\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 25, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2005/25\hfil On positive solutions]
{Positive solutions to a semilinear higher-order ODE on the half-line}
\author[M. I. Gil'\hfil EJDE-2005/25\hfilneg]
{Michael I. Gil'}

\address{Department of Mathematics \\
 Ben Gurion University of the Negev \\
P.0. Box 653, Beer-Sheva 84105, Israel}
\email{gilmi@cs.bgu.ac.il}

\date{}
\thanks{Submitted September 28, 2004. Published March 6, 2005.}
\thanks{Supported by the Kamea fund of the Israel}
\subjclass[2000]{34C10, 34C11}
\keywords{Ordinary differential equations; nonlinear equations; \hfill\break\indent
 positive solutions on the half-line}

\begin{abstract}
 We study a semilinear non-autonomous ordinary differential
 equation (ODE) of order $n$.  Explicit conditions for the
 existence of $n$ linearly independent and positive solutions
 on the positive half-line  are obtained. Also we establish
 lower solution estimates.
\end{abstract}

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


\section{Introduction and statement of main result}

The problem of existence of
positive of solutions to  a higher order nonlinear nonautonomous
ordinary differential equations (ODEs)
 continues to attract the attention of many specialists,
despite its long history, cf. \cite{ag,ag1,el,kr,krl,sw}
and references therein. It is still one of the most burning problems of
 theory of ODEs, because of the absence of its complete solution.
Let $p_k(t)$ ($t\geq 0$ $k=1,\dots  n$) be real
 continuous scalar-valued functions defined and bounded on $[0,\infty)$ and
$p_0\equiv 1$. Let
$F: [0,\infty)\times \mathbb{R}\to \mathbb{R}$ be a continuous function.
In the present paper we investigate  the semilinear equation
\begin{equation}
\sum_{k=0}^n p_k(t) \frac{d^{n-k}x(t)}{dt^{n-k}}=F(t, x)\quad (t>0,\;x=x(t))
\label{e1.1}
\end{equation}
with the initial conditions
\begin{equation}
x^{(j)}(0)=x_j\in \mathbb{R} \quad (j=0, \dots,  n-1).
\label{e1.2}
\end{equation}
A solution of problem \eqref{e1.1}, \eqref{e1.2} is a function $x(\cdot)$ defined on
$[0, \infty)$, having continuous derivatives
up to the $n$-th order.
In addition,  $x(\cdot)$ satisfies \eqref{e1.2} and  \eqref{e1.1} for all  $t> 0$.
The existence of solutions is assumed.

As it is well-known, the existence of positive solutions on the half-line for
such equations is proved mainly in the case when $p_k$ are constants,
cf. \cite{kr,krl,gi83}. In \cite{gi04}
 the positivity conditions were derived for
a class of semilinear nonautonomous equations in the divergent form.
In \cite{le},  the following remarkable result is established:
Solutions to the equation
\begin{equation}
\sum_{k=0}^n p_k(t) \frac{d^{n-k}v(t)}{dt^{n-k}}=0\quad (t>0)
\label{e1.3}
\end{equation}
do not oscillate, if the  roots
of the polynomial
$$
P(t,z)=\sum_{k=0}^n p_{n-k}(t)z^k\quad (z\in\mathbb{C}, t\geq 0)
$$
are real and  not intersecting. In the present
paper, under some ``close" conditions we prove that the nonlinear
equation \eqref{e1.1} has $n$ linearly independent positive solutions.
 Besides, we generalize the corresponding result from \cite{gi04}.

Let  polynomial $P(z,t)$ have the purely real roots $\rho_k(t)$ ($k=1,\dots, n$)
with the property
\begin{equation}
\rho_k(t)\geq -\mu \;\;(t\geq 0;\; k=1,\dots, n)
\label{e1.4}
\end{equation}
for some $\mu>0$. Put
\begin{equation}
q_m(t)=\sum_{k=0}^m p_k(t) C_{n-k}^{m-k} (-1)^k\mu^{m-k}\quad
(m=1, \dots, n ),\; q_0\equiv 1.
\label{e1.5}
\end{equation}
and
$$
d_0=1, d_{2k}=\sup_k q_{2k}(t)\quad\mbox{and}\quad
d_{2k-1}=\inf_k q_{2k-1}(t)\;(k= 1, \dots, [n/2]),
$$
where $[x]$ is the integer part of $ x > 0 $ and $C_n^k=\frac{n!}{k! (n-k)!}$.

