\documentclass[twoside]{article} \usepackage{amsfonts, amsmath} % used for R in Real numbers \pagestyle{myheadings} \markboth{\hfil Nonlinear initial-value problems \hfil EJDE/Conf/10} {EJDE/Conf/10 \hfil John V. Baxley \& Cynthia G. Enloe \hfil} \begin{document} \setcounter{page}{71} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent Fifth Mississippi State Conference on Differential Equations and Computational Simulations, \newline Electronic Journal of Differential Equations, Conference 10, 2003, pp 71--78. \newline http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \vspace{\bigskipamount} \\ % Nonlinear initial-value problems with positive global solutions % \thanks{ {\em Mathematics Subject Classifications:} 34A12, 34B15. \hfil\break\indent {\em Key words:} Nonlinear initial-value problems, positive global solutions, Carath\'eodory. \hfil\break\indent \copyright 2003 Southwest Texas State University. \hfil\break\indent Published February 28, 2003. } } \date{} \author{John V. Baxley \& Cynthia G. Enloe} \maketitle \begin{abstract} We give conditions on $m(t)$, $p(t)$, and $f(t,y,z)$ so that the nonlinear initial-value problem \begin{gather*} \frac{1}{m(t)} (p(t)y')' + f(t,y,p(t)y') = 0,\quad\mbox{for }t>0,\\ y(0)=0,\quad \lim_{t \to 0^+} p(t)y'(t) = B, \end{gather*} has at least one positive solution for all $t>0$, when $B$ is a sufficiently small positive constant. We allow a singularity at $t=0$ so the solution $y'(t)$ may be unbounded near $t=0$. \end{abstract} \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \section{Introduction} We consider the initial-value problem \begin{gather} \label{ode} \frac{1}{m(t)}(p(t) y')' + f(t, y, p(t) y') = 0 , \quad t>0,\\ \label {ic} y(0)=0, \quad \lim_{t \to 0^+}p(t) y'(t) = B, \quad B > 0. \end{gather} We allow a singularity at $t=0$, and so $y'(t)$ may not be bounded near $t=0$. However, we require of a solution that it be continuous at $t=0$, satisfy (\ref{ode}) a.e. on some interval $(0,\delta)$, and satisfy (\ref{ic}). The singularity may be caused by the behavior of $m$ or $p$ or $f$ near $t=0$ or by some combination of them. This problem was considered earlier by Zhao \cite{Z} and by Maagli and Masmoudi \cite{MM}. In particular, \cite{Z} considered the case that $m \equiv p \equiv 1$ while \cite{MM} required that $m \equiv p$. In each of these papers, only one of the initial conditions ($y(0)=0$) was imposed and conditions were specified which guaranteed that this incomplete'' initial-value problem has infinitely many positive solutions existing on the entire interval $(0,\infty)$. Both papers viewed the problem as a boundary-value problem by imposing a condition at $\infty$, namely $$\label{infcond} \lim_{t \to \infty} \frac{y(t)}{r(t)} = c > 0 ,$$ where $r(t) = \int_0^t (p(s))^{-1} \, ds$. In \cite{Z}, $r(t)$ reduces to $r(t) = t$. In both \cite{MM} and \cite{Z}, the Schauder fixed point theorem is the main tool and the hypotheses imposed allow the authors to prove existence of at least one solution of the boundary-value problem for $c$ sufficiently small. Here, we