\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2012 (2012), No. 20, pp. 1--10.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2012 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2012/20\hfil Nonexistence of solutions to BVP] {Nonexistence of solutions to some boundary-value problems for second-order ordinary differential equations} \author[G. L. Karakostas \hfil EJDE-2012/20\hfilneg] {George L. Karakostas} \address{George L. Karakostas \newline Department of Mathematics, University of Ioannina, 451 10 Ioannina, Greece} \email{gkarako@uoi.gr} \thanks{Submitted December 9, 2011. Published February 2, 2012.} \subjclass[2000]{34B15, 34B99, 34K10} \keywords{Boundary value problems; non-existence of solutions} \begin{abstract} We present a method to obtain concrete sufficient conditions which guarantee non-existence of solutions lying into a prescribed domain of six two- or three-point boundary value problems for second-order ordinary differential equations. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \allowdisplaybreaks \section{Introduction} Let $I$ be the interval [0,1] of the real line $\mathbb{R}$ and let $C(I)$ be the Banach space of all continuous functions $x: I\to\mathbb{R}$, endowed with the sup-norm $\|\cdot\|$. In this article, we investigate the non-existence of a solution $x$ of the equation $$\label{e7} x''(t)+(Fx)(t)=0,\quad \text{a. a. }t\in{I},$$ satisfying one of the following conditions: \begin{gather} \label{e1} x(0)=0,\quad x'(1)=\alpha x'(0),\\ \label{e2} x(0)=0,\quad x(1)=\alpha x'(0),\\ \label{e3} x'(0)=0,\quad x(1)=\alpha x(\eta),\\ \label{e4} x'(0)=0,\quad x'(1)=\alpha x(0),\\ \label{e5} x(0)=0,\quad x(1)=\alpha x(\eta), \\ \label{e6} x(0)=\alpha x'(0),\quad x(1)=0, \end{gather} and lying into a prescribed domain of the space $C(I)$. Here assume that $\alpha\geq 0$ and $\eta\in[0,1]$. The dependence of the response $F$ on the function $x$ might be in a moment or a functional way. Some publications which deal with the existence of positive solutions of equations of the form \eqref{e7} lying in a certain domain, associated with the conditions (or the multi-point version of them) respectively, are, for instance, \cite{bf, ma, kt1, kt2, kt3} for \eqref{e1}; \cite{LG1} for \eqref{e2}, \cite{aoy1, JGZ} for \eqref{e3}; \cite{cz1} for \eqref{e4}; \cite{bf1, chen, chen1, CR1, DG1, db1, HG1, KT1, LP1, MW, YR, SW1, xx, YS, w1, WL1, wi1, ZS1} for \eqref{e5}; \cite{mt} for \eqref{e6}. (See, also, the references therein) In most of these cases the response factor $(Fx)(t)$ has the Hammerstein form $q(t)f(x(t))$ and the quotient ${f(u)}/{u}$ plays a central role. In particular, its least and upper limiting values at $0+$ and at $+\infty$ are combined with some arguments related to the Green's function, in order to ensure the applicability of a method leading to the existence of a fixed point of an appropriate integral operator. But, as the sufficient conditions are important for the existence of solutions, the necessary conditions are (more or less) equally important. Working in this direction, in this paper we provide some rather simple sufficient conditions for the nonexistence of (positive) solutions, which lie in an angular domain of the origin. Our discussion