\documentclass[reqno]{amsart} \usepackage{amsfonts} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 106, pp. 1--18.\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 2005 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2005/106\hfil Three-point boundary-value problem] {Positive solutions of three-point boundary-value problems for p-Laplacian singular differential equations} \author[G. N. Galanis, A. P. Palamides\hfil EJDE-2005/106\hfilneg] {George N. Galanis, Alex P. Palamides} \address{George N. Galanis \hfill\break Naval Academy of Greece, Piraeus, 185 39, Greece} \email{ggalanis@math.uoa.gr} \address{Alex P. Palamides \hfill\break Department of Communication Sciences, University of Peloponnese, 22100 Tripolis, Greece} \email{palamid@uop.gr} \date{} \thanks{Submitted May 13, 2005. Published October 7, 2005.} \subjclass[2000]{34B15, 34B18} \keywords{Three-point singular boundary-value problem; p-Laplacian; \hfill\break\indent positive and negative solutions; vector field; Nonlinear alternative of Leray-Schauder} \begin{abstract} In this paper we prove the existence of positive solutions for the three-point singular boundary-value problem \begin{equation*} -[\phi _{p}(u')]'=q(t)f(t,u(t)),\quad 01)$is the well known$p$-Laplacian operator,$0<\eta <1$;$0<\alpha ,\beta <1$are fixed points and$g$is a monotone continuous function defined on the real line$\mathbb{R}$with$g(0)=0$and$ug(u)\geq 0$. The inhomogeneous term in \eqref{1.1} is allowed to be singular at$u=0$and$q(t)$may be singular at$t=0$or/and$t=1$and finally we suppose that$q(0)>0$. Further assumptions concerning the nonlinearity$f(t,u)$will be clarified later. Let$B$be the Banach space$C[0,1]$endowed with the norm$\Vert x\Vert =\max_{t\in [ 0,1]}|x(t)|$. A solution$u(t)$of \eqref{1.1} subject to \eqref{1.2} or \eqref{1.3} means that$u(t)\in C^{1}[0,1]$, is positive on$(0,1)$,$\phi _{p}(u'(.))\in C(0,1)\cap L^{1}[0,1]$and satisfies the differential equation as well as the corresponding boundary conditions. It is well known that when$p>1$,$\phi _{p}(s)$is strictly increasing on$\mathbb{R}$and so its inverse$\phi _{p}^{-1}$exists and further$\phi _{p}^{-1}=\phi _{q}$, where$1/p+1/q=1$. In \cite{EW}, Erbe and Wang by using Green's functions and Krasnoselskii's fixed point theorem in cones proved existence of a positive solution of the boundary-value problem studied the Sturm-Liouville boundary-value problem \begin{gather*} x^{\prime \prime }(t)=-f(t,x(t)), \\ \alpha x(0)-\beta x'(0)=0,\quad \gamma x(1)+\delta x'(1)=0, \end{gather*}% where$\alpha ,\beta ,\gamma ,\delta \geq 0$and$\rho :=\beta \gamma +\alpha \gamma +\alpha \delta >0$, mainly under the assumptions: \begin{gather*} f_{0}:=\lim_{x\rightarrow 0+}\max_{0\leq t\leq 1}\frac{f(t,x)}{x}=0, \\ f_{\infty }:=\lim_{x\rightarrow +\infty }\min_{0\leq t\leq 1} \frac{f(t,x)}{x}=+\infty \end{gather*}% i.e.,$f$is \emph{supelinear} at both ends points$x=0$and$x=\infty $or under \begin{gather*} f_{0}:=\lim_{x\rightarrow 0+}\min_{0\leq t\leq 1}\frac{f(t,x)}{x}=+\infty \\ f_{\infty }:=\lim_{x\rightarrow +\infty }\max_{0\leq t\leq 1} \frac{f(t,x)}{x}=0, \end{gather*}% i.e.,$f$is \emph{sublinear} at both$x=0$and$x=\infty $. The study of multi-point boundary-value problems was initiated by Il'in and Moiseev in \cite{im1,im2}. Many authors since then considered nonlinear 3-point boundary-value problems (see e.g., \cite{da, fe, fw, ghm, ek, ma1, ma2, We, w2} and the references therein). In particular, Ma in \cite{ma2} proved the existence of a positive solution to the three-point nonlinear boundary-value problem \begin{gather*} -u''(t)=q(t)f(u(t)),\quad 00$ continuous non-increasing on $(0,+\infty )$ and $% \int_{0}^{L}f_{1}(u)du<+\infty$ for any fixed $L>0$; $f_{2}\geq 0$ and continuous on $[0,+\infty )$ \item[(H4)] For any $K>0$ there exists $\psi _{K}(t):(0,1)\to (0,+\infty )$ such that $f(t,y)\geq \psi _{K}(t),\ t\in (0,1)$ for any $y(t)\in C[0,1]$ with $0\leq