\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2006(2006), No. 12, 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 2006 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2006/12\hfil Focal differential equation] {Green's function and existence of solutions for a functional focal differential equation} \author[D. R. Anderson, T. O. Anderson, M. M. Kleber\hfil EJDE-2006/12\hfilneg] {Douglas R. Anderson, Tyler O. Anderson, Mathew M. Kleber} % in alphabetical order \address{Douglas R. Anderson \hfill\break Department of Mathematics and Computer Science\\ Concordia College \\ Moorhead, MN 56562 USA} \email{andersod@cord.edu} \address{Tyler O. Anderson \hfill\break Department of Mathematics and Computer Science\\ Concordia College \\ Moorhead, MN 56562 USA} \email{toanders@cord.edu} \address{Mathew M. Kleber \hfill\break Department of Mathematics and Computer Science\\ Concordia College \\ Moorhead, MN 56562 USA} \email{mmkleber@cord.edu} \date{} \thanks{Submitted November 28, 2005. Published January 26, 2006.} \subjclass[2000]{34B10, 34B18, 34B27} \keywords{Multiple solutions; boundary value problems; Green's function;\hfill\break\indent third order differential equation} \begin{abstract} We determine Green's function for a third-order three-point bound\-ary-value problem of focal type and determine conditions on the coefficients and boundary points to ensure its positivity. We then apply this in the determination of the existence of positive solutions to a related higher-order functional differential equation. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \section{finding green's function} Since at least the time of Chazy's attempt \cite{chazy} to completely classify all third-order differential equations of certain form, analysts have been fascinated by the study of third-order differential equations in the pure sense, but also in the applied sense, as in Gamba and J\"{u}ngel \cite{gamba}. Here we will be concerned initially with a certain class of third-order differential equations, namely the homogeneous three-point mixed boundary-value type given by \begin{gather} x'''(t)=0, \quad t_1\le t\le t_3 \label{b0} \\ \alpha x(t_1)-\beta x'(t_1)=0 \label{b1}\\ \gamma x(t_2)+\delta x'(t_2)=0 \nonumber \\ x''(t_3)=0. \nonumber \end{gather} Here we assume \begin{enumerate} \item[$(i)$] $t_1 < t_2 < t_3$ are real numbers; \item[$(ii)$] $\alpha, \beta, \gamma \ge 0$; \item[$(iii)$] $k:=\alpha\delta+\beta\gamma+\alpha\gamma(t_2-t_1)\ne 0$; \item[$(iv)$] $\delta > \max\{\gamma(t_3-t_2), \frac{k(t_3-t_1)^2}{2(t_2-t_1)[\alpha(t_3-t_1)+\beta]}-\frac{\gamma}{2}(t_2-t_1)\}$; see Lemma \ref{lemmak} and Theorem \ref{positive}. \end{enumerate} This is a generalization of the third-order, three-point, right-focal boundary value problem found in \cite{An1,An2,anddavis,gy}, where $\alpha=\delta=1$ and $\beta=\gamma=0$. We prove the existence of and find an explicit formula for Green's function associated with \eqref{b0}, \eqref{b1}; for more on Green's functions and their applications, see \cite{kelley}. \begin{lemma}\label{lemmak} The number $k$ satisfies $$\label{k} k=\alpha\delta+\beta\gamma+\alpha\gamma(t_2-t_1)\ne 