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\markboth{\hfil Basis properties of eigenfunctions \hfil EJDE--2000/28}
{EJDE--2000/28\hfil P. E. Zhidkov \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc Electronic Journal of Differential Equations},
Vol.~{\bf 2000}(2000), No.~28, pp.~1--13. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)}
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Basis properties of eigenfunctions of nonlinear Sturm-Liouville problems
\thanks{ {\em Mathematics Subject Classifications:} 34L10, 34L30, 34L99.
\hfil\break\indent
{\em Key words and phrases:} Riesz basis, nonlinear eigenvalue problem,
\hfill\break\indent Sturm-Liouville operator, completeness, basis.
\hfil\break\indent
\copyright 2000 Southwest Texas State University and University of
North Texas. \hfil\break\indent
Submitted November 17, 1999. Published April 13, 2000.} }
\date{}
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\author{ P. E. Zhidkov }
\maketitle
\begin{abstract}
We consider three nonlinear eigenvalue problems that consist of
$$-y''+f(y^2)y=\lambda y$$
with one of the following boundary conditions:
$$\displaylines{ y(0)=y(1)=0 \quad y'(0)=p \,,\cr y'(0)=y(1)=0
\quad y(0)=p\,, \cr y'(0)=y'(1)=0 \quad y(0)=p\,, }$$
where $p$ is a positive constant. Under smoothness
and monotonicity conditions on $f$,
we show the existence and uniqueness of a sequence of eigenvalues
$\{\lambda _n\}$ and corresponding eigenfunctions $\{y_n\}$ such
that $y_n(x)$ has precisely $n$ roots in the interval $(0,1)$,
where $n=0,1,2,\dots$.
For the first boundary condition, we show that $\{y_n\}$ is a basis and
that $\{y_n/\|y_n\|\}$ is a Riesz basis in the space $L_2(0,1)$.
For the second and third boundary conditions, we show that $\{y_n\}$
is a Riesz basis.
\end{abstract}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\section{Introduction}
We consider three eigenvalue problems which consist of the
nonlinear equation
$$ -y''+f(y^2)y=\lambda y \quad\mbox{on }(0,1) \eqno (1)$$
with one of the following three boundary conditions:
$$\displaylines{
\hfill y(0)=y(1)=0 \quad y'(0)=p\,,
\hfill\llap{(2a)} \cr \hfill y'(0)=y(1)=0 \quad y(0)=p\,,
\hfill\llap{(2b)} \cr \hfill y'(0)=y'(1)=0 \quad
y(0)=p\,.\hfill\llap{(2c)} }$$
Hereafter all quantities are real, including
$\lambda$ which is a spectral parameter and $p$ which is an arbitrary
constant. However, we consider only $p>0$, because when $p<0$ the
substitution of $y(x)$ by $-y(x)$ leads to a positive $p$.
We assume that $f$ is a continuously differentiable function on
${\mathbb R}$; so that the standard theorems, such as
local existence, uniqueness, and
continuous dependence on parameters are valid for (1).
A pair $(\lambda,y)$ with $\lambda$ real and $y=y(x)$ twice continuously
differentiable for all $x\in [0,1]$, that satisfies (1) and one of the
boundary conditions (2) is called an {\it eigenvalue} and corresponding
{\it eigenfunction}. Since each of the boundary conditions (2)
contains Cauchy data and an additional condition, by the standard uniqueness
theorem to each value $\lambda $ there corresponds at most one
eigenfunction.
By the {\it spectrum} we mean the set of all eigenvalues for each problem
defined by (1)-(2). For any $g\in L_2=L_2(0,1)$, by the
{\it normalized function} $g$ we mean the normalization of this function
to $1$ in the above space.
In the present paper, we investigate the properties of the eigenvalues and
eigenfunctions of each problem defined by (1)-(2).
In particular, we prove that for problems (1)-(2b) and (1)-(2c) the
eigenvectors form Riesz bases, and that the normalized eigenvectors of
problem (1)-(2a) also form a Riesz basis. For problem (1)-(2b), these
questions have already been considered in \cite{z1}. The proof that eigenfunctions
form a Riesz basis was done using Bary's theorem.
In the present paper the author establishes a direct proof (not based on
Bary's theorem) of the result in \cite{z1}, and then studies
problems (1)-(2a) and (1)-(2c).
According to Bary's theorem \cite{b1,b2,g1,m1}, a linearly
independent system of functions $\{h_n\}$ which
is $L_2(a,b)$ quadratically close to a Riesz basis in this space is a
Riesz basis in $L_2(a,b)$. This result in \cite{b1,b2} is presented
in a weaker form. However the proof in \cite{b2} applies without
modifications to the above statement.
Note that an orthonormal basis in $L_2(a,b)$ is also a Riesz basis in $L_2$
(see definitions in Section 2). A Riesz basis that is quadratically
close to an orthonormal basis is called Bary basis \cite{g1}.
