\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
$0<p(t)\le 1$ on $\mathbb{T}$; the boundary points from $\mathbb{T}$ satisfy
$t_1<t_2<t_3$,  with $t_2/\alpha\in\mathbb{T}$ such that
\begin{enumerate}
 \item[(H1)] the constants $d$ and $\alpha$ satisfy
 $$ d:=\int_{\rho(t_1)}^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}
-\alpha\int_{\rho(t_1)}^{t_2}\frac{\Delta\tau}{p(\tau)}>0
\quad \text{and}\quad
1<\alpha<\frac{\int_{\rho(t_1)}^{\sigma(t_3)}
\frac{\Delta\tau}{p(\tau)}}{\int_{\rho(t_1)}^{t_2}\frac{\Delta\tau}{p(\tau)}};
 $$
 \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
\begin{equation}\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{equation}
\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
\begin{equation}\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).
\end{equation}


\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 $0<p(t)\le 1$ on $\mathbb{T}$ by assumption. Thus $j(t)$ is nonnegative,
guaranteeing overall that $v(t,s)$ is nonnegative, so that ultimately
$G(t,s)$ is nonnegative as claimed.

Now we show that $G(t,s)\le g(s)$ on
$[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}\times[t_1,\sigma(t_3)]_\mathbb{T}$, where $g$
is given in \eqref{gdef}. For any fixed $s\in[t_1,\sigma(t_3)]_\mathbb{T}$,
 a delta derivative of $G(t,s)$ with respect to $t$ yields
$$
G^{\Delta_t}(t,s)=
    \begin{cases}
 \frac{1}{d}\left(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}
 -\alpha\int_s^{t_2}\frac{\Delta\tau}{p(\tau)}\right)\int_{\rho(t_1)}^t
 \frac{\Delta\tau}{p(\tau)} - \int_{s}^t\frac{\Delta\tau}{p(\tau)}
&: s\le\min\{t_2,t\} \\[5pt]
        \frac{1}{d}\left(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}
 -\alpha\int_s^{t_2}\frac{\Delta\tau}{p(\tau)}\right)\int_{\rho(t_1)}^t
 \frac{\Delta\tau}{p(\tau)}
&: t\le s\le t_2 \\[5pt]
        \frac{1}{d}\left(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}\right)
 \int_{\rho(t_1)}^t\frac{\Delta\tau}{p(\tau)}
 - \int_{s}^t\frac{\Delta\tau}{p(\tau)}
&:t_2\le s\le t \\[5pt]
        \frac{1}{d}\left(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}\right)
        \int_{\rho(t_1)}^t\frac{\Delta\tau}{p(\tau)}
&:\max\{t_2,t\}\le s.
    \end{cases}
$$
Then rewriting we see that
\begin{align*}
0&\le G^{\Delta_t}(t,s)\\
&=\frac{1}{d}
    \begin{cases}
    \left(\int_{\rho(t_1)}^{s}\frac{\Delta\tau}{p(\tau)}\right)
 \left[d+(\alpha-1)\int_{\rho(t_1)}^{t}\frac{\Delta\tau}{p(\tau)}\right]
&: s\le\min\{t_2,t\} \\[5pt]
    \left(\int_{\rho(t_1)}^{t}\frac{\Delta\tau}{p(\tau)}\right)
 \left[d+(\alpha-1)\int_{\rho(t_1)}^{s}\frac{\Delta\tau}{p(\tau)}\right]
&: t\le s\le t_2 \\[5pt]
    \left(\int_t^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}\right)
 \int_{\rho(t_1)}^s\frac{\Delta\tau}{p(\tau)}
 + \alpha \left(\int_{s}^t\frac{\Delta\tau}{p(\tau)}\right)
 \int_{\rho(t_1)}^{t_2}\frac{\Delta\tau}{p(\tau)}
&:t_2\le s\le t \\[5pt]
    \left(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}\right)
        \int_{\rho(t_1)}^t\frac{\Delta\tau}{p(\tau)} &:\max\{t_2,t\}\le s,
    \end{cases}
\end{align*}
and rewriting again we obtain
\begin{equation}\label{Gdel}
\begin{aligned}
  0&\le G^{\Delta_t}(t,s)
\le
\begin{cases}
 \frac{\alpha}{d}\left(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}\right)
 \int_{\rho(t_1)}^s\frac{\Delta\tau}{p(\tau)}
&: s\le\min\{t_2,t\} \\[5pt]
 \frac{\alpha}{d}\left(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}\right)
 \int_{\rho(t_1)}^s\frac{\Delta\tau}{p(\tau)}
&: t\le s\le t_2 \\[5pt]
 \frac{\alpha+1}{d}\left(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}\right)
 \int_{\rho(t_1)}^s\frac{\Delta\tau}{p(\tau)}
&:t_2\le s\le t \\[5pt]
 \frac{\alpha}{d}\left(\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}\right)
 \int_{\rho(t_1)}^s\frac{\Delta\tau}{p(\tau)} &:\max\{t_2,t\}\le s
\end{cases}\\
&\le \frac{g(s)}{\big(\sigma^2(t_3)-\rho(t_1) \big)}.
\end{aligned}
\end{equation}
Delta integration from $\rho(t_1)$ to $t$ yields $G(t,s)\le g(s)$ for
$g(s)$ given in \eqref{gdef}.
\end{proof}

