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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 97, 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 2008 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2008/97\hfil Positivity  of the Green functions]
{Positivity  of the Green functions for higher order
ordinary differential equations}

\author[M. I. Gil'\hfil EJDE-2008/97\hfilneg]
{Michael I. Gil'}

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

\thanks{Submitted May 8, 2008. Published July 25, 2008.}
\thanks{Supported by  the Kamea Fund of Israel}
\subjclass[2000]{34C10, 34A40}
\keywords{Linear ODE; Green function; fundamental  solution;
 positivity; \hfill\break\indent comparison of solutions}

\begin{abstract}
We consider the equation
$$
\sum_{k=0}^n a_k(t)x^{(n-k)}(t)=0,\quad t\geq 0,
$$
where $a_0(t)\equiv 1$, $a_k(t)$ ($k=1, \dots, n$)
are  real  bounded functions.
Assuming that   all the roots  of the polynomial
$z^n+a_1(t)z^{n-1}+ \dots +a_n(t)$ ($t\geq 0$)  are real,
we derive positivity  conditions for the Green function for the
Cauchy problem.  We also establish
 a lower estimate for the  Green function
and a comparison theorem for solutions.
\end{abstract}

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

\section{Introduction and statement of the main result}

In this paper we establish  positivity conditions of the Green function
for the Cauchy problem  (the fundamental solution) for
 the scalar equation
\begin{equation}
\sum_{k=0}^n a_k(t)x^{(n-k)}(t)=0,\quad t>0,
\label{e1.1}
\end{equation}
where $a_0(t)\equiv 1$; $a_k(t)$ ($k=1, \dots, n$) are real continuous
functions bounded on $[0,\infty)$.


The literature on the positive and nonoscillating  solutions of ordinary
differential  equations is very rich, cf. \cite{ag2,el,na1,na2,sw} and
references therein. In particular, Yu and Levin  \cite[Section 5]{lea}
among other remarkable results,
proved the following result: Suppose that, the roots
$r_1(t), \dots, r_n(t)$ of the polynomial
$$
P(z, t):=\sum_{k=0}^n a_k(t)z^{n-k},\quad z\in\mathbb{C}.
$$
for each $t\geq 0$ are  real and satisfy the inequalities
$$
\nu_0\leq r_1(t)< \nu_1\leq r_2(t)< \nu_2\leq  \dots
< \nu_{n-1}\leq r_n(t)\leq \nu_{n},\quad t\geq 0,
$$
where $\nu_j$ are  constants.  Then equation \eqref{e1.1}
has non-oscillating solutions.
That result is very useful, see for instance \cite{gi04,gi05} and
references therein.
It should be noted that  the existence of non-oscillating solutions
does not guarantee the positivity of the Green function.
Obtaining the  positivity conditions for the Green function
requires additional restrictions. On the other hand such conditions are
very important for various applications, cf. \cite{gi07,kr}.
To the best of our knowledge, the  positivity conditions for the
  Green function  were established
only in the cases of the second order equations, cf. \cite{kr}, and
equations with constant coefficients \cite{gi05};
the nonautonomous higher order differential equations were not
found in the available literature.

A solution of \eqref{e1.1} is a function $x(t)$ having continuous
derivatives up to $n$-order satisfying  \eqref{e1.1} for all $t>0$
and  given initial conditions.
Recall  that   the Green function $G(t,\tau)$ for \eqref{e1.1}
 is a function defined for $t\geq \tau\geq 0$, satisfying \eqref{e1.1}
 for $t> \tau\geq 0$,
 and the initial conditions
\begin{equation}
\lim_{t\downarrow \tau}\frac{\partial^k G(t,\tau)}{\partial t^k}=0\quad(k=0,\dots, n-2);\quad
\lim_{t\downarrow \tau}\frac{\partial^{n-1} G(t,\tau)}{\partial t^{n-1}}=1.
\label{e1.2}
\end{equation}
Assume that
\begin{equation}
a_k(t)\leq b_k ,\quad t\geq 0;\;k=1, \dots, n,
\label{e1.3}
\end{equation}
where $b_k$ are constant, and introduce the polynomial
$$
Q(\lambda)=\lambda^n+ b_1\lambda^{n-1}+b_2\lambda^{n-2}+ \dots+b_n.
$$
The aim of this paper is to prove the following theorem.