Now we are in a position to formulate the main result of the paper.

\begin{theorem}\label{thm1.1}
 Let all the roots of the polynomial
\begin{equation}
\tilde Q(z):=\sum_{k=0}^n (-1)^k d_k z^{n-k} \label{e1.6}
\end{equation}
be real and nonnegative. In addition, let
\begin{equation}
F(y, t)\geq 0\quad (y, t\geq 0). \label{e1.7}
\end{equation}
Then  \eqref{e1.1} has on $(0,\infty)$ $n$ linearly independent positive
  solutions $x_1, \dots, x_n$, satisfying the inequalities
$$
x_j(t)\geq  \mathrm{const}\; e^{(-\mu+\tilde r_1) t}\geq 0\quad
(j=1, \dots, n; \;\; t\geq t_0>0),
\label{e1.8}
$$
where $\tilde r_1\geq 0$ is the smallest root of $\tilde Q(z)$.
\end{theorem}

This theorem is proved in the next two sections.

\subsection*{Example}
Consider the equation
\begin{equation}
\frac{d^{2}x}{dt^{2}}+p_1(t)\frac{dx}{dt}+p_2(t)x=F(t,x)\quad (t>0).
\label{e1.9}
\end{equation}
Assume that  $p_1(t), p_2(t)\geq 0$ and $p_1^2(t)> 4p_2(t)$ ($t\geq 0$).
Put
$$
p_1^+=\sup_{t\geq 0} \;p_1(t).
$$
Since $\rho_1(t)+\rho_2(t)=-p_1(t)$, we can take $\mu=p_1^+$. Hence,
$$
q_1(t)= 2p_1^+ - p_1(t),\; q_2(t)=p_1^+\;(p_1^+-p_1(t))+ p_2(t)
$$
and
$$
d_1=\inf_t q_1(t)=p_1^+,\quad d_2=\sup_t q_2(t).
$$
If, in addition,  $(p_1^+)^2>4d_2$ and \eqref{e1.7} holds, then due to
Theorem \ref{thm1.1}, equation \eqref{e1.9} has 2 positive linearly
independent solutions satisfying
 inequalities \eqref{e1.8}  with $n=2$ and
$$
-\mu+\tilde r_1=-p_1^+/2-\sqrt{(p_1^+)^2/4-d_2}\,.
$$

\section{Preliminaries}

Let $a_k(t)$ ($t\geq 0;\;k=1, \dots, n$) be continuous scalar-valued
functions defined and bounded on $[0,\infty)$, and $a_0\equiv 1$.
Consider the equation
\begin{equation}
\sum_{k=0}^n (-1)^k a_k(t) \frac{d^{n-k}u(t)}{dt^{n-k}}=0\quad (t>0).
\label{e2.1}
\end{equation}
Put
$$
c_{2k}:=\sup_{t\geq 0} a_{2k}(t),\quad
c_{2k-1}:=\inf_{t\geq 0} a_{2k-1}(t)\quad (k=1, \dots, [n/2]).
$$

\begin{lemma} \label{lem2.1}
Assume all the roots of  the polynomial
$$
Q(z)=\sum_{k=0}^n (-1)^kc_k z^{n-k}\;\;(c_0=1,\;z\in\mathbb{C})
$$
 be real and nonnegative. Then a solution $u$ of
 \eqref{e2.1} with the initial  conditions
\begin{equation}
u^{(j)}(0)=0, j=0, \dots,  n-2;\;u^{(n-1)}(0)=1
\label{e2.2}
\end{equation}
satisfies the inequalities
$$
u^{(j)}(t)\geq e^{r_1 t}\sum_{k=0}^j C^j_k
\frac{ r_1^{j-k} t^{n-1-k}}{(n-1-k)!}\geq   0\quad (j=0, \dots, n-1;\;t>0),
$$
where $r_1\geq 0$ is the smallest root of $Q(z)$.
\end{lemma}

