\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2004(2004), No. 119, 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 2004 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2004/119\hfil Semipositone $m$-point boundary-value problems] {Semipositone $m$-point boundary-value problems} \author[Nickolai Kosmatov\hfil EJDE-2004/119\hfilneg] {Nickolai Kosmatov} \address{Department of Mathematics and Statistics\\ University of Arkansas at Little Rock\\ Little Rock, AR 72204-1099, USA} \email{nxkosmatov@ualr.edu} \date{} \thanks{Submitted April 23, 2004. Published October 10, 2004.} \subjclass[2000]{34B10, 34B18} \keywords{Green's function; fixed point theorem; positive solutions; \hfill\break\indent multi-point boundary-value problem} \begin{abstract} We study the $m$-point nonlinear boundary-value problem $$\displaylines{ -[p(t)u'(t)]' = \lambda f(t,u(t)), \quad 0 < t < 1, \cr u'(0) = 0, \quad \sum_{i=1}^{m-2}\alpha_i u(\eta_i) = u(1), }$$ where $0 < \eta_1 < \eta_2 < \dots < \eta_{m-2} < 1$, $\alpha_i > 0$ for $1 \leq i \leq m-2$ and $\sum_{i=1}^{m-2}\alpha_i < 1$, $m \geq 3$. We assume that $p(t)$ is non-increasing continuously differentiable on $(0,1)$ and $p(t) > 0$ on $[0,1]$. Using a cone-theoretic approach we provide sufficient conditions on continuous $f(t,u)$ under which the problem admits a positive solution. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} \section{Introduction} In this note we consider the nonlinear $m$-point eigenvalue problem \begin{gather} -[p(t)u'(t)]' = \lambda f(t,u(t)), \quad 0 < t < 1, \label{e}\\ u'(0) = 0, \quad \sum_{i=1}^{m-2}\alpha_i u(\eta_i) = u(1), \label{bc} \end{gather} where $0 < \eta_1 < \eta_2 < \dots < \eta_{m-2} < 1$, $\alpha_i > 0$ for $1 \leq i \leq m-2$, $\sum_{i=1}^{m-2}\alpha_i < 1$. We also assume that the function $p(t)$ is non-increasing continuously differentiable on $(0,1)$ and $p(t) > 0$ on $[0,1]$. The inhomogeneous term in (\ref{e}) is allowed to change its sign. Other assumptions on $f(t,u(t))$ will be made later. The study of multi-point boundary-value problems was initiated by Il'in and Moiseev in \cite{im1,im2}. Many authors since then considered nonlinear multi-point boundary-value problems (see, e.g., \cite{da, fe, fw, ghm, ek, ma1, ma2, w1, w2} and the references therein). In particular, Ma studied in \cite{ma2} positive solutions to the three-point nonlinear boundary-value problem \begin{gather*} -u''(t) = a(t)f(u(t)), \quad 0 < t < 1, \\ u(0) = 0, \quad \alpha u(\eta) = u(1), \end{gather*} where $0 < \alpha$, $0 < \eta < 1$ and $\alpha\eta < 1$. The results of \cite{ma2} were complemented in the works of Webb \cite{w2}, Kaufmann \cite{ek}, Kaufmann and Kosmatov \cite{kk}, and Kaufmann and Raffoul \cite{kar}. Among the studies dealing with semipositone multi-point boundary-value problems, we mention the papers by Cao and Ma \cite{cm} and Liu \cite{liu}. Cao and Ma considered the boundary-value problem \begin{gather*} -u''(t) = \lambda a(t)f(u(t),u'(t)), \quad 0 < t < 1, \\ u(0) = 0, \quad \sum_{i=1}^{m-2}\alpha_i u(\eta_i) = u(1). \end{gather*} The authors applied the Leray-Schauder fixed point theorem