\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2004(2004), No. 63, pp. 1--6.\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 2004 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2004/63\hfil Existence of $\psi$-bounded solutions] {Existence of $\Psi$-bounded solutions for a system of differential equations} \author[A. Diamandescu\hfil EJDE-2004/63\hfilneg] {Aurel Diamandescu} \address{Department of Applied Mathematics, University of Craiova, 13 Al. I. Cuza'' st., 200.585 Craiova, Rom\^{a}nia} \email{adiamandescu@central.ucv.ro} \date{} \thanks{Submitted March 5, 2004. Published April 23, 2004.} \subjclass[2000]{34D05, 34C11} \keywords{$\Psi$-bounded, Lebesgue $\Psi$-integrable function} \begin{abstract} In this article, we present a necessary and sufficient condition for the existence of solutions to the linear nonhomogeneous system $x'=A(t)x + f(t)$. Under the condition stated, for every Lebesgue $\Psi$-integrable function $f$ there is at least one $\Psi$-bounded solution on the interval $(0,+\infty)$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \section{Introduction} We give a necessary and sufficient condition for the nonhomogeneous system $$x' = A(t)x + f(t) \label{1}$$ to have at least one $\Psi$-bounded solution for every Lebesgue $\Psi$-integrable function $f$, on the interval $\mathbb{R}_{+} = [0,+\infty )$. Here $\Psi$ is a continuous matrix function, instead of a scalar function, which allows a mixed asymptotic behavior of the components of the solution. The problem of $\Psi$-boundedness of the solutions for systems of ordinary differential equations has been studied by many authors; see for example Akinyele \cite{a1}, Constantin \cite{c1}, Avramescu \cite{a2}, Hallam \cite {h1}, and Morchalo \cite{m1}. In these papers, the function $\Psi$ is a scalar continuous function: Increasing, differentiable, and bounded in \cite {a1}; nondecreasing with $\Psi (t) \geq 1$ on $\mathbb{R}_{+}$ in \cite{c1}). Let $\mathbb{R}^d$ be the Euclidean $d$-space. Elements in this space are denoted by $x = (x_1, x_2,\dots x_d)^T$ and their norm by $\|x\| =\max \{ |x_1|,|x_2|,\dots |x_d|\}$. For $d\times d$ real matrices, we define the norm $|A| = \sup_{\|x\| \leq 1}\|Ax\|$. Let $\Psi _{i}:\mathbb{R}_{+}\to (0,\infty )$, $i=1,2,\dots d$, be continuous functions, and let \begin{equation*} \Psi =\mathop{\rm diag} [\Psi _1,\Psi _2,\dots \Psi _d]. \end{equation*} Then the matrix $\Psi (t)$ is invertible for each $t\geq 0$. \smallskip \noindent \textbf{Definition.} % 1.1. A function $\varphi :\mathbb{R}_{+}\to R^{d}$ is said to be $\Psi$-bounded on $\mathbb{R}_{+}$ if $\Psi (t)\varphi (t)$ is bounded on $\mathbb{R}_{+}$. \smallskip \noindent \textbf{Definition.