\documentstyle{amsart} \newtheorem{theorem}{Theorem} \newtheorem{lemma}{Lemma} \newtheorem{corollary}{Corollary} \newtheorem{proposition}{Proposition} \newtheorem{remark}{Remark} \newcommand{\refp}[1]{(\ref{#1})} \begin{document} {\noindent\small {\sc Electronic Journal of Differential Equations}\newline Vol. 1995(1995), No. 03, pp. 1-8. Published March 2, 1995.\newline ISSN 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp (login: ftp) 147.26.103.110 or 129.120.3.113 } \thanks{\copyright 1995 Southwest Texas State University and University of North Texas.} \vspace{1.5cm} \title[\hfilneg EJDE--1995/03\hfil Positive Solutions]{Positive Solutions for Higher Order Ordinary Differential Equations} \author[P.W. Eloe \& J. Henderson\hfil EJDE--1995/03\hfilneg] {Paul W. Eloe \& Johnny Henderson } \address{Department of Mathematics \\ University of Dayton\\ Dayton, OH 45469-2316 USA} \email{eloe@@udavxb.oca.udayton.edu} \address{Discrete and Statistical Sciences\\ Auburn University\\ Auburn, AL 36849-5307 USA} \email{hendej2@@mail.auburn.edu} \date{} \thanks{Submitted on December 4, 1994.} \subjclass{34B15} \keywords{Boundary value problems, positive solutions, superlinear and sublinear, operators on a cone} \begin{abstract} Solutions that are positive with respect to a cone are obtained for the boundary value problem, $u^{(n)} + a(t)f(u) = 0$, $u^{(i)}(0) = u^{(n-2)}(1) = 0$, $0 \leq i \leq n-2$, in the cases that $f$ is either superlinear or sublinear. The methods involve application of a fixed point theorem for operators on a cone. \end{abstract} \maketitle \section{Introduction} We are concerned with the existence of solutions for the two-point boundary value problem, $$\label {e1} u^{(n)} + a(t) f(u) = 0, \quad 0 < t < 1,$$ $$\label {e2} u^{(i)}(0) = u^{(n-2)}(1) = 0, \quad 0 \leq i \leq n-2,$$ where \begin{itemize} \item [(A)] $f: [0, \infty) \to [0, \infty)$ is continuous, and \item [(B)] $a: [0, 1] \to [0, \infty)$ is continuous and does not vanish identically on any subinterval. \end{itemize} We remark that, if $u(t)$ is a nonnegative solution of \refp {e1}, \refp {e2}, then $u^{(n-2)}(t)$ is concave on $[0, 1]$. Specifically, our aim is to extend the work of Erbe and Wang \cite {LH} to obtain solutions of \refp {e1}, \refp {e2}, that are positive with respective to a cone, in the cases when, either (i) $f$ is superlinear, or (ii) $f$ is sublinear; that is, in the respective cases when, either (i) $f_0 = 0$ and $f_{\infty} = \infty$, or (ii) $f_0 = \infty$ and $f_{\infty} = 0$, where $$f_0 = \lim _{x \to 0^+} \frac {f(x)}{x} \; \mbox { and } \; f_{\infty} = \lim_{x \to \infty} \frac {f(x)}{x}.