y(t)\leq K$ \item[(H5)] $\int_{0}^{\eta }\phi _{p}^{-1}(\int_{s}^{\eta }q(r)\psi (r)dr)ds>0$, $\int_{\eta }^{1}\phi _{p}^{-1}(\int_{\eta }^{s}q(r)\psi (r) dr)ds>0$ and for any $k_{1}>0$ and $k_{2}>0$, $\int_{0}^{% \eta}f_{1}(k_{1}s)q(s)ds +\int_{\eta }^{1}f_{1}(k_{2}(1-s))q(s)ds<+\infty$ and mainly \begin{equation*} \sup_{c>0}\frac{c}{\phi _{p}^{-1}(I^{-1}[G_{0}(c) ])(\frac{p}{p-1})^{\frac{1% }{p}}[\frac{1}{ 1-\beta }(\int_{0}^{\eta }[q(s)]^{\frac{1}{p} }ds+\int_{\eta }^{1}[q(s)]^{\frac{1}{p}}ds) ]}>1, \end{equation*} where \begin{equation*} I(c):=\int_{0}^{c}\phi _{p}^{-1}(z)dz=\frac{p-1}{p} c^{\frac{p}{p-1}}\text{ \ \ and \ }G_{0}(c) =\int_{0}^{c}[f_{1}(u)+f_{2}(u)]du\text{\ }. \end{equation*} \end{itemize} It is not difficult to prove the next useful properties of $I(c)$: \begin{equation*} I^{-1}(uv)\leq I^{-1}(u)I^{-1}(v) \quad\text{for }u\geq 0,v\geq 0 \end{equation*} and whenever $c<0$, we have $I(-c)=I(c)$. Indeed, since $-c>0$, \begin{equation*} I(-c)=\int_{0}^{-c}\phi _{p}^{-1}(z) dz=\int_{0}^{-c}\phi _{q}(z) dz=\int_{0}^{-c}|z|^{q-2}zdz=\int_{0}^{-c}z^{q-1}dz=\frac{(-c) ^{q}}{q} \end{equation*} and \begin{equation*} I(c) =\int_{0}^{c}|z|^{q-2}zdz=\int_{0}^{c}(-z)^{q-2}zdz=- \int_{0}^{c}(-z)^{q-1}d(-z)=\frac{(-c)^{q}}{q}. \end{equation*} In this work, mainly motivated by the above mentioned paper of Ma and Ge \cite{MG}, we combine the properties of the vector field at the face $(u,u')$ plane and sublinearity of $f(t,u)$ at the origin $u=0$ with the alternative continuation principle of Leray-Shauder, proving the existence of a positive solution for the boundary-value problem \eqref{1.1}-\eqref{1.3} and eliminating several of the assumptions (H1)-(H5). \section{Preliminaries} We now proceed with the auxiliaries. Consider the boundary-value problem \begin{gather} -[\phi _{p}(u')]'=q(t)f(t,u(t)),\quad 00$and nondecreasing on$(0,1)$\item[(A2)]$f\in C([0,1]\times (0,+\infty),(0,+\infty ))$\item[(A3)]$\int_{0}^{L}\max_{0\leq t\leq 1}f(t,u)du<+\infty $, for any fixed$L>0$. \item[(A4)]$g\in C(\mathbb{R},\mathbb{R})$, is a nondecreasing function with$ug(u)>0$,$u\neq 0$. \end{itemize} \begin{remark} \label{rmk1} \rm Note that the differential equation \eqref{2.1} defines a vector field whose properties will be crucial for our study. More specifically, working at the$(u,u')$\textit{face semi-plane}$(u>0)$, the sign condition on$f$(see assumption (A2)), immediately gives (since$\phi _{p}'(u')>0$for all$u'\in \mathbb{R}$) that$u''<0$. Thus, any trajectory$(u(t),u'(t))$,$t\geq 0$, emanating from the semi-line \begin{equation*} E:=\{(u,u'):u-g(u')=0,\quad \;u>0\} \end{equation*} trends'' in a natural way, (when$u'(t)>0$) toward the positive$u$-semi-axis and then (when$u'(t)<0$) turns toward the negative$u'$-semi-axis. Finally, by setting a certain growth rate on$f$(say sublinearity) we can control the vector field, so that all trajectories with$u(0)$small enough satisfy the relation \begin{equation*} u(1)-\beta u(\eta )\neq 0. \end{equation*} So, all solutions of the given boundary-value problem cannot have their initial values arbitrary small, avoiding in this way the singular point$u=0$of the nonlinearity. \end{remark} Namely we have the next result. \begin{lemma} \label{Le1} Let$0<\beta <1$. If$u\in C[0,1]$is a solution of \eqref{2.1}-\eqref{2.2}, then$u=u(t)$is concave. Furthermore for every solution with$u(0)>0$, it follows that \begin{itemize} \item[(i)] There exists a$t_{0}\in [0,1)$such that$u(t_{0})=\max_{0\leq t\leq 1}\|u(t)\|=\|u\|$, \item[(ii)]$u(t)>0$,$t\in [0,1]$and \item[(iii)]$\inf_{t\in [\eta ,1]}u(t)\geq \gamma u(t_{0})=\gamma \|u\|$, where$\gamma =\min \{ \beta \eta ,\;\frac{\beta (1-\eta ) }{1-\beta \eta }\} $. \end{itemize} \end{lemma} \begin{proof} Let$u(t)$be a solution to \eqref{2.1}-\eqref{2.2}. Then, since$[\phi _{p}(u')]'=-q(t)f(t,u(t))\leq 0$,$\phi _{p}(u')$is non-increasing. Consequently$u'(t)$is non-increasing which implies the concavity