shall treat the problem in the initial-value form (\ref{ode}), (\ref{ic}). We shall impose conditions rather close to those of \cite{MM} and \cite{Z}, and prove that our initial-value problem has at least one positive solution for $B$ sufficiently small. Our methods use initial-value techniques, similar to those used already in \cite{B1,B2}, and are completely different from the previous papers discussed above. To get started, we must have a local solution on some interval $(0,\delta)$ and for that purpose, we need a slight generalization of the classical theorem of Carath\'eodory \cite{CL}, which we provide in Section 2. In Section 3, we prove our main result, which we state below. Let $r(t) = \int_0^t (p(s))^{-1} \, ds$ and assume that \begin{itemize} \item[M1:] $p(t)$ and $m(t)$ are positive and continuous on $(0,\infty)$; \item[M2:] $\frac{1}{p(t)} \in L^1(0,1)$; \item[M3:] for some positive number, $D<\infty$, $f:(0,\infty) \times (0,Dr(\infty)) \times (0,D) \to {\mathbb R}$ is a measurable function on $(0,\infty) \times (0,Dr(\infty)) \times (0,D)$ and $f(t,\cdot,\cdot)$ is continuous on $(0,Dr(\infty)) \times (0,D)$ for each fixed $t \in (0,\infty)$; \item[M4:] $|f(t,y,z)| \leq h_1(t,y,z)y + h_2(t,y,z)z$ where $h_1(t,y,z) \to 0$ and $h_2(t,y,z) \to 0$ as $(y,z) \to (0,0)$, $h_1$ and $h_2$ are nonnegative, and for $\alpha > 0$, let $h(t,y,z) = h_1(t,y,z)r(t) + h_2(t,y,z)$, $g_\alpha(s) = \sup\{h(s,y,z): 0 < y < \alpha r(s), 0 < z < \alpha\}, s>0,$ and $m(s)g_\alpha(s) \in L^1(0,\infty)$ for sufficiently small $\alpha > 0$. \end{itemize} \begin{theorem} Under assumptions M1--M4, there exists $\gamma > 0$ so that $B \in (0,\gamma)$ implies that the initial-value problem (\ref{ode}), (\ref{ic}) has at least one solution existing for $0 < t < \infty$ and satisfying \begin{gather*} \frac{B}{2} < p(t) y'(t) < \frac{3B}{2} ,\\ \frac{B r(t)}{2} < y(t) < \frac{3Br(t)}{2} , \end{gather*} for $0 < t < \infty$. Moreover, the two limits $\lim_{t \to \infty} \frac{y(t)}{r(t)} , \quad \lim_{t \to \infty} p(t) y'(t)$ exist, and if $r(\infty) = \infty$, the two limits are equal. \end{theorem} Other than the fact that \cite{MM} requires that $m \equiv p$, the only substantive difference in our hypotheses is that we do not require that $h_1$, $h_2$ be nondecreasing with respect to $y$ and $z$, as they do. Of course, we prove existence for an initial-value problem, not a boundary-value problem as they do. The key to our proof is that our local existence theorem in Section 2 is formulated carefully to provide a lower bound on the length of the interval of existence. In applying it in Section 3, we show that this lower bound gives us a uniform lower bound on the length of the interval of existence, regardless of where in the interval $[0,\infty)$ we start the solution. Thus, we are able to step from $0$ to $\infty$ inductively, without fear that the sum of the lengths of our intervals will converge, to complete the proof. \section{Local Solutions} In this section, we consider the initial-value problem \begin{gather} \label{locode} \frac{1}{m(t)}(p(t) y')' + f(t, y, p(t) y') = 0 , \quad t>t_0,\\ \label{locic} y(t_0)=A, \quad \lim_{t \to t_0^+}p(t) y'(t) = B . \end{gather} We use $x_1 = y$, $x_2 = p(t) y'$ to transform to the two-dimensional system $$\label{system} \begin{gathered} x'_1 = \frac{x_2}{p(t)} \\ x'_2 = -m(t) f(t,x_1,x_2) \end{gathered}$$ with initial conditions $$\label{systemic} \lim_{t \to t_0^+} x_1 (t) = A, \quad \lim_{t \to t_0^+} x_2 (t) = B .