} % 1.2. A function $\varphi : \mathbb{R}_+ \to R^d$ is said to be Lebesgue $\Psi$% -integrable on $\mathbb{R}_+$ if $\varphi (t)$ is measurable and $\Psi (t)\varphi(t)$ is Lebesgue integrable on $\mathbb{R}_+$. \smallskip By a solution of \eqref{1}, we mean an absolutely continuous function satisfying the system for almost all $t \geq 0$. Let $A$ be a continuous $d\times d$ real matrix and the associated linear differential system be $$y'= A(t)y. \label{2}$$ Also let $Y$ be the fundamental matrix of \eqref{2} with $Y(0) = I_d$, the identity $d\times d$ matrix. Let $X_1$ denote the subspace of $\mathbb{R}^d$ consisting of all vectors which are values of $\Psi$-bounded solutions of \eqref{2} at $t = 0$. Let $% X_2$ be an arbitrary closed subspace of $\mathbb{R}^d$, supplementary to $% X_1$. Let $P_1$, $P_2$ denote the corresponding projections of $\mathbb{R}^d$ onto $X_1$, X$_2$. \section{The Main Results} In this section, we give the main results of this Note. \begin{theorem} \label{thm2.1} If $A$ is a continuous $d\times d$ real matrix, then \eqref{1} has at least one $\Psi$-bounded solution on $\mathbb{R}_+$ for every Lebesgue $\Psi$-integrable function $f$ on $\mathbb{R}_+$ if and only if there is a positive constant $K$ such that $$\label{3} \begin{gathered} |\Psi (t)Y(t)P_1 Y^{-1}(s)\Psi ^{-1}(s)|\leq K,\quad\text{for } 0 \leq s \leq t, \\ |\Psi (t)Y(t)P_2 Y^{-1}(s)\Psi ^{-1}(s)|\leq K,\quad\text{for }0 \leq t \leq s. \end{gathered}$$ \end{theorem} \begin{proof} First, we prove the only if'' part. We define the sets: \\ $C_{\Psi} = \{ x : \mathbb{R}_+ \to \mathbb{R}^d : x$ is $\Psi$-bounded and continuous on $\mathbb{R}_+ \}$, \\ $B = \{ x : \mathbb{R}_+ \to \mathbb{R}^d : x \mbox{ is Lebesgue$\Psi$-integrable on }\mathbb{R}_+ \}$, \\ $D = \{ x : \mathbb{R}_+ \to \mathbb{R}^d : x$ is absolutely continuous on all intervals $J \subset \mathbb{R}_+$, $\Psi$-bounded on $\mathbb{R}_+$, $x(0)$ in $X_2$, $x'(t) - A(t)x(t)$ in $B \}$. It is well-known that $C_\Psi$ is a real Banach space with the norm $\| x\| _{C_\Psi } = \sup_{t\geq 0} \| \Psi (t)x(t)\| .$ Also, it is well-known that $B$ is a real Banach space with the norm $\| x\| _B=\int_0^\infty \| \Psi(t)x(t)\| dt.$ The set $D$ is obviously a real linear space and \begin{equation*} \| x\| _D=\underset{t\geq 0}{\sup }\| \Psi(t)x(t)\| +\| x' - A(t)x \| _{B} \end{equation*} is a norm on $D$. Now, we show that ($D, \|\cdot\|_D$) is a Banach space. Let $(x_n)_n$ be a fundamental sequence in $D$. Then, $(x_n)_n$ is a fundamental sequence in $C_\Psi$. Therefore, there exists a continuous and bounded function $x : \mathbb{R}_+ \to \mathbb{R}^d$ such that \begin{equation*} \lim_{n\to \infty } \Psi (t)x_n (t) = x(t), \quad\mbox{uniformly on } \mathbb{R}_+. \end{equation*} Denote $\bar{x}(t) = \Psi ^{-1}(t)x(t) \in C_\Psi$. From \begin{equation*} \|x_n(t) - \bar{x}(t)\| \leq |\Psi ^{-1}(t)|\| \Psi (t)x_n(t) - x(t) \| , \end{equation*} it follows that $\lim_{n\to \infty } x_n(t) = \bar{x}(t)$, uniformly on every compact of $\mathbb{R}_{+}$. Thus, $\bar{x}(0) \in X_2$. On the other hand, ($f_n(t)$), where $f_n(t) = \Psi (t)(x_n'(t) - A(t)x_n(t))$, is a fundamental sequence in $L$, the Banach