\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{amssymb}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 215, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/215\hfil A third-order three-point BVP]
{Iterative technique for a third-order three-point BVP with
sign-changing Green's function}

\author[J. -P. Sun and J. Zhao \hfil EJDE-2013/215\hfilneg]
{Jian-Ping Sun, Juan Zhao}  % in alphabetical order

\address{Jian-Ping Sun \newline
Department of Applied Mathematics, Lanzhou University of Technology,
Lanzhou, Gansu 730050,  China}
\email{jpsun2012@163.com}

\address{Juan Zhao \newline
Department of Applied Mathematics, Lanzhou University of Technology,
Lanzhou, Gansu 730050,  China}
\email{jzhao79@163.com}

\thanks{Submitted July 5, 2013. Published September 30, 2013.}
\subjclass[2000]{34B10, 34B18}
\keywords{Boundary value problem; Green's function;
 positive solution; \hfill\break\indent 
existence of solutions; iterative technique}

\begin{abstract}
 In this article, by applying iterative technique,
 we study the third-order three-point boundary value problem
 \begin{gather*}
 u'''(t)=f(t,u(t)),\quad t\in [0,1], \\
 u'(0)=u''(\eta)=u(1)=0,
 \end{gather*}
 where $f\in C([0,1]\times[0,+\infty),[0,+\infty))$ and
 $\eta\in[2-\sqrt{2},1)$. The emphasis is mainly that
 although the corresponding Green's function is sign-changing, the
 solution obtained is still positive. Moreover, our iterative scheme
 starts off with zero function, which implies that the iterative
 scheme is feasible. An example is also included to illustrate the
 main results.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

Third-order differential equations arise from a variety of areas of
applied mathematics and physics, e.g., in the deflection of a curved
beam having a constant or varying cross section, a three-layer beam,
electromagnetic waves or gravity driven flows and so on \cite{6}.

Recently, the existence of single or multiple positive solutions to
some third-order three-point boundary value problems (BVPs for
short) has received much attention from many authors. For example,
in 1998, by using the Leggett-Williams fixed point theorem, Anderson
\cite{2} proved the existence of at least three positive solutions
to the BVP
\begin{gather*}
-x'''(t)+f(x(t))=0,\quad t\in [0,1],\\
x(0)=x'(t_2)=x''(1)=0,
\end{gather*}
where $t_2\in [1/2,1)$. In 2003, Anderson
\cite{1} obtained some existence results of positive solutions for
the BVP
\begin{gather*}
x'''(t)=f(t,x(t)),\quad t_1\leq t\leq t_3,\\
x(t_1)=x'(t_2)=0,\quad \gamma x(t_3)+\delta x''(t_3)=0.
\end{gather*}
The main tools used were the Guo-Krasnosel'skii and Leggett-Williams
fixed point theorems. In 2005, Sun \cite{15} studied the existence
of single and multiple positive solutions for the singular BVP
\begin{gather*}
u'''(t)-\lambda a(t)F(t,u(t))=0,\ t\in (0,1),\\
u(0)=u'(\eta)=u''(1)=0,
\end{gather*}
where $\eta\in [1/2,1)$, $\lambda$ was a positive
parameter and $a(t)$ was a nonnegative continuous function defined
on $(0, 1)$. His main tool was the Guo-Krasnosel'skii fixed point
theorem. In 2008, by using the Guo-Krasnosel'skii fixed point
theorem, Guo, Sun and Zhao \cite{7} obtained the existence of at
least one positive solution for the BVP
\begin{gather*}
u'''(t)+h(t)f(u(t))=0,\quad t\in (0,1),\\
u(0)=u'(0)=0,\quad u'(1)=\alpha u'(\eta ),
\end{gather*}
where $0<\eta <1$ and $1<\alpha <\frac 1\eta $. For more results
concerning the existence of positive solutions to third-order
three-point BVPs, one can refer to \cite{3,4,5,9,11,14,16}.

We want to point out that all the above-mentioned works are
achieved when the corresponding Green's functions are nonnegative,
which is a very important condition. A natural question is that
whether we can obtain the existence of positive solutions to some
third-order three-point BVPs when the corresponding Green's
functions are sign-changing. It is worth mentioning that Palamides
and Smyrlis \cite{8} discussed the existence of at least one
positive solution to the singular third-order three-point BVP with
an indefinitely signed Green's function
\begin{gather*}
u'''(t)=a(t)f(t,u(t)),\quad t\in (0,1),\\
u(0)=u(1)=u''(\eta)=0,\quad \eta\in (\frac{17}{24},1).
\end{gather*}
Their technique was a combination of the Guo-Krasnosel'skii fixed
point theorem and properties of the corresponding vector field. For
some related results, one can refer to \cite{10}.