0$$ if and only if the boundary value problem \eqref{b0}, \eqref{b1} has only the trivial solution. \end{lemma} \begin{proof} A general solution of \eqref{b0} is $x(t)=k_1t^2+k_2t+k_3$. The condition at $t=t_3$ implies that $k_1=0$. The mixed boundary conditions at $t_1$ and $t_2$ lead to the two equations \begin{align*} \beta k_2-\alpha(t_1k_2+k_3)=0 \\ (t_2\gamma+\delta)k_2+\gamma k_3=0. \end{align*} The determinant of the coefficients for this system is $k$. It follows that $k_2=k_3=0$ if and only if $k\ne 0$. This implies the given boundary value problem \eqref{b0}, \eqref{b1} has only the trivial solution if and only if $k\ne 0$. \end{proof} \begin{theorem}\label{green} Assume for $k$ given in \eqref{k} that $k > 0$. Then Green's function for the homogeneous problem \eqref{b0} satisfying the boundary conditions \eqref{b1} is given via $$\label{greensfunction} g(t,s)= \begin{cases} s\in [t_1,t_2] &: \begin{cases} u_1(t,s) &:t\le s \\ v_1(t,s) &:t\ge s \end{cases}\\[3pt] s\in [t_2,t_3] &: \begin{cases} u_2(t,s) &:t\le s \\ v_2(t,s) &:t\ge s \end{cases} \\ \end{cases}$$ for $t,s\in[t_1,t_3]$, where \begin{align*} u_1(t,s) &:= \frac{1}{k}(s-t_1)[\alpha(t-t_1)+\beta] \left[\delta+\frac{\gamma}{2}(2t_2-t_1-s)\right] -\frac{1}{2}(t-t_1)^2, \\ v_1(t,s) &:= u_1(t,s)+\frac{1}{2}(t-s)^2 =\frac{1}{2k}(s-t_1)[\alpha(s-t_1)+2\beta][\gamma(t_2-t) +\delta], \\ u_2(t,s) &:= \frac{1}{k}[\alpha(t-t_1)+\beta] \left[\delta(t_2-t_1)+\frac{\gamma}{2}(t_2-t_1)^2\right] -\frac{1}{2}(t-t_1)^2, \\ v_2(t,s) &:= u_2(t,s)+\frac{1}{2}(t-s)^2. \end{align*} \end{theorem} \begin{proof} Note that $g(t,s)$ is well defined for all $(t,s)\in[t_1,t_3]\times[t_1,t_3]$. First, check that $g$ satisfies the boundary conditions \eqref{b1}. For convenience we note that $$\frac{\partial}{\partial t}g(t,s)= \begin{cases} s\in [t_1,t_2] &: \begin{cases} \frac{\alpha}{k}(s-t_1)\left[\delta+\frac{\gamma}{2}(2t_2-t_1-s)\right] -t+t_1 & \\ \frac{\alpha}{k}(s-t_1)\left[\delta+\frac{\gamma}{2}(2t_2-t_1-s)\right] -s+t_1 & \end{cases}\\[3pt] s\in [t_2,t_3] &: \begin{cases} \frac{\alpha}{k}\left[\delta(t_2-t_1)+\frac{\gamma}{2}(t_2-t_1)^2\right] -t+t_1 & \\ \frac{\alpha}{k}\left[\delta(t_2-t_1)+\frac{\gamma}{2}(t_2-t_1)^2\right] -s+t_1 & \end{cases} \\ \end{cases}$$ and $$\frac{\partial^2}{\partial t^2}g(t,s)= \begin{cases} s\in [t_1,t_2] &: \begin{cases} -1 & : t < s \\ 0 & : t > s \end{cases}\\[3pt] s\in [t_2,t_3] &: \begin{cases} -1 & : t < s \\ 0 & : t > s, \end{cases} \\ \end{cases}$$ for fixed $s$; in the rest of this proof we will employ the shorthand $g'$ and $g''$ for these two expressions. For $t=t_1$ and $s\in[t_1,t_2]$: \begin{align*} \alpha g(t_1,s)-\beta g'(t_1,s) & = \alpha u_1(t_1,s)-\beta u_1'(t_1,s) \\ & = \frac{\alpha\beta}{k}(s-t_1)\left[\delta+\frac{\gamma}{2}(2t_2-t_1-s)\right]\\ & \quad -\frac{\beta\alpha}{k}(s-t_1)\left[\delta+\frac{\gamma}{2}(2t_2-t_1-s)\right] \\ & = 0. \end{align*} For $t=t_1$ and $s\in[t_2,t_3]$: \begin{align*} \alpha g(t_1,s)-\beta g'(t_1,s) & = \alpha u_2(t_1,s)-\beta u_2'(t_1,s) \\ & = \frac{\alpha\beta}{k}\left[\delta(t_2-t_1)+\frac{\gamma}{2}(t_2-t_1)^2\right]\\ & \quad -\frac{\beta\alpha}{k}\left[\delta(t_2-t_1)+\frac{\gamma}{2}(t_2-t_1)^2\right] \\ & = 0. \end{align*} For $t=t_2$ and $s\in[t_1,t_2]$: \begin{align*} \gamma g(t_2,s)+\delta g'(t_2,s) & = \gamma v_1(t_2,s)+\delta v_1'(t_2,s) \\ & = \frac{\gamma}{2}[(t_2-s)^2-(t_2-t_1)^2] \\ & \quad +\frac{\gamma}{k}(s-t_1)[\alpha(t_2-t_1)+\beta]\left[\delta+\frac{\gamma}{2}(2t_2-t_1-s)\right] \\ & \quad +\delta(t_1-s)+\frac{\delta\alpha}{k}(s-t_1) \left[\delta+\frac{\gamma}{2}(2t_2-t_1-s)\right] \\ & = \frac{\gamma}{2}(2t_2-t_1-s)(t_1-s)+\delta(t_1-s) \\ & \quad +(s-t_1)\left[\delta+\frac{\gamma}{2}(2t_2-t_1-s)\right]\frac{k}{k} \\ & = 0. \end{align*} For $t=t_2$ and $s\in[t_2,t_3]$: \begin{align*} \gamma g(t_2,s)+\delta g'(t_2,s) & = \gamma u_2(t_2,s)+\delta u_2'(t_2,s) \\ & = \frac{\gamma}{k}[\alpha(t_2-t_1)+\beta] \left[\delta(t_2-t_1)+\frac{\gamma}{2}(t_2-t_1)^2\right] \\ & \quad -\frac{\gamma}{2}(t_2-t_1)^2+\delta(t_1-t_2) \\ & \quad +\frac{\delta\alpha}{k}\left[\delta(t_2-t_1)+\frac{\gamma}{2}(t_2-t_1)^2\right] \\ & = 0. \end{align*} Finally, differentiating the expression $g'(t,s)$ with respect to $t$ shows that $g''(t_3,s)=0$ for any $s\in[t_1,t_3]$. Thus, $g$ as in \eqref{greensfunction} satisfies the boundary conditions \eqref{b1}. Now, for any function $f$ continuous on $[t_1,t_3]$, define $$x(t):=\int_{t_1}^{t_3}g(t,s)f(s)ds.$$ As shown above, this $x$ satisfies the boundary conditions \eqref{b1} via $g$. We will show that $x'''(t)=f(t)$. Note that for $t\in[t_1,t_2]$, \begin{align*} x''(t) & = \left(\int_{t_1}^{t_2} + \int_{t_2}^{t_3}\right)g''(t,s)f(s)ds \\ & = \left(\int_{t_1}^t + \int_t^{t_2}\right)g''(t,s)f(s)ds+\int_{t_2}^{t_3}(-1)f(s)ds \\ & = \int_{t_1}^t(0)f(s)ds + \int_t^{t_2}(-1)f(s)ds-\int_{t_2}^{t_3}f(s)ds \\ & = \int_{t_3}^tf(s)ds, \end{align*} so that $x'''(t)=f(t)$ using the Fundamental Theorem of Calculus. Likewise for $t\in[t_2,t_3]$, $$x''(t)=\int_{t_1}^{t_2}(0)f(s)ds + \int_{t_2}^t(0)f(s)ds -\int_t^{t_3}f(s)ds$$ again implies that $x'''(t)=f(t)$. Therefore $g$ as given in \eqref{greensfunction} is Green's function for \eqref{b0}, \eqref{b1}. \end{proof} \section{positivity of green's function} \begin{theorem}\label{positive} Assume $k > 0$. If $$\delta > \max\big\{\gamma(t_3-t_2), \;\; \frac{k(t_3-t_1)^2}{2(t_2-t_1)[\alpha(t_3-t_1)+\beta]} -\frac{\gamma}{2}(t_2-t_1)\big\},$$ then Green's function as given in \eqref{greensfunction} satisfies $g(t,s) > 0$ on $(t_1,t_3]\times(t_1,t_3]$. \end{theorem} \begin{proof} Note that $g(t,t_1)=0$ for all $t\in[t_1,t_3]$. We proceed by cases on the two branches of Green's function \eqref{greensfunction}. \noindent\emph{Case $I$:} Let $s\in(t_1,t_2]$. Then $$g(t_1,s) = u_1(t_1,s) = \frac{\beta}{k}(s-t_1)\left[\delta+\frac{\gamma}{2}(2t_2-t_1-s)\right] \ge 0,$$ and $$\frac{\partial}{\partial t} g(t,s)=\frac{\partial}{\partial t} u_1(t,s) = \frac{\alpha}{k}(s-t_1)\left[\delta+\frac{\gamma}{2}(2t_2-t_1-s)\right]-t+t_1 \ge 0$$ for $t\in[t_1,\tau(s)]$; here $$\label{tau} \tau(s):= \frac{\alpha}{k}(s-t_1)\left[\delta+\frac{\gamma}{2}(2t_2-t_1-s)\right]+t_1 \le s$$ for $s\in(t_1,t_2]$, since $\tau(s)=s$ only if $s=t_1-\frac{2\beta}{\alpha}$ or $s=t_1$, and $\beta\ge 0$. For $t\ge s$, $$\frac{\partial}{\partial t} g(t,s) = \frac{\partial}{\partial t} v_1(t,s) = \tau(s)-s < 0,$$ so that $g$ is increasing in $t$ on $[t_1,\tau(s)]$, decreasing in $t$ on $[\tau(s),t_3]$, $\tau$ as defined in \eqref{tau}. It follows that $g(t,s)>0$ on $(t_1,t_3]\times(t_1,t_2]$ if $g(t_3,s)>0$ for these $s$: \begin{align*} g(t_3,s) & = v_1(t_3,s) \\ & = \frac{1}{k}(s-t_1)[\alpha(t_3-t_1)+\beta]\left[\delta+\frac{\gamma}{2}(2t_2-t_1-s)\right] \\ & \quad -\frac{1}{2}(t_3-t_1)^2+\frac{1}{2}(t_3-s)^2 \\ & = \frac{1}{2k}(s-t_1)[\alpha(s-t_1)+2\beta][\delta+\gamma(t_2-t_3)]. \end{align*} From this expression we see that $$\frac{\partial}{\partial s} v_1(t_3,s) = \frac{1}{k}[\alpha(s-t_1)+\beta][\delta+\gamma(t_2-t_3)] >0$$ if $\delta > \gamma(t_3-t_2)$; this is the first condition on $\delta$ mentioned in the theorem. Since $v_1(t_3,t_1)=0$ and $v_1(t_3,s)$ is increasing in $s$, $v_1(t_3,s) > 0$ for all $s\in(t_1,t_2]$. Consequently, $g(t,s) > 0$ for $(t,s)\in(t_1,t_3]\times(t_1,t_2]$. In addition, we note for later use that $$0 < g(t,s) \le g(\tau(s),s)$$ for $(t,s)\in(t_1,t_3]\times(t_1,t_2]$. %%%%%%%%%%% % Case II % %%%%%%%%%%% \noindent \emph{Case $II$:} Now let $s\in[t_2,t_3]$. For $t\le s$, \begin{align*} \frac{\partial}{\partial t} g(t,s) & = \frac{\partial}{\partial t} u_2(t,s) \\ & = \frac{\alpha}{k}\left[\delta(t_2-t_1)+\frac{\gamma}{2}(t_2-t_1)^2\right]+t_1-t \\ & = \tau(t_2)-t \ge 0 \end{align*} if $t \le \tau(t_2)$, $\tau$ as in \eqref{tau}. Note that $\tau(t_2) < t_2\le s$ here. As a result we have that $$0\le \frac{\beta}{k}\left[\delta(t_2-t_1)+\frac{\gamma}{2}(t_2-t_1)^2\right]=u_2(t_1,s)\le g(t,s)$$ for all $(t,s)\in[t_1,\tau(t_2)]\times[t_2,t_3]$. For $t\in[\tau(t_2),s]$, $g$ is then decreasing in $t$, and for $t\ge s$, \begin{align*} \frac{\partial}{\partial t} g(t,s) & = \frac{\partial}{\partial t} v_2(t,s) \\ & = \frac{\alpha}{k}\left[\delta(t_2-t_1)+\frac{\gamma}{2}(t_2-t_1)^2\right]+t_1-s \\ & = \tau(t_2)-s < 0 \end{align*} as mentioned previously. Therefore $g$ is increasing in $t$ on $[t_1,\tau(t_2)]$ and decreasing in $t$ on $[\tau(t_2),t_3]$, with a maximum at $g(\tau(t_2),s)$. Again we check to see that $g(t_3,s)>0$ for $s\in[t_2,t_3]$: \begin{align*} g(t_3,s) & = v_2(t_3,s) \\ & = \frac{1}{k}[\alpha(t_3-t_1)+\beta]\left[\delta(t_2-t_1)+\frac{\gamma}{2}(t_2-t_1)^2\right] \\ & \quad -\frac{1}{2}(t_3-t_1)^2 + \frac{1}{2}(t_3-s)^2. \end{align*} As a function of $s$ we have $$\frac{\partial}{\partial s} g(t_3,s) = \frac{\partial}{\partial s} v_2(t_3,s) = s-t_3 \le 0$$ for $s\in[t_2,t_3]$; in other words, $g(t_3,t_3)\le g(t_3,s)$ for these $s$. To ensure that $$g(t_3,t_3) = \frac{1}{k}[\alpha(t_3-t_1)+\beta]\left[\delta(t_2-t_1) +\frac{\gamma}{2}(t_2-t_1)^2\right]-\frac{1}{2}(t_3-t_1)^2$$ is positive, take $$\delta > \frac{k(t_3-t_1)^2}{2(t_2-t_1)[\alpha(t_3-t_1)+\beta]}-\frac{\gamma}{2}(t_2-t_1);$$ this is the second condition on $\delta$ mentioned in the theorem. (The fraction in this last expression is a finite real number, since by \eqref{k} $\alpha$ and $\beta$ cannot both be zero.) \end{proof} \section{functional focal