$$ In the case that $n= 2$, the boundary value problem \refp {e1}, \refp {e2} arises in applications involving nonlinear elliptic problems in annular regions; see \cite {CB}, \cite {BC}, \cite {XG}, \cite {HW}. Applications of \refp {e1}, \refp {e2} can also be made to singular boundary value problems as in \cite {fb}, \cite {CJ}, \cite {EP}, \cite {JA}, \cite {CD}, \cite {ST}, as well as to extremal point characterizations for boundary value problems in \cite {PW}, \cite {DH}, \cite {KS}. In these applications, frequently, only solutions that are positive are useful. The results herein are also somewhat related to those obtained in \cite {CC} and \cite {AF}. Our arguments for establishing the existence of solutions of \refp {e1}, \refp {e2} involve concavity properties of solutions that are used in defining a cone on which a positive integral operator is defined. A fixed point theorem due to Krasnosel'skii \cite {MK} is applied to yield a positive solution of \refp {e1}, \refp {e2}. In Section 2, we present some properties of a Green's function which will be used in defining the positive operator. We also state the fixed point theorem from \cite {MK}. In Section 3, we provide an appropriate Banach space and cone in order to apply the fixed point theorem yielding solutions of \refp {e1}, \refp {e2} in both the superlinear and sublinear cases. \section{Some preliminaries} In this section, we state a theorem due to Krasnosel'skii, an application of which will yield in the next section a positive solution of \refp {e1}, \refp {e2}. The mapping to which we apply this fixed point theorem will include an integral whose kernel, $G(t, s)$, is the Green's function for $$\label {e3} \begin{gathered} -y^{(n)} = 0,\\ y^{(i)}(0) = y^{(n-2)}(1) = 0, \quad 0 \leq i \leq n -2. \end{gathered}$$ Eloe \cite {PE} has shown that, for $0 \leq i \leq n-2$, $$\label {e4} \frac {\partial ^i}{\partial t^i} G(t, s) > 0 \mbox { on } (0, 1) \times (0, 1),$$ as well as the fact that the function $$\label {e5} K(t, s) = \frac {\partial ^{n-2}}{\partial t^{n-2}} G(t, s)$$ is the Green's function for $$\label {e6} \begin{gathered} -y'' = 0,\\ y(0) = y(1) = 0. \end{gathered}$$ We note that $$\label {e7} K(t, s) = \left \{ \begin{array}{ll} t(1-s), & 0 \leq t < s \leq 1,\\ s(1-t), & 0 \leq s < t \leq 1, \end{array} \right.$$ from which it is straightforward that $$\label {e8} K(t, s) \leq K(s, s), 0 \leq t, s \leq 1,$$ and a nice argument in \cite {LH} shows that $$\label {e9} K(t, s) \geq \frac {1}{4} K(s, s), \; \frac {1}{4} \leq t \leq \frac {3}{4}, \; 0 \leq s \leq 1.$$ The existence of solutions of \refp {e1}, \refp {e2} is based on an application of the following fixed point theorem \cite {MK}. \begin{theorem} \label {t21} Let ${\cal B}$ be a Banach space, and let ${\cal P} \subset {\cal B}$ be a cone in ${\cal B}$. Assume $\Omega _1, \Omega _2$ are open subsets of ${\cal B}$ with $0 \in \Omega _1 \subset \bar {\Omega}_1 \subset \Omega _2$, and let $$T: {\cal P} \cap (\bar {\Omega}_2 \backslash \Omega _1) \to {\cal P}$$ be a completely continuous operator such that, either \begin{itemize} \item [(i)] $\|Tu\| \leq \|u\|, u \in {\cal P} \cap \partial \Omega _1$, and $\|Tu\| \geq \|u\|$, $u \in {\cal P} \cap \partial \Omega _2$, or \item [(ii)] $\|Tu\| \geq \|u\|$, $u \in {\cal P} \cap \partial \Omega _1$, and $\|Tu\| \leq \|u\|$, $u \in {\cal P} \cap \partial \Omega _2$. \end{itemize} Then $T$ has a fixed point in ${\cal P} \cap (\bar {\Omega}_2 \backslash \Omega _1)$. \end{theorem} \section{Existence of solutions} We are now ready to apply Theorem \ref {t21}. We remark that $u(t)$ is a solution of \refp {e1}, \refp {e2} if, and only if, $$u(t) = \int_0^1 G(t, s) a(s) f(u(s))ds, \quad 0 \leq t \leq 1.