refers to the existence of a real number $\rho>0$ such that no solution of the problem exists satisfying the inequality $$|(Fx)(t)|<\rho\|x\|,$$ for almost all $t\in I$. This fact implies that, if we have the response $q(t)f(x(t))$, then we can proceed further to obtain a more concrete domain not containing solutions. Indeed, assume that $(Fx)(t)$ is of the form $q(t)f(x(t))$, where, in the simplest case, $q$ is measurable and essentially bounded and $f$ is nondecreasing. Then, the general result is formulated as follows: \emph{There is no solution $x$ of equation \eqref{e7} satisfying the conditions $(1,j)$ such that} $$f(\|x\|)<\frac{\rho_j}{\|q\|_{\infty}}\|x\|,$$ where $j=2,\dots,7$. \section {Problem \eqref{e1}-\eqref{e7}} We start with the following theorem. \begin{theorem}\label{th1} Assume that $F$ is a function defined in a domain ${\mathcal{D}}(F)$ of the space $C(I)$ such that for each $x\in{\mathcal{D}}(F)$ the value $Fx:I\to{\mathbb{R}}$ is a Lebesgue measurable function. \begin{itemize} \item[(i)] If $\alpha>1$, then there is no positive solution $x$ of the problem \eqref{e1}-\eqref{e7} lying in ${\mathcal{D}}(F)$ and such that $(Fx)(t)\geq 0$, a.e. on $I$. \item[(ii)] If $\alpha\in[0,1]$, then there is no solution $x$ of problem \eqref{e1}-\eqref{e7} lying in ${\mathcal{D}}(F)$ and such that $$\label{e01} \operatorname{ess\,sup}\frac{(Fx)(t)}{x^2(t)}<+\infty.$$ \end{itemize} \end{theorem} \begin{proof} (i) Assume that $\alpha>1$ and let $x$ be a positive solution of the problem in ${\mathcal{D}}(F)$. Consider a real number $\lambda\neq 0$ and write equation \eqref{e7} in the form $$x''(t)+\lambda x'(t)=\lambda x'(t)-(Fx)(t).$$ Multiply both sides with $e^{\lambda t}$ and take $$\big(x'(t)e^{\lambda t}\big)'=[x''(t)+\lambda x'(t)]e^{\lambda t} =[\lambda x'(t)-(Fx)(t)]e^{\lambda t}.$$ Integrating from 0 to $t$, we obtain \begin{align*} x'(t)e^{\lambda t}&=x'(0)+{\lambda}\int_0^tx'(s)e^{\lambda s}ds -\int_0^t(Fx)(s)e^{\lambda s}ds.\\ &=x'(0)+\lambda x(t)e^{\lambda t}-\lambda x(0) -\int_0^tz(s)e^{\lambda s}ds, \end{align*} where $z(s):=\lambda^2x(s)+(Fx)(s)$. Thus we have $$\label{e8} x'(t)-\lambda x(t)=x'(0)e^{-\lambda t} -\int_0^tz(s)e^{-\lambda (t-s)}ds.$$ Multiplying both sides by $e^{-\lambda t}$, we obtain $\big(x(t)e^{-\lambda t}\big)' =[x'(t)-\lambda x(t)]e^{-\lambda t} =x'(0)e^{-2\lambda t}-\int_0^tz(s)e^{-\lambda (2t-s)}ds.$ Integrate both sides from 0 to $t$ and take \begin{align*} x(t)e^{-\lambda t} &=x'(0)\int_0^te^{-2\lambda s}ds-\int_0^t\int_0^rz(s)e^{-\lambda (2r-s)}dsdr\\ &=\frac{x'(0)}{2\lambda}(1-e^{-2\lambda t}) -\int_0^tz(s)e^{\lambda s}\int_s^te^{-2\lambda r}drds\\ &=\frac{x'(0)}{2\lambda}(1-e^{-2\lambda t}) -\frac{1}{2\lambda}\int_0^tz(s)e^{\lambda s}(e^{-2\lambda s}-e^{-2\lambda t}), \end{align*} because of \eqref{e1}. Finally we obtain $$\label{e9} x(t)=\frac{x'(0)}{\lambda}\sinh({\lambda t}) -\frac{1}{\lambda}\int_0^tz(s)\sinh(\lambda(t-s))ds.