of$u(t)$. (i) Since$u(0)>0$, by the first condition in \eqref{2.2} and the assumption (A4), we get$u'(0)>0$. If$u'(t)\geq 0$,$t\in [0,1]$then$% u(1)\geq u(\eta )>\beta u(\eta )$, a contradiction. Hence there exists$t_{0}>0$such that$u(t_{0}) =\max_{0\leq t\leq 1}\|u(t)\|=\|u\|$. (ii) If$\eta \in (0,t_{0})$, then$u(\eta )>u(0)>0$and so$u(1)=\beta u(\eta )>0$. If$\eta \in (t_{0},1)$, then$u(\eta )>u(1)$and so \begin{equation*} 0=u(1)-\beta u(\eta )0$. Finally, the concavity of $u(t)$ yields $u(t)>0$, $t\in [0,1]$. (iii) The proof follows the concavity of the solution. Indeed, since $u(1)=\beta u(\eta )u(1)$, we have again \begin{equation*} \min_{t\in [ \eta ,1]}u(t)=u(1). \end{equation*}% >From the concavity of $u$, we know that \begin{equation*} \frac{u(\eta )}{\eta }\geq \frac{u(t_{0})}{t_{0}}. \end{equation*}% Combining the above and the boundary condition $u(1)=\beta u(\eta )$, we conclude that \begin{equation*} \frac{u(\eta )}{\beta \eta }\geq \frac{u(t_{0})}{t_{0}}\geq u(t_{0})=\Vert u\Vert , \end{equation*} that is $\min_{t\in [ \eta ,1]}u(t)\geq \beta \eta \Vert u\Vert$. \end{proof} \section{Existence for the first boundary-value problem} In this section we consider the boundary-value problem \eqref{2.1}-\eqref{2.2} and prove the next result. \begin{lemma} Suppose that conditions \eqref{A2S}-\eqref{G} hold. Then, there exists an $\eta _{0}>0$ such that for any $\eta \leq \eta _{0}$ any solution of \eqref{2.1} with $u(0)=\frac{\eta }{2}$, satisfies the inequality 00$\ there exists$\eta _{0}$such that \min_{0\leq t\leq 1}f(t,u)>Ku,\;0\max \{2\mu ^{2},\;2\frac{1+2\mu }{\min \{\gamma ,1\}}\}$. We examine first the case $p>2$. Taking into account\ \eqref{G}, we may chose $\eta _{0}$ small enough so that $$\frac{g^{-1}(\frac{\eta _{0}}{2})}{\frac{\eta _{0}}{2}}\leq 2\mu \quad \text{and}\quad (p-1)\frac{[g^{-1}(\eta )]^{p-2}}{\min_{0\leq t\leq 1}q(t)} <1,\quad \eta \in (0,\eta _{0}] \label{2.40}$$ (if $\lim_{u\rightarrow 0}\frac{g^{-1}(u)}{u}=0$, then we may find $\mu >0$ so that \eqref{2.40} still holds true). Assume that the boundary-value problem \eqref{2.1}-\eqref{2.2} has a solution $u(t)$, $t\in [0,1]$ with initial value $u(0)$ arbitrary small. Then, we may assume that $u(0)=\eta /2$ for some $\eta \in (0,\eta _{0}]$ with \begin{equation*} \frac{\min_{0\leq t\leq 1}q(t)}{(p-1)[ g^{-1}(\frac{\eta }{2})]^{p-2}}\geq 1. \end{equation*} We demonstrate first that \eqref{2.4} holds true. If not, by Lemma \ref{Le1}, there exist $t^{\ast }\in (0,1]$ such that $\frac{\eta }{2}\leq u(t)<\eta$, $0\leq t2\mu ^{2}$, yields $\phi (t)>0$ for all $t\in [ 0,1]$. As a result, noticing Lemma \ref{Le1}, we obtain $0\frac{1}{2\mu}>1$ (in view of \eqref{G}), a contradiction due to the initial choice of $t^{\ast }\in [ 0,1]$. \end{proof} \begin{lemma} \label{lem3} Suppose that conditions \eqref{A2S}-\eqref{G} hold. Then for any $\eta \in (0,\eta _{0})$\ ($\eta _{0}$ as above) there exists an $\alpha _{0}=\alpha _{0}(\eta )>0$ such that for any (possible) solution $u=u(t)$ of \eqref{2.1}-\eqref{2.2}, with $u(0)=\frac{\eta }{2}$, the following inequality holds: $u(t)\geq \alpha _{0},\quad t\in [0,1]\,.$ \end{lemma} \begin{proof} By the concavity of $u(t)$, it is obvious that \begin{equation*} \min_{t\in [0,1]}u(t)=\min \{ u(0),\;u(1)\} . \end{equation*} However, in view of Lemma \ref{Le1}, $\min_{t\in [\eta ,1]}u(t)\geq \gamma u(t_{0})=\gamma \|u\|\geq \gamma u(0)=\gamma \frac{\eta }{2}$ and so \begin{equation*} u(t)\geq \min_{t\in [0,1]}u(t)\geq \frac{\eta }{2}\min \{ 1,\gamma \} :=\alpha _{0}(\eta). \end{equation*} \end{proof} \begin{proposition} \label{Pr1} Suppose that conditions \eqref{A2S}-\eqref{G} hold. Then, there exists an $\eta _{0}^{\ast }>0$ such that any solution of \eqref{2.1}-\eqref{2.2} satisfies the inequality $u(0)\geq \eta _{0}^{\ast }$ and so, by the previous Lemma, \begin{equation*} u(t)\geq \alpha _{0}(\eta _{0}^{\ast }):=\alpha _{0}^{\ast },\quad t\in [0,1], \end{equation*} $\alpha _{0}^{\ast }$ being a positive constant. \end{proposition} \begin{proof} Supposing the opposite, we may find a solution $u(t)$ of \eqref{2.1}-\eqref{2.2}, such that $u(0)=\frac{\eta }{2}$, $\eta$ being the same as in \eqref{2.4}, i.e. $\eta$ is arbitrarily small. Then, noticing \eqref{2.400}, as above by Taylor formula we can find a $t\in [0,1]$ such that \begin{align*} u(1)-\beta u(\eta ) & \leq \frac{\eta }{2} +g^{-1}(\frac{\eta }{2}) +\frac{1}{2}u''(t)-\beta u(\eta ) \\ & <\frac{\eta }{2}+g^{-1}(\frac{\eta }{2})-\frac{1}{2}Ku(t) \\ & \leq \frac{\eta }{2}+g^{-1}(\frac{\eta }{2})-\frac{K}{2}\min \{ \alpha _{0},\frac{\eta }{2}\} \\ & \leq \frac{\eta }{2}+g^{-1}(\frac{\eta }{2})-\frac{K}{2} \frac{\eta }{2} \min \{ 1,\gamma \} <0, \end{align*} due to the choice $K>2\frac{1+2\mu }{\min \{ \gamma ,1\} }$. This contradiction completes the proof. \end{proof} We give now an existence principle, which is crucial for the proof of our results. \begin{lemma} \label{Le4}(Nonlinear Alternative of Leray-Shauder Type) \cite{AOW} Let $V$ be a Banach space and $C\subset V$ a convex set. Assume that $U$ is a relative open subset of $C$ with $u_{0}\in U$ and $T:\bar{U}\rightarrow C$ a\ completely continuous\ (continuous and compact) map. Then either (I)\qquad $T$ has a fixed point, or (II)\qquad there exists $u\in \partial U$ and $\lambda \in \left( 0,1\right)$ with $u=\lambda T\left( u\right) +\left( 1-\lambda \right) u_{0}$. \end{lemma} \begin{theorem} \label{Th1} Assume (A1)-(A4) hold and $$\sup_{c>0,\;0\leq t\leq 1}\frac{c^{p}}{\int_{0}^{c}f(t,u)du}> \frac{p}{p-1}\Big[\frac{1}{1-\beta }\int_{\eta }^{1}[q(t) ]^{1/p}dt+\int_{0}^{\eta }[q(t)]^{1/p}dt\Big]^{p}. \label{2.70}$$ Then the 3-point boundary-value problem \eqref{2.1}-\eqref{2.2} has at least a positive solution. \end{theorem} \begin{proof} In order to show that \eqref{2.1}-\eqref{2.2} has a solution, we consider the boundary-value problem \begin{gathered} -[\phi _{p}(u')]'=q(t)F(t,u(t)),\quad 0\alpha _{0}^{\ast }$such that$u(0)\leq A_{0}^{\ast }$. Indeed, setting$% (u(0),u'(0))=(u_{0},u_{0}')\in Eand since \begin{gather*} u'(t)=\phi _{p}^{-1}[\phi _{p}(u_{0}')-\lambda \int_{0}^{t}q(s)F(s,u(s))ds], \\ u(t)=u_{0}+\int_{0}^{t}\phi _{p}^{-1}[\phi _{p}(u_{0}')-\lambda \int_{0}^{t}q(s)F(s,u(s))ds]dt, \end{gather*}% the initial values must be chosen so that \begin{aligned} Q(u_{0}') & :=u(1)-\beta u(\eta ) \\ & =g(u_{0}')+\int_{0}^{1}\phi _{p}^{-1}[\phi _{p}(u_{0}')-\lambda \int_{0}^{t}q(s) F(s,u(s))ds]dt \\ & -\beta [g(u_{0}')+\int_{0}^{\eta }\phi _{p}^{-1}[\phi _{p}(u_{0}')-\lambda \int_{0}^{t}q(s)F(s,u(s))ds]dt]=0. \end{aligned} \label{2.12}% By Proposition \ref{Pr1} and its proof, there is an\eta >0$such that$Q(g^{-1}(\frac{\eta }{2}))<0$, and moreover by the definition of$Q$, \begin{equation*} Q\big\{\phi _{p}^{-1}\big(\lambda \int_{0}^{t}q(s)F(s,u(s))ds\big)\big\}>0. \end{equation*} Hence$u_{0}'$is upper bounded and similarly$u_{0}=g(u_{0}')$, i.e.$(u_{0},u_{0}')\in E_{0}\subset E$,$E_{0}$being a compact subset of$\mathbb{R}^{2}$. We consider now the Banach space$B=C[0,1]$and for any$x\in B, $let$u=u(t)$be a solution of the boundary-value problem \begin{gather*} -[\phi _{p}(u')]'=q(t)F(t,x(t)),\quad 00$ such that $\Vert u\Vert \leq M$ for any solution of \eqref{2.9}. We set \begin{equation*} G(c)=\int_{0}^{c}\max_{0\leq t\leq 1}f(t,u)ds \end{equation*}% Noting \eqref{2.70}, we may indeed find a $M>0$ such that \begin{equation*} \frac{M}{\Big(\int_{0}^{M}\sup_{\;0\leq t\leq 1}f(t,u)du\Big)^{\frac{1}{p}}(% \frac{p}{p-1})^{\frac{1}{p}}\Big[\frac{1}{1-\beta }\int_{\eta }^{1}[q(t)]^{1/p}dt+\int_{0}^{\eta }[q(t)]^{1/p}dt\Big]}>1. \end{equation*}% Also by Proposition \ref{Pr1}, any solution of the boundary-value problem \eqref{2.9} is convex and there exists a point $t_{0}\in (0,1)$ such that \begin{equation*} u'(t)\geq 0,\;t\in [ 0,t_{0}),\ u'(t_{0})=0\quad \text{and}\quad u'(t)\leq 0,\;t\in (t_{0},1]. \end{equation*}% Working