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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 102, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/102\hfil Behavior at infinity]
{Behavior at infinity of $\psi$-evanescent solutions to linear
 differential equations}

\author[P. N. Boi\hfil EJDE-2010/102\hfilneg]
{Pham Ngoc Boi}  % in alphabetical order

\address{Pham Ngoc Boi \newline
Department of Mathematics, 
Vinh University, Vinh City, Vietnam}
\email{pnboi\_vn@yahoo.com}

\thanks{Submitted April 29, 2010. Published July 28, 2010.}
\subjclass[2000]{34A12, 34C11, 34D05}
\keywords{$\psi$-bounded solutions;
$\psi$-ordinary dichotomy; $\psi$-evanescent solutions}

\begin{abstract}
 In this article we present some necessary and sufficient
 conditions for the existence of $\psi$-evanescent solution of the
 nonhomogeneous linear differential equation $x'=A(t)x+f(t)$, which
 is related to  the notion of  $\psi$-ordinary dichotomy for the
 equation $x'=A(t)x$. We associate that with the condition of
 $\psi$-ordinary dichotomy for the homogeneous linear differential
 equation $x'=A(t)x$.
\end{abstract}

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

\section{Introduction}

The existence of $\psi$-bounded and $\psi$-stable solutions
on $\mathbb{R}_+$ for systems of ordinary differential equations
has been studied by many authors; such as
Akinyele \cite{a1}, Avramescu \cite{a2},
Boi \cite{b1,b2}, Constantin \cite{c1}, Diamandescu \cite{d2,d3,d4}.
Also, in \cite{b2,d2,d3,d4} the authors prove several sufficient
conditions of the $\psi$-evanescence at $\infty$, $-\infty$ for the solutions
of linear differential equations.

 The purpose of this paper is to provide a condition for the existence
of  $\psi$- evanescent solution of the equations  $x'=A(t)x +f(t)$,
which is concerned with the notion of  $\psi$-ordinary dichotomy
for the equation $x'=A(t)x$. We shall deal with the existence of
$\psi$-evanescent solution of nonhomogeneous equations,
which have been studied in recent works, such as \cite{b2,d2,d4}.

Denote by $\mathbb{R}^d$ the $d$-dimensional Euclidean space.
Elements in this space are denoted by $x=(x_1,x_2,\dots , x_d)^T$
and their norm by $\Vert x\Vert =\max \{ |x_1|, |x_2|,\dots ,
|x_d|\}$. For real $d\times d$  matrices $A$, we define the norm
$|A|= \sup_{\Vert x\Vert \leqslant 1}\Vert Ax\Vert$.
Let $\mathbb{R}_+ =[0, \infty)$,
$\mathbb{R}_-=(- \infty,0]$, $J=\mathbb{R}_-$,$J=\mathbb{R}_+$
or $J=\mathbb{R}$. Let $\psi_i :J \to (0,\infty)$,
$i=1,2,\dots,d$ be continuous functions and let
$$
\psi =\mathop{\rm diag}\{\psi_1, \psi_2,\dots ,\psi_d\}.
$$

\begin{definition} \label{def1.1}  \rm
 A function $f:J \to \mathbb{R}^d$ is said to be
\begin{itemize}
\item $\psi$-bounded on $J$ if $\psi f$ is
bounded on $J$.
\item $\psi$-integrable on $J$ if $f$ is measurable
and $\psi f$ is Lebesgue integrable on $J$.
\end{itemize}
\end{definition}

 In $\mathbb{R}^d$, consider  the following equations on $J$.
\begin{gather}
 x'=A(t)x+f(t),\label{e1.1} \\
 x'=A(t)x. \label{e1.2}
\end{gather}
where $A(t)$ is a continuous $d\times d$ matrix function and
$f(t)$ is a continuous function for $t \in J$.

 By a solution of \eqref{e1.1}, we mean a continuous function
satisfying \eqref{e1.1} for almost $t$ in $J$.
Let $Y(t)$ be the fundamental matrix of  \eqref{e1.2} with
$Y(0)=I_d$, the identity $d \times d$ matrix.
  A $d\times d$ matrix $P$ is said to be a projection matrix if $P^2=P$.
If $P$ is a projection, then so is $I_d-P$. Two  projections
$P, I_d-P$ are called supplementary.

\begin{definition} \label{def1.2} \rm
 Equation \eqref{e1.2} is said to have a $\psi$-ordinary dichotomy on $J$
 if there exist  positive constants $K, L$ and two supplementary
projections $P_1, P_2 $ such that
\begin{gather}
|\psi(t) Y(t) P_1 Y^{-1}(s)\psi^{-1}(s) |\leqslant
 K \quad \text{for }    s\leqslant   t; s,t\in J, \label{e1.3}
\\
|\psi(t) Y(t) P_2 Y^{-1}(s)\psi^{-1}(s)|\leqslant
L \quad \text{for }    t\leqslant   s; s,t \in J. \label{e1.4}
\end{gather}
Also we say that  \eqref{e1.2} has a $\psi$-ordinary dichotomy
on $J$ with two supplementary projections $P_1, P_2 $.
\end{definition}

\begin{remark} \label{rmk1.3} \rm
It is easily verified that if \eqref{e1.2} has a $\psi$-ordinary
dichotomy on $\mathbb{R}_+$ and on $\mathbb{R}_-$ with two
supplementary projections $P_1, P_2 $ then \eqref{e1.2} has a
$\psi$-ordinary dichotomy on $\mathbb{R}$ with two supplementary
projections $P_1, P_2$.
Note that for $\psi = I_d$, we obtain the notion of ordinary
dichotomy (see \cite{c2,d1})
\end{remark}

\begin{theorem}[\cite{b1,d2}] \label{thm1.4}
(a) Equation \eqref{e1.1} has at least one $\psi$-bounded solution
on $\mathbb{R}_+$ for every  $\psi$-integrable function $f$ on
$\mathbb{R}_+$ if and only if  \eqref{e1.2} has a
$\psi$-ordinary dichotomy on $\mathbb{R}_+$.

