\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2017 (2017), No. 35, pp. 1--10.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2017 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2017/35\hfil Existence of solutions] {Existence of solutions to nonlinear problems with three-point boundary conditions} \author[D. P. D. Santos \hfil EJDE-2017/35\hfilneg] {Dionicio Pastor Dallos Santos} \address{Dionicio Pastor Dallos Santos \newline Department of Mathematics, IME-USP, Cidade Universit\'aria, CEP 05508-090, S\~ao Paulo, SP, Brazil} \email{dionicio@ime.usp.br} \thanks{Submitted September 21, 2016. Published January 30, 2017.} \subjclass[2010]{34B15, 34B16, 47H10, 47H11} \keywords{Boundary value problems; Schauder fixed point theorem; \hfill\break\indent Leray-Schauder degree; lower and upper solutions} \begin{abstract} Using Leray-Schauder degree theory and the method of upper and lower solutions, we obtain a solution for nonlinear boundary-value problem \begin{gather*} \big(\varphi(u' )\big)'= f(t,u,u') \\ l(u,u')=0, \end{gather*} where $l(u,u')=0$ denotes the three-point boundary conditions on $[0,T]$, and $\varphi$ is a homeomorphism such that $\varphi(0)=0$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} The purpose of this article is to obtain a solution for the nonlinear problem \begin{equation}\label{equ} \begin{gathered} \big(\varphi(u' )\big)' = f(t,u,u') \\ l(u,u')=0, \end{gathered} \end{equation} where $l(u,u')=0$ denotes the boundary conditions $u(T) = u'(0)= u'(T)$ or $u(0) = u(T)= u'(0)$ on the interval $[0,T]$, $\varphi$ is a singular or classic homeomorphism such that $\varphi(0)=0$, and $f: [0, T]\times \mathbb{R} \times \mathbb{R}\to \mathbb{R}$ is continuous. Solvability of two-point boundary value problems can be investigated by various methods: fixed point theorems, topological degree arguments, variational methods, lower and upper functions, etc., see for example, \cite{man4, garcia, grae, huang, maw, oregan} and the reference therein. In particular, the author in \cite{oregan} proved the existence of solutions for the Dirichlet and mixed problems, assuming $f$ and $\varphi$are continuous and that $\varphi$ is strictly increasing and satisfies $\varphi(\mathbb{R})=\mathbb{R}$ and $\varphi^{-1}\in C^{1}(\mathbb{R})$. Bereanu and Mawhin \cite{ma1} proved the existence of solutions for the periodic bound\-ary-value problem \begin{gather*} \big(\varphi(u' )\big)' = f(t,u,u') \\ u(0)=u(T), \quad u'(0)=u'(T), \end{gather*} assuming that $f : [0, T]\times \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is a continuous function and $\varphi: \mathbb{R} \to (-a,a) \ (01$. Using the barrier strip argument and topological transversality theorem the authors in \cite{chinos} obtained the existence of solutions for nonlinear boundary-value problems \begin{gather*} \big(\varphi(u' )\big)' = f(t,u,u') \\ u(0)=A, \quad u'(1)=B, \end{gather*} where $f: [0, T]\times \mathbb{R} \times \mathbb{R}\to \mathbb{R}$ is continuous and $\varphi:\mathbb{R} \to \mathbb{R}$ is an increasing homeomorphism. Inspired by these results, the main aim of this paper is to study the existence of solutions for \eqref{equ} using topological methods based upon Leray-Schauder degree. The main contribution of this paper is the extension of some results above cited to a more general type of boundary conditions. Such problems do not seem to have been studied in the literature. This article is organized as follows. In Section 2, we introduce some notations and preliminaries, which will be crucial in the proofs of our results. Section 3 is devoted to the study of existence of solutions for boundary-value problems \begin{gather*} \big(\varphi(u' )\big)' = f(t,u,u') \\ u(T)=u'(0)=u'(T), \end{gather*} where $\varphi:(-a,a)\to \mathbb{R}$ (we call it singular). We call \emph{solution } of this problem any function $u: [0, T]\to \mathbb{R}$ of class $C^{1}$ such that $ \max_{ [0, T]}|u'(t)|0$ such that \begin{equation}\label{dp4} \| u\|_1\leq \rho. \end{equation} Next, we show that $\overline{M_1(\Lambda)}\subset C^{1}$ is a compact set. Let $(v_n)_n $ be a sequence in $M_1(\Lambda)$, and let $(u_n)_n$ be a sequence in $\Lambda$ such that $v_n=M_1(u_n)$. Using \eqref{dp4}, we have that there exists a constant $L_1>0$ such that, for all $n\in \mathbb{N}$, \[ \| N_{f}(u_n)\|_{\infty}\leq L_1, \] which implies \[ \| H(N_{f}(u_n)-Q(N_f(u_n))) \|_{\infty}\leq 2L_1T. \] Hence the sequence $ (H(N_{f}(u_n)-Q(N_f(u_n))))_n$ is bounded in $C$. Moreover, for $t, t_1\in [0, T]$ and for all $n\in \mathbb{N}$, we have \begin{align*} &| H(N_{f}(u_n)-Q(N_f(u_n)))(t) - H(N_{f}(u_n)-Q(N_f(u_n)))(t_1) |\\ &\leq \big|\int_{t_1}^{t}N_f(u_n)(s) ds\big| + \big| \int_{t_1}^{t}Q(N_f(u_n))(s) ds\big|\\ &\leq L_1|t-t_1| + |t-t_1|\| Q(N_f(u_n))\|_{\infty}\\ &\leq 2L_1|t-t_1|, \end{align*} which implies that $ \big(H(N_{f}(u_n)-Q(N_f(u_n)))\big)_n$ is equicontinuous. Thus, by the Arzel\`a-Ascoli theorem there is a subsequence of $(H(N_{f}(u_n)-Q(N_f(u_n))))_n$, which we call $(H(N_{f}(u_{n_{j}})-Q(N_f(u_{n_{j}}))))_{j}$, which is convergent in $C$. Then, passing to a subsequence if necessary, we obtain that the sequence $$ (H(N_{f}(u_{n_{j}})-Q(N_f(u_{n_{j}}))) + \varphi(S(u_{n_{j}})))_{j} $$ is convergent in $C$. Using the fact that $\varphi^{-1}: C \to B_{a}(0)\subset C$ is continuous it follows from \[ M_1(u_{n_{j}})'=\varphi^{-1} [(H(N_{f}(u_{n_{j}})-Q(N_f(u_{n_{j}}))) + \varphi(S(u_{n_{j}}))) ] \] that the sequence $(M_1(u_{n_{j}})')_{j}$ is convergent in $C$. Therefore, passing if necessary to a subsequence, we have that $(v_{n_{j}})_{j}=( M_1(u_{n_{j}}))_{j}$ is convergent in $C^{1}$. Finally, let $(v_n)_n$ be a sequence in $\overline{M_1(\Lambda)}$. Let $(z_n)_n\subseteq M_1(\Lambda)$ be such that \[ \lim_{n \to \infty}\| z_n-v_n\|_1=0. \] Let $(z_{n_{j}})_{j}$ be a subsequence of $(z_n)_n$ such that converge to $z$. It follows that $z\in \overline{M_1(\Lambda)}$ and $(v_{n_{j}})_{j}$ converge to $z$. This completes the proof. \end{proof} To apply the Leray-Schauder