in the interval $[t_{0},t]\subset [ t_{0},1]$, we have \begin{equation*} 0\leq -(\phi _{p}(u'))'=\lambda q(t)F(t,u)\leq q(t)\max_{0\leq t\leq 1}f(t,u). \end{equation*} Multiplying by $-u'>0$, we get \begin{equation*} (\phi _{p}(u'))'\phi _{p}^{-1}(\phi _{p}(u'))\leq q(t)\max_{0\leq t\leq 1}f(t,u),\quad t\in [ t_{0},1] \end{equation*} and then integrating on $[t_{0},t]$, we obtain \begin{align*} \int_{0}^{\phi _{p}(u'(t))}\phi _{p}^{-1}(u'(t))u^{\prime }(t)dt& \leq q(t)\int_{u(t)}^{u(t_{0})}\max_{0\leq t\leq 1}f(t,u)du \\ & \leq q(t)\int_{0}^{u(t_{0})}\max_{0\leq t\leq 1}f(t,u)du=q(t)G(u(t_{0})), \end{align*} hence \begin{equation*} I(-\phi _{p}(u'(t)))=I(\phi _{p}(u'(t)))\leq q(t)G(u(t_{0})) \end{equation*} and so $$0\leq -u'(t)\leq \phi _{p}^{-1}\{[I^{-1}(q(t))]I^{-1}(G(u(t_{0})))\},\quad t\in [ t_{0},t]. \label{2.11}$$ If $\eta \in (t_{0},1]$, an integration over $[\eta ,1]$ yields \begin{equation*} u(\eta )-u(1)\leq \phi _{p}^{-1}\{I^{-1}[(G(u(t_{0})))]\}\int_{\eta }^{1}\phi _{p}^{-1}\{I^{-1}((q(t))\}. \end{equation*} If $\eta \in (0,t_{0}]$, we integrate over $[t_{0},1]$ to obtain \begin{align*} u(t_{0})-u(1)& \leq \phi _{p}^{-1}\{I^{-1}(G(u(t_{0})))\}\int_{t_{0}}^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt \\ & \leq \phi _{p}^{-1}\{I^{-1}(G(u(t_{0})))\}\int_{\eta }^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt. \end{align*} Then clearly it follows that \begin{equation*} u(\eta )-u(1)\leq u(t_{0})-u(1)\leq \phi _{p}^{-1}\{I^{-1}(G(u(t_{0})))\}\int_{\eta }^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt. \end{equation*} Moreover, since $u(1)=\beta u(\eta )$, we get \begin{equation*} u(1)\leq \frac{\beta }{1-\beta }\phi _{p}^{-1}\{I^{-1}(G(u(t_{0})))\}\int_{\eta }^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt \end{equation*}% and so a new integration from $t_{0}$ to $1$ of (\ref{2.11}) yields \begin{align*} u(t_{0})& =u(1)+\phi _{p}^{-1}\{I^{-1}(G(u(t_{0})))\}\int_{t_{0}}^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt \\ & \leq u(1)+\phi _{p}^{-1}\{I^{-1}(G(u(t_{0})))\}\int_{0}^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt \\ & \leq \phi _{p}^{-1}\{I^{-1}(G(u(t_{0})))\}[\frac{\beta }{1-\beta } \int_{\eta }^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt+\int_{0}^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt] \\ & \leq \phi _{p}^{-1}\{I^{-1}(G(u(t_{0})))\}\big[\frac{1}{1-\beta } \int_{\eta }^{1}\phi _{p}^{-1}[I^{-1}(q(t))\big]dt+\int_{0}^{\eta }\phi _{p}^{-1}[I^{-1}(q(t))]dt] \\ & =\phi _{p}^{-1}\{I^{-1}(G(u(t_{0})))\}(\frac{p}{p-1})^{\frac{1}{p}}\big[ \frac{1}{1-\beta }\int_{\eta }^{1}q^{\frac{1}{p}}(t)dt+\int_{0}^{\eta }q^{ \frac{1}{p}}(t)dt\big]. \end{align*} Consequently \begin{equation*} \frac{u(t_{0})}{\phi _{p}^{-1}\{I^{-1}(G(u(t_{0})))\}(\frac{p}{p-1})^{\frac{1 }{p}}\big[\frac{1}{1-\beta }\int_{\eta }^{1}q^{\frac{1}{p} }(t)dt+\int_{0}^{\eta }q^{\frac{1}{p}}(t)dt\big]}<1 \end{equation*}% which by the assumption \eqref{2.70} implies that $u(t_{0}) \frac{p}{p-1}\Big[\frac{1}{1-\alpha }\int_{1-\eta }^{1}[q(t) ]^{1/p}dt+\int_{0}^{1-\eta }[q(t)]^{1/p}dt\Big]^{p}. \label{2.1400} Then, the 3-point boundary-value problem \eqref{2.13}-\eqref{2.14} has at least one negative solution. \end{theorem} To prove the above Theorem, we give some lemmas symmetrical to the previous case, with similar proofs which are partly omitted. Note that the differential equation \eqref{2.13} defines also a vector field. So if we focus on the$(y,y')$face semi-plane$(y<0)$, then by \eqref{2.1400}, we see that$y^{\prime \prime }>0$. Thus, any trajectory$(y(t),y'(t))$,$t\geq 0$, emanating from the semi-line \begin{equation*} E^{\ast }:=\{(y,y'):y-g(y')=0,\quad \;y<0\} \end{equation*}% \textquotedblleft trends\textquotedblright\ in a natural way, (when$y'(t)<0$) toward the negative$y$-semi-axis and then (when$y'(t)>0$) trends toward the positive$y'$-semi-axis. As a result, we may control the vector field, so that$y(1)+\alpha y(\eta )=0$. \begin{lemma}\label{Le11} Let$0<\alpha <1$. If$y\in C[0,1]$is a solution of the boundary-value problem \eqref{2.13}-\eqref{2.14}, then$y$is convex. Furthermore for every solution with$y(0)<0$, it follows that \begin{itemize} \item[(i)]$y(t)<0$,$t\in [0,1]$\item[(ii)] There exists a$t_{0}\in [0,1)$such that \begin{equation*} \sup_{t\in [\eta ,1]}y(t)\leq \delta y(t_{0})=-\delta \|y\|, \end{equation*} where$\delta =\min \{ \alpha (1-\eta ),\;\frac{\alpha (1-\eta )}{1-\alpha (1-\eta )}\}$\item[(iii)] There exists a$t_{0}\in [0,1)$such that$y( t_{0})=-\max_{0\leq t\leq 1}y(t)=-\|y\|$\end{itemize} \end{lemma} \begin{proof} Let$y(t)$be a solution of \eqref{2.13}-\eqref{2.14}. Then since, \begin{equation*} [ \phi _{p}(y')]'=-q(t)f(t,y(t))\geq 0, \end{equation*}$[\phi _{p}(y')]$is nondecreasing and so is$y'(t)$, a fact that implies the convexity of$y(t)$. (i) Since$y(0)<0$, by the first condition in \eqref{2.14} we get$y'(0)<0$. If$y'(t)\leq 0$,$t\in [ 0,1]$, then$y(1)\leq y(\eta )<\alpha y(\eta )$, a contradiction. Hence, there exists a$t_{0}>0$such that$y(t_{0})=\min_{0\leq t\leq 1}y(t)=-\Vert y\Vert $. (ii) If$1-\eta \in (0,t_{0})$, then$y(1-\eta )y(1)-\alpha y(1). \end{equation*} Hence, $y(1)<0$. Finally, the convexity of $y(t)$ yields $y(t)<0$, $t\in [ 0,1]$. (iii) The proof follows by the convexity of the solution and since it is analogous to the given one at Lemma \ref{Le1}, we omit it. \end{proof} \begin{lemma} \label{lem6} Suppose that conditions \eqref{A2S} and \eqref{G} hold. Then, there exists an $\eta _{0}<0$ such that for any $\eta \in (\eta _{0},0)$, any solution of \eqref{2.13} with $y(0)=\eta / 2$, satisfies the inequality \begin{equation*} \eta _{0}\leq \eta \leq y(t)<0,\quad t\in [0,1]. \end{equation*} Furthermore, there exists an $\alpha _{0}=\alpha _{0}(\eta )<0$ such that any (possible) solution $y=y(t)$ of \eqref{2.13}-\eqref{2.14} with $y(0)=\frac{\eta }{2}$, satisfies the inequality $$y(t)\leq \alpha _{0}(\eta ),\quad t\in [0,1]. \label{2.510}$$ \end{lemma} \begin{proof} By the sublinearity of the function $f(t,y)$ at the point $y=0$, for every $K>\max \{2\mu ^{2},2\frac{1+2\mu }{\min \{\delta ,1\}}\}$ there exists an $\eta _{0}<0$ such that \max_{0\leq t\leq 1}f(t,y)1. \end{equation*}% We shall prove first that $\eta \leq y(t)<0$, $t\in [ 0,1]$. If not, by Lemma \ref{Le11}, there exists a $t^{\ast }\in (0,1]$ such that $\eta \leq y(t)<\frac{\eta }{2}$, 0\leq t0, \end{align*} where, noticing the monotonicity of\phi _{p}^{\prime }(s)=(p-1)(-s)^{p-2}>0 $,$s<0$and of$y'(t)$,$0\leq t<1$, \begin{equation*} M_{3}=\max \big\{\phi _{p}'(y')= \begin{cases} \phi _{p}'(g^{-1}[\eta /2]), & \mbox{if }p>2 \\ \phi _{p}'(g^{-1}[\eta ]), & \mbox{if }p\in (1,2)% \end{cases} \big\}>0 \end{equation*} Consequently, by \eqref{2.40} and Taylor's formula we conclude that for some$t\in [ 0,t^{\ast }], \begin{align*} \eta =y(t^{\ast })& =\frac{\eta }{2}+t^{\ast }g^{-1}(\frac{\eta }{2}) +\frac{(t^{\ast })^{2}}{2}y^{\prime \prime }(t) \\ & \geq \frac{\eta }{2}+t^{\ast }g^{-1}(\frac{\eta }{2})-\frac{(t^{\ast })^{2} }{2}K\frac{\eta }{2} \\ & \geq \frac{\eta }{2}+t^{\ast }2\mu \frac{\eta }{2}-\frac{(t^{\ast })^{2}}{2 }K\frac{\eta }{2}. \end{align*}% Considering now the map \begin{equation*} \phi (t^{\ast }):=Kt^{\ast }{}^{2}+4\mu t^{\ast }+2, \end{equation*}% the above inequality yields\phi (t^{\ast })\leq 0$, given that$\eta <0$. This is a contradiction, since the above choice of$K>2\mu ^{2}$, yields$\phi (t)>0$for all$t\in [ 0,1]$. Consequently, noticing Lemma \ref{Le11}, we obtain$\eta \leq y(t)<0$,$t\in [ 0,1]$. We proceed now with the proof of inequality \eqref{2.510}. By the convexity of$y(t)$, it is obvious that \begin{equation*} \max_{t\in [ 0,1]}y(t)=\max \{y(0),\;y(1)\}. \end{equation*} However, in view of the same Lemma \ref{Le11}, \begin{equation*} \sup_{t\in [ \eta ,1]}y(t)\leq \delta y(t_{0})=-\delta \Vert y\Vert \leq \delta y(0)=\delta \frac{\eta }{2} \end{equation*} and