space of all vector functions which are Lebesgue integrable on $\mathbb{R}_+$ with the norm \begin{equation*} \| f\| =\int_0^{\infty }\| \Psi (t)f(t)\| dt. \end{equation*} Thus, there is a function $f$ in $L$ such that \begin{equation*} \lim_{n\to \infty }\int_0^{\infty }\|f_n(t) - f(t)\| dt = 0. \end{equation*} Putting $\bar{f}(t) = \Psi ^{-1}(t)f(t)$, it follows that $\bar{f}(t) \in B$ For a fixed, but arbitrary, $t \geq 0$, we have \begin{align*} \bar{x}(t) - \bar{x}(0) &= \lim_{n\to \infty } (x_n(t) - x_n(0)) \\ &= \lim_{n\to \infty} \int_0^t x_n'(s)ds \\ &= \lim_{n\to \infty } \int_0^t [ (x_n'(s) - A(s)x_n(s) ) +A(s)x_n(s) ]ds \\ &= \lim_{n\to\infty}\int_0^t \{\Psi ^{-1}(s)[ f_n(s) - f(s) ] + \bar{f}(s) + A(s)x_n(s) \}ds \\ &=\int_0^t [ \bar{f}(s) + A(s)\bar{x}(s) ]ds. \end{align*} It follows that $\bar{x}'(t) - A(t)\bar{x}(t) = \bar{f}(t) \in B$ and $\bar{x}(t)$ is absolutely continuous on all intervals $J \subset \mathbb{R}_+$. Thus, $\bar{x}(t) \in D$. From $\lim_{n\to \infty } \Psi (t)x_n (t) = \Psi (t)\bar{x}(t)$, uniformly on $\mathbb{R}_{+}$ and \begin{equation*} \lim_{n\to\infty}\int_0^{\infty }\| \Psi (t)[( x_n'(t) - A(t)x_n (t) ) - ( \bar{x}'(t) - A(t)\bar{x}(t) )]\| dt = 0 , \end{equation*} it follows that $\lim_{n\to\infty}\| x_n-\bar{x} \| _D= 0$. Thus, $(D,\| \cdot \| _D)$ is a Banach space. Now, we define \begin{equation*} T: D\to B, \quad Tx = x' - A(t)x. \end{equation*} Clearly, $T$ is linear and bounded, with $\| T \| \leq 1$. Let $Tx = 0$. Then, $x'= A(t)x, x \in D$. This shows that $x$ is a $\Psi$-bounded solution of \eqref{2}. Then, $x(0) \in X_1 \cap X_2 = \{0\}$. Thus, $x = 0$, such that the operator $T$ is one-to-one. Now, let $f \in B$ and let $x(t)$ be the $\Psi$-bounded solution of the system \eqref{1}. Let $z(t)$ be the solution of the Cauchy problem \begin{equation*} z'= A(t)z + f(t), \quad z(0) = P_2x(0). \end{equation*} Then, $x(t) - z(t)$ is a solution of \eqref{2} with $P_2(x(0) - z(0)) = 0$, i.e. $x(0) - z(0)\in X_1$. It follows that $x(t) - z(t)$ is $\Psi$-bounded on $R_{+}$. Thus, $z(t)$ is $\Psi$-bounded on $\mathbb{R}_{+}$. It follows that $z(t) \in D$ and $Tz = f$. Consequently, the operator $T$ is onto. From a fundamental result of Banach: If $T$ is a bounded one-to-one linear operator from Banach space onto another, then the inverse operator $T^{-1}$ is also bounded, we have that there is a positive constant $K = \| T^{-1}\| - 1$ such that, for $f\in B$ and for the solution $x \in D$ of \eqref{1}, \begin{equation*} \sup_{t\geq 0}\| \Psi(t)x(t)\| \leq K\int_0^{\infty } \| \Psi (t)f(t)\| . \end{equation*} For $s \geq 0$, $\delta > 0$, $\xi \in \mathbb{R}^d$, we consider the function $f : \mathbb{R}_+ \to \mathbb{R}^d$, \begin{equation*} f(t) =\begin{cases} \Psi ^{-1}(t)\xi ,&\text{for } s \leq t \leq s + \delta \\ 0, &\text{elsewhere.