to obtain an interval of eigenvalues for which at least one positive solution exists. Liu applied a fixed point index method to obtain such an interval for \begin{gather*} -u''(t) = \lambda a(t)f(u(t)), \quad 0 < t < 1, \\ u'(0) = 0, \quad \alpha u(\eta) = u(1). \end{gather*} Our approach is based on Krasnosel'ski\u{\i}'s cone-theoretic theorem \cite{kra} and enables us to show the existence of a positive solution for the semipositone problem (\ref{e}), (\ref{bc}). Other applications of Krasnosel'ski\u{\i}'s fixed point theorem to semipositone problems can, for example, be found in \cite{ag}. \section{Preliminaries} We now proceed with the auxiliaries. Consider the equation $$\label{eq1} -[p(t)u'(t)]' = g(t), \quad 0 < t < 1,$$ with the boundary conditions (\ref{bc}). For convenience we set $\alpha = \sum_{i=1}^{m-2}\alpha_i$. Recall that $\alpha < 1$. \begin{lemma}\label{solutn} If $g \in C[0,1]$ and $g(t) \geq 0$ on $[0,1]$, then \label{sol1} \begin{aligned} u(t) &=-\int_0^t \Big( \int_s^t \frac{d \tau}{p(\tau)} \Big) g(s) \, ds +\frac{1}{1-\alpha}\int_0^1 \Big( \int_s^1 \frac{d \tau}{p(\tau)} \Big) g(s) \, ds \\ &\quad -\frac{1}{1-\alpha} \sum_{i=1}^{m-2}\alpha_i \int_0^{\eta_i} \left ( \int_s^{\eta_i} \frac{d \tau}{p(\tau)} \right ) g(s)\, ds \end{aligned} is the unique nonnegative solution on $[0,1]$ of the problem (\ref{eq1}), (\ref{bc}). \end{lemma} \begin{proof} Integration of (\ref{eq1}) from $0$ to $t$ with the use of the boundary condition (\ref{bc}) at $0$ yields $u'(t) = - \frac{1}{p(t)} \int_0^t g(s) \, ds \leq 0.$ Integrating again we get $u(t) = - \int_0^t \frac{1}{p(s)} \Big( \int_0^s g(\tau) \, d\tau \Big) \,ds + A = - \int_0^t \Big( \int_s^t \frac{d \tau}{p(\tau)} \Big) g(s) \, ds + A.$ Using the multi-point condition in (\ref{bc}) we determine $A$ and obtain (\ref{sol1}). Since $u'(t) \leq 0$, \begin{align*} u(t) &\geq u(1) \\ &= \frac{\alpha}{1-\alpha} \int_0^1 \Big( \int_s^1 \frac{d \tau}{p(\tau)} \Big) g(s) \, ds -\frac{1}{1-\alpha} \sum_{i=1}^{m-2}\alpha_i \int_0^{\eta_i} \Big( \int_s^{\eta_i} \frac{d \tau}{p(\tau)} \Big) g(s) \, ds \\ &= \frac{1}{1-\alpha} \sum_{i=1}^{m-2}\alpha_i \Big[ \int_0^1 \Big( \int_s^1 \frac{d \tau}{p(\tau)} \Big) g(s) \, ds - \int_0^{\eta_i} \Big( \int_s^{\eta_i} \frac{d \tau}{p(\tau)} \Big) g(s) \, ds \Big] \geq 0 \end{align*} on $[0,1]$ and the proof is complete. \end{proof} For $g(t) = 1$ on $[0,1]$, we denote by $u_0(t)$ the unique solution (\ref{sol1}). Then we have \begin{align*} C &= \max_{t\in[0,1]} u_0(t) = u_0(0) \\ &=\frac{1}{1-\alpha}\int_0^1 \Big( \int_s^1 \frac{d \tau}{p(\tau)} \Big) g(s)ds - \frac{1}{1-\alpha} \sum_{i=1}^{m-2}\alpha_i \int_0^{\eta_i} \Big( \int_s^{\eta_i} \frac{d \tau}{p(\tau)} \Big) g(s)ds. \end{align*} The Green's function for $-[p(t)u'(t)]' = 0$ with (\ref{bc}) is given by \begin{align*} G(t,s) &= \frac{1}{1-\alpha} \int_s^1 \frac{d \tau}{p(\tau)} \nonumber \\ &\quad - \begin{cases} \int_s^t \frac{d \tau}{p(\tau)}, & s \leq t\\ 0, & s > t \end{cases} - \begin{cases} \frac{1}{1-\alpha} \sum_{i=1}^{m-2} \alpha_i \chi_i(s) \int_s^{\eta_i} \frac{d \tau}{p(\tau)}, & s \leq \eta_{m-2}\\ 0, & s > \eta_{m-2}, \end{cases} \end{align*} where $\chi_i(s) = \begin{cases} 1, & s \leq \eta_i\\ 0, & s > \eta_i. \end{cases}$ Note that $$\label{est11} \max_{t\in[0,1]}\int_{0}^1 G(t,s) \, ds = C.