\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 99, pp. 1--10.\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 2008 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2008/99\hfil Third-order three-point BVP] {Positive solutions of a third-order three-point boundary-value problem} \author[B. Yang\hfil EJDE-2008/99\hfilneg] {Bo Yang} \address{Bo Yang\newline Department of Mathematics and Statistics, Kennesaw State University, GA 30144, USA} \email{byang@kennesaw.edu} \thanks{Submitted June 18, 2008. Published July 25, 2008.} \subjclass[2000]{34B18, 34B10} \keywords{Fixed point theorem; cone; multi-point boundary-value problem; \hfill\break\indent upper and lower estimates} \begin{abstract} We obtain upper and lower estimates for positive solutions of a third-order three-point boundary-value problem. Sufficient conditions for the existence and nonexistence of positive solutions for the problem are also obtained. Then to illustrate our results, we include an example. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{example}[theorem]{Example} \section{Introduction} Recently third-order multi-point boundary-value problems have attracted a lot of attention. In 2003, Anderson \cite{A} considered the third-order boundary-value problem \begin{gather} u'''(t) = f(t,u(t)), \quad 0\le t\le 1, \label{ae1} \\ u(t_1) = u'(t_2) = \gamma u(t_3)+ \delta u''(t_3) = 0. \label{ab1} \end{gather} In 2008, Graef and Yang \cite{GY} studied the third-order nonlocal boundary-value problem \begin{gather} u'''(t) = g(t)f(u(t)), \quad 0\le t\le 1, \label{ee1} \\ u(0) = u'(p) = \int_q^1 w(t)u''(t)dt = 0.\label{ee2} \end{gather} For more results on third-order boundary-value problems we refer the reader to \cite{AAK,CMS,GSZ,HT,M,WZ,W}. In this paper, we consider the third-order three-point nonlinear boundary-value problem \begin{gather} u'''(t) = g(t)f(u(t)), \quad 0\le t\le 1, \label{bhu} \\ u(0)-\alpha u'(0)= u'(p) = \beta u'(1)+\gamma u''(1) = 0. \label{rfv} \end{gather} % To our knowledge, the problem \eqref{bhu}-\eqref{rfv} has not been considered before. Note that the set of boundary conditions \eqref{rfv} is a very general one. For example, if we let $\alpha=\beta=0$ and $p=\gamma=1$, then \eqref{rfv} reduces to $$u(0) = u'(1) = u''(1) = 0, \label{rfvvv}$$ which are often referred to as the $(1,2)$ focal boundary conditions. If we let $\alpha=0$, $\beta=0$, and $\gamma=1$, then \eqref{rfv} reduces to $$u(0) = u'(p) = u''(1) = 0. \label{rfvv}$$ The boundary-value problem that consists of the equation \eqref{bhu} and the boundary conditions \eqref{rfvv} has been considered by Anderson and Davis in \cite{AD} and Graef and Yang in \cite{GY-2}. Our goal in this paper is to generalize some of the results from \cite{AD,GY-2} to the problem \eqref{bhu}-\eqref{rfv}. In this paper, we are interested in the existence and nonexistence of positive solutions of the problem \eqref{bhu}-\eqref{rfv}. By