\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 139, pp. 1--9.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2010 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2010/139\hfil Solvability] {Solvability of a three-point nonlinear boundary-value problem} \author[A. Guezane-Lakoud, S. Kelaiaia \hfil EJDE-2010/139\hfilneg] {Assia Guezane-Lakoud, Smail Kelaiaia} % in alphabetical order \address{Assia Guezane-Lakoud \newline Department of Mathematics\\ Faculty of Sciences \\ University Badji Mokhtar\\ B.P. 12, 23000, Annaba, Algeria} \email{a\_guezane@yahoo.fr} \address{Smail Kelaiaia \newline Department of Mathematics\\ Faculty of Sciences \\ University Badji Mokhtar\\ B.P. 12, 23000, Annaba, Algeria} \email{kelaiaiasmail@yahoo.fr} \thanks{Submitted March 19, 2010. Published September 27, 2010.} \subjclass[2000]{34B10, 34B15} \keywords{Fixed point theorem; three-point boundary-value problem; \hfill\break\indent non trivial solution} \begin{abstract} Using the Leray Schauder nonlinear alternative, we prove the existence of a nontrivial solution for the three-point boundary-value problem \begin{gather*} u''+f(t,u)= 0,\quad 01$such that$T(x)=\lambda x$, or there exists a fixed point$x^{\ast }\in \overline{\Omega }$. \end{lemma} \section{Main Results} In this section, we present and prove our main results. \begin{theorem} \label{thm3} We assume that$f(t,0)\neq 0$,$1+\alpha \neq \beta $and there exist nonnegative functions$k,h\in L^{1} [0,1] $such that \begin{gather} | f(t,x)| \leq k(t)| x| +h(t),\quad \text{a.e. }(t,x)\in [0,1 ] \times\mathbb{R}, \label{e5} \\ (1+\frac{| 1+\alpha | }{| 1+\alpha -\beta | })\int_{0}^{1}(1-s)k(s)ds+| \beta \frac{1+\alpha }{1+\alpha -\beta }| \int_{0}^{\eta }k(s)ds<1 \label{e6} \end{gather} Then BVP \eqref{e1}-\eqref{e2} has at least one nontrivial solution$u^{\ast }\in C[0,1] $. \end{theorem} \begin{proof} Setting \begin{gather*} M=(1+\frac{| 1+\alpha | }{| 1+\alpha -\beta | })\int_{0}^{1}(1-s)k(s)ds+| \beta \frac{1+\alpha }{1+\alpha -\beta }| \int_{0}^{\eta }k(s)ds, \\ N=(1+\frac{| 1+\alpha | }{| 1+\alpha -\beta | })\int_{0}^{1}(1-s)h(s) ds+| \beta \frac{1+\alpha }{1+\alpha -\beta }| \int_{0}^{\eta }(\eta -s)h(s)ds. \end{gather*} By \eqref{e6}, we have$M<1$. Since$f(t,0)\neq 0$, then there exists an interval$[\sigma ,\tau ] \subset [0,1] $such that$\underset{\sigma \leq t\leq r}{\min }| f(t,0)| >0$and as$h(t)\geq |f(t,0)| $, for all$t\in [0,1] $then$N>0$. Let$m=\frac{N}{(1-M)}$,$\Omega =\{ u\in C[0,1] :\| u\| 1$such$Tu=\lambda u, then \begin{align*} \lambda m&=\lambda \|u\| =\| Tu\| =\max_{0\leq t\leq 1}| (Tu)(t)|\\ &\leq \| u\| \Big[(1+\frac{| 1+\alpha| }{| 1+\alpha -\beta | }) \int_{0}^{1}(1-s)k(s)ds+| \beta \frac{1+\alpha }{ 1+\alpha -\beta }| \int_{0}^{\eta }k(s)ds \\ &\quad + (1+\frac{| 1+\alpha | }{| 