\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 $$\sum_{k=0}^n p_k(t) \frac{d^{n-k}x(t)}{dt^{n-k}}=F(t, x)\quad (t>0,\;x=x(t)) \label{e1.1}$$ with the initial conditions $$x^{(j)}(0)=x_j\in \mathbb{R} \quad (j=0, \dots, n-1). \label{e1.2}$$ 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 $$\sum_{k=0}^n p_k(t) \frac{d^{n-k}v(t)}{dt^{n-k}}=0\quad (t>0) \label{e1.3}$$ 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 $$\rho_k(t)\geq -\mu \;\;(t\geq 0;\; k=1,\dots, n) \label{e1.4}$$ for some $\mu>0$. Put $$q_m(t)=\sum_{k=0}^m p_k(t) C_{n-k}^{m-k} (-1)^k\mu^{m-k}\quad (m=1, \dots, n ),\; q_0\equiv 1. \label{e1.5}$$ 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 $$\tilde Q(z):=\sum_{k=0}^n (-1)^k d_k z^{n-k} \label{e1.6}$$ be real and nonnegative. In addition, let $$F(y, t)\geq 0\quad (y, t\geq 0). \label{e1.7}$$ 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 $$\frac{d^{2}x}{dt^{2}}+p_1(t)\frac{dx}{dt}+p_2(t)x=F(t,x)\quad (t>0). \label{e1.9}$$ 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 $$\sum_{k=0}^n (-1)^k a_k(t) \frac{d^{n-k}u(t)}{dt^{n-k}}=0\quad (t>0). \label{e2.1}$$ 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 $$u^{(j)}(0)=0, j=0, \dots, n-2;\;u^{(n-1)}(0)=1 \label{e2.2}$$ 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 $$\sum_{k=0}^n (-1)^k c_k \frac{d^{n-k}u}{dt^{n-k}}=\sum_1^n b_k(t) \frac{d^{n-k}u}{dt^{n-k}}\,. \label{e2.3}$$ 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 $$\sum_{k=0}^n (-1)^k c_k\frac{d^{n-k}w(t)}{dt^{n-k}} =0. \label{e2.4}$$ Put $$y(t)\equiv \sum_{k=0}^n c_k (-1)^k\frac{d^{n-k}u(t)}{dt^{n-k}}. \label{e2.5}$$ 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, $$G^{(j)}(t)= \sum_{k=0}^j \frac{j!e^{\theta t}\theta^{j-k} t^{n-1-k}}{(j-k)!(n-1-k)!k!} \geq \sum_{k=0}^j \frac{j! e^{r_1 t}r_1^{j-k} t^{n-1-k}}{(j-k)!(n-1-k)!k!}\geq 0\,. \label{e2.6}$$ According to the initial conditions \eqref{e2.2}, we can write $w(t)=G(t)$. So $$u(t) = G(t)+ \int_0^t G(t-s)y(s)\,ds\,. \label{e2.7}$$ 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{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} 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 $$P(t, \frac{d}{dt}-\mu)u\equiv \sum_{k=0}^n p_{k}(t)(\frac{d}{dt} -\mu)^{n-k}u=0\,. \label{e3.1}$$ 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, $$\tilde W(t, \tau)\geq e^{(-\mu+\tilde r_1) (t-\tau)} \frac{(t-\tau)^{n-1}}{(n-1)!}\quad (t> \tau\geq 0). \label{e3.2}$$ \end{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} \begin{thebibliography}{00} \bibitem{ag} Agarwal, R., O'Regan, D. and Wong, P. J. Y.; {\em Positive solutions of differential, difference and integral equations.} Kluwer Academic Publishers, Dordrecht (1999). \bibitem{ag1} Agarwal, R. P.; positive solutions of nonlinear problems. A tribute to the academician M. A. Krasnosel'skij, {\em Comput. Appl. Math.} {\bf 88} (1998), No.1, 238 p. \bibitem{el} Eloe, P.W. and J. Henderson; Positive solutions for higher order ordinary differential equations. {\em Electron. J. Differ. Equ.} {\bf 1995} (1995) No. 03, 1-?. \bibitem{gi83} Gil', M. 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