\begin{theorem} \label{thm1.1}
Assume  \eqref{e1.3}, and
let all the roots  of polynomial $Q(z)$  be real
and non-negative.  Then the  Green function for  \eqref{e1.1} is positive.
Moreover,
\begin{equation}
\frac{\partial^k G(t,s)}{\partial t^k}\geq 0,\quad t>s\geq 0,\;
k=1,\dots, n-1
\label{e1.4}
\end{equation}
\end{theorem}

This theorem is proved in the next section.
Below we also consider the case when $Q$ has negative roots.
Theorem \ref{thm1.1} supplements the very interesting recent
investigations of  higher order differential equations,
cf. \cite{car,de,tu}.


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

Denote by  $C(\mathbb{R}_+)$ the Banach space of functions
continuous and bounded on $\mathbb{R}_+:=[0, \infty)$ and
consider the nonhomogeneous equation
\begin{equation}
\sum_{k=0}^n a_k(t)D^{n-k}v(t)=f(t)
\label{e2.1}
\end{equation}
with a positive $f\in C(\mathbb{R}_+)$,
$D^kx(t): =\frac{d^k v}{dt^k}, t>0$,
and the zero initial conditions
\begin{equation}
v^{(k)}(0)=0, \quad k=0, 1, \dots, n-1.
\label{e2.2}
\end{equation}
Since the coefficients of  \eqref{e2.1} are bounded on $\mathbb{R}_+$,
a solution $v(t)$ of problem \eqref{e2.1}--\eqref{e2.2} satisfies
the conditions
$$
|v^{(k)}(t)|\leq Me^{\nu t},\quad t\geq 0, \; k=0, 1, \dots, n
$$
with  constants $M\geq 1$ and $\nu$. So $v(t)$ admits the Laplace transform.
Let $\tilde v(\lambda)$ be
the Laplace transform to $v(t)$, $\lambda$ the  dual variable.
Put  $\tilde y(\lambda)=Q(\lambda)\tilde v(\lambda)$.
Then
\begin{equation}
v(t)=\frac 1{2\pi i} \int_{c_0-i\infty}^{c_0+i\infty}
\frac{ e^{\lambda t} \tilde y(\lambda)}{Q(\lambda)}\,d\lambda\quad
(c_0=\mathop{\rm const}).
\label{e2.3}
\end{equation}
We can write as
\begin{equation}
f(t)=P(D, t)v(t)=\frac 1{2\pi i} \int_{c_0-i\infty}^{c_0+i\infty}
\frac{ e^{\lambda t} P(\lambda, t)\tilde y(\lambda)d\lambda}{Q(\lambda)}.
\label{e2.4}
\end{equation}
Hence,
\begin{equation}
f(t)=\frac 1{2\pi i} \int_{c_0-i\infty}^{c_0+i\infty} e^{\lambda t}
[1-\frac{Q(\lambda)- P(\lambda, t) }{Q(\lambda)}]\tilde y(\lambda)d\lambda= y(t)-Z(t)
\label{e2.5}
\end{equation}
where $y(t)$ is the Laplace original  to  $\tilde y(\lambda)$ and
$$
Z(t)=
\frac 1{2\pi i} \int_{c_0-i\infty}^{c_0+i\infty}
e^{\lambda t} \frac{Q(\lambda)- P(\lambda, t) }{Q(\lambda)}\tilde y(\lambda)d\lambda.
$$
By the convolution property,
$$
Z(t)=\int_0^t K(t, t-s)y(s) ds,
$$
where
$$
K(\nu, t)=\frac 1{2\pi i} \int_{c_0-i\infty}^{c_0+i\infty}
 e^{\lambda t} \frac{Q(\lambda)- P(\lambda, \nu) }{Q(\lambda)}d\lambda,
\quad \nu\geq 0.
$$
So
$$
y(t)-\int_0^t K(t, t-s)y(s) ds =f(t).
$$
Take into account that
$$
K(\nu, t)=\sum_{k=1}^n (b_k-a_k(\nu)) \mu_k(t),
$$
where
\begin{equation}
\mu_k(t)=
\frac 1{2\pi i} \int_{c_0-i\infty}^{c_0+i\infty} e^{\lambda t}
\frac{\lambda^{n-k}}{Q(\lambda)}d\lambda, \quad k=1, \dots, n.
\label{e2.6}
\end{equation}
By \cite[Lemma 1.11.2, p. 23]{gi05}
\begin{equation}
\mu_k(t)= \frac 1{(n-1)!} \frac{d^{n-1}e^{st}s^{n-k}}{ds^{n-1}}
\big|_{s\in [z_1, z_n]}\geq 0
\label{e2.7}
\end{equation}
where $z_1$ is the smallest (nonnegative) root of $Q$ and
$z_n$ is the largest root of $Q$.

