\documentclass[12pt,thmsa]{article} \usepackage{amssymb} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%% \usepackage{sw20lart} %TCIDATA{TCIstyle=Article/art4.lat,lart,article} %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{LastRevised=Fri Aug 22 10:21:59 2003} %TCIDATA{} %TCIDATA{Language=American English} %TCIDATA{CSTFile=article.cst} \input tcilatex \begin{document} \title{On the Riesz Basisness of the Root \ Functions of the Nonselfadjoint Sturm-Liouville Operator } \author{Ne\c{s}e Dernek, O.A. Veliev \\ %EndAName {\small \ Depart. of Math, Fen-Ed. Fak, Marmara Univ., }\\ {\small \ G\"{o}ztepe Kamp\"{u}s\"{u}, 81040, Kadik\"{o}y, }\\ {\small \ Istanbul, Turkey.}\\ {\small \ e-mail: ndernek@marmara.edu.tr}\\ \ {\small oveliev@marmara.edu.tr}} \date{} \maketitle \begin{abstract} In this article we obtain the asymptotic formulas of arbitrary order for eigenfunctions and eigenvalues of the nonselfadjoint Sturm-Liouville operators with \ periodic and antiperiodic boundary conditions, when the potential is a summable function. Then using these asymptotic formulas, we find the conditions on Fourier coefficients of the potential for which the eigenfunctions and associated functions of these operators form a Riesz basis in $L_{2}(0,1)$. \end{abstract} Let $L_{t}(q)$ be the operator generated in $L_{2}[0,1]$ by the expression \begin{equation} -y^{^{\prime \prime }}+q(x)y, \end{equation} and the boundary conditions \begin{equation} y(1)=e^{it}y(0),\text{ }y^{^{\prime }}(1)=e^{it}y^{^{\prime }}(0), \end{equation} where $t=0,\pi$ and $q(x)$ is a complex-valued summable function. In this article we obtain the asymptotic formulas \ of order $O(n^{-l})$ ( for all $l>0$ ) for $n$-th\ eigenvalue and corresponding eigenfunction of the operator $L_{t}(q)$ \ with $q(x)\in L_{1}(0,1),$ for $t=0,\pi$ , that is, for periodic and antiperiodic boundary conditions. Then using these asymptotic formulae, we find the conditions on Fourier coofficients $% q_{n}=(q(x),e^{i2\pi nx})$, where $(.,.)$ denotes inner product in $% L_{2}(0,1),$ of $q(x)$ for which the root functions ( the eigenfunctions and associated functions) of $L_{t}(q),$ for $t=0,\pi ,$ form a Riesz basis in $% L_{2}(0,1)$. Note that the periodic and antiperiodic boundary conditions are regular boundary conditions, but are not strongly regular boundary conditions. Therefore, in general, the eigenfunctions and associated functions of $L_{0}(q)$ and $L_{\pi }(q)$ \ do not form a Riesz basis, they form a basis with brachet \ ( see ,). In this paper we prove that if \begin{equation} \lim_{n\rightarrow \infty }\frac{ln\left| n\right| }{nq_{2n}}=0,\text{ \ }% q_{2n}\sim q_{-2n}, \end{equation} where \ $a_{n}\sim b_{n}$ means that $c_{1}\mid b_{n}\mid <\mid a_{n}\mid &\left| 2(n-k)\pi \right| \left| 2(n+k)\pi \right| -c_{3}n^{\frac{1}{2}}>c_{4}n, \nonumber \\ \forall \text{ }j &=&1,2,\text{\ }\forall k\neq n,\text{ }k=0,1,..., \end{eqnarray} \begin{eqnarray} \left| \mu _{n,j}-(2\pi k+\pi )^{2}\right| &>&\left| 2(n-k)\pi \right| \left| 2(n+k+1)\pi \right| -c_{5}n^{\frac{1}{2}}>c_{6}n\text{ } \nonumber \\ \forall \text{\ }j &=&1,2,\text{\ }\forall k\neq n,,\forall k=0,1,.... \end{eqnarray} for$n\geq N.