\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. $$\begin{definition}[\cite{b2}] \label{def2.2}\rm Let f\in C([0,1]\times \mathbb{R}^4,\mathbb{R}), \phi, \psi\in C([0,1],\mathbb{R}) and \phi(t)\le \psi(t), t\in [0,1]. We say that f(t,x_1,x_2,x_3,x_4) satisfies a Nagumo-type condition with respect to \phi,\psi if there exists a positive continuous function h(s) on [0,\infty) satisfying $$|f(t,x_1,x_2,x_3,x_4)|\le h(|x_4|),\label{e2.4}$$ for all (t,x_1,x_2,x_3,x_4)\in[0,1]\times[-M, M]^2 \times[\phi(t),\psi(t)]\times \mathbb{R}, and $$\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}$$ where \lambda=\max\{|\psi(1)-\phi(0)|,|\psi(0)-\phi(1)|\}. \end{definition} \begin{lemma}\label{lem2.3} Suppose f satisfies the Nagumo-type condition with respect to \phi,\psi\in C^2[0,1] and \phi\le \psi. If BVP (\ref{e2.3}) has a solution v(t) such that \phi(t)\le v(t)\le \psi(t), then there exists N>0 such that |v'(t)|\le N, for t\in[0,1]. \end{lemma} The proof of the above lemma is similar to that in \cite{b2}, therefore, we omit it. \section{Main results} \begin{theorem}\label{thm3.1} Suppose \alpha,\beta are lower and upper solutions to BVP \eqref{e1.1} such that \alpha''(t)\ge\beta''(t) and f satisfies a Nagumo-type condition with respect to \alpha'',\beta''. In addition, we assume that g is odd, continuous and increasing on \mathbb{R}, \theta is increasing on [0,1] and \theta(0)=0. Then BVP \eqref{e1.1} has a solution u(t) such that$$ \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 $$\int^N_\lambda \frac{s}{h(s)}ds>\max\limits_{0\le t\le 1}\psi(t) -\min\limits_{0\le t\le 1}\phi(t).\label{e3.1}$$ 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{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