$$ Let $R(t) = \int_{t_0}^t (p(s))^{-1} \, ds$. We shall assume that \begin{itemize} \item[L1:] There exists $b>t_0$ such that $p(t)$ and $m(t)$ are positive and continuous on $(t_0,b)$. \item[L2:] $\frac{1}{p(t)} \in L^1(t_0,b)$. \item[L3:] $f:S \to {\mathbb R}$, where $S = \{t_0 < t \leq b, A + cR(t) < y < A + dR(t), c < z < d\}$, and $f$ is measurable in $t$ for each fixed $(y,z)$ and continuous in $(y,z)$ for each fixed $t$. \item[L4:] There exists $h(t) \in L^1(t_0,b)$ such that $m(t)|f(t,y,z)| \leq h(t)$, almost everywhere on the set $S$. \end{itemize} We shall prove the following generalization of Carath\'eodory's local existence theorem. The proof follows the same general lines as the well-known proof in \cite{CL}. \begin{theorem} \label{cara} Suppose hypotheses L1-L4 are satisfied. Let $0 < d^* < \min\{d-B,B-c\}$ and suppose $\beta \in (0,b)$ and satisfies $\int_{t_0}^{t_0+\beta} h(s)ds < d^*$. Then, the initial-value problem (\ref{locode}), (\ref{locic}) has a solution existing on the interval $[t_0, t_0 + \beta]$ and satisfies \begin{gather*} A+cR(t) < x_1(t) < A+dR(t)\\ c < x_2(t) < d \end{gather*} for $t_0 < t \leq t_0 + \beta$. \end{theorem} \paragraph{Proof:} Choose a fixed integer $n >1$. Let $h_n = \beta/n$ and let $t_k = t_0 + k h_n$ for $k = 1,2,\cdots,n$. Define $u_{2,n} (t) = B, \quad \mbox{for} \quad t_0 \leq t \leq t_1 .$ Note that $B - d^* < u_{2,n} (t) < B + d^*$ for $t_0 \leq t \leq t_1$. Also define $u_{1,n} (t) = A + \int_{t_0}^t \frac{u_{2,n} (s)}{p(s)} \, ds, \quad \mbox{for} \quad t_0 \leq t \leq t_1 .$ It follows that $(B-d^*) R(t) < u_{1,n} (t) - A < (B+d^*) R(t)$ and so $A + (B-d^*) R(t) < u_{1,n} (t) < A + (B+d^*) R(t).$ Thus, $(t,u_{1,n} (t),u_{2,n} (t)) \in S$ for $t_0 \leq t \leq t_1$. We extend the pair $(u_{1,n} , u_{2,n} )$ to the entire interval $[t_0 ,t_0 + \beta]$ by recursively defining the pair on the subintervals $[t_{j-1} ,t_j]$. Thus, for each $j=2,3,\cdots,n$, we define \begin{gather*} u_{2,n} (t) = B - \int_{t_0}^{t-h_n} m(s) f(s,u_{1,n} (s),u_{2,n} (s)) \, ds, \quad \mbox{for} \quad t_{j-1} \leq t \leq t_j ,\\ u_{1,n} (t) = A + \int_{t_0}^t \frac{u_{2,n} (s)}{p(s)} \, ds, \quad \mbox{for} \quad t_{j-1} \leq t \leq t_j . \end{gather*} (The measurability of the integrand in the integral for $u_{2,n}$ follows from L3 by approximating with simple functions.) Using L4, we have \begin{eqnarray*} |u_{2,n}(t)-B| &\leq& \int_{t_0}^{t-h_n}m(s)|f(s,u_{1,n}(s),u_{2,n}(s))|ds \\ &\le& \int_{t_0}^{t_0 + \beta}h(s)ds < d^* , \end{eqnarray*} and therefore, $B-d^* < u_{2,n}(t) < B + d^*$. Further, $(B-d^*)R(t) < u_{1,n}(t) - A < (B+d^*)R(t),$ and so $A + (B-d^*)R(t) < u_{1,n}(t) < A + (B+d^*)R(t).