Important properties of Riesz and Bary bases are presented in
\cite{b1,b2,g1}.
In the present paper, we shall prove linear independence and
quadratic closeness of the normalized eigenfunctions of all problems (1)-(2)
to orthonormal bases in the space $L_2$.
We also obtain estimates of the form
$\overline c<\|y_n\|_{L_2}<\overline C$ for the eigenfunctions
of problems (1)-(2b) and (1)-(2c) with constants
$0<\overline c<\overline C$ independent of $n$. Thus,
eigenfunctions of problems (1)-(2b)
and (1)-(2c) and normalized eigenfunctions
of the problem (1)-(2a) are Riesz bases. Moreover, the latter system is a
Bary basis in $L_2$. We present a direct simple and short
proof, not based on Bary's theorem, of these systems being Riesz
bases in $L_2$.
Several papers have presented proofs of completeness, of being a basis,
and of other properties for eigenfunctions of nonlinear boundary-value
problems; see for example \cite{m1}.
In \cite{z2}, the main result consists of proving that the eigenfunctions of a
nonlinear Sturm-Liouville-type problem form a basis in $L_2$.
These results are also announced (without proofs) in \cite{z3}.
However, \cite{z2} contains errors which have been corrected
in \cite{z4}, where also an analog of the Fourier transform over eigenfunctions
of nonlinear Sturm-Liouville-type problems on a half-line is considered.
An independent (and shorter) proof based on the Bary's theorem
of the result from \cite{z2} is done in \cite{z5}.
Lately, in \cite{z6} it is shown that the eigenfunctions form a
basis in $H^s$, where $s$ is a negative constant. This is done for
a boundary-value similar to (1), without the spectral parameter, and
with zero Dirichlet boundary conditions.
Concerning the applications of our results, the author hopes
that once developed our results be used in Galerkin and Fourier methods.
In addition, our results could be applied in those numerous areas of
quantum physics where various modifications of the nonlinear
Schr\"odinger equation arise.
\section{Main results}
First we introduce some notation. Let $L_2=L_2(0,1)$ (or $L_2(a,b)$
where $a**0$ has the points
$1/(2n) + k/n $ as roots; while $y_0$ is identically equal to $p$, thus
it has no roots. In each of these three cases $k=0,\pm 1,\pm 2,\dots $.
\item{(f)} For any eigenfunction $y$ of problems (1)-(2)
and for all $x\in{\mathbb R}$, $y(c+x)=-y(c-x)$ if
$y(c)=0$, and $y(d-x)=y(d+x)$ if $y'(d) =0$.
\item{(g)} The eigenvalues of problems (1)-(2) satisfy
$\lambda _0<\lambda _1<\lambda _2<\dots $.
\end{description}
\end{theorem}
Parts of Theorem 1 are known (see, for example, \cite{k1}) but the author does
not know a reference for all of this theorem. The author does not
know of any function $g$ that is an eigenfunction of a problem (1)-(2)
and is representable as a superposition of elementary functions.
The main result of the paper is stated as follows.
\begin{theorem} \label{thm2} Under Assumption {\rm (F)}
the eigenfunctions of problems (1)-(2b) and (1)-(2c)
form Riesz bases in $L_2$. Also under Assumption {\rm (F)}, the
eigenfunctions of problem (1)-(2a) form a basis in $L_2$ and they form a
Riesz basis after being normalized.
\end{theorem}
\paragraph{Remark 1.} It follows from (17) in the proof of Theorem 2
that for problem (1)-(2a), $\|y_n\|\to 0$
as $n\to \infty$. Therefore, this sequence of eigenfunctions is not a
Riesz basis in $L_2$.
\section{Proofs}
First we prove the statement of Theorem 1 for problem (1)-(2a).
Consider the Cauchy problem
$$\displaylines{
\hfill -y''+f(y^2)y=\lambda y \hfill\llap{(3)} \cr
\hfill y(0)=0, \quad y'(0)=p\,. \hfill\llap{(4)}
}$$
Since $p>0$, for any $\lambda \in {\mathbb R}$ the solution $y$ satisfies
$y'(x)>0$ in a neighborhood of zero.
It is well known that equation (3) can be solved by quadratures
and our proof of Theorem 1 is based on this fact (see below).
Clearly, if a solution $y(x)$ of the problem (3)-(4) can be continued onto a
segment $[0,d]$, and if $y'(x)>0$ for $x\in [0,d)$, then the
inverse function $x(y)$ with domain $[0,y(d)]$ has the form
$$
x(y)=\int_0^y \frac {dr}{\sqrt {U(\lambda ,r^2)}}\,, \eqno (5)
$$
where $U(\lambda ,r^2)=p^2+F(r^2)-\lambda r^2$, $F(r)=\int_0^r f(t)\, dt$,
$y\in [0,y(d)]$, and $x(y(d))=d$.