$\ To obtain the asymptotic formula for eigenvalues$\lambda _{n,j}$and corresponding normalized eigenfunctions$\Psi _{n,j}(x)$of$% L_{0}(q)$we use (7) and the following well-known relation \begin{equation} (\lambda _{N,j}-(2\pi n)^{2})(\Psi _{N,j}(x),e^{i2\pi nx})=(q(x)\Psi _{N,j}(x),e^{i2\pi nx}). \end{equation} Moreover, we use the following relations, which can be easily proved arguing as in the proof of \ Lemma 1 of , \begin{equation} (q(x)\Psi _{N,j}(x),e^{i2\pi nx})=\sum_{m=-\infty }^{\infty }q_{-m}(\Psi _{N,j}(x),e^{i2\pi (n+m)x}) \end{equation} \begin{equation} \left| (q(x)\Psi _{N,j}(x),e^{i2\pi nx})\right| <4M,\text{ }\forall n,\text{ }\forall N\gg 1 \end{equation} where$M=\stackunder{n\in Z}{sup}\left| q_{n}\right| .$Note that$% q_{n}\rightarrow 0$as$\left| n\right| \rightarrow \infty $and, without loss of generality, we assume that$q_{0}=0.$Using (10) in (9) we get \begin{equation} (\lambda _{n,j}-(2\pi n)^{2})(\Psi _{n,j}(x),e^{i2\pi nx})=\sum\Sb % n_{1}=-\infty , \\ n_{1}\neq 0 \endSb ^{\infty }q_{n_{1}}(\Psi _{n,j}(x),e^{i2\pi (n-n_{1})x}). \end{equation} Now isolating the term in right-hand side of (12) containing the multiplicant$(\Psi _{n,j}(x),e^{-i2\pi nx})$(i.e., case$n_{1}=2n$), replacing \bigskip\$(\Psi _{n,j}(x),e^{i2\pi (n-n_{1})x})$by$\frac{(q(x)\Psi _{n,j}(x),e^{i2\pi (n-n_{1})x}))}{\lambda _{n,j}-(2\pi (n-n_{1}))^{2}},$for$n_{1}\neq 2n$( this replacement can be obtained from (9) by taking$n$and$n-n_{1}$instead of$N$and$n$) and using (10) for the numerator of the last fraction we obtain $(\lambda _{n,j}-(2\pi n)^{2})(\Psi _{n,j}(x),e^{i2\pi nx})-q_{2n}(\Psi _{n,j}(x),e^{-i2\pi nx})=$ \begin{equation} \sum\Sb n_{1},n_{2}=-\infty \\ n_{1}\neq 2n \endSb ^{\infty }\frac{% q_{n_{1}}q_{n_{2}}(\Psi _{n,j}(x),e^{i2\pi (n-n_{1}-n_{2})x})}{\lambda _{n,j}-(2\pi (n-n_{1}))^{2}}. \end{equation} Note that, since$q_{0}=0,$here and in the forthcoming relations the sums are taken under conditions$n_{1},n_{2},...,\neq 0$. Now we isolate the terms in right-hand side of (13) containing one of the multiplicants$(\Psi _{n,j}(x),e^{i2\pi nx})$,$(\Psi _{n,j}(x),e^{-i2\pi nx})$(i.e., cases$% n_{1}+n_{2}=0,2n$) and apply the above replacement to other terms (i.e., cases$n_{1}+n_{2}\neq 0,2n$) to get $(\lambda _{n,j}-(2\pi n)^{2})(\Psi _{n,j}(x),e^{i2\pi nx})-q_{2n}(\Psi _{n,j}(x),e^{-i2\pi nx})-$ $\sum\Sb n_{1}=-\infty \\ n_{1}\neq 2n \endSb ^{\infty }\frac{% q_{n_{1}}q_{-n_{1}}(\Psi _{n,j}(x),e^{i2\pi nx})}{\lambda _{n,j}-(2\pi (n-n_{1}))^{2}}-\sum\Sb n_{1}=-\infty \\ n_{1}\neq 2n \endSb ^{\infty }% \frac{q_{n_{1}}q_{2n-n_{1}}(\Psi _{n,j}(x),e^{-i2\pi nx})}{\lambda _{n,j}-(2\pi (n-n_{1}))^{2}}=$ \begin{equation} \sum\Sb n_{1},n_{2}=-\infty \\ n_{1}\neq 2n,n_{1}+n_{2}\neq 0,2n \endSb % ^{\infty }\frac{q_{n_{1}}q_{n_{2}}(q(x)\Psi _{n,j}(x),e^{i(2\pi (n-n_{1}-n_{2})x})}{\lambda _{n,j}-(2\pi (n-n_{1}))^{2}(\lambda _{n,j}-(2\pi (n-n_{1}-n_{2}))^{2})} \end{equation} Repeating this process$m-$times (i.e. applying (10) to the numerator of the fractions in the rigth-hand side of (14) and isolating the terms containing one of the multiplicants$(\Psi _{n,j}(x),e^{i2\pi nx})$,$(\Psi _{n,j}(x),e^{-i2\pi nx})$etc.) we get \begin{equation} (\lambda _{n,j}-(2\pi n)^{2}-A_{m}(\lambda _{n,j}))(\Psi _{n,j}(x),e^{i2\pi nx})- \end{equation} $(q_{2n}+B_{m}(\lambda _{n,j})(\Psi _{n,j}(x),e^{-i2\pi nx})=R_{m},$ where$A_{m}(\lambda _{n,j})=\sum_{k=1}^{m}a_{k}(\lambda _{n,j}),% B_{m}(\lambda _{n,j})=\sum_{k=1}^{m}b_{k}(\lambda _{n,j})$% $a_{k}(\lambda _{n,j})=\sum_{n_{1},n_{2},...,n_{k}}\frac{% q_{n_{1}}q_{n_{2}}...q_{n_{k}}q_{-n_{1}-n_{2}-...-n_{k}}}{[\lambda _{n,j}-(2\pi (n-n_{1}))^{2}]...[\lambda _{n,j}-(2\pi (n-n_{1}-...-n_{k}))^{2}]},$ $b_{k}(\lambda _{n,j})=\sum_{n_{1},n_{2},...,n_{k}}\frac{% q_{n_{1}}q_{n_{2}}...q_{n_{k}}q_{2n-n_{1}-n_{2}-...-n_{k}}}{[\lambda _{n,j}-(2\pi (n-n_{1}))^{2}]...[\lambda _{n,j}-(2\pi (n-n_{1}-..-n_{k}))^{2}]% },$ $R_{m}=\sum_{n_{1},n_{2},...,n_{m+1}}\frac{% q_{n_{1}}q_{n_{2}}...q_{n_{m}}q_{n_{m+1}}(q(x)\Psi _{n,j}(x),e^{i2\pi (n-n_{1}-...-n_{m+1})x})}{[\lambda _{n,j}-(2\pi (n-n_{1}))^{2}]...[\lambda _{n,j}-(2\pi (n-n_{1}-...-n_{m+1}))^{2}]},$ Here the sums are taken under the conditions $n_{j}\neq 0,\forall j\text{ ; }\sum\limits_{j=1}^{k}n_{j}\neq 0,2n,\text{ }% \forall k=1,2,...,m+1.$ Using (7) and arguing as in the proof of equalities (27), (28) of  we get the following relations \begin{equation} a_{k}=O\left( (\frac{ln\left| n\right| }{n})^{k}\right) ,b_{k}=O\left( (% \frac{ln\left| n\right| }{n})^{k}\right) ,R_{m}=O\left( (\frac{ln\left| n\right| }{n})^{m+1}\right) . \end{equation} In the same way one can obtain the relations \begin{equation} (\lambda _{n,j}-(2\pi n)^{2}-A_{m}^{^{\prime }}(\lambda _{n,j}))(\Psi _{n,j}(x),e^{-i2\pi nx})- \end{equation} $(q_{-2n}+B_{m}^{^{\prime }}(\lambda _{n,j}))(\Psi _{n,j}(x),e^{i2\pi nx})=R_{m}^{^{\prime }},$ where$A_{m}^{^{\prime }}(\lambda _{n,j})=\sum_{k=1}^{m}a_{k}^{\prime }(\lambda _{n,j}),B_{m}^{^{\prime }}(\lambda _{n,j})=\sum_{k=1}^{m}b_{k}^{^{\prime }}(\lambda _{n,j}),$% $a_{k}^{^{\prime }}(\lambda _{n,j})=\sum_{n_{1},n_{2},...,n_{k}}\frac{% q_{n_{1}}q_{n_{2}}...q_{n_{k}}q_{-n_{1}-n_{2}-...-n_{k}}}{[\lambda _{n,j}-(2\pi (n+n_{1}))^{2}]...[\lambda _{n,j}-(2\pi (n+n_{1}+...+n_{k}))^{2}]},$ $b_{k}^{^{\prime }}(\lambda _{n,j})=\sum_{n_{1},n_{2},...,n_{k}}\frac{% q_{n_{1}}q_{n_{2}}...q_{n_{k}}q_{-2n-n_{1}-n_{2}-...-n_{k}}}{[\lambda _{n,j}-(2\pi (n+n_{1}))^{2}]...[\lambda _{n,j}-(2\pi (n+n_{1}+...+n_{k}))^{2}]},$ \begin{equation} a_{k}^{^{\prime }}=O\left( (\frac{ln\left| n\right| }{n})^{k}\right) ,b_{k}^{^{\prime }}=O\left( (\frac{ln\left| n\right| }{n})^{k}\right) ,R_{m}^{^{\prime }}=O\left( (\frac{ln\left| n\right| }{n})^{m+1}\right) . \end{equation} Here the sums are taken under the conditions $n_{j}\neq 0,\forall j\text{ ; }\sum\limits_{j=1}^{k}n_{j}\neq 0,-2n,\forall k=1,2,...,m+1.$ It follows from (7), (9) and (11) that $\sum_{k\in Z,k\neq n,-n}\left| (\Psi _{n,j}(x),e^{i2\pi kx})\right| ^{2}=\sum_{k\in Z,k\neq n,-n}\frac{(4M)^{2}}{[\lambda _{n,j}-(2\pi k)^{2}]^{2}}=O(\frac{1}{n^{2}}).