$$ For our construction, we let $${\cal B} = \{x \in C^{(n-2)}[0, 1] \mid x^{(i)}(0) = 0, \quad 0 \leq i \leq n-3\},$$ with norm, $\|x\| = |x^{(n-2)}|_{\infty}$, where $|\cdot|_{\infty}$ denotes the supremum norm on $[0, 1]$. Then $(\cal {B}, \| \cdot \|)$ is a Banach space. \begin{remark} \label {r1} We note that, for each $x \in {\cal B}$, $$\label {e10} |x^{(i)}|_{\infty} \leq \|x\|, \quad 0 \leq i \leq n-2.$$ We will seek solutions of \refp {e1}, \refp {e2} which lie in a cone, ${\cal P}$, defined by $${\cal P} = \{x \in {\cal B} \mid x^{(n-2)}(t) \geq 0 \mbox { on } [0, 1], \mbox { and } \min _{\frac {1}{4} \leq t \leq \frac {3}{4}} x^{(n-2)}(t) \geq \frac {1}{4} \|x\|\}.$$ \end{remark} \begin{remark} \label {r2} We note here that, if $x \in {\cal P}$, then $x^{(i)}(t) \geq 0$ on $[0, 1]$ and $$x^{(i)}(t) \geq \frac {1}{4} \|x\| \frac {(t- \frac {1}{4})^{n-i-2}}{(n-i-2)!}$$ on $[\frac {1}{4}, \frac {3}{4}]$, $0 \leq i \leq n-2$. As a consequence $$x^{(i)}(t) \geq \frac {1}{(n-i-2)! 4^{n-i-1}} \|x\|$$ on $[\frac {1}{2}, \frac {3}{4}]$, $0 \leq i \leq n-2$. \end{remark} \begin{theorem} \label {t31} Assume that conditions (A) and (B) are satisfied. If, either \begin{itemize} \item [(i)] $f_0 = 0$ and $f_{\infty} = \infty$ (i.e., $f$ is superlinear), or \item [(ii)] $f_0 = \infty$ and $f_{\infty} = 0$ (i.e., $f$ is sublinear), \end{itemize} then \refp {e1}, \refp {e2} has at least one solution in ${\cal P}$. \end{theorem} \medskip \noindent {\bf Proof.} We begin by defining an integral operator $T: {\cal P} \to {\cal B}$ by $$\label {e11} Tu(t) = \int_0^1G(t, s) a(s) f(u(s))\, ds, \quad u \in {\cal P},$$ and we seek a fixed point of $T$ in the cone ${\cal P}$ for the respective cases of $f$ superlinear and $f$ sublinear. Before dealing with these cases, we make a few observations. First, if $u \in {\cal P}$, it follows from \refp {e8} that \begin{eqnarray*} (Tu)^{(n-2)}(t) & = & \int_0^1 \frac {\partial ^{n-2}}{\partial t^{n-2}} G(t, s) a(s) f(u(s))\, ds\\ & = & \int_0^1 K(t, s)a(s)f(u(s))\, ds\\ & \leq & \int_0^1 K(s, s)a(s)f(u(s))\, ds, \end{eqnarray*} so that $$\|Tu\| = |(Tu)^{(n-2)}|_{\infty} \leq \int_0^1 K(s,s) a(s)f(u(s))\, ds.$$ In fact, $$\label {E12} \|Tu\| = \int_0^1 K(s,s)a(s) f(u(s))\, ds.