(b) Suppose that \eqref{e1.2} has a $\psi$-ordinary dichotomy
and $lim_{t\to\infty}|\psi(t)Y(t)P_1|=0$. Let $f$ be a
$\psi$-integrable function on $\mathbb{R}_+$. Then every
$\psi$-bounded solution x(t) of \eqref{e1.1} on $\mathbb{R}_+$
is such that $lim_{t\to\infty}\|\psi(t)x(t)\| =0$.
 \end {theorem}

\section{Preliminaries}

\begin{lemma} \label{lem2.1}
Equation  \eqref{e1.2} has a $\psi$-ordinary dichotomy on $J$
with two supplementary projections $P_1, P_2 $
 if and only if two following conditions are satisfied for
 all $\xi \in \mathbb{R}^d$:
\begin{gather}
 \|\psi(t)Y(t)P_1\xi\|\leqslant K \|\psi(s)Y(s)\xi\|\quad\text{for }
  s \leqslant t ; s,t \in J \label{e2.1}\\
 \|\psi(t)Y(t)P_2\xi\|\leqslant  L\|\psi(s)Y(s)\xi\|\quad \text{for }
  t \leqslant  s ; s , t \in J  \label{e2.2}
 \end{gather}
 \end{lemma}

\begin{proof}
 If \eqref{e1.2} has a $\psi$-ordinary dichotomy on $J$ then
  \begin{gather}
 \|\psi(t)Y(t)P_1Y^{-1}(s)\psi^{-1}(s)y\|\leqslant K\|y\| \quad \text{for }
 s \leqslant  t; s,t \in J  \label{e2.3}\\
 \|\psi(t)Y(t)P_2Y^{-1}(s)\psi^{-1}(s)y\|\leqslant L\|y\| \quad\text{for }
 t \leqslant  s ; s,t \in J \label{e2.4}
 \end{gather}
for any vector $y\in \mathbb{R}^d$. Choose $y=\psi(s)Y(s)\xi$,
we obtain \eqref{e2.1}, \eqref{e2.2}. Conversely, if
\eqref{e2.1} \eqref{e2.2} are true, for any vector $y\in \mathbb{R}^d$,
 putting $\xi=Y^{-1}(s)\psi^{-1}(s)y$ we get \eqref{e2.3}, \eqref{e2.4}.
This implies that \eqref{e1.2} has a $\psi$-ordinary dichotomy on $J$.
% $\mathbb{R}_+$.
The proof is complete.
 \end{proof}

\begin{remark} \label{rmk2.2}\rm
 If \eqref{e1.2} has a $\psi$-ordinary dichotomy on $\mathbb{R}_+$
with two supplementary projections $P_1, P_2 $ then there exist
positive constants $K_P , L_P $ such that
 \[
 \|\psi(t)Y(t)P_1\xi\|\leqslant K_P\| \xi\| , \quad
 \|\psi(t)Y(t)\xi\| \geqslant L_P\|P_2\xi\|
\]
 for all $\xi \in \mathbb{R}^d$, all $t\geqslant 0$.
\end{remark}

 Indeed, let $s=0$, we deduce from \eqref{e2.1} that
$\|\psi(t)Y(t)P_1\xi\| \leqslant K\|\psi(0)\xi\|\leqslant K_P\|\xi\|$ for
all $t\geqslant 0$, where $K_P=K|\psi(0)|$.
 Let $t=0$, we deduce from \eqref{e2.2} that
$\|\psi(0)P_2\xi\|\leqslant L\|\psi(s)Y(s)\xi\|$, for all $s\geqslant0$.
Then $\|\psi(s)Y(s)\xi\|\geqslant L_P\|P_2\xi\|$, for all $s\geqslant0$,
where $L_P=[L|\psi^{-1}(0)|]^{-1}$.

 Now, let $X_1=\{ u\in \mathbb{R}^d | u=x(0), x(t)
 \text{ is a $\psi$-bounded solution  of \eqref{e1.2} on $\mathbb{R}_+$}
 \}$
 and let $X_0=\{ u\in \mathbb{R}^d | u=x(0), x(t)$ is a solution
of \eqref{e1.2} on $\mathbb{R}_+$ such that
$\psi(t)x(t)\to 0, $ as $ t \to \infty$ \}.