1+\alpha -\beta | })\int_{0}^{1}(1-s) h(s)ds+| \beta \frac{1+\alpha }{1+\alpha -\beta }| \int_{0}^{\eta }h(s)ds\Big] \\ &= M\| u\| +N \end{align*} From this we obtain $\lambda \leq M+\frac{N}{m}=M+\frac{N}{N(1-M)^{-1}}=M+(1-M)=1,$ this contradicts\lambda >1$. By Lemma \ref{lem2} we conclude that operator$T$has a fixed point$u^{\ast }\in \overline{\Omega }$and then BVP \eqref{e1}-\eqref{e2} has a nontrivial solution$u^{\ast }\in C[0,1]$. \end{proof} \begin{theorem} \label{thm4} Assume \eqref{e5} and one of the following four conditions: \begin{itemize} \item[(1)] There exists a constant$p>1$such that $$\int_{0}^{1}k^{p}(s)ds<\Big[\frac{(1+q)^{1/q}}{1+| \frac{ 1+\alpha }{1+\alpha -\beta }| +(| \beta \frac{ 1+\alpha }{1+\alpha -\beta }| )(\eta (1+q))^{1/q}} \Big] ^{p}\quad (\frac{1}{p}+\frac{1}{q}=1); \label{e7}$$ \item[(2)] There exists constant$\mu >-1$such that $$k(s)<\frac{(\mu +1)(\mu +2)}{1+| \frac{ 1+\alpha }{1+\alpha -\beta }| +(| \beta \frac{ 1+\alpha }{1+\alpha -\beta }| )(\mu +2)\eta ^{\mu +1}}s^{\mu } \label{e8}$$ and $\operatorname{meas}\big\{ s\in [0,1] : k(s)<\frac{(\mu +1)( \mu +2)}{1+| \frac{1+\alpha }{1+\alpha -\beta }| +| \beta \frac{1+\alpha }{1+\alpha -\beta }| (\mu +2)\eta ^{\mu +1}}s^{\mu }\big\} >0;$ \item[(3)] The function$k(s)$satisfies $$k(s)<\frac{| 1+\alpha -\beta | }{| 1+\alpha | +| 1+\alpha -\beta | +| \beta ( 1+\alpha )| \eta } \label{e9}$$ and $\operatorname{meas}\big\{ s\in [0,1]:k(s)<\frac{| 1+\alpha -\beta | }{| 1+\alpha | +| 1+\alpha -\beta | +| \beta (1+\alpha )| \eta } \big\} >0;$ \item[(4)] The function$f(t,x)$satisfies $$\omega =\limsup_{| x| \to \infty} \max_{t\in [0,1] } | \frac{f(t,x)}{x}| <\frac{1}{2}(\frac{| 1+\alpha -\beta | }{| 1+\alpha | +| 1+\alpha -\beta | +| \beta (1+\alpha )| \eta }). \label{e10}$$ \end{itemize} Then BVP \eqref{e1}-\eqref{e2} has at least one nontrivial solution$u^{\ast }\in C[0,1]$. \end{theorem} \begin{proof} Let$M$be defined as in the proof of Theorem \ref{thm3}. To prove Theorem \ref{thm4}, we only need to prove that$M<1. (1) By using H\"{o}lder inequality, we obtain \begin{align*} M &\leq \big(1+| \frac{1+\alpha }{1+\alpha -\beta }|\big) \Big(\int_{0}^{1}k^{p}(s)ds\Big)^{1/p} \Big(\int_{0}^{1}(1-s)^{q}ds\Big)^{1/q}\\ &\quad + | \beta \frac{1+\alpha }{1+\alpha -\beta }| \Big(\int_{0}^{\eta }k^{p}(s)ds\Big)^{1/p} \Big(\int_{0}^{\eta}ds\Big)^{1/q}\,. \end{align*} Then \begin{align*} M &\leq \Big(\int_{0}^{1}k^{p}(s)ds\Big)^{1/p} \Big[\big(1+| \frac{1+\alpha }{1+\alpha -\beta }| \big) \Big(\int_{0}^{1}(1-s)^{q}ds\Big)^{1/q}\\ &\quad + |\beta \frac{1+\alpha }{1+\alpha -\beta }| \Big(\int_{0}^{\eta }ds\Big)^{1/q}\Big]. \end{align*} Integrating, it yields \begin{align*} M&<\frac{(1+q)^{1/q}}{(1+| \frac{1+\alpha }{1+\alpha -\beta }| )+|\beta \frac{1+\alpha }{1+\alpha -\beta }|(\eta (1+q))^{1/q}}\\ &\quad \times \Big[(1+\frac{|1+\alpha |}{|1+\alpha -\beta |}) (\frac{1}{1+q})^{1/q}+|\beta \frac{1+\alpha }{1+\alpha -\beta }| \eta ^{1/q}\Big] =1. \end{align*} (2) Using the same reasoning as in the proof of the first statement we obtain \begin{align*} M &<\frac{(\mu +1)(\mu +2)}{1+| \frac{ 1+\alpha }{1+\alpha -\beta }| +(| \beta \frac{ 1+\alpha }{1+\alpha -\beta }| )(\mu +2)\eta ^{\mu +1}}\\ &\quad \times \Big[\big(1+| \frac{1+\alpha }{1+\alpha -\beta }| \big)\int_{0}^{1}(1-s)s^{\mu }ds+| \beta \frac{1+\alpha }{ 1+\alpha -\beta }| \int_{0}^{\eta }s^{\mu }ds\Big] \\ &=\frac{(\mu +1)(\mu +2)}{1+| \frac{ 1+\alpha }{1+\alpha -\beta }| +| \beta \frac{1+\alpha }{ 1+\alpha -\beta }| (\mu +2)\eta ^{\mu +1}} \big(1+| \frac{1+\alpha }{1+\alpha -\beta }| \big) \big(\frac{1}{(\mu +1)(\mu +2)}\big)\\ &\quad + \big| \beta \frac{1+\alpha }{1+\alpha -\beta }\big| \big(\frac{\eta ^{\mu +1}}{\mu +1}\big)= 1 \end{align*} (3) we have $M = (1+\frac{| 1+\alpha | }{| 1+\alpha -\beta | })\int_{0}^{1}(1-s)k(s)ds +| \beta \frac{1+\alpha }{1+\alpha -\beta }| \int_{0}^{\eta }k(s)ds.$ Then \begin{align*} M &<\big(\frac{| 1+\alpha -\beta | }{| ( 1+\alpha )| +| 1+\alpha -\beta | +| \beta (1+\alpha )| \eta }\big)\\ &\quad \times \big(1+| \frac{1+\alpha }{1+\alpha -\beta }| \big) \int_{0}^{1}(1-s)ds+| \beta \frac{1+\alpha }{1+\alpha -\beta } | \int_{0}^{\eta }ds\\ &=\frac{1}{2}(\tfrac{| 1+\alpha -\beta | }{ | (1+\alpha )| +| 1+\alpha -\beta | +| \beta (1+\alpha )| \eta } )(1+| \tfrac{1+\alpha }{1+\alpha -\beta }| )+| \beta \tfrac{1+\alpha }{1+\alpha -\beta }| \eta =1 \end{align*} (4) From\omega =\limsup_{| x| \to \infty } \max_{t\in [0,1] } | \frac{ f(t,x)}{x}| $we deduce that there exists$c>0$such that for$| x| >c$we have $| f(t,x)| \leq (\omega +\varepsilon )| x| \quad \forall \varepsilon >0.$ Set $h(t)=\max \{ | f(t,x)| :( t,x)\in [0,1] \times (-c,c)\}.$ Then for$(t,x)\in [0,1] \times\mathbb{R}$, with$\varepsilon =\omega, we obtain \begin{align*} | f(t,x)| &\leq 2\omega |x| +h(t)\\ &\leq \frac{| 1+\alpha -\beta | }{| (1+\alpha )| +| 1+\alpha -\beta | +| \beta (1+\alpha )| \eta }| x| +h(t). \end{align*} Setting $k(t)<\frac{| 1+\alpha -\beta | }{| (1+\alpha )| +| 1+\alpha -\beta | +| \beta (1+\alpha )| \eta },$ by applying the above statement we complete the proof. \end{proof} \begin{corollary} \label{coro5} Assume the conditions of Theorem \ref{thm4}, and one of the following four conditions: \begin{itemize} \item[(1)] There exists a constantp>1$such that $\int_{0}^{1}k^{p}(s)ds<\Big[\frac{(1+q)^{1/q}}{1+| \frac{ 1+\alpha }{1+\alpha -\beta }| +| \beta \frac{1+\alpha }{ 1+\alpha -\beta }| (1+q)^{1/q}}\Big] ^{p},\quad (\frac{1}{p}+\frac{1}{q}=1).