Very recently, inspired greatly by \cite{8}, the authors
\cite{12,13} studied the following third-order three-point BVP
\begin{equation}
\begin{gathered}
u'''(t)=f(t,u(t)),\quad t\in [0,1], \\
u'(0)=u''(\eta)=u(1)=0,
\end{gathered}\label{e1.2}
\end{equation}
where $ \eta\in(1/2,1)$. Although the corresponding Green's
function was sign-changing, the existence of single or multiple
positive solutions for the BVP \eqref{e1.2} was still obtained. The
main tools used were the Guo-Krasnosel'skii and Leggett-Williams
fixed point theorems.

In this article, we continue to study the BVP \eqref{e1.2}. Throughout
this paper, we always assume that 
$f\in C([0,1]\times[0,+\infty),[0,+\infty))$ and
$\eta\in[2-\sqrt{2},1)$. Although the corresponding
Green's function is sign-changing, we still obtain the existence of
a decreasing positive solution under some suitable conditions on
$f$. Our main method is iterative technique. It is worth mentioning
that the iterative scheme starts off with zero function, which is
feasible for computational purpose. An example is also included to
illustrate our main results.


\section{Main results}

Let the space $E=C[0,1]$ be equipped with the norm
$\| u\|= \max_{t\in [0,1]}| u(t)|$. To obtain the existence of a 
positive solution for \eqref{e1.2}, we need to construct a suitable
cone in the Banach $E$. Let $u$ be
a solution of  \eqref{e1.2}. Then it is easy to verify that
$u(t)\geq 0$  for $t\in [0,1]$ provided that $u'(1)\leq 0$.
In fact, since $f$ is nonnegative, we know that 
$u'''(t)\geq 0$  for $t\in [0,1]$, which together with $u''(\eta)=0$
implies that
\begin{equation}
u''(t)\leq 0 \text{ for  }t\in [0,\eta] \quad\text{and}\quad
u''(t)\geq 0 \text{ for  }t\in [\eta,1]. \label{e1.001}
\end{equation}
In view of \eqref{e1.001} and $u'(0)=0$, we have
\begin{equation}
u'(t)\leq 0 \text{ for  }t\in [0,\eta] \quad\text{and}\quad
u'(t)\leq u'(1) \text{ for  }t\in [\eta,1]. \label{e2}
\end{equation}
If $u'(1)\leq 0$, then it follows from \eqref{e2} that 
$u'(t)\leq 0\text{ for  }t\in [0,1]$, which together with 
$u(1)=0$ implies that $u(t)\geq 0\ \text{\ for \ }t\in [0,1]$. 
Therefore, we define a cone in $E$ as follows:
\[
K=\{u\in E: u(t) \text{ is nonnegative and decreasing on }[0,1]\}.
\]
Note that this induces an order relation $\lesssim$ in $E$ by
defining $u \lesssim v$ if and only if $v-u\in K$.

For any $y\in E$, we consider the boundary value problem
\begin{equation}
\begin{gathered}
u'''(t)=y(t),\quad t\in [0,1], \\
u'(0)=u''(\eta)=u(1)=0.
\end{gathered} \label{e2.1}
\end{equation}

After a simple computation, we may obtain the expression of Green's 
function $G(t,s)$ for \eqref{e2.1} as follows:
 for $s\geq\eta$,
\[
G(t,s)=\begin{cases}
-\frac{(1-s)^{2}}{2},&  t\leq s, \\
\frac{t^{2}-2st+2s-1}{2},& s\leq t
\end{cases}
\]
and for $s<\eta$,
\[
G(t,s)=\begin{cases}
\frac{-t^{2}-s^{2}+2s}{2},& t\leq s, \\
(1-t)s, & s\leq t.
\end{cases}
\]
Obviously,
\[
G(t,s)\geq 0 \text{ for  }0\leq s< \eta\quad \text{and}\quad
G(t,s)\leq 0 \text{ for  }\eta\leq s\leq 1.
\]
Moreover, for $s\geq\eta$,
\begin{gather*}
\max\{G(t,s):t\in[0,1]\}=G(1,s)=0, \\
\min\{G(t,s):t\in[0,1]\}=G(s,s)=-\frac{(1-s)^{2}}{2}\geq-\frac{(1-\eta)^{2}}{2}
\end{gather*}
and for $s<\eta$,
\begin{gather*}
\max\{G(t,s):t\in[0,1]\}=G(0,s)=s-\frac{s^{2}}{2}\leq\eta-\frac{\eta^{2}}{2},\\
\min\{G(t,s):t\in[0,1]\}=G(1,s)=0.
\end{gather*}
So,
\[
\max\{|G(t,s)|:t,s\in[0,1]\}
=\max\big\{\frac{(1-\eta)^{2}}{2},\eta-\frac{\eta^{2}}{2}\big\}
=\eta-\frac{\eta^{2}}{2} =: M.
\]
It is obvious that $\sqrt{2}-1\leq M<1/2$.