\begin{lemma}\label{lemma23}
Assume {\rm (H1)}. The Green function \eqref{greenf} corresponding to the
problem \eqref{b1}, \eqref{b2} satisfies
\begin{equation}\label{gammadef}
 G(t,s) \ge \gamma g(s), \quad \gamma
:=\frac{\min\{\alpha-1,\alpha\}\int_{\rho(t_1)}^{t_2/\alpha}
\int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u}{(\alpha+1)
\big(\sigma^2(t_3)-\rho(t_1) \big)\int_{\rho(t_1)}^{\sigma(t_3)}
\frac{\Delta\tau}{p(\tau)}}\in(0,1)
\end{equation}
for $(t,s)\in[t_2/\alpha,t_2]_\mathbb{T}\times[t_1,\sigma(t_3)]_\mathbb{T}$, where
$g(s)$ is given in \eqref{gdef}.
\end{lemma}

\begin{proof}
If $s=\sigma(t_3)$, or if $t_1$ is a left-dense point and $s=t_1$,
then the result follows from \eqref{gdef}. Thus consider the cases
where $(t,s)\in[t_2/\alpha,t_2]_\mathbb{T}\times(\rho(t_1),\sigma(t_3))_\mathbb{T}$.
For $s\le t\le t_2$,
\begin{align*}
\frac{G(t,s)}{g(s)}
&=  \frac{\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 - d\int_{s}^t\int_{s}^{u}
   \frac{\Delta\tau}{p(\tau)}\Delta u}{dg(s)} \\
&=  \frac{d\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)}{dg(s)} \\
&\quad +\frac{ (\alpha-1)
  \Big(\int_{\rho(t_1)}^{s}\frac{\Delta\tau}{p(\tau)}\Big)
  \int_{\rho(t_1)}^{t}\int_{\rho(t_1)}^{u}
  \frac{\Delta\tau}{p(\tau)}\Delta u}{dg(s)} \\
&\ge  \frac{(\alpha-1)\int_{\rho(t_1)}^{t}\int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u}{(\alpha+1)\big(\sigma^2(t_3)-\rho(t_1)\big)\int_s^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}} \ge \gamma.
\end{align*}
For $t\le s\le t_2$,
\begin{align*}
\frac{G(t,s)}{g(s)}
&=  \frac{\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}{dg(s)} \\
 &=  \frac{d\int_{\rho(t_1)}^{t}\int_{\rho(t_1)}^{u}
 \frac{\Delta\tau}{p(\tau)}\Delta u +(\alpha-1)\left(\int_{\rho(t_1)}^{s}\frac{\Delta\tau}{p(\tau)}\right)\int_{\rho(t_1)}^{t}\int_{\rho(t_1)}^{u}\frac{\Delta\tau}{p(\tau)}\Delta u}{dg(s)} \\
 &\geq   \frac{(\alpha-1)\int_{\rho(t_1)}^{t}\int_{\rho(t_1)}^{u}
\frac{\Delta\tau}{p(\tau)}\Delta u}{(\alpha+1)
\big(\sigma^2(t_3)-\rho(t_1)\big)\int_s^{\sigma(t_3)}
\frac{\Delta\tau}{p(\tau)}} \ge \gamma.
\end{align*}
For $t\le t_2\le s$,
$$
\frac{G(t,s)}{g(s)} = \frac{\int_{\rho(t_1)}^t\int_{\rho(t_1)}^{u}
\frac{\Delta\tau}{p(\tau)}\Delta u}{(\alpha +1)\big(\sigma^2(t_3)
- \rho(t_1)\big)\int^s_{\rho(t_1)}\frac{\Delta\tau}{p(\tau)}} \ge \gamma.
$$
In all cases the statement holds.
\end{proof}