$ Therefore$\Psi _{n,t}(x)$has an expansion of the form \begin{equation} \Psi _{n,j}(x)=u_{n,j}e^{i2\pi nx}+v_{n,j}e^{-i2\pi nx}+h(x), \end{equation} where$u_{n,j}=(\Psi _{n,j}(x),e^{i2\pi nx}),v_{n,j}=(\Psi _{n,j}(x),e^{-i2\pi nx}),\left\| h(x)\right\| =O(\frac{1}{n}),$% \begin{equation} \left| u_{n,j}\right| ^{2}+\left| v_{n,j}\right| ^{2}=1+O(\frac{1}{n^{2}}) \end{equation} Using (16), (18), (20) from the system of equations (15), (17) we can find asymptotic formulas \ of order$O(n^{-l})$( for all$l>0$) for eigenfunctions$\Psi _{n,j}(x)$and eigenvalues$\lambda _{n,j}$\ of the operator$L_{0}(q)$\ with$q(x)\in L_{1}(0,1).$We do it under conditions (3), since in these cases the asymptotic formulas give the Riesz basisness. \bigskip \begin{theorem} Let the conditions (3) hold. Then:$(a)$The big eigenvalues of the operator$L_{0}$are simple. They consists of two sequences \$\{\lambda _{n,1}:n>N\}$and$\{\lambda _{n,2}:n>N\}$\ satisfying \begin{equation} \lambda _{n,j}=(2\pi n)^{2}+(-1)^{j}p_{n}+O\left( \frac{ln\left| n\right| }{n% }\right) ,\forall j=1,2, \end{equation} where$p_{n}=(q_{2n}q_{-2n})^{\frac{1}{2}}.$The corresponding eigenfunction$\varphi _{n,j}(x)$satisfies \begin{equation} \varphi _{n,j}(x)=e^{i2\pi nx}+\alpha _{n,j}e^{-i2\pi nx}+O(\frac{1}{n}% ),\forall j=1,2, \end{equation} where \begin{eqnarray} \alpha _{n,j} &\sim &1,\forall j=1,2, \\ \alpha _{n,j} &=&\frac{(-1)^{j}p_{n}}{q_{2n}}+O\left( \frac{ln\left| n\right| }{nq_{2n}}\right) ,\forall j=1,2, \end{eqnarray}$(b)$\ The root functions of$L_{0}(q)$form a Riesz basis in$L_{2}(0,1).$\end{theorem} %TCIMACRO{ %\TeXButton{Proof}{\proof% %} }% %BeginExpansion \proof% % %EndExpansion First we prove that if the conditions (3) hold then \begin{equation} u_{n,j}\sim v_{n,j}\sim 1,\forall j=1,2 \end{equation} ( for the definitions of$u_{n,j}$and$v_{n,j}$see (19)) and \begin{equation} \lambda _{n,j}-(2\pi n)^{2}=a_{j}p_{n}+O\left( \frac{ln\left| n\right| }{n}% \right) . \end{equation} where$a_{j}$\ is either$1$or$-1.$From (15)- (18) we obtain the system of equations \begin{equation} (\lambda _{n,j}-(2\pi n)^{2})u_{n,j}=q_{2n}v_{n,j}+O\left( \frac{ln\left| n\right| }{n}\right) , \end{equation} \begin{equation} (\lambda _{n,j}-(2\pi n)^{2})v_{n,j}=q_{-2n}u_{n,j}+O\left( \frac{ln\left| n\right| }{n}\right) . \end{equation} Without loss of generality we assume that \$\left| v_{n,j}\right| >\frac{1}{% 2}$(see (20)) and hence$v_{n,j}\sim 1.$Then from (28) and (3) we obtain$% \left| \lambda _{n,j}-(2\pi n)^{2}\right| \frac{1}{2},$and (27) imply that$\left| \lambda _{n,j}-(2\pi n)^{2}\right| >c_{8}\left| q_{2n}\right| .$\ Therefore using (3) \ from these inequalities we get \begin{equation} \lambda _{n,j}-(2\pi n)^{2}\sim q_{2n}\sim q_{-2n}\sim p_{n},\text{ }% \lim_{n\rightarrow \infty }\frac{ln\left| n\right| }{np_{n}}=0. \end{equation} This with (27) implies that$u_{n,j}\sim v_{n,j}.