$$ The integral operator $T \colon \mathcal{B} \to \mathcal{B}$ associated with (\ref{e}), (\ref{bc}) is defined by $Tu(t) = \int_0^1 G(t,s)f(s,u(s)) \, ds$ A routine argument shows that $T$ is completely continuous. \begin{definition} \rm Let $\mathcal{B}$ be a Banach space and let $\mathcal{C} \subset \mathcal{B}$ be closed and nonempty. Then $\mathcal{C}$ is said to be a cone if \begin{enumerate} \item $\alpha u + \beta v \in \mathcal{C}$ for all $u, v \in \mathcal{C}$ and for all $\alpha, \beta \geq 0$, and \item $u, -u \in \mathcal{C}$ implies $u \equiv 0$. \end{enumerate} \end{definition} Our Banach space, $\mathcal{B}$, is the space $C[0,1]$ with the norm $\|u\| = \max_{t \in [0,1]} |u(t)|$. We will show now that the unique solution (\ref{sol1}) satisfies $$\label{coneconst1} \min_{t\in[0,1]} u(t) \geq {\gamma} \| u \|,$$ where $\gamma = \max_{1 \leq i \leq m-2} \frac{\alpha_i (1-\eta_i)}{1-\alpha_i \eta_i}.$ To this end, note that the solution (\ref{sol1}) is concave, since $g(t) \geq 0$ and $u'(t),p'(t) \leq 0$ on $[0,1]$. By concavity and since $u(1) > \alpha_i u(\eta_i)$ for each $1 \leq i \leq m-2$, \begin{align*} \|u\| &= u(0) \\ &\leq u(1) + \frac{u(1) - u(\eta_i)}{1 - \eta_i}(0 - 1) \\ & < u(1) \frac{1 - \alpha_i \eta_i}{\alpha_i (1 - \eta_i)} \\ & = \frac{1-\alpha_i \eta_i}{\alpha_i (1-\eta_i)} \min_{t \in [0,1]} u(t) \end{align*} and hence (\ref{coneconst1}) holds. The estimate (\ref{coneconst1}) is used for defining our cone $\mathcal{C} \subset \mathcal{B}$ by $$\label{cone} {\mathcal{C}} = \{u(t) \in \mathcal{B}: u(t) \geq 0 \; \text{on} \; [0,1], \; \min_{t \in [0,1]} u(t) \geq \gamma \| u \| \}.$$ It turns out that our operator $T$ is cone-preserving. Fixed points of $T$ are solutions of (\ref{e}), (\ref{bc}). The existence of a fixed point of $T$ follows from a fixed point theorem due to Krasnosel'ski\u{\i} \cite{kra}, which we now state. \begin{theorem}\label{Kr} Let $\mathcal{B}$ be a Banach space and let $\mathcal{C}\subset \mathcal{B}$ be a cone in $\mathcal{B}$. Assume that $\Omega_{1}$, $\Omega_{2}$ are open with $0 \in \Omega_1$, ${\overline{\Omega}}_1\subset \Omega_2$, and let $$T\colon {\mathcal{C}}\cap({\overline{\Omega}}_2 \setminus \Omega_1)\to \mathcal{C}$$ be a completely continuous operator such that either \begin{itemize} \item[(i)] $\| Tu \| \leq \| u\|$, $u \in {\mathcal{C}} \cap \partial \Omega_1$, and $\| Tu \| \geq \| u\|$, $u \in {\mathcal{C}} \cap \partial \Omega_2$, or \item[(ii)] $\| Tu \| \geq \| u\|$, $u \in {\mathcal{C}} \cap \partial \Omega_1$, and $\| Tu \| \leq \| u\|$, $u \in {\mathcal{C}} \cap \partial \Omega_2$. \end{itemize} Then $T$ has a fixed point in ${\mathcal{C}}\cap({\overline{\Omega}}_2 \setminus \Omega_1)$. \end{theorem} The following assumptions will stand throughout the remainder of this note: \begin{itemize} \item[(A1)] $f(t,z)$ is a continuous function on $[0,1] \times [0,\infty)$ \item[(A2)] There exists $M > 0$ such that $f(t,z) + M \geq 0$ on $[0,1] \times [0,\infty)$ \item[(A3)] There exist continuous nonnegative nondecreasing on $[0,\infty)$ functions $\psi_a(z)$ and $\psi_b(z)$ with $\psi_b(z) \leq f(t,z) + M \leq \psi_a(z)$ on $[0,1] \times [0,\infty)$. \end{itemize} \section{Positive solutions} We now state our main results. \begin{theorem}\label{pos1} Let the assumptions (A1)-(A3) be satisfied. Assume, in addition, that $\lim_{z \to 0^+} \frac{\psi_a(z)}{z} = 0 \quad\mbox{and}\quad \lim_{z \to \infty} \frac{\psi_b(z)}{z} = \infty.$ Then, for a sufficiently small $\lambda > 0$, the problem (\ref{e}), (\ref{bc}) has a positive solution. \end{theorem} \begin{proof} Consider the equation $$\label{aue} -[p(t)u'(t)]' = \lambda f_p(t,u(t)-u_{\lambda}(t)), \quad 0 < t < 1,$$ with the boundary conditions (\ref{bc}), where $f_p(t,z) = \begin{cases} f(t,z) + M, & z \geq 0\\ f(t,0) + M, & z \leq 0 \end{cases}$ and $u_{\lambda}(t)=\lambda M u_0(t)$ ($u_0(t)$ is given by (\ref{sol1}) for $g \equiv 1$). Our objective is to show that the problem (\ref{aue}), (\ref{bc}) has a positive solution. Our completely continuous and cone-preserving operator associated with (\ref{aue}), (\ref{bc}) is defined by $T_{\lambda}u(t) = \lambda \int_0^1 G(t,s)f_p(s,u(s)-u_{\lambda}(s)) \, ds$ Since $\lim_{z \to 0^+} \frac{\psi_a(z)}{z} = 0$, there exists $R_1 > 0$ such that $\psi_a(z) \leq \frac{1}{\lambda C} z$ for all $z \leq R_1$. Define $\Omega_1 = \{ u \in \mathcal{B} : \|u\| < R_1 \}$, then for $u \in {\mathcal{C}} \cap \partial \Omega_1$ we have $$\label{nearzero} \psi_a(u(s)) \leq \psi_a(\|u\|) \leq \frac{1}{\lambda C} R_1$$ for all $s \in [0,1]$, since $\psi_a(z)$ is nondecreasing. Now, if $u(s) \geq u_{\lambda}(s)$ for $s \in [0,1]$, then $f_p(s,u(s)-u_{\lambda}(s)) = f(s,u(s)-u_{\lambda}(s)) + M \leq {\psi}_a(u(s)-u_{\lambda}(s)) \leq {\psi}_a(u(s)).$ If $u(s) \leq u_{\lambda}(s)$, then $f_p(s,u(s)-u_{\lambda}(s)) = f(s,0) + M \leq {\psi}_a(0) \leq {\psi}_a(u(s))$ (we know that $u(s) \geq 0$ as an element of $\mathcal{C}$). Combining both cases and using (\ref{nearzero}) and (\ref{est11}), we get \begin{align*} \|T_{\lambda}u\| &= \max_{t \in [0,1]} \lambda \, \int_0^1 \! G(t, s) f_p(s,u(s)-u_{\lambda}(s)) \, ds\\ &\leq \max_{t \in [0,1]} \lambda \, \int_0^1 \! G(t, s) {\psi_a}(u(s)) \, ds \\ &\leq \lambda \max_{t \in [0,1]} \int_0^1 \! G(t, s) \, ds \, \frac{1}{\lambda C} R_1 = R_1, \end{align*} that is, $\|T_{\lambda}u\| \leq \|u\|$ on ${\mathcal{C}} \cap \partial \Omega_1$. Since $\lim_{z \to \infty} \frac{\psi_b(z)}{z} = \infty$, then also $\lim_{z \to \infty} \frac{\psi_b(\gamma z - \lambda M C)}{z} = \infty$. Thus, there exists $R_2 > 0$ large enough (so that $R_2 > \frac{\lambda M C}{\gamma}$ and $R_2 > R_1$) such that $\psi_b(\gamma z - \lambda M C) \geq \frac{1}{\lambda C} z$ for all $z \geq R_2$. In fact, $$\label{nearinf} \psi_b(\gamma R_2 - \lambda M C) \geq \frac{1}{\lambda C} R_2.$$ Define $\Omega_2 = \{ u \in \mathcal{B} : \|u\| < R_2 \}$, then for $u \in {\mathcal{C}} \cap \partial \Omega_2$ we have $u(s)-u_{\lambda}(s) \geq \gamma \|u\| - \lambda M u_0(s) \geq \gamma R_2 - \lambda M C > 0.