$ These inequalities show that $(t,u_{1,n} (t),u_{2,n} (t))$ remains in $S$ on each subinterval and the recursive definition is allowed. Moreover, the two sequences $\{u_{1,n}\}$, $\{u_{2,n}\}$ are uniformly bounded on $t_0 \leq t \leq t_0 + \beta$. We shall show that these sequences are equicontinuous so that Ascoli's theorem may be applied. Suppose $t_0 \leq t \leq t^* \leq t_0 + \beta$. Then $|u_{1,n}(t) - u_{1,n}(t^*)| = \Big|\int_{t}^{t^*} \frac{u_{2,n}(s)}{p(s)} \, ds \Big| \leq \int_{t}^{t^*} \frac{Q}{p(s)} \, ds,$ where $Q = \max\{|B-d^*|,|B+d^*|\}$. Moreover, \begin{eqnarray*} |u_{2,n}(t) - u_{2,n}(t^*)| &=& \Big|\int_{t-h_n}^{t^*-h_n} m(s)f(s,u_{1,n}(s),u_{2,n}(s)) \, ds\Big| \\ &\leq& \int_{t-h_n}^{t^*-h_n} h(s) \, ds. \end{eqnarray*} The desired equicontinuity follows from absolute continuity of the integral. Using Ascoli's theorem, we may assume without loss of generality that both sequences converge uniformly on $[t_0,t_0+\beta]$ to limit functions $u_1 (t)$, $u_2 (t)$. We may use the Lebesgue dominated convergence theorem (for $u_{2,n}$ the dominating function is $h(s)$; for $u_{1,n}$, the dominating function is $(p(s))^{-1}$) to take limits under each integral sign as $n \to \infty$ to show that \begin{gather*} u_1 (t) = \int_{t_0}^t \frac{u_2 (s)}{p(s)} \, ds ,\\ u_2(t) = B - \int_{t_0}^{t} m(s)f(s,u_1(s),u_2(s)) \, ds , \end{gather*} for $t_0 \leq t \leq t_0 + \beta$, from which we obtain \begin{gather*} u_2'(t) = -m(t)f(t,u_1(t),u_2(t)),\\ u_1'(t) = \frac{u_2(t)}{p(t)} , \end{gather*} almost everywhere on $[t_0 ,t_0 + \beta]$. \medskip The specific size of $\beta$ provided by the hypotheses of this last theorem is crucial for our main proof in the next section. \section{Proof of Main Theorem} First note that the hypotheses M1-M4 imply that the earlier hypotheses L1-L4 hold on any interval $(t_0,b)$ with $0 \leq t_0 < b < \infty$, so we may apply Theorem \ref{cara} as needed. From hypothesis M4, we have $m(s) g_{\alpha_0} (s) \in L^1 (0,\infty)$ if $\alpha_0 > 0$ is sufficiently small. Further, M4 implies that $m(s) g_\alpha (s) \leq m(s) g_{\alpha_0} (s)$ whenever $0<\alpha<\alpha_0$ and also that $m(s) g_\alpha (s) \to 0$ as $\alpha \to 0$, for all $s > 0$. Thus, by the Lebesgue Dominated Convergence Theorem, $\int_{0}^{\infty}m(t)g_{\alpha}(t)dt \to 0 \quad \mbox{as } \alpha \to 0 .$ Hence, there exists $\delta \in (0,\alpha_0 ]$ such that $0 < \alpha < \delta$ implies $\int_{0}^{\infty}m(t)g_{\alpha}(t)dt < \frac{1}{4}.$ We shall show that $\gamma = \frac 12 \min\{D,\delta\}$, where $D$ is the number from our hypothesis M3, satisfies the requirements of our theorem. To apply Theorem \ref{cara}, we pick $0 < C \leq \gamma$, $d^* =C/2$, $c = 0$, $d = 2C$, $t_0 = 0$, $b = 1$, $B = C$, and $A = 0$. Note that $d^* = \frac{d}{4} < d - d/2 = d - C = d - B$ and $d^* < C = B - c$. So $d^* < \min\{d-B,B-c\}$. By absolute continuity of the integral, there exists $\beta \in (0,b) = (0,1)$ so that for $k = 0,1,\cdots$, $$\label{beta} \int_{k \beta}^{(k+1)\beta} \alpha m(s)g_{\alpha}(s) \, ds < d^* .