$\ Thus (25) is proved. Now dividing both sides of (27) and (28) by$u_{n,j}q_{2n}$\ and$% (\lambda _{n,j}-(2\pi n)^{2})u_{n,j}$respectively we get \begin{equation} \frac{v_{n,j}}{u_{n,j}}=\frac{\lambda _{n,j}-(2\pi n)^{2}}{q_{2n}}+O\left( \frac{ln\left| n\right| }{nq_{2n}}\right) =\frac{q_{-2n}}{\lambda _{n,j}-(2\pi n)^{2}}+O\left( \frac{ln\left| n\right| }{nq_{2n}}\right) , \end{equation} from which using (29) by simple calculation we get (26). The proof of$(a).$Suppose$\lambda _{n,j},$where$n>N,N\gg 1,$is multiple. There are two cases: 1. There is associated functions$\Psi _{n,j}^{1}(x)$corresponding to the eigenfunctions$\Psi _{n,j}(x),$that is, \begin{equation} (L_{0}-\lambda _{n,j})\Psi _{n,j}^{1}(x)=\Psi _{n,j}(x). \end{equation} Since the boundary condition (2) for$t=0$is selfadjoint$\overline{\lambda _{n,j}}$and$\overline{\Psi _{n,j}}$are eigenvalue and eigenfunction of adjoint operator$L_{0}^{\ast }.$Therefore multiplying both sides of (31) by$\overline{\Psi _{n,j}}$we get$(\Psi _{n,j}(x),\overline{\Psi _{n,j}}% )=0.$This with (19) give the equality$u_{n,j}v_{n,j}=O(\frac{1}{n})$which contradicts (25). 2. There are two eigenfunctions corresponding$\lambda _{n,j}.$Then all solutions of the equation $-y^{^{\prime \prime }}+q(x)y=\lambda _{n,j}y$ are eigenfunctions. In particular, the solution $y(x,\lambda _{n,j})=e^{\rho _{n,j}x}+O(\frac{1}{n})=e^{2\pi nx}+O(\frac{1}{n}% ),$ where$\rho _{n,j}=(\lambda _{n,j})^{\frac{1}{2}}=2\pi n+(\frac{1}{n}),$( see page 52 of ), and (26)) is eigenfunction. In (27) instead of using$% \Psi _{n,j}$taking this eigenfunction we get $\lambda _{n,j}=(2\pi n)^{2}+O\left( \frac{ln\left| n\right| }{n}\right) ,$ which contradicts (29). Thus the eigenvalues$\lambda _{n,j}$for$n>N,j=1,2$are simple and satisfy (26). Now, in order to obtain (21) from (26) it remains to prove that the numbers$a_{1},a_{2}$in (26) are different, say$% a_{1}=-1,a_{2}=1.$Suppose both$a_{1}$and$a_{2}$are$-1$. Then (26), (27) yield \begin{equation} -p_{n}u_{n,1}=q_{2n}v_{n,1}+O\left( \frac{ln\left| n\right| }{n}\right) ,-p_{n}u_{n,2}=q_{2n}v_{n,2}+O\left( \frac{ln\left| n\right| }{n}\right) , \end{equation} Multiplying both sides of f\i rst and second equalities by$-v_{n,2}$and$% v_{n,1}$respectively and then summing the obtained equalities we get \begin{equation} u_{n,1}v_{n,2}-u_{n,2}v_{n,1}=O\left( \frac{ln\left| n\right| }{np_{n}}% \right) . \end{equation} On the other hand, since$\overline{\lambda _{n,j}}$and$\overline{\Psi _{n,j}}$are eigenvalue and eigenfunction of adjoint operator$L_{0}^{\ast }$and$\lambda _{n,1}\neq \lambda _{n,2}$\ we have \begin{equation} 0=(\Psi _{n,1},\overline{\Psi _{n,2}})=u_{n,1}v_{n,2}+u_{n,2}v_{n,1}+O(\frac{% 1}{n}) \end{equation} (see (19)). It follows from (33), (34), (29) that$u_{n,1}v_{n,2}=O\left( \frac{ln\left| n\right| }{np_{n}}\right) =o(1)$which contradicts (25). If we suppose that both$a_{1}$and$a_{2}$are$1$then, in the same way, we obtain the same contradiction. Thus one of these numbers, say$a_{1}$is$-1$and \ the other$a_{2}$is$1$. The formulas (21) are proved. Since$\lambda _{n,1}$satisfies (21), \ for$j=1,$first equality in (32) holds. Therefore \begin{equation} \frac{v_{n,1}}{u_{n,1}}=\frac{-p_{n}}{q_{2n}}+O\left( \frac{ln\left| n\right| }{nq_{2n}}\right) \end{equation} Denoting$\alpha _{n,j}=\frac{v_{n,j}}{u_{n,j}},\varphi _{n,j}(x)=\frac{% \Psi _{n,j}(x)}{u_{n,j}}$\ from (19), (25) and (35) we get (22), (23), (24) for$j=1$. Taking into account that$\lambda _{n,2}$satisfies (21) for$j=2$in the same way we obtain (22), (23), (24) for$j=2$. The proof of$(b).