mid \frac{1}{\lambda _{n},_{i}-\lambda _{k_{1},j_{1}}^{0}}\mid 
<\frac{%
c_{23}}{n^{2}}$ ( see (8), (9)). So by (32) we have
\begin{equation}
\sum_{j_{1}=1}^{2}\sum_{k_{1}=1}^{n_{0}-1}\mid \frac{1}{\lambda
_{n}{}_{,i}-\lambda _{k_{1},j_{1}}^{0}}\mid <\frac{c_{24}}{n^{2}}
\end{equation}
Therefore from (34) we obtain $S_{1,2}(\lambda _{n,i})=O(\frac{1}{n^{2}}),$
\begin{equation}
A_{2}(\lambda _{n}{}_{,i})=S_{1,1}(\lambda _{n,i})+O(\frac{1}{n^{2}}).
\end{equation}
Using $\lambda _{n,i}-\lambda _{n,i}^{0}=O(1)$ and (41) one can \ easily
verify that
\begin{equation}
\sum\limits_{(k,j)\neq (n,i)}\mid \frac{1}{\lambda _{n,i}-\lambda 
_{k,j}^{0}}%
 -\frac{1}{\lambda _{n,i}^{0}-\lambda _{k,j}^{0}}\mid =O\left( (\frac{\ln 
n}{n%
})^{2}\right)
\end{equation}
and hence
\begin{equation}
S_{1,1}(\lambda _{n,i})=S_{1,1}(\lambda _{n,i}^{0})+O\left( (\frac{\ln 
n}{n}%
)^{2}\right)
\end{equation}
Therefore from (59), (61) and (57) for $m=2$ we get the proof of the last
formula in the theorem. 

Now we prove (54) by induction with respect to $m.$ Assume that (54) holds
for $m=j-1$, that is, \ $\lambda _{n,i}=\lambda
_{n,i}^{0}+c(n,i)+F_{j-1}+O\left( (\frac{\ln n}{n})^{j-1}\right) .$
Substituting this value of $\lambda _{n,i}$ into $A_{j}(\lambda _{n,j})$ in
(57) for $m=j$ we get
\[
\lambda _{n,i}=\lambda _{n,i}^{0}+c(n,i)+A_{j}(\lambda
_{n,i}^{0}+c(n,i)+F_{j-1}+O((\frac{\ln n}{n})^{j-1}))+O\left( (\frac{\ln 
n}{n%
})^{j}\right) .
\]
Arguing as in the proof of (61) one can verify that
\begin{eqnarray}
&&A_{j}(\lambda _{n,i})=A_{j}(\lambda _{n,i}^{0}+c(n,i)+F_{j-1}+O((\frac{\ln
n}{n})^{j-1}))= \\
&&A_{j}(\lambda _{n,i}^{0}+c(n,i)+F_{j-1})+O\left( (\frac{\ln n}{n}%
)^{j}\right) .  \nonumber
\end{eqnarray}
The last two formulas imply (54) for $m=j$. The theorem is proved
\end{proof} 

Now we prove the asymptotic formulas of arbitrary order for eigenfunctions 
$%
\Psi _{n,_{i}}(x)$ of the operator $L(q).$ As we noted above the normalized
eigenfunctions $\Psi _{n,i}(x)$ satisfy the formula (56). Since the
eigenfunctions is deternined up to constant there exists eigenfunction, (for
simplisity of notation we denote it again by $\Psi _{n,i}(x))$ satisfying
\begin{equation}
(\Psi _{n,i},\Phi _{n,i}^{0})=1
\end{equation}
The asymptotic formulas of arbitrary order for this eigenfunctions accord
the well-known formulas Rellich-Shrodinger pertubation theory. Finding the
norm of this eigenfunctions one can find the asymptotic formulas of
arbitrary order for normalized eigenfunctions. 

\begin{theorem}
For the eigenfunction $\Psi _{n,_{i}}(x)$ satisfying (63) the following
holds
\begin{equation}
\Psi _{n,_{i}}(x)=\Psi _{n,i_{{}}}^{0}(x)+A_{m}^{\ast }(\lambda
_{n,i}^{0}+c(n,i)+F_{m-1})+O((\frac{\ln n}{n})^{m}),
\end{equation}
where $F_{m-1}$ is defined in Theorem 3, $A_{m}^{\ast }(\lambda
_{n}{}_{,i})=\sum_{l=1}^{m-1}S_{l}^{\ast }(\lambda _{n},_{i})$,
\[
S_{l}^{\ast }(\lambda
_{n}{}_{,i})=\sum\limits_{j_{1},j_{2},...,j_{l}}\sum%
\limits_{k_{1},k_{2},...,k_{l}}\sum\limits_{p_{1},p_{2},...,p_{l}}a^{\ast
}(n,i,0,k_{1},j_{1},p_{1})a(k_{1},j_{1},p_{1},k_{2},j_{2},p_{2})\times
\]
\[
a(k_{l-1},j_{l-1},p_{l-1},k_{l},j_{l},p_{l})c(k_{l},j_{l},p_{l},n,i,0),
\]
\[
a^{\ast 
}(n,i,0,k_{1},j_{1},p_{1})=\sum_{m_{1}=p_{1}}^{t(k_{1},j_{1})}\frac{%
\Psi _{k_{1},j_{1},p_{1}}^{0}(x)}{(\lambda _{n},_{i}-\lambda
_{k_{1},j_{1}}^{0})^{m_{1}+1-p_{1}}},
\]
that is, $a^{\ast }(n,i,0,k_{1},j_{1},p_{1}),$ $S_{l}^{\ast }(\lambda
_{n},_{i}),A_{m}^{\ast }(\lambda _{n}{}_{,i})$ are obtained from $%
a(n,i,0,k_{1},j_{1},p_{1}),$ $S_{l}(\lambda _{n},_{i}),$ $A_{m}(\lambda
_{n}{}_{,i})$ respectively by replasing $c(n,i,0,k_{1},j_{1},p_{1})$ by $%
\Psi _{k_{1},j_{1},p_{1}}^{0}(x)$ ( see (45), (49), (51)). In case $m=2$ the
formula (64) has the form
\begin{equation}
\Psi _{n,_{i}}(x)=\Psi _{n,_{i}}^{0}(x)+\sum\limits_{j=1}^{2}(\sum\limits\Sb 
%
(k,j)\neq (n,i),  \\ k_{1}\geq n_{0}  \endSb \frac{c(n,i,0,k,j,0)}{\lambda
_{n}^{0},_{i}-\lambda _{k,,j}^{0}}\Psi _{k,j}^{0}(x)+O((\frac{\ln n}{n}%
)^{2})).  \label{asym1}
\end{equation}
\end{theorem} 