\begin{proof} We have
$$
b_k(t):=(-1)^k (c_k-a_k(t))\geq 0\quad (k=1, \dots, n).
$$
Rewrite equation  \eqref{e2.1} in the form
\begin{equation}
\sum_{k=0}^n (-1)^k c_k \frac{d^{n-k}u}{dt^{n-k}}=\sum_1^n b_k(t)
\frac{d^{n-k}u}{dt^{n-k}}\,.
\label{e2.3}
\end{equation}
Denote
$$
G(t)=\frac{1}{2i\pi} \int_C  \frac{e^{zt}dz}{Q(z)},
$$
where $C$ is a smooth contour  surrounding all the zeros of $Q(z)$. That is, $G$
is the Green functions for the autonomous equation
\begin{equation}
\sum_{k=0}^n (-1)^k c_k\frac{d^{n-k}w(t)}{dt^{n-k}} =0.
\label{e2.4}
\end{equation}
Put
\begin{equation}
y(t)\equiv \sum_{k=0}^n c_k (-1)^k\frac{d^{n-k}u(t)}{dt^{n-k}}.
\label{e2.5}
\end{equation}
Then thanks to the variation of constants formula,
$$
u(t)=w(t)+ \int_0^t G(t-s)y(s)ds,
$$
where $w(t)$ is a solution of  \eqref{e2.3}. Since
$$
G^{(j)}(t)=\frac{1}{2i\pi} \int_C  \frac{z^j e^{zt}dz}{(z-r_1) \dots (z-r_n)}\,,
$$
where $r_1\leq \dots \leq r_n$ are the roots of $Q(z)$ with their multiplicities,
due to \cite[Lemma 1.11.2 ]{gi03}, we get
$$
G^{(j)}(t)= \frac{1}{(n-1)!} \big[ \frac{d^{n-1}z^je^{zt}}{dz^{n-1}}\big ]_{z=\theta}
$$
with  $\theta\in [r_1, r_n]$. Hence,
\begin{equation}
G^{(j)}(t)=
\sum_{k=0}^j \frac{j!e^{\theta t}\theta^{j-k} t^{n-1-k}}{(j-k)!(n-1-k)!k!} \geq
\sum_{k=0}^j \frac{j! e^{r_1 t}r_1^{j-k} t^{n-1-k}}{(j-k)!(n-1-k)!k!}\geq   0\,.
\label{e2.6}
\end{equation}
According to the initial conditions \eqref{e2.2}, we can write $w(t)=G(t)$. So
\begin{equation}
u(t) = G(t)+ \int_0^t G(t-s)y(s)\,ds\,. \label{e2.7}
\end{equation}
For $j\leq n-2$ we have  $G^{(j)}(0)=0$ and
\begin{align*}
\frac{d^j}{dt^j}\int_0^t G(t-s)y(s)ds
&=\frac{d}{dt}\int_0^t G^{(j-1)}(t-s)y(s)ds\\
&=\int_0^t G^{(j)}(t-s)y(s)ds\quad (j=0, \dots, n-1)\,.
\end{align*}
Hence thanks to \eqref{e2.3} and \eqref{e2.5},
\begin{equation}
\begin{aligned}
y(t) &= \sum_1^n b_k(t)[G^{(n-k)}(t) + \int_0^t G^{(n-k)}(t-s)y(s)ds]\\
&=K(t, t)+\int_0^t K(t, t-s)y(s)\,ds\,,
\end{aligned} \label{e2.8}
\end{equation}
where
$$
K(t, \tau)=\sum_1^n b_k(t)G^{(n-k)}(\tau)\quad (t,\tau\geq 0).
$$
According to \eqref{e2.6}, $K(t, \tau)\geq 0$ ($t,\tau\geq 0$).
Put $h(t)=K(t,t)$. Let $V$ be the Volterra operator with the kernel $K(t, t-s)$.
Then thanks to \eqref{e2.8} and the Neumann series,
$$
y(t)=h(t)+\sum_1^\infty (V^k h)(t)\geq h(t)\geq 0.
$$
Hence  \eqref{e2.7} yields,
\begin{align*}
u^{(j)}(t)&= G^{(j)}(t)+ \int_0^t G^{(j)}(t-s)y(s)ds\\
&\geq G^{(j)}(t)+ \int_0^t G^{(j)}(t-s)K(s, s)ds\\
&\geq G^{(j)}(t)\quad (j=0, \dots, n-1).
\end{align*}
This inequality and \eqref{e2.6} prove the lemma.
\end{proof}

