\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 107, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/107\hfil Zeros of solutions to second order ODE]
{Sums of zeros of solutions to second order ODE
with non-polynomial coefficients}

\author[M. I. Gil' \hfil EJDE-2012/107\hfilneg]
{Michael I. Gil'} 

\address{Michael I. Gil' \newline
Department of Mathematics \\
Ben Gurion University of the Negev \\
P.0. Box 653, Beer-Sheva 84105, Israel}
\email{gilmi@bezeqint.net}

\thanks{Submitted September 8, 2011.  Published June 25, 2012.}
\subjclass[2000]{34C10, 34A30}
\keywords{Complex differential equation; zeros of solutions}

\begin{abstract}
 We consider the equation $y''=F(z)y$ ($z\in\mathbb{C}$)
 with an entire function $F$  satisfying the condition
 $$
 |F(z)|\le A \exp \big(\frac{|z|^\rho}{\rho}\big)\quad (\rho\ge 1,\; A=\text{const}>0).
 $$
 Let  $z_k(y)$, $k=1, 2, \dots $  be the  zeros of a solution $y(z)$
 to the above equation. Bounds for the sums
 $$
 \sum_{k=1}^j  \frac{1}{|z_k(y)|} \quad (j=1, 2, \dots)
 $$
 are established. Some applications of these bounds are also considered.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks

\section{Introduction and statement of the main result}


In the present article, we consider linear
differential equations with non-polynomial coefficients in the complex domain.
The literature devoted to the zeros of solutions of such equations is very rich.
Here the main tool  is the Nevanlinna theory.
An excellent exposition of
the Nevanlinna theory  and its applications to differential equations 
is given in the book \cite{lai}.
In that book, in particular,  the well-known
results of Bank, Br\H{u}ggemann,  Hellerstein,  Rossi,
Huang  and other mathematicians are featured.
In connection with recent  results see the
very interesting papers \cite{bela}-\cite{chen}, \cite{far}, \cite{li}-\cite{xu}.
In particular, in the paper \cite{tu}, the authors study the  
convergence  of the zeros of a   non-trivial (entire) solution  to the linear
 differential equation
$$
f''+\bigl\{Q_1(z)e^{P_1(z)}+
  Q_2(z)e^{P_2(z)}+Q_3(z)e^{P_3(z)}\big \}f=0
$$
 where  $P_j$ are
    polynomials of degree $n\geq 1$
     and  $Q_j$ ($ Q_j\not\equiv 0)$ are entire functions of
    order less than $n$ $(j=1,2,3)$.
The  remarkable results on the zeros of
a wide class of ordinary differential equations with polynomial coefficients
whose solutions are classical orthogonal polynomials was established by
 Anghel \cite{ang}. Besides, he had derived the  important results
connected with the  equations of
mathematical physics.


Certainly we could not survey the whole subject here and we refer
the reader to the above listed publications and references given therein.

In the  above cited works  mainly the asymptotic distributions of zeros
 and  counting functions of zeros are investigated.
At the same time, bounds for    the zeros of solutions are very important
in various applications.
But to the best of our knowledge, they  have been investigated considerably less
than the asymptotic distributions.
In the paper \cite{gi11},
bounds for the sums of the zeros of solutions are established
for the second order equations with polynomial coefficients.
In this paper, we obtain such bounds in
the case of non-polynomial coefficients.
Besides, below we estimate  the zero free domains.
That estimation supplements the well-known results of
Eloe   and Henderson \cite{eloe} on
the positivity of solutions for higher order ordinary differential equations, 
since the positivity of solutions implies the absence of zeros.
Note that, the proof of the main result of the present paper is
considerably different from   the proof
of the paper \cite{gi11}.

Consider the equation
\begin{equation}
\frac{d^2y(z)}{dz^2}=F(z)y(z) \label{e1.1}
\end{equation}
with an entire function $F$, satisfying the condition
\begin{equation}
|F(z)|\le A\exp  \big(\frac{|z|^\rho}{\rho}\big)\quad (\rho\ge 1;\;
A=\text{const}>0;\;z\in\mathbb{C}).
\label{e1.2}
\end{equation}
In Section 3 below we check  that to inequality \eqref{e1.2} can be reduced
the formally more general inequality
\begin{equation}
|F(z)|\le A \exp [B|z|^\rho]\quad (B=\text{const}>0) \label{e1.3}
\end{equation}
by the substitution
\begin{equation}
z=\frac{w}{(\rho B)^{1/\rho}} \label{e1.4}
\end{equation}
into \eqref{e1.1}.  Everywhere below, $y(z)$ is a solution of \eqref{e1.1} with
$y(0)=1$. Enumerate the zeros $z_k(y)$ of $y(z)$,
with multiplicities taken into account,  in  order of
increasing  modulus:
$|z_{k}(y)|\leq |z_{k+1}(y)|$  $(k=1, 2, \dots )$.
Put
$$
v_n= n\; \frac{e^{1+2/\rho}}{\rho}.
$$

