\documentclass[reqno]{amsart} \usepackage{amssymb} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2004(2004), No. 25, pp. 1--14.\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 2004 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2004/25\hfil Multi point boundary-value problems] {Multi point boundary-value problems at resonance for n-order differential equations: Positive and monotone solutions} \author[Panos K. Palamides\hfil EJDE-2004/25\hfilneg]{Panos K. Palamides} \address{Naval Academy of Greece, Piraeus, 185 39, Greece} \email{ppalam@otenet.gr \quad ppalam@snd.edu.gr} \urladdr{http://ux.snd.edu.gr/\symbol{126}maths-ii/pagepala.htm} \date{} \thanks{Submitted January 12, 2004. Published February 24, 2004.} \subjclass[2000]{34B10, 34B18, 34B15, 34G20} \keywords{Focal boundary value problem, multi-point, resonance, vector field, \hfill\break\indent positive monotone solution, Sperner's lemma, Knaster-Kuratowski-Mazurkiewicz's principle} \begin{abstract} In this article, we study a complete $n$-order differential equation subject to the $(p,n-p)$ right focal boundary conditions plus an additional nonlocal constrain. We establish sufficient conditions for the existence of a family of positive and monotone solutions at resonance. The emphasis in this paper is not only that the nonlinearity depends on all higher-order derivatives but mainly that the obtaining solution satisfies the above extra condition. Our approach is based on the Sperner's Lemma, proposing in this way an alternative to the classical methodologies based on fixed point or degree theory and results the introduction of a new set of quite natural hypothesis. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \section{Introduction} Consider the differential equation x^{(n)}(t)=f(t,x(t),x'(t),\dots ,x^{(n-1)}(t)),\quad 01) \label{11} subject to the multi-point boundary conditions $$\begin{gathered} x^{(i)}(0)=0\quad \mbox{for }i=0,1,\dots ,p-1, \\ x^{(i)}(1)=0\quad \mbox{for }i=p,p+1,\dots ,n-1, \\ \sum_{i=1}^{m}\alpha_{i}x^{(j)}(\xi _{i})=0, \end{gathered} \label{12}$$ where $p\leq j\leq n-1$ and $m\in \mathbb{N}$ are fixed numbers. The boundary value problem \begin{gathered} x^{(n)}(t)=f(t,x(t),x'(t),\dots ,x^{(n-1)}(t)),\quad 0k_{0}$. Chyan and Henderson \cite{CH}, consider the (between conjugate and focal) BVP \begin{gather*} (-1)^{n-k}x^{(n)}=\lambda a(t)f(t,x), \\ x^{(i)}(0)=0,\;0\leq i\leq k-1 \\ x^{(l)}(1)=0,\;j\leq l\leq j+n-k-1 \end{gather*}% for a (fixed)$1\leq j\leq k-1$and established values of$\lambda $to get positive solutions to that problem, assuming similar conditions. A general overview off much of the work which has been done on these subjects and the methods used is given in the book of Agarwal, O'Regan and Wong \cite{aow}. Consider the$2n$-order nonlinear scalar