degree to the equivalent fixed point operator $M_1$, for $\lambda \in [0,1]$, we introduce the family of boundary-value problems \begin{equation}\label{misto3} \begin{gathered} (\varphi(u' ))' = \lambda N_{f}(u)+(1-\lambda)Q(N_{f}(u)) \\ u(T)=u'(0)=u'(T). \end{gathered} \end{equation} Note that \eqref{misto3} coincide with \eqref{equa1} for $\lambda =1$. So, for each $ \lambda \in [0,1]$, the nonlinear operator associated with \eqref{misto3} by Lemma \ref{mate2} is the operator $M(\lambda,\cdot)$, where $M$ is defined on $[0,1]\times C^{1}$ by \begin{equation}\label{eme} M(\lambda,u)=S(u)+Q(N_{f}(u)) + K(\varphi^{-1} [\lambda H( N_f (u) -Q(N_{f}(u)))+\varphi(S(u))]). \end{equation} Using the same arguments as in the proof of Lemma \ref{mate3} we show that the operator $M$ is completely continuous. Moreover, using the same reasoning as in Lemma \ref{mate2}, system \eqref{misto3} is equivalent to the problem \[ u=M(\lambda, u). \] The following lemma gives a priori bounds for the possible fixed points of $M$. \begin{lemma}\label{cota} Let $f: [0,T]\times \mathbb{R}\times \mathbb{R}\to \mathbb{R}$ be continuous. If $(\lambda,u)\in [0,1]\times C^{1} $ is such that $u=M(\lambda, u)$, then \begin{center} $\|u\|_1\leq a(2+T)$. \end{center} \end{lemma} \begin{proof} Let $ [0,T]\times C^{1}$ be such that $u=M(\lambda, u)$. Then \[ u =M(\lambda, u)= S(u)+Q(N_{f}(u)) + K(\varphi^{-1} [\lambda H( N_f (u)-Q(N_{f}(u)))+\varphi(S(u))]). \] Differentiating, we obtain \[ u' = [M(\lambda, u)]' = \varphi^{-1} [\lambda H( N_f (u)-Q(N_{f}(u)))+\varphi(S(u))], \] so that $\|u'\|_{\infty}\leq a$. Because $u \in C^{1}$ is such that $u(T)=u'(0)$, we have \[ |u(t)|\leq |u(T)|+ \int_ 0 ^T |u'(s)| ds \leq a +aT,\quad t\in[0,T], \] and hence $\|u\|_1 =\|u\|_{\infty}+ \|u'\|_{\infty}\leq a+aT +a=a(2+T)$. This completes the proof \end{proof} \subsection{Existence result} %\label{S:0} We can now prove an existence theorem for \eqref{equa1}. We denote by $\deg_{B}$ the Brouwer degree and for $\deg_{LS}$ the Leray-Schauder degree, and define the mapping $G:\mathbb{R}^2\to \mathbb{R}^2$ by \[ G (x,y)= (xT+yT^2-yT-\frac{1}{T}\int_0^T f(t,x+yt,y)dt, y-x-yT). \] \begin{theorem}\label{prin1} Let $f: [0,T]\times \mathbb{R}\times \mathbb{R}\to \mathbb{R}$ be continuous. Then for all $\rho>a(2+T)$ \[ \deg_{LS}(I-M(1,\cdot), B_{\rho}(0),0) = \deg_{B}(G, B_{\rho}(0) \cap \mathbb{R}^2,0). \] If furthermore \[ \deg_{B}(G,B_{\rho}(0) \cap \mathbb{R}^2,0)\neq 0, \] then problem \eqref{equa1} has at least one solution. \end{theorem} \begin{proof} Let $M$ be the operator given by \eqref{eme} and let $\rho>a(2+T)$. Using Lemma \ref{cota}, we have that, for each $\lambda \in [0,T]$, the Leray-Schauder degree $\deg_{LS}(I-M(\lambda,\cdot), B_{\rho}(0),0)$ is well defined, and by the homotopy invariance, one has \[ \deg_{LS}(I-M(0,\cdot),B_{\rho}(0),0)= \deg_{LS}(I-M(1,\cdot),B_{\rho}(0),0). \] On the other hand, \[ \deg_{LS}(I-M(0,\cdot),B_{\rho}(0),0) = \deg_{LS}(I-(S+QN_{f}+KS) ,B_{\rho}(0),0). \] But the range