$$ Next, if $u \in {\cal P}$, it follows from \refp {e9} and \refp {E12} that \begin{eqnarray*} \min _{\frac {1}{4} \leq t \leq \frac {3}{4}} (Tu)^{(n-2)}(t) & = & \min _{\frac {1}{4} \leq t \leq \frac {3}{4}} \int_0^1 K(t, s) a(s) f(u(s)) \, ds\\ & \geq & \frac {1}{4} \int_0^1 K(s, s) a(s) f(u(s))\, ds\\ & \geq & \frac {1}{4} \|Tu\|. \end{eqnarray*} Moreover, properties of $G(t, s)$ give that $(Tu)^{(n-2)}(t) \geq 0$, so that $Tu \in {\cal P}$, and in particular $T: {\cal P} \to {\cal P}$. Also, the standard arguments yield that $T$ is completely continuous. We now turn to the cases of the theorem. \begin{itemize} \item [(i)] Assume $f_0 = 0$ and $f_{\infty} = \infty$. First, dealing with $f_0 = 0$, there exist $\eta > 0$ and $H_1 > 0$ such that $f(x) \leq \eta x$, for $0 < x \leq H_1$, and $$\eta \int_0^1 K(s, s) a(s)\, ds \leq 1.$$ So, if we choose $u \in {\cal P}$ with $\|u\| = H_1$, and if we recall from Remark \ref {r1} that $|u|_{\infty} \leq \|u\|$, we have from \refp {e8}, \begin{eqnarray*} (Tu)^{(n-2)}(t) & = & \int_0^1 K(t, s) a(s) f(u(s))\, ds\\ & \leq & \int_0^1 K(s,s) a(s) f(u(s))\, ds\\ & \leq & \int_0^1 K(s, s) a(s) \eta u(s)\, ds\\ & \leq & \eta \int_0^1 K(s, s) a(s)\, ds \|u\|\\ & \leq & \|u\|, \quad 0 \leq t \leq 1. \end{eqnarray*} As a consequence $\|Tu\| = |(Tu)^{(n-2)}(t)|_{\infty} \leq \|u\|$. Thus, if we set $$\Omega _1 = \{x \in {\cal B} \mid \|x\| < H_1\},$$ then $$\label {e13} \|Tu\| \leq \|u\|, \mbox { for } u \in {\cal P} \cap \partial \Omega _1.$$ Next, dealing with $f_{\infty} = \infty$, there exist $\lambda > 0$ and $\bar {H}_2 > 0$ such that $f(x) \geq \lambda x$, for $x \geq \bar {H}_2$, and $$\frac {\lambda}{(n-2)! 4^{n-1}} \int_{\frac {1}{2}}^{\frac {3}{4}} K(\frac {1}{2}, s) a(s)\, ds \geq 1.$$ Now, let $H_2 = \max \{2H_1, (n-2)! 4^{n-1} \bar {H}_2\}$ and set $$\Omega _2 = \{x \in {\cal B} \mid \|x\| < H_2\}.$$ So, if $u \in {\cal P}$ with $\|u\| = H_2$, and if we recall from Remark \ref {r2} that $u(t) \geq \frac {1}{(n-2)! 4^{n-1}} \|u\| \geq \bar {H}_2$ on $[\frac {1}{2}, \frac {3}{4}]$, we have \begin{eqnarray*} (Tu)^{(n-2)} (\frac {1}{2}) & = & \int_0^1 K(\frac {1}{2}, s) a(s) f(u(s))\, ds\\ & \geq & \int_{\frac {1}{2}}^{\frac {3}{4}} K(\frac {1}{2}, s) a(s) \lambda u(s)\, ds\\ & \geq & \lambda \int_{\frac {1}{2}}^{\frac {3}{4}} K(\frac {1}{2}, s) a(s) \frac {1}{(n-2)! 4^{n-1}} \|u\|\, ds\\ & = & \frac {\lambda}{(n-2)! 4^{n-1}} \int_{\frac {1}{2}}^{\frac {3}{4}} K(\frac {1}{2}, s) a(s)\, ds\|u\|\\ & \geq & \|u\|, \end{eqnarray*} so that $\|Tu\| \geq \|u\|$. Consequently, $$\label {e14} \|Tu\| \geq \|u\|, \mbox { for } u \in {\cal P} \cap \partial \Omega _2.$$ Therefore, by part (i) of Theorem \ref {t21} applied to \refp {e13} and \refp {e14}, $T$ has a fixed point $u(t) \in {\cal P} \cap (\bar {\Omega}_2 \backslash \Omega _1)$ such that $H_1 \leq \|u \| \leq H_2$, and as such, $u(t)$ is a desired solution of \refp {e1}, \refp {e2}. (We remark that the arguments carry through, if we had set $H_2 = \max \{H_1, (n-2)!4^{n-1} \bar {H}_2\}$ and if $H_2 = H_1$, then there is a solution $u \in {\cal P}$ with $\|u\| = H_1$.) This completes the case when $f$ is superlinear. \item [(ii)] Now, assume $f_0 = \infty$ and $f_{\infty} = 0$. Dealing with $f_0 = \infty$, there exist $\bar {\eta} > 0$ and $J_1 > 0$ such that $f(x) \geq \bar {\eta} x$, for $0 < x \leq J_1$, and $$\frac {\bar {\eta}}{(n-2)! 