\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small Seventh Mississippi State - UAB Conference on Differential Equations and Computational Simulations, {\em Electronic Journal of Differential Equations}, Conf. 17 (2009), pp. 81--94.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2009 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \setcounter{page}{81} \title[\hfilneg EJDE-2009/Conf/17\hfil A third-order m-point boundary-value problem] {A third-order m-point boundary-value problem of Dirichlet type involving a p-Laplacian type operator} \author[C. P. Gupta\hfil EJDE/Conf/17 \hfilneg] {Chaitan P. Gupta} \address{Chaitan P. Gupta \newline Department of Mathematics, 084\\ University of Nevada, Reno, NV 89557, USA} \email{gupta@unr.edu} \thanks{Published April 15, 2009.} \subjclass[2000]{34B10, 34B15, 34L30} \keywords{m-point boundary value problems; $p$-Laplace type operator; non-resonance; resonance; topological degree} \begin{abstract} Let $\phi$, be an odd increasing homeomorphisms from $\mathbb{R}$ onto $\mathbb{R}$ satisfying $\phi (0)=0$, and let $f:[0,1]\times \mathbb{R}\times \mathbb{R}\times \mathbb{R}\mapsto \mathbb{R}$ be a function satisfying Caratheodory's conditions. Let $\alpha _{i}\in {\mathbb{R}}$, $\xi _{i}\in (0,1)$, $i=1,\dots ,m-2$, $0<\xi _{1}<\xi _{2}<\dots <\xi _{m-2}<1$ be given. We are interested in the existence of solutions for the $m$-point boundary-value problem: \begin{gather*} (\phi (u''))'=f(t,u,u',u''), \quad t\in (0,1), \\ u(0)=0,\quad u(1)=\sum_{i=1}^{m-2}\alpha _{i}u(\xi _{i}), \quad u''(0)=0, \end{gather*} in the resonance and non-resonance cases. We say that this problem is at \emph{resonance} if the associated problem $(\phi (u''))'=0, \quad t\in (0,1),$ with the above boundary conditions has a non-trivial solution. This is the case if and only if $\sum_{i=1}^{m-2}\alpha _{i}\xi _{i}=1$. Our results use topological degree methods. In the non-resonance case; i.e., when $\sum_{i=1}^{m-2}\alpha_{i}\xi _{i}\neq 1$ we note that the sign of degree for the relevant operator depends on the sign of $\sum_{i=1}^{m-2}\alpha _{i}\xi _{i}-1$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \section{Introduction} In this paper we consider the boundary-value problem $$\begin{gathered} (\phi (u''))'=f(t,u,u',u''), \quad t\in (0,1), \\ u(0)=0, \quad u(1)=\sum_{i=1}^{m-2}\alpha _{i}u(\xi _{i}), \quad u''(0)=0, \end{gathered} \label{P}$$ where $\phi$ is an odd increasing homeomorphism from $\mathbb{R}$ onto $\mathbb{R}$ with $\phi (0)=0$ and the function $f:[0,1]\times \mathbb{R}\times \mathbb{R}\times \mathbb{R}\mapsto \mathbb{R}$ is Caratheodory. Also $\alpha_{i}\in {\mathbb{R}}$, $\xi _{i}\in (0,1)$, for $i=1,2,\dots m-2$, are such that $0<\xi _{1}<\xi _{2}<\dots <\xi _{m-2}<1$. We say that \eqref{P} is at \emph{resonance}, if the associated multi-point boundary-value problem $$\begin{gathered} (\phi (u''))'=0, \quad t\in (0,1), \\ u(0)=0, u(1)=\sum_{i=1}^{m-2}\alpha _{i}u(\xi _{i}),\quad u''(0)=0, \end{gathered}\label{HP}$$ has a non-trivial solution. We are interested here in the existence of solutions for the $m$-point boundary-value problem \eqref{P} in the resonance and in the non-resonance cases. The study of multipoint second-order boundary-value problems for $\phi (u)\equiv u$ was initiated by Il'in and Moiseev in \cite{I1,I2} and has been the subject of many papers, see for example \cite{FW1, FW2,GU1,GU2,GU3,GU4,GNT1,GNT,GT,L1,L2,L3,L4,rayun}. More recently multipoint second-order boundary-value problems containing the $p$-Laplace operator or the more general operator $-(\phi (u'))'$ complemented with linear boundary conditions, have been studied in \cite{B,GGM1,GM,LG,WA1,WA2}. Problem \eqref{P} is at resonance if and only if $\sum_{i=1}^{m-2}\alpha _{i}\xi _{i}=1$, having $u(t)=\rho t$ as a non-trivial solution, where $\rho \in \mathbb{R}$ is an arbitrary constant. Our aim in this paper is to obtain existence of solutions for problem (\ref {P}), by using topological degree arguments. Thus, in section 2, we first derive a deformation lemma that is needed when problem \eqref{P} is at resonance. In section 3 an existence theorem for problem \eqref{P} is derived from this lemma. Finally in section 4 we consider problem \eqref{P} when it is non-resonant. The crucial point here is to prove that the Leray Schauder degree of a certain