of the mapping \[ u \mapsto S(u)+Q(N_{f}(u))+K(S(u)) \] is contained in the subspace of related functions, isomorphic to $\mathbb{R}^2$. Using homotopy invariance and reduction properties of Leray-Schauder degree \cite{man7}, we obtain \begin{align*} & \deg_{LS}(I-(S+QN_{f}+KS) ,B_{\rho}(0),0)\\ & = \deg_{B}\big(I-(S+QN_{f}+KS)\big|_{\overline{B_{\rho}(0) \cap \mathbb{R}^2}} ,B_{\rho}(0)\cap \mathbb{R}^2, 0\big)\\ &=\deg_{B}(G,B_{\rho}(0)\cap \mathbb{R}^2,0)\neq 0. \end{align*} Then, $\deg_{LS}(I-M(1,\cdot),B_{\rho}(0),0)\neq 0$. Hence, there exists $u\in B_{\rho}(0)$ such that $M_1(u)=u$, which is a solution for \eqref{equa1}. \end{proof} Let us give now an application of Theorem \ref{prin1}. \begin{example} \label{examp3.6}\rm We consider the boundary-value problem \begin{equation}\label{exemplo} \begin{gathered} ( \varphi(u'))' = e^{u'} +e \\ u(T)=u'(0)=u'(T), \end{gathered} \end{equation} where $\varphi(s)=s/\sqrt{1-s^2}$. It is not difficult to verify that $\varphi:(-1,1) \to \mathbb{R}$ is a homeomorphism and $f(t,x,y)=e^{y}+e$ is a continuous function. If we choose $\rho> 2+T$, then the equation \begin{align*} G(x,y) &=\Big(xT+yT^2-yT-\frac{1}{T}\int_{0}^{T} f(t,x+yt,y)dt, y-x-yT\Big)=(0,0)\\ &= \Big(xT+yT^2-yT-\frac{1}{T}\int_{0}^{T}(e^{y}-e)dt, y-x-yT\Big)=(0,0)\\ &= \Big(xT+yT^2-yT-e^{y}+e, y-x-yT\Big)=(0,0) \end{align*} does not have solutions on $\partial B_{\rho}(0)\cap\mathbb{R}^2$. Then we have that the Brouwer degree $\deg_{B}(G,B_{\rho}(0) \cap \mathbb{R}^2,(0,0))$ is well defined and, by the properties of that degree, we have that \[ \deg_{B}(G, B_{\rho}(0)\cap\mathbb{R}^2, (0,0)) = \sum_{(x,y) \in G^{-1}(0,0)}\operatorname{sgn}J_{G}(x,y) = \operatorname{sgn}(-e)=-1, \] where $(0,0)$ is a regular value of $G$ and $J_{G}(x,y)=$det$ G'(x,y)$ is the Jacobian of $G$ at $(x,y)$. So, using Theorem \ref{prin1}, we obtain that the boundary-value problem \eqref{exemplo} has at least one solution. \end{example} \begin{remark}\label{obs} \rm Using the family of boundary-value problems \begin{equation}\label{equa9} \begin{gathered} (\varphi(u' ))' = \lambda N_{f}(u)+(1-\lambda)Q(N_{f}(u)) \\ u(0) = u'(0) = u'(T) \end{gathered} \end{equation} which gives the completely continuous homotopy $\widetilde{M}$ defined on $[0,1]\times C^{1}$ by \[ \widetilde{M}(\lambda,u)= P(u)+Q(N_{f}(u)) + H(\varphi^{-1} [\lambda H( N_f (u)-Q(N_{f}(u)))+\varphi(P(u))]), \] and similar a priori bounds as in the Lemma \ref{cota}, it is not difficult to see that \eqref{equa9} has a solution for $\lambda=1$. \end{remark} \section{Boundary value problems with $\varphi$-Laplacian} \label{S4} In this section we study the existence of solutions for the boundary-value problem \begin{equation}\label{equa2} \begin{gathered} (\varphi(u' ))' = f(t,u,u') \\ u(0) = u(T) = u'(0), \end{gathered} \end{equation} where $\varphi: \mathbb{R} \to \mathbb{R}$ is an increasing homeomorphism, $\varphi(0)=0$ and $f: [0, T]\times \mathbb{R} \times \mathbb{R}\to \mathbb{R}$ is continuous. Let us consider the operator $M_1:C^{1} \to C^{1}$, \[ M_1(u) \varphi^{-1}(-Q_{\varphi}(H(N_{f}(u)))) + H(\varphi^{-1} [H(N_{f}(u))-Q_{\varphi}(H(N_{f}(u)))]). \] As in the previous section, here $\varphi^{-1}$ with an abuse of notation is understood as the operator $\varphi^{-1}: C \to C$ defined for $\varphi^{-1}(v)(t)=\varphi^{-1}(v(t))$. It is clear that $\varphi^{-1}$ is continuous and sends bounded sets into bounded sets. To transform problem \eqref{equa2} to a fixed point problem we use Lemma \ref{lebema1}. \begin{lemma}\label{dallos11} A map $u \in C^{1}$ is a solution of \eqref{equa2} if and only if $u$ is a fixed point of the operator $M_1$. \end{lemma} \begin{proof} If $u\in C^{1}$ is solution of \eqref{equa2}, then \[ (\varphi(u'(t)))' = N_f (u)(t)=f(t,u(t),u'(t)), \quad u(0)=u(T),\quad u(0)=u'(0) \] for all $t\in [0, T]$. Applying $H$ to both members and using the fact that $u(0)=u'(0)$, we deduce that \[ \varphi(u'(t)) = \varphi(u(0)) + H(N_{f}(u))(t). \] Composing with the function $\varphi^{-1}$, we obtain \[ u'(t) =\varphi^{-1} [ H(N_{f}(u))(t)+c], \] where $c=\varphi (u(0))$. Integrating from 0 to $t\in [0, T]$, we have \[ u(t)=u(0)+ H( \varphi^{-1} [H(N_{f}(u))+c ])(t). \] Because $u(0)=u(T)$, we have \[ \int_0^T \varphi^{-1} [H(N_{f}(u))(t)+c]dt=0. \] Using Lemma \ref{lebema1}, it follows that $c=-Q_{\varphi}(H(N_{f}(u)))$. Hence, \[ u=\varphi^{-1}(-Q_{\varphi}(H(N_{f}(u)))) + H ( \varphi^{-1} [H(N_{f}(u))-Q_{\varphi}(H(N_{f}(u)))] ) . \] Now suppose that $u\in C^{1}$ be such that $ u=M_1(u)$. It follows that \begin{equation}\label{pof} u(t)=\varphi^{-1}(-Q_{\varphi}(H(N_{f}(u)))) + H( \varphi^{-1} [H(N_{f}(u))-Q_{\varphi}(H(N_{f}(u)))])(t) \end{equation} for all $t\in [0, T]$. Since \[ \int_0^T \varphi^{-1} [H(N_{f}(u))(t)-Q_{\varphi}(H(N_{f}(u)))]dt=0, \] we have $u(0)=u(T)$. Differentiating \eqref{pof}, we obtain \begin{align*} u'(t) &=\varphi^{-1} [H(N_{f}(u))-Q_{\varphi}(H(N_{f}(u)))](t)\\ &= \varphi^{-1} [H(N_{f}(u))(t)-Q_{\varphi}(H(N_{f}(u)))].%\\ \end{align*} Applying $\varphi$ to both of its members, and differentiating we have \[ (\varphi(u'(t)))' = N_f (u)(t), \quad u(0)=u(T), \quad u(0)=u'(0) \] for all $t\in [0, T]$. This completes the proof. \end{proof} Using an argument similar to the one introduced in the proof of \cite[Lemma 4.2]{dallos}, it is not difficult to see that $M_1:C^{1} \to C^{1} $ is well defined and completely continuous. \subsection{Upper and lower solutions} \label{S:0} The functions considered as lower and upper solutions for the initial problem \eqref{equa2} are defined as follows. \begin{definition} \label{def4.2} \rm A lower solution $\alpha$ (resp. upper solution $\beta$) of \eqref{equa2} is a function $\alpha \in C^{1}$ such that $\varphi(\alpha')\in C^{1}, \ \alpha'(0)\geq \alpha(0)=\alpha(T)$ (resp. $\beta \in C^{1}, \ \varphi(\beta')\in C^{1}, \ \beta'(0)\leq \beta(0)=\beta(T))$ and \begin{equation}\label{pas1} (\varphi(\alpha'(t)))' \geq f(t,\alpha(t),\alpha'(t)) \quad (\text{resp. } (\varphi(\beta'(t)))' \leq