In the remaining of this article, we  assume that 
$f:[0,1]\times [0,+\infty)\to [0,+\infty)$
is continuous and satisfies the following two conditions:
\begin{itemize}
\item[(C1)] For each $u\in [0,+\infty)$, the mapping $t\mapsto f(t,u)$ 
is decreasing;
\item[(C2)] For each $t\in [0,1]$, the mapping $u\mapsto
f(t,u)$ is increasing.
\end{itemize}
Now, we define an operator $T$ on $K$ by
\[
(Tu)(t)=\int_0^1G(t,s)f(s,u(s))ds,\quad t\in [0,1].
\]
Obviously, if $u$ is a fixed point of $T$ in $K$, then $u$ is a
decreasing nonnegative solution of \eqref{e1.2}.

\begin{lemma} \label{lem2.1}
The operator $T:K\to K$
is completely continuous.
\end{lemma}

\begin{proof}
Let $u\in K$. Then for $0\leq t\leq\eta$, we obtain
\begin{align*}
(Tu)(t) &= (1-t)\int_0^tsf(s,u(s))ds+\int_t^\eta
\frac{-t^{2}-s^{2}+2s}{2}f(s,u(s))ds\\
&\quad -\int_\eta^1\frac{(1-s)^{2}}{2}f(s,u(s))ds,
\end{align*}
which shows that
\[
(Tu)'(t) =-\Big(\int_0^t
sf(s,u(s))ds+t\int_t^\eta f(s,u(s))ds\Big) \leq0.
\]
For $\eta\leq t\leq1$, we have
\begin{align*}
(Tu)(t) &=(1-t)\int_0^\eta sf(s,u(s))ds
+\int_\eta^t \frac{t^{2}-2st+2s-1}{2}f(s,u(s))ds\\
&\quad-\int_t^1\frac{(1-s)^{2}}{2}f(s,u(s))ds,
\end{align*}
which together with (C1) and (C2) imply that
\[
(Tu)'(t) =-\int_0^\eta
sf(s,u(s))ds+\int_\eta^t(t-s)f(s,u(s))ds
\leq-f(\eta,u(\eta))(\eta-\frac{t}{2})t\leq0.
\]
So, $(Tu)(t)$ is decreasing on $[0,1]$. At the same time, since
$(Tu)(1)=0$, $(Tu)(t)$ is nonnegative on $[0,1]$. This indicates
that $Tu\in K$.