differential equation $$x^{(2n)}(t)=f(t,x(t),x'(t),\dots x^{(2n-1)}(t)),\quad 0\leq t\leq \gamma , \label{14}$$ and further the associated$(2k,2(n-k))$multi-point focal boundary value problem defined by: $$\begin{gathered} x^{(i)}(0)=0,\;0\leq j\leq 2k-1 \\ x^{(j)}(1)=0,\;2k\leq j\leq 2n-1, \\ \sum_{i=1}^{m}a_{i}x^{(2p) }(\xi _{j}) =0 \end{gathered} \label{15}$$ where$f:[0,1]\times \mathbb{R}^{2n}\rightarrow \mathbb{R}$is continuous,$% m\geq 2$,$n\geq 2$are integers,$1\leq k\leq n-1$,$p\in \{k,k+1,\dots ,n-1\}$,$\alpha _{i}\in \mathbb{R}(i=1,2,\dots ,m)$with$% \sum_{i=1}^{m}a_{i}=0$and$0\leq \xi _{1}<\xi _{2}<\dots <\xi _{m}\leq 1are fixed. Consider the cone \begin{align*} \mathbb{K}_{0}=\big\{& (x,x',\dots ,x^{(2n-1)})\in \mathbb{R}^{2n}/\{0\}:x^{(i)} \geq 0,\;0\leq i\leq 2k-1 \\ & \text{and }(-1)^{j}x^{(j)}\geq 0,\;2k\leq j\leq 2n-1\big\}\,. \end{align*} We assume throughout this paper that the functionf:[0,1]\times \mathbb{R}% ^{2n}\rightarrow \mathbb{R}$is continuous and positive: $$f(t,x,x',x^{\prime \prime },\dots ,x^{(2n-1)})>0,\text{ on the cone }\mathbb{K}_{0}\,. \label{17}$$ Further assume that$f$is: \begin{enumerate} \item Nondecreasing on every of its last$2(n-k)$variables and (strictly) increasing in (at least) one of$n-k-1$even-order derivatives$x^{(2k) },x^{(2k+2) },\dots,x^{(2n-2)}$\item Bounded at$-\infty $on every of its last$n-k-1$odd-order derivatives$x^{(2k+1) }$, \dots,$x^{(2n-1) }$, uniformly for \begin{equation*} \big(t,x,x',\dots,x^{(2k-2) },x^{(2k-1)},x^{(2k) },x^{(2k+2) },x^{(2k+4) },\dots,x^{(2n-2)}\big) \in W, \end{equation*} where$W$is any compact subset of$[0,1] \times \mathbb{R}^{n-k}$: $$\lim_{x^{(2\rho +1) \rightarrow -\infty }}f(t,x(t),x'(t),\dots x^{(2n-1)}(t)) \leq K,\;(\rho =k,k+1,\dots,n-1). \label{18}$$ \end{enumerate} In a recent paper (published in this journal) Liu and Ge \cite{gl} based on the \emph{coincidence degree method} of Gaines and Mawhin \cite{gm} proved that the boundary-value problem at resonance (without any extra condition) (% \ref{11})--(\ref{19}), where $$\begin{gathered} x^{(i)}(0)=0\quad \mbox{for }i=0,1,\dots ,p-1, \\ x^{(i)}(1)=0\quad \mbox{for }i=p+1,\dots ,n-1, \\ \sum_{i=1}^{m}\alpha _{i}x^{(p)}(\xi _{i})=0, \end{gathered} \label{19}$$ has at least one solution, under the following assumptions: \begin{itemize} \item[(A1)] There is$M>0$such that for any$x\in \mathop{\rm dom}L/\ker L$% , with$|x^{(p)}(t)|>M$for all$t\in (0,\frac{1}{2})$, it follows that \begin{equation*} \sum_{i=1}^{m}\alpha _{i}\int_{\xi _{i}}^{1}\frac{(s-\xi _{i})^{n-p-1}}{% (n-p-1)!