\begin{lemma} \label{lem2.3}
 If \eqref{e1.2} has a $\psi$-ordinary dichotomy on $\mathbb{R}_+$
 and $Q_1, Q_2$ are two supplementary projections, then
\eqref{e1.2} has a $\psi$-ordinary dichotomy on $\mathbb{R}_+$
with two supplementary projections $Q_1, Q_2$ if and only if
\begin{equation}
X_0 \subset Q_1\mathbb{R}^d \subset X_1   \label{e2.5}
\end{equation}
\end{lemma}

\begin{proof}
  The ``only if'' part. Suppose that \eqref{e1.2} has a
$\psi$-dichotomy with two supplementary projections $Q_1, Q_2$,
we show that \eqref{e2.5} holds. First, we prove
$Q_1\mathbb{R}^d \subset X_1$. For any $u\in Q_1\mathbb{R}^d$,
there exists $v\in \mathbb{R}^d$ such that $u=Q_1v$.  Let $y(t)$
be a solution of \eqref{e1.2} such that $y(0)=u$. It follows from
Remark \ref{rmk2.2} that
$$
\|\psi(t)y(t)\|=\|\psi(t)Y(t)u\|=\|\psi(t)Y(t)Q_1v\|\leqslant{K}_Q\|v\| \quad
\text{for } t\geqslant 0,
$$
where ${K}_Q$ is a positive constant.
This implies that $u\in X_1$. Hence $Q_1\mathbb{R}^d \subset X_1$.
 We prove $X_0\subset Q_1\mathbb{R}^d $.
For $u\in X_0$, let $x(t)$ be a solutions of \eqref{e1.2} such that
$x(0)=u$. It implies that
\begin{equation}
\|\psi(t)x(t)\| \to 0, \text{ as } t\to \infty \label{e2.6}
\end{equation}
 From  Remark \ref{rmk2.2}, we have
\begin{equation}
 \|\psi(t)x(t)\|=\|\psi(t) Y(t)u\|\geqslant {L}_Q\|Q_2u\|,\quad
 \text{ for } t\geqslant 0 \label{e2.7}
\end{equation}
where  ${L}_Q$ is a positive constant.
The relations \eqref{e2.6} and \eqref{e2.7} imply $Q_2u=0$,
then $u\in Q_1\mathbb{R}^d$. Thus \eqref{e2.5} holds.

 We prove the ``if'' part. Suppose that \eqref{e1.2} has a
$\psi$-ordinary dichotomy on $\mathbb{R}_+$ with two supplementary
projections $P_1, P_2$.
  Let $Q_1, Q_2$  be two supplementary projections such that
 \eqref{e2.5} holds. We will  prove that \eqref{e1.2} has a
$\psi$-ordinary dichotomy on $\mathbb{R}_+$ with two
supplementary projections $Q_1, Q_2$.
Let $\widetilde{Q}_1 ,\widetilde{Q}_2 $ be two supplementary
projections such that $\widetilde{Q}_1 \mathbb{R}^d=X_0$.
Applying $\eqref{e2.5}$ to $P_1, P_2$ we get
$\widetilde{Q}_1\mathbb{R}^d=X_0 \subset P_1\mathbb{R}^d \subset X_1 $.
 The set  $X_0' = (P_1-\widetilde{Q}_1 )\mathbb{R}^d$  is a subset
of  $P_1 \mathbb{R}^d$, supplementary to $X_0$. We will show that
there exists a positive constant number $N$ such that
\begin{equation}
 \|\psi(t)Y(t)u\| \geqslant N\|u\|, \quad \text{for all } u\in X_0',\;
t\geqslant 0 \label{e2.8}
\end{equation}
In fact, otherwise there exists a sequence of unit vectors
$\{v_n\}\subset X_0', n=1,2,\dots $ and a sequence of numbers
$t_n\geqslant 0$ such that $\|\psi(t_n)Y(t_n)v_n\|\to 0$.
By the compactness of the unit sphere in $X_0'$, we may assume
that $v_n\to v \in X_0'$ as $n\to \infty$, where $v$ is
 a unit vector. By  Remark \ref{rmk2.2} and
$(v-v_n)\in X_0'\subset P_1\mathbb{R}^d$, we obtain
$$
\|\psi(t_n)Y(t_n)(v-v_n)\|=\|\psi(t_n)Y(t_n)P_1(v-v_n)\| \leqslant K_P\|v-v_n\|
$$
Letting $n \to \infty $, we obtain $\|\psi(t_n)Y(t_n)(v-v_n)\|\to 0$.
Then $ \|\psi(t_n)Y(t_n)v_n\|+\|\psi(t_n)Y(t_n)(v-v_n)\|\to 0$
as $t_n\to \infty$. Then $\|\psi(t_n)Y(t_n)v\|\to 0$ as $t_n\to \infty$.
Hence $v\in X_0$. On the other hand, $v\in X_0'$, we have $v=0$,
which is a contradiction to the unit of $v$.
Thus \eqref{e2.8} holds.