Now, we assume that $D\subset K$ is a bounded set. Then there exists
a constant $M_1>0$ such that $\| u\| \leq M_1$ for any
$u\in D$. In what follows, we will prove that $T(D)$ is relatively
compact. Let
\[
M_2=\sup \{f( t,u):( t,u) \in [0,1] \times[0,M_1]\}.
\]
Then for any $y \in T(D)$, there exists $u \in D$ such that $y =Tu$,
and so,
\begin{align*}
| y(t)|&=| (Tu)(t)| 
=\Big|\int_0^1G(t,s)f(s, u(s))ds\Big|\\
&\leq\int_0^1|G(t,s)|f(s,u(s))ds\\
&\leq M\int_0^1f(s,u(s))ds
\leq MM_2,\quad t\in [0,1],
\end{align*}
which implies that $T(D)$ is uniformly bounded. On the other hand,
when $\varepsilon>0$, if we choose 
$0<\xi<\min\{1-\eta, \frac{\varepsilon}{12M_2(M+1)}\}$, then, for any
 $u \in D$,
\begin{equation}
\int_{\eta-\xi}^{\eta+\xi}f(s,u(s))ds \leq 2M_2\xi<
\frac{\varepsilon}{6(M+1)}. \label{e4.2}
\end{equation}
Since $G(t,s)$ is uniformly continuous on $[0,1]\times[0,\eta-\xi]$
and $[0,1]\times[\eta+\xi,1]$, there exists $\delta>0$ such that for
any $t_{1},t_2\in[0,1]$ with $| t_1-t_2|<\delta$,
\begin{equation}
|G(t_1,s)-G(t_2,s)|<\frac{\varepsilon}{3(M_2+1)(\eta-\xi)},\quad
s\in[0,\eta-\xi]{\label{e4.3}}
\end{equation}
and
\begin{equation}
|G(t_1,s)-G(t_2,s)|<\frac{\varepsilon}{3(M_2+1)(1-\eta-\xi)},\quad
s\in[\eta+\xi,1].{\label{e4.4}}
\end{equation}
In view of \eqref{e4.2}, \eqref{e4.3} and \eqref{e4.4}, for any $y \in
T(D)$ and $t_{1}, t_2\in[0,1]$ with\ $| t_1-t_2|
<\delta $, we have
\begin{align*}
| y(t_1) -y(t_2)|
&= |(Tu)(t_1)-(Tu)(t_2)|  \\
&= |\int_0^1 ( G(t_1,s)-G(t_2,s))f(s,u(s))ds|\\
&\leq \int_0^1| G(t_1,s)-G(t_2,s)|f(s,u(s))ds\\
&= \int_0^{\eta-\xi}|
G(t_1,s)-G(t_2,s)|f(s,u(s))ds\\
&\quad +\int_{\eta-\xi}^{\eta+\xi}|
G(t_1,s)-G(t_2,s)|f(s,u(s))ds\\
&\quad +\int_{\eta+\xi}^1| G(t_1,s)-G(t_2,s)|f(s,u(s))ds\\
&\leq \frac{M_2\varepsilon}{3(M_2+1)}+
\frac{M\varepsilon}{3(M+1)}+\frac{M_2\varepsilon}{3(M_2+1)}
<\varepsilon,
\end{align*}
which implies that $T(D)$ is equicontinuous. By Arzela-Ascoli
theorem, we know that
$T(D)$ is relatively compact. Thus, we have shown that $T$ is a compact operator.

 Finally, we prove that $T$ is continuous. Suppose that
$u_m$ $(m=1,2,\dots)$, $u_{0}\in K$ and 
$\| u_m-u_{0}\| \to 0( m\to \infty ) $. 
Then there exists
$M_3>0$ such that for any $m$, $\| u_m\| \leq M_3$. Let
\[
M_4=\sup \{ f(t,u) :(t,u)\in [0,1] \times[0,M_3] \}.
\]
Then for any $m$ and $t\in [0,1]$, we have
\begin{align*}
G(t,s)f(s,u_m(s))&\leq MM_4,\ s\in [0,1].
\end{align*}
By applying Lebesgue Dominated Convergence theorem, we obtain
\begin{align*}
\lim_{m\to \infty }(Tu_m)(t)
&= \lim_{m\to \infty } \int_0^1G(t,s)f(s,u_m(s))ds\\
&=\int_0^1G(t,s)\lim_{m\to \infty }f(s,u_{m}(s))ds\\
&= \int_0^1G(t,s)f(s,u_{0}(s))ds=(Tu_{0})(t),\ t\in [0,1],
\end{align*}
which indicates that $T$ is continuous.
Therefore, $T:K\to K$ is completely continuous.
\end{proof}

\begin{theorem} \label{thm2.2}
Assume that $f(t,0)\not\equiv 0$ for $t\in[0,1]$ and there exist two positive 
constants $a$ and $b$ such that the
following conditions are satisfied:
\begin{itemize}
\item[(C3)] $f(0,a)\leq 2a$;
\item[(C4)] $b(u_2-u_{1})\leq f(t, u_2)-f(t, u_{1})\leq 2b(u_2-u_{1})$,
$0\leq t\leq 1$, $0\leq u_1\leq u_2\leq a$.
\end{itemize}
If we construct a iterative sequence $v_{n+1}=Tv_n$,
$n=0,1,2,\dots$, where $v_0( t) \equiv 0$ for $t\in [ 0,1]$,
then $\{ v_n\} _{n=1}^\infty$ converges to $v^{*}$ in $E$
and $v^{*}$ is a decreasing positive solution of the BVP
\eqref{e1.2}.
\end{theorem}

\begin{proof} 
Let $K_a=\{ u\in K:\| u\| \leq a\}$.
 Then we may assert that $T:K_a\to K_a$. In fact, if 
$u\in K_a$, then it follows from Lemma \ref{lem2.1} that $Tu \in K$. 
In view of (C3) and $0\leq u(s)\leq a$ for $s\in [0,1]$, we have
\begin{equation*}
0\leq ( Tu) ( t) =\int_0^1G( t,s)
f( s,u( s))ds\leq 2aM\leq a,\ t\in [
0,1],
\end{equation*}
which shows that $\|Tu\|\leq a$. So, $T:K_a\to K_a$.