a positive solution, we mean a solution $u(t)$ to the boundary-value problem such that $u(t)>0$ for $00$, $0< p\le 1$, and $2p(1+\alpha)\ge 1$. \item[(H3)] If $p=1$, then $\gamma>0$. \end{itemize} To prove some of our results, we will use the following fixed point theorem, which is due to Krasnosel'skii \cite{K}. \begin{theorem} \label{tt1} Let $(X,\|\cdot\|)$ be a Banach space over the reals, and let $P\subset X$ be a cone in $X$. Assume that $\Omega_1,\Omega_2$ are bounded open subsets of $X$ with $0\in\Omega_1\subset \overline{\Omega_1}\subset \Omega_2$, and let $$L:P\cap( \overline{\Omega_2}-\Omega_1 )\to P$$ be a completely continuous operator such that, either \begin{itemize} \item[(K1)] $\|Lu\|\leq\|u\|$ if $u\in P\cap\partial \Omega_1$, and $\|Lu\|\ge\|u\|$ if $u\in P\cap\partial \Omega_2$; or \item[(K2)] $\|Lu\|\geq\|u\|$ if $u\in P\cap\partial \Omega_1$, and $\|Lu\|\le\|u\|$ if $u \in P\cap\partial \Omega_2$. \end{itemize} Then $L$ has a fixed point in $P\cap( \overline{\Omega_2}-\Omega_1)$. \end{theorem} Before the Krasnosel'skii fixed point theorem can be used to obtain any existence result, we have to find some nice estimates to positive solutions to the problem \eqref{bhu}--\eqref{rfv} first. These a priori estimates are essential to a successful application of the Krasnosel'skii fixed point theorem. It is based on these estimates that we can define an appropriate cone, on which Theorem~\ref{tt1} can be applied. Better estimates will result in sharper existence and nonexistence conditions. We now fix some notation. Throughout we let $X=C[0,1]$ with the supremum norm $$\|v\|=\max_{t\in [0,1]} |v(t)|,\quad \forall v\in X.$$ Obviously $X$ is a Banach space. Also we define the constants \begin{gather*} F_0=\limsup_{x\to 0^+}\frac{f(x)}{x},\quad f_0=\liminf_{x\to 0^+}\frac{f(x)}{x}, \\ F_{\infty}=\limsup_{x\to + \infty} \frac{f(x)}{x},\quad f_{\infty}=\liminf_{x\to + \infty} \frac{f(x)}{x}. \end{gather*} These constants will be used later in our statements of the existence theorems. This paper is organized as follows. In Section 2, we obtain some {a priori} estimates to positive solutions to the problem \eqref{bhu}-\eqref{rfv}. In Section 3, we define a positive cone of the Banach space $X$ using the estimates obtained in Section 2, and apply Theorem \ref{tt1} to establish some existence results for positive solutions of the problem \eqref{bhu}-\eqref{rfv}. In Section 4, we present some nonexistence results. An example is given at the end of the paper to illustrate the existence and nonexistence results. \section{Green's Function and Estimates of Positive Solutions} In this section, we shall study Green's function for the problem \eqref{bhu}-\eqref{rfv}, and prove some estimates for positive solutions of the problem. Throughout the section, we define the constant $M=\beta+\gamma-\beta p$. By conditions (H2) and (H3), we know that $M$ is a positive constant. \begin{lemma} \label{dfg} If $u\in C^3[0,1]$ satisfies the boundary conditions \eqref{rfv} and $u'''(t)\equiv~0$ on $[0,1]$, then $u(t)\equiv 0$ on $[0,1]$. \end{lemma} \begin{proof} Since $u'''(t)\equiv 0$ on $[0,1]$, there exist constants $a_1$, $a_2$, and $a_3$ such that $u(t)=a_1+a_2t+a_3t^2,\quad 0\le t\le 1.