problem} Letting $\gamma=0$ and $\delta=1$, we now apply Green's function and its properties from the first two sections to an investigation of the existence of positive solutions to the higher-order, three-point functional problem \begin{gather} x^{(n)}(t)=f(t,x(t+\theta)), \quad t_1\le t\le t_3, \quad -\tau\le \theta\le 0 \label{b0ak}\\ x^{(i)}(t_1)=0, \quad 0\le i\le n-4, \quad n\ge 4 \nonumber \\ \alpha x^{(n-3)}(t)-\beta x^{(n-2)}(t)=\sigma(t), \quad t_1-\tau\le t\le t_1 \nonumber \\ x^{(n-2)}(t_2)=x^{(n-1)}(t_3)=0. \label{b2ak} \end{gather} Here we assume \begin{enumerate} \item[$(i)$] $t_1 0$, \; $t_3-t_1\ge \tau\ge 0$, and $\theta\in[-\tau,0]$ is constant; \item[$(iii)$] $\sigma:[t_1-\tau,t_1]\rightarrow\mathbb{R}$ is continuous with $\sigma(t_1)=0$; \item[$(iv)$] $f:\mathbb{R}^2\rightarrow\mathbb{R}$ is continuous and nonnegative for $x\ge 0$. \end{enumerate} For the rest of this paper we also have the assumptions \begin{enumerate} \item[(A1)] $G(t,s)$ is Green's function for the differential equation $$u^{(n)}(t)=0, \;\; t\in(t_1,t_3)$$ subject to the boundary conditions \eqref{b2ak} with $\tau=0$. \item[(A2)] $g(t,s)$ is Green's function for the differential equation $$u'''(t)=0, \;\; t\in(t_1,t_3)$$ subject to the boundary conditions $$\begin{gathered} \alpha u(t_1) - \beta u'(t_1) = 0 \nonumber\\ u'(t_2) = u''(t_3) = 0 \end{gathered}\label{gbc}$$ for $\alpha, \beta$ as in $(ii)$. \item[(A3)] $\|y\|_{[u,v]}:=\displaystyle\sup_{u\le x\le v}|y^{(n-3)}(x)|$. \item[(A4)] For $\Xi:=\{s\in[t_1,t_3]:t_1\le s+\theta\le t_3\}$, the set $$\Xi_h:=\{s\in\Xi: t_2-h\le s+\theta\le t_2+h\}$$ has nonzero measure for some $h\in(0,t_3-t_2)$. \end{enumerate} The corresponding Green's function for the homogeneous problem $u'''(t)=0$ satisfying the boundary conditions \eqref{gbc} is given in \eqref{greensfunction}, rewritten here for convenience as $$\label{ngreen} g(t,s)= \begin{cases} s\in [t_1,t_2] &: \begin{cases} \frac{1}{2}(t-t_1)(2s-t-t_1)+\frac{\beta}{\alpha}(s-t_1) &:t\leq s \\ \frac{1}{2}(s-t_1)^2+\frac{\beta}{\alpha}(s-t_1) &:s\leq t \end{cases}\\[3pt] s\in [t_2,t_3] &: \begin{cases} \frac{1}{2}(t-t_1)(2t_2-t-t_1) +\frac{\beta}{\alpha}(t_2-t_1) & \\ \frac{1}{2}(t-t_1)(2t_2-t-t_1)+\frac{\beta}{\alpha}(t_2-t_1)+\frac{1}{2}(t-s)^2 & \end{cases} \\ \end{cases}$$ \begin{remark} \label{rmk1} \rm As in Theorem \ref{positive}, if $$\frac{\beta}{\alpha}(t_2-t_1) > \frac{1}{2}(t_3-t_1)(t_3+t_1-2t_2),$$ then $g(t,s)>0$ for all $t\in(t_1,t_3]$, $s\in(t_1,t_3]$. Note that if the boundary points satisfy $$\label{boundarydistance} t_3-t_2 < t_2-t_1,$$ then the above inequality holds for any choice of $\alpha,\beta > 0$. Thus throughout this section we assume that \eqref{boundarydistance} holds. Moreover, as in \cite[Lemma 3]{An1} or \cite[Lemma 1]{gy}, we have the following boundedness result. \end{remark} \begin{lemma} \label{lem1} For all $t,s\in[t_1,t_3]$, $$\label{greenbounds} \ell(t)g(t_2,s)\leq g(t,s)\leq g(t_2,s)$$ where $$\label{g} \ell(t):=\frac{\alpha(t-t_1)(2t_2-t-t_1)+2\beta(t_2-t_1)}{\alpha(t_2-t_1)^2+2\beta(t_2-t_1)}.