  From \eqref{e2.8} and \eqref{e2.1} we obtain
\begin{equation}
\begin{aligned}
N\|(P_1-\widetilde{Q}_1)u\|
&\leqslant \|\psi(t)Y(t)(P_1-\widetilde{Q}_1)u\| \\
&\leqslant \|\psi(t)Y(t)P_1u\|+\|\psi(t)Y(t)\widetilde{Q}_1u\|  \\
&\leqslant K\|\psi(s)Y(s)u\|+\|\psi(t)Y(t)\widetilde{Q}_1u\|
\end{aligned} \label{e2.9}
\end{equation}
 for $u\in \mathbb{R}^d$, $0\leqslant s\leqslant t$.
Let $t\to \infty$, we get $\|\psi(t)Y(t)\widetilde{Q}_1u\|\to0$.
 From \eqref{e2.9}, we have
\begin{equation}
N\|(P_1-\widetilde{Q}_1)u\| \leqslant \|\psi(s)Y(s)u\| \quad
\text{for } s\geqslant0 \label{e2.10}
\end{equation}
 From Remark \ref{rmk2.2} and \eqref{e2.10}, we have
\begin{equation}
\|\psi(t)Y(t)(P_1-\widetilde{Q}_1)u\|
\leqslant K_P\|(P_1-\widetilde{Q}_1)u\| \leqslant K_PN^{-1}\|\psi(s)Y(s)u\|\quad
\text{ for } t,s \geqslant 0 \label{e2.11}
\end{equation}
Consequently,
\begin{equation}
\begin{aligned}
\|\psi(t)Y(t)\widetilde{Q}_1u\|
&\leqslant \|\psi(t)Y(t)P_1u\|+\|\psi(t)Y(t)(P_1-\widetilde{Q}_1)u\|   \\
&\leqslant (K+K_PN^{-1})\|\psi(s)Y(s)u\| \quad \text{for } 0\leqslant s\leqslant t
\end{aligned}    \label{e2.12}
\end{equation}
 From $\widetilde{Q}_2=P_2+P_1-\widetilde{Q}_1$ and \eqref{e2.11},
 we obtain
\begin{equation}
\begin{aligned}
\|\psi(t)Y(t)\widetilde{Q}_2u\|
&\leqslant \|\psi(t)Y(t)P_2u\|+\|\psi(t)Y(t)(P_1-\widetilde{Q}_1)u\|   \\
&\leqslant (L+K_PN^{-1})\|\psi(s)Y(s)u\| \quad\text{for } 0\leqslant t\leqslant s
\end{aligned}  \label{e2.13}
\end{equation}
  From $\widetilde{Q}_1\mathbb{R}^d=X_0\subset
 \mathbb{Q}_1\mathbb{R}^d\subset X_1$, we obtain
 $Q_2\widetilde{Q}_1\mathbb{R}^d\subset Q_2Q_1\mathbb{R}^d=0$
then $Q_1\widetilde{Q}_1=(I_d-Q_2)\widetilde{Q}_1=\widetilde{Q}_1$.
Thus
\begin{equation}
Q_1\widetilde{Q}_2=Q_1(I_d-\widetilde{Q}_1)=Q_1-\widetilde{Q}_1   \label{e2.14}
\end{equation}
By the definition of $X_1$, there exists $N' >0$ such that
\begin{equation}
 \|\psi(t)Y(t)u\| \leqslant N'\|u\| \text{, for } t \geqslant 0  \label{e2.15}
\end{equation}
It follows from Lemma \ref{lem2.1}, \eqref{e2.12}, \eqref{e2.13} that
 \eqref{e2.2} has a $\psi$-ordinary dichotomy on $\mathbb{R}_+$
with two supplementary projections $\widetilde Q_1, \widetilde Q_2$.
By Remark \ref{rmk2.2} we have
$$
\|\psi(s)Y(s)u\| \geqslant \widetilde L_Q \|\widetilde Q_2 u\|\quad
 \text{for } s \geqslant 0\,.
$$
Combining this inequality, \eqref{e2.14} and \eqref{e2.15} we obtain
\begin{equation}
\begin{aligned}
\|\psi(t)Y(t)(Q_1-\widetilde{Q}_1)u\|
&\leqslant N'\|(Q_1-\widetilde{Q}_1)u\| \\
& \leqslant N'\|Q_1 \widetilde{Q}_2u\|   \leqslant N'|Q_1|\| \widetilde{Q}_2u\| \\
&\leqslant K_2\|\psi(s)Y(s)u\|, \quad\text{for }t,  s\geqslant 0
\end{aligned}  \label{e2.16}
\end{equation}
where $K_2$ is a positive constant.
 From \eqref{e2.12}, \eqref{e2.16}, we have
\begin{equation}
\begin{aligned}
\|\psi(t)Y(t)Q_1u\|
&\leqslant\|\psi(t)Y(t) \widetilde{Q}_1u\|+\|\psi(t)Y(t)(Q_1-\widetilde{Q}_1)u\|\\
&\leqslant (K+K_PN^{-1}+K_2)\|\psi(s)Y(s)u\|, \quad\text{ for }
0\leqslant s\leqslant t
\end{aligned}  \label{e2.17}
\end{equation}
 From $Q_2=\widetilde{Q}_2+\widetilde{Q}_1-Q_1$, \eqref{e2.13}
and \eqref{e2.16}, we obtain
\begin{equation}
\begin{aligned}
\|\psi(t)Y(t)Q_2u\|
&\leqslant\|\psi(t)Y(t) \widetilde{Q}_2u\|+\|\psi(t)Y(t)(\widetilde{Q}_1-Q_1)u\|\\
&\leqslant (L+K_PN^{-1}+K_2)\|\psi(s)Y(s)u\|, \quad\text{for }
 0\leqslant t\leqslant s
\end{aligned} \label{e2.18}
\end{equation}
Lemma \ref{lem2.1} and \eqref{e2.17}, \eqref{e2.18} follow that \eqref{e1.2}
 has a $\psi$-ordinary dichotomy on $\mathbb{R}_+$ with two
supplementary projections $Q_1,Q_2$. The proof is complete.
\end{proof}