so \begin{equation*} y(t)\leq \sup_{t\in [ \eta ,1]}y(t)\leq \frac{\eta }{2}\min \{1,\delta \}:=\alpha _{0}(\eta )<0. \end{equation*} \end{proof} \begin{proposition}\label{Pr11} Suppose that conditions \eqref{A2S} and \eqref{G} hold. Then there exists an$\eta _{0}^{\ast }<0$such that any solution of \eqref{2.13}-\eqref{2.14} satisfies the inequality$y(0)\leq \eta_{0}^{\ast }$and furthermore, \begin{equation*} y(t)\leq \alpha _{0}(\eta _{0}^{\ast }):=\alpha_{0}^{\ast },\quad t\in [0,1], \end{equation*}$\alpha _{0}^{\ast }$being a negative constant. \end{proposition} \begin{proof} Supposing the opposite, we may find a solution$y(t)$of the\ boundary-value problem \eqref{2.13}-\eqref{2.14}, such that$y(0)=\frac{\eta }{2}$, with$\eta \in (\eta _{0},0),\;\eta _{0}$being the same as in the previous Lemma. Then, noticing \eqref{2.140}, we may find a$t\in [ 0,1]such that \begin{align*} y(1)-\alpha y(1-\eta )& \geq \frac{\eta }{2}+g^{-1}(\frac{\eta }{2})+\frac{1% }{2}y^{\prime \prime }(t)-\alpha y(1-\eta ) \\ & >\frac{\eta }{2}+g^{-1}(\frac{\eta }{2})-\frac{1}{2}Ky(t) \\ & \geq \frac{\eta }{2}+g^{-1}(\frac{\eta }{2})-\frac{K}{2}\min \{\alpha _{0}(\eta ),\frac{\eta }{2}\} \\ & \geq \frac{\eta }{2}+g^{-1}(\frac{\eta }{2})-\frac{K}{2}\frac{\eta }{2} \min \{1,\delta \}>0, \end{align*}% due to the choiceK>2\frac{1+2\mu }{\min \{\delta ,1\}}$and since$y(1-\eta )<0$. This contradiction completes the proof. \end{proof} \begin{proof}[Proof of Theorem] \ref{Th2}. To show that \eqref{2.13}-\eqref{2.14} has a solution, we consider the problem \begin{gathered} -[\phi _{p}(y')]'=q(t)F(t,y(t)),\quad 0\alpha _{0}^{\ast }. \end{cases} \end{equation*} Then, clearly$F\in C([0,1]\times (-\infty ,0],(-\infty ,0))$. Consider now the family of problems \begin{gathered} -[\phi _{p}(y')]'=q(t)\lambda F(t,y(t)),\quad 00$ and moreover by the definition of $Q^{\ast }$ and the fact that $g(y_{0}')<0$, \begin{equation*} Q^{\ast }\Big\{\phi _{p}^{-1}\Big(\lambda \int_{0}^{t}q(s)F(s,y(s))ds\Big) \Big\}<0. \end{equation*} Hence $y_{0}'$ is lower bounded and similar $y_{0}=g(y_{0}')$, i.e. $(y_{0},y_{0}')\in E_{0}^{\ast }$ with $E_{0}^{\ast }$ a compact subset of $\mathbb{R}^{2}$. Furthermore, by the monotonicity of the functions $g$ and $Q^{\ast }$, for each solution $y(t)$ of the boundary-value problem \eqref{221}, its initial value $(y_{0},y_{0}')$ is uniquely determined and continuous. Consider now the operator \begin{equation*} T_{\lambda }y(t)=y_{0}+\int_{0}^{t}\phi _{p}^{-1}\Big[\phi _{p}[g^{-1}(y_{0})]-\lambda \int_{0}^{s}q(r)F(r,y(r))dr\Big]ds, \end{equation*}% where $y_{0}$ is the unique constant corresponding to the function $y(t)$ satisfying (\ref{222}). It is easily verified that $y(t)$ is a solution of \eqref{220} if and only if $y$ is a fixed point of $T_{1}$ in $C[0,1]$. (i) We consider the Banach space $B=C[0,1]$ and we may easily, as above, prove that $T:=T_{1}:B\rightarrow B$ is completely continuous. (ii) We will show that there exists a $M>0$ such that $\Vert y\Vert \leq M$ for any solution of \eqref{221}. We set \begin{equation*} G(c)=\int_{c}^{0}\min_{0\leq t\leq 1}f(t,y)ds>0,\quad c\leq 0. \end{equation*}% Noting \eqref{2.1400}, we may find a $M>0$ such that \begin{equation*} \frac{-M}{\big(\int_{-M}^{0}\min_{0\leq t\leq 1}f(t,y)dy\big)^{\frac{1}{p}} (\frac{p}{p-1})^{\frac{2}{p}}\big[\frac{1}{1-\alpha }\int_{1-\eta }^{1}[q(t)]^{1/p}dt+\int_{0}^{1-\eta }[q(t)]^{1/p}dt\big]}<-1. \end{equation*} Also by the previous, any solution of \eqref{221} being convex, satisfies \begin{equation*} y(t)\leq \alpha _{0}(\eta _{0}^{\ast }):=\alpha _{0}^{\ast }<0,\quad t\in [ 0,1], \end{equation*} and further there is a $t_{0}\in (0,1)$ such that \begin{equation*} y'(t)\leq 0,\quad t\in [ 0,t_{0}),\quad y^{\prime }(t_{0})=0\quad \text{and}\quad y'(t)\geq 0,\;t\in (t_{0},1]. \end{equation*} Working in the interval $[t_{0},t]\subset [ t_{0},1]$, we have \begin{equation*} -(\phi _{p}(y'))'=\lambda