$ Because $u(t)$ satisfies the boundary conditions \eqref{rfv}, we have $$\begin{pmatrix} 1 & -\alpha & 0 \\ 0 & 1 & 2p \\ 0 & \beta & 2(\beta+\gamma) \end{pmatrix} \begin{pmatrix}a_1\cr a_2\\ a_3\end{pmatrix} =\begin{pmatrix}0\\ 0\\ 0 \end{pmatrix}. \label{cvb}$$ The determinant of the coefficient matrix for the above linear system is $2M>0$. Therefore, the system \eqref{cvb} has only the trivial solution $a_1=a_2=a_3=0$. Hence $u(t)\equiv 0$ on $[0,1]$. The proof is complete. \end{proof} We need the indicator function $\chi$ to write the expression of Green's function for the problem \eqref{bhu}-\eqref{rfv}. Recall that if $[a,b]\subset R:=(-\infty,+\infty)$ is a closed interval, then the indicator function $\chi$ of $[a,b]$ is given by $\chi_{[a,b]}(t)= \begin{cases} 1, & \mbox{if }t\in [a,b], \\ 0, & \mbox{if }t\not\in [a,b]. \end{cases}$ Now we define the function $G:[0,1]\times[0,1]\to [0,\infty)$ by \begin{align*} G(t,s)&=\frac{\beta +\gamma-\beta s}{2( \beta +\gamma -p\beta)} (2\alpha p + 2pt - t^2) + \frac{(t-s)^2}{2}\chi_{[0,t]}(s)\\ &\quad -\frac{p-s}{2(\beta +\gamma -p\beta)} (2(\alpha + t)(\beta+\gamma)-\beta t^2) \chi_{[0,p]}(s). \end{align*} We are going to show that $G(t,s)$ is Green's function for the problem \eqref{bhu}-\eqref{rfv}. \begin{lemma}\label{seed} Let $h\in C[0,1]$. If $y(t)=\int_0^1 G(t,s)h(s)ds, \quad 0\le t\le 1,$ then $y(t)$ satisfies the boundary conditions \eqref{rfv} and $y'''(t)=h(t)$ for $0\le t\le 1$. \end{lemma} \begin{proof} If $y(t)=\int_0^1 G(t,s)h(s)ds,$ then \begin{aligned} y(t)&=-\int_0^p \frac{p-s}{2M} (2(\alpha + t)(\beta+\gamma)-\beta t^2) h(s)ds + \int_0^t\frac{(t-s)^2}{2}h(s)ds \\ &\quad + \int_0^1\frac{\beta +\gamma-\beta s}{2M} (2\alpha p + 2pt - t^2) h(s)ds. \end{aligned} \label{gg} Differentiating the above expression, we have \begin{aligned} y'(t)&=-\int_0^p \frac{p-s}{M} (\beta+\gamma-\beta t) h(s)ds +\int_0^t (t-s) h(s)ds \\ &\quad + \int_0^1\frac{\beta +\gamma-\beta s}{M} (p - t) h(s)ds, \end{aligned}\label{ggg} $$y''(t)= \beta \int_0^p \frac{p-s}{M} h(s)ds +\int_0^t h(s)ds - \int_0^1\frac{\beta +\gamma-\beta s}{M} h(s)ds, \label{gggg}$$ and $y'''(t)=h(t),\quad 0\le t\le 1.$ It is easy to see from \eqref{ggg} that $y'(p)=0$. By making the substitution $t=0$ in \eqref{gg} and \eqref{ggg}, we get $$y(0)= -\int_0^p \frac{p-s}{M} \alpha (\beta+\gamma) h(s)ds + \int_0^1\frac{\beta +\gamma-\beta s}{M} \alpha p h(s)ds\label{gg1}$$ and $$y'(0)=-\int_0^p \frac{p-s}{M} (\beta+\gamma) h(s)ds + \int_0^1\frac{\beta +\gamma-\beta s}{M} p h(s)ds.