 Let $\widetilde{X}_1=\{ u\in \mathbb{R}^d | u=x(0), x(t) \text{ is a
$\psi$-bounded solution  of \eqref{e1.2} on $\mathbb{R}_-$ }\}$,
and let $\widetilde{X}_0=\{ u\in \mathbb{R}^d | u=x(0), x(t)$ is a solution of $\eqref{e1.2}$ on $\mathbb{R}_-$ such that
$\psi(t)x(t)\to 0, $ as $ t \to -\infty$ \}.
  From Theorem \ref{thm1.4} and Lemma \ref{lem2.3},
we obtain the following results
 on half-line $\mathbb{R}_-$.

\begin{lemma} \label{lem2.4}
 (a) Equation \eqref{e1.1} has at least one $\psi$-bounded
solution on $\mathbb{R}_-$
 for every  $\psi$-integrable function $f$ on $\mathbb{R}_-$ if
and only if  \eqref{e1.2} has a
$\psi$-ordinary dichotomy on $\mathbb{R}_-$.

 (b) If \eqref{e1.2} has a $\psi$-ordinary dichotomy on $\mathbb{R}_-$
and $\widetilde{Q}_1, \widetilde{Q}_2$ are two supplementary
projections, then \eqref{e1.2} has a $\psi$-ordinary dichotomy
on $\mathbb{R}_-$ with two supplementary projections
$\widetilde{Q}_1, \widetilde{Q}_2$ if and only if
\begin{equation}
\widetilde{X}_0 \subset \widetilde{Q}_2\mathbb{R}^d
\subset \widetilde{X}_1   \label{e2.19}
\end{equation}
\end{lemma}

\begin{proof}
The proof of this Lemma is similar to that of
Theorem \ref{thm1.4} and Lemma \ref{lem2.3} with the corresponding
replacement
($t\geqslant s \geqslant 0 $ with
$ 0\geqslant  s\geqslant t$, $P_1$ with $-P_2, P_2$ with $-P_1$,
$\infty$
with $-\infty$, $- \infty$ with $\infty$ \dots ).
\end{proof}

\begin{definition} \label{def2.5}  \rm
 A function $x(t)$ is said to be
\begin{itemize}
\item  $\psi$-evanescent at
$ \infty$ if $\lim_{t\to  \infty}\|\psi(t)x(t)\|=0$.
\item  $\psi$-evanescent at $ -\infty$ if
 $\lim_{t\to  -\infty}\|\psi(t)x(t)\|=0$.
\item  $\psi$-evanescent at $ \pm\infty$ if
 $\lim_{t\to  \pm\infty}\|\psi(t)x(t)\|=0$.
\end{itemize}
\end{definition}

  Note that for $\psi = I_d$, we obtain the notion of evanescent
solution of \eqref{e1.1} at $\pm\infty$ (see \cite{a3})

\begin{lemma} \label{lem2.6}
 If  \eqref{e1.1} has at least one solution on $\mathbb{R}$,
$\psi$-evanescent  at $\infty$ for every $\psi$-integrable function
$f$ on $\mathbb{R}$ then every solution of  \eqref{e1.2} is the
sum of  two solution of \eqref{e1.2}, one of which is $\psi$-bounded
on $\mathbb{R}_-$, and the other is defined on $\mathbb{R}_+$,
 $\psi$-evanescent at $\infty$.
\end{lemma}

\begin{proof} Set
\[
h(t)=\begin{cases}
0 & \text{for } | t|\geqslant 1\\
1 &\text{for } t=0\\
\text{linear} &\text{for } t\in[-1,0],\;t\in[0,1]
\end{cases}
\]
 Fix a solution $x(t)$ of  \eqref{e1.2}. Then $h(t)x(t)$ is a
$\psi$-integrable  function on $\mathbb{R}$.
Set $y(t) =x(t)\int_0^th(s)ds$ , we have
 $$
y'(t) = A(t)x(t)\int_0^th(s)ds +h(t)x(t)  = A(t)y(t)+h(t)x(t).
$$
By hypothesis, the equation
$$
y'(t)=A(t)y(t)+h(t)x(t)
$$
has a solution $\widetilde y(t)$ on $\mathbb{ R}$,
$\psi$-evanescent at $\infty$.
Set $x_1(t) = \widetilde y(t)-y(t)+\frac{1}{2}x(t)$   and
$x_2(t)=  -\widetilde y(t)+y(t)+\frac{1}{2}x(t)$. It follows from
$\int_{-1}^0h(t)dt =\int_0^1h(t)dt=\frac{1}{2}$  that
$x_1(t) = \widetilde y(t)$  for $t \geqslant 1$;
$x_2(t) =-\widetilde y(t)$
for $t\leqslant  -1$. Then $x_2$ is the solution of \eqref{e1.2},
$\psi$-bounded on $\mathbb{R}_-$, $x_1$ is the solution
of  \eqref{e1.2} on $\mathbb{ R}_+$, $\psi$-evanescent at $\infty$.
The solution $x(t)$ is the sum of  two solutions $x_1(t)$
and $ x_2(t)$ of \eqref{e1.2}, these solutions satisfy the
conditions of Lemma.
The proof is complete.
\end{proof}