\begin{theorem} \label{thm1.1}
Let $y(z)$ be a solution of \eqref{e1.1} with
$y(0)=1$ and let condition \eqref{e1.2} hold. Then
$$
\sum_{k=1}^j  \frac{1}{|z_k(y)|}<
\theta_0 + \zeta_0\sum_{k=1}^j  \frac{1}{\sqrt[\rho]{\ln  v_k}}\quad (j=1, 2, \dots),
$$
where $\theta_0$ and $\zeta_0$ are positive constants
defined by
$$
\theta_0=2\sqrt{\frac e3}(1+|y'(0)|)\exp [ A^2e^{2/\rho}] \quad
\text{and}\quad \zeta_0=2e^{1/2}(3/\rho)^{1/\rho}.
$$
\end{theorem}

The proof of this theorem is presented in the next  section.
Let us point some corollaries of Theorem \ref{thm1.1}.

Denote by $\nu(y, r)$ $(r>0)$ the counting function of the zeros of $y$ in
$|z|\le r$. Theorem \ref{thm1.1} implies

\begin{corollary} \label{cor1.2}
With the notation
$$
\eta_j(y):=
\frac j {\theta_0 + \zeta_0\sum_{k=1}^j  \frac{1}{\sqrt[\rho]{\ln  v_k}}}\quad
(j=1, 2, \dots),
$$
the inequality  $|z_j(y)|>\eta_j(y)$ holds and thus $\nu(u, r)\leq j-1$  
 for any $r \leq \eta_j(y)$ $(j\ge 2)$.
\end{corollary}

Furthermore, put
$$
\vartheta_1=\theta_0+\frac{\zeta_0}{\ln ^{1/\rho} v_1},\quad
\vartheta_k=\frac{\zeta_0}{\ln ^{1/\rho} v_k}
\quad ( k=2, 3, \dots).
$$
Theorem \ref{thm1.1} and
\cite[Lemma 1.2.1]{gi09} yield the following result.

\begin{corollary} \label{cor1.3} 
Under the hypothesis of Theorem \ref{thm1.1},
  let  $\phi(t)$ $(0\leq t<\infty)$ be a continuous convex
 scalar-valued function, such that $\phi(0)=0$. Then
$$
\sum_{k=1}^j  \phi (|z_k(y)|^{-1})\leq  \sum_{k=1}^j \phi(\vartheta_k)\quad (j=1, 2, \dots).
$$
\end{corollary}

In particular, take
$$
\phi(t)=t^{\rho+1}\exp [-\frac{\zeta_0^{\rho}}{t^{\rho}}].
$$
Then
$$
\phi(\vartheta_k)=\frac{\zeta_0^{\rho+1}}{\ln ^{1+1/\rho} v_k}\exp [-\ln  v_k]\le
\frac{\text{const}}{(\ln  k)^{1+1/\rho}\;k}\quad (k>1).
$$
By the previous corollary we get the following result.

\begin{corollary} \label{cor1.4}
Under the hypothesis of Theorem \ref{thm1.1},
we have
$$
\sum_{k=1}^\infty
\frac 1{|z_k(y)|^{\rho+1}} e^{-(\zeta_0|z_k(y)|)^{\rho}}<\infty.
$$
\end{corollary}

In addition, in light of  Theorem \ref{thm1.1} and \cite[Lemma 1.2.2]{gi09}
we obtain our next result.