Since $b_k-a_k(t)\geq 0, t\geq 0,$ we can assert the
$K(\nu, t)\geq 0$ for all $\nu, t\geq 0$.

Furthermore, denote by $C_\tau$ the Banach space of functions
continuous  on $[0, \tau]$ with a positive $\tau<\infty$.
In addition $C_\tau^+$ denotes the cone of positive functions from $C_\tau$.
Introduce on $C_\tau$ the Volterra operator $V$ by
$$
(Vw)(t)= \int_0^t K(t, t-s)w(s) ds.
$$
Then $y-Vy=f$. By the Neumann series,
$$
(I-V)^{-1} f=\sum_{k=0}^\infty V^k f\geq f\geq 0.
$$
Here $I$ is the unit operator.
Note that the Neumann series of any Volterra operator with a
continuous kernel converges in the sup-norm on each finite segment,
since the spectral radius of that  operator
in a space of continuous functions defined on a finite segment
is equal to zero, cf. \cite{dal}.
So $y(t)\geq f(t), t\in [0, \tau]$.
But $\tau$ is an arbitrary positive number. So we obtain
$y(t)\geq f(t), t\in \mathbb{R}_+$. Recall that $y(t)$ is the
Laplace original to $\tilde y(\lambda)$; so according to \eqref{e2.3} and
the convolution property, we get
\begin{equation}
v(t)=\int_0^t \mu_n(t-s)y(s)ds
\label{e2.8}
\end{equation}
where $\mu_n$ is defined by \eqref{e2.6}.
According to \eqref{e2.7},
\begin{equation}
\mu_n(t)= \frac 1{(n-1)!} \frac{d^{n-1}e^{st}}{ds^{n-1}}
\big|_{s\in [z_1, z_n]}\geq    e^{z_1t}\frac{t^{n-1}}{(n-1)!}.
\label{e2.9}
\end{equation}
Now the inequality  $y(t)\geq f(t), t\geq 0$, yields
$$
v(t)\geq \int_0^t \mu_n (t-s)f(s)ds\geq 0, \quad t\geq 0.
$$
Thus the solution of problem  \eqref{e2.1}--\eqref{e2.2} is positive,
provided $f$ is positive.
But
\begin{equation}
v(t)=\int_0^t G(t,s)f(s)ds.
\label{e2.10}
\end{equation}
Hence it follows that $G(t,s)\geq 0$.
Furthermore, by \eqref{e1.2}, \eqref{e2.3} and the convolution property
$$
v^{(k)}(t)=\frac 1{2\pi i} \int_{c_0-i\infty}^{c_0+i\infty}
\frac{ e^{\lambda t} \lambda^k \tilde y(\lambda)}
{Q(\lambda)}\;d\lambda=\int_0^t \mu_{n-k}(t-s)y(s)ds.
$$
But as it was above shown, $\mu_k(t)\geq 0$, $k=1,\dots, n$. Thus
$v^{(k)}(t)\geq 0$.
So  by \eqref{e2.10},
$$
v^{(k)}(t)=\int_0^t \frac{\partial^k G(t,s)}{\partial t^k} f(s)ds\geq 0,
\quad k=1,\dots, n-1.
$$
Hence \eqref{e1.4}  follows.
As claimed.