$ \item[(2)] There exists a constant$\mu >-1$such that \begin{gather*} k(s)<\frac{(\mu +1)(\mu +2)}{1+| \frac{ 1+\alpha }{1+\alpha -\beta }| +(| \beta \frac{ 1+\alpha }{1+\alpha -\beta }| )(\mu +2)}s^{\mu}, \\ \operatorname{meas}\big\{ s\in [0,1] : k(s)<\frac{(\mu +1)( \mu +2)}{1+| \frac{1+\alpha }{1+\alpha -\beta }| +| \beta \frac{1+\alpha }{1+\alpha -\beta }| (\mu +2)}s^{\mu }\big\} >0 \end{gather*} \item[(3)] The function$k(s)$satisfies \begin{gather*} k(s)<\frac{| 1+\alpha -\beta | }{| ( 1+\alpha )| +| 1+\alpha -\beta | +| \beta (1+\alpha )| \eta }, \\ \operatorname{meas}\left\{ s\in [0,1] ,k(s)<\frac{| 1+\alpha -\beta | }{| (1+\alpha )| +| 1+\alpha -\beta | +| \beta (1+\alpha ) | \eta }\right\} >0 \end{gather*} \item[(4)] The function$f(t,x)satisfies \begin{align*} \omega &= \limsup_{| x| \to \infty } \max_{t\in [0,1]} | \frac{f(t,x)}{x}| \\ &< 1/2(\frac{| 1+\alpha -\beta | }{| ( 1+\alpha )| +| 1+\alpha -\beta | +| \beta (1+\alpha )| }) \end{align*} \end{itemize} Then the BVP \eqref{e1}-\eqref{e2} has at least one nontrivial solutionu^{\ast }\in C[0,1]. \end{corollary} \begin{proof} Taking into account \begin{align*} M &= \big(1+\frac{| 1+\alpha | }{| 1+\alpha -\beta | }\big)\int_{0}^{1}(1-s)k(s)ds+| \beta \frac{1+\alpha }{1+\alpha -\beta }| \int_{0}^{\eta }k(s)ds \\ &\leq \big(1+\frac{| 1+\alpha | }{| 1+\alpha -\beta | }\big)\int_{0}^{1}(1-s)k(s)ds+| \beta \frac{1+\alpha }{1+\alpha -\beta }| \int_{0}^{1}k(s)ds. \end{align*} the proof follows. \end{proof} \section{Examples} To illustrate our results, we give the following examples. \begin{example} \label{exa6} \rm Consider the three-point BVP \begin{gathered} u''+\frac{u}{2}(t\sin \sqrt{u}-e^{-t}\cos u) +te^{t}=0,\quad 00\,. \end{align*} Hence, by Theorem \ref{thm4}, BVP \eqref{e12} has at least one nontrivial solutionu^{\ast }$in$C[0,1]. \end{example} \begin{example} \label{exa8} \rm Consider the three-point BVP $$\begin{gathered} u''+\frac{u^{3}}{2(1+e^{-t^{2}})^{4}(1+u^{2})}-\sqrt{t}\sin t=0 \\ u(0)=\frac{1}{3}u'(0),\quad u(1)=\frac{-1}{6}u'(\frac{1}{4}). \end{gathered} \label{e13}$$ We have \begin{gather*} f(t,x)=\frac{x^{3}}{2(1+e^{-t^{2}})^{4}( 1+x^{2})}-\sqrt{t}\sin t, \\ | f(t,x)| \leq | x| \frac{(1+e^{-t^{2}})^{-4}}{2}+\sqrt{t}\sin t = k(t)| x| +h(t) \\ | \frac{f(t,x)}{x}| \leq \frac{(1+e^{-t^{2}})^{-4}}{2}+\frac{\sqrt{t}\sin t}{| x| } \leq 0.142+\frac{1}{| x|}. \end{gather*} Applying the fourth statement in Theorem \ref{thm4} we obtain \begin{align*} \omega &= \limsup_{| x| \to \infty } \max_{t\in [0,1] } | \frac{f(t,x)}{x}| =\limsup_{| x| \to\infty } (0.142+\frac{1}{| x| })\\ &=0.142< \frac{1}{2}(\frac{\frac{13}{9}}{\frac{4}{3} +\frac{13}{9}+\frac{1}{27}})=0.25658 \end{align*} Hence, BVP \eqref{e13} has at least one nontrivial solutionu^{\ast }$in$C[0,1]\$. \end{example} \begin{example} \label{exa9} \rm Consider the three-point BVP \begin{gathered} u''+\frac{\sin u}{2\sqrt{4+t}}-t^{2}\cos t+t\cos (t^2)=0, \quad 0