\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 107, pp. 1--15.\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 2007 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2007/107\hfil Third-order Three-Point BVP] {Existence of solutions to a third-order multi-point problem on time scales} \author[D. R. Anderson, J. Hoffacker\hfil EJDE-2007/107\hfilneg] {Douglas R. Anderson, Joan Hoffacker} % in alphabetical order \address{Douglas R. Anderson \newline Department of Mathematics and Computer Science, Concordia College, Moorhead, MN 56562 USA} \email{andersod@cord.edu} \address{Joan Hoffacker \newline Department of Mathematical Sciences, Clemson University, Clemson, SC 29634 USA} \email{johoff@clemson.edu} \thanks{Submitted July 8, 2007. Published August 7, 2007.} \subjclass[2000]{34B18, 34B27, 34B10, 39A10} \keywords{Boundary value problem; time scale; third order; Green function} \begin{abstract} We are concerned with the existence and form of solutions to nonlinear third-order three-point and multi-point boundary-value problems on general time scales. Using the corresponding Green function, we prove the existence of at least one positive solution using the Guo-Krasnosel'skii fixed point theorem. Moreover, a third-order multi-point eigenvalue problem is formulated, and eigenvalue intervals for the existence of a positive solution are found. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \section{introduction} We will establish the corresponding Green function whereby conditions can be given such that a positive solution exists for the following nonlinear third-order three-point boundary value problem on arbitrary time scales \begin{gather} (px^{\Delta\Delta})^\nabla(t)+ a(t)f(x(t))=0, \quad t\in[t_1,t_3]_\mathbb{T}, \label{bvp} \\ x(\rho(t_1))=x^\Delta(\rho(t_1))= 0, \quad x^\Delta(\sigma(t_3))-\alpha x^\Delta(t_2)=0, \label{bvpbc} \end{gather} where: $p$ is a right-dense continuous, real-valued function with 00 \quad \text{and}\quad 1<\alpha<\frac{\int_{\rho(t_1)}^{\sigma(t_3)} \frac{\Delta\tau}{p(\tau)}}{\int_{\rho(t_1)}^{t_2}\frac{\Delta\tau}{p(\tau)}}; $$\item[(H2)] the continuous function f:[0,\infty)\rightarrow[0,\infty) is such that the following exist:$$ f_0:=\lim_{x\rightarrow 0^+}\frac{f(x)}{x} \quad f_{\infty}:=\lim_{x\rightarrow\infty}\frac{f(x)}{x}. $$\item[(H3)] the left-dense continuous function a:[\rho(t_1),\sigma(t_3)]_\mathbb{T} \rightarrow[0,\infty) is such that a is not identically zero on [t_2/\alpha,t_2]_\mathbb{T}. \end{enumerate} If \mathbb{T}=\mathbb{R}, then \eqref{bvp}, \eqref{bvpbc} is the ordinary third-order three-point boundary value problem \begin{gather*} (px'')'(t)+ a(t)f(x(t))=0, \quad t\in[t_1,t_3]_\mathbb{R}, \\ x(t_1)=x'(t_1)= 0, \quad x'(t_3)-\alpha x'(t_2)=0. \end{gather*} If \mathbb{T}=\mathbb{Z}, then \eqref{bvp}, \eqref{bvpbc} is the discrete third-order three-point boundary value problem \begin{gather*} \nabla\left(p\Delta^2 x)(t)+ a(t)f(x(t)\right)=0, \quad t\in[t_1,t_3]_\mathbb{Z}, \\ x(t_1-1)=\Delta x(t_1-1)= 0, \quad \Delta x(t_3+1)-\alpha \Delta x(t_2)=0, \end{gather*} where \Delta y(t)=y(t+1)-y(t) and \nabla y(t)=y(t)-y(t-1). As a final illustration, if \mathbb{T} is a quantum time scale for some real q>1, then \eqref{bvp}, \eqref{bvpbc} is the third-order three-point quantum boundary value problem \begin{gather*} D^q\left(pD_q(D_q x)\right)(t)+ a(t)f(x(t))=0, \quad t\in[t_1,t_3]_\mathbb{T}, \\ x(t_1/q)=D_q x(t_1/q)= 0, \quad D_q x(qt_3)-\alpha D_q x(t_2)=0, \end{gather*} where the quantum derivatives are given by the difference quotients$$ D_q y(t)=\frac{y(qt)-y(t)}{(q-1)t} \quad\text{and}\quad D^q y(t)=\frac{y(t)-y(t/q)}{(1-1/q)t}. $$Third-order differential equations, though less common in applications than even-order problems, nevertheless do appear, for example in the study of quantum fluids; see Gamba and J\"{u}ngel \cite{gamba}. Here we approach a third-order three-point problem on general time scales, namely on any nonempty closed subset of the real line, to include the discrete, continuous, and quantum