$Now we prove that the eigenfunctions$\varphi _{n,j}\equiv \varphi _{n,j}^{0}$and associated functions$\varphi _{n,j}^{k}, $where$n=1,2,...,$and$\ k=1,2,,,,s(n,j),$of$L_{0}(q)$form a Riesz basis in$L_{2}(0,1)$. For every$f(x)\in L_{2}(0,1)$, the asymptotic formulas (22), (23) yield \begin{equation} \sum_{n=1}^{N}(\sum_{k=0}^{s(n,j)}\sum_{j=1}^{2}\left| (f,\varphi _{n,j}^{k})\right| ^{2})+\sum_{j=1}^{2}\sum\limits_{n=N+1}^{\infty }\left| (f,\varphi _{n,j})\right| ^{2}<\infty . \end{equation} Let$\chi _{n,j}\equiv \chi _{n,j}^{0}$and$\chi _{n,j}^{k},$where$% k=1,2,,,,t(n,j),$be the biorthonormal system of eigenfunctions and associated functions of$L_{0}^{\ast }.$Clearly$\chi _{n,j}(x)=\frac{% \overline{\varphi _{n,j}(x)}}{(\varphi _{n,j},\overline{\varphi _{n,j}})}$for$n>N$. Therefore using (22), (23) we get \begin{equation} \sum_{n=1}^{N}(\sum_{k=0}^{t(n,j)}\sum_{j=1}^{2}\left| (f,\chi _{n,j}^{k})\right| ^{2})+\sum_{j=1}^{2}\sum\limits_{n=N+1}^{\infty }\left| (f,\chi _{n,j})\right| ^{2}<\infty . \end{equation} Since$\{\varphi _{n,j}^{k}:k=0,1,2,,,,s(n,j);j=1,2;n=1,2,...\}$, and the biorthonormal system is total (see chap.1, sec.3 of ) by the well-known theorem of Bari (see , chap. 6) for the Riesz basis, (36) and (37) imply that the system of the eigenfunctions and associated functions of$L_{0}(q)$forms a Riesz basis in$L_{2}(0,1)$.The theorem is proved$\blacksquare $Instead of (3), (7), and (9) using (4), (8) and \begin{equation} (\mu _{N,j}-(2\pi n+\pi )^{2})(\Phi _{N,j}(x),e^{i\pi (2n+1)x})=(q(x)\Phi _{N,j}(x),e^{i\pi (2n+1)x}), \end{equation} where$\Phi _{N,j}(x)$is normalized eigenfunction of$L_{\pi }(q),$and arguing as in proof of Theorem 1 we obtain \bigskip \begin{theorem} Let the conditions (4) hold. Then:$(a)$The big eigenvalues of the operator$L_{\pi }$are simple. They consist of two sequences \$\{\mu _{n,1}:n>N\}$and$\{\mu _{n,2}:n>N\}$\ satisfying \begin{equation} \mu _{n,j}=(2\pi n+\pi )^{2}+(-1)^{j}p_{n}^{^{\prime }}+O\left( \frac{% ln\left| n\right| }{n}\right) ,\text{ }\forall j=1,2, \nonumber \end{equation} where$p_{n}^{^{\prime }}=(q_{2n+1}q_{-2n-1})^{\frac{1}{2}}.$The corresponding eigenfunction$\phi _{n,j}(x)$satisfies \begin{equation} \phi _{n,j}(x)=e^{i(2\pi n+\pi )x}+\alpha _{n,j}^{^{\prime }}e^{-i(2\pi n+\pi )x}+O(\frac{1}{n}),\text{ }\forall j=1,2, \end{equation} where$\alpha _{n,j}^{^{\prime }}\sim 1,\alpha _{n,j}^{^{\prime }}=\frac{% (-1)^{j}p_{n}^{^{\prime }}}{q_{2n+1}}+O\left( \frac{ln\left| n\right| }{% nq_{2n+1}}\right) ,\forall j=1,2.(b)$\ The root functions of$L_{\pi }(q)$form a Riesz basis in$L_{2}(0,1)$% . \end{theorem} \bigskip Now we prove asymptotic formulas \ of arbitrary order for eigenfunctions and eigenvalues of the operators$L_{0}(q),L_{\pi }(q)$\ with$q(x)\in L_{1}(0,1).