$$ By using \eqref{e1} and \eqref{e8} it follows that \begin{align*} \alpha x'(0)=x'(1) &=\lambda x(1)+x'(0)e^{-\lambda}-\int_0^1z(s)e^{-\lambda (1-s)}ds\\ &=x'(0)\cosh({\lambda})-\int_0^1z(s)\cosh({\lambda(1-s)})ds \end{align*} and therefore the solution $x$ must satisfy $$\label{e10} x'(0)(\cosh(\lambda)-\alpha)=\int_0^1z(s)\cosh({\lambda(1-s)})ds,$$ for all $\lambda\neq 0$. Observe that the right side is a positive quantity, while the left side depends on $\lambda$. Hence, if it holds $x'(0)>0$, choose $\lambda$ such that $\cosh(\lambda)<\alpha$ and if $x'(0)<0$, choose $\lambda$ such that $\cosh(\lambda)>\alpha$, to get a contradiction. \smallskip (ii) Next assume that $\alpha\in[0,1]$. Let $x$ be a solution satisfying relation \eqref{e01}. Choose $\lambda$ negative large enough such that $$z(t):=(Fx)(t)+\lambda x^2(t)<0,$$ for a.a. $t\in I$. This and \eqref{e10} imply that $x'(0)<0$ and so, due to the fact that $x(0)=0$, the solution $x$ can not be positive. The proof is complete. \end{proof} In the sequel we shall assume that $0<\alpha<1$ and moreover the function $F$ satisfies the following condition: \begin{itemize} \item[(C)] $F$ is a function defined in a domain ${\mathcal{D}}(F)$ of the space $C(I)$ and for each $x\in{\mathcal{D}}(F)$ the value $Fx:I\to{\mathbb{R}}$ is an element of ${\mathcal{L}}_{\infty}$. \end{itemize} For each $\rho>0$ we shall denote by ${\mathcal{A}}_{\rho}$ the set of all functions $x\in{\mathcal{D}}(F)$ satisfying the inequality $$\|Fx\|_{\infty}<\rho\|x\|.$$ \begin{theorem} \label{thm2.2} Assume that $F$ satisfies condition $(C)$. Then there is none $x\in {\mathcal{A}}_{\rho_1}$ which solves problem \eqref{e1}-\eqref{e7}, where $$\rho_1:=1-\alpha.$$ \end{theorem} \begin{proof} Assume that a solution exists satisfying the requirements of the theorem and take $\rho'$ such that $\rho'<\rho_1$ and $\|Fx\|_{\infty}< \rho'\|x\|$. Since $1-\alpha$ is the maximum of the function $$\psi(\lambda):=\frac{\cosh(\lambda)-\sinh^2(\lambda)-\alpha}{\sinh^2(\lambda)} \lambda^2,\quad \lambda\geq 0,$$ we can choose $\lambda\geq 0$ such that $$\label{e11} \rho'<\psi(\lambda).$$ Next, as in Theorem \ref{th1}, we obtain relation \eqref{e9}. By using relation \eqref{e8} and the boundary condition \eqref{e1} we obtain $$\alpha x'(0)=x'(1)=\lambda x(1)+x'(0)e^{-\lambda}-\int_0^1z(s)e^{-\lambda(1-s)}ds,$$ and due to \eqref{e9} it follows that $$x'(0)=\frac{1}{\cosh(\lambda)-\alpha}\int_0^1z(s)\cosh(\lambda(1-s))ds.$$ Hence we have $$\label{e12} x(t) =\frac{\sinh({\lambda t})}{\lambda[\cosh(\lambda) -\lambda\alpha]}\int_0^1z(s)\cosh(\lambda(1-s))ds -\frac{1}{\lambda}\int_0^tz(s)\sinh(\lambda(t-s))ds.$$ Assume that $\|x\|=x(t_0)$, for some $t_0\in[0,1]$. Dividing both sides of \eqref{e12} by $x(t_0)$, we obtain $\lambda[\cosh(\lambda)-\alpha] \leq\sinh(\lambda)\int_0^1\Big[\lambda^2\frac{x(s)}{\|x\|} +\frac{|(Fx)(s)|}{\|x\|}\Big]\cosh({\lambda(1-s)})ds.$ From this relation we obtain $$\lambda(\cosh(\lambda)-\alpha)< \sinh({\lambda})\int_0^1[\lambda^2+\rho'] \cosh({\lambda(1-s)})ds$$ and so $$\lambda^2(\cosh(\lambda)-\alpha)< \sinh^2({\lambda})[\lambda^2+\rho'].$$ The latter contradicts to \eqref{e11} and so there is no solution of the problem. \end{proof} \section {Problem \eqref{e2}-\eqref{e7}} Before we will discuss the problem, we need some information about the function defined by $$\phi(\lambda):=\frac{\sinh(\lambda)}{\lambda}(2-\cosh(\lambda)), \quad \lambda\in[0,1].