\section{the main results}

\begin{theorem} \label{thm3.1}
Suppose that $f$ is a $\psi$-integrable function on $\mathbb{R}_+$.
Then  \eqref{e1.1} has at least one solution on
$\mathbb{ R}_+$, $\psi$-evanescent at $\infty$ if and only
if  \eqref{e1.2} has a $\psi$-ordinary dichotomy on $\mathbb{R}_+$.
\end{theorem}

\begin{proof}
First, we prove the ``if'' part.
By Lemma \ref{lem2.3}, we can consider \eqref{e1.2} has a $\psi$-ordinary
dichotomy on $\mathbb{R}_+$ with two supplementary projections
$P_1,P_2$ such that $P_1\mathbb{R}^d =X_o$. Let
$$
g(t) = \int_0^tY(t)P_1Y^{-1}(s)f(s)ds
-\int_t^{\infty}Y(t)P_2Y^{-1}(s)f(s)ds.
$$
It is easy to see that $g(x)$ is a solution of \eqref{e1.1} on
$\mathbb{ R}_+$. We shall prove that $g(x)$ is $\psi$-evanescent
at $\infty$ on $\mathbb{ R}_+$. Since $f$ is  $\psi$-integrable
on $\mathbb{R}_+$, it follows that for a given  $\varepsilon>0$,
there exists $T > 0$ such that
 $$
(K+L)\int_T^{\infty} \| \psi(s)f(s)\|ds<\varepsilon/2.
$$
By
$P_1\mathbb{R}^d=X_o$, there exists $t_1>T$ such that, for $t\geqslant t_1$,
$$
|\psi(t)Y(t)P_1|\int_0^T\|Y^{-1}(s)f(s)\|ds<\varepsilon/2.
$$
Then for $t\geqslant t_1$, we have
 \begin{align*}
\| \psi(t)g(t)\|
&\leqslant \int_0^T| \psi(t)Y(t)P_1|. \|Y^{-1}(s)f(s)\|ds\\
& \quad+\int_T^t| \psi(t)Y(t)P_1Y^{-1}(s)\psi^{-1}(s)|.\|\psi(s)f(s)\|ds\\
& \quad+ \int ^{\infty}_t| \psi(t)Y(t)P_2Y^{-1}(s)\psi^{-1}(s)|.
 \| \psi(s)f(s)\|ds \\
 &\leqslant  |\psi(t)Y(t)P_1|\int_0^T\|Y^{-1}(s)f(s)\|ds
 +(K+L)\int_T^{\infty}\|\psi(s)f(s)\|ds\\
& < \varepsilon/2+\varepsilon/2=\varepsilon
\end{align*}
This shows that $g(x)$ is $\psi$-evanescent at $\infty$.
The ``only if'' part evidently holds, by Theorem \ref{thm1.4}(a).
\end{proof}

  Similarly, we have the following Theorem.

\begin{theorem} \label{thm3.2}
Suppose that $f$ is a $\psi$-integrable function on $\mathbb{R}_-$.
Then  \eqref{e1.1} has at least one solution on $\mathbb{ R}_-$,
$\psi$-evanescent at $-\infty$ if and only if  \eqref{e1.2}
has a $\psi$-ordinary dichotomy on $\mathbb{R}_-$.
\end{theorem}

\begin{theorem} \label{thm3.3}
 Suppose that \eqref{e1.2} has a $\psi$-ordinary dichotomy on
$\mathbb{R}_+$ and $f$ is a $\psi$-integrable function on
$\mathbb{R}_+$. Then  following statements are equivalent
\begin{itemize}
\item[(a)] every $\psi$-bounded solution of \eqref{e1.2} on
 $\mathbb{R}_+$ is  $\psi$- evanescent at $\infty$.

\item[(b)] every  $\psi$-bounded solution of \eqref{e1.1} on
 $\mathbb{R}_+$ is  $\psi$-evanescent at $\infty$.
\end{itemize}
\end{theorem}

\begin{proof}
By Lemma \ref{lem2.3}, we consider \eqref{e1.2} has a $\psi$-ordinary
dichotomy on $\mathbb{R}_+$ with two supplementary projections
$P_1,P_2$ such that $P_1\mathbb{R}^d =X_o$.
 Let $S_1$ be the set of all  $\psi$-bounded solutions
of \eqref{e1.1} on $\mathbb{R}_+$ and let $S_2$ be the set of
all $\psi$-bounded solutions of \eqref{e1.2} on $\mathbb{R}_+$.
We establish a mapping $h$ from $S_2$ to $S_1$:
$$
(hx)(t) = x(t) + g(t),
$$
 where $g(t)$ as in the proof of Theorem \ref{thm3.1}. We obtain
$$
\lim_{t\to\infty}\|\psi(t)(hx)(t)-\psi(t)x(t)\|
=\lim_{t\to\infty}\|\psi(t)g(t)\|=0
$$
Thus $h(x)$ is  $\psi$-bounded on $\mathbb{R}_+$. Hence $h(x)$
belongs to $S_1$. It is easily to verify that $h$ is  one-to-one
 mapping  between $S_2$ and $S_1$.