$ Now, for all $s \in [0,1]$, $f_p(s,u(s)-u_{\lambda}(s)) = f(s,u(s)-u_{\lambda}(s)) + M \geq {\psi}_b(u(s)-u_{\lambda}(s)) \geq {\psi}_b(\gamma R_2 - \lambda M C),$ since $\psi_b(z)$ is nondecreasing. Therefore, by (\ref{nearinf}) and (\ref{est11}), \begin{align*} \|T_{\lambda}u\| &= \max_{t \in [0,1]} \lambda \, \int_0^1 \! G(t, s) f_p(s,u(s)-u_{\lambda}(s)) \, ds\\ &\geq \max_{t \in [0,1]} \lambda \, \int_0^1 \! G(t, s) {\psi}_b(\gamma R_2 - \lambda M C) \, ds \\ &\geq \lambda \max_{t \in [0,1]} \int_0^1 \! G(t, s) \, ds \, \frac{1}{\lambda C} R_2 = R_2, \end{align*} that is, $\|T_{\lambda}u\| \leq \|u\|$ on ${\mathcal{C}} \cap \partial \Omega_2$. Since the assumptions of Theorem \ref{Kr} are satisfied, we conclude that the problem (\ref{aue}), (\ref{bc}) has a positive solution in ${\mathcal{C}}\cap({\overline{\Omega}}_2 \setminus \Omega_1)$, which we denote by $u_p$. Let $\lambda$ be small enough so that $R_1 > \frac{\lambda M C}{\gamma}$. Now we have $u_p(t) \geq \gamma \|u_p\| \geq \gamma R_1 > \lambda M C \geq u_{\lambda}(t)$ for all $t \in [0,1]$. Set $u(t) = u_p(t) - u_{\lambda}(t)$, then \begin{align*} -[p(t)u'(t)]' &= -[p(t)u_p'(t)]' - \lambda M \\ &= \lambda f_p(t,u_p(t) - u_{\lambda}(t)) - \lambda M \\ &= \lambda (f(t,u_p(t) - u_{\lambda}(t)) + M) - \lambda M \\ &= \lambda f(t,u(t)), \end{align*} which shows that $u(t)$ is a positive solution of (\ref{e}), (\ref{bc}). The proof is complete. \end{proof} \subsection*{Example} To illustrate our main result, we consider the inhomogeneous term in the form of the function $f(t,z) = -1 + z^2(2+\sin{(4 \pi z (1+t^3))}).$ The function $f(t,z)$ is continuous and, setting $M = 1$, we get $f(t,z)+ M \geq 0$ on $[0,1] \times [0,\infty)$. In addition, for $\psi_b(z)= z^2$ and $\psi_a(z)= 3z^2$, we have $\psi_b(z) \leq f(t,z)+ M \leq\psi_a(z)$ and $\lim_{z \to 0^+} \frac{\psi_a(z)}{z} = 0 \quad \text{and} \quad \lim_{z \to \infty} \frac{\psi_b(z)}{z} = \infty.$ Thus, Theorem \ref{pos1} applies. With only minor adjustments to the argument above one can prove our next theorem. \begin{theorem}\label{pos2} Let the assumptions (A1)-(A3) be satisfied. Assume, in addition, that $\lim_{z \to 0^+} \frac{\psi_a(z)}{z} = \infty \quad \text{and} \quad \lim_{z\to \infty} \frac{\psi_b(z)}{z} = 0.$ Then, for a sufficiently small $\lambda > 0$, the problem (\ref{e}), (\ref{bc}) has a positive solution. \end{theorem} \subsection*{Remark} If problem (\ref{e}), (\ref{bc}) has a positive solution for some $\lambda_1 > 0$, there is also a positive solution for each $\lambda \in (0,\lambda_1]$. We say that a function $\psi(z)$ is sublinear if $\lim_{z \to 0^+} \frac{\psi(z)}{z} = \infty \quad \text{and} \quad \lim_{z \to \infty} \frac{\psi(z)}{z} = 0.$ On the other hand, if $\lim_{z \to 0^+} \frac{\psi(z)}{z} = 0 \quad \text{and} \quad \lim_{z \to \infty} \frac{\psi(z)}{z} = \infty,$ then the function $\psi$ is called superlinear. If in the assumption (A3) we take $\psi_a(z) = \psi_b(z)$, then the following corollary to Theorems \ref{pos1} and \ref{pos2} becomes immediate. \begin{corollary}\label{pos4} Let the assumptions (A1)-(A3) be satisfied. 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