Note that if $y'(d)=0$, then $x'(y(d)-0)=+\infty$.
If there exists $\overline \lambda$ such that for $\lambda >\overline \lambda$
the function $U(\lambda ,r^2)$ is positive in a half-open interval
$r\in [0,b)$, where $b>0$, then the solution $y$ is defined on the closed
interval $[0,x(b)]$.
In this case $y(x(b))=b$, $y'(x)>0$ for $x\in [0,x(b))$ and the function $x(y)$
with the domain $[0,b]$ is inverse of $y(x)$ on $[0,x(b)]$.
In addition, $y'_x(x(b))=0$ if $x(b)<+\infty$ and $x'(b-0)=+\infty$.
We denote by $\Lambda$ the set of values $\lambda$ for
which there exists $r>0$ such that $U(\lambda ,r^2)=0$.
In view of condition {\rm (F)}, $\inf \Lambda \geq f(0)$. It is clear
that if $\lambda \in \Lambda$ and $\tilde {\lambda} > \lambda $
then $\tilde {\lambda}$ also belongs to $\Lambda$.
Therefore, either $\Lambda =(\overline \lambda,+\infty)$
(the case A) or $\Lambda =[\overline \lambda,+\infty)$ (the case B)
for some $\overline \lambda\geq f(0)$.
For each $\lambda \in \Lambda $ we denote by $z(\lambda)$ the greatest
lower bound of the set of values $r>0$ for which
$U(\lambda ,r^2)=0$.
Clearly, $z(\lambda)>0$ and $U(\lambda ,z^2(\lambda ))=0$.
Furthermore, $z(\lambda )$ is a monotonically decreasing
function. Therefore, if $\lambda >\overline \lambda$, then
$$[U(\lambda ,r^2)]'_r\big|_{r=
z(\lambda)}=2z(\lambda)f(z^2(\lambda))-2\lambda z(\lambda)<
2z(\lambda )(f(z^2(\lambda '))-\lambda ')\leq 0
$$
for any $\lambda '\in (\overline \lambda,\lambda )$
(because $0\geq U'_r(\lambda ',r^2)\big|_{r=z
(\lambda ')}=2z(\lambda ')(f(z^2(\lambda '))-\lambda ')$,
where $z(\lambda ')>0$); hence by the implicit function theorem
$z(\lambda)$ is a continuously differentiable function
and $z'(\lambda)<0$ for $\lambda >\overline \lambda$.
Also, in the case B $\lim_{\lambda \to \overline {\lambda}+0}
z(\lambda )=z(\overline {\lambda} )$.
Let us prove now that for any $d>0$ there exists
$\lambda >\overline \lambda$ such that
$$
J(\lambda)=\int_0^{z(\lambda)}\frac {dr}{\sqrt {U(\lambda ,r^2)}}=d\,,
\eqno (6)
$$
where the integral in the right-hand side is understood as improper in a
neighborhood of the point $r=z(\lambda )$.
Note that as indicated above,
$[U(\lambda ,r^2)]'_r\big|_{r=z(\lambda)}<0$ for
$\lambda >\overline \lambda$; thus for $\lambda >\overline \lambda$
the improper integral in the right-hand side
of (6) converges, and $J(\lambda)$ is a continuous function.
Let us consider the case A: when $\Lambda =(\overline \lambda,+\infty)$.
It is clear that
$z(\overline \lambda+0)=+\infty$, because otherwise in view of the continuity
of the function
$U(\lambda ,r^2)$ of the argument $(\lambda,r)$ we should get that
$\overline \lambda \in \Lambda$, and that $z(+\infty)=0$.
This easily implies that $\overline \lambda=
\lim_{r\to \infty}f(r^2)<+\infty$ and the latter fact yields that
$$
\lim_{\lambda \to \overline \lambda+0} J(\lambda)=+\infty \quad {\rm and}
\quad \lim_{\lambda \to +\infty} J(\lambda)=0\,.
\eqno (7)
$$
Therefore, in case A the existence of $\lambda >\overline \lambda$ such that
(6) holds is proved.
Let us consider the case B: when $\Lambda =[\overline \lambda,+\infty)$.
We have as in the case A
$\lim_{\lambda \to + \infty} J(\lambda )=0$. Hence, in view of the continuity
of the function $J(\lambda)$, we have only to prove that $\lim_{\lambda \to
\overline \lambda+0} J(\lambda)=+\infty$.