\begin{corollary} \label{cor1.5}
Let
 $\Phi(t_1, t_2, \dots, t_j)$
be a function with an integer $j$  defined on the domain
$$
0< t_j\leq t_{j-1}\dots \leq t_2\leq t_1<\infty
$$
and satisfying the condition
$$
\frac{\partial \Phi}{\partial t_1}>\frac{\partial \Phi}{\partial t_2}> \dots
>\frac{\partial \Phi}{\partial t_j}>0\quad \text{for } t_1> t_2 > \dots > t_j>0.
$$
Then
$$
\Phi \Big(\frac{1}{|z_1(y)|},  \dots, \frac{1}{|z_j(y)|} \Big)
\leq \Phi(\vartheta_1,  \dots, \vartheta_j).
$$
\end{corollary}

In particular, let $\{d_k\}_{k=1}^\infty$ be a decreasing sequence of
positive numbers.
Then the previous corollary  yields the inequality
$$
\sum_{k=1}^j   \frac{d_k}{|z_k(y)|}\leq
\sum_{k=1}^j    d_k\vartheta_k  \quad (j= 1, 2,   \dots ).
$$

\section{Proof of Theorem \ref{thm1.1}}

Consider the entire function
\begin{equation}
f(z)=\sum_{k-0}^\infty c_kz^k\quad (c_0=1) \label{e2.1}
\end{equation}
satisfying the condition
\begin{equation}
|f(z)|\leq  (1+qr)\exp[ A \exp (r^\rho/\rho)]\quad (q=\text{const}>0;
z\in\mathbb{C}, \;r=|z|).
\label{e2.2}
\end{equation}
Put
$$
C=(1+q) \exp[\frac 12 e^{2/\rho} A^2].
$$

\begin{lemma} \label{lem2.1}
Let condition \eqref{e2.2} hold. Then
the Taylor coefficients of $f$ are subjected to the inequality
$$
|c_n|\leq C e^{n/2}\Big(\frac{3}{\rho\ln  v_n }\Big)^{(n-1)/\rho}.
$$
\end{lemma}

\begin{proof}
Let
$$
M_f(r)=\max_{|z|=r}|f(z)|.
$$
By the well-known inequality for the coefficients of a power series,
$$
|c_n|\leq \frac{M_f(r)}{r^n}\quad (r>0).
$$
Take into account that
$$
ab\le \frac{a^2}{4c}+b^2c\quad (a,b,c \text{positive constants}).
$$
Then for a constant $\mu>0$,
$$
A \exp (r^\rho/\rho)\le  \frac{A^2}{4\mu}+
\mu\exp (2r^\rho/\rho).
$$
Due to \eqref{e2.2},
\begin{equation}
|c_n|\leq (1+qr)\exp[\frac{A^2}{4\mu}]h(r)\text{ where }h(r)
:=\frac{\exp[\mu  e^{2r^\rho/\rho}]}{r^n}\quad (n=1, 2, \dots).
\label{e2.3}
\end{equation}
Let us  use the usual method for finding extrema. Clearly,
$$
r^{2n}h'(r)=e^{\mu e^{2r^\rho/\rho}}
[2\mu  e^{2r^\rho/\rho}r^{n+\rho-1}- nr^{n-1}].
$$
Thus the zero
$r_0=r_0(n)$ of $h'(r)$ is defined by
\begin{equation}
2\mu   r_0^{\rho} e^{2r_0^\rho/\rho}=n. \label{e2.4}
\end{equation}
Take
$$
\mu  =\frac 12e^{-2/\rho}.
$$
Then
\begin{equation}
r_0^{\rho} e^{2(r_0^\rho-1)/\rho}=n. \label{e2.5}
\end{equation}
So for $n\ge 1$ we have $r_0\ge 1$. Hence by \eqref{e2.4},
\begin{equation}
\mu   e^{2r_0^\rho/\rho}\le n/2. \label{e2.6}
\end{equation}
Since $x\le e^{x-1}\;\;(x>0)$, by
\eqref{e2.5}
we have $e^{3r_0^\rho/\rho}\ge ne^{1+2/\rho}/\rho=v_n$, and therefore,
\begin{equation}
r_0\ge (\frac{\rho}3  \ln  v_n)^{1/\rho}. \label{e2.7}
\end{equation}
Since $r_0\ge 1$, we obtain $\frac{1+qr_0}{r_0}\le 1+q$.
Now \eqref{e2.3} and   \eqref{e2.6}  imply
$$
|c_n|\leq \exp[\frac{A^2}{4\mu  }](1+q) \frac{e^{n/2}}{r_0^{n-1}}.
$$
Hence, \eqref{e2.7} proves the lemma.
\end{proof}