operator is different from zero which is shown to be an explicit consequence of the non-resonance condition, i.e., $\sum_{i=1}^{m-2}\alpha _{i}\xi _{i}\neq 1$. In addition we obtain the interesting property that the degree of the operator changes sign when $\sum_{i=1}^{m-2}\alpha _{i}\xi _{i}$ goes from being less than one to being greater than one. We shall denote by $C[0,1]$ (resp. $C^{1}[0,1]$, $C^{2}[0,1]$) the classical space of continuous (resp. continuously differentiable, twice continuously differentiable) real-valued functions on the interval $[0,1]$. The norm in $C[0,1]$ is denoted by $| \cdot | _{\infty }$. Also, we shall denote by $L^{1}(0,1)$ the space of real-valued (equivalence classes of) functions whose absolute value is Lebesgue integrable on $(0,1)$. The Brouwer and Leray-Schauder degree shall be respectively denoted by $\deg _{B}$ and $\deg _{LS}$. \section{\label{defor-lemma}A deformation lemma for the resonance case} We begin this section by formulating a general deformation lemma for the solvability of the boundary-value problem \eqref{P} in the resonance case. Let $f^{\ast }:[0,1]\times \mathbb{R}\times \mathbb{R\times R}\times [ 0,1]\mapsto \mathbb{R}$ be a function satisfying Caratheodory's conditions; i.e., (i) for all $(s,r,q,\lambda )\in \mathbb{R}\times \mathbb{R} \times \mathbb{R}\times [ 0,1]$ the function $f^{\ast }(\cdot ,s,r,q,\lambda )$ is measurable on $[0,1]$, (ii) for a.e. $t\in [ 0,1]$ the function $f^{\ast }(t,\dots ,\cdot )$ is continuous on $\mathbb{R}\times \mathbb{R\times R}\times [ 0,1]$, and (iii) for each $R>0$ there exists a Lebesgue integrable function $\rho _{R}:[0,1]\mapsto \mathbb{R}$ such that $|f^{\ast }(t,s,r,q,\lambda )|\leq \rho _{R}(t)$ for a.e. $t\in [ 0,1]$ and all $(s,r,q,\lambda )\in \mathbb{R}\times \mathbb{R\times R}\times [ 0,1]$ with $| s| \leq R$, $| r| \leq R$, and $| q| \leq R$. We suppose that $f(t,s,r,q)=f^{\ast }(t,s,r,q,1)$ is the given function in problem \eqref{P}. We, now, introduce an operator ${\mathfrak{B}}(u,\lambda ):C^{2}[0,1]\times [ 0,1]\mapsto \mathbb{R}$ defined for $(u,\lambda )\in C^{2}[0,1]\times [ 0,1]$ by \begin{aligned} \mathfrak{B}(u,\lambda ) &=\lambda \Big(u(1)-\sum_{i=1}^{m-2}\alpha _{i}u(\xi _{i})\Big)\\ &\quad+(1-\lambda )\Big(\int_{0}^{1}\int_{0}^{s}f^{\ast }(\tau ,u(\tau ),u'(\tau ),u''(\tau ),\lambda )d\tau ds \\ &\quad -\sum_{i=1}^{m-2}\alpha _{i}\int_{0}^{\xi _{i}}\int_{0}^{s}f^{\ast }(\tau ,u(\tau ),u'(\tau ),u''(\tau ),\lambda )d\tau ds \Big). \end{aligned} \label{eqB} For $\lambda \in [ 0,1]$ we consider the family of boundary-value problems: $$\begin{gathered} (\phi (u''))'=\lambda f^{\ast }(t,u,u',u'',\lambda ), \quad t\in (0,1), \\ u(0)=0, \quad u''(0)=0, \quad {\mathfrak{B}}(u,\lambda )=0. \end{gathered} \label{Plambda}$$ Let $\Omega \subset C^{2}[0,1]$ be a bounded open set. Let us set for $\rho \in \mathbb{R}$, $i_{\rho }(t)=\rho t$, for $t\in [ 0,1]$, and \begin{equation*} X=\{i_{\rho }: \rho \in \mathbb{R}\}, \end{equation*} then $X$ is a one dimensional subspace of $C^{2}[0,1]$. Defining $i:{\mathbb{R}}\mapsto X$ by $i(\rho )=i_{\rho }$ it is clear that $i$ is an isomorphism from ${\mathbb{R}}$ onto $X$. Next let us define $F:X\mapsto \mathbb{R}$ by \begin{equation*} F(i_{\rho })=\int_{0}^{1}\int_{0}^{s}f^{\ast }(\tau ,\rho \tau ,\rho ,0,0)d\tau ds-\sum_{i=1}^{m-2}\alpha _{i}\int_{0}^{\xi _{i}}\int_{0}^{s}f^{\ast }(\tau ,\rho \tau ,\rho ,0,0)d\tau ds, \end{equation*} and set $\mathcal{F}=F\circ i$, then $\mathcal{F}:{\mathbb{R}}\mapsto { \mathbb{R}}$ is continuous, and is given by \begin{equation*} \mathcal{F}(\rho )=\int_{0}^{1}\int_{0}^{s}f^{\ast }(\tau ,\rho \tau ,\rho ,0,0)d\tau ds-\sum_{i=1}^{m-2}\alpha _{i}\int_{0}^{\xi _{i}}\int_{0}^{s}f^{\ast }(\tau ,\rho \tau ,\rho ,0,0)d\tau ds. \end{equation*} We have the following lemma. \begin{lemma} \label{DefLemma} Assume that \begin{itemize} \item[(i)] for $\lambda \in (0,1)$ the boundary-value problem (\ref{Plambda}) has no solution $u\in \partial \Omega$, \item[(ii)] the equation $\mathcal{F}(\rho)= 0$ has no solution for any $\rho$ with $i_{\rho }(t)\in \partial \Omega \cap X$, and \item[(iii)] the Brouwer degree $\deg _{B}(F,\Omega \cap X,0)\neq 0$. \end{itemize} Then the boundary-value problem \eqref{P} has at least one solution in $\overline{\Omega }$. \end{lemma} \begin{proof} If the boundary-value problem \eqref{P} has