It follows from the implicit function theorem that
$[U(\overline \lambda ,r^2)]'_r\big|_{r=z(\overline \lambda)}=0$ (because otherwise
by this theorem there exist values of $\lambda$ smaller than
$\overline \lambda$
and belonging to $\Lambda$). This fact implies the required property because
$\lim_{\lambda \to \overline \lambda+0}J(\lambda)
\geq J(\overline \lambda)$ (indeed,
$U(\overline \lambda,r^2)>0$ for $r\in (0,z(\overline \lambda))$ and
it is clear that for any $y\in (0,z(\overline \lambda ))$
$$\int_0^y \frac {dr}{\sqrt {U(\lambda ,r^2)}} \to
\int_0^y \frac {dr}{\sqrt {U(\overline \lambda ,r^2)}}$$ as $\lambda \to
\overline \lambda +0$ and also, $U(\overline \lambda,r^2)=
{\rm O}((z(\overline \lambda)-r)^2)$ as $r\to z(\overline \lambda)-0$).
In view of the above arguments, it is proved that for any $d>0$ there exists
$\lambda \in {\mathbb R}$ such that the corresponding solution $y(x)$
of the problem (3)-(4) can be continued
on the segment $[0,d]$, $y'(x)>0$ for $x\in [0,d)$ and $y'(d)=0$
where $y(d)=z(\lambda )$.
Let us prove that for any $d>0$ there exists precisely one value
$\lambda$ possessing this property. On the contrary, suppose that for some
$d>0$ there exist two values $\lambda _1<\lambda _2$ such that the
corresponding solutions $y_1(x)$ and $y_2(x)$ of the problem (3)-(4) can be
continued on the segment $[0,d]$,
$y'_i(x)>0$ for $x\in [0,d)$ and $y'_i(d)=0$ ($i=1,2$).
In view of the above arguments $\lambda _1>\overline \lambda$.
Multiply equation (3), with $y(x)=y_1(x)$, by $y_2(x)$.
Multiply equation (3), with $y(x)=y_2(x)$, by $y_1(x)$. Then subtract these
equalities from each other and integrate over the segment $[0,d]$.
In view of the initial data (4), applying the integration by parts, we get
$$
0=\int_0^d y_1(x) y_2(x) [\lambda _1-\lambda _2-f(y_1^2(x))+f(y_2^2
(x))] dx\,. \eqno (8)
$$
By (5) the inverse functions $x_1(y)$ and $x_2(y)$
of $y_1(x)$ and $y_2(x)$ respectively, satisfy $x_1(y)y_2(x)$ for all $x\in (0,d]$.
Then, in view of the assumption {\rm (F)}, the right-hand side in (8)
is negative, which is a contradiction.
Thus, it is proved that for any $d>0$ there exists
a unique $\lambda$ such that the corresponding solution $y(x)$
of the problem (3)-(4) satisfies $y'(x)>0$ if $x\in [0,d)$ and $y'(d)=0$.
Finally, in view of the autonomy of the equation (3) and its invariance
with respect to
the changes of variables $y(x)\to -y(x)$ and $x\to c-x$, where $c$
is an arbitrary constant, we obtain that if a solution $y(x)$ of the problem
(3)-(4) can be continued
onto the segment $[0,d]$ and satisfies the conditions
$y'(x)>0$ for $x\in [0,d)$ and
$y'(d)=0$, then it can be continued onto the whole real line, and
$y(d+x)=y(d-x)$ (in particular $y(2d)=0$) and $y(2kd+x)=-y(2kd-x)$ for
all $x\in{\mathbb R}$ and for $k=0,\pm 1, \pm 2,\dots $.
Therefore, Theorem 1, except the statements (d) and (g),
for the problem (1)-(2a) is proved.
The statement (d) follows from the fact that the function
$z(\lambda)$ for $\lambda >\overline \lambda$, coinciding with $y(d)$,
decreases monotonically. Since, as it is proved earlier,
for any $d >0$ there exists a unique $\lambda \in{\mathbb R}$
(where $\lambda >\overline \lambda $)
such that $y'(x)>0$ for $x\in [0,d)$ and $y'(d)=0$, where $y(x)$ is
the solution of the problem (3)-(4) corresponding to this value of the
parameter $\lambda$, and
since $d>0$ continuously depends on $\lambda$ (because $d=J(
\lambda)$), by the properties (7), which, as it is proved earlier, hold
in each of the cases A and B,
the value $d>0$ is a monotonically decreasing function of the argument
$\lambda \in (\overline \lambda,+\infty)$. The statement (g) of Theorem 1
for the problem (1)-(2a) immediately follows from this fact and
the above arguments.
For problems (1)-(2b) and (1)-(2c) the proof of the statements of Theorem 1
can be made by analogy. Thus, Theorem 1 is proved.\hfill $\diamondsuit$
\smallskip
Next, we turn to the proof of Theorem 2.