4^{n-1}} \int_{\frac {1}{2}}^{\frac {3}{4}} K(\frac {1}{2}, s) a(s) \, ds \geq 1.$$ This time, we choose $u \in {\cal P}$ with $\|u\| = J_1$. Since $|u|_{\infty} \leq \|u\| = J_1$, we have $f(u(s)) \geq \bar {\eta} u(s)$, $0 \leq s \leq 1$. Also, we know $u(s) \geq \frac {1}{(n-2!) 4^{n-1}} \|u\|$, $\frac {1}{2} \leq s \leq \frac {3}{4}$. Thus, \begin{eqnarray*} (Tu)^{(n-2)} (\frac {1}{2}) & = & \int_0^1 K(\frac {1}{2}, s) a(s) f(u(s))\, ds\\ & \geq & \int_{\frac {1}{2}}^{\frac {3}{4}} K(\frac {1}{2}, s) a(s) \bar {\eta} u(s)\, ds\\ & \geq & \frac {\bar {\eta}}{(n-2)! 4^{n-1}} \int_{\frac {1}{2}}^{\frac {3}{4}} K(\frac {1}{2}, s)a(s)\, ds \|u\|\\ & \geq & \| u\|, \end{eqnarray*} and in particular, $\|Tu\| \geq \|u\|$. Setting $$\Omega _1 = \{x \in {\cal B} \mid \|x\| < J_1\},$$ we conclude $$\label {e15} \|Tu\| \geq \|u\|, \mbox { for } u \in {\cal P} \cap \partial \Omega _1.$$ \end{itemize} For the final part of this case, we deal with $f_{\infty} = 0$. There exist $\bar {\lambda} > 0$ and $\bar {J}_2 > 0$ such that, $f(x) \leq \bar {\lambda} x$, for $x \geq \bar {J}_2$, and $$\bar {\lambda} \int_0^1 K(s, s) a(s) \, ds \leq 1.$$ There are two further sub-cases to be considered: (I) We suppose first that $f$ is bounded. Then, there exists $N > 0$ such that $f(x) \leq N$, for all $0 < x < \infty$. Let $J_2 = \max \{2 J_1, N \int_0^1 K(s, s) a(s)\, ds\}$. Then, for $u \in {\cal P}$ with $\|u\| = J_2$, since $|u|_{\infty} \leq \|u\|$ and $K(t, s) \leq K(s, s)$, $0 \leq s, t \leq 1$, we have \begin{eqnarray*} (Tu)^{(n-2)} (t) & = & \int_0^1 K(t, s) a(s) f(u(s))\, ds\\ & \leq & N \int_0^1 K(s, s) a(s) \, ds\\ & \leq & J_2\\ & = & \|u\|, \quad 0 \leq t \leq 1. \end{eqnarray*} Consequently, $\|Tu\| \leq \|u\|$. (II) For the second sub-case, suppose that $f$ is unbounded. Then, there exists $J_2 > \max \{2J_1, \bar {J}_2\}$ such that $f(x) \leq f(J_2)$, for $0 < x \leq J_2$. We now choose $u \in {\cal P}$ with $\|u\| = J_2$. Again, recalling $|u|_{\infty} \leq \|u\|$ and $K(t, s) \leq K(s, s)$ leads to \begin{eqnarray*} (Tu)^{(n-2)}(t) & = & \int_0^1 K(t, s) a(s) f(u(s))\, ds\\ & \leq & \int_0^1 K(s, s) a(s) f(J_2) \, ds\\ & \leq & \bar {\lambda} \int_0^1 K(s, s) a(s) \, ds J_2\\ & \leq & \|u\|, \quad 0 \leq t \leq 1. \end{eqnarray*} Thus, $\|Tu\| \leq \|u\|$. We conclude from each sub-case, (I) and (II), if we set $$\Omega _2 = \{x \in {\cal B} \mid \|x\| < J_2\},$$ then $$\label {e16} \|Tu\| \leq \|u\|, \mbox { for } u \in {\cal P} \cap \partial \Omega _2.$$ Therefore, by part (ii) of Theorem \ref {t21} applied to \refp {e15} and \refp {e16}, $T$ has a fixed point $u(t) \in {\cal P} \cap (\Omega _2 \backslash \Omega _1)$ such that $J_1 \leq \|u\| \leq J_2$, and $u(t)$ is a sought solution of \refp {e1}, \refp {e2}. This completes the argument for the case of $f$ sublinear. The proof is complete. \quad$\Box$ \begin{thebibliography}{99} \bibitem{CB} C. Bandle, C. V. Coffman and M. 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