$$ \end{lemma} \begin{remark}\label{discussion} \rm The following discussion is similar to that found in \cite{hong} for a two-point problem on the unit interval. If $x$ is a solution of \eqref{b0ak}, \eqref{b2ak}, it can be written as $$x(t)=\begin{cases} x(-\tau;t) & t_1-\tau\leq t\leq t_1 \\[2pt] \int_{t_1}^{t_3} G(t,s)f(s,x(s+\theta))ds & t_1\leq t\leq t_3 \end{cases}$$ where $x(-\tau;t)$ satisfies $$x^{(n-3)}(-\tau;t) =e^{\frac{\alpha}{\beta}(t-t_1)}x^{(n-3)}(t_1) + \frac{1}{\beta}\int_t^{t_1} e^{\frac{\alpha}{\beta}(t-s)}\sigma(s)ds$$ for $t\in[t_1-\tau,t_1]$. Now assume that $u_0$ is the solution of \eqref{b0ak}, \eqref{b2ak} with $f\equiv 0$. Then $u_0$ satisfies $$\label{unot} u_0^{(n-3)}(t)=\begin{cases} \frac{1}{\beta}\int_t^{t_1} e^{\frac{\alpha}{\beta}(t-s)}\sigma(s)ds & t_1-\tau\leq t\leq t_1 \\ 0 & t_1\leq t\leq t_3. \end{cases}$$ If $x$ is any solution of \eqref{b0ak}, \eqref{b2ak} set $u(t):=x(t)-u_0(t)$. Then $u(t)\equiv x(t)$ on $[t_1,t_3]$, and $u$ satisfies $$u^{(n-3)}(t) =\begin{cases} e^{\frac{\alpha}{\beta}(t-t_1)}u^{(n-3)}(t_1) & t_1-\tau\leq t\leq t_1 \\ \int_{t_1}^{t_3} g(t,s)f(s,u(s+\theta)+u_0(s+\theta))ds & t_1\leq t\leq t_3. \end{cases}$$ But this implies $$u(t) =\begin{cases} \big(\frac{\beta}{\alpha}\big)^{n-3}e^{\frac{\alpha}{\beta}(t-t_1)} u^{(n-3)}(t_1) & t_1-\tau\leq t\leq t_1 \\[2pt] \int_{t_1}^{t_3} G(t,s)f(s,u(s+\theta)+u_0(s+\theta))ds & t_1\leq t\leq t_3. \end{cases}$$ \end{remark} \section{Existence of at Least One Positive Solution} As mentioned in the previous section, assume $(i)-(iv)$ and $(A1)-(A4)$ hold. We are concerned with proving the existence of positive solutions of the higher-order nonlinear boundary value problem \eqref{b0ak}, \eqref{b2ak}; for related work on the existence of positive solutions, see \cite{davis1,davis2,He}. In light of the above discussion in Remark \ref{discussion}, the solutions of \eqref{b0ak}, \eqref{b2ak} can be found using the fixed points of the operator $\mathcal{A}$ with domain $C^{n-3}[t_1-\tau,t_3]$ defined by $$\mathcal{A} u(t) =\begin{cases} \big(\frac{\beta}{\alpha}\big)^{n-3}e^{\frac{\alpha}{\beta}(t-t_1)} u^{(n-3)}(t_1) & t_1-\tau\leq t\leq t_1 \\[2pt] \int_{t_1}^{t_3} G(t,s)f(s,u(s+\theta)+u_0(s+\theta))ds & t_1\leq t\leq t_3. \end{cases}$$ If $u=\mathcal{A} u$, then a solution $x$ of \eqref{b0ak}, \eqref{b2ak} is given by $x=u+u_0$, where $u_0$ satisfies \eqref{unot}. \begin{remark} \label{rmk3}\rm In the following discussion we will need an $h\in(0,t_3-t_2)$ to satisfy $(A4)$; note that $$\label{uofh} \ell(t_2+h) = \ell(t_2-h)=\frac{\alpha(t_2+h-t_1)(t_2-h-t_1)+2\beta(t_2-t_1)}{\alpha(t_2-t_1)^2+2\beta(t_2-t_1)}$$ for all $h\in(0,t_3-t_2)$, where $\ell$ is given in \eqref{g}, and $\ell(t)\ge\ell(t_2+h)$ for all $t\in[t_2-h,t_2+h]$. Moreover, let $k,m>0$ such that \begin{aligned} k^{-1}&:=\int_{t_1}^{t_3}g(t_2,s)ds \\ &=\frac{1}{6}(t_2-t_1)^2(3t_3-2t_2-t_1) +\frac{\beta}{2\alpha}(t_2-t_1)(2t_3-t_2-t_1)\nonumber \end{aligned} \label{kak} and $$\label{m} m^{-1}:=\int_{\Xi_h}g(t_2,s)\,ds.