We associate with three problems (1)-(2), respectively, the following three
linear eigenvalue problems:
$$ \displaylines{
\hfill -u''=\mu u, \quad x\in (0,1), \quad u=u(x),\quad u(0)=u(1)=0,
\hfill\llap{(9a)}\cr
\hfill -u''=\mu u, \quad x\in (0,1), \quad u=u(x),\quad u'(0)=u(1)=0,
\hfill\llap{(9b)}\cr
\hfill -u''=\mu u, \quad x\in (0,1), \quad u=u(x), \quad u'(0)=u'(1)=0,
\hfill\llap{(9c)}
}$$
where all quantities are real and $\mu$ is the spectral parameter.
For each of the problems (9) by $\{e_n\}$ ($n=0,1,2,\dots$) we denote the
orthonormal basis consisting of its normalized eigenfunctions.
More precisely,
$e_n(x)=\sqrt {2}\sin \pi (n+1)x$ for problem (1)-(2a),
$e_n(x)=\sqrt {2}\cos{\pi (2n+1)x \over 2}$ for problem (1)-(2b)
(with $n=0,1,2,\dots$), and
$e_0(x)\equiv 1$, $e_n(x)=\sqrt {2}\cos \pi nx$ (with $n=1,2,3,\dots $) for
the problem (1)-(2c).
The corresponding eigenvalues
are the numbers $\mu _n=(\pi (n+1))^2$ for problem (9a), the numbers
$\mu _n=(\pi (2n+1)/2)^2$ for problem (9b),
and the numbers $\mu _n=(\pi n)^2$ for problem (9c) ($n=0,1,2,\dots $).
For each of the problems (1)-(2) we set $v_n(x)=y_n(x)/ \|y_n\|$.
\begin{lemma} \label{Lemma1}
For each of the problems (1)-(2) and an arbitrary integer $n\geq 0$
the following expansion in the Fourier series, understood in the sense
of the space $L_2$, takes place
$$
v_n(x)=\sum_{k=0}^\infty d_{n,k}e_k,
$$
where coefficients $d_{n,k}$ are real,
$d_{n,0}=\dots =d_{n,n-1}=0$ and $d_{n,n}>0$.
\end{lemma}
\paragraph{Proof.} For arbitrary integers $n\geq 0$ consider the spaces
$L_2(0,I_n)$, where $I_n=1/(n+1)$ for problem (1)-(2a), and $I_n=1/(2n+1)$
for problem (1)-(2b) (i.\ e., $I_n$ is the smallest positive root of $v_n(x)$,
see Theorem 1(e)). It is clear that in each of these two cases functions from
the system $\{e_k\}$ equal to zero at the point $x=I_n$ ($k=0,1,\dots$)
form an orthogonal basis of the space $L_2(0,I_n)$.
We also note that for any integer $n\geq 0$ the minimal integer $k\geq 0$,
for which $e_k(I_n)=0$, is $k=n$. Hence,
$$
v_n(x)=\sum_{k=0}^\infty d_{n,k}e_k, \quad n=0,1,2,\dots
\eqno (10)
$$
in $L_2(0,I_n)$, where $d_{n,k}=0$ if $e_k(I_n)\not =0$,
and in view of the above arguments $d_{n,0}=\dots =d_{n,n-1}=0$. In addition,
$d_{n,n}>0$ because $v_n(x)>0$ and $e_n(x)>0$ for
$x\in (0,I_n)$. We also note that in the case of the problem (1)-(2b)
the functions $v_n(x)$ and
$e_k(x)$ are even with respect to the point $x=0$, therefore the expansions
(10) also hold in the space $L_2(-I_n,I_n)$. Further, one can
easily verify that if $e_k(I_n)=0$ for some value of the index $k$, then
this function $e_k(x)$, continued for all $x\in{\mathbb R}$,
is equal to zero at all points of the real line
which are roots of the function $v_n(x)$. Therefore, since clearly
any function $e_k(x)$ is odd with respect to an arbitrary its root, by
Theorem 1(f) we easily get that expansions (10) of the functions $v_n(x)$
($n=0,1,2,\dots$) take place in each space $L_2(I)$, too, where $I$ is
the interval between two arbitrary nearest roots of the function $v_n(x)$.
Hence, the expansions (10) are also valid in the space $L_2$.
For problem (1)-(2c) using the same procedure, one can obtain
$v_n(x)=\sum_{k=0}^\infty d_{n,k}e_k(x)$, $n=1,2,3,\dots $, in
the space $L_2$, where $d_{n,0}=\dots =d_{n,n-1}=0$ and $d_{n,n}>0$,
and, in addition, that $v_0(x)=pe_0(x)$. Thus, Lemma 1 is proved.