$$ Observe that $2-\cosh(\lambda)>0$ for all $\lambda\in[0,1]$. Also, since $\phi(0)=1$ we conclude that for each $\alpha\in(0,1)$ there is a $\lambda\in (0,1)$ such that $\alpha<\phi(\lambda)$. Thus the set $$D_{\alpha}:=\{\lambda\in[0,1): \phi(\lambda)>\alpha \}$$ is nonempty and it contains a right neighborhood of 0. \begin{theorem} \label{thm3.1} Assume that $F$ satisfies condition {\rm (C)}. Then there is none $x\in {\mathcal{A}}_{\rho_2}$ which solves problem \eqref{e2}-\eqref{e7}, where $$\rho_2:=2(1-\alpha).$$ \end{theorem} \begin{proof} Assume that a solution exists satisfying the requirements of the theorem and take $\rho'$ such that $\rho'<\rho_2$ and $\|Fx\|_{\infty}< \rho'\|x\|$. Since the number $2(1-\alpha)$ is the maximum of the quantity $$\psi_1(\lambda):= \frac{\phi(\lambda)-\alpha}{[\cosh(\lambda)-1] \sinh(\lambda)} \lambda^3,\quad \lambda\geq 0,$$ we can choose $\lambda>0$ such that $$\label{e13} \rho'<\psi_1(\lambda).$$ Next, following the same method as in Theorem \ref{th1}, we obtain relation \eqref{e9}. By using the boundary condition \eqref{e2} we obtain $$\alpha x'(0)=x(1)=\frac{x'(0)}{\lambda}\sinh({\lambda}) -\frac{1}{\lambda}\int_0^1z(s)\sinh(\lambda(1-s))ds,$$ from which it follows that $$x'(0)=\frac{1}{\sinh(\lambda)-\lambda\alpha}\int_0^1z(s)\sinh(\lambda(1-s))ds.$$ Hence we have $$\label{e14} x(t) =\frac{\sinh({\lambda t})}{\lambda[\sinh(\lambda) -\lambda\alpha]}\int_0^1z(s)\sinh(\lambda(1-s))ds -\frac{1}{\lambda}\int_0^tz(s)\sinh(\lambda(t-s))ds.$$ From here and our assumptions we conclude that $\sinh(\lambda)>\lambda\alpha$. Next assume that $\|x\|=x(t_0)$, for some $t_0\in[0,1]$. Dividing both sides of \eqref{e14} by $x(t_0)$, we obtain $\lambda[\sinh(\lambda)-\lambda\alpha] \leq\sinh(\lambda)\int_0^1\Big[\lambda^2\frac{x(s)}{\|x\|} +\frac{|(Fx)(s)|}{\|x\|}\Big]\sinh({\lambda(1-s)})ds.$ From this relation we obtain $$\lambda(\sinh(\lambda)-\lambda\alpha) < \sinh({\lambda})\int_0^1[\lambda^2+\rho']\sinh({\lambda(1-s)})ds$$ and so $$\lambda^2(\sinh(\lambda)-\lambda\alpha) < \sinh({\lambda})[\lambda^2+\rho'](\cosh({\lambda)-1)}.$$ The latter contradicts to \eqref{e13} and so there is no solution of the problem. \end{proof} \section {Problem \eqref{e3}-\eqref{e7}} \begin{theorem}\label{th2} Assume that $F$ satisfies condition {\rm (C)}. Then there is no $x\in {\mathcal{A}}_{\rho_3}$ that solves problem \eqref{e22}-\eqref{e1}, where $$\rho_3:=\sup_{\lambda>0}\frac{2-\cosh(\lambda)-\alpha e^{\lambda(\eta-1)}}{\cosh(\lambda)-1} \lambda^2.$$ \end{theorem} \begin{proof} Assume that a solution exists satisfying the requirements of the theorem and take $\rho'$ such that $\rho'<\rho_3$ and $\|Fx\|_{\infty}< \rho'\|x\|$. Choose $\lambda>0$ such that $$\label{e15} \rho'<\frac{2-\cosh(\lambda)-\alpha e^{\lambda(\eta-1)}}{\cosh(\lambda)-1} \lambda^2.$$ Following the same method as in Theorem \ref{th1}, we obtain relation \eqref{e8}, which due to \eqref{e3} becomes $$\label{e16} x'(t)-\lambda x(t)=-\int_0^tz(s)e^{-\lambda (t-s)}ds.$$ Multiplying both sides with $e^{\lambda t}$ we obtain $$\label{e17} x(t)=x(0)e^{\lambda t}-\int_0^tz(s)\sinh(\lambda(t-s))ds.