}f(s,x(s),x'(s),\dots ,x^{(n-1)}(s))ds\neq 0 \end{equation*} \item[(A2)] There is a function$a\in C^{0}[0,1]$and positive numbers$% a_{i} $and$\beta _{i}\in \lbrack 0,1)(i=0,1,\dots ,n-1)$such that \begin{equation*} |f(t,x_{0},x_{1},\dots ,x_{n-1})|\leq a(t)+\sum_{i=0}^{n-1}a_{i}|x_{i}|^{\beta _{i}} \end{equation*} for$t\in \lbrack 0,1]$and$(x_{0},x_{1},\dots ,x_{n-1})\in \mathbb{R}^{n}$\item[(A3)] There is$M^{\ast }>0$such that for any$c\in \mathbb{R}$then either \begin{equation*} c\sum_{i=1}^{m}\alpha _{i}\int_{\xi _{i}}^{1}\frac{(s-\xi _{i})^{n-p-1}}{ (n-p-1)!}f(s,cs^{p},cps^{p-1},\dots ,cp!,0,\dots ,0)ds<0, \end{equation*} or \begin{equation*} c\sum_{i=1}^{m}\alpha _{i}\int_{\xi _{i}}^{1}\frac{(s-\xi _{i})^{n-p-1}}{ (n-p-1)!}f(s,cs^{p},cps^{p-1},\dots ,cp!,0,\dots ,0)ds>0, \end{equation*} for any$c\in \mathbb{R}$with$|c|>M^{\ast }.$\end{itemize} Motivated and inspired by \cite{gl} see also \cite{eh1,fw2,ht,PA}, we establish in this paper sufficient conditions for the existence of at least one solution of (\ref{11})--\ref{12} at resonance. Our principal tool for the analysis of trajectories of (\ref{14}) will be a theorem of combinatorial topology, namely the Knaster-Kuratowski-Mazurkiewicz's principle or (as it is known) Sperner's Lemma (cf. \cite{A1}), as we did in \cite{PAL}. It is worth noticing that the use of Sperner's Lemma in this study, gives an alternative to the usual considerations of topological methods, such as fixed point theories, and results to more strongly conclusions under possibly weaker but in any case much different assumptions. In addition, the most known results on focal or conjugate BVP are based on conditions which are expressed in terms of Green's functions a fact making the resulting criteria too complicated for practical use. In this paper, conditions are of simple form which contain no explicit reference to the Green's functions. Moreover we get a whole$(n-p-1)-$parametric family of solutions of BVP (\ref{14})-(\ref{15}) any member of which satisfies properties as the above (see Remark \ref{Re2} at the end of paper). We study first two special cases of (\ref{14}). More precisely, for the$% (2n) $-order differential equation (\ref{14}) and the$(2k,2(n-k))$multi-point focal boundary value conditions (\ref{15}) \emph{we need only natural assumptions like monotonicity, a kind of boundednes of }$f$\emph{\ and a sign property }(see (\ref{17})-(\ref{18}), so we do not assume any growth or separation constraint on$f$). Then the obtaining solution$x(t)$is not only positives, but it has its firth$2k$derivatives also positive on the interval$(0,1]$, i.e. \begin{equation*} x^{(i)}(t)>0,\quad 00,\quad x^{(2\rho +1)}(t)<0,\quad 00,\quad 01 \label{21} associated wit the ($2k,2(n-k)$multipoint focal value problem $$\begin{gathered} x^{(i)}(0)=0, \quad 0\leq i\leq 2k-1, \\ x^{(j)}(1)=0, \quad 2k\leq j\leq 2n-1,\\ \sum_{i=1}^{m}a_{i}x^{(2p) }(\xi _{j}) =0. \end{gathered} \label{22}$$ It will be convenient to represent (\ref{21}) as a second order system of the form \begin{equation*} X''(t)=f(t,X,X') \end{equation*} where, for