Put $D=\sqrt e C$, and
$$
\tau_n=2e^{1/2}\left(\frac{3}{\rho\ln  v_n }\right )^{1/\rho}.
$$
Then according to Lemma \ref{lem2.1},
$$
|c_n|\leq D\frac{\tau_n^{n-1}}{2^{n-1}}.
$$
Denote
$\psi_1=1, \psi_n=\tau_n^{n-1}\;\;(n>1)$;
$a_n=c_n/\psi_n$, and $m_{n+1}=\psi_{n+1}/\psi_{n}$.
As it is proved in  \cite[Theorem 5.1.1]{gi09}, the inequality
$$
\sum_{k=1}^j  \frac 1{|z_k(f)|}\le \theta(f)+ \sum_{k=1}^j  m_{k+1}
$$
is valid, where $z_k(f)$ are  the zeros  of $f(z)$,
with multiplicities taken into account,
enumerated  in  order of increasing  modulus, and
$$
\theta(f):=\Big[\sum_{k=1}^\infty |a_k|^2\Big]^{1/2}.
$$
But  $|a_1|=|c_1|\le D$; $|a_n|\le D/2^{n-1}, n\ge 2$.
Moreover, since $\tau_{n+1}\le \tau_n$, we get
$m_{n+1}\le \tau_n$ $(n=1, 2, \dots)$, and
$\theta(f)\le \tilde\theta_0$,  where
$$
\tilde\theta^2_0=D^2\sum_{k=0}^\infty \frac 1{4^k}=4D^2/3.
$$
We thus have proved the following result.

\begin{lemma} \label{lem2.2}
Let an entire function $f$ satisfy condition \eqref{e2.2}.
Then
$$
\sum_{k=1}^j  \frac 1{|z_k(f)|}\le \tilde\theta_0+ \sum_{k=1}^j  \tau_{n}=
\tilde\theta_0 + \zeta_0\sum_{k=1}^j  \frac{1}{\ln ^{1/\rho} v_k}\quad (j=1, 2, \dots).
$$
\end{lemma}

\begin{lemma} \label{lem2.3}
A solution $y$ of \eqref{e1.1} with the conditions \eqref{e1.2}
and $y(0)=1$ is an entire function
satisfying the inequality
$$
|y(z)|\le (1+|y'(0)|r) \exp[A e^{r^\rho/\rho}]\quad (z\in \mathbb{C}).
$$
\end{lemma}

\begin{proof}
From \eqref{e1.1} for a  $z=re^{it}$ with a fixed argument $t$  we
 have $e^{-2it} d^2y(z)/dr^2=F(z)y(z)$.
Hence, putting $q=|y'(0)|$,
$g(r)=|y(re^{it})|$, and taking into account \eqref{e1.2}
we obtain
$$
g(r)\le 1+qr+ \int_0^r (r-s)|F(se^{it})|g(s)ds
\le 1+qr+A\int_0^r (r-s)\exp[s^\rho/\rho] g(s)ds.
$$
By \cite[Lemma III.2.1]{dal} we have $g(r)\le m(r)$, where $m(r)$ 
is a solution of the equation
$$
m(r)=1+qr+   A\int_0^r (r-s)\exp[s^\rho/\rho] m(s)ds.
$$
However,
$$
\int_0^r (r-s)f(s)ds=\int_0^r \int_0^s f(\tau)d\tau\; ds
$$
for any integrable function $f$. Thus,
$$
m(r)= 1+qr+ A\int_0^r \int_0^s \exp[\tau^\rho/\rho]m(\tau)d\tau\; ds .
$$
Clearly the derivative of $m$ is positive. So
$$
m(r)\le 1+ qr+ A\int_0^r m(\tau) \int_0^\tau \exp[s^\rho/\rho]ds \;d\tau.
$$
But for $r\le 1$,
$$
\int_0^r  \exp[s^\rho/\rho]ds \le \exp[r^\rho/\rho],
$$
and for an $r\ge 1$,
$$
\int_0^r  e^{s^\rho/\rho} ds\le \int_0^1
e^{s^\rho/\rho} ds
+ \int_1^r s^{\rho-1} e^{s^\rho/\rho} ds \le  e^{1/\rho}+ (e^{r^\rho/\rho}-e^{1/\rho})= e^{r^\rho/\rho}.
$$
Thus,
\begin{equation}
\int_0^r  \exp[s^\rho/\rho]ds \le \exp[r^\rho/\rho]. \label{e2.8}
\end{equation}
Consequently, $m(r)\le 1+qr+A\int_0^r m(s)\exp[s^\rho/\rho]ds$.
By the Gronwall inequality,
$$
m(r)\le (1+qr)\exp\big[A \int_0^r   e^{s^\rho/\rho}ds\big].
$$
Now \eqref{e2.8} implies the required result.
\end{proof}

 Then the assertion of Theorem \ref{thm1.1}  follows from
Lemmas \ref{lem2.2} and \ref{lem2.3}.