\section{Lower solution estimates and comparison of Green's functions}

\begin{lemma} \label{lem3.1}
Under the hypothesis of Theorem \ref{thm1.1},
for any nonnegative $f$ a solution of problem \eqref{e2.1}--\eqref{e2.2}
satisfies the inequality
$$
v(t)\geq \frac{1}{(n-1)!}\int_0^t e^{z_1 (t-s)}(t-s)^{n-1}f(s)ds
$$
where $z_1\geq 0$ is the smallest root of $Q(\lambda)$.
\end{lemma}

Indeed, this result immediately follows
from \eqref{e2.8} and \eqref{e2.9}.
Recall also that   $G(t,\tau)$ is a solution of the equation
$$
P(D,t)y=\delta(t-\tau),\quad t > 0
$$
where $\delta(t)$ is the Dirac Delta function.
Hence thanks to the previous lemma  we easily get the inequality
$$
G(t, \tau)\geq \frac{1}{(n-1)!} (t-\tau)^{n-1}e^{z_1 (t-\tau)},\quad t>\tau.
$$
Furthermore, together with \eqref{e1.1}, let us consider the equation
\begin{equation}
\sum_{k=0}^{n-1} c_k(t)x^{(n-k)}(t)=0,\quad t>0,
\label{e3.1}
\end{equation}
where $c_k(t)$ are bounded real functions satisfying the
conditions
\begin{equation}
c_k(t)\leq a_k(t),\quad t\geq 0;\; k=1, \dots, n.
\label{e3.2}
\end{equation}

\begin{lemma} \label{lem3.2}
Let the  Green function  $G(t,s)$ for \eqref{e1.1} be positive
and the inequalities \eqref{e1.4} and \eqref{e3.2} hold.
Then
 the Green function  $W(t,s)$ for  \eqref{e3.1} satisfies the inequalities
\begin{equation}
W(t,s)\geq G(t,s)\geq 0;\quad
 \frac{\partial^k W(t,s)}{\partial t^k}\geq \frac{\partial^k G(t,s)}{\partial t^k}\geq 0
\label{e3.3}
\end{equation}
for all $t>s\geq 0$ and $k=1,\dots, n-1$.
\end{lemma}

\begin{proof} Rewrite \eqref{e3.1} as
$$
\sum_{k=0}^n a_k(t)x^{(n-k)}(t)=\sum_{k=1}^n (a_n(t)-c_n(t))x^{(n-k)}(t),
\quad t\geq 0.
$$
Then with the notation $w(t)=W(t,0)$, we have
\begin{equation}
w(t)=G(t,0)+ \int_0^t G(t,s) \sum_{k=1}^n (a_{n-k}(s)-c_{n-k}(s))w^{(k)}(s)ds.
\label{e3.4}
\end{equation}
Hence, according to \eqref{e1.2},
\begin{equation}
w^{(k)}(t)=\frac{\partial^k G(t,0)}{\partial t^k}+ \int_0^t
\frac{\partial^k G(t,s)}{\partial t^k}\sum_{k=0}^n (a_{n-k}(s)-c_{n-k}(s))w^{(k)}(s)ds.
\label{e3.5}
\end{equation}
Rewrite \eqref{e3.4} and \eqref{e3.5} as the $n$-vector equation
$$
\widehat w=\widehat G+ \tilde V\widehat w
$$
where $\tilde V$ is a Volterra equation with a positive continuous
matrix kernel
\begin{gather*}
\widehat G(t)=\mathop{\rm column}
\big[G(t,0), \frac{\partial G(t,0)}{\partial t},  \dots,
\frac{\partial^{n-1} G(t,0)}{\partial t^{n-1}}\big],\\
\widehat w(t)=\mathop{\rm column}\big[w(t), w'(t), \dots, w^{(n-1)}(t)\big]
\end{gather*}.
Hence  by the Neumann series
$$
\widehat w=\sum_{k=0}^\infty \tilde V^k \widehat G\geq \widehat G.
$$
So for $s=0$ the inequalities \eqref{e3.3} are proved.
 But the case $s>0$ can be similarly proved.
As it was above mentioned,  the Neumann series
of any Volterra operator with a continuous kernel
converges in the sup-norm on each finite segment, since the
spectral radius of that  operator
in a space of continuous functions defined on a finite segment
is equal to zero. This proves the lemma.
\end{proof}

Note that a relatively special but related comparison result is
due to MacKenna and Reichel \cite{ma}.