$$ From this relation and \eqref{e3} we derive $$x(0)=\frac{1}{\lambda(e^{\lambda}-\alpha e^{\lambda\eta})} \Big[\int_0^1z(s)\sinh(\lambda(1-s))ds -\alpha\int_0^{\eta}z(s)\sinh(\lambda(\eta-s))ds\Big],$$ and therefore we have \label{e18}\begin{aligned} x(t)&=\frac{e^{\lambda t}}{\lambda(e^{\lambda} -\alpha e^{\lambda\eta})}\Big[\int_0^1z(s)\sinh(\lambda(1-s))ds\\ &\quad-\alpha\int_0^{\eta}z(s)\sinh(\lambda(\eta-s))ds\Big] -\frac{1}{\lambda}\int_0^tz(s)\sinh(\lambda(t-s))ds. \end{aligned} Assume that $\|x\|=x(t_0)$, for some $t_0\in[0,1]$. Dividing both sides of \eqref{e18} by $x(t_0)$, we obtain $$\lambda^2[e^{\lambda}-\alpha e^{\lambda\eta}]\leq e^{\lambda} [\lambda^2+\rho'](\cosh({\lambda)-1)}).$$ The latter contradicts to \eqref{e15} and so there is no solution of the problem. \end{proof} The following table shows the values of the bound $\rho_3$ for some values of the coefficient $\alpha$ and the argument $\eta$. \begin{center} \begin{tabular}{|l||l|l|l|l|l|l|l|l|l|l|l|} \hline $\displaystyle{}_\eta\quad ^{\alpha}_{\rho_3}$ &0.1&0.2&0.3&0.4&0.5&0.6&0.7&0.8&0.9 \\ \hline\hline $0$& 1.808&1.628&1.464&1.307&1.160&1.020&0.888&0.761&0.639\\ \hline $0.1$& 1.806&1.625&1.453&1.291&1.137&0.990&0.849&0.714&0.585 \\ \hline $0.2$& 1.805&1.620&1.444&1.275&1.114&0.959&0.811&0.667&0.529 \\ \hline $0.3$& 1.804&1.615&1.434&1.260&1.092&0.929&0.772&0.620&0.472 \\ \hline $0.4$& 1.803&1.611&1.426&1.246&1.071&0.900&0.735&0.573&0.416 \\ \hline $0.5$& 1.802&1.608&1.418&1.233&1.051&0.873&0.699&0.528&0.361 \\ \hline $0.6$& 1.801&1.605&1.412&1.221&1.034&0.849&0.667&0.487&0.310 \\ \hline $0.7$& 1.800&1.603&1.407&1.212&1.019&0.828&0.639&0.451&0.265 \\ \hline $0.8$& 1.800&1.599&1.403&1.205&1.009&0.813&0.618&0.423&0.230 \\ \hline $0.9$& 1.800&1.600&1.400&1.201&1.002&0.803&0.604&0.379&0.207 \\ \hline \end{tabular} \end{center} \section {Problem \eqref{e4}-\eqref{e7}} \begin{theorem} \label{thm5.1} Assume that $F$ satisfies condition {\rm (C)}. Then there is none $x\in {\mathcal{A}}_{\rho_4}$ which solves problem \eqref{e4}-\eqref{e7}, where $$\rho_4:=\sup_{\lambda>0}\frac{\lambda[e^{\lambda}-\alpha]}{\lambda\cosh(\lambda) -\lambda+1-e^{-\lambda}}-\lambda^2.$$ \end{theorem} \begin{proof} Assume that a solution $x$ exists satisfying the requirements of the theorem and take $\rho'$ such that $\rho'<\rho_4$ and $\|Fx\|_{\infty}< \rho'\|x\|$. Choose $\lambda>0$ such that $$\label{e19} \rho'<\frac{\lambda[e^{\lambda}-\alpha]}{\lambda\cosh(\lambda)-\lambda+1-e^{-\lambda}} -\lambda^2.$$ Next, following the same method as in Theorem \ref{th2}, we obtain relation \eqref{e17}, which because of \eqref{e4} gives $$\alpha x(0)=\frac{1}{\lambda e^{\lambda}-\alpha}\int_0^1z(s)\Big[\sinh(\lambda(1-s)) +e^{-\lambda(1-s)}\Big]ds.$$ Therefore, $$\label{e20} x(t)=\frac{e^{\lambda t}}{\lambda e^{\lambda}-\alpha)}\int_0^1z(s) \Big[\sinh(\lambda(1-s))+e^{-\lambda(1-s)}\Big]ds -\int_0^tz(s)e^{\lambda(t-s)}ds.$$ Next assume that $\|x\|=x(t_0)$, for some $t_0\in[0,1]$. Dividing both sides of \eqref{e20} by $x(t_0)$, we, finally, obtain $$\lambda[\lambda- e^{-\lambda}\alpha]\leq[\lambda^2+\rho'] (\lambda\cosh(\lambda)-\lambda+1-e^{-\lambda}).