notational reasons, we set$X=(Y,Z)$, if \begin{gather*} Y =(y_{0},y_{1},\dots,y_{2n_{1}-1})=(x,x^{''},\dots,x^{(2k-2)})% \in \mathbb{R}^{k}, \\ Z =(z_{0},z_{1},\dots,z_{2(n-k)})=(x^{(2k)},x^{(2k+2)},\dots,x^{(2n-2)})\in \mathbb{R}^{(n-k) }. \end{gather*} Then the boundary conditions at (\ref{22}) take the form \begin{equation*} Y(0)=Y'(0)=0\quad \text{and}\quad Z(1)=Z'(1)=0. \end{equation*} For a (fixed)$\alpha >0$, solutions of (\ref{21}) are defined by trajectories of the initial value problems $$\begin{gathered} x^{(i)}(0)=0,\quad 0\leq i\leq 2k-1, \\ x^{(2j)}(0)=\alpha, \\ x^{(2j+1)}(0)=-\lambda _{j+1}\leq 0,\quad k\leq j\leq n-1; \end{gathered} \label{23}$$ i.e. (\ref{23}) can be written in vector notation as $$Y(0)=0=Y'(0)\quad \text{and}\quad Z(0)=\alpha(1,1,\dots,1),\;Z'(0)=v, \label{24}$$ where$v=-(\lambda _{k},\lambda _{k+1},\dots,\lambda _{n-k})\in \mathbb{R} ^{n-k}$. A solution of the initial value problem (\ref{21})-(\ref{24}) will be denoted by \begin{equation*} X=X(t;v)=(Y(t,v),Z(t,v))\quad \text{or simply }x=x(t;v). \end{equation*} Assume that$\mathbb{K}$denotes the closed positive cone of$\mathbb{R}^{n-k}$and let$\partial \mathbb{K}$be its boundary, which consists of the hyperplanes \begin{equation*} H_{i}=\{x\in \mathbb{R}^{n-k}: x_{i}=0,\; x_{j}\geq 0,\; j\neq i\} \quad (i=k,k+1,\dots,n-1). \end{equation*} \begin{definition} \label{De1} \rm The trajectory$X(t,v)$egresses from$\mathbb{K}$through$H_{i}$, whenever there exists$00$for$0\leq t\leq t_{1}\;\;(j\neq i)$. If moreover there exists an$\varepsilon >0$such that$z_{i}(t,v)<0$,$t_{1}0$such that for any$\lambda >\lambda _{\rho }$, any trajectory$X(t,v)$egresses from$\mathbb{K}$through$H_{\rho }$, for some$ t_{1}\leq 1$. \end{lemma} \begin{proof} Taking into account the assumption (\ref{17}) we may write for all$0\leq t\leq \gamma$, $$F(t,x,x',\dots x^{(2n-1)})>0,\quad Y\geq 0,\quad Y'\geq 0,\quad Z>0, \quad Z'\leq 0\,. \label{26}$$ Consider now, for any (fixed)$\rho =k,\dots,n-1$, a solution$X(t,\lambda e^{\rho })$of the differential equation (\ref{25}), satisfying the initial conditions $$\begin{gathered} y_{j}(0,\lambda e^{\rho })=0,\quad y_{j}'(0,\lambda e^{\rho})=0\quad (j=0,1,\dots,2k-1), \\ z_{j}(0,\lambda e^{\rho })=\alpha ,\quad \;z_{j}'(0,\lambda e^{\rho })=0,\;k\leq j\leq n-1,\;(j\neq \rho ),\\ z_{\rho }(0,\lambda e^{\rho })=\alpha ,\quad z_{\rho }'(0,\lambda e^{\rho })=-\lambda . \end{gathered} \label{27}$$ Since$z_{\rho }(0;\lambda e^{\rho })=\alpha >0$, by the above representation of$X=(Y,Z)$, there exists a$\bar{t}>0$such that for$00$and$% y_{j}'(0,\lambda e^{\rho })>0$when$j=0,1,\dots,k-1$and$z_{j}(t)>0 $and$z_{j}'(t)>0$for$j=k,\dots,\rho -1$. By the modification$F$of$f$and (\ref{26}), it follows also that$z_{j}(t)>0$and$% z_{j}'(t)>0$,$00$. As a