\hfill$\diamondsuit$
\paragraph{Remark 2.} An example of expansions of the form as in Lemma
1 is given in \cite{z6}. In that example the matrix
$D=(d_{n,k})$ is upper-triangular with positive elements on the main diagonal,
which shows that, generally speaking, the completeness
(in particular, the property of being a basis) of the system of functions
$\{v_n\}$ does not follow from the indicated properties of
the matrix $D$.
\begin{lemma} \label{Lemma2} There exists a positive constant $C$ such that
$$
\|v_n-e_n\|\leq C(n+1)^{-1}, \quad n=0,1,2,\dots
$$
for each of the problems (1)-(2).
\end{lemma}
\paragraph{Proof.} Let $t_n(x)=p e_n(x)/(\sqrt {2} \pi (n+1))$
for problem (1)-(2a), \\ $t_n(x)=p e_n(x)/\sqrt {2}$ for the problem
(1)-(2b) (with $n=0,1,2,\dots $), and $t_0(x)=pe_0(x)$,
$t_n(x)=p e_n(x)/\sqrt {2}$ (with $n=1,2,3,\dots $) for problem (1)-(2c).
We note that by the standard comparison theorem \cite{l1},
Theorem 1(d) immediately implies the existence of $C_1>0$ such that
$$
|\lambda _n-\mu _n|\leq C_1 \eqno (11)
$$
for $n=0,1,2,\dots $ for each of three problems (1)-(2). Therefore, by
Theorem 1(d), for each of three problems (1)-(2) we get
for $u_n(x)=y_n(x)-t_n(x)$
$$
-u''_n=W_n(x)+\mu _n u_n,\quad x\in (0,1)\,. \eqno (12)
$$
In addition, for problems (1)-(2a) and (1)-(2c) by Theorem 1(f),
$$
u_n(0)=u'_n(0)=u_n(1)=u'_n(1)=0\,\eqno (13)
$$
where $W_n(x)=(\lambda _n-\mu _n)y_n(x)-f(y_n^2(x))y_n(x)$
is a sequence of continuous functions uniformly bounded with respect to
$x\in [0,1]$.
For problem (1)-(2a) or (1)-(2c) for each number $n$ we multiply
the equality (12) by $2xu'_n(x)$ and integrate over the segment $[0,1]$
with the application of the integration by parts.
Then, in view of (13) and the obvious inequality $2ab\leq a^2+b^2$
we get
$$
\mu _n\|u_n\|^2=2\int_0^1xW_n(x)u'_n(x)dx-\int_0^1[u'_n(x)]
^2 dx \leq \int_0^1W_n^2(x) dx. \eqno (14)
$$
Hence, there exists $C_2>0$ such that
$$
\mu _n\|u_n\|^2\leq C_2 \eqno (15)
$$
for all integer $n\geq 0$. Since in problem (1)-(2c) we have
$\|t_0\|=p, \quad \|t_n\|={p\over \sqrt {2}}$, where $n=1,2,3,\dots $, it
follows from (15)
that $\left| \quad \|y_n\|-{p\over \sqrt {2}}\right|\leq C'_2(n+1)^{-1}$
for some $C'_2>0$ independent of $n$. Therefore, taking into account
(15), we get the statement of Lemma 2 for problem (1)-(2c).
For problem (1)-(2a) we have
$$
\|t_n\|={p\over \sqrt {2} \pi (n+1)} \quad (n=0,1,2,\dots ); \eqno (16)
$$
therefore, (15) implies the existence of $C_3>0$ such that
$$
\|y_n\|\leq C_3(n+1)^{-1} \eqno (17)
$$
for all integer $n\geq 0$. In view of Theorem 1(d),
(17) yields the existence of $C_4>0$
such that $\|W_n\|\leq C_4(n+1)^{-1}$ for all $n=0,1,2,\dots $.
So, in problem (1)-(2a) we get from the latter estimate and (14)
that
$$
\|u_n\|\leq C_5(n+1)^{-2}
$$
with a constant $C_5>0$ independent of the number $n=0,1,2,\dots $.
The statement of Lemma 2 for problem (1)-(2a) follows from
this estimate together with (16) as for problem (1)-(2c).
For problem (1)-(2b) the proof of the statement in Lemma 2 can be done
as for problem (1)-(2c). One should only take into
account the estimate for the eigenfunctions of the problem (1)-(2b)
$$
\left| y'_n(1)-t'_n(1) \right| \leq C_6
$$
with a constant $C_6>0$ independent of $n=0,1,2,\dots $ (this estimate
follows from (11), the properties of the function $t_n(x)$, Theorem 1(e), and
the identities
$$
-[y'_n(x)]^2+F(y^2_n(x))-F(p^2)=\lambda _ny_n^2(x)-\lambda _np^2 \quad
(n=0,1,2,\dots )\,.
$$
These identities can be obtained by multiplying (1),
with $y(x)=y_n(x)$, by $2y'_n(x)$, and then integrating from $0$
to $x$). Thus, Lemma 2 is proved.\hfill$\diamondsuit$
\begin{lemma} \label{Lemma3} For each of the problems (1)-(2) the system of
functions $\{v_n\}$ is linearly independent in the space $L_2$.