$$ The latter contradicts \eqref{e19} and so there is no solution of the problem. \end{proof} The parameter $\rho_4$ for various values of the coefficient $\alpha$ is given in the following table. \begin{center} \begin{tabular}{|l||l|l|l|l|l|l|l|l|l|} \hline $\alpha$&0.1&0.2&0.3&0.4&0.5&0.6&0.7&0.8&0.9 \\ \hline $\rho_4$& 1.480&1.375&1.271&1.167&1.063&0.961&0.859&0.757&0.657 \\ \hline \end{tabular} \end{center} \section {Problem \eqref{e5}-\eqref{e7}} \begin{theorem} Assume that $F$ satisfies condition {\rm (C)} and, moreover, assume that $0<\alpha<1$ and $0<\eta<1$. Then there is none $x\in {\mathcal{A}}_{\rho_5}$ which solves the problem \eqref{e5}-\eqref{e7}, where $$\rho_5:=\sup_{\lambda>0}\frac{\lambda^2[\sinh(\lambda) -\alpha\sinh(\eta\lambda)]}{\sinh(\lambda)\big(\cosh(\lambda)-1\big)}-\lambda^2.$$ \end{theorem} \begin{proof} Assume that a solution $x$ exists satisfying the requirements of the theorem and take $\rho'$ such that $\rho'<\rho_5$ and $\|Fx\|_{\infty}< \rho'\|x\|$. Choose $\lambda>0$ such that $$\label{e21} \rho'<\frac{\lambda^2[\sinh(\lambda)-\alpha\sinh(\eta\lambda)]}{\sinh(\lambda) \big(\cosh(\lambda)-1\big)}-\lambda^2.$$ Next, following the same method as in Theorem \ref{th1}, we obtain relation \eqref{e9}, which because of \eqref{e5} gives $x'(0)=\frac{1}{\sinh(\lambda)-\alpha\sinh(\lambda\eta)} \Big[\int_0^1z(s)\sinh(\lambda(1-s))ds -\int_0^{\eta}z(s)\sinh(\lambda(\eta-s))ds\Big].$ Therefore, \label{e22} \begin{aligned} x(t)&=\frac{\sinh(\lambda t)}{\lambda\big(\sinh(\lambda) -\alpha\sinh(\lambda\eta)\big)}\Big[\int_0^1z(s)\sinh(\lambda(1-s))ds\\ &\quad -\int_0^{\eta}z(s)\sinh(\lambda(\eta-s))ds\Big]-\frac{1}{\lambda} \int_0^tz(s)\sinh({\lambda(t-s)})ds. \end{aligned} Next assume that $\|x\|=x(t_0)$, for some $t_0\in[0,1]$. Dividing both sides of \eqref{e22} by $x(t_0)$, we, finally, obtain $$\lambda^2[\sinh(\lambda)-\alpha\sinh(\lambda\eta)] \leq[\sinh(\lambda)[\lambda^2+\rho'](\cosh(\lambda)-1).$$ The latter contradicts \eqref{e21} and so there is no solution of the problem. \end{proof} The following table shows the values of the bound $\rho_5$ for some values of the coefficient $\alpha$ and the argument $\eta$. \begin{center} \begin{tabular}{|l||l|l|l|l|l|l|l|l|l|l|} \hline $\displaystyle{_\alpha\quad ^{\eta}_{\rho_5}}$ &0.1&0.2&0.3&0.4&0.5&0.6&0.7&0.8&0.9 \\ \hline\hline $0.1$& 1.978&1.959&1.939&1.919&1.899&0.878&0.858&0.839&0.818 \\ \hline $0.2$& 1.958&1.919&1.878&1.839&1.798&1.759&1.718&1.679&1.637 \\ \hline $0.3$& 1.938&1.919&1.878&1.839&1.798&1.759&1.718&1.518&1.458 \\ \hline $0.4$& 1.917&1.838&1.758&1.679&1.598&1.519&1.439&1.355&1.267 \\ \hline $0.5$& 1.897&1.797&1.697&1.599&1.498&1.398&1.298&1.195&1.097 \\ \hline $0.6$& 1.879&1.759&1.638&1.519&1.399&1.279&1.159&1.037&0.918 \\ \hline $0.7$& 1.858&1.719&1.578&1.439&1.296&1.159&1.017&0.879&0.736 \\ \hline $0.8$& 1.839&1.678&1.518&1.359&1.195&1.039&0.879&0.719&0.559 \\ \hline $0.9$& 1.819&1.637&1.458&1.278&1.099&0.917&0.738&0.559&0.379 \\ \hline \end{tabular} \end{center} \section {Problem \eqref{e6}-\eqref{e7}} \begin{theorem} Assume that $F$ satisfies condition {\rm (C)}. Then there is no $x\in {\mathcal{A}}_{\rho_6}$ which solves problem \eqref{e6}-\eqref{e7}, where $$\rho_6:=\sup_{\lambda>0}\frac{\lambda^2[\alpha\lambda +\cosh(\lambda)]}{\sinh(\lambda)\big(\lambda\alpha e^{\lambda} +\cosh(\lambda)\big)}-\lambda^2.$$ \end{theorem} \begin{proof} Assume that a solution $x$ exists satisfying the requirements of the theorem and take $\rho'$ such that $\rho'<\rho_5$ and $\|Fx\|_{\infty}< \rho'\|x\|$. Choose $\lambda>0$ such that $$\label{e23} \rho'<\frac{\lambda^2[\sinh(\lambda)-\alpha\sinh(\eta\lambda)]}{\sinh(\lambda) \big(\cosh(\lambda)-1\big)}-\lambda^2.