Recall that a  scalar valued function $W(t, \tau)$ defined for $t\geq \tau\geq 0$
is {\em the Green function to equation \eqref{e2.1}} if it satisfies
that equation for $t>\tau$ and the initial conditions
\begin{gather*}
\lim_{t\downarrow \tau}\frac{\partial^{k} W(t, \tau)}{\partial t^k}=0\quad (k=0, \dots, n-2)\\
\lim_{t\downarrow \tau}\frac{\partial^{n-1} W(t, \tau)}{\partial t^{n-1}}=1\,.
\end{gather*}

\begin{lemma} \label{lem2.2}
Assume all the roots of  polynomial $Q(z)$
are real and nonnegative. Then the  Green function to equation \eqref{e2.1} and
its derivatives up to $(n-1)$ order are nonnegative. Moreover,
$$
\frac{\partial^{j} W(t, \tau)}{\partial t^j}\geq
e^{r_1 (t-\tau)}\;\sum_{k=0}^{j} C_j^k \frac{r_1^{j-k} (t-\tau)^{n-1-k}}{(n-1-k)!}
\geq   0
\quad (j=0, \dots, n-1;\;t> \tau\geq 0),
$$
\end{lemma}

\begin{proof} For a $\tau>0$, take the initial conditions
$$
u^{(j)}(\tau)=0, \quad j=0, \dots,  n-2;\quad u^{(n-1)}(\tau)=1.
$$
Then the corresponding solution $u(t)$ to \eqref{e2.1} is equal to $W(t,\tau)$.
Repeating  the argument in the proof of Lemma \ref{lem2.1}, we have
$$
\frac{\partial^{j} W(t, \tau)}{\partial t^j}
\geq G^{(j)}(t-\tau)+ \int_\tau^t G^{(j)}(t-\tau- s)K(s,s-\tau)ds\geq G^{(j)}(t-\tau).
$$
According to \eqref{e2.6} this proves the lemma.
\end{proof}


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

In \eqref{e1.3} put $v(t)=e^{-\mu t} u(t)$. Then
$$
0=e^{\mu t} \sum_{k=0}^n p_{k}(t)\frac{d^{n-k}e^{-\mu t} u}{dt^{n-k}}=
\sum_{k=0}^n p_{k}(t)(\frac{d}{dt} -\mu)^{n-k}u.
$$
That is, \eqref{e1.3} is reduced to the  equation
\begin{equation}
P(t, \frac{d}{dt}-\mu)u\equiv \sum_{k=0}^n p_{k}(t)(\frac{d}{dt} -\mu)^{n-k}u=0\,.
\label{e3.1}
\end{equation}
However,
\begin{align*}
P(t, z-\mu)&=\sum_{k=0}^n p_{k}(t)(z -\mu)^{n-k}\\
&=\sum_{k=0}^n p_k(t)\sum_{j=0}^{n-k} C_{n-k}^j (-\mu)^{j}z^{n-k-j}\\
&=\sum_{k=0}^n p_k(t)\sum_{m=k}^{n} C_{n-k}^{m-k} (-\mu)^{m-k}z^{n-m}\\
&=\sum_{m=0}^{n}z^{n-m}\sum_{k=0}^m p_k(t) C_{n-k}^{m-k} (-\mu)^{k-m}\,.
\end{align*}
So
$$
P(t, z-\mu)=\sum_{m=0}^{n}(-1)^m q_m(t)z^{n-m},
$$
where $q_m(t)$ are defined by \eqref{e1.5}.
Take into account that
$$
P(t, z-\mu)=\prod_{k=1}^n (z-\rho_k(t)-\mu)= \prod_{k=1}^n (z-\tilde \rho_k(t)),
$$
where according to \eqref{e1.4}, $\tilde \rho_k(t)\equiv \rho_k(t)+\mu\geq 0$.
Hence it follows that  $q_m(t)$ are nonnegative and we can apply
Lemma \ref{lem2.1} to \eqref{e3.1}. Due to  Lemma \ref{lem2.1} and
the substitution  $v(t)=e^{-\mu t} u(t)$, we have the following statement.