result, if some component has a zero in$(0,1]$, the first such zero must be in$z_{\rho }(t,\lambda e^{\rho })$. Therefore, we must show that for$\lambda $sufficiently large the tranjectory$X(t;\lambda e^{\rho })$egresses (strictly) from the positive cone$\mathbb{K}$for some$t_{1}\leq 1$. Assume that this is not the case, i.e. for every$\lambda \in \mathbb{R}$, x^{(2\rho ) }(t;\lambda e^{\rho })\geq 0,\quad 00$ independent of $\hat{t}$ and $\lambda$ such that $$|F(\hat{t},X(\hat{t}),X'(\hat{t}))|\leq M, \label{210}$$% then letting $\lambda \rightarrow +\infty$, we obviously obtain $x^{(2\rho )}(t)<0,$ i.e. a contradiction to (\ref{28}). By the above analysis, the solution $x(t)$ egresses from the cone $\mathbb{K}$ through $H_{\rho }$ and this certainly means that \begin{gather*} z_{\rho }(t,\lambda e^{\rho })>0,\quad 0\leq t0$. So let us set $$\lambda _{\rho }:=\max \{\lambda \geq 0:z_{\rho }(t,\lambda e^{\rho })=x^{(2\rho )}(t,\lambda e^{\rho })\geq 0,\;0\leq t\leq 1\}. \label{211}$$% Now, to show (\ref{210}), we note first that for$\lambda >\lambda _{0}\geq 0 $and since \begin{gather*} x^{(2\rho )}(0,\lambda e^{\rho })=\alpha \quad \text{and}\quad x^{(2\rho +1)}(0,\lambda e^{\rho })=-\lambda \\ x^{(2\rho )}(0,\lambda _{0}e^{\rho })=\alpha \ \quad \and\quad x^{(2\rho +1)}(0,\lambda _{0}e^{\rho })=-\lambda _{0}, \end{gather*}% by continuity of$z_{\rho }(.,\lambda e^{\rho })=x^{(2\rho )}(.,\lambda e^{\rho })$, we get a number$0<\tau \leq 1$such that \begin{equation*} x^{(2\rho )}(t,\lambda _{0}e^{\rho })>x^{(2\rho )}(t,\lambda e^{\rho }),\quad 0x^{(i)}(t,\lambda e^{\rho }),\quad 0x^{(j)}(t,\lambda e^{\rho }),\quad 0x^{(i)}(t,\lambda e^{\rho }),\quad 00,\quad 0\lambda _{0}. \end{gathered} \label{216} \noindent(2) Assume that$j>2\rho $. Then, we also get the contradiction \begin{equation*} 0=\psi ^{(2j) }(\mathbb{\hat{\tau}},\lambda ) =\frac{t^{2n}}{(2n) !} [F(\bar{t},X(\bar{t};\lambda e^{\rho }),X'(\bar{t};\lambda e^{\rho })) -F(\bar{t},X(\bar{t};\lambda _{0}e^{\rho }),X'( \bar{t};\lambda e^{\rho }))], \end{equation*} by noting (\ref{216}) and the (strictly) monotonicity of$F$. Thus, (\ref{214}) can not also be true and so we get (\ref{216}) once again. We set now \begin{equation*} \widehat{K}=\max \{x^{(j)}(t,\lambda _{0}e^{\rho }):j=0,1,\dots,2n-1,\;0\leq t\leq 1,\;j\neq 2\rho +1\}>0 \end{equation*} and consider the rectangle$R=[0,1]\times \lbrack 0,\widehat{K}]^{2\rho }\times (-\infty ,0]\times \lbrack 0,\widehat{K}]^{2(n-\rho ) }$. By (\ref{18}), (\ref{216}) and the continuity of$F$we get \begin{equation*} \max \{F(t,X,X'): (t,X,X')\in R\}=M<+\infty \end{equation*} and thus \begin{equation*} \big| F(t,x(t,\lambda e^{\rho }),x'(t,\lambda e^{\rho }),\dots,x^{(2n-1)}(t,\lambda e^{\rho }))\big| \leq M, \quad 00,\quad 00$, $\;00,\quad z_{j}'(t;v_{0})<0,\quad 00$. In this