\label{ggg1}$$ It is clear from \eqref{gg1} and \eqref{ggg1} that $y(0)-\alpha y'(0)=0$. By making the substitution $t=1$ in \eqref{ggg} and \eqref{gggg}, we get $y'(1)=-\gamma\int_0^p \frac{p-s}{M} h(s)ds +\int_0^1 (1-s) h(s)ds + \int_0^1\frac{\beta +\gamma-\beta s}{M}\cdot (p-1) h(s)ds$ and $y''(1)= \beta \int_0^p \frac{p-s}{M} h(s)ds +\int_0^1 h(s)ds -\int_0^1\frac{\beta +\gamma-\beta s}{M} h(s)ds.$ Simplifying the last two equations, we get \begin{gather} y'(1)= -\gamma\int_0^p \frac{p-s}{M} h(s)ds +\gamma \int_0^1 \frac{p-s}{M} h(s)ds, \label{ggglt}\\ y''(1)=\beta \int_0^p \frac{p-s}{M} h(s)ds -\beta \int_0^1\frac{p-s}{M} h(s)ds.\label{gggg1} \end{gather} It is easily seen from \eqref{ggglt} and \eqref{gggg1} that $\beta y'(1)+\gamma y''(1)=0$. The proof is now complete. \end{proof} \begin{lemma} \label{lem2.3} Let $h\in C[0,1]$ and $y\in C^3[0,1]$. If $y(t)$ satisfies the boundary conditions \eqref{rfv} and $y'''(t)=h(t)$ for $0\le t\le 1$, then $y(t)=\int_0^1 G(t,s)h(s)ds, \quad 0\le t\le 1.$ \end{lemma} \begin{proof} Suppose that $y(t)$ satisfies the boundary conditions \eqref{rfv} and $y'''(t)=h(t)$ for $0\le t\le 1$. Let $k(t)=\int_0^1 G(t,s)h(s)ds, \quad 0\le t\le 1.$ By Lemma \ref{seed} we have $k'''(t)=h(t)$ for $0\le t\le 1$, and $k(t)$ satisfies the boundary conditions \eqref{rfv}. If we let $m(t)=y(t)-k(t)$, $0\le t\le 1$, then $m'''(t)=0$ for $0\le t\le 1$ and $m(t)$ satisfies the boundary conditions \eqref{rfv}. By Lemma \ref{dfg}, we have $m(t)\equiv 0$ on $[0,1]$, which implies that $y(t)=\int_0^1 G(t,s)h(s)ds, \quad 0\le t\le 1.$ The proof is complete. \end{proof} We see from the last two lemmas that $y(t)=\int_0^1 G(t,s)h(s)ds \quad \mbox{for } 0\le t\le 1$ if and only if $y(t)$ satisfies the boundary conditions \eqref{rfv} and $y'''(t)=h(t)$ for $0\le t\le 1$. Hence the problem \eqref{bhu}-\eqref{rfv} is equivalent to the integral equation $$u(t) =\int_0^1 G(t,s)g(s)f(u(s))ds,\quad 0\le t\le 1, \label{ee3}$$ and $G(t,s)$ is Green's function for the problem \eqref{bhu}-\eqref{rfv}. Now we investigate the sign property of $G(t,s)$. We start with a technical lemma. \begin{lemma} \label{lem2.4} If {\rm (H2)} holds, then $2p\alpha + 2pt - t^2\ge 0, \quad 0\le t\le 1.$ \end{lemma} \begin{proof} Assuming that (H2) holds; if $0\le t\le 1$, then $2p\alpha + 2pt - t^2 = t(1-t)+2p\alpha(1-t)+(2p(1+\alpha)-1)t\ge 0.$ The proof is complete. \end{proof} \begin{lemma} \label{lem2.5} If {\rm (H2)--(H3)} hold, then we have \begin{enumerate} \item If $0\le t\le 1$ and $0\le s\le 1$, then $G(t,s)\ge 0$. \item If $0< t<1$ and $0< s< 1$, then $G(t,s)>0$. \end{enumerate} \end{lemma} \begin{proof} We shall prove (1) only. We take four cases to discuss the sign property of $G(t,s)$. \begin{itemize} \item[(i)] If $s\ge p$ and $s\ge t$, then $G(t,s)=\frac{\beta+\gamma-\beta s}{2M} (2p\alpha + 2pt - t^2)\ge 0.$ \item[(ii)] If $s\ge p$ and $s\le t$, then $G(t,s)=\frac{\beta+\gamma-\beta s}{2M} (2p\alpha + 2pt - t^2)+\frac{(t-s)^2}{2}\ge 0.