a solution in $\partial \Omega$, then there is nothing to prove. Accordingly, let us assume that the boundary-value problem \eqref{P} has no solution in $\partial \Omega$. This assumption combined with assumption (i) implies that the boundary-value problem (\ref{Plambda}) has no solution $u\in \partial \Omega$ for $\lambda \in (0,1]$. Let us define an operator $\Psi ^{\ast }:C^{2}[0,1]\times [ 0,1]\mapsto C^{2}[0,1]$ by setting for $(u,\lambda )\in C^{2}[0,1]\times [ 0,1]$ \begin{aligned} \Psi ^{\ast }(u,\lambda )(t) &=\int_{0}^{t}\Big(u'(0)+\int_{0}^{s}\phi ^{-1}\Big(\lambda \int_{0}^{r}f^{\ast }(\tau ,u(\tau ),u'(\tau ),u''(\tau ),\lambda )d\tau \Big)dr\Big)ds\\ &\quad +t{\mathfrak{B}}(u,\lambda ), \end{aligned} \label{Eq2} where ${\mathfrak{B}}(u,\lambda )$ is as defined in equation (\ref{eqB}). We note from our assumptions that the function $f^{\ast }$ satisfies Caratheodory's conditions so that for $(u,\lambda )\in$ $C^{2}[0,1]\times [ 0,1]$, $f^{\ast }(t,u(t),u'(t),u''(t),\lambda )\in L^{1}(0,1)$. Accordingly, the function $s\in [ 0,1]\mapsto \int_{0}^{s}f^{\ast }(\tau ,u(\tau ),u'(\tau ),u''(\tau ),\lambda )d\tau$ is absolutely continuous on $[0,1]$. Since, now, the integrand in (\ref{Eq2}) is continuous on $[0,1]$ we see that the operator $\Psi ^{\ast }$ is well defined. Next, let us suppose that $u(t)$ be a solution to the boundary-value problem (\ref{Plambda}) for some $\lambda \in [ 0,1]$. We, then, see by integrating the equation in (\ref{Plambda}) and using the boundary conditions in (\ref{Plambda}) that $u(t)$ satisfies the equation \begin{equation*} u(t)=\Psi ^{\ast }(u,\lambda )(t), t\in [ 0,1], \end{equation*} along with \begin{equation*} u(0)=0, u''(0)=0, {\mathfrak{B}}(u,\lambda )=0. \end{equation*} Conversely, let us suppose that for some $\lambda \in [ 0,1]$, $u(t)$, $t\in [ 0,1]$, satisfies the equation $$u(t)=\Psi ^{\ast }(u,\lambda )(t). \label{Eq3}$$ We first see from the equation (\ref{Eq3}) and the definition of $\Psi ^{\ast }(u,\lambda )$ that \begin{equation*} u(0)=0. \end{equation*} Next, we obtain, by differentiating the equation (\ref{Eq3}) that $$u'(t)=u'(0)+\int_{0}^{t}\phi ^{-1}\Big(\lambda \int_{0}^{r}f^{\ast }(\tau ,u(\tau ),u'(\tau ),u''(\tau ),\lambda )d\tau \Big)dr +{\mathfrak{B}}(u,\lambda ), t\in [ 0,1]. \label{Eq4}$$ Evaluating (\ref{Eq4}) at $t=0$ we see that \begin{equation*} {\mathfrak{B}}(u,\lambda )=0. \end{equation*} Again, we obtain, by differentiating (\ref{Eq4}) that $$u''(t)=\phi ^{-1}\Big(\lambda \int_{0}^{t}f^{\ast }(\tau ,u(\tau ),u'(\tau ),u''(\tau ),\lambda )d\tau \Big). \label{eq41}$$ Evaluating the equation (\ref{eq41}) at $t=0$ we see that \begin{equation*} u''(0)=0. \end{equation*} Also, equation (\ref{eq41}) further implies that $\phi (u''(t))$ is absolutely continuous on $[0,1]$ and \begin{equation*} (\phi (u''(t)))'=\lambda f^{\ast }(t,u(t),u'(t),u''(t),\lambda ), t\in [ 0,1]. \end{equation*} Thus $u(t)$, $t\in (0,1)$, is a solution to the boundary-value problem (\ref {Plambda}). We have, accordingly, proved that $u(t)$, $t\in (0,1)$, is a solution to the boundary-value problem (\ref{Plambda}) if and only if $u(t)$, $t\in [ 0,1]$, is a solution to the equation (\ref{Eq3}). We observe that it is easy to show, using standard arguments, that $\Psi^{\ast }:C^{2}[0,1]\times [ 0,1]\mapsto C^{2}[0,1]$ is a completely continuous operator. If, now, $u(t)\in \partial \Omega$ is a solution to the boundary-value problem \eqref{P} then we are done. Accordingly, let us assume that the boundary-value problem \eqref{P} has no solution on $\partial \Omega$. Since, now, $f^{\ast }(t,s,r,q,1)=f(t,s,r,q)$ for all $(t,s,r,q)\in [ 0,1]\times \mathbb{R}\times \mathbb{R}\times \mathbb{R}$ we see that the assumption (i) of the lemma implies that \begin{equation*} u\neq \Psi ^{\ast }(u,\lambda )\quad \text{for all }u\in \partial \Omega \text{ and }\lambda \in (0,1]. \end{equation*} We, next, assert that $u\neq \Psi ^{\ast }(u,0)$ for all $u\in \partial \Omega$. Indeed, let $u\in \partial \Omega$ be such that $u=\Psi ^{\ast }(u,0)$. It then follows from the definition of $\Psi ^{\ast }$, as given in (\ref{Eq2}), that $u(t)=\rho t=i_{\rho }(t)$, with $\rho =u'(0)+{\mathfrak{B}}(u,0)$, $u'(t)=\rho +{\mathfrak{B}}(u,0)$, $u''(0)=0$, ${\mathfrak{B}}(u,0)=0$, $u\in \partial \Omega \cap X$, and \begin{align*} {\mathfrak{B}}(u,0) &= \int_{0}^{1}\int_{0}^{s}f^{\ast }(\tau ,u(\tau ),u'(\tau ),u''(\tau ),0)d\tau ds \\ &\quad -\sum_{i=1}^{m-2}\alpha _{i}\int_{0}^{\xi _{i}}\int_{0}^{s}f^{\ast }(\tau ,u(\tau ),u'(\tau ),u''(\tau ),0)d\tau ds \\ &= \int_{0}^{1}\int_{0}^{s}f^{\ast }(\tau ,\rho \tau ,\rho ,0,0)d\tau ds-\sum_{i=1}^{m-2}\alpha _{i}\int_{0}^{\xi _{i}}\int_{0}^{s}f^{\ast }(\tau ,\rho \tau ,\rho ,0,0)d\tau ds \\ &= \mathcal{F}(\rho )=0. \end{align*} But this contradicts the assumption (ii) of the lemma. We thus get that \begin{equation*} u\neq \Psi ^{\ast }(u,\lambda )\quad \text{for all }u\in \partial \Omega \text{ and }\lambda \in [ 0,1]. \end{equation*} Thus $\deg _{LS}(I-\Psi ^{\ast }(\cdot ,\lambda ),\Omega ,0)$ is well defined for all $\lambda \in [ 0,1]$. By the homotopy invariance property of Leray-Schauder degree we obtain immediately that $$\deg _{LS}(I-\Psi ^{\ast }(\cdot ,1),\Omega ,0) = \deg _{LS}(I-\Psi ^{\ast}(\cdot ,0),\Omega ,0) = \deg _{B}(I-\Psi ^{\ast }(\cdot ,0)| _{X},\Omega _{0},0), \label{eq5}$$ where, $\Omega _{0}=\Omega \cap X$. Now since for $v\in X$ \begin{equation*} \big(I-\Psi ^{\ast }(\cdot ,0)\big)v=-i_{F(v)}, \end{equation*} we have \begin{equation*} \deg _{LS}(I-\Psi ^{\ast }(\cdot ,1),\Omega ,0)=\deg _{B}(-i_{F(\cdot )},\Omega _{0},0)=-\deg _{B}(i_{F(\cdot )},\Omega _{0},0). \end{equation*} Since, $i^{-1}\circ i_{F(\cdot )}\circ i=\mathcal{F}$, we obtain by using a standard formula in degree theory that \begin{equation*} \deg _{B}(i_{F(\cdot )},\Omega _{0},0))=\deg _{B}(\mathcal{F},i^{-1}(\Omega _{0}),0)). \end{equation*} Hence, by assumption $(iii)$ of the lemma, it follows that $\deg _{LS}(I-\Psi ^{\ast }(\cdot ,1),\Omega ,0)\neq 0$. Thus, the mapping $\Psi \equiv \Psi ^{\ast }(\cdot ,1):C^{2}[0,1]\mapsto C^{2}[0,1]$ has at least one fixed-point in $\overline{\Omega }$ and hence the boundary value problem \eqref{P} has at least one solution in $\overline{\Omega }$. This completes the proof of the lemma. \end{proof} \section{Existence Theorems} We shall assume that for any constants $\Lambda \geq 0$, $A>0$ with $\Lambda0$ (resp. $<0$) for every $t\in (0,1)$. Moreover, given $\alpha _{i}\geq 0$, $\xi _{i}\in (0,1)$, $i=1,2,\cdot \cdot \cdot ,m-2$ with $\sum_{i=1}^{m-2}\alpha _{i}\xi _{i}=1$ we have $\int_{0}^{1}g(s)ds-\sum_{i=1}^{m-2}\alpha _{i}\int_{0}^{\xi _{i}}g(s)ds>0$ (resp. $<0$). \end{lemma} \begin{proof} Let us suppose that $g$ is a strictly increasing function on $[0,1]$. Now we see that $G'(t) = \frac{g(t)}{t}-\frac{1}{t^{2}}\int_{0}^{t}g(s)ds = \frac{1}{t^{2}}(\int_{0}^{t}(g(t)-g(s))ds >0,$ for every $t\in (0,1]$. Accordingly, $G$ is strictly increasing on $(0,1]$ and $\int_{0}^{1}g(s)ds-\frac{1}{t}\int_{0}^{t}g(s)ds>0$ for every $t\in (0,1)$. Finally, we see that \begin{align*} & \int_{0}^{1}g(s)ds-\sum_{i=1}^{m-2}\alpha _{i}\int_{0}^{\xi _{i}}g(s)ds \\ &= \sum_{i=1}^{m-2}\alpha _{i}\xi _{i}(\int_{0}^{1}g(s)ds-\frac{1}{\xi _{i}} \int_{0}^{\xi _{i}}g(s)ds)>0. \end{align*} Similarly $G$ is strictly decreasing on $(0,1]$ and $\int_{0}^{1}g(s)ds- \sum_{i=1}^{m-2}\alpha _{i}\int_{0}^{\xi _{i}}g(s)ds<0$ when $g$ is a strictly decreasing function on $[0,1]$. This completes the proof of the lemma. \end{proof} \begin{theorem} \label{thm3} Let $f:[0,1]\times \mathbb{R}\times \mathbb{R}\times \mathbb{R} \mapsto \mathbb{R}$ in the boundary-value problem \eqref{P} be a continuous function and satisfies the following conditions: \begin{itemize} \item[(i)] there exist non-negative functions $d_{1}(t)$, $d_{2}(t)$, $d_{3}(t)$, and $r(t)$ in $L^{1}(0,1)$ such that $|f(t,u,v,w)|\leq d_{1}(t)\phi (|u|)+d_{2}(t)\phi (|v|) +d_{3}(t)\phi (|w|)+r(t),$ for all $t\in [ 0,1]$, $u$, $v$, $w\in \mathbb{R}$, \item[(ii)] there exist constants $\Lambda \geq 0$, $B\geq 0$, $A>0$ with $\Lambda0$ such that for all $v$ with $|v|>v_{0}$, all $t\in[ 0,1]$ and all $u$, $w\in \mathbb{R}$ one has \begin{equation*} |f(t,u,v,w)|\geq -\Lambda |u|+A|v|-\Lambda |w|-B, \end{equation*} \item[(iii)] there exists an $R>0$ such that for all $\rho$, with $|\rho |>R$, either \begin{gather*} \rho f(t,\rho t,\rho ,0)>0,\text{ for all }t\in [ 0,1],\quad\text{or}\\ \rho f(t,\rho t,\rho ,0)<0,\text{ for all }t\in [ 0,1]. \end{gather*} \end{itemize} Suppose, further, that $$\tilde{\alpha}(A,\Lambda)(\|d_{1}\|_{L^{1}(0,1)}+\|d_{2}\|_{L^{1}(0,1)}) +\|d_{3}\|_{L^{1}(0,1)}<1. \label{cond1}$$ Then, given $\alpha _{i}\geq 0$, $\xi _{i}\in (0,1)$, $i=1,2,\cdot \cdot \cdot ,m-2$ with $\sum_{i=1}^{m-2}\alpha _{i}\xi _{i}=1$ the boundary value problem \eqref{P} has at least one solution in $u(t)\in C^{2}[0,1]$. \end{theorem} \begin{proof} We first choose an $\varepsilon >0$ be such that \begin{equation*} (\tilde{\alpha}(A,\Lambda )+\varepsilon )(\|d_{1}\|_{L^{1}(0,1)}+\|d_{2}\|_{L^{1}(0,1)})+\|d_{3}\|_{L^{1}(0,1)}<1, \end{equation*} which is possible to do, in view of (\ref{cond1}). We consider the family of boundary-value problems: $$\begin{gathered} (\phi (u''(t)))'=\lambda f(t,u(t),u'(t),u''(t)), \quad t\in (0,1), \lambda \in [0,1], \\ u(0)=0, \quad {\mathfrak{B}}(u,\lambda )=0, \quad u''(0)=0, \end{gathered} \label{EQ1}$$ where ${\mathfrak{B}}(u,\lambda )$ is as defined in (\ref{eqB}). Let $u(t)$ be a solution to the boundary-value problem (\ref{EQ1}) for some $\lambda \in (0,1)$. Then either there exists a $t_{0}\in [0,1]$ such that $$|u'(t_{0})|\leq v_{0} \label{Eq31}$$ or $|u'(t)|>v_{0}$ for all $t\in [ 0,1]$. In case, $|u'(t)|>v_{0}$ for all $t\in [ 0,1]$, we claim that there exists a $\tau _{0}\in [ 0,1]$ such that $f(\tau _{0},u(\tau _{0}),u'(\tau _{0}),u''(\tau _{0}))=0$. Indeed, let us suppose that $f(t,u(t),u'(t),u''(t))\neq 0$ for all $t\in [ 0,1]$. It then follows from the continuity of $f(t,u(t),u'(t),u''(t))$ on the interval $[0,1]$ either $f(t,u(t),u'(t),u''(t))>0$ for all $t\in [ 0,1]$ or $f(t,u(t),u'(t),u''(t))<0$ for all $t\in [ 0,1]$. Let us first suppose that $f(t,u(t),u'(t),u''(t))>0$ for all $t\in [ 0,1]$. It then follows from the boundary condition in (\ref{Eq3}) that \begin{aligned} &\lambda \Big[\int_{0}^{1}\Big(u'(0)+\int_{0}^{s}\phi ^{-1} \Big(\lambda \int_{0}^{r}f(\tau ,u(\tau ),u'(\tau ),u''(\tau ))d\tau \Big)dr\Big) ds \\ &-\sum_{i=1}^{m-2}\alpha _{i}\int_{0}^{\xi _{i}}\Big(u'(0)+\int_{0}^{s}\phi ^{-1}\Big(\lambda \int_{0}^{r}f(\tau ,u(\tau ),u'(\tau ),u''(\tau ))d\tau \Big)dr\Big)ds \Big] \\ &+(1-\lambda )\Big[\int_{0}^{1}\int_{0}^{r}f(\tau ,u(\tau ),u'(\tau ),u''(\tau )) d\tau ds\\ & -\sum_{i=1}^{m-2}\alpha _{i}\int_{0}^{\xi _{i}}\int_{0}^{r}f(\tau ,u(\tau ),u'(\tau ),u''(\tau ))d\tau dr\Big]\\ &=0. \end{aligned} \label{EQ4} We, next, see that the functions \begin{gather*} \int_{0}^{t}\Big(u'(0)+\int_{0}^{s}\phi ^{-1}\Big(\lambda \int_{0}^{r}f(\tau ,u(\tau ),u'(\tau ),u''(\tau ))d\tau \Big)dr\Big)ds, \\ \int_{0}^{s}\int_{0}^{r}f(\tau ,u(\tau ),u'(\tau ),u''(\tau ))d\tau dr \end{gather*} are strictly increasing functions on $(0,1]$, in view of our assumption \begin{equation*} f(t,u(t),u'(t),u''(t))>0 \end{equation*} for all $t\in [ 0,1]$. We then get from Lemma \ref{Lemma1} and (\ref {EQ4}) that $0>0$, a contradiction. Similarly, the supposition $f(t,u(t),u'(t),u''(t))<0$ for all $t\in [ 0,1]$ leads to the contradiction $0<0$. Hence, there must exist a $\tau _{0}\in[ 0,1]$ such that $$f(\tau _{0},u(\tau _{0}),u'(\tau _{0}),u''(\tau _{0}))=0, \label{Eq5}$$ proving the claim. We next see from (\ref{Eq5}) and assumption (ii) that $$|u'(\tau _{0})|\leq \frac{B}{A}+\frac{\Lambda }{A}\|u\|_{\infty }+ \frac{\Lambda }{A}\|u''\|_{\infty }. \label{Eq32}$$ Thus we see from (\ref{Eq31}) and (\ref{Eq32}) that there exists a $\tau _{1}\in [ 0,1]$ (either $t_{0}$ or $\tau _{0}$) such that $$|u'(\tau _{1})|\leq v_{0}+\frac{B}{A}+\frac{\Lambda }{A} \|u\|_{\infty }+\frac{\Lambda }{A}\|u''\|_{\infty }. \label{Eq33}$$ It then follows from the equation $u'(t)=u'(\tau _{1})+\int_{\tau _{1}}^{t}u''(s)ds$ and (\ref{Eq33}) that $$\|u'\|_{\infty }\leq \frac{A+\Lambda }{A-\Lambda }\|u''\|_{\infty } +\frac{Av_{0}+B}{A-\Lambda }. \label{Eq34}$$ Next, we see by integrating the equation in (\ref{EQ1}) from $0$ to $t\in [ 0,1]$ and noting $u''(0) =0$, that $$\phi (u''(t))=\lambda \int_{0}^{t}f(\tau ,u(\tau ),u'(\tau ),u''(\tau ))d\tau . \label{Eq51}$$ It now follows from equations (\ref{Eq51}), (\ref{Eq33}) using assumption (i), the fact that $u(0)=0$ implies $\|u\|_{\infty }\leq \|u'\|_{\infty }$ that \begin{align*} &\phi (|u''(t)|) \\ &\leq \phi (\|u\|_{\infty})\|d_{1}\|_{L^{1}(0,1)} +\phi (\|u'\|_{\infty })\|d_{2}\|_{L^{1}(0,1)}+\phi (\|u''\|_{\infty })\|d_{3}\|_{L^{1}(0,1)}+\|r\|_{L^{1}(0,1)} \\ &\leq (\|d_{1}\|_{L^{1}(0,1)}+\|d_{2}\|_{L^{1}(0,1)})\phi (\frac{A+\Lambda }{A-\Lambda }\|u''\|_{\infty }+\frac{Av_{0}+B}{A-\Lambda }) \\ &\quad +\|d_{3}\|_{L^{1}(0,1)})\phi (\|u''\|_{\infty })+\|r\|_{L^{1}(0,1)} \\ &\leq ((\tilde{\alpha}(A,\Lambda )+\varepsilon )(\|d_{1}\|_{L^{1}(0,1)}+\|d_{2}\|_{L^{1}(0,1)})+\|d_{3}\|_{L^{1}(0,1)})\phi (\|u''\|_{\infty }) \\ &\quad +C_{\varepsilon }(\|d_{1}\|_{L^{1}(0,1)}+\|d_{2}\|_{L^{1}(0,1)})+\|r\|_{L^{1}(0,1)}, \end{align*} and hence \begin{aligned} \phi (\|u''\|_{\infty }) &\leq ((\tilde{\alpha}(A,\Lambda )+\varepsilon )(\|d_{1}\|_{L^{1}(0,1)}+\|d_{2}\|_{L^{1}(0,1)}) +\|d_{3}\|_{L^{1}(0,1)})\phi (\|u''\|_{\infty }) \\ &\quad +C_{\varepsilon}(\|d_{1}\|_{L^{1}(0,1)} +\|d_{2}\|_{L^{1}(0,1)})+\|r\|_{L^{1}(0,1)}. \end{aligned} \label{Eq6} It now follows from (\ref{cond1}), the estimates (\ref{Eq6}), (\ref{Eq34}) and $\|u\|_{\infty }\leq \|u'\|_{\infty }$ that there exists an $R_{0}>R$ , where $R$ is as in assumption (iii), such that the family of boundary value problems (\ref{EQ1}) have no solution on the boundary of a bounded open set $\Omega =B(0,\widetilde{R})\subset C^{2}[0,1]$, for every $\widetilde{R}\geq R_{0}$. Accordingly, we see that the family of boundary value problems (\ref{EQ1}) satisfy condition (i) of Lemma \ref{DefLemma}. Next, we see from assumption (iii) and Lemma \ref{Lemma1} for all $\rho$, $|\rho |>R$, that \begin{equation*} \int_{0}^{1}\int_{0}^{s}f(\tau ,\rho \tau ,\rho ,0)d\tau ds-\sum_{i=1}^{m-2}\alpha _{i}\int_{0}^{\xi _{i}}\int_{0}^{s}f(\tau ,\rho \tau ,\rho ,0)d\tau ds \end{equation*} is strictly positive or strictly negative. Accordingly, we see that $f^{\ast }(t,u,v,w,\lambda )=f(t,u,v,w)$ satisfies the condition (ii) of Lemma \ref{DefLemma}. Finally, we again see from assumption (iii), the continuity in $\rho \in \mathbb{R}$ of the function \begin{equation*} \psi (\rho )=\int_{0}^{1}\int_{0}^{s}f(\tau ,\rho \tau ,\rho ,0)d\tau ds-\sum_{i=1}^{m-2}\alpha _{i}\int_{0}^{\xi _{i}}\int_{0}^{s}f(\tau ,\rho \tau ,\rho ,0)d\tau ds \end{equation*} and the assumption that $\widetilde{R}>R$, that $F(i_{\widetilde{R}}(t))$ and $F(i_{-\widetilde{R}}(t))$ have opposite signs. It follows immediately that $F(i_{\rho }(t))=0$ for an odd number of $\rho \in (-\widetilde{R}, \widetilde{R})$ which implies that the Brouwer degree $\deg _{B}(F,\Omega \cap X,0)\neq 0$. Thus the condition (iii) of Lemma \ref{DefLemma} is also satisfied. \ Thus it follows from Lemma \ref{DefLemma} that the boundary value problem \eqref{P} has at least one solution in $\overline{\Omega }$. This completes the proof of the theorem. \end{proof} \section{A result for the non-resonance case} In this section we will consider problem \eqref{P} in the non-resonance case. Problem \eqref{P} is in the non-resonance case if problem (\ref{HP}) has only the trivial solution. This holds if and only if the $\alpha _{i}$, $\xi _{i}$ satisfy $\sum_{i=1}^{m-2}\alpha _{i}\xi _{i}\neq 1$. We assume henceforth that $\alpha _{i}$, $\xi _{i}$ satisfy this condition. Notice that we do not assume a sign condition on the $\alpha _{i}'s$. In addition, we shall assume that for any $\sigma$, $0<\sigma <1$, it holds that $$\tilde{\alpha}(\sigma )=\limsup_{z\rightarrow \infty }\frac{\phi (\frac{1}{ 1-\sigma }z)}{\phi (z)}<\infty . \label{EQ8}$$ Let us set $\xi _{m-1}=1$, $\alpha _{m-1}=-1$, $\sigma _{ij}=\alpha _{i}(\xi _{i}-\xi _{j})$ for $i\neq j$ and $\sigma _{jj}=\sum_{i=1}^{m-1}\alpha _{i}\xi _{j}$ for $i, j = 1, 2, \cdot,\cdot,\cdot,m-1$. We note that the assumption $\sum_{i=1}^{m-2}\alpha _{i}\xi _{i}\neq 1$ is equivalent to $\sum_{i=1}^{m-1}\alpha _{i}\xi _{i}\neq 0$. Also, for each $j = 1, 2, \cdot,\cdot,\cdot,m-1$ we have \begin{equation*} \sum_{i=1}^{m-1}\sigma _{ij}=\sum_{i=1,i\neq j}^{m-1}\sigma _{ij}+\sigma _{jj}=\sum_{i=1,i\neq j}^{m-1}\alpha _{i}(\xi _{i}-\xi _{j})+\sum_{i=1}^{m-1}\alpha _{i}\xi _{j}=\sum_{i=1}^{m-1}\alpha _{i}\xi _{i}\neq 0. \end{equation*} It follows that $\sum_{i=1}^{m-1}(\sigma _{ij})^{+} \neq \sum_{i=1}^{m-1}(\sigma _{ij})^{-},$ for $j = 1, 2, \cdot,\cdot,\cdot,m-1$, where for $\alpha \in \mathbb{R}$, $\alpha^{+}= \max(\alpha,0)$ and $\alpha^{-} = \max ( -\alpha, 0)$. Let us set $$\sigma ^{\ast }= \begin{cases} \min \{\frac{\sum_{i=1}^{m-1}(\sigma _{ij})^{+}}{\sum_{i=1}^{m-1}(\sigma _{ij})^{-}},\frac{\sum_{i=1}^{m-1}(\sigma _{ij})^{-}}{\sum_{i=1}^{m-1}( \sigma _{ij})^{+}}\}& \text{if }\sum_{i=1}^{m-1}(\sigma _{ij})^{+}\neq 0 \text{ and}\\ &\quad \sum_{i=1}^{m-1}(\sigma _{ij})^{-}\neq 0 \text{ for all }j, \\ 0,& \text{otherwise.