\begin{lemma} \label{lem3.1}
Assume condition  \eqref{e1.4} and that all the roots of the polynomial
$\tilde Q(z)$ defined by \eqref{e1.6} are real and nonnegative.
Then the  Green function $\tilde W(t,\tau)$ for equation \eqref{e1.3} is positive
and
$$
\frac{\partial^{j}  e^{\mu (t-\tau)}\tilde W(t, \tau)}{\partial t^j}\geq
e^{\tilde r_1 (t-\tau)}
\sum_{k=0}^j C_j^k \frac{ \tilde r_1^{j-k} (t-\tau)^{n-1-k}}{(n-1-k)!}\geq   0
$$
for $j=0, \dots, n-1;\;t> \tau\geq 0$.
In particular,
\begin{equation}
\tilde W(t, \tau)\geq e^{(-\mu+\tilde r_1) (t-\tau)}
\frac{(t-\tau)^{n-1}}{(n-1)!}\quad (t> \tau\geq 0).
\label{e3.2}
\end{equation}
\end{lemma}

\begin{lemma} \label{lem3.2}
Assume the hypothesis of Theorem \ref{thm1.1}. Let $v(t)$
be a positive solution of the linear non-autonomous problem
\eqref{e1.2}--\eqref{e1.3}.
Then a solution $x(t)$ of problem \eqref{e1.1}--\eqref{e1.2} is also positive.
Moreover, $x(t)\geq v(t)$, $t\geq 0$.
\end{lemma}

\begin{proof}
Thanks to the Variation of Constants Formula, \eqref{e1.1} can be rewritten as
$$
x(t)=v(t)+ \int_0^t \tilde W(t,s)F(s, x(s))ds.
$$
Since $\tilde W(t,s)$ is positive due to the previous lemma, there is a
sufficiently small $t_0\geq 0$, such that
$x(t)\geq 0$, $t\leq t_0$. Hence, $x(t)\geq v(t), \;t\leq t_0$.
Extending this inequality to
all $t\geq 0$, we prove the lemma.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.1}]
Take $n$ solutions $x_k(t)$ ($k=1, \dots, n$) of \eqref{e1.1} satisfying the
conditions
$$
x_k^{(j)}(\epsilon k)=0, \quad (j=0, \dots,  n-2),\quad
 x_k^{(n-1)}(\epsilon k)=1
$$
with an arbitrary $\epsilon>0$.
It can be directly checked that these solutions are linearly independent.

Now take $n$ solutions $v_k(t)$ ($k=1, \dots, n$) of \eqref{e1.3} satisfying
the same conditions
$$
v_k^{(j)}(\epsilon k)=0, \quad (j=0, \dots,  n-2),\quad  v_k^{(n-1)}(\epsilon k)=1.
$$
Then due to Lemma  \ref{lem3.1},
$$
v_k(t)\equiv \tilde W(t, \epsilon k)\geq 0\quad (t\geq \epsilon k).
$$
Now the required result is due to Lemma \ref{lem3.2}.
\end{proof}


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\end{document}