\section{an existence result on cones}

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
$$
\mathcal{P}=\{x\in\mathcal{B}: x(t)\ge 0 \;\text{for}\;
t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}, x(t)\ge \gamma\|x\| \;\text{on}\;
[t_2/\alpha,t_2]_\mathbb{T}\}.
$$
For $x\in\mathcal{P}$, define
$$
Ax(t):=\int_{\rho(t_1)}^{\sigma(t_3)}G(t,s)a(s)f(x(s))\nabla s, \quad
t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}. $$
 From Lemma \ref{lemma22}, for  $t\in[\rho(t_1),\sigma^2(t_3)]_\mathbb{T}$,
$$
0\le Ax(t)=\int_{\rho(t_1)}^{\sigma(t_3)}G(t,s)a(s)f(x(s))\nabla s\le
\int_{\rho(t_1)}^{\sigma(t_3)}g(s)a(s)f(x(s))\nabla s,
 $$
so that
\begin{equation}\label{3bd}
  \|Ax\|\le \int_{\rho(t_1)}^{\sigma(t_3)}g(s)a(s)f(x(s))\nabla s.
\end{equation}
By Lemma \ref{lemma23} and \eqref{3bd}, for $t\in[t_2/\alpha,t_2]_\mathbb{T}$,
$$
Ax(t)=\int_{\rho(t_1)}^{\sigma(t_3)}G(t,s)a(s)f(x(s))\nabla s
\ge \gamma \int_{\rho(t_1)}^{\sigma(t_3)}g(s)a(s)f(x(s))\nabla s
\ge \gamma \|Ax\|,
$$
giving us
$$
Ax(t)\ge \gamma\|Ax\|, \quad [t_2/\alpha,t_2]_\mathbb{T},
$$
and $A\mathcal{P}\subset\mathcal{P}$. Furthermore, it is straightforward
to verify that $A:\mathcal{P}\rightarrow\mathcal{P}$ is a completely
continuous operator whose fixed points are solutions of
\eqref{bvp}, \eqref{bvpbc}.

To establish an existence result we will employ the following fixed point
theorem due to Guo and Krasnosel'skii
\cite{gu}, and seek a fixed point of $T$ in $\mathcal{P}$.

\begin{theorem}\label{fixedpt}
Let $E$ be a Banach space, $P\subseteq E$ be a cone, and suppose that
$\mathcal{S}_1$, $\mathcal{S}_2$ are bounded open balls of $E$ centered
at the origin with
$\overline{\mathcal{S}}_1\subset\mathcal{S}_2$.
Suppose further that
$L:P\cap(\overline{\mathcal{S}}_2\setminus\mathcal{S}_1)\rightarrow P$
is a completely continuous operator such that either
\begin{enumerate}
 \item[(i)] $\|Ly\| \le \|y\|$, $y\in P\cap\partial\mathcal{S}_1$ and $\|Ly\| \ge \|y\|$,
      $y\in P\cap\partial\mathcal{S}_2$, or
 \item[(ii)] $\|Ly\| \ge \|y\|$, $y\in P\cap\partial\mathcal{S}_1$ and $\|Ly\| \le \|y\|$,
      $y\in P\cap\partial\mathcal{S}_2$
\end{enumerate}
holds.  Then $L$ has a fixed point in $P\cap(\overline{\mathcal{S}}_2\setminus\mathcal{S}_1)$.
\end{theorem}

\begin{theorem}
Assume {\rm (H1), (H2)}, and {\rm (H3)} hold. Then the boundary value
problem \eqref{bvp}, \eqref{bvpbc} has at least one positive solution if
\begin{enumerate}
 \item[(i)] $f_0=0$ and $f_\infty=\infty$ ($f$ is superlinear), or
 \item[(ii)] $f_0=\infty$ and $f_\infty=0$ ($f$ is sublinear).
\end{enumerate}
\end{theorem}

\begin{proof}
The proof in the general time-scale setting is similar to that given in
the case $\mathbb{T}=\mathbb{R}$ in \cite{guo} and is omitted.
\end{proof}