\section{The case of negative roots}

Let  $Q(\lambda)$ have at least one negative root. With a positive number $r$,
substitute in \eqref{e1.1} $x(t)=e^{-rt}w(t)$.
After simple calculations we get we equation
$$
P(D-r, t)w(t)=0.
$$
Take into account that
\begin{align*}
P(z-r, t)&=\sum_{k=0}^n  a_{n-k}(t)(z-r)^k\\
&= \sum_{k=0}^n  a_{n-k}(t)\sum_{j=0}^k C_k^j (-r)^{k-j}z^j \\
&=\sum_{j=0}^n z^j \sum_{k=j}^n  a_{n-k}(t) C_k^j (-r)^{k-j}\\
&=\sum_{j=0}^n z^j \tilde a_{n-j}(t, r)
\end{align*}
where $C_k^j=\frac {k!}{j!(k-j)!}$ and
$$
\tilde a_{n-j}(t,r)=\sum_{k=j}^n  a_{n-k}(t) C_k^j (-r)^{k-j}.
$$
Thus we have
\begin{equation}
\sum_{k=0}^n \tilde a_{n-k}(t,r)w^{(k)}(t)=0,\quad t>0,.
\label{e4.1}
\end{equation}
Assume that
\begin{equation}
\tilde a_j(t, r)\leq \tilde b_j(r)\;\;(\tilde b_j(r)=\mathop{\rm const};
\;t\geq 0;\; j=1, \dots, n,
\label{e4.2}
\end{equation}
and introduce the polynomial
$$
\tilde Q(\lambda, r)=\lambda^n+ \tilde b_1(r)\lambda^{n-1}+ \tilde b_2(r)\lambda^{n-2}
+ \dots+\tilde b_n(r).
$$
Then applying Theorem \ref{thm1.1} to equation \eqref{e4.1}, we obtain the
following result.

\begin{corollary} \label{coro4.1}
Under condition \eqref{e4.2}, for a positive number $r$, let all the
roots  of polynomial $\tilde Q(\lambda, r)$  be real
and non-negative.  Then the  Green function for \eqref{e1.1}
is positive. Moreover,
$$
\frac{\partial^k (e^{rt} G(t,s))}{\partial t^k}\geq 0, \quad 
t>s\geq 0;\;k=1,\dots, n-1.
$$
\end{corollary}

In particular, consider the equation
\begin{equation}
x''+ a_1(t)x'+a_2(t)x=0 \label{e4.3}
\end{equation}
assuming that
\begin{equation}
0< m_k\leq a_k(t)\leq M_k, \quad  t\geq 0;\;k=1, 2,
\label{e4.4}
\end{equation}
where $m_k$ and $M_k$ are constant.
Then
$$
\tilde a_{1}(t,r)=-2r+a_{1}(t),  \tilde a_{2}(t,r)=r^2-a_1r+ a_{2}(t).
$$
Hence
$$
\tilde a_{1}(t,r)\leq \tilde b_1(r)=-2r+M_1,  \tilde a_{2}(t,r)\leq
\tilde b_2(r)=r^2-m_1r+ M_2.
$$
If $\tilde b_1(r)<0, \tilde b_2(r)>0$ and $\tilde b^2_1(r)\geq 4\tilde b_2(r)$,
then $\tilde Q(z,r)=z^2+\tilde b_1(r)z+\tilde b_2(r)$ has two non-negative roots.
Let  $m_1^2> 4M_2$. Take $r\geq M_1$.
Then
$$
r\geq M_1> m_1/2+ \sqrt{ m_1^2/4-M_2}
$$
and therefore $\tilde b_2(r)> 0$. We also should have the inequality
$$
(-2r+M_1)^2\geq 4(r^2-m_1r+ M_2).
$$
Hence $M_1^2- 4rM_1^2\geq -4m_1r+ 4M_2$. So we get
\begin{equation}
M_1\leq r\leq (M_1^2- 4M_2)/4(M_1 -m_1).
\label{e4.5}
\end{equation}
Now the previous corollary implies the following result.

\begin{corollary} \label{coro4.2}
Let the conditions \eqref{e4.4}, $m_1^2> 4M_2$ and
$$
1\leq \frac{(M_1^2- 4M_2)}{4M_1(M_1 -m_1)}
$$
hold,  then the  Green function $G(t,s)$ for \eqref{e4.1} is positive.
Moreover,
$$
\frac{\partial (e^{rt} G(t,s))}{\partial t}\geq 0,\quad t>s\geq 0,
$$
for any $r$ satisfying \eqref{e4.5}. In particular for $r=M_1$.
\end{corollary}

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\end{document}