calculus as special cases. Of late there have been several papers on third-order boundary value problems. Hopkins and Kosmatov \cite{hopkins}; Li \cite{li}; Liu, Ume, and Kang \cite{liu,liu2}; and Minghe and Chang \cite{minghe} have all recently considered third-order problems. All of these papers, however, were two-point problems with \mathbb{T}=\mathbb{R}. Graef and Yang \cite{gy2}, Sun \cite{sun}, and Wong \cite{wong} consider three-point focal problems, while Palamides and Smyrlis consider the three-point boundary conditions$$ x(0)=x''(\eta)=x(1)=0, \quad \mathbb{T}=[0,1]_{\mathbb{R}}. On general time scales there are also a few papers on third-order problems. Sun \cite{jpsun} considers a third-order two-point boundary value problem; a couple of papers on third-order three-point boundary value problems considered on general time scales are \cite{and,and2} in the right-focal case. Note that boundary value problems on time scales that utilize both delta and nabla derivatives, such as the one here, were first introduced by Atici and Guseinov \cite{ag}. For more on existence of solutions to boundary value problems, see \cite[Chapters 4 and 6-9]{bp}, the text by Guo and Lakshmikantham \cite{gu}, and Zhang and Liu \cite{zl}. Problem \eqref{bvp}, \eqref{bvpbc} is an extension of the unit interval boundary value problem \begin{gather*} x'''(t)+a(t) f(x(t))=0, \quad t\in(0,1)_{\mathbb{R}}, \\ x(0)=x'(0)= 0, \quad \alpha x'(\eta)=x'(1), \end{gather*} to arbitrary time scales \cite{guo}; in other words, take \mathbb{T}=\mathbb{R}, p\equiv 1, t_1=0, t_2=\eta and t_3=1 in \eqref{bvp}, \eqref{bvpbc} to get the results in \cite{guo}. One could also consider a third-order problem with derivatives in the order of nabla, nabla, delta, but the results would be similar; other permutations of nablas and/or deltas lead to a Green function that is less easy to calculate. \section{preliminary lemmas} Underlying our technique will be the Green function for the homogeneous, third-order, three-point boundary-value problem \begin{gather} -(px^{\Delta\Delta})^\nabla(t)=0, \quad t\in[t_1,t_3]_\mathbb{T}, \label{b1}\\ x(\rho(t_1))=x^\Delta(\rho(t_1))= 0, \quad \alpha x^\Delta(t_2)=x^\Delta(\sigma(t_3)).\label{b2} \end{gather} The Green function for \eqref{b1}, \eqref{b2} will be defined, nonnegative, and bounded above on [\rho(t_1),\sigma^2(t_3)]_\mathbb{T}\times[t_1,\sigma(t_3)]_\mathbb{T}, as will be shown in the following lemmas. \begin{lemma} \label{lem2.1} For y\in C_{ld}[\rho(t_1),\sigma(t_3)]_\mathbb{T}, the boundary value problem \begin{gather} (px^{\Delta\Delta})^\nabla(t)+y(t)=0, \quad t\in[t_1,t_3]_\mathbb{T}, \label{lemeq}\\ x(\rho(t_1))=x^\Delta(\rho(t_1))= 0, \quad \alpha x^\Delta(t_2)=x^\Delta(\sigma(t_3))\label{lembc} \end{gather} has a unique solution x(t)=\int_{\rho(t_1)}^{\sigma(t_3)}G(t,s)y(s)\nabla s, where the Green function corresponding to the problem \eqref{b1}, \eqref{b2} is given by $$\label{greenf} G(t,s)= \begin{cases} \frac{1}{d}\left(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)} -\alpha\int_s^{t_2}\frac{\Delta\tau}{p(\tau)}\right) \int_{\rho(t_1)}^t\int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)} \Delta u \\ - \int_{s}^t\int_{s}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u &: s\le\min\{t_2,t\} \\[5pt] \frac{1}{d}\left(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)} -\alpha\int_s^{t_2}\frac{\Delta\tau}{p(\tau)}\right)\int_{\rho(t_1)}^t \int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u &: t\le s\le t_2 \\[5pt] \frac{1}{d}\left(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}\right) \int_{\rho(t_1)}^t\int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u - \int_{s}^t\int_{s}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u &:t_2\le s\le t \\[5pt] \frac{1}{d}\left(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}\right) \int_{\rho(t_1)}^t\int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u &:\max\{t_2,t\}\le s. \end{cases}$$ \end{lemma} \begin{proof} We follow the approach given in the case \mathbb{T}=\mathbb{R} in \cite{guo}. If \rho(t_1)\le t\le t_2, then \begin{align*} x(t) &= \int_{\rho(t_1)}^{\sigma(t_3)}G(t,s)y(s)\nabla s \\ &= \int_{\rho(t_1)}^{t} \Big[\frac{1}{d}\Big(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)} -\alpha\int_s^{t_2}\frac{\Delta\tau}{p(\tau)}\Big) \int_{\rho(t_1)}^t\int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u\\ &\quad - \int_{s}^t\int_{s}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u \Big] y(s)\nabla s \\ &\quad + \int_{t}^{t_2} \Big[\frac{1}{d}\Big(\int_s^{\sigma(t_3)} \frac{\Delta\tau}{p(\tau)} -\alpha\int_s^{t_2}\frac{\Delta\tau}{p(\tau)}\Big) \int_{\rho(t_1)}^t\int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)} \Delta u \Big]y(s)\nabla s \\ &\quad + \int_{t_2}^{\sigma(t_3)} \Big[\frac{1}{d}\Big(\int_s^{\sigma(t_3)} \frac{\Delta\tau}{p(\tau)}\Big) \int_{\rho(t_1)}^t\int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u \Big]y(s)\nabla s \\ &= \frac{1}{d}\Big[\int_{\rho(t_1)}^{t_2} \Big(\int_s^{\sigma(t_3)} \frac{\Delta\tau}{p(\tau)}-\alpha\int_s^{t_2}\frac{\Delta\tau}{p(\tau)} \Big)y(s)\nabla s + \int_{t_2}^{\sigma(t_3)} \int_s^{\sigma(t_3)} \frac{\Delta\tau}{p(\tau)}y(s)\nabla s \Big] \\ &\quad \times\Big[\int_{\rho(t_1)}^t \int_{\rho(t_1)}^u \frac{\Delta\tau}{p(\tau)}\Delta u \Big] - \int_{\rho(t_1)}^t\Big(\int_s^t\int_s^u \frac{\Delta\tau}{p(\tau)} \Delta u\Big)y(s)\nabla s. \end{align*} If \sigma^2(t_3)\ge t\ge t_2, then \begin{align*} x(t) &= \int_{\rho(t_1)}^{\sigma(t_3)}G(t,s)y(s)\nabla s \\ &= \int_{\rho(t_1)}^{t_2} \Big[\frac{1}{d}\Big(\int_s^{\sigma(t_3)} \frac{\Delta\tau}{p(\tau)} -\alpha\int_s^{t_2}\frac{\Delta\tau}{p(\tau)}\Big) \int_{\rho(t_1)}^t\int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u\\ &\quad - \int_{s}^t\int_{s}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u \Big] y(s)\nabla s \\ &\quad + \int_{t_2}^{t} \Big[\frac{1}{d}\Big(\int_s^{\sigma(t_3)} \frac{\Delta\tau}{p(\tau)}\Big) \int_{\rho(t_1)}^t\int_{\rho(t_1)}^{u} \frac{\Delta\tau}{p(\tau)}\Delta u - \int_{s}^t\int_{s}^{u} \frac{\Delta\tau}{p(\tau)}\Delta u \Big]y(s)\nabla s \\ &\quad + \int_{t}^{\sigma(t_3)} \Big[\frac{1}{d}\Big(\int_s^{\sigma(t_3)} \frac{\Delta\tau}{p(\tau)}\Big) \int_{\rho(t_1)}^t\int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u \Big]y(s)\nabla s \\ &= \frac{1}{d}\Big[\int_{\rho(t_1)}^{t_2} \Big(\int_s^{\sigma(t_3)} \frac{\Delta\tau}{p(\tau)}-\alpha\int_s^{t_2}\frac{\Delta\tau}{p(\tau)} \Big)y(s)\nabla s + \int_{t_2}^{\sigma(t_3)} \int_s^{\sigma(t_3)} \frac{\Delta\tau}{p(\tau)}y(s)\nabla s \Big] \\ &\quad \times\Big[\int_{\rho(t_1)}^t \int_{\rho(t_1)}^u \frac{\Delta\tau}{p(\tau)}\Delta u \Big] - \int_{\rho(t_1)}^t \Big(\int_s^t\int_s^u \frac{\Delta\tau}{p(\tau)}\Delta u\Big)y(s)\nabla s. \end{align*} For the remainder of the proof let k(s):=\frac{1}{d}\Big[\int_{\rho(t_1)}^{t_2} \Big(\int_s^{\sigma(t_3)} \frac{\Delta\tau}{p(\tau)}-\alpha\int_s^{t_2} \frac{\Delta\tau}{p(\tau)} \Big)y(s)\nabla s + \int_{t_2}^{\sigma(t_3)} \int_s^{\sigma(t_3)} \frac{\Delta\tau}{p(\tau)}y(s)\nabla s \Big]. $$Thus for all t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T},$$ x(t) = k(s)\int_{\rho(t_1)}^t \int_{\rho(t_1)}^u \frac{\Delta\tau}{p(\tau)}\Delta u - \int_{\rho(t_1)}^t \left(\int_s^t\int_s^u \frac{\Delta\tau}{p(\tau)}\Delta u\right) y(s)\nabla s. $$Note that x(\rho(t_1))=0. Taking a delta derivative,$$ x^\Delta(t) = k(s)\int_{\rho(t_1)}^t\frac{\Delta\tau}{p(\tau)} - \int_{\rho(t_1)}^t\left(\int_s^t \frac{\Delta\tau}{p(\tau)}\right) y(s)\nabla s. Again it is easy to see that x^\Delta(\rho(t_1))=0. To verify the third boundary condition, check that \begin{align*} & x^\Delta(\sigma(t_3))-\alpha x^\Delta(t_2) \\ &= dk(s) - \int_{\rho(t_1)}^{\sigma(t_3)}\Big(\int_s^{\sigma(t_3)} \frac{\Delta\tau}{p(\tau)}\Big)y(s)\nabla s + \alpha\int_{\rho(t_1)}^{t_2}\Big(\int_s^{t_2} \frac{\Delta\tau}{p(\tau)} \Big)y(s)\nabla s\\ & = 0. \end{align*} It follows that the boundary conditions \eqref{lembc} are satisfied. To finish the proof, another delta derivative yields x^{\Delta\Delta}(t) = \frac{k(s)}{p(t)} - \int_{\rho(t_1)}^{t} \frac{y(s)}{p(t)}\nabla s, $$which results in (px^{\Delta\Delta})^\nabla(t) = -y(t), so that \eqref{lemeq} is satisfied as well. \end{proof} We now seek bounds on the Green function given in \eqref{greenf}. For later reference, set $$\label{gdef} g(s):=\frac{1}{d} (\alpha+1)\big(\sigma^2(t_3)-\rho(t_1)\big) \Big(\int_{\rho(t_1)}^{s}\frac{\Delta\tau}{p(\tau)}\Big) \Big(\int^{\sigma(t_3)}_{s}\frac{\Delta\tau}{p(\tau)}\Big).