q(t)F(t,y)\geq q(t)\min_{t\in [ 0,1]}F(t,y). \end{equation*} Multiplying by $-y'<0$, integrating on $[t_{0},t]$, and given that $q(t)$ is non-increasing, we obtain \begin{align*} \int_{0}^{\phi _{p}[y'(t)]}\phi _{p}^{-1}(z)dz& \leq -q(t)\int_{y(t_{0})}^{y(t)}\min_{0\leq t\leq 1}F(t,y)dy \\ & \leq -q(t)\int_{y(t_{0})}^{0}\min_{0\leq t\leq 1}F(t,y)dy \\ & =-q(t)G(y(t_{0}))<0. \end{align*} Hence, \begin{equation*} I(\phi _{p}^{-1}(y'(t)))\leq -q(t)G(y(t_{0}))\leq q(t)G(y(t_{0})) \end{equation*}% and so $$0\leq y'(t)\leq \phi _{p}^{-1}\{I^{-1}(q(t))I^{-1}[G(y(t_{0}))]\},\quad t\in [ t_{0},t]. \label{2.110}$$ If $1-\eta \in (t_{0},1]$, an integration over $[1-\eta ,1]$ yields \begin{equation*} y(1)-y(1-\eta )\leq \phi _{p}^{-1}[I^{-1}(G(y(t_{0})))]\int_{1-\eta }^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt. \end{equation*} If $1-\eta \in (0,t_{0}]$, we integrate over $[t_{0},1]$ to obtain \begin{align*} y(1)-y(t_{0})& \leq \phi _{p}^{-1}[I^{-1}(G(y(t_{0})))]\int_{t_{0}}^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt \\ & \leq \phi _{p}^{-1}[I^{-1}(G(y(t_{0})))]\int_{1-\eta }^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt. \end{align*} Since $\ y(t_{0})\leq y(1-\eta )$, it follows that \begin{equation*} y(1)-y(1-\eta )\leq y(1)-y(t_{0})\leq \phi _{p}^{-1}[I^{-1}(G(y(t_{0})))]\int_{1-\eta }^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt. \end{equation*} Moreover, since $y(1)=\alpha y(1-\eta )$, we get \begin{equation*} y(1)\geq \frac{\alpha }{\alpha -1}\phi _{p}^{-1}[I^{-1}(G(y(t_{0})))]\int_{1-\eta }^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt, \end{equation*} and so a new integration from $t_{0}$ to $1$ of \eqref{2.110} yields \begin{align*} & y(t_{0}) \\ & =y(1)-\phi _{p}^{-1}[I^{-1}(G(y(t_{0})))]\int_{t_{0}}^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt \\ & \geq y(1)-\phi _{p}^{-1}[I^{-1}(G(y(t_{0})))]\int_{0}^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt \\ & \geq \phi _{p}^{-1}[I^{-1}(G(y(t_{0})))][\frac{\alpha }{\alpha -1} \int_{1-\eta }^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt-\int_{0}^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt] \\ & =-\phi _{p}^{-1}[I^{-1}(G(y(t_{0})))][\frac{1}{1-\alpha }\int_{1-\eta }^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt+\int_{0}^{1-\eta }\phi _{p}^{-1}[I^{-1}(q(t))]dt] \\ & =-\phi _{p}^{-1}[I^{-1}(G(y(t_{0})))](\frac{p}{p-1})^{\frac{1}{p}} [\frac{1}{1-\alpha }\int_{1-\eta }^{1}[q(t)]^{\frac{1}{p}}(t)dt+\int_{0}^{1-\eta }[q(t)]^{\frac{1}{p}}(t)dt]. \end{align*} Consequently, (recall that $y(t_{0})<0$) \begin{equation*} \frac{y(t_{0})}{[G(y(t_{0}))]^{\frac{1}{p}}(\frac{p}{p-1})^{\frac{1}{p}} [\frac{1}{1-\alpha }\int_{1-\eta }^{1}[q(t)]^{\frac{1}{p}}(t)dt+\int_{0} ^{1-\eta }[q(t)]^{\frac{1}{p}}(t)dt]}>-1, \end{equation*} which in turn, by the assumption \eqref{2.1400} and the choice of $M$, implies $y(t_{0})>-M$. Hence we obtain $\Vert y\Vert \leq M$. Finally, in view of Lemma \ref{Le4}, we may set \begin{equation*} C:=\{y\in B=C[0,1]:\Vert y\Vert \leq M\}\quad \text{and}\quad U:=\{y\in C:\Vert y\Vert 0$is a constant. Comparing to Theorem \ref{Th1}, we have chosen$g(v)=v^{1/3}$,$q(t)=a/\sqrt{(1-t)}$,$\beta =1/2$and$\eta =1/3$. It is trivial to verify that assumptions (A1)-(A4) hold true for the system \eqref{41}. Furthermore, since \begin{equation*} \frac{c^{3}}{\int_{0}^{c}[u^{-1/2}+\sin ^{2}u^{-1/4}]}\geq \frac{c^{3}}{c+2\sqrt{c}} \end{equation*}% and \begin{equation*} \frac{3}{3-1}\Big[\frac{1}{1-1/2}\int_{1/3}^{1}\frac{a^{1/3}}{(\sqrt{1-t} )^{1/3}}dt+\int_{0}^{1/3}\frac{a^{1/3}}{(\sqrt{1-t})^{1/3}}dt\Big] =\frac{18}{5}a^{1/3}, \end{equation*} it follows that \eqref{2.70} is fulfilled for every$a>0$. Hence the boundary-value problem \eqref{41} admits a positive solution. \begin{remark} \rm The results in \cite{MG} can not be applied to \eqref{41}, since the assumption (H5) is not satisfied. 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