f(t,\beta(t),\beta'(t)) \end{equation} for all $t\in [0,T]$. \end{definition} We can now prove some existence results for \eqref{equa2}. These results are inspired on works by Bereanu and Mawhin \cite{ma1} and Carrasco and Minh\'os \cite{minhos}. \begin{theorem}\label{gato} Suppose that \eqref{equa2} has a lower solution $\alpha$ and an upper solution $\beta$ such that $\alpha(t)\leq \beta(t)$ for all $t\in [0,T]$. If there exists a continuous function $g(t,x)$ on $ [0,T]\times \mathbb{R}$ such that \begin{equation}\label{pas2} |f(t,x,y)|\leq |g(t,x)|, \quad\text{for all } (t, x,y)\in [0,T]\times \mathbb{R}\times \mathbb{R}, \end{equation} then \eqref{equa2} has a solution $u$ such that $\alpha(t)\leq u(t)\leq \beta(t)$ for all $t\in [0,T]$. \end{theorem} \begin{proof}\label{gato1} Let $\alpha$, $\beta$ be, respectively, lower and upper solutions of \eqref{equa2}. Let $\gamma: [0,T]\times \mathbb{R}\to \mathbb{R}$ be the continuous function defined by $$ \gamma(t,x) = \begin{cases} \beta(t), & x>\beta(t)\\ x, & \alpha(t)\leq x\leq \beta(t) \\ \alpha(t), & x <\alpha(t), \end{cases} $$ and define $F: [0,T]\times \mathbb{R}\times \mathbb{R} \to \mathbb{R}$ by $F(t,x,y)= f(t,\gamma (t,x),y) +\frac{x-\gamma(t,x)}{1+|x-\gamma(t,x)|}$. We consider the modified problem \begin{equation}\label{equa3} \begin{gathered} (\varphi(u' ))' = F(t,u,u') \\ u(0)= u(T)=u'(0). \end{gathered} \end{equation} For clearness, the proof will follow several steps. \smallskip \noindent\textbf{Step 1} We show that if $u$ is a solution of \eqref{equa3}, then $ \alpha(t)\leq u(t)\leq \beta(t)$ for all $t\in [0,T]$ and hence $u$ is a solution of \eqref{equa2}. Let $u$ be a solution of the modified problem \eqref{equa3} and suppose by contradiction that there is some $t_{0}\in [0,T]$ such that \begin{equation}\label{equa4} \max_{[0,T]}(\alpha(t)-u(t)) = \alpha(t_{0})-u(t_0)>0. \end{equation} If $t_{0}\in (0,T)$, there are sequences $(t_{k})$ in $ [t_{0}-\epsilon, t_{0})$ and $(t'_{k})$ in $(t_{0}, t_{0}+\epsilon ]$ converging to $t_{0}$ such that $ \alpha'(t_{k})-u'(t_{k})\geq 0$ and $\alpha'(t'_{k})-u'(t'_{k})\leq 0$. Therefore $\alpha'(t_{0})=u'(t_{0})$. Using the fact that $\varphi$ is an increasing homeomorphism, this implies $(\varphi(\alpha'(t_{0}) ))'\leq (\varphi(u'(t_{0}) ))'$. By \eqref{pas1} we get the contradiction \begin{align*} (\varphi(\alpha'(t_{0}) ))' &\leq (\varphi(u'(t_{0})))' = F(t_{0},u(t_{0}),u'(t_{0})) \\ &\leq f(t_{0},\alpha(t_{0}), \alpha'(t_{0}))) +\frac{u(t_{0}) -\alpha(t_{0})}{1+|u(t_{0})-\alpha(t_{0}) |} \\ &< f(t_{0},\alpha(t_{0}), \alpha'(t_{0}))) \leq (\varphi(\alpha'(t_{0}) ))'. \end{align*} So $\alpha(t)\leq u(t)$ for all $t\in (0,T)$. If the maximum is attained at $t_{0}=0$ then \[ \max_{[0,T]}(\alpha(t)-u(t)) = \alpha(0)-u(0)>0. \] Using the fact that $u$ is solution of \eqref{equa3} and $\alpha'(0)\leq u'(0)$, we obtain the following contradiction \[ \alpha(0)\leq \alpha'(0)\leq u'(0)=u(0)<\alpha(0). \] If \[ \max_{[0,T]}(\alpha(t)-u(t)) = \alpha(T)-u(T)>0, \] then, since $\alpha(0)=\alpha(T)$ and $u(0)=u(T)$ we obtain again a contradiction. In consequence we have that $\alpha(t)\leq u(t)$ for all $t\in [0,T]$. In a similar way we can prove that $u(t)\leq \beta(t)$ for all $t\in [0,T]$. \smallskip \noindent\textbf{Step 2} We show that problem \eqref{equa3} has at least one solution. Let $u\in C^{1}$ and define the operator $M_{F}: C^{1} \to C^{1}$ by \[ M_{F}(u)=\varphi^{-1}(-Q_{\varphi}(H(N_{F}(u)))) + H(\varphi^{-1} [H(N_{F}(u))-Q_{\varphi}(H(N_{F}(u)))]), \] with $F(t,u,u')=f(t,\gamma (t,u),u') +\frac{u-\gamma(t,u)}{1+|u-\gamma(t,u)|}$, by Lemma \ref{dallos11} it is enough to prove that $M_{F}$ has a fixed point. Under the hypothesis of the theorem, the operator $M_{F}$ is bounded. Indeed, if $v=M_{F}(u)$ then \begin{equation}\label{pepa15} \varphi(v') = [H(N_F(u))- Q_{\varphi}(H(N_F(u)))], \end{equation} where \begin{align*} |H(N_F(u))(t)| &\leq \int_0^T |f(s,\gamma(s,u(s)),u'(s)) +\frac{u(s)-\gamma(s,u(s))}{1+|u(s)-\gamma(s,u(s))|}|ds \\ &\leq \int_0^T |f(s,\gamma(s,u(s)),u'(s))|ds + T \\ & \leq \int_0^T |g(s,\gamma(s,u(s)))|ds + T\\ & \leq \sigma T +T, \end{align*} with $\sigma:= \sup_{s\in[0,T]}|g(s,\gamma(s,u(s)))|$. Using \eqref{pepa15}, we have that \begin{equation}\label{pepa} |\varphi(v'(t))|\leq 2(\sigma T +T):= \delta \quad t\in [0,T], \end{equation} and hence \begin{equation}\label{pepa1} \|v'\|_{\infty}\leq \omega, \end{equation} where $\omega= \max \{|\varphi^{-1}(\delta)|, \ |\varphi^{-1}(-\delta)|\}$. Because $v \in C^{1}$ is such that $v(0) = v'(0)$, we have \[ |v(t)|\leq |v(0)| + \int_0^T |v'(s)|ds \leq \omega+ T \omega \quad t\in [0,T], \] and hence \[ \|v\|_1=\|v\|_{\infty}+ \|v'\|_{\infty}\leq \omega + T\omega + \omega =\omega(2+T). \] As the operator $M_{F}$ is completely continuous and bounded, we can use Schauder's fixed point theorem to deduce the existence of at least one fixed point in $\overline{B_{\omega(2+T)}(0)}$. The proof is complete. \end{proof} \begin{corollary}\label{pepa3} Let $f(t,x,y)=f(t,x)$ be a continuous function. If \eqref{equa2} has a lower solution $\alpha$ and a upper solution $\beta$ such that $\alpha(t)\leq \beta(t)$ for all $t\in [0,T]$, then problem \eqref{equa2} has a solution such that $\alpha(t)\leq u(t)\leq \beta(t)$ for all $t\in [0,T]$. \end{corollary} \begin{example} \rm We consider the boundary-value problem \begin{equation}\label{pepa4} \begin{gathered} \big(|u'|^{p-2}u'\big)'= \frac{\sin(u+1)-1 +4ue^{u^2t}}{1+t^2}, \\ u(0)=u(T)=u'(0), \end{gathered} \end{equation} where $p\in (1,\infty)$. As $f(t,x)$ is a continuous function, and the functions $\alpha(t)=-1$ and $\beta(t)=1$ are lower and upper solutions of \eqref{pepa4}, respectively, then, by Corollary \eqref{pepa3}, we obtain that \eqref{pepa4} has at least one solution $u$ such that $-1\leq u(t)\leq 1$ for all $t\in [0,T]$. \end{example} \subsection*{Acknowledgements} This research was supported by CAPES and CNPq/Brazil. \begin{thebibliography}{99} \bibitem{man4} C. Bereanu, J. 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