$$ \begin{lemma}\label{lemma22} Assume {\rm (H1)}. The Green function \eqref{greenf} corresponding to the problem \eqref{b1}, \eqref{b2} satisfies$$ 0 \le G(t,s) \le g(s) $$for (t,s)\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}\times[t_1,\sigma(t_3)]_\mathbb{T}. \end{lemma} \begin{proof} First we show that G(t,s) is nonnegative. For t\le s\le t_2, consider the coefficient$$ c_1(s):=\frac{1}{d}\Big(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)} -\alpha\int_s^{t_2}\frac{\Delta\tau}{p(\tau)}\Big). $$Since c_1(\rho(t_1))=1 and c_1^\Delta(s)=\frac{\alpha-1}{dp(s)}>0, c_1(s)\ge 1 for all s\ge t_1. It follows that branches 1,2, and 4 of G(t,s) in \eqref{greenf} are nonnegative, so we consider the third branch of G(t,s):$$ \frac{1}{d}\Big[\Big(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}\Big) \int_{\rho(t_1)}^t\int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u - d\int_{s}^t\int_{s}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u\Big]. $$For t_2\le s\le t let$$ v(t,s):= \Big(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}\Big) \int_{\rho(t_1)}^t\int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u - d\int_{s}^t\int_{s}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u. Then \begin{align*} v(t,s) &= \Big(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}\Big) \Big[\int_{\rho(t_1)}^s\int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)} \Delta u +\int_{s}^t\int_{\rho(t_1)}^{s} \frac{\Delta\tau}{p(\tau)}\Delta u\Big] \\ &\quad + \Big(\int_{s}^t\int_{s}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u\Big) \Big(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}-d\Big) \\ &= \Big(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}\Big) \Big[\int_{\rho(t_1)}^s\int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)} \Delta u +\int_{s}^t\int_{\rho(t_1)}^{s}\frac{\Delta\tau}{p(\tau)} \Delta u\Big] \\ &\quad + \alpha\Big(\int_{\rho(t_1)}^{t_2}\frac{\Delta\tau}{p(\tau)}\Big) \Big(\int_{s}^t\int_{s}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u\Big) - \Big(\int_{\rho(t_1)}^{s}\frac{\Delta\tau}{p(\tau)}\Big) \Big(\int_{s}^t\int_{s}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u\Big) \\ &= \Big(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}\Big) \Big(\int_{\rho(t_1)}^s\int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)} \Delta u\Big) + \alpha\Big(\int_{\rho(t_1)}^{t_2} \frac{\Delta\tau}{p(\tau)}\Big) \Big(\int_{s}^t\int_{s}^{u} \frac{\Delta\tau}{p(\tau)}\Delta u\Big) \\ &\quad + \Big(\int^s_{\rho(t_1)}\frac{\Delta\tau}{p(\tau)}\Big) \Big[\Big(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}\Big) \Big(\int_s^{t}\frac{\Delta\tau}{p(\tau)}\Big) -\int_{s}^t\int_{s}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u\Big]. \end{align*} Clearly this is nonnegative if the last term in brackets is nonnegative. For the remainder of the proof set j(t):=\Big(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}\Big) \Big(\int_s^{t}\frac{\Delta\tau}{p(\tau)}\Big) -\int_{s}^t\int_{s}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u $$for t_2\le s\le t. Then j(s)=0, and$$ j^\Delta(t) = \frac{1}{p(t)}\int_{s}^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)} -\int_{s}^{t}\frac{\Delta\tau}{p(\tau)} \ge 0 $$since 00 is a real scalar; the boundary points from \mathbb{T} satisfy t_10 \quad \text{and}\quad 1<\alpha<\frac{\int_{\rho(t_1)}^{\sigma(t_3)} \frac{\Delta\tau}{p(\tau)}}{\int_{\rho(t_1)}^{t_2} \frac{\Delta\tau}{p(\tau)}};$$ \item[(K2)] the coefficients satisfy\alpha_i\ge 0$for$i=1,2,\cdots,n$and the points$\xi_i\in(\rho(t_1),\sigma^2(t_3))_\mathbb{T}$are such that $$\xi_1<\xi_2<\cdots<\xi_n \quad\text{and}\quad d -\sum_{i=1}^{n}\alpha_i\int_{\rho(t_1)}^{\xi_i}\frac{\Delta\tau}{p(\tau)}>0;$$ \item[(K3)] the continuous function$f:[0,\infty)\rightarrow[0,\infty)$is such that the following exist: $$f_0:=\lim_{x\rightarrow 0^+}\frac{f(x)}{x}, \quad f_{\infty}:=\lim_{x\rightarrow\infty}\frac{f(x)}{x};$$ \item[(K4)] the left-dense continuous function$a:[\rho(t_1),\sigma(t_3)]_\mathbb{T}\rightarrow[0,\infty)$is such that $$\label{c5} \exists t_*\in[t_2/\alpha,t_2]_\mathbb{T} \ni a(t_*) > 0.