} \end{cases} \end{equation*} Then, $f \in B$ and $\|f \| _B = \delta \| \xi \|$. The corresponding solution $x \in D$ is \begin{equation*} x(t) = \int_s^{s + \delta } G(t,u)du, \end{equation*} where \begin{equation*} G(t,u) = \begin{cases} Y(t)P_1 Y^{-1}(u), &\text{for } 0 \leq u \leq t \\ -Y(t)P_2Y^{-1}(u), &\text{for } 0 \leq t \leq u. \end{cases} \end{equation*} Clearly, $G$ is continuous except on the line $t = u$, where it has a jump discontinuity. Therefore, \begin{equation*} \| \Psi(t)x(t)\| =\| \int_s^{s + \delta }\Psi (t)G(t,u)\Psi ^{-1}(u)\xi du \| \leq K\delta \| \xi \| . \end{equation*} It follows that \begin{equation*} \| \Psi (t)G(t,s) \Psi ^{-1}(s) \xi \|\leq K\| \xi \| . \end{equation*} Hence, \begin{equation*} |\Psi (t)G(t,s)\Psi ^{-1}(s)|\leq K, \end{equation*} which is equivalent with \eqref{3}. By continuity, \eqref{3} remains true also in the case $t = s$. Now, we prove the if'' part. We consider the function \begin{equation*} x(t) = \int_0^t Y(t)P_1Y^{-1}(s)f(s)ds - \int_t^{\infty } Y(t)P_2Y^{-1}(s)f(s)ds, t \geq 0, \end{equation*} where $f$ is a Lebesgue $\Psi$-integrable function on $\mathbb{R}_+$ It is easy to see that $x(t)$ is a $\Psi$-bounded solution on $\mathbb{R}_+$ of \eqref{1}. The proof is now complete. \end{proof} \noindent\textbf{Remark.}\; %2.1 By taking $\Psi (t) = I_d$ in Theorem \ref{thm2.1}, the conclusion in \cite[Theorem 2, Chapter V]{c2} follows. \begin{theorem} \label{thm2.2} Suppose that: \begin{enumerate} \item The fundamental matrix $Y(t)$ of \eqref{2} satisfies the conditions: \begin{itemize} \item[(a)] $\lim_{t\to \infty }\Psi (t)Y(t)P_1 = 0$; \item[(b)] $|\Psi (t)Y(t)P_1Y^{-1}(s)\Psi ^{-1}(s)\leq K$, for $0 \leq s \leq t$,\\ $|\Psi (t)Y(t)P_2Y^{-1}(s)\Psi ^{-1}(s)| \leq K$, for $0 \leq t \leq s$, \end{itemize} where $K$ is a positive constant and $P_1$ and $P_2$ are as in the Introduction \item The function $f : \mathbb{R}_{+} \to \mathbb{R}^d$ is Lebesgue $\Psi$-integrable on $\mathbb{R}_{+}$. \end{enumerate} Then, every $\Psi$-bounded solution $x(t)$ of \eqref{1} is such that \begin{equation*} \lim_{t\to \infty }\| \Psi(t)x(t)\| =0. \end{equation*} \end{theorem} \begin{proof} Let $x(t)$ be a $\Psi$-bounded solution of \eqref{1}. There is a positive constant $M$ such that $\|\Psi (t)x(t) \| \leq M$, for all $t \geq 0$. We consider the function \begin{equation*} y(t) = x(t) - Y(t)P_1x(0) -\int_0^t Y(t)P_1 Y^{-1}(s)f(s)ds +\int_t^{\infty }Y(t)P_2Y^{-1}(s)f(s)ds \end{equation*} for all t $\geq$ 0. From the hypotheses, it follows that the function $y(t)$ is a $\Psi$-bounded solution of \eqref{2}. Then, $y(0) \in X_1$. On the other hand, $P_1y(0) = 0$. Therefore, $y(0) = P_2 y(0) \in X_2$. Thus, $y(0) = 0$ and then $y(t) = 0$ for $t\geq 0$. Thus, for $t \geq 0$ we have \begin{equation*} x(t) = Y(t)P_1 x(0) + \int_0^t Y(t)P_1 Y^{-1}(s)f(s)ds -\int_t^{\infty } Y(t)P_2 Y^{-1}(s)f(s)ds. \end{equation*} Now, for a given $\varepsilon >0$, there exists $t_1\geq 0$ such that \begin{equation*} \int_t^{\infty}\| \Psi (s)f(s)\| ds < \frac{\varepsilon }{2K},\quad \text{for }t \geq