Now, we prove that $\{ v_n\} _{n=1}^\infty$ converges to
$v^{*}$ in $E$ and $v^{*}$ is a decreasing positive solution of 
\eqref{e1.2}. Indeed, in view of $v_0\in K_a$ and
$T:K_a\to K_a$, we have $v_n\in K_a,\ n=1,2,\dots $.
 Since the set $\{ v_n\} _{n=0}^\infty $ is bounded and $T$ is
completely continuous, we know that the set $\{ v_n\}_{n=1}^\infty $ 
is relatively compact. In what follows, we prove
that $\{ v_n\} _{n=0}^\infty $ is monotone by induction.
First, it is obvious that $v_1 -v_0 =v_1\in K$, which shows that
$v_0\lesssim v_1$. Next, we assume that $v_{k-1}\lesssim v_k$. Then
it follows from (C4) that for $0 \leq t\leq\eta $,
\begin{align*}
&v_{k+1}'( t) -v_k'( t) \\
&=  (Tv_{k})'(t)-(Tv_{k-1})'(t)\\
&= \int_0^1 \frac{\partial G( t,s) }{\partial t} [ f(
s,v_k( s) ) -f(s,v_{k-1}( s) )] ds\\
&= -\big\{\int_0^t s[ f(s,v_k( s) ) -f(s,v_{k-1}( s) )] ds+
t\int_t^\eta [ f(s,v_k( s) ) -f(s,v_{k-1}( s) )] ds\big\} \\
&\leq 0,
\end{align*}
and for $\eta \leq t\leq 1 $,
\begin{align*}
&v_{k+1}'( t) -v_k'( t)\\
&=  (Tv_{k})'(t)-(Tv_{k-1})'(t)\\
&= \int_0^1 \frac{\partial G( t,s) }{\partial t}[ f(
s,v_k( s) ) -f(s,v_{k-1}( s) )] ds\\
&= -\int_0^\eta s[ f(s,v_k( s) ) -f(s,v_{k-1}( s) )] ds+
\int_\eta^t (t-s)[ f(s,v_k( s) ) -f(s,v_{k-1}( s) )] ds \\
&\leq b[ v_k(\eta)-v_{k-1}(\eta) ]\big(t^{2}-2\eta t+\frac{\eta^{2}}{2}\big)
\leq  0,
\end{align*}
hence
\begin{equation}
v_{k+1}'( t) -v_k'( t)\leq 0,\ t\in [ 0,1]; \label{e3.3}
\end{equation}
that is, $v_{k+1}(t) -v_k(t)$ is decreasing on $[0,1]$. At the same
time, it is easy to see that
\[
v_{k+1}(1) -v_k(1) =\int_0^1 G( 1,s) [ f(s,v_k( s)) -
f(s,v_{k-1}( s))] ds= 0,
\]
therefore,
\begin{equation}
v_{k+1}(t) -v_k(t) \geq v_{k+1}(1) -v_k(1)=0,\quad t\in [ 0,1].\label{e3.4}
\end{equation}
It follows from \eqref{e3.3} and \eqref{e3.4} that $v_{k+1}-v_k\in K$,
which indicates that $v_k\lesssim v_{k+1}$. Thus, we have shown that
$v_n \lesssim v_{n+1}$, $n=0,1,2\dots$.
Since $\{ v_n\}_{n=1}^\infty $ is relatively compact and monotone,
there exists a $v^{*}\in K_a$ such that
$\| v_n-v^{*}\| \to 0( n\to \infty ) $, which together with the
continuity of $T$ and the fact that $v_{n+1}=Tv_n$ implies that
$v^{*}=Tv^{*}$. This indicates that $v^{*}$ is a decreasing
nonnegative solution of  \eqref{e1.2}. Moreover, in view of
$f(t,0)\not\equiv0$, $t\in[0,1]$, we know that zero function is not a
solution of  \eqref{e1.2}, which shows that $v^{*}$ is a
positive solution of \eqref{e1.2}.
\end{proof}