 Suppose that statement (a) is satisfied. Let $z$ be arbitrary
$\psi$-bounded solution of \eqref{e1.1} on $\mathbb{R}_+$.
The foregoing follow that there exists $\psi$-bounded solution
$x$ of \eqref{e1.2} on $\mathbb{R}_+$ such that $h(x)=z$ and
$$
\lim_{t\to\infty}\|\psi(t)z(t)-\psi(t)x(t)\|=0
$$
By hypothesis, $x$ is $\psi$-evanescent at $\infty$.
Thus $z$ is $\psi$-evanescent at $\infty$.
 Suppose that statement (b) is satisfied, the  proof is similarly.
 The proof is complete.
\end{proof}

  Note that the above Theorem is a supplement to
Theorem \ref{thm1.4}(b).
 Similarly, we have the following Theorem.

\begin{theorem} \label{thm3.4}
 Suppose that \eqref{e1.2} has a $\psi$-ordinary dichotomy on
$\mathbb{R}_-$ and $f$ is a $\psi$-integrable function on
$\mathbb{R}_-$. Then  following statements are equivalent
\begin{itemize}
\item[(a)] every $\psi$-bounded solution of \eqref{e1.2} on $\mathbb{R}_-$ is  $\psi$- evanescent at $-\infty$. \par

\item[(b)] every  $\psi$-bounded solution of \eqref{e1.1} on
$\mathbb{R}_-$ is  $\psi$-evanescent at $-\infty$.
\end{itemize}
\end{theorem}

\begin{corollary}\label{coro3.5}
 Suppose that \eqref{e1.2} has a $\psi$-ordinary dichotomy on
$\mathbb{R}$ and $f$ is a $\psi$-integrable function on
$\mathbb{R}$. Then  following statements are equivalent
\begin{itemize}
\item[(a)] every $\psi$-bounded solution of \eqref{e1.2} on
 $\mathbb{R}_+$ is  $\psi$- evanescent at $\infty$ and  every
$\psi$-bounded solution of \eqref{e1.2} on $\mathbb{R}_-$ is
$\psi$- evanescent at $-\infty$.

\item[(b)] every  $\psi$-bounded solution of $\eqref{e1.1}$ on
$\mathbb{R}$ is  $\psi$-evanescent at $ \pm\infty$.
\end{itemize}
\end{corollary}

  Note that the above corollary is a supplement to
\cite[Theorem 3.3]{d4}.


\begin{theorem} \label{thm3.6}
Suppose that \eqref{e1.2} has no non-trivial solution on
$\mathbb{R}$, $\psi$-evanescent  at $\infty$. Then \eqref{e1.1}
has a unique solution on $\mathbb{ R}$, $\psi$-evanescent  at
$\infty$  for every $\psi$-integrable function f  on
 $\mathbb{ R}$ if and only if \eqref{e1.2} has a
 $\psi$-ordinary dichotomy on $\mathbb{ R}$.
\end{theorem}

\begin{proof}
 First, we prove the ``if'' part. By Lemma \ref{lem2.3}, we can consider
\eqref{e1.2} has a $\psi$-ordinary dichotomy on $\mathbb{R}_+$
with two supplementary projections $P_1,P_2$ such that
$P_1\mathbb{R}^d =X_o$. Let
 $$
x(t) = \int_{-\infty}^tY(t)P_1Y^{-1}(s)f(s)ds
-\int_t^{\infty}Y(t)P_2Y^{-1}(s)f(s)
$$
 Then the function $x(t)$ is a $\psi$-bounded solution of
\eqref{e1.1} on $\mathbb{R}$. We shall prove that $x(t)$ is
$\psi$-evanescent at $\infty$. We have, for $t > 0$,
 $$
\psi(t)x(t) = \psi(t)Y(t)P_1\int_{-\infty}^0P_1Y^{-1}(s)f(s)ds+\psi(t)g(t),
$$
where $g(t)$ as in the proof of Theorem \ref{thm3.1}.
Since
$$
\|P_1Y^{-1}(s)f(s)\|\leqslant|Y^{-1}(0)|.|\psi^{-1}(0)|.
|\psi(0)Y(0)P_1Y^{-1}(s)\psi^{-1}(s)|.\|\psi(s)f(s)\|
$$
and $f$ is $\psi$-integrable on $\mathbb{R}$, we have that
$P_1Y^{-1}(s)f(s)$ is integrable on $\mathbb{R}_-$.
Let $a=\int_{-\infty}^0P_1Y^{-1}(s)f(s)ds$. It follows from
$P_1\mathbb{R}^d=X_0$ that
$$
\lim_{t\to\infty}\|\psi(t)Y(t)P_1a\|=0.
$$
On the other hand, as in the proof of Theorem \ref{thm3.1},
we have
$$\lim_{t\to\infty}\|\psi(t)g(t)\|=0.
$$
Consequently $x(t)$ is defined on $\mathbb{ R}$,
$\psi$-evanescent at $\infty$.
The uniqueness of solution $x(t)$ result from \eqref{e1.2}
has no non-trivial on $\mathbb{ R}$, $\psi$-evanescent solution
at $\infty$. Indeed, suppose that y is a solution on $\mathbb{ R}$
of  \eqref{e1.1}, $\psi$-evanescent at $\infty$  then $x - y$
is a solution solution on $\mathbb{ R}$ of \eqref{e1.2},
$\psi$-evanescent at $\infty$. We conclude $x = y$ since
 $x - y$ is the trivial solution of \eqref{e1.2}.