$$ \end{enumerate} By the novelty of the multi-point boundary conditions, problem \eqref{signbvp}, \eqref{signbc} is introduced for the first time on any time scale, including$\mathbb{R}$,$\mathbb{Z}$, and the quantum time scale. We now turn our attention to the problem $$\label{bvp2} (px^{\Delta\Delta})^\nabla(t)+\lambda y(t)=0, \quad t\in[t_1,t_3]_\mathbb{T},$$ with multi-point boundary conditions \eqref{signbc}, where$y:[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}\rightarrow(0,\infty)$is a left-dense continuous function, and$\lambda>0$. \begin{lemma} \label{lem4.1} Assume {\rm (K1)} and {\rm (K2)}. If$y\in C_{ld}[\rho(t_1),\sigma^2(t_3)]$with$y\ge 0$, then the nonhomogeneous dynamic equation \eqref{bvp2} with boundary conditions \eqref{signbc} has a unique solution$x^*$on$t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$given by $$\label{form} x^*(t)=\lambda\Big(\int_{\rho(t_1)}^{\sigma(t_3)} G(t,s)y(s) \nabla s+B(y)\int_{\rho(t_1)}^{t}\int_{\rho(t_1)}^{u} \frac{\Delta\tau}{p(\tau)}\Delta u\Big),$$ where:$G(t,s)$is the Green function \eqref{greenf} of the boundary value problem \eqref{b1}, \eqref{b2} and the functional$B$is defined by $$\label{B} B(y):=\Big(d-\sum_{i=1}^{n}\alpha_i\int_{\rho(t_1)}^{\xi_i} \frac{\Delta\tau}{p(\tau)}\Big)^{-1}\sum_{i=1}^{n}\alpha_i \int_{\rho(t_1)}^{\sigma(t_3)}G^{\Delta_t}(\xi_i,s)y(s)\nabla s.$$ \end{lemma} \begin{proof} Let$y\in C_{ld}[\rho(t_1),\sigma^2(t_3)]$with$y\ge 0$; we show that the function$x^*$given in \eqref{form} is a solution of \eqref{bvp2} with conditions \eqref{signbc} only if$B(y)$is given by \eqref{B}. If$x^*$is a solution of \eqref{bvp2}, \eqref{signbc}, then $$x^*(t)=\lambda\int_{\rho(t_1)}^t G(t,s)y(s)\nabla s + \lambda\int_{t}^{\sigma(t_3)} G(t,s)y(s)\nabla s + B\int_{\rho(t_1)}^{t} \int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u$$ for some constant$B$. Taking the delta derivative with respect to$t$yields $$x^{*\Delta}(t) = \lambda\int_{\rho(t_1)}^t G^\Delta(t,s)y(s)\nabla s + \lambda\int_{t}^{\sigma(t_3)} G^\Delta(t,s)y(s)\nabla s + B\int_{\rho(t_1)}^{t}\frac{\Delta\tau}{p(\tau)};$$ clearly$x^*(\rho(t_1))=x^{*\Delta}(\rho(t_1))=0$as$G(t,s)$satisfies \eqref{bvpbc}. Since$ptimes the delta derivative of this expression is \begin{align*} (px^{*\Delta\Delta})(t) &= \lambda\int_{\rho(t_1)}^t \Big[\frac{1}{d} \Big(\int_{s}^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}-\alpha\int_{s}^{t_2} \frac{\Delta\tau}{p(\tau)}\Big)-1\Big]y(s)\nabla s \\ &\quad + \lambda\int_{t}^{t_2} \Big[\frac{1}{d} \Big(\int_{s}^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)} -\alpha\int_{s}^{t_2}\frac{\Delta\tau}{p(\tau)}\Big)\Big]y(s)\nabla s + B \\ &\quad + \frac{\lambda}{d}\int_{t_2}^{\sigma(t_3)} \Big(\int_{s}^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}\Big)y(s)\nabla s, \end{align*} we see that \begin{align*} (px^{*\Delta\Delta})^\nabla(t) &= \lambda \Big[\frac{1}{d}\Big(\int_{t}^{\sigma(t_3)} \frac{\Delta\tau}{p(\tau)}-\alpha\int_{t}^{t_2} \frac{\Delta\tau}{p(\tau)}\Big)-1\Big]y(t) \\ &\quad - \lambda \Big[\frac{1}{d}\Big(\int_{t}^{\sigma(t_3)} \frac{\Delta\tau}{p(\tau)}-\alpha\int_{t}^{t_2} \frac{\Delta\tau}{p(\tau)}\Big)\Big]y(t)= - \lambda y(t) \end{align*} and \eqref{bvp2} holds; this works regardless of the placement oft$in$[t_1,t_3]_\mathbb{T}$. To meet the other boundary condition in \eqref{signbc}, we must have that $$Bd = \sum_{i=1}^{n}\alpha_i\Big[\int_{\rho(t_1)}^{\sigma(t_3)} G^\Delta(\xi_i,s)y(s)\nabla s +B\int_{\rho(t_1)}^{\xi_i} \frac{\Delta\tau}{p(\tau)}\Big],$$ from which \eqref{B} follows. \end{proof} \begin{corollary} \label{coro4.2} Assume {\rm (K1)} and {\rm (K2)}. If$y\in C_{ld}[\rho(t_1),\sigma^2(t_3)]$with$y\ge 0$, then the unique solution$x^*$as in \eqref{form} of the problem \eqref{bvp2}, \eqref{signbc} satisfies$x^*(t)\ge 0$for$t\in[\rho(t_1),\sigma^2(t_3)]$. \end{corollary} \begin{proof} From Lemma \ref{lemma22} we know that on$[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}\times[t_1,\sigma(t_3)]_\mathbb{T}$the Green function \eqref{greenf} satisfies$G(t,s)\ge 0$. >From equation \eqref{Gdel} and (K2) we have that$B(y)\ge 0$for$y\ge 0$. \end{proof} \begin{lemma}\label{lemma43} Assume {\rm (K1)} and {\rm (K2)}. If$y\in C_{ld}[\rho(t_1),\sigma^2(t_3)]$with$y\ge 0$, then the unique solution$x^*$as in \eqref{form} of the problem \eqref{bvp2}, \eqref{signbc} satisfies $$\gamma\|x^*\| \le x^*(t),\quad t\in[t_2/\alpha,t_2]_\mathbb{T}$$ for$\gamma$given in \eqref{gammadef}. \end{lemma} \begin{proof} Let$y\in C_{ld}[\rho(t_1),\sigma^2(t_3)]$with$y\ge 0$. From previous work in Lemma \ref{lemma22}, it is clear that for all$t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$and$B(y)$given in \eqref{B}, $$\label{xless} x^*(t) \le \lambda\Big(\int_{\rho(t_1)}^{\sigma(t_3)}g(s)y(s)\nabla s + B(y)\int_{\rho(t_1)}^{\sigma^2(t_3)}\int_{\rho(t_1)}^{u} \frac{\Delta\tau}{p(\tau)}\Delta u \Big).$$ For$t\in[t_2/\alpha,t_2]_\mathbb{T}$, from Lemma \ref{lemma23} and the definition of$\gammain \eqref{gammadef} we have \begin{aligned} x^*(t) &= \lambda\Big(\int_{\rho(t_1)}^{\sigma(t_3)} G(t,s)y(s)\nabla s +B(y)\int_{\rho(t_1)}^{t}\int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)} \Delta u\Big) \\ &\geq \lambda\Big(\int_{\rho(t_1)}^{\sigma(t_3)} \gamma g(s)y(s)\nabla s +B(y)\int_{\rho(t_1)}^{t_2/\alpha}\int_{\rho(t_1)}^{u} \frac{\Delta\tau}{p(\tau)}\Delta u\Big) \\ & = \gamma\lambda\Big(\int_{\rho(t_1)}^{\sigma(t_3)} g(s)y(s)\nabla s +B(y)\frac{(\alpha+1)\int_{\rho(t_1)}^{\sigma^2(t_3)} \int_{\rho(t_1)}^{\sigma(t_3)} \frac{\Delta\tau}{p(\tau)}\Delta u}{\min\{\alpha-1,\alpha\}}\Big) \\ &\geq \gamma\|x^*\|. \end{aligned}\label{xmore} The proof is complete. \end{proof} \section{eigenvalue intervals} \noindent To establish eigenvalue intervals for the eigenvalue problem \eqref{signbvp}, \eqref{signbc} we will employ Theorem \ref{fixedpt}. To that end, let\mathcal{B}$denote the Banach space$C[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$with the norm$\|x\|=\sup_{t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}}|x(t)|$. Define the cone$\mathcal{P}\subset\mathcal{B}$by $$\mathcal{P}=\{x\in\mathcal{B}: x(t)\ge 0 \;\text{for}\; t\in[\rho(t_1),\; \sigma^2(t_3)]_\mathbb{T},\; x(t)\ge \gamma\|x\| \;\text{on}\; [t_2/\alpha,t_2]_\mathbb{T}\},$$ where$\gamma$is given in \eqref{gammadef}. When$x\in\mathcal{P}$define the operator$T:\mathcal{P}\rightarrow \mathcal{B}$for$t\in [\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$by $$\label{T} (Tx)(t):=\lambda\Big(\int_{\rho(t_1)}^{\sigma(t_3)} G(t,s)a(s)f(x(s))\nabla s +B(af(x))\int_{\rho(t_1)}^{t} \int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u\Big),$$ using \eqref{B}. We seek a fixed point of$T$in$\mathcal{P}$by establishing the hypotheses of Theorem \ref{fixedpt}. \begin{lemma}\label{lemma51} Assume {\rm (K1)} through {\rm (K4)}. Then$T:\mathcal{P}\rightarrow\mathcal{P}$is completely continuous. \end{lemma} \begin{proof} Consider the integral operator$T$in \eqref{T}. If$x\in\mathcal{P}$, then by Lemma \ref{lemma22} we have, as in \eqref{xless} and \eqref{xmore}, $$(Tx)(t) \le \lambda\Big(\int_{\rho(t_1)}^{\sigma(t_3)} g(s)a(s)f(x(s)) \nabla s +B(af(x))\int_{\rho(t_1)}^{\sigma^2(t_3)}\int_{\rho(t_1)}^{u} \frac{\Delta\tau}{p(\tau)}\Delta u\Big),$$ so that for$t\in[t_2/\alpha,t_2]_\mathbb{T}, \begin{align*} (Tx)(t) & \ge \lambda\Big(\int_{\rho(t_1)}^{\sigma(t_3)} \gamma g(s)a(s)f(x(s))\nabla s+B(af(x))\int_{\rho(t_1)}^{t_2/\alpha} \int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u\Big) \\ & = \gamma\lambda\Big(\int_{\rho(t_1)}^{\sigma(t_3)} g(s)a(s) f(x(s))\nabla s+B(af(x))\frac{(\alpha+1)\int_{\rho(t_1)}^{\sigma^2(t_3)} \int_{\rho(t_1)}^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)} \Delta u}{\min\{\alpha-1,\alpha\}}\Big) \\ & \ge \gamma\|Tx\|. \end{align*} Therefore,T:\mathcal{P}\rightarrow\mathcal{P}$. Moreover,$T$is completely continuous by a typical application of the Ascoli-Arzela Theorem. \end{proof} For$G(t,s)$in \eqref{greenf} and$B$\eqref{B}, define the constant $$\label{Jdef} J:=\int_{\rho(t_1)}^{\sigma(t_3)}g(s)a(s)\nabla s+B(a) \int_{\rho(t_1)}^{\sigma^2(t_3)}\int_{\rho(t_1)}^{u} \frac{\Delta\tau}{p(\tau)}\Delta u.$$ \begin{theorem}\label{ethm1} Assume {\rm (K1)} through {\rm (K4)}. Then for each$\lambda$satisfying $$\label{lambda1} \frac{1}{f_{\infty}\gamma\int_{t_2/\alpha}^{t_2} G(\sigma^2(t_3),s)a(s) \nabla s} < \lambda < \frac{1}{f_0 J}$$ there exists at least one positive solution of \eqref{signbvp}, \eqref{signbc} in$\mathcal{P}$. \end{theorem} \begin{proof} Let$J$be as in \eqref{Jdef},$\lambda$as in \eqref{lambda1}, and let$\epsilon > 0$be such that $$\label{lambdae1} \frac{1}{(f_{\infty}-\epsilon)\gamma\int_{t_2/\alpha}^{t_2} G(\tau,s)a(s)\nabla s} \le \lambda \le \frac{1}{(f_0+\epsilon)J}.$$ First consider$f_0$. There exists an$R_1>0$such that$f(x)\le (f_0+\epsilon)x$for$0From \eqref{B} we have $|B(af(x))| \le B(a)\|f(x)\|$. Using Lemma \ref{lemma22} we have \begin{align*} (Tx)(t) & = \lambda \Big(\int_{\rho(t_1)}^{\sigma(t_3)} G(t,s)a(s)f(x(s))\nabla s +B(af(x)) \int_{\rho(t_1)}^{t}\int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u\Big) \\ & \le \lambda\|f(x)\|\Big(\int_{\rho(t_1)}^{\sigma(t_3)} g(s)a(s) \nabla s +B(a)\int_{\rho(t_1)}^{\sigma^2(t_3)}\int_{\rho(t_1)}^{u} \frac{\Delta\tau}{p(\tau)}\Delta u\Big) \\ & \le \lambda(f_0+\epsilon)\|x\|J \le \|x\| \end{align*} from the right side of \eqref{lambdae1}. As a result, $\|Tx\| \le \|x\|$. Thus, take $$\Omega_1:=\{x\in\mathcal{B}:\|x\| < R_1\}$$ so that $\|Tx\|\le\|x\|$ for $x\in\mathcal{P}\cap\partial\Omega_1$. Next consider $f_{\infty}$. Again by definition there exists an $R_2' > R_1$ such that $f(x)\ge(f_{\infty}-\epsilon)x$ for $x\ge R_2'$; take $R_2=\max\{2R_1,R_2'/\Gamma\}$. If $x\in\mathcal{P}$ with $\|x\|=R_2$, then for $s\in[t_2/\alpha,t_2]_\mathbb{T}$ we have $$\label{xandH2} x(s)\ge \gamma\|x\|=\gamma R_2.$$ Define $\Omega_2:=\{x\in\mathcal{B}:\|x\| < R_2\}$; using \eqref{xandH2} for $s\in[t_2/\alpha,t_2]$ we get \begin{align*} &(Tx)(\sigma^2(t_3)) \\ & = \lambda\Big(\int_{\rho(t_1)}^{\sigma(t_3)} G(\sigma^2(t_3),s) a(s)f(x(s))\nabla s +B(af(x))\int_{\rho(t_1)}^{\sigma^2(t_3)} \int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u\Big) \\ & \ge \lambda \int_{t_2/\alpha}^{t_2} G(\sigma^2(t_3),s)a(s)f(x(s)) \nabla s \ge \lambda (f_{\infty}-\epsilon)\int_{t_2/\alpha}^{t_2} G(\sigma^2(t_3),s)a(s)x(s)\nabla s \\ & \ge \lambda(f_{\infty}-\epsilon)\gamma R_2 \int_{t_2/\alpha}^{t_2} G(\sigma^2(t_3),s)a(s) \nabla s \ge R_2 = \|x\|, \end{align*} where we have used the left side of \eqref{lambdae1}. Hence we have shown that $$\|Tx\| \ge \|x\|, \quad x\in\mathcal{P}\cap\partial\Omega_2.$$ An application of Theorem \ref{fixedpt} yields the conclusion of the theorem; in other words, $T$ has a fixed point $x$ in $\mathcal{P}\cap(\overline{\Omega}_2\setminus\Omega_1)$ with $R_1 \le \|x\| \le R_2$. \end{proof} \begin{theorem}\label{ethm2} Assume {\rm (K1)} through {\rm (K4)}. Then for each $\lambda$ satisfying $$\label{lambda2} \frac{1}{f_0\gamma\int_{t_2/\alpha}^{t_2} G(\sigma^2(t_3),s)a(s)\nabla s} < \lambda < \frac{1}{f_{\infty} J}$$ there exists at least one positive solution of \eqref{signbvp}, \eqref{signbc} in $\mathcal{P}$. \end{theorem} \begin{proof} Let $J$ be as in \eqref{Jdef}, $\lambda$ as in \eqref{lambda2}, and let $\epsilon > 0$ be such that $$\label{lambdae2} \frac{1}{(f_0-\epsilon)\gamma\int_{t_2/\alpha}^{t_2} G(\tau,s)a(s)\nabla s} \le \lambda \le \frac{1}{(f_{\infty}+\epsilon)J}.