} \end{cases} \label{EQ9}$$ Note that $0\leq \sigma ^{\ast }<1$. The main result of this section is the following theorem. \begin{theorem}\label{nonresonant} Let $f:[0,1]\times \mathbb{R}\times\mathbb{R}\times\mathbb{R}\mapsto \mathbb{R}$ be a function satisfying Caratheodory's conditions such that the following condition holds: \\ there exist non-negative functions $d_{1}(t)$, $d_{2}(t)$, $d_{3}(t)$, and $r(t)$ in $L^{1}(0,1)$ such that $| f(t,u,v,w)| \leq d_{1}(t)\phi (| u| )+d_{2}(t)\phi (| v|) \\ +d_{3}(t)\phi (| w| )+r(t),$ for a. e. $t\in [ 0,1]$ and all $u,v,w\in \mathbb{R}$. Suppose, further, $$\tilde{\alpha}(\sigma ^{\ast })(\|d_{1}\|_{L^{1}(0,1)}+\| d_{2}\| _{L^{1}(0,1)})+\| d_{3}\| _{L^{1}(0,1)}<1\text{, } \label{cond1m}$$ where $\sigma ^{\ast }$ is as defined in (\ref{EQ9}) and $\tilde{\alpha}$ is as defined in (\ref{EQ8}). Then, the boundary-value problem \eqref{P} has at least one solution $u\in C^{2}[0,1]$. \end{theorem} We need the following variant of an a priori estimate from \cite{GT0} in the proof of Theorem \ref{nonresonant} and present this in the following lemma. \begin{lemma}\label{APestimate} Let $u\in C^{1}[0,1]$, be such that $u''\in L^{\infty }(0,1)$ and satisfies \begin{equation*} u(0)=0, \quad u(1)=\sum_{i=1}^{m-2}\alpha _{i}u(\xi _{i}), \end{equation*} with $\sum \alpha _{i}\xi _{i}\neq 1$. If $\sum_{i=1}^{m-1}(\sigma_{ij})^{+}\neq 0$, and $\sum_{i=1}^{m-1}(\sigma _{ij})^{-}\neq 0$ for all $j$, then $$\| u'\| _{\infty }\leq \frac{1}{1-\sigma ^{\ast }} \| u''\| _{\infty }. \label{APEstimate}$$ If one of $\sum_{i=1}^{m-1}(\sigma _{ij})^{+}$, $\sum_{i=1}^{m-1}(\sigma _{ij})^{-}$ is zero for some $j=1,2,\dots ,m-1$, then $u'(\eta _{0})=0$ for some $\eta _{0}\in [ 0,1]$, and $$\| u'\| _{\infty }\leq \| u''\| _{\infty }. \label{mana10}$$ \end{lemma} \begin{proof} We first, note, that the assumption \begin{equation*} u(1)=\sum_{i=1}^{m-2}\alpha _{i}u(\xi _{i}) \end{equation*} is equivalent to \begin{equation*} \sum_{i=1}^{m-1}\alpha _{i}u(\xi _{i})=0, \end{equation*} with $\xi _{m-1}=1$, $\alpha _{m-1}=-1$ and the non-resonant condition $\sum_{i=1}^{m-2}\alpha _{i}\xi _{i}\neq 1$ is equivalent to $\sum_{i=1}^{m-1}\alpha _{i}\xi _{i}\neq 0$. Next, for each $j = 1, 2, \cdot,\cdot,\cdot,m-1$ we have $u(\xi _{j})=\xi _{j}u'(\eta _{jj})$ for some $\eta _{jj}\in [ 0,1]$. Also for $i, j = 1, 2, \cdot,\cdot,\cdot,m-1$ with $i\neq j$ we have $u(\xi _{i})-u(\xi _{j})=u'(\eta _{ij})(\xi _{i}-\xi _{j})$ for some $\eta _{ij}\in [ 0,1]$. Accordingly, \begin{align*} \sum_{i=1,i\neq j}^{m-1}\alpha _{i}u'(\eta _{ij})(\xi _{i}-\xi _{j}) &=\sum_{i=1,i\neq j}^{m-1}\alpha _{i}(u(\xi _{i})-u(\xi_{j}))\\ &=-\sum_{i=1}^{m-1}\alpha _{i}u(\xi _{j})=-\sum_{i=1}^{m-1}\alpha _{i}\xi _{j}u'(\eta _{jj}), \end{align*} using the mean-value theorem and the assumptions $u(0)=0$, $\sum_{i=1}^{m-1}\alpha _{i}u(\xi _{i})=0$ (equivalently, $u(1)=\sum_{i=1}^{m-2}\alpha _{i}u(\xi _{i})$). We thus get $\sum_{i=1}^{m-1}\sigma _{ij}u'(\eta _{ij})=0$, and hence $\sum_{i=1}^{m-1}(\sigma _{ij})^{+}u'(\eta _{ij})=\sum_{i=1}^{m-1}(\sigma _{ij})^{-}u'(\eta _{ij})$. So there must exist $\chi _{j}^{1}$ and $\chi _{j}^{2}$ in $[0,1]$ such that $$\big(\sum_{i=1}^{m-1}(\sigma _{ij})^{+}\Big)u'(\chi _{j}^{1})=\Big(\sum_{i=1}^{m-1}(\sigma _{ij})^{-}\Big)u'(\chi _{j}^{1}) . \label{EQ100}$$ If one of $\sum_{i=1}^{m-1}(\sigma _{ij})^{+}$, $\sum_{i=1}^{m-1}(\sigma _{ij})^{-}$ is zero for some $j=1,2,\dots ,m-1$ then it follows from (\ref{EQ100}) that there is an $\eta _{0}\in [ 0,1]$ (indeed one of $\chi _{j}^{1}$ or $\chi _{j}^{2}$) such that $u'(\eta _{0})=0$ and the estimate (\ref{mana10}) is immediate. Next, suppose that $\sum_{i=1}^{m-1}(\sigma _{ij})^{+}\neq 0$ and $\sum_{i=1}^{m-1}(\sigma _{ij})^{-}\neq 0$ for every $j=1,2,\dots ,m-1$. Then either $u'(\chi _{j}^{1})=u'(\chi _{j}^{1})=0$ for some $j=1,2,\dots ,m-1$, in which case the estimate (\ref{mana10}) is immediate, or $u'(\chi _{j}^{1})\neq u'(\chi _{j}^{1})$ for every $j=1,2,\dots ,m-1$. It follows that there exist $\eta _{1},\eta _{2}\in [ 0,1]$ with $u'(\eta _{1})\neq u'(\eta _{2})$ such that $$u'(\eta _{1})=\sigma ^{\ast }u'(\eta _{2}). \label{EQ101}$$ The estimate (\ref{APEstimate}) is now immediate from (\ref{EQ8}), (\ref {EQ101}) and the equation \begin{equation*} u'(t)=u'(\eta _{1})+\int_{\eta _{1}}^{t}u''ds. \end{equation*} This completes the proof of the lemma. \end{proof} \begin{proof}[Proof