way, we get a whole $(n-k)-$parametric family of solutions of the modified differential equation (\ref{25}) satisfying the above focal boundary conditions. We assert that there is an $a>0$ such that the obtaining trajectory $X(t,v_{0})$ satisfies further the additional condition $$\sum_{i=1}^{m}\alpha _{i}x^{(p)}(\xi _{i})=0,\quad p=2\rho ,\;(\rho =k,k+1,\dots ,n-1). \label{218}$$% Recalling that $\sum_{i=1}^{m}\alpha _{i}=0$, we set \begin{equation*} \alpha _{i}=% \begin{cases} a_{i}^{+}, & \mbox{if }\alpha _{i}\geq 0 \\ a_{i}^{-}, & \mbox{if }\alpha _{i}<0,% \end{cases}% \end{equation*} $I_{+}=\left\{ i:\alpha _{i}\geq 0\right\}$, $I_{-}=\left\{ i:\alpha _{i}<0\right\}$ and \begin{equation*} A=\sum_{i\in I_{+}}\alpha _{i}^{+}=\sum_{i\in I_{-}}\alpha _{i}^{-}. \end{equation*} Since the solution $x=x^{(2\rho )}(t)$, $0\leq t\leq 1$ is decreasing, we get \begin{aligned} \sum_{i=1}^{m}\alpha _{i}x^{(2\rho )}(\xi _{i}) &=\sum_{i\in I_{+}}\alpha_{i}^{+}x^{(2\rho )}(\xi _{i}) -\sum_{i\in I_{-}}(-\alpha _{i}^{-})x^{(2\rho )}(\xi _{i}) \\ &\leq \sum_{i\in I_{+}}\alpha _{i}^{+}x^{(2\rho )}(\xi _{1})-\sum_{i\in I_{-}}(-\alpha _{i}^{-})x^{(2\rho )}(\xi _{m}) \\ &=A\big[ x^{(2\rho )}(\xi _{1})-x^{(2\rho )}(\xi _{m})\big] . \end{aligned} \label{219}% Suppose that (\ref{218}) is not fulfilled and so there exists an $% \varepsilon _{0}>0$ such that for every $a>0$, \begin{equation*} \sum_{i=1}^{m}\alpha _{i}x^{(2\rho )}(\xi _{i})\geq \varepsilon _{0}. \end{equation*}% If we choose $a\leq \varepsilon _{0}/(2A)$, then by positivity and monotonicity of $x=x^{(2\rho )}(t)$, $0\leq t\leq 1$ and noticing (\ref{219}% ), we get the contradiction \begin{equation*} \sum_{i=1}^{m}\alpha _{i}x^{(2\rho )}(\xi _{i})\leq Ax^{(2\rho )}(\xi _{1})\leq A\frac{\varepsilon _{0}}{2A}=\frac{\varepsilon _{0}}{2}. \end{equation*} It is worth noticing (see the following Remark) that the solution, $x=x(t;v_{0})$, of the modified differential equation (\ref{25}) is actually a solution of the original equation (\ref{21}). \end{proof} \begin{remark} \label{Re1} \rm The solution $x(t;v_{0})$ of (\ref{25}) fulfilling the boundary conditions (\ref{22}), satisfies further the inequalities: \[ x^{(2\rho )}(t,v_{0})>0,\quad x^{(2\rho +1)}(t,v_{0})<0,\quad 00,\quad 00,\quad 00, \quad x^{(2\rho +2)}(t)<0,\;00$,$00,\quad 00,\;\lambda >0,\;m\geq 0)$$$\begin{gathered} x^{(i)}(0)=0,\quad 0\leq i\leq 2k-1,\quad x^{(2k) }(0) =m, \\ x^{(2j+1)}(0)=-\alpha,\quad x^{(2j+2)}(0)=0,\quad k\leq j\leq n-1.