$$ Again let $T$ be the operator defined in \eqref{T}. We once more seek a fixed point of $T$ in $\mathcal{P}$ by establishing the hypotheses of Theorem \ref{fixedpt}. First consider $f_0$. There exists an $R_1>0$ such that $f(x)\ge (f_0-\epsilon)x$ for $0 R_1$ such that $f(x)\le(f_{\infty}+\epsilon)x$ for $x\ge R_2'$; take $R_2=\max\{2R_1,R_2'/\Gamma\}$. If $f$ is bounded, there exists $M > 0$ with $f(x) \le M$ for all $x\in(0,\infty)$. Let $$R_2:=\max\Big\{2R_2', \;\lambda M\Big(\int_{\rho(t_1)}^{\sigma(t_3)} g(s)a(s)\nabla s +B(a)\int_{\rho(t_1)}^{\sigma^2(t_3)} \int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u\Big)\Big\}.$$ If $x\in\mathcal{P}$ with $\|x\|=R_2$, then we have $$(Tx)(t) \le \lambda M\Big(\int_{\rho(t_1)}^{\sigma(t_3)} g(s)a(s)\nabla s +B(a)\int_{\rho(t_1)}^{\sigma^2(t_3)}\int_{\rho(t_1)}^{u} \frac{\Delta\tau}{p(\tau)}\Delta u\Big) \le R_2 = \|x\|.$$ As a result, $\|Tx\| \le \|x\|$. Thus, take $$\Omega_2:=\{x\in\mathcal{B}:\|x\| < R_2\}$$ so that $\|Tx\|\le\|x\|$ for $x\in\mathcal{P}\cap\partial\Omega_2$. If $f$ is unbounded, take $R_2:=\max\{2R_1,R_2'\}$ such that $f(x) \le f(R_2)$ for $0 < x \le R_2$. If $x\in\mathcal{P}$ with $\|x\|=R_2$, then we have \begin{align*} (Tx)(t) & = \lambda\Big(\int_{\rho(t_1)}^{\sigma(t_3)} G(t,s)a(s)f(x(s)) \nabla s +B(af(x))\int_{\rho(t_1)}^{t}\int_{\rho(t_1)}^{u} \frac{\Delta\tau}{p(\tau)}\Delta u\Big) \\ & \le \lambda f(R_2) \Big(\int_{\rho(t_1)}^{\sigma(t_3)} g(s)a(s)\nabla s +B(a)\int_{\rho(t_1)}^{\sigma^2(t_3)}\int_{\rho(t_1)}^{u} \frac{\Delta\tau}{p(\tau)}\Delta u\Big) \\ & \le \lambda(f_\infty+\epsilon)R_2J \le R_2=\|x\|, \end{align*} where we have used the left side of \eqref{lambdae2}. Hence we have shown that $$\|Tx\| \le \|x\|, \quad x\in\mathcal{P}\cap\partial\Omega_2$$ if we take $$\Omega_2:=\{x\in\mathcal{B}:\|x\| < R_2\}.$$ As before an application of Theorem \ref{fixedpt} yields the conclusion that $T$ has a fixed point $x$ in $\mathcal{P}\cap(\overline{\Omega}_2\setminus\Omega_1)$ with $R_1 \le \|x\| \le R_2$. \end{proof} \begin{corollary} \label{coro5.4} Assume {\rm (K1)} through {\rm (K4)}. If $f$ is sublinear (i.e., $f_0=\infty$ and $f_{\infty}=0$), or if $f$ is superlinear (i.e., $f_0=0$ and $f_{\infty}=\infty$), then for any $\lambda > 0$ the boundary value problem \eqref{signbvp}, \eqref{signbc} has at least one positive solution in $\mathcal{P}$. \end{corollary} \begin{proof} For the superlinear claim, use \eqref{lambda1} of Theorem \ref{ethm1}; for the sublinear claim, use \eqref{lambda2} of Theorem \ref{ethm2}. \end{proof} As remarked earlier, the results in this section are new for ordinary differential equations (when $\mathbb{T} = \mathbb{R}$) and for difference equations (when $\mathbb{T} = \mathbb{Z}$). We now provide an example to illustrate that conditions (K1) through (K4) are naturally satisfied. \begin{example} \label{exa5.5} \rm Consider for $\mathbb{T} = \mathbb{R}$ and the following choices: $t_1 = 0$, $t_2 = 1/2$, $t_3 = 1$; $p \equiv 1$; continuous $f(x)$ such that $f_0$ and $f_{\infty}$ exist; $\alpha = 3/2$; $\alpha_1=1/2$; $\xi_1=1/4$; $a(t)=t$. Then the boundary value problem \eqref{signbvp}, \eqref{signbc} has at least one positive solution in $\mathcal{P}$ for any $$\frac{1718.45}{f_\infty} < \lambda < \frac{0.824}{f_0}.$$ With these choices, \eqref{signbvp}, \eqref{signbc} reduces to a third-order four-point boundary value problem \begin{gather*} x'''(t)+ \lambda tf(x(t))=0, \quad t\in[0,1]_\mathbb{R}, \\ x(0)=x'(0)= 0, \\ x'(1)-3 x'(1/2)/2=x'(1/4)/2. \end{gather*} It is not difficult to verify that conditions (K1) through (K4) are satisfied. Some calculations lead to $\gamma=1/90$, $g(s)=10s(1-s)$, $B(s)=73/96$, $J=233/192$, and $\int_{1/3}^{1/2}sG(1,s)ds=181/3456$, so that we can find the interval in \eqref{lambda1} to be $$\frac{311040}{181 f_\infty}<\lambda <\frac {192}{233 f_0},$$ which is approximately $$\frac{1718.45}{f_\infty} < \lambda < \frac{0.824}{f_0}.$$ \end{example} \begin{thebibliography}{99} \bibitem{and} D. R. Anderson and A. Cabada, Third-order right-focal multi-point problems on time scales, \emph{J. 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