of Theorem \ref{nonresonant}] We consider the family of boundary-value problems: $$\begin{gathered} (\phi (u''(t)))'=\lambda f(t,u(t),u'(t),u''(t)), \quad t\in (0,1), \lambda \in [0,1], \\ u(0)=0, \quad u(1)=\sum_{i=1}^{m-2}\alpha _{i}u(\xi _{i}), \quad u''(0)=0. \end{gathered} \label{NRlambda}$$ Also, we define an operator $\Psi ^{\ast }:C^{2}[0,1]\times [0,1]\mapsto C^{2}[0,1]$ by setting for $(u,\lambda )\in C^{2}[0,1]\times[ 0,1]$ \begin{align*} \Psi ^{\ast }(u,\lambda ) &= \int_{0}^{t}\Big(u'(0)+\int_{0}^{s}\phi ^{-1}\Big(\lambda \int_{0}^{r}f^{\ast }(\tau,u(\tau ),u'(\tau ),u''(\tau ),\lambda )d\tau \Big)dr\Big)ds \\ &\quad +t\Big(u(1)-\sum_{i=1}^{m-2}\alpha _{i}u(\xi _{i})\Big). \end{align*} Following standard arguments, it can be proved that $\Psi ^{\ast }$ is a completely continuous operator. Furthermore reasoning in an entirely similar way as we did in the proof of Lemma \ref{DefLemma} it can be proved that $u$ is a solution to the family of boundary-value problems (\ref{NRlambda}) if and only if $u$ is a fixed point for the operator $\Psi ^{\ast }(\cdot,\lambda )$; i.e., $u$ satisfies \begin{equation*} u=\Psi ^{\ast }(u,\lambda ). \end{equation*} We will show next that there is a constant $R>0$ independent of $\lambda \in [ 0,1]$ such that if $u$ satisfies (\ref{NRlambda}) for some $\lambda \in [ 0,1]$ then $\|u\|_{C^{2}[0,1]}0$ since the proof for the case $\sigma ^{\ast }=0$ is simpler. Let us choose $\varepsilon >0$ such that $$(\tilde{\alpha}(\sigma ^{\ast })+\varepsilon )(\|d_{1}\|_{L^{1}(0,1)}+\| d_{2}\| _{L^{1}(0,1)})+\| d_{3}\| _{L^{1}(0,1)}<1, \label{EQ11}$$ which can be done in view of the assumption (\ref{cond1m}). Next, we have from the definition of $\tilde{\alpha}$, as given in (\ref{EQ8}), that there exists a constant $C_{\varepsilon }^{1}$ such that $$\phi (\frac{1}{1-\sigma ^{\ast }}z)\leq (\tilde{\alpha}(\sigma ^{\ast })+\varepsilon )\phi (z)+C_{\varepsilon }^{1}\text{, for all }z. \label{EQ12}$$ Let, now, $u$ be a solution of the family of boundary-value problems (\ref {NRlambda}). Then $u\in C^{2}[0,1]$ with $\phi (u''(t))$ absolutely continuous on $[0,1]$ and satisfies \begin{equation*} u(0)=0, u(1)=\sum_{i=1}^{m-2}\alpha _{i}u(\xi _{i}), u''(0)=0. \end{equation*} We, now, use the estimates $$\|u\|_{\infty }\leq \|u'\|_{\infty }, \| u'\| _{\infty } \leq \frac{1}{1-\sigma ^{\ast }}\| u''\| _{\infty }, \phi (\| u''\| _{\infty })\leq \| (\phi (u''))'\| _{L^{1}(0,1)} \label{EQ13n}$$ and the inequality (\ref{EQ12}) to get \begin{align*} &\| (\phi (u''))'\| _{L^{1}(0,1)}\\ &\leq \phi (\| u\| _{\infty })\| d_{1}\| _{L^{1}(0,1)}+\phi (\| u'\| _{\infty })\| d_{2}\| _{L^{1}(0,1)} \\ &\quad +\phi (\| u''\| _{\infty })\| d_{3}\| _{L^{1}(0,1)}+\| r\| _{L^{1}(0,1)} \\ &\leq (\| d_{1}\| _{L^{1}(0,1)}+\| d_{2}\|_{L^{1}(0,1)}) \phi (\frac{1}{1-\sigma ^{\ast }}\|u''\|_{\infty }) \\ &\quad +\phi (\| u''\| _{\infty })\| d_{3}\| _{L^{1}(0,1)}+\| r\| _{L^{1}(0,1)} \\ &\leq \Big(\tilde{\alpha}(\sigma ^{\ast })+\varepsilon \Big) (\| d_{1}\| _{L^{1}(0,1)}+\| d_{2}\| _{L^{1}(0,1)})\phi (\|u''\|_{\infty })+\| d_{3}\| _{L^{1}(0,1)}\phi (\| u''\| _{\infty })+C_{\varepsilon } \\ &\leq { [(}\tilde{\alpha}(\sigma ^{\ast })+\varepsilon \Big) (\| d_{1}\| _{L^{1}(0,1)}+\| d_{2}\| _{L^{1}(0,1)})+\| d_{3}\| _{L^{1}(0,1)}{ ]}\| (\phi (u''))'\| _{L^{1}(0,1)}+C_{\varepsilon }, \end{align*} %\label{EQ15} where \begin{equation*} C_{\varepsilon }=\| r\| _{L^{1}(0,1)}+C_{\varepsilon }^{1}(\| d_{1}\| _{L^{1}(0,1)}+\| d_{2}\| _{L^{1}(0,1)}). \end{equation*} It, now, follows from (\ref{EQ11}) that there exists a constant $R_{0}>0$, independent of $\lambda \in (0,1]$ such that if $u$ is a solution of the family of boundary-value problems (\ref{NRlambda}) then \begin{equation*} \| (\phi (u''))'\| _{L^{1}(0,1)})\leq R_{0}. \end{equation*} This, combined with (\ref{EQ13n}) gives that there exist a constant $R>0$ such that \begin{equation*} \|u\|_{C^{2}[0,1]}1 \\ -1&\text{if }\sum_{i=1}^{m-2}\alpha _{i}\xi _{i}<1. \end{cases} \end{equation*} Hence if $\sum_{i=1}^{m-2}\alpha _{i}\xi _{i}\neq 1$ we have that $\deg _{LS}(I-\Psi ^{\ast }(\cdot ,1),B(0,R),0)\neq 0$ and there is a $u\in B(0,R)$ that satisfies \begin{equation*} u=\Psi ^{\ast }(\cdot ,1), \end{equation*} equivalently $u$ is a solution to the boundary-value problem (\ref {nonresonant}). 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