\section{Example}

In this section we consider an example that illustrates Theorem \ref{thm1.1}.
First  substitute \eqref{e1.4} into \eqref{e1.1}. Then we arrive at the equation
$$
\frac{d^2x(w)}{dw^2}=F_1(w),\quad \text{where }F_1(w)=\frac{1}{(\rho B)^{2/\rho}}
F\big(\frac{w}{(\rho B)^{1/\rho}}\big)
$$
 and $x(w)=y(w/(\rho B)^{1/\rho})$. If condition \eqref{e1.3} holds, then
$$
|F_1(w)|\le A_1\exp \big(\frac{|w|^\rho}{\rho}\big),
$$
where
$A_1=A/(\rho B)^{2/\rho}$.
By Theorem \ref{thm1.1},
\begin{equation}
\sum_{k=1}^j  \frac{1}{|z_k(x)|}<
\theta_1 + \zeta_0\sum_{k=1}^j  \frac{1}{\ln ^{1/\rho} v_k}\quad (j=1, 2, \dots),
\label{e3.1}
\end{equation}
where
$$
\theta_1=2\sqrt{\frac e3}\Big(1+|\frac{dx(0)}{dw}|\Big)
\exp[ A_1^2e^{2/\rho}]
= 2\sqrt{\frac e3}\Big(1+\frac{1}{(B\rho)^{1/\rho}}
 |\frac{dy(0)}{dz}|\Big)\exp[ A_1^2e^{2/\rho}].
$$
But  $z_k(y)=z_k(x)(\rho B)^{1/\rho}$.
Now \eqref{e3.1} implies
\begin{equation}
\sum_{k=1}^j  \frac{1}{|z_k(y)|}<(\rho B)^{1/\rho}\Big[
\theta_1 + \zeta_0\sum_{k=1}^j  \frac{1}{\ln ^{1/\rho} v_k}\Big]\quad (j=1, 2, \dots).
\label{e3.2}
\end{equation}
So the following result is holds.

\begin{corollary} \label{cor3.1}
Let $y(z)$ be a solution of \eqref{e1.1} with
$y(0)=1$ and condition \eqref{e1.3} hold. Then inequality \eqref{e3.2} is valid.
\end{corollary}

Furthermore,  it can be directly checked that the function
\begin{equation}
y(z)=c e^{-z/2}\sin(e^z) \label{e3.3}
\end{equation}
with
$c=1/\sin(1)$ is a solution of the equation
\begin{equation}
y''(z)=(\frac 14-e^{2z})y(z) \label{e3.4}
\end{equation}
Besides, $y(0)=1$.
Clearly, the zeros of $y$ are $\ln  \pi k$ $(k=0, \pm 1, \pm 2,\dots)$.
Hence, for a sufficiently large $j$ we have
\begin{equation}
\sum_{k=1}^{2j} \frac 1{|z_k(y)|}=\sum_{k=1}^j
\big[\frac 1{|z_{2k-1}(y)|}+ \frac 1{|z_{2k}(y)|}\big]
= \sum_{k=1}^j  \big[\frac 1{\ln  \pi k} + \frac 1{|\ln (-\pi k)|}\big].
\label{e3.5}
\end{equation}
On the other hand, due to \eqref{e3.4},$F(z)=\frac 14-e^{2z}$ and therefore,
 $|F(z)|\le (1+\frac 14)e^{2|z|}$.
By \eqref{e3.2}  with $B=2,   A=1+1/4$, we have
$$
\sum_{k=1}^j  \frac{1}{|z_k(y)|}< 2
\theta_1 + 2\zeta_0\sum_{k=1}^j  \frac{1}{\ln (k e^3 )}\quad(j=1, 2, \dots).
$$
This result is rather close to \eqref{e3.5}.

Note that, if $F(z)$ is of infinite order, then
the problem  considered in this paper is much more complicated.

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\end{document}