$$ Finally, set $$\label{Mnot} M_0:= \|u_0\|_{[t_1-\tau,t_3]}$$ for $u_0$ as in \eqref{unot}. \end{remark} We will employ the following fixed point theorem due to Krasnoselskii \cite{Kr}. \begin{theorem}\label{fixedpt} Let $E$ be a Banach space, $P\subseteq E$ be a cone, and suppose that $\Omega_1$, $\Omega_2$ are bounded open balls of $E$ centered at the origin with $\overline{\Omega}_1\subset\Omega_2$. Suppose further that $\mathcal{A}:P\cap(\overline{\Omega}_2\setminus\Omega_1)\to P$ is a completely continuous operator such that either \begin{enumerate} \item[$(i)$] $\|\mathcal{A} u\| \leq \|u\|$, $u\in P\cap\partial\Omega_1$ and $\|\mathcal{A} u\| \geq \|u\|$, $u\in P\cap\partial\Omega_2$, or \item[$(ii)$] $\|\mathcal{A} u\| \geq \|u\|$, $u\in P\cap\partial\Omega_1$ and $\|\mathcal{A} u\| \leq \|u\|$, $u\in P\cap\partial\Omega_2$ \end{enumerate} holds. Then $\mathcal{A}$ has a fixed point in $P\cap(\overline{\Omega}_2\setminus\Omega_1)$. \end{theorem} \begin{theorem}\label{theorem4} Assume $(i)-(iv)$ and $(A1)-(A4)$ hold. Let $k, m, M_0$ be as in \eqref{kak}, \eqref{m}, \eqref{Mnot}, respectively, and suppose the following conditions are satisfied. \begin{itemize} \item[(C1)] There exists a $p > 0$ such that $f(t,w)\le kp$ for $t \in [t_1,t_3]$ and $0\leq \|w\| \le p+M_0$. \item[(C2)] There exists a $q > p$ such that $f(t,w)\ge mq$ for $t \in \Xi_h$ and $q\ell(t_2+h)\le \|w\| \leq q$, for $h\in(0,t_3-t_2)$ and $\Xi_h$ as in $(A4)$. \end{itemize} Then system \eqref{b0ak}, \eqref{b2ak} has a positive solution $x$ such that $\|x\|_{[t_1-\tau, t_3]}$ lies between $\max\{0,p-M_0\}$ and $q+M_0$. \end{theorem} \begin{proof} Many of the techniques employed here are as in \cite{He, hong}. Let $\mathcal{B}$ denote the Banach space $C^{n-3}[t_1-\tau, t_3]$ with the norm $$\|u\|_{[t_1-\tau, t_3]}=\sup_{t\in[t_1-\tau,t_3]}|u^{(n-3)}(t)|.$$ Define the cone $\mathcal{P}\subset\mathcal{B}$ by $$\mathcal{P}=\{u\in\mathcal{B}:\displaystyle\min_{t\in[t_2-h,t_2+h]}u^{(n-3)}(t) \geq \ell(t_2+h)\|u\|_{[t_1-\tau, t_3]}\}.$$ Consider the mapping $\mathcal{A}:\mathcal{P}\rightarrow\mathcal{B}$ via $$\mathcal{A} u(t) =\begin{cases} \big(\frac{\beta}{\alpha}\big)^{n-3}e^{\frac{\alpha}{\beta}(t-t_1)}u^{(n-3)}(t_1) & t_1-\tau\leq t\leq t_1 \\ \int_{t_1}^{t_3} G(t,s)f(s,u(s+\theta)+u_0(s+\theta))ds & t_1\leq t\leq t_3. \end{cases}$$ Then $$\label{Aun-3} (\mathcal{A} u)^{(n-3)}(t)=\begin{cases} e^{\frac{\alpha}{\beta}(t-t_1)} \int_{t_1}^{t_3}g(t_1,s) f(s,u(s+\theta)+u_0(s+\theta))ds \\[2pt] \int_{t_1}^{t_3} g(t,s)f(s,u(s+\theta)+u_0(s+\theta))ds, \end{cases}$$ so that $(\mathcal{A} u)^{(n-3)}(t)\leq (\mathcal{A} u)^{(n-3)}(t_1)$ for $t_1-\tau\le t\le t_1$. In other words, $\|\mathcal{A} u\|_{[t_1-\tau, t_3]} = \|\mathcal{A} u\|_{[t_1, t_3]}$. It follows for $h\in(0,t_3-t_2)$ and $t\in[t_2-h,t_2+h]$ that \begin{align} (\mathcal{A} u)^{(n-3)}(t) & = \int_{t_1}^{t_3} g(t,s)f(s,u(s+\theta)+u_0(s+\theta))ds \nonumber \\ & \ge \ell(t)\int_{t_1}^{t_3} g(t_2,s)f(s,u(s+\theta)+u_0(s+\theta))ds \nonumber \\ & \ge \ell(t_2+h)\|\mathcal{A} u\|_{[t_1-\tau, t_3]}\label{Auge} \end{align} by properties of Green's function \eqref{greenbounds}, so that $\mathcal{A}:\mathcal{P}\rightarrow\mathcal{P}$. For $0c$ for all $u\in \partial P(\chi,c)$; \item[(ii)] $\psi(Au)a$ for all $u\in\partial P(\eta,a)$. \end{itemize} Then $A$ has