$\begin{theorem} For the eigenvalue$\lambda _{n,j}$satisfying (21), the following formulas hold \begin{eqnarray} \lambda _{n,j} &=&(2\pi n)^{2}+F_{m,j}+(-1)^{j}G_{m,j}+O\left( (\frac{% ln\left| n\right| }{n})^{m+1}\right) , \\ \forall m &=&0,1,2,...,\forall j=1,2, \nonumber \end{eqnarray} where$F_{0,j}=0,G_{0,j}=p_{n},F_{k,j}=\frac{1}{2}(A_{k}((2\pi n)^{2}+F_{k-1,j}+(-1)^{j}G_{k-1,j})+A_{k}^{^{\prime }}((2\pi n)^{2}+F_{k-1,j}+(-1)^{j}G_{k-1,j})),G_{k,j}=[\frac{1}{4}(A_{k}((2\pi n)^{2}+F_{k-1,j}+(-1)^{j}G_{k-1,j})-A_{k}^{^{\prime }}((2\pi n)^{2}+F_{k-1,j}+(-1)^{j}G_{k-1,j}))^{2}+(B_{k}((2\pi n)^{2}+F_{k-1,j}+(-1)^{j}G_{k-1,j})+q_{2n})(B_{k}^{^{\prime }}((2\pi n)^{2}+F_{k-1,j}+(-1)^{j}G_{k-1,j})+q_{-2n})]^{\frac{1}{2}},\forall k=1,2,...,$For the eigenfunction$\varphi _{n,j}(x)$satisfying (23) the following formulas hold \begin{eqnarray} \varphi _{n,j}(x) &=&e^{i2\pi nx}+\alpha _{n,j}e^{-i2\pi nx}+A_{m}^{\ast }((2\pi n)^{2}+F_{m,j}+(-1)^{j}G_{m,j}) \\ + &&(B_{m}^{\ast }((2\pi n)^{2}+F_{m,j}+(-1)^{j}G_{m,j}))\alpha _{n,j}+O((% \frac{\ln n}{n})^{m+1}), \nonumber \end{eqnarray} where$A_{m}^{\ast }$,$B_{m}^{\ast }$are obtained from$A_{m},B_{m}$respectively by replacing$q_{n_{1}}$with$e^{i2\pi (n-n_{1})x}$and \begin{eqnarray*} \alpha _{n,j} &=&\frac{v_{n,j}}{u_{n,j}}=\frac{% F_{m,j}+(-1)^{j}G_{m,j}-A_{m}((2\pi n)^{2}+F_{m,j}+(-1)^{j}G_{m,j})}{% q_{2n}+B_{m}((2\pi n)^{2}+F_{m,j}+(-1)^{j}G_{m,j})}+ \\ &&O\left( \frac{1}{q_{2n}}(\frac{ln\left| n\right| }{n})^{m+1}\right) . \end{eqnarray*} For the eigenvalue$\mu _{n,j}$and eigenfunction$\phi _{n,j}(x)$\ of$% L_{\pi }(q)$satisfying (39), (40) similar formulas hold. \end{theorem} \bigskip %TCIMACRO{ %\TeXButton{Proof}{\proof% %} }% %BeginExpansion \proof% % %EndExpansion Arguing as in the proof of (30), using (15), (17) in place of (27), (28), and denoting$\lambda =\lambda _{n,j}-(2\pi n)^{2}$we get \begin{eqnarray} \frac{v_{n,j}}{u_{n,j}} &=&\frac{\lambda -A_{m}(\lambda _{n,j})}{% (q_{2n}+B_{m}(\lambda _{n,j}))}+O\left( \frac{1}{q_{2n}}(\frac{ln\left| n\right| }{n})^{m+1}\right) \nonumber \\ &=&\frac{(q_{-2n}+B_{m}^{^{\prime }}(\lambda _{n,j}))}{\lambda -A_{m}^{^{\prime }}(\lambda _{n,j})}+O\left( \frac{1}{q_{2n}}(\frac{ln\left| n\right| }{n})^{m+1}\right) , \end{eqnarray} Solving this square equation with respect to$\lambda $we obtain $\lambda _{n,j}-(2\pi n)^{2}=\frac{1}{2}(A_{m}(\lambda _{n,j})+A_{m}^{^{\prime }}(\lambda _{n,j}))\mp \lbrack \frac{1}{4}% (A_{m}(\lambda _{n,j})-A_{m}^{^{\prime }}(\lambda _{n,j}))^{2}+$ \begin{equation} (B_{m}(\lambda _{n,j})+q_{2n})(B_{m}^{^{\prime }}(\lambda _{n,j})+q_{-2n}))+O\left( q_{2n}(\frac{ln\left| n\right| }{n})^{m+1}\right) ]^{\frac{1}{2}}. \end{equation} Here upper and lower sign is taken for$j=1,$and$j=2$respectively, since the expression in the square bracket is $q_{2n}q_{-2n}+O(q_{2n}\frac{ln\left| n\right| }{n})=q_{2n}q_{-2n}(1+o(1))$ (see (3), (16), (18)), and (21) holds. Thus in (44) the$\mp $can be replaced by$(-1)^{j}.