t_1. \end{equation*} Moreover, there exists $t_2>t_1$ such that, for $t \geq t_2$, \begin{equation*} |\Psi (t)Y(t)P_1|\leq \frac{\varepsilon }{2} \Big[ \| x(0)\| +\int_0^{t_1} \| Y^{-1}(s)f(s)\| ds\Big]^{-1}. \end{equation*} Then, for $t \geq t_2$ we have \begin{align*} \| \Psi(t)x(t)\| &\leq |\Psi (t)Y(t)P_1|\| x(0)\| +\int_0^{t_1}|\Psi (t)Y(t)P_1|\| Y^{-1}(s)f(s)\| ds\\ &\quad + \int_{t_1}^{t}|\Psi (t)Y(t)P_1 Y ^{-1}(s)\Psi ^{-1}(s)|\| \Psi (s)f(s)\| ds\\ &\quad +\int_{t}^{\infty }|\Psi (t)Y(t)P_2 Y^{-1}(s) \Psi ^{-1}(s)|\| \Psi (s)f(s)\| ds \\ &\leq |\Psi (t)Y(t)P_1|\Big[ \| x(0)\| +\int_0^{t_1}\| Y^{-1}(s)f(s)\| ds\Big] \\ &\quad + K\int_{t_1}^{\infty }\| \Psi (s)f(s)\| ds <\varepsilon . \end{align*} This shows that $\lim{t\to \infty }\|\Psi (t)x(t) \| =0$. The proof is now complete. \end{proof} \noindent\textbf{Remark.}% 2.2. Theorem \ref{thm2.2} generalizes a result in Constantin \cite{c1}. Note that Theorem \ref{thm2.2} is no longer true if we require that the function $f$ be $\Psi$-bounded on $\mathbb{R}_{+}$, instead of condition (2) of the Theorem. Even if the function $f$ is such that \begin{equation*} \lim {t\to \infty }\Vert \Psi (t)f(t)\Vert =0, \end{equation*} Theorem \ref{thm2.2} does not apply. This is shown by the next example. \noindent\textbf{Example.} % 2.1. Consider the linear system \eqref{2} with $A(t) = O_2$. Then $Y(t) = I_2$ is a fundamental matrix for \eqref{2}. Consider \begin{equation*} \Psi(t) = \begin{pmatrix} \frac{1}{t + 1} & 0 \\ 0 & t + 1 \end{pmatrix} \end{equation*} We have $\Psi (t)Y(t) = \Psi (t)$, such that \begin{equation*} P_1= \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} , \quad p_2= \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \end{equation*} It follows that the first hypothesis of the Theorem is satisfied with $K=1$. When we take $f(t) = (\sqrt{t + 1},( t + 1 )^{-2} )^T$, then $\lim_{t\to \infty }\|\Psi (t)f(t)\| = 0$. On the other hand, the solutions of the system \eqref{1} are \begin{equation*} x(t) = \begin{pmatrix} \frac{2}{3} ( t + 1 )^{3/2} + c_1 \\ -\frac{1}{t + 1} + c_2 \end{pmatrix} \end{equation*} It follows that the solutions of the system \eqref{1} are $\Psi$-unbounded on $\mathbb{R}_{+}$. \smallskip \noindent\textbf{Remark.}\; % 2.3.\ } When in the above example we consider \begin{equation*} f(t) = \big( ( t + 1 )^{-1} , (t+1)^{-3}\big)^T , \end{equation*} then we have \begin{equation*} \int_0^{\infty }\| \Psi (t)f(t)\| dt = 1. \end{equation*} On the other hand, the solutions of the system \eqref{1} are \begin{equation*} x(t) = \begin{pmatrix} \ln ( t + 1 ) + c_1 \\ -\frac{1}{2} ( t + 1 )^{-2} + c_2 \end{pmatrix} \end{equation*} It is easy to see that these solutions are $\Psi$-bounded on $\mathbb{R}_+$ if and only if $c_2 = 0$. In this case, $\lim_{t\to \infty }\| \Psi (t)x(t)\| = 0$. Note that the asymptotic properties of the components of the solutions are not the same. 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