at least two fixed points $u_1$ and $u_2$ such that $$a<\eta(u_1) \text{ with } \psi(u_1) am for all s\in\Xi_h and \|w\|\in[a,\frac{a}{\ell(t_2+h)}+M_0], \item[(iii)] f(s,w) < \frac{b}{\int_{t_1}^{t_3}g(t_2+h,s)ds} for all s\in[t_1,t_3] and \|w\|\in[0,\frac b{\ell(t_2+h)}+M_0], \item[(iv)] f(s,w) > \frac{cm}{\ell(t_2+h)} for s\in\Xi_h and \|w\|\in[c,\frac{c}{\ell(t_2+h)}+M_0]. \end{enumerate} Then, the higher-order boundary value problem \eqref{b0ak}, \eqref{b2ak}, has at least two positive solutions x_1 and x_2 such that$$ \max_{t\in[t_2-h,t_2+h]}x_1^{(n-3)}(t)>a \quad \mbox{with}\quad \max_{t\in[t_1,t_2-h]\cup[t_2+h,t_3]}x_1^{(n-3)}(t)b \quad \mbox{with} \quad \min_{t\in[t_2-h,t_2+h]}x_2^{(n-3)}(t)From Lemma \ref{lemma5.1} it follows that $$(\mathcal{A} u)^{(n-3)}(t_2+h)\ge (\mathcal{A} u)^{(n-3)}(t_2-h),$$ and $$\min_{t\in[t_2-h,t_2+h]}(\mathcal{A} u)^{(n-3)}(t)\ge\ell(t_2+h)\|\mathcal{A} u\|_{[t_1-\tau, t_3]}$$ as in \eqref{Auge}. Therefore $\mathcal{A}:\mathcal{P}\to\mathcal{P}$. For any $u\in\mathcal{P}$, \eqref{gpd} and \eqref{normgam} imply that \begin{gather*} \chi(u)\le\psi(u)\le\eta(u), \\ \|u\|\le\frac{1}{\ell(t_2+h)}\chi(u). \end{gather*} It is clear that $\psi(0)=0$, and for all $u\in\mathcal{P}$, $\lambda\in[0,1]$ we have \begin{align*} \psi(\lambda u) &= \max_{t\in[t_1,t_2-h]\cup[t_2+h,t_3]}(\lambda u)^{(n-3)}(t) \\ &= \lambda\max_{t\in[t_1,t_2-h]\cup[t_2+h,t_3]} u^{(n-3)}(t) =\lambda\psi(u). \end{align*} Since $0\in\mathcal{P}$ and $a>0$, $P(\eta,a)\ne\emptyset$. In the following claims, we verify the remaining conditions of Theorem \ref{averyhen}. \noindent {\bf Claim 1.} If $u\in\partial P(\eta,a)$, then $\eta(\mathcal{A} u)>a$: Note that $u\in\partial P(\eta,a)$ and \eqref{del} yield $a=\|u\|\le \frac{a}{\ell(t_2+h)}$. By hypothesis $(ii)$, \begin{align*} \eta(\mathcal{A} u) & = \max_{t\in[t_2-h,t_2+h]} \int_{t_1}^{t_3} g(t,s)f(s,u(s+\theta) +u_0(s+\theta))ds \\ & = \int_{t_1}^{t_3} g(t_2,s)f(s,u(s+\theta)+u_0(s+\theta))ds \\ &\ge \int_{\Xi_h} g(t_2,s)f(s,u(s+\theta)+u_0(s+\theta))ds \\ & > am\int_{\Xi_h}g(t_2,s)ds = a. \end{align*} \noindent {\bf Claim 2.} If $u\in\partial P(\psi,b)$, then $\psi(\mathcal{A} u)c$: Since $u\in\partial P(\chi,c)$, from \eqref{normgam} we have that $\displaystyle\min_{t\in[t_2-h,t_2+h]}u^{(n-3)}(t)=c$ and $c\le \|u\|\le \frac{c}{\ell(t_2+h)}$. Thus, \begin{align*} \chi (\mathcal{A} u) & = \min_{t\in[t_2-h,t_2+h]} \int_{t_1}^{t_3} g(t,s)f(s,u(s+\theta)+u_0(s+\theta))ds \\ & \ge \min_{t\in[t_2-h,t_2+h]} \ell(t) \int_{t_1}^{t_3} g(t_2,s)f(s,u(s+\theta)+u_0(s+\theta))ds \\ & \ge \ell(t_2+h) \int_{\Xi_h} g(t_2,s)f(s,u(s+\theta)+u_0(s+\theta))ds \\ & > \ell(t_2+h) \frac{cm}{\ell(t_2+h)} \int_{\Xi_h} g(t_2,s)ds = c \end{align*} by hypothesis $(iv)$, using arguments as in Claim 1. Therefore the hypotheses of Theorem \ref{averyhen} are satisfied and there exist at least two positive fixed points $u_1$ and $u_2$ of $\mathcal{A}$ in $\overline{P(\chi,c)}$. Thus, the higher-order boundary value problem \eqref{b0ak}, \eqref{b2ak}, has at least two positive solutions $x_1$ and $x_2$ such that \begin{gather*} a<\eta(x_1) \quad \mbox{with} \quad \psi(x_1)