$$ This last inequality, for $k=0$ allows us to apply Theorem \ref{cara} to get a solution $y_1(t)$ on $[0,\beta]$ so that for $0 < t \leq \beta$, $$\label{first} 0 < y_1(t) < 2Cr(t), \quad 0 < p(t)y'_1(t) < 2C\,.$$ Integrating (\ref{ode}) from $0$ to $t$ and using M4, we obtain \begin{eqnarray*} |p(t)y_1'(t) - C| &\leq& \int_{0}^{t} m(s)|f(s,y_1(s),p(s)y_1'(s))| \, ds \\ &=& 2C \int_{0}^{t} m(s) h(s,y_1(s),p(s)y_1'(s)) \, ds \\ &<& 2C\int_{0}^{t} m(s)g_{2C}(s) \, ds < \frac{C}{2}, \end{eqnarray*} if $t \in [0,\beta]$. Hence, $\frac{C}{2} < p(t)y_1'(t) < \frac{3C}{2}.$ Then $y_1(t) = \int_{0}^{t}\frac{p(s)y_1'(s)}{p(s)}ds$ and so $\frac{C}{2}r(t) < y_1(t) < \frac{3C}{2}r(t).$ We claim that for $k=2,3,\dots,$ there exists a solution $y_k(t)$ of (\ref{ode}) on the interval $(k-1)\beta \leq t \leq k \beta$ so that \begin{gather*} y_{k+1}(k\beta) = y_k(k\beta),\\ y_{k+1}'(k\beta) = y_k'(k\beta), \end{gather*} for $k \geq 1$, and $$\label{ineqC} \begin{gathered} \frac{C}{2}r(t) < y_k(t) < \frac{3C}{2}r(t), \\ \frac{C}{2} < p(t)y_k'(t) < \frac{3C}{2}, \end{gathered}$$ for $(k-1)\beta \leq t \leq k\beta$. Noting that $y_1 (t)$ has already been constructed, we continue by induction and assume that $y_1(t), y_2(t),\dots,y_n(t)$ have been constructed. Next, we construct $y_{n+1} (t)$. To use Theorem \ref{cara}, we keep $C$, $d^*$, $c$, $d$, $b$, and $\beta$ as before and let $t_0 = n\beta$, $A = A_n = y_n(n\beta)$, and $B = B_n = p(n\beta)y_n'(n\beta)$. Inequality (\ref{beta}) for $k=n$ allows us to apply Theorem \ref{cara} to get a solution $y_{n+1}(t)$ of (\ref{ode}) on $[t_0,t_0 + \beta] = [n\beta, (n+1)\beta]$ so that $y_{n+1} (n \beta) = y_n (n \beta)$ and \begin{gather*} A_n \leq y_{n+1}(t) < A_n + 2C\int_{t_0}^{t} \frac{1}{p(s)}ds, \\ 0 < p(t)y_{n+1}'(t) < 2C. \end{gather*} To complete the induction, we must verify that $y_{n+1} (t)$ satisfies (\ref{ineqC}). Define $y(t)$ for $0 \leq t \leq (n+1)\beta$ by $y(t) = y_k(t)$ for $(k-1)\beta \leq t \leq k\beta$. Since $\frac{C}{2} < p(s)y'(s) < \frac{3C}{2}, \quad \mbox{for } 0 \leq s \leq n\beta$ and $0 < p(s)y'(s) < 2C, \quad \mbox{for } n\beta \leq s \leq (n+1)\beta ,$ it follows that $0 < p(s)y'(s) < 2C$ for the larger interval, $0 \leq s \leq (n+1)\beta$, and it follows by integrating that $0 < y(t) < 2Cr(t), \quad \mbox{for } 0 \leq t \leq (n+1)\beta.$ The calculation appearing just after (\ref{first}) may now be repeated to show that $\frac{C}{2} < p(t)y'(t) < \frac{3C}{2}, \quad \mbox{for } 0 \leq t \leq (n+1)\beta ,$ which implies, as before, that $\frac{C}{2} r(t) < y(t) < \frac{3C}{2}r(t), \quad \mbox{for } 0 \leq t \leq (n+1)\beta.$ Finally, we define $y(t)$ for $0 \leq t < \infty$ by $y(t) = y_k(t)$ for $(k-1)\beta \leq t < k\beta$, and each $k = 1,2,\dots$. Clearly, $y(t)$ is the desired solution. To investigate the limit of $p(t)y'(t)$ at infinity, we examine \begin{eqnarray*} p(t)y'(t) &=& p(t_0)y'(t_0) + \int_{t_0}^{t}(p(s)y'(s))'\,ds \\ &=& p(t_0)y'(t_0) - \int_{t_0}^{\infty} (m(s)f(s,y(s),p(s)y'(s)){\mathcal X}_{[t_0,t]}(s) \, ds. \end{eqnarray*} Since \$(m(s)f(s,y(s),p(s)y'(s)){\mathcal X}_{[t_0,t]}(s)