$ \item[(iii)] If $s\le p$ and $s\ge t$, then $G(t,s)=\frac{1}{2}(2 \alpha s +2st-t^2)\ge 0.$ \item[(iv)] If $s\le p$ and $s\le t$, then $G(t,s)=\alpha s+\frac{s^2}{2}\ge 0.$ \end{itemize} Therefore $G(t,s)$ is nonnegative in all four cases. The proof of (1) is complete. If we take a closer look at the expressions of $G(t,s)$ in the four cases, then we will see easily that (2) is also true. We leave the details to the reader. \end{proof} \begin{lemma} \label{tt2} If $u\in C^3[0,1]$ satisfies \eqref{rfv}, and $$u'''(t)\ge 0 \quad \text{for } 0\le t\le 1, \label{ee5}$$ then \begin{enumerate} \item $u(t)\ge 0$ for $0\le t\le 1$. \item $u'(t)\ge 0$ on $[0,p]$ and $u'(t)\le 0$ on $[p,1]$. \item $u(p) = \|u\|$. \end{enumerate} \end{lemma} \begin{proof} Note that $G(t,s)\ge 0$ if $t,s\in[0,1]$. If $u'''(t)\ge 0$ on $[0,1]$, then for each $t\in[0,1]$ we have $u(t)=\int_0^1 G(t,s)u'''(s)\,ds\ge 0.$ The proof of (1) is complete. It follows from \eqref{ggg} that $u'(t)= \int_t^p (s-t)u'''(s)\,ds +\int_p^1\frac{\beta +\gamma-\beta s}{M}\cdot(p-t)u'''(s)ds.$ If $0\le t\le p$, then it is obvious that $u'(t)\ge 0$. If $t\ge p$, then we put \eqref{ggg} into an equivalent form $u'(t)= -\int_p^t \frac{(s-p)(\gamma+\beta(1-t))}{M}u'''(s)\,ds -\int_t^1 \frac{\beta +\gamma-\beta s}{M}(t-p)u'''(s)\,ds,$ from which we can see easily that $u'(t)\le 0$. Part (3) of the lemma follows immediately from parts (1) and (2). The proof is now complete. \end{proof} Throughout the remainder of the paper, we define the continuous function $a:[0,1]\to[0,+\infty)$ by $a(t)=\frac{2p\alpha+2pt-t^2}{2p\alpha + p^2},\quad 0\le t\le 1.$ It can be shown that $a(t)\geq \min\{t,1-t\},\quad 0\le t\le 1.$ The proof of the last inequality is omitted. \begin{lemma} \label{lem2.7} If $u\in C^3[0,1]$ satisfies \eqref{ee5} and the boundary conditions \eqref{rfv}, then $u(t)\ge a(t)u(p)$ on $[0,1]$. \end{lemma} \begin{proof} If we define $h(t) = u(t) - a(t)u(p),\quad 0\le t\le 1,$ then $h'''(t)=u'''(t)\geq 0,\quad 0\le t\le 1.$ Obviously we have $h(p)=h'(p)=0$. To prove the lemma, it suffices to show that $h(t)\ge 0$ for $0\le t\le 1$. We take two cases to continue the proof. \noindent Case I: $h'(0)\le 0$. We note that $h'(p)=0$ and $h'$ is concave upward on $[0,1]$. Since $h'(0)\le 0$, we have $h'(t)\le 0$ on $[0,p]$ and $h'(t)\ge 0$ on $[p,1]$. Since $h(p)=0$, we have $h(t)\ge 0$ on $[0,1]$. \noindent Case II: $h'(0)>0$. It is easy to see from the definition of $h(t)$ that $h(0)=\alpha h'(0).$ Since $\alpha\ge 0$, we have $h(0)\ge 0$. Because $h'(0)>0$ and $h(0)\ge 0$, there exists $\delta\in(0,p)$ such that $h(\delta)>0$. By the mean value theorem, since $h(\delta)>h(p)=0$, there exists $r_1\in(\delta,p)$ such that $h'(r_1)<0$. Now we have $h'(0)>0$, $h'(r_1)<0$, and $h'(p)=0$. Because $h'(t)$ is concave upward on $[0,1]$, there exists $r_2\in (0,r_1)$ such that $h'(t)>0\quad \text{on } [0,r_2), \quad h'(t)\le 0 \quad \text{on } [r_2,p],\quad h'(t)\ge 0 \quad \text{on } (p,1].