\end{lemma}
\paragraph{Proof.} On the contrary suppose that
$$
\sum_{n=0}^\infty c_nv_n=0 \eqno (18)
$$
in the space $L_2$ where $c_n$ are real coefficients, $c_0=\dots =c_{l-1}
=0$ and $c_l\not =0$ for a number $l$. Multiply (18)
in the space $L_2$ by the function $e_l(x)$. Then, in view of Lemma 1, we get
$d_{l,l}c_l=0$. But according to Lemma 1 $d_{l,l}>0$, which is a contradiction
that proves the present lemma. \hfill$\diamondsuit$
\paragraph {Remark 3.} In view of Lemmas 2 and 3, it follows from Bary's
theorem that for each of the problems (1)-(2) the system of functions
$\{v_n\}$ is a Riesz basis in the space $L_2$. Moreover,
since $\{e_n\}$ is an orthonormal basis in $L_2$,
the functions $\{v_n\}$ form a Bary basis in $L_2$ for each case.
As stated in the introduction, we shall establish a direct proof
of this fact without using Bary's theorem.
In what follows, unless otherwise stated, all three problems (1)-(2)
are considered simultaneously.
For an arbitrary number $N\geq 0$, denote by $L^N$ the closure in $L_2$
of the set of all finite linear combinations of the functions
$\{e_n\}_{n\geq N}$.
Denote by $A$ the linear operator defined
by the rule $Ae_n={(d_{n,n})}^{-1}v_n$. Then $A$ is defined for
the linear combinations of $\{e_n\}$. In view of Lemma 1, it is clear that
$A$ maps the subspace $L^N$ into itself, and that $A$ is defined on dense
subsets of $L_2$ and of the subspaces $L^N$. Let $A_N$ denote
the restriction of the operator $A$ to the subspace $L^N$
($N=0,1,2,\dots $). Then, in view of Lemma 1
$A_N=\Lambda _N+G_N$ where $\Lambda _N$ is the identity
operator in $L^N$ and $G_N$ is an operator defined on all finite
linear combinations of $\{e_n\}_{n\geq N}$. In addition,
$G_Ne_n={(d_{n,n})}^{-1}\sum_{k>n}d_{n,k}e_k$
so that $A_Ne_n={(d_{n,n})}^{-1}v_n$ for $n=N,N+1,N+2,\dots $.
In view of Lemma 2, the inequality $d_{n,n}>0$, and the condition
$\|v_n\|=1$ for $n=0,1,2,\dots $, there exists $\overline d\in (0,1)$
such that for $n\geq 0$,
$$
\overline d\leq d_{n,n}\leq 1\,. \eqno (19)
$$
Let $\|G_N\|=\sup\{\|G_Ng\| :g=\sum_{n=N}^M c_ne_n\mbox {and }\|g\|=1\}$,
where supremum is taken over all numbers $M=N,N+1,N+2,\dots $ and
all real coefficients $c_n$.
\begin{lemma} \label{Lemma4} For each of three problems (1)-(2),
$\|G_N\|<\infty$ for all numbers $N\geq 0$ and
$\lim_{N\to \infty}\|G_N\|=0$.
\end{lemma}
\paragraph{Proof.} Let $g=\sum_{n=N}^\infty c_ne_n$ where
only finitely many coefficients $c_n$ are non-zero,
and $\|g\|=1$. It follows from Lemma 2 and (19) that
$\sum_{n,k=0 \atop n
\not = k}^\infty {(d_{n,n})}^{-2}d_{n,k}^2 <\infty$. Therefore,
\begin{eqnarray*}
\|G_Ng\|^2&=&\sum_{k=N+1}^\infty \big[\sum_{n=N}^{k-1}
c_n{(d_{n,n})}^{-1}d_{n,k}\big]^2 \\
&\leq& \sum_{k=N+1}^\infty \big[\sum_{n=N}^{k-1}c_n^2\big]
\times \big[\sum_{n=N}^{k-1}{(d_{n,n})}^{-2}d^2_{n,k}\big]\\
&\leq& \|g\|^2 \sum_{n,k=N \atop n\not =k}^\infty {(d_{n,n})}^{-2}
d^2_{n,k} \to 0 \quad \mbox{as } N\to \infty,
\end{eqnarray*}
and the proof of Lemma 4 is complete.