\section{nonlinear multi-point problem}

In the next two sections we consider the related multi-point boundary
value problem
\begin{gather}
  (px^{\Delta\Delta})^\nabla(t)+ \lambda a(t)f(x(t))=0,
\quad t\in[t_1,t_3]_\mathbb{T},  \label{signbvp} \\
 x(\rho(t_1))=x^\Delta(\rho(t_1))= 0,   \quad
 x^\Delta(\sigma(t_3))-\alpha x^\Delta(t_2)
=\sum_{i=1}^{n}\alpha_ix^\Delta(\xi_i), \label{signbc}
\end{gather}
where: $p$ is a right-dense continuous, real-valued function with
$0<p(t)\le 1$ on $\mathbb{T}$; $\lambda>0$ is a real scalar; the boundary points
from $\mathbb{T}$ satisfy $t_1<t_2<t_3$, with $t_2/\alpha\in\mathbb{T}$ such that
\begin{enumerate}
 \item[(K1)] the constants $d$ and $\alpha$ satisfy
 $$
d:=\int_{\rho(t_1)}^{\sigma(t_3)}\frac{\Delta\tau}{p(\tau)}
 -\alpha\int_{\rho(t_1)}^{t_2}\frac{\Delta\tau}{p(\tau)}>0
 \quad \text{and}\quad
 1<\alpha<\frac{\int_{\rho(t_1)}^{\sigma(t_3)}
  \frac{\Delta\tau}{p(\tau)}}{\int_{\rho(t_1)}^{t_2}
 \frac{\Delta\tau}{p(\tau)}};
$$
 \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
\begin{equation}\label{c5}
    \exists t_*\in[t_2/\alpha,t_2]_\mathbb{T} \ni a(t_*) > 0.
\end{equation}
\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
\begin{equation}\label{bvp2}
 (px^{\Delta\Delta})^\nabla(t)+\lambda y(t)=0,  \quad t\in[t_1,t_3]_\mathbb{T},
\end{equation}
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
\begin{equation}\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),
\end{equation}
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
\begin{equation}\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{equation}
\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 $p$ times 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 of $t$
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},
\begin{equation}\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).
\end{equation}
For $t\in[t_2/\alpha,t_2]_\mathbb{T}$, from Lemma \ref{lemma23} and the definition
of $\gamma$ in \eqref{gammadef} we have
\begin{equation}
\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}
\end{equation}
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
\begin{equation}\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),
\end{equation}
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
\begin{equation}\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.
\end{equation}

\begin{theorem}\label{ethm1}
Assume {\rm (K1)} through {\rm (K4)}. Then for each $\lambda$ satisfying
\begin{equation}\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}
\end{equation}
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
\begin{equation}\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}.
\end{equation}
First consider $f_0$.  There exists an $R_1>0$ such that
$f(x)\le (f_0+\epsilon)x$ for $0<x\le R_1$ by the
definition of $f_0$.  Pick $x\in\mathcal{P}$ with $\|x\|=R_1$.
>From \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
\begin{equation}\label{xandH2}
  x(s)\ge \gamma\|x\|=\gamma R_2.
\end{equation}
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
\begin{equation}\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}
\end{equation}
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
\begin{equation}\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}.
\end{equation}
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 $0<x\le R_1$ by the
definition of $f_0$.  Pick $x\in\mathcal{P}$ with $\|x\|=R_1$.
For $s\in[t_2/\alpha,t_2]_\mathbb{T}$ we have
\begin{equation}\label{xandH1}
  x(s)\ge \gamma\|x\|=\gamma R_1.
\end{equation}
Using the left side of \eqref{lambdae2} and \eqref{xandH1} we get,
for $s\in[t_2/\alpha,t_2]_\mathbb{T}$,
\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 (f_0-\epsilon)\int_{t_2/\alpha}^{t_2} G(\sigma^2(t_3),s)
 a(s)x(s)\nabla s \\
& \ge  \lambda (f_0-\epsilon)R_1 \gamma\int_{t_2/\alpha}^{t_2}
 G(\sigma^2(t_3),s)a(s)\nabla s \ge R_1 = \|x\|,
\end{align*}
Therefore $\|Tx\| \ge \|x\|$.  This motivates us to define
$$
\Omega_1:=\{x\in\mathcal{B}:\|x\| < R_1\},
$$
whereby our work above confirms
$$
\|Tx\|\ge\|x\|, \quad x\in\mathcal{P}\cap\partial\Omega_1.
$$
Next consider $f_{\infty}$. Again by definition there exists an $R_2' > R_1$
such that $f(x)\le(f_{\infty}+\epsilon)x$ 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}


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\end{document}