$ Since $h(0)\ge 0$ and $h(p)=0$, we have $h(t)\geq 0$ on $[0,1]$. We have shown that $h(t)\geq 0$ on $[0,1]$ in both cases. The proof is complete. \end{proof} In summary, we have \begin{theorem}\label{uhj} Suppose that {\rm (H1)--(H3)} hold. If $u\in C^3[0,1]$ satisfies \eqref{ee5} and the boundary conditions \eqref{rfv}, then $u(p)=\|u\|$ and $u(t)\ge a(t)u(p)$ on $[0,1]$. In particular, if $u\in C^3[0,1]$ is a nonnegative solution to the boundary-value problem \eqref{bhu}-\eqref{rfv}, then $u(p)=\|u\|$ and $u(t)\ge a(t)u(p)$ on $[0,1]$. \end{theorem} \section{Existence of Positive Solutions} Now we give some notation. Define the constants $A = \int_0^1 G(p,s) g(s) a(s)\,ds,\quad B = \int_0^1 G(p,s) g(s) \,ds$ and let $$P=\{v\in X : v(p)\ge 0 , \, a(t)v(p)\le v(t)\le v(p) \text{ on \ } [0,1]\}.$$ Obviously $X$ is a Banach space and $P$ is a positive cone of $X$. Define an operator $T:P\to X$ by $$Tu(t)= \int_0^1 G(t,s)g(s)f(u(s))ds,\quad 0\leq t\leq 1,\; u\in X.$$ It is well known that $T:P\to X$ is a completely continuous operator. And by the same argument as in Theorem \ref{uhj} we can prove that $T(P)\subset P$. Now the integral equation \eqref{ee3} is equivalent to the equality $$Tu=u,\quad u\in P.$$ To solve problem \eqref{bhu}-\eqref{rfv} we need only to find a fixed point of $T$ in $P$. \begin{theorem} \label{tt6} If $BF_0 <1< A f_{\infty}$, then the problem \eqref{bhu}-\eqref{rfv} has at least one positive solution. \end{theorem} \begin{proof} Choose $\epsilon>0$ such that $(F_0+\epsilon)B\le 1$. There exists $H_1 >0$ such that $$f(x)\leq (F_0+\epsilon)x \quad\text{for } 00 such that$$ (f_\infty-\delta)\int_c^{1-c} G(p,s) g(s) a(s)\,ds >1. $$There exists  H_3 >0  such that f(x)\geq (f_\infty-\delta)x for x\geq H_3. Let H_2=H_1 + {H_3}/{c}. If u\in P with \|u\| = H_2, then for c\le t\le 1-c, we have$$ u(t)\geq\min\{t,1-t\}\|u\| \ge c H_2\ge H_3. So, if u\in P with \|u\| = H_2, then \begin{align*} (Tu)(p) & \geq \int_c^{1-c} G(p,s) g(s) f(u(s))\,ds\\ &\ge \int_c^{1-c} G(p,s) g(s) (f_\infty-\delta)u(s)ds\\ & \geq \int_c^{1-c} G(p,s) g(s) a(s)\,ds\cdot (f_\infty-\delta) \|u\| \ge \|u\|, \end{align*} which implies  \|Tu\|\geq \|u\| . So, if we let \Omega_2=\{ u\in X\,|\ \|u\|0 for 0x for all x\in(0,+\infty), then \eqref{bhu}-\eqref{rfv} has no positive solution. \end{theorem} We conclude this paper with an example. \begin{example}\rm Consider the third-order boundary-value problem \begin{gather} u'''(t)=g(t)f(u(t)), \quad 0\frac{1}{A}\approx 10.613, then problem \eqref{rfv2}-\eqref{rfv3} has no positive solution. This example shows that our existence and nonexistence conditions are quite sharp. \end{example} \subsection*{Acknowledgment} The author is grateful to the anonymous referee for his/her careful reading of the manuscript and valuable suggestions for its improvement. \begin{thebibliography}{00} \bibitem{AAK}{D. R. 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