\hfill$\diamondsuit$\smallskip
Let $N$ be a positive integer such that $\|G_N\|<1/2$. As shown in
\cite{l2}, the linear operator $G_N$ can be uniquely extended to the whole
space $L^N$. For this extension we keep the same symbol $G_N$, and we
have $\|G_N\| < 1/2$. In view of the latter fact,
the linear operator $A_N=\Lambda _N+G_N$, with
$G_N\in {\cal L}(L^N;L^N)$) belongs to ${\cal L}(L^N;L^N)$
and possesses a bounded inverse:
$$
A_N^{-1}=\Lambda _N+\sum_{m=1}^\infty (-1)^m(G_N)^m\,.
$$
The proof of this statement can be found in \cite{l2}.
The following statement is in fact proved in \cite{b2,g1}. However, since its
proof is simple and short, we present it here.
\begin{lemma} \label{Lemma5} Let $N$ be the number fixed above. Then
for each of the problems (1)-(2), the functions $\{v_n\}_{n\geq N}$ form
a Riesz basis in $L^N$.
\end{lemma}
\paragraph{Proof.} Obviously, the functions $\{e_n\}_{n\geq N}$
form an orthonormal basis in $L^N$. Take an arbitrary function
$u\in L^N$ and let $v=A^{-1}_Nu=\sum_{n=N}^\infty c_ne_n$ in the space
$L^N$ where $c_n$ are real coefficients.
We have $u=A_Nv=\sum_{n=N}^\infty c_nA_Ne_n=\sum
_{n= N}^\infty c_n{(d_{n,n})}^{-1}v_n$ where all series converge in $L^N$.
Therefore, in view of Lemma 3,
the system of functions $\{v_n\}_{n\geq N}$ is a basis in the space $L^N$.
Further, with the same notation, $\sum_{n=N}^\infty c_n^2<\infty$
because the series $\sum_{n= N}^\infty c_ne_n$ converges in the space
$L^N$.
Hence, if $\sum_{n=N}^\infty c_n^2<\infty$, then the series
$\sum _{n=N}^\infty c_n{(d_{n,n})}^{-1}v_n$ converges in the space
$L^N$. Conversely, let the series $\sum
_{n=N}^\infty c_nv_n$ converge in $L^N$. Then, $A_N^{-1}
\sum_{n=N}^\infty c_nv_n=\sum_{n=N}^\infty c_nA_N^{-1}v_n=
\sum_{n=N}^\infty c_nd_{n,n}e_n$ in the space $L^N$, therefore, $\sum
_{n=N}^\infty c_n^2d_{n,n}^2<\infty$. So, in view of the estimates
(19), Lemma 5 is proved.\hfill$\diamondsuit$
\begin{lemma} \label{Lemma6} For each of the three problems (1)-(2)
the functions $\{v_n\}_{n=0,1,2,\dots }$ form a Riesz basis in $L_2$.
\end{lemma}
\paragraph{Proof.} Let $N$ be the number fixed above, and
$L^N_\bot$ be the finite-dimensional subspace in
$L_2$ spanned by $e_0,\dots ,e_{N-1}$, and $P_N$ be
the orthogonal projector in the space $L_2$ onto the subspace $L^N_\bot$.
We set $w_n=P_Nv_n$, with $0\leq n \leq N-1$. Then
$$
w_n=v_n-\sum_{k=N}^\infty d_{n,k}e_k\,.
\eqno (20)
$$
Further, the dimension of the subspace $L^N_\bot$ is obviously equal to $N$
and, in view of Lemma 1, the system of functions $\{w_n\}_{0 \leq n \leq N-1}$
is linearly independent in $L_2$, therefore, it is a basis in $L^N_\bot$.
Hence, in view of Lemma 5, the system of functions
$\{w_n\}_{0 \leq n \leq N-1} \cup \{v_n\}_{n\geq N}$ is a basis in $L_2$.
For an arbitrary $u\in L_2$,
$$
u=\sum_{n=0}^{N-1}c_nw_n+\sum_{n=N}^\infty c_nv_n
\eqno (21)
$$
in $L_2$ where $c_n$ are real coefficients.
Substituting the expressions for functions $w_n$ from (20) into (21), we get
$$
u=\sum_{n=0}^{N-1}c_nv_n+\sum_{n=N}^\infty \left(c_n-\sum
_{m=0}^{N-1}c_md_{m,n}\right)e_n
$$
in $L_2$, hence the functions $\{v_n\}
_{n=0,1,2,\dots }$ form a basis for $L_2$. Finally, that these
functions form a Riesz basis in $L_2$ follows from Lemma 5, and concludes
the present proof. \hfill$\diamondsuit$\smallskip
For problem (1)-(2c), the estimate (15)
and the arguments following the estimate imply the existence of constants
$0from Lemma 6 and these estimates.
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\noindent{\sc Peter E. Zhidkov} \\
Bogoliubov Laboratory of Theoretical Physics \\
Joint Institute for Nuclear Research \\
141980 Dubna (Moscow region), Russia \\
email: zhidkov@thsun1.jinr.ru
\end{document}
**