$Now we prove (41) by induction. It is proved for$% m=1 $( see (21) ). Assume that it is true for$m=k-1$. Substituting the value of$\lambda _{n,j},$given by this formula for$m=k-1,$in (44) for$% m=k$and taken into account that \begin{eqnarray} &&A_{k}((2\pi n)^{2}+F_{k-1,j}+(-1)^{j}G_{k-1,j}+O\left( (\frac{ln\left| n\right| }{n})^{k}\right) ) \nonumber \\ &=&A_{k}((2\pi n)^{2}+F_{k-1,j}+(-1)^{j}G_{k-1,j})+O\left( (\frac{ln\left| n\right| }{n})^{k+1}\right) , \end{eqnarray} and the same formulas hold for$A_{k}^{^{\prime }},B_{k},B_{k}^{^{\prime }}$we get the proof of (41) for$m=k.$Note that the validity of (45) can be easily verified by using the obvious relation $\sum\Sb n_{1}=-\infty , \\ n_{1}\neq 0,2n \endSb ^{\infty }\mid \frac{1}{% (2\pi n)^{2}+F_{k-1,j}+(-1)^{j}G_{k-1,j}+O\left( (\frac{ln\left| n\right| }{n% })^{k}\right) -(2\pi (n-n_{1})^{2}}-$ $\frac{1}{(2\pi n)^{2}+F_{k-1,j}+(-1)^{j}G_{k-1,j}-(2\pi (n-n_{1})^{2}}\mid =O\left( (\frac{ln\left| n\right| }{n})^{k+1}\right) .$ Now we prove (42). Writing the decomposition of$\Psi _{n,j}(x)$by$\{e^{i2\pi (n-n_{1})x}:n_{1}\in Z\}$we obtain \begin{equation} \Psi _{n,j}(x)-u_{n,j},e^{i2\pi nx}=\sum\Sb n_{1}=-\infty , \\ n_{1}\neq 0 \endSb ^{\infty }(\Psi _{n,j}(x),e^{i2\pi (n-n_{1})x})e^{i2\pi (n-n_{1})x}. \end{equation} The right side of this formula can be obtained from the right side of (12) by replacing$q_{n_{1}}$with$e^{i2\pi (n-n_{1})x}.$Therefore doing the iteration which was dane in order to obtain (15) from (12) ( namely, isolate the terms in right-hand side of (46) containing the multiplicant$(\Psi _{n,j}(x),e^{-i2\pi nx})$(i.e., case$n_{1}=2n$) and apply (9), (10) to other terms (i.e., cases$n_{1}\neq 2n$), then isolate the terms containing one of the multiplicants$(\Psi _{n,j}(x),e^{-i2\pi nx})$,$(\Psi _{n,j}(x),e^{-i2\pi nx})$and apply (9), (10) to other terms , ets.) and the estimation as in (16) we get \begin{eqnarray} \Psi _{n,j}(x) &=&u_{n,j}e^{i2\pi nx}+v_{n,j}e^{-i2\pi nx}+ \\ &&u_{n,j}A_{m}^{\ast }(\lambda _{n,j})+v_{n,j}B_{m}^{\ast }(\lambda _{n,j})+O\left( (\frac{ln\left| n\right| }{n})^{m+1}\right) \nonumber \end{eqnarray} Dividing both sides of (47) by$u_{n,j}$and using (43), (41), (45) we get (42). Instead of (3), (7), (9) and Theorem 1 using (4), (8), (38) and Theorem 2 in the same way we get the similar asymptotic formulas for eigenvalues and eigenfunctions of$L_{\pi }(q)\blacksquare \$ \begin{thebibliography}{9} \bibitem{} M. A. Naimark, Linear Differential Operators, George G. Harap\&Company, 1967. \bibitem{} V. A. Marchenko, \ Sturm-Liouville Operators and Applications, Birkhauser Verlag, Basel, 1986. \bibitem{} I. T. Goghberg and M. G. Krein, Introduction to Theory of Linear Nonselfadjoint operators in Hilbert Space, Nauka, Moscow, 1965. \bibitem{} A. A. Shkalikov, The basis property of eigenfunctions of an ordinary differential operator, Uspekhi Mat.Nauk 34 (1979), no. 5(209), 235-236. \bibitem{} A. A. Shkalikov, Boundary value problem for ordinary differential equations with a parameter in the boundary conditions. Trudy Sem. Petrovsk, 9 (1983), 190-229. \bibitem{} O. A .Veliev, \ M. 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