\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 $$X_0 \subset Q_1\mathbb{R}^d \subset X_1 \label{e2.5}$$ \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 $$\|\psi(t)x(t)\| \to 0, \text{ as } t\to \infty \label{e2.6}$$ From Remark \ref{rmk2.2}, we have $$\|\psi(t)x(t)\|=\|\psi(t) Y(t)u\|\geqslant {L}_Q\|Q_2u\|,\quad \text{ for } t\geqslant 0 \label{e2.7}$$ 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 $$\|\psi(t)Y(t)u\| \geqslant N\|u\|, \quad \text{for all } u\in X_0',\; t\geqslant 0 \label{e2.8}$$ 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{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} 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 $$N\|(P_1-\widetilde{Q}_1)u\| \leqslant \|\psi(s)Y(s)u\| \quad \text{for } s\geqslant0 \label{e2.10}$$ From Remark \ref{rmk2.2} and \eqref{e2.10}, we have $$\|\psi(t)Y(t)(P_1-\widetilde{Q}_1)u\| \leqslant K_P\|(P_1-\widetilde{Q}_1)u\| \leqslant K_PN^{-1}\|\psi(s)Y(s)u\|\quad \text{ for } t,s \geqslant 0 \label{e2.11}$$ Consequently, \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} From $\widetilde{Q}_2=P_2+P_1-\widetilde{Q}_1$ and \eqref{e2.11}, we obtain \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} 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 $$Q_1\widetilde{Q}_2=Q_1(I_d-\widetilde{Q}_1)=Q_1-\widetilde{Q}_1 \label{e2.14}$$ By the definition of $X_1$, there exists $N' >0$ such that $$\|\psi(t)Y(t)u\| \leqslant N'\|u\| \text{, for } t \geqslant 0 \label{e2.15}$$ 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{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} where $K_2$ is a positive constant. From \eqref{e2.12}, \eqref{e2.16}, we have \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} From $Q_2=\widetilde{Q}_2+\widetilde{Q}_1-Q_1$, \eqref{e2.13} and \eqref{e2.16}, we obtain \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} 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 $$\widetilde{X}_0 \subset \widetilde{Q}_2\mathbb{R}^d \subset \widetilde{X}_1 \label{e2.19}$$ \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 $$\label{e3.1} \mathbb{R}^d =X_0 + \widetilde{X}_1 .$$ 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 $$\label{e3.2} X_0\cap \widetilde{X}_1 =0$$ 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. 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