  Now, we prove the ``only if'' part. Suppose that
\eqref{e1.1} has a unique $\psi$-bounded solution on $\mathbb{R}$
for every $\psi$- integrable function $f$ on $\mathbb{R}$.
For each $u \in\mathbb{R}^d$, denote by $ x=x(t)$ the solution
of \eqref{e1.2}, $x(0) =u$. By Lemma \ref{lem2.6}, we get $x =x_1 + x_2$,
where $x_2$ is a $\psi$-bounded solution of \eqref{e1.2}
on $\mathbb{R}_-$, $x_1$ is a solutions of \eqref{e1.2}
on $\mathbb{ R}_+$ and $\psi$-evanescent at $\infty$.
Thus $x_1(0) \in X_0$ and $x_2(0) \in \widetilde{X}_1$.
 It follows from $u = x_1(0) + x_2(0)$ that
\begin{equation} \label{e3.1}
\mathbb{R}^d =X_0 + \widetilde{X}_1 .
\end{equation}
 For any $v \in X_0\cap \widetilde{X}_1$ , denote by $x(t)$
the solution of \eqref{e1.2} such that $x(0) = v$.
Thus $x(t)$  is a solution on $\mathbb{ R}$ of \eqref{e1.2},
$\psi$-evanescent at $\infty$.
By hypothesis,  \eqref{e1.2} has no non-trivial solution on
$\mathbb{R}$, $\psi$-evanescent at $\infty$, then $x(t)$
is the trivial solution. This implies $v = 0$. Consequently
\begin{equation} \label{e3.2}
X_0\cap \widetilde{X}_1 =0
\end{equation}
    The relations \eqref{e3.1} and \eqref{e3.2} imply that
$\mathbb{R}^d$  is the direct sum of $X_0$ and $\widetilde{X}_1$.
 Every $\psi$-integrable function $f$ on $\mathbb{R}_+$,
or on $\mathbb{R}_- $ is the restriction of a $\psi$-integrable
function $f$ on $\mathbb{R}$, it follows that \eqref{e1.2}
satisfies Theorem \ref{thm1.4}(a) and Lemma \ref{lem2.4}(a). Hence \eqref{e1.2}
has a $\psi$-ordinary dichotomy on $\mathbb{R}_+$ and has
a $\psi$-ordinary dichotomy on  $\mathbb{R}_ -$.
Let $P_1,P_2$ be two projections  such that
$\operatorname{Im}P_1=X_0$, $\operatorname{Im}P_2=\widetilde{X}_1$.
Lemmas \ref{lem2.3} and \ref{lem2.4}(b) follow that \eqref{e1.2} has a
$\psi$-ordinary dichotomy on $\mathbb{R}_+$ and has a
$\psi$-ordinary dichotomy on $\mathbb{R}_-$ with  two
supplementary projections $P_1,P_2$. Remark \ref{rmk1.3}
  follows that \eqref{e1.2} has a $\psi$-ordinary dichotomy on
$\mathbb{R}$ with two supplementary projections $P_1, P_2$.
 The proof  is complete.
\end{proof}

 Similarly, we have the following Theorem.

\begin{theorem} \label{thm3.7}
Suppose that \eqref{e1.2} has no non-trivial solution on
$\mathbb{R}$, $\psi$-evanescent  at $-\infty$. Then \eqref{e1.1}
 has a unique solution on $\mathbb{ R}$, $\psi$-evanescent
 at $-\infty$  for every $\psi$-integrable function $f$
on $\mathbb{ R}$ if and only if \eqref{e1.2} has a
$\psi$-ordinary dichotomy on $\mathbb{ R}$.
\end{theorem}

Now, consider the equations
\begin{gather}
{x'}(t)=[A(t)+B(t)]x(t), \label{e3.3}\\
{x'}(t)=[A(t)+B(t)]x(t)+f(t)\label{e3.4}
\end{gather}
where $B(t)$ is a $d\times d$ continuous matrix function on
 $\mathbb{R}_{+}$ and  $f$ is a $\psi$-integrable function
on $\mathbb{R}_+$.
We have the following result.

\begin{theorem} \label{thm3.8}
Suppose that  \eqref{e1.2} has a $\psi$-ordinary dichotomy on
$\mathbb{R}_+$.
If $\delta=\sup_{t\geqslant0}|\psi(t)B(t)\psi^{-1}(t)|$ is
sufficiently small, then following statements are equivalent
\begin{itemize}
\item[(a)] every $\psi$-bounded solution of \eqref{e3.3} on
$\mathbb{R}_+$ is  $\psi$- evanescent at $+\infty$.

\item[(b)] every  $\psi$-bounded solution of \eqref{e3.4}
 on $\mathbb{R}_+$ is  $\psi$-evanescent at $+\infty$.
\end{itemize}
\end{theorem}

\begin{proof}
 By \cite[Theorem 3.7]{b1}, equation \eqref{e3.3} has a
$\psi$-ordinary dichotomy on $\mathbb{R}_+$. By Theorem \ref{thm3.3},
 we have the conclusion.
\end{proof}

With similar  proof, we can conclude that $J=\mathbb{R}_-$.

\begin{theorem} \label{thm3.9}
Suppose that  \eqref{e1.2} has a $\psi$-ordinary dichotomy on
$\mathbb{R}_-$ and
$\delta=\sup_{t\leqslant0}|\psi(t)B(t)\psi^{-1}(t)|$ is
sufficiently small. Then the following statements are equivalent
\begin{itemize}
\item[(a)] every $\psi$-bounded solution of \eqref{e3.3}
on $\mathbb{R}_-$ is  $\psi$- evanescent at $-\infty$.

\item[(b)] every  $\psi$-bounded solution of \eqref{e3.4}
on $\mathbb{R}_-$ is  $\psi$-evanescent at $-\infty$.
\end{itemize}
\end{theorem}

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\end{document}