$$ Next, following the same method as previously, we obtain $x'(0)=\frac{1}{\alpha\lambda+\cosh(\lambda)}\int_0^1z(s)\sinh(\lambda(1-s))ds.$ Therefore, $$\label{e24} x(t)=\frac{\lambda\alpha e^{\lambda t}+\sinh(\lambda t)}{\lambda \big(\alpha\lambda+\cosh(\lambda)\big)}\int_0^1z(s)\cosh(\lambda(1-s))ds -\frac{1}{\lambda}\int_0^tz(s)\cosh({\lambda(t-s)})ds.$$ Next assume that $\|x\|=x(t_0)$, for some $t_0\in[0,1]$. Dividing both sides of \eqref{e24} by $x(t_0)$, we, finally, obtain $$\lambda^2[\alpha\lambda+\cosh(\lambda)]\leq\sinh(\lambda) [\lambda^2+\rho'](\lambda\alpha e^{\lambda}+\sinh(\lambda)).$$ The latter contradicts \eqref{e23} and so there is no solution of the problem. \end{proof} The following tableshows the values of the bound $\rho_6$, for some values of the coefficient $\alpha$. \begin{center} \begin{tabular}{|l||l|l|l|l|l|l|l|l|l|} \hline $\alpha$&0.1&0.2&0.3&0.4&0.5&0.6&0.7&0.8&0.9 \\ \hline $\rho_6$& 0.223&0.216&0.211&0.206&0.202&0.198&0.195&0.192&0.190 \\ \hline \end{tabular} \end{center} \begin{thebibliography}{00} \bibitem{aoy1} Ravi P. Agarwal, Donal O'Regan, Baoqiang Yan; Positive solutions for singular three-point boundary-value problems, \emph{Electron. J. Differential Equations} \textbf{2008}(2008), No. 116, 1-20. \bibitem {bf1} Chuanzhi Bai, Jinxuan Fang; Existence of multiple positive solutions for nonlinear m-point boundary value problems, \emph{J. Math. Anal. Appl.} \textbf{281} (2003) 76-85. \bibitem{bf} Donglong Bai, Hanying Feng; Three positive solutions for m-point boundary value problems with one-dimensional $p$-Laplacian, \emph{Electron. J. Differential Equations}, \textbf{2011} (2011), No. 75, 1-10. \bibitem{ma} R. Ma, D. Cao; Positive solutions to an m-point boundary value problem, \emph{Appl. Math. J. Chinese Univ. Ser. B} \textbf{17} (2002), 24-30. \bibitem{chen} Haibo Chen; Positive solutions for the nonhomogeneous three-point boundary value problem of second-order differential equations, \emph{Math. Comp. Modelling} \textbf{45} (2007), 844-852. \bibitem{chen1} Shihua Chen, Qunkiao Zhang, Li Chen; Positive Solutions for an $n$-Point Nonhomogeneous Boundary Value Problem, \emph{Math. Comp. Modelling} \textbf{40} (2004), 1405-1412. \bibitem{cz1} Yun Chen, Baoqiang Yan, Lili Zhang; Positive solutions for singular three-point boundary-value problems on $x'$, \emph{Electron. J. Differential Equations} \textbf{2007}(2007), No. 63, 1-9. \bibitem{CR1} Wing-Sum Cheung, Jingli Ren; Positive solution for $m$-point boundary value problems, \emph{J. Math. Anal. Appl.} \textbf{303} (2005), 565-575. \bibitem{DG1} Shijie Dong, Weigao Ge; Positive Solutions of an $m$-Point Boundary Value Problem with Sign Changing Nonlinearities, \emph{Comput. Math. Appl.} \textbf{49} (2005), 589-598. \bibitem{db1} Zengji Du, Zhanbing Bai; Asymptotic solutions for a second-order differential equation with three-point boundary conditions, \emph{Appl. Math. Comput.} \textbf{186} (2007), 469-473. \bibitem{HG1} Xiaoming He, Weigao Ge; Triple solutions for second-order three-point boundary value problems. \emph{J. Math. Anal. Appl.