\section{An example}

Consider the boundary value problem
\begin{equation}
\begin{gathered}
u'''(t)=\frac{1}{4}u^{2}(t)+u(t)+(1-t),\quad t\in [0,1], \\
u'(0)=u''(\frac{2}{3})=u(1)=0.
\end{gathered}\label{e4.1}
\end{equation}
If we let $\eta=2/3$ and $f(t,u)=\frac{1}{4}u^{2}+u+(1-t)$,
$(t,u)\in [ 0,1] \times [ 0,+\infty )$, then all the hypotheses of
Theorem \ref{thm2.2}
are fulfilled with $a=2$ and $b=1$. It follows from Theorem \ref{thm2.2} that
 \eqref{e4.1} has a decreasing positive solution $v^{*}$.
Moreover, the iterative scheme is
$v_{0}( t) \equiv 0$ for $t\in [ 0,1]$,
\[
v_{n+1}(t) =\begin{cases}
\int_0^t s(1-t)\big[\frac{1}{4}(v_{n}(s))^{2}+v_{n}(s)+(1-s)\big]ds\\
+\int_t^{2/3} \frac{-t^{2}-s^{2}+2s}{2}
\big[\frac{1}{4}(v_{n}(s))^{2}+v_{n}(s)+(1-s)\big]ds\\
-\int_{2/3}^1\frac{(1-s)^{2}}{2}[\frac{1}{4}(v_{n}(s))^{2}
+v_{n}(s)+(1-s)]ds,\\
\quad\text{if } t\in [ 0,2/3],\; n=0,1,2,\dots, 
\\[5pt]
\int_0^{2/3} s(1-t)[\frac{1}{4}(v_{n}(s))^{2}+v_{n}(s)+(1-s)]ds\\
+\int_{2/3}^t \frac{t^{2}-2st+2s-1}{2}[\frac{1}{4}(v_{n}(s))^{2}
 +v_{n}(s)+(1-s)]ds\\
-\int_t^1\frac{(1-s)^{2}}{2}[\frac{1}{4}(v_{n}(s))^{2}+v_{n}(s)+(1-s)]ds,\\
\quad\text{if } t\in [2/3,1],\; n=0,1,2,\dots.
\end{cases}
\]
The first, second, third and fourth terms of the scheme are as
follows:
\begin{gather*}
v_{0}(t)\equiv 0, \quad 
v_1( t) =-\frac 29t^2+\frac 16t^3-\frac
1{24}t^4+\frac 7{72}, 
\\
\begin{aligned}
v_2( t) &=-\frac{18305}{62208}t^2+\frac{26425}{124416}t^3+%
\frac{1915}{62208}t^4-\frac{779}{7776}t^5+\frac{1291}{31104}t^6
 +\frac{23}{62208}t^7\\
&\quad -\frac{95}{13824}t^8+\frac 1{324}t^9 -\frac 5{6912}t^{10}
+\frac 1{13824}t^{11}+\frac{14161}{124416},
\end{aligned} \\
\begin{aligned}
v_3( t) &=-\frac{58154370371}{185752092672}t^2
+\frac{27688166033}{123834728448}t^3+\frac{1099507181}{15479341056}t^4
-\frac{4628786297}{30958682112}t^
5\\
&\quad +\frac{3441194273}{123834728448}t^6
 +\frac{223703119}{3869835264}t^7-\frac{169069657}{3869835264}t^8
 +\frac{1231643521}{371504185344}t^9
\\
&\quad +\frac{89491399}{7739670528}t^{10}
-\frac{1210466063}{185752092672}t^{11}-\frac{824305}{92876046336}t^{12}
+\frac{52150655}{30958682112}t^{13}
\\
&\quad -\frac{153634915}{185752092672}t^{14}
 +\frac{3356471}{92876046336}t^{15}+\frac{18718607}{123834728448}t^{16}\\
&\quad -\frac{15792949}{185752092672}t^{17}
 +\frac{1212829}{61917364224}t^{18}
 +\frac{223969}{123834728448}t^{19}
\\
&\quad -\frac{10753}{3439853568}t^{20}
 +\frac{1087}{859963392}t^{21}
 -\frac{157}{509607936}t^{22}
+\frac{169}{3439853568}t^{23}\\
&\quad -\frac{11}{2293235712}t^{24}
 +\frac 1{4586471424}t^{25}
+\frac{43366416289}{371504185344}.
\end{aligned}
\end{gather*}

\subsection*{Acknowledgments}
This research was supported by grant 1208RJZA240
from the NSF of Gansu Province of China.

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\end{document}