\begin{proof}
\bigskip The decomposition of $\Psi _{n,i}(x)$ relative to the basis $\{\Psi
_{k,j,m}^{0}(x)\}$ is
\begin{equation}
\Psi _{n,i}(x)=\sum\limits_{j_{1}=1}^{2}\sum\limits_{k_{1}=0}^{\infty
}\sum\limits_{p_{1}=0}^{t(k_{1},j_{1})}(\Psi _{n,i}(x),\Phi
_{k_{1},j_{1},p_{1}}^{0}(x))\Psi _{k_{1},j_{1},p_{1}}^{0}(x)
\end{equation}
The right side of this formula can be obtained from the right side of (42)
by replasing $c(n,i,0,k_{1},j_{1},p_{1})$ by $\Psi
_{k_{1},j_{1},p_{1}}^{0}(x).$ Therefore doing the iteration which was done
in order to obtain (51) from (42) ( naimly, isolate the term with
multiplicant $(\Psi _{n,i}(x),\Phi _{n,i}^{0}(x))=1$ ( see (63) ) then use
(30) and (31) for $(\Psi _{n,i}(x),\Phi _{k_{1},j_{1},m_{1}}^{0}(x))$ with 
$%
m_{1}=0,(k_{1},j_{1})\neq (n,i)$ and $m_{1}\neq 0$ respectively, then use
(33), and again isolate the term with multiplicant $(\Psi _{n,i}(x),\Phi
_{n,i}^{0}(x)),$ use (30) and (31) for $(\Psi _{n,i}(x),\Phi
_{k_{1},j_{1},m_{1}}^{0}(x))$ with $m_{1}=0,(k_{1},j_{1})\neq (n,i)$ and $%
m_{1}\neq 0$ respectively, then use (33), and etc. ) we get
\begin{equation}
\Psi _{n,_{i}}(x)=\Psi _{n,i_{{}}}^{0}(x)+A_{m}^{\ast }(\lambda
_{n,i})+C_{m}^{\ast },
\end{equation}
where $C_{m}^{\ast }$ is obtained \ from $C_{m}$ ( see (53), (45), (46)) by
replasing $c(n,i,0,k_{1},j_{1},p_{1})$ by $\Psi 
_{k_{1},j_{1},p_{1}}^{0}(x).$
Therefore instead of (34) using (38) and doing the estimation which was done
for $C_{m},A_{m}$ ( see (55), (62)) we obtain
\begin{equation}
C_{m}^{\ast }=O\left( (\frac{\ln n}{n})^{m}\right) ,\forall m=1,2,...
\end{equation}
\[
A_{m}^{\ast }(\lambda _{n,i})=A_{m}^{\ast }(\lambda
_{n,i}^{0}+c(n,i)+F_{m-1}))+O\left( (\frac{\ln n}{n})^{m}\right) .
\]
Using these in (67) we obtain (64). 

In case $m=2$ we have $A_{2}^{\ast }(\lambda _{n,i})=S_{1}^{\ast }(\lambda
_{n,i})=S_{1,1}^{\ast }(\lambda _{n,i})+S_{1,2}^{\ast }(\lambda _{n,i}),$
where $S_{1,1}^{\ast }(\lambda _{n,i})$ and $S_{1,2}^{\ast }(\lambda 
_{n,i})$
is obtained from $S_{1,1}(\lambda _{n,i})$ and $S_{1,2}(\lambda _{n,i})$
respectively by replasing $c(n,i,0,k_{1},j_{1},p_{1})$ by $\Psi
_{k_{1},j_{1},p_{1}}^{0}(x).$ Arguing as above from (59), (61) we obtain
\[
A_{2}^{\ast }(\lambda _{n,i})=S_{1,1}^{\ast }(\lambda _{n,i})+O(\frac{1}{%
n^{2}})=S_{1,1}^{\ast }(\lambda _{n,i}^{0})+O\left( (\frac{\ln n}{n}%
)^{2}\right) .
\]
Therefore (67) and (68) for $m=2$ imply (65). The theorem is proved
\end{proof} 

\[
REFERENCES
\] 

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\end{document}