\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 52, 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 2008 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2008/52\hfil Fourth-order boundary-value problem] {Existence of solutions for a fourth-order boundary-value problem} \author[Y. Liu\hfil EJDE-2008/52\hfilneg] {Yang Liu} \address{Yang Liu\newline Department of Mathematics\\ Yanbian University \\ Yanji, Jilin 133000, China. \newline Department of Mathematics \\ Huaiyin Teachers College\\ Huaian, Jiangsu 223300, China} \email{liuyang19830206@yahoo.com.cn} \thanks{Submitted October 19, 2007. Published April 10, 2008.} \subjclass[2000]{34B15} \keywords{Fourth-order boundary-value problem; upper and lower solution; \hfill\break\indent Riemann-Stieltjies integral; Nagumo-type condition} \begin{abstract} In this paper, we use the lower and upper solution method to obtain an existence theorem for the fourth-order boundary-value problem \begin{gather*} u^{(4)}(t)=f(t,u(t),u'(t),u''(t),u'''(t)),\quad 00. $$ \begin{definition}[\cite{b2}] \label{def2.2}\rm Let $f\in C([0,1]\times \mathbb{R}^4,\mathbb{R})$, $\phi, \psi\in C([0,1],\mathbb{R})$ and $\phi(t)\le \psi(t), t\in [0,1]$. We say that $f(t,x_1,x_2,x_3,x_4)$ satisfies a Nagumo-type condition with respect to $\phi,\psi$ if there exists a positive continuous function $h(s)$ on $[0,\infty)$ satisfying \begin{equation} |f(t,x_1,x_2,x_3,x_4)|\le h(|x_4|),\label{e2.4} \end{equation} for all $(t,x_1,x_2,x_3,x_4)\in[0,1]\times[-M, M]^2 \times[\phi(t),\psi(t)]\times \mathbb{R}$, and \begin{equation} \int^\infty_\lambda \frac{s}{h(s)}ds>\max\limits_{0\le t\le 1}\psi(t) -\min\limits_{0\le t\le 1}\phi(t),\label{e2.5} \end{equation} where $\lambda=\max\{|\psi(1)-\phi(0)|,|\psi(0)-\phi(1)|\}$. \end{definition} \begin{lemma}\label{lem2.3} Suppose $f$ satisfies the Nagumo-type condition with respect to $\phi,\psi\in C^2[0,1]$ and $\phi\le \psi$. If BVP (\ref{e2.3}) has a solution $v(t)$ such that $\phi(t)\le v(t)\le \psi(t)$, then there exists $N>0$ such that $|v'(t)|\le N$, for $t\in[0,1]$. \end{lemma} The proof of the above lemma is similar to that in \cite{b2}, therefore, we omit it. \section{Main results} \begin{theorem}\label{thm3.1} Suppose $\alpha,\beta$ are lower and upper solutions to BVP \eqref{e1.1} such that $\alpha''(t)\ge\beta''(t)$ and $f$ satisfies a Nagumo-type condition with respect to $\alpha'',\beta''$. In addition, we assume that $g$ is odd, continuous and increasing on $\mathbb{R}$, $\theta$ is increasing on $[0,1]$ and $\theta(0)=0$. Then BVP \eqref{e1.1} has a solution $u(t)$ such that $$ \alpha(t)\le u(t)\le \beta(t),\quad\alpha''(t)\ge u''(t)\ge \beta''(t). $$ \end{theorem} \begin{proof} Since $f$ satisfies the Nagumo-type condition with respect to $\phi=-\alpha'',\psi=-\beta''$, there exists a constant $N>0$ depending on $\phi,\psi,h$ such that \begin{equation} \int^N_\lambda \frac{s}{h(s)}ds>\max\limits_{0\le t\le 1}\psi(t) -\min\limits_{0\le t\le 1}\phi(t).\label{e3.1} \end{equation} Take $C> \max\{N,\|\phi'\|,\|\psi'\|\}$ and $p(v')=\max\{-C,\min\{v',C\}\}$. By modifying $\hat{f}$ and $g$ with respect to $\phi, \psi$, we aim at obtaining a second-order boundary-value problem and reformulating the new problem as an integral equation. We show that solutions of the modified problem lie in the region where $\hat{f},g$ are unmodified and hence are solutions of problem (\ref{e2.3}). Let $\varepsilon>0$ be a fixed small number and $F(v(t),v'(t)), G(\int_0^1v(t)d\theta(t))$ are the modifications of $\hat{f}(v(t),v'(t))$ and $g(\int_0^1v(t)d\theta(t))$ as follows \begin{align*} &F(v(t),v'(t))\\ &= \begin{cases} \hat{f}(\psi(t),\psi'(t))+\frac{v(t)-\psi(t)}{1+|v(t)-\psi(t)|}, &\text{if } v(t)\ge \psi(t)+\varepsilon,\\[3pt] \hat{f}(\psi(t),p(v'))+[\hat{f}(\psi(t),\psi'(t))-\hat{f}(\psi(t),p(v'(t)))\\ +\frac{v(t)-\psi(t)}{1+|v(t)-\psi(t)|}]\times \frac{v(t)-\psi(t)}{\varepsilon}, &\text{if } \psi(t)\le v(t)<\psi(t)+\varepsilon,\\[3pt] \hat{f}(v(t),p(v'(t))),& \text{if } \phi(t)\le v(t)\le \psi(t),\\[3pt] \hat{f}(\phi(t),p(v'(t)))+[\hat{f}(\phi(t),\phi'(t)) -\hat{f}(\phi(t),p(v'(t)))\\ +\frac{\phi(t)-v(t)}{1+|\phi(t)-v(t)|}]\times \frac{\phi(t)-v(t)}{\varepsilon}, & \text{if } \phi(t)-\varepsilon< v(t)\le \phi(t),\\[3pt] \hat{f}(\phi(t),\phi'(t))+\frac{\phi(t)-v(t)}{1+|\phi(t)-v(t)|}, & \text{if } v(t)\le \phi(t)-\varepsilon, \end{cases} \end{align*} and \begin{align*} &G\big(\int_0^1v(t)d\theta(t)\big)\\ &=\begin{cases} g(\int^1_0\psi(t)d\theta(t))+ \frac{\int_0^1v(t)d\theta(t)-\int^1_0\psi(t)d\theta(t) }{1+|\int_0^1v(t)d\theta(t)-\int^1_0\psi(t)d\theta(t)|}, & \text{if } v(t)>\psi(t),\\ g(\int_0^1v(t)d\theta(t)), & \text{if } \phi(t)\le v(t)\le\psi(t),\\ g(\int^1_0\phi(t)d\theta(t))+ \frac{\int^1_0\phi(t)d\theta(t)-\int_0^1v(t)d\theta(t)} {1+|\int^1_0\phi(t)d\theta(t)-\int_0^1v(t)d\theta(t)|}, & \text{if } v(t)<\phi(t). \end{cases} \end{align*} Obviously, $F: \mathbb{R}\times \mathbb{R}\to \mathbb{R}$ and $G : \mathbb{R}\to \mathbb{R}$ are continuous and bounded. Consider the modified problem \begin{equation} \begin{gathered} -v''(t)=F(v(t),v'(t)),\quad 0C$, $m>0$ such that \begin{gather*} |F(v(t),v'(t))|0$. We divide the proof into three cases. \smallskip \noindent\textbf{Case 1.} $t_0=0$. Then we have $w(0)=\phi(0)- v_*(0)=\phi(0)>0$. It contradict the definition of $\phi$. \smallskip \noindent\textbf{Case 2.} $t_0=1$. Then $w(1)>0$ and $w'(1)\ge 0$. The boundary value conditions of \eqref{e3.2} imply $$ w'(1)=\phi'(1)-{v_*}'(1)\le g(\int^1_0\phi(t)d\theta(t)) -G(\int^1_0v_{*}(t)d\theta(t)). $$ If $v_*(t)<\phi(t)$, then \begin{align*} G(\int^1_0v_*(t)d\theta(t)) &=g(\int^1_0\phi(t)d\theta(t))+ \frac{\int^1_0\phi(t)d\theta(t)-\int^1_0v_*(t)d\theta(t)} {1+\int^1_0\phi(t)d\theta(t)-\int^1_0v_*(t)d\theta(t)}\\ &>g(\int^1_0\phi(t)d\theta(t)), \end{align*} which implies $w'(1)<0$. It is a contradiction. If $v_*(t)>\psi(t)$, then \begin{align*} G(\int^1_0v_*(t)d\theta(t)) &=g(\int^1_0\psi(t)d\theta(t))+ \frac{\int^1_0v_*(t)d\theta(t)-\int^1_0\psi(t)d\theta(t)} {1+\int^1_0v_*(t)d\theta(t)-\int^1_0\psi(t)d\theta(t)}\\ &>g(\int^1_0\psi(t)d\theta(t))\\ &\ge g(\int^1_0\phi(t)d\theta(t)), \end{align*} we can also get $w'(1)<0$, which is a contradiction. Hence, $\phi(t)\le v_*(t)\le\psi(t).$ So $$ G(\int^1_0v_*(t)d\theta(t))=g(\int^1_0v_*(t)d\theta(t)) \ge g(\int^1_0\phi(t)d\theta(t)), $$ which implies $w'(1)\le 0$. If $w'(1)<0$, it is a contradiction. So we have $w'(1)=0$. Since $t_0\neq 0$, there exists $t_1\in[0,1)$ such that $w(t_1)=0$ and $w(t)>0$ on $(t_1,1]$. Then for each $t\in[t_1,1]$, we have $$ w''(t)=\phi''(t)-{v_*}''(t)\ge-\hat{f}(\phi(t),\phi'(t)) +\Big[\hat{f}(\phi(t),\phi'(t))+ \frac{w(t)}{1+w(t)}\Big]>0. $$ Thus, by $w'(1)=0$, we get $w'(t)\le0\quad {\rm on} \quad[t_1,1]$, which implies that $w$ is decreasing on $[t_1,1]$ and hence $w(1)\le 0$, it is a contradiction. \smallskip \noindent\textbf{Case 3.} $t_0\in(0,1)$. Then, we have $w'(t_0)=0$ and $w''(t_0)\le 0$. However, for $00, \end{align*} a contradiction. For $w(t_0)\ge \varepsilon$, we obtain $$ w''(t_0)=\phi''(t_0)-{v_*}''(t_0)\ge\frac{w(t_0)}{1+w(t_0)}>0, $$ it is also a contradiction. Thus, $\phi(t)\le v_*(t), t\in [0,1]$. By the similar discussion, we can get $v_*(t)\le \psi(t)$. According to the Lemma \ref{lem2.3} and the choice of $C$, for the solution $v_*$ of \eqref{e3.2} with $\phi(t)\le v_*(t)\le \psi(t)$, $t\in [0,1]$, we have $$ |{v_*}'(t)|\le N