} \textbf{268} (2002), 256-265. \bibitem{mt} Andrei Horvat-Marc, Cristina Ticala; Localization of solutions for a problem arizing in the theory of adiabatic tubular chemical reactor, \emph{Carpathian J. Math.} \textbf{20} (2004), 187-192. \bibitem{JGZ} Yude Ji, Yanping Guo, Jiehua Zhang; Positive solutions for second-order quasilinear multi-point boundary value problems, \emph{Appl. Math. Comput.} \textbf{189} (2007) , 725-731. \bibitem{kt1} G. L. Karakostas, P. Ch. Tsamatos; Positive solutions of a boundary-value problem for second order ordinary differentiual equation, \emph{Electron. J. Differential Equations}, \textbf{2000} (2000), No. 49, 1-9. \bibitem{KT1} G. L. Karakostas, P. Cr. Tsamatos; Existence of multiple positive solutions for a nonlocal boundary value problem, \emph{Topol. Methods Nonlinear Anal.} \textbf{19}, 2002, 109-121. \bibitem{kt2} G. L. Karakostas, P. Ch. Tsamatos; Sufficient Conditions for the Existence of Nonnegative Solutions of a Nonlocal Boundary Value Problem, \emph{Appl. Math. Lett.} \textbf{15} (2002), 401-407. \bibitem{kt3} G. L. Karakostas, P. Ch. Tsamatos; Positive solutions for a nonlocal boundary value problem with increasing response, \emph{Electron. J. Differential Equations}, \textbf{2000} (2000), No. 73, 1-8. \bibitem{LG1} Hairong Lian, Weigao Ge; Positive solutions for a four-point boundary value problem with the p-Laplacian, \emph{Nonl. Analysis} \textbf{68} (2008), 3493-3503. \bibitem{LP1} Xuezhe Lv, Minghe Pei; Existence and Uniqueness of Positive Solution for a Singular Nonlinear Second-Order $m$-Point Boundary Value Problem, \emph{Bound. Value Probl.} \textbf{2010}, Article ID 254928, (16 pages), doi:10.1155/2010/254928. \bibitem{MW} Ruyun Ma, Haiyan Wang; Positive solutions of nonlinear three-point boundary-value problems, \emph{J. Math. Anal. Appl.} \textbf{279} (2003), 216-227. \bibitem{YR} Youssef N. Raffoul; Positive Solutions of Three-Point Nonlinear Second Order Boundary Value Problem, \emph{Electron. J. Qual. Theory Differ. Equ.}, \textbf{5} (2002), 1-11. \bibitem{SW1} Jian-Ping Sun, Jia Wei; Existence of positive solution for semipositone second-order three-point boundary-value problem, \emph{Electron. J. Differential Equations} \textbf{2008} (2008), No. 41, 1-7. \bibitem{YS} Yongping Sun; Positive solutions of nonlinear second-order $m$-point boundary value problem, \emph{Nonlinear Anal.} \textbf{61} (2005), 1283-1294. \bibitem{WL1} Yong Wan, Yuji Liu; On Nonlinear Boundary Value Problems for Functional Difference Equations with p-Laplacian, \emph{Discrete Dyn. Nat. Soc.} \textbf{2010}, Article ID 396840, (12 pages), doi:10.1155/2010/396840. \bibitem{w1} J. R. L. Webb; Uniqueness of the principal eigenvalue in nonlocal boundary value problems, \emph{Discrete Contin. Dyn. Syst. Ser. S} \textbf{1} (1), (2008), 177-186. \bibitem{wi1} J. R. L. Webb, Gennaro Infante; Positive solutions of nonlocal boundary value problems: A unified approach, \emph{J. London Math. Soc.} \textbf{74} (2) (2006), 673-693. \bibitem{xx} Xu Xian; Positive solutions for singular m-point boundary value problems with positive parameter, \emph{J. Math. Anal. Appl.} 291 (2004) 352-367. \bibitem{ZS1} Guowei Zhang, Jingxian Sun; Positive solutions of $m$-point boundary value problems, \emph{J. Math. Anal. Appl.} \textbf{291} (2004), 406-418. \end{thebibliography} \end{document}