\;j\neq \rho \\ x^{(2\rho +1)}(0)=-\alpha ,\quad x^{(2\rho +2)}(0)=\lambda _{\rho+1}, \end{gathered} \label{224}$$% Consider the modification (replace now by$0$only the negative coordinates of$V^{\ast }$and/or$Z^{\ast \prime }$) \begin{equation*} F^{\ast }(t,Y^{\ast },V^{\ast },Z^{\ast },Z^{\ast \prime }):= \begin{cases} f(t,Y^{\ast },V_{0}^{\ast },Z^{\ast },Z_{0}^{\ast \prime }) & \mbox{if } V_{0}^{\ast }\ngtr 0\mbox{ or/and }Z^{\ast \prime }\ngtr 0 \\ f(t,Y^{\ast },V^{\ast },Z^{\ast },Z^{\ast \prime }) & \mbox{otherwise} \end{cases} \end{equation*} and the differential equation $$x^{(2n+1)}(t)=F^{\ast }(t,Y^{\ast },V^{\ast },Z^{\ast },Z^{\ast \prime }). \label{225}$$ For the moment, we fix the initial value$x^{(2k)}(0)=m>0$, and then we may follow the lines of Lemma \ref{lm3} and Theorem \ref{thm1}, under the obvious symmetrical alterations to get a solution \begin{equation*} x=x_{m}(t;v_{0}),\quad 0\leq t\leq 1 \end{equation*}% of (\ref{224}) such that for$k\leq \rho \leq n-1$, \begin{gather*} x_{m}^{(2\rho +1)}(t;v_{0})<0,\quad x_{m}^{(2\rho +2)}(t;v_{0})>0,\quad 0\leq t\leq 1 \\ x_{m}^{(2\rho +1)}(1;v_{0})=0,\quad x_{m}^{(2\rho +2)}(1;v_{0})=0, \end{gather*}% Especially, since$x_{m}^{(2k+1)}(t;v_{0})<0$,$0\leq t\leq 1$, the map$x_{m}^{(2k)}(t;v_{0})$is decreasing on$0\leq t\leq 1$and we may show that there is an$m_{0}>0$such that $$x_{m_{0}}^{(2k)}(t;v_{0})>0,\quad 0\leq t<1\quad \text{and}\quad x_{m_{0}}^{(2k)}(1;v_{0})=0. \label{226}$$% Indeed, suppose that for an$m_{1}$(for example$m_{1}=0$) \begin{equation*} x_{m_{1}}^{(2k)}(t;v_{0})>0,\;0\leq t<\hat{\tau}\quad \text{and}\quad x_{m_{1}}^{(2k)}(t;v_{0})\leq 0,\;\hat{\tau}\leq t\leq 1. \end{equation*}% Taking now$m\rightarrow +\infty $and noticing that the function$f$is bounded, we may get (following a procedure similar to the given one in the proof of Lemma \ref{lm3}) an$m_{2}>m_{1}$such that \begin{equation*} x_{m_{2}}^{(2k)}(t;v_{0})>0,\quad 0\leq t\leq 1. \end{equation*}% Hence by the continuity (Knesser's property) of solutions upon their initial values (see for details \cite{PA} and the references therein), we obtain the requesting in (\ref{226})$m_{0}\in (m_{1},m_{2})$. Finally, the obtaining solution$x(t)=x_{m_{0}}(t;v_{0})$clearly satisfies, for$00,\quad i=0,1,\dots ,2k, \\ x_{m_{0}}^{(2\rho +1)}(t;v_{0})<0, \\ x_{m_{0}}^{(2\rho +2)}(t;v_{0})>0,\quad \rho =k,k+1,\dots ,n-1, \end{gather*} and thus noticing the definition of the modification $F$, we conclude that it is actually a solution of the initial differential equation. \end{proof} \begin{remark} \label{Re2} \rm By the construction of the simplex $S$ (see Theorem \ref{thm1}) and especially the choice of initial conditions (\ref{27}), we clearly get a whole $(n-k-1)$-parametric family of solutions of BVP (\ref{220})-(\ref{221}). Indeed, the only restriction of the constant $\alpha$ comes from the inequality $a>0$. \end{remark} \begin{thebibliography}{99} \bibitem{a2} R. P. Agarwal, \emph{Focal Boundary Value Problems for Differential and Difference Equations}, Kluwer, Dordrecht, 1998. \bibitem{ao1} R. P. Agarwal, D. 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