\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 142, 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  (login: ftp)}
\thanks{\copyright 2004 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE-2004/142\hfil Uniqueness of positive solutions]
{Uniqueness of positive solutions for a class of ODE's with
Dirichlet boundary conditions}

\author[Yulian An \& Ruyun Ma\hfil EJDE-2004/142\hfilneg]
{Yulian An, Ruyun Ma} % in alphabetical order


\address{Yulian An \hfill\break
School of Mathematics, Physics \& Software Engineering,
 Lanzhou Transportation University,
  Lanzhou 730070, China}
\email{yulian\_an@tom.com}

\address{Ruyun Ma \hfill\break
Department of Mathematics, Northwest Normal University,
Lanzhou 730070,  China}
\email{mary@nwnu.edu.cn}

\date{}
\thanks{Submitted August 18, 2004. Published November 29, 2004.}
\thanks{Supported by grants: 10271095 form the NSFC, and  GG-110-10736-1003,
 NWNU-KJCXGC-\hfill\break\indent
 212 from the Foundation of Excellent Young
Teacher of the Chinese Education Ministry}
\subjclass[2000]{34B15}
\keywords{Boundary value problem; positive solutions; uniqueness;\hfill\break\indent
shooting method; Sturm comparison theorem}

\begin{abstract}
 We study the uniqueness of positive solutions of the boundary-value
 problem
 \begin{gather*}
 u''+a(t)u'+f(t,u)=0 ,\quad t\in (0,b)\\
 u(0)=0,u(b)=0\,,
 \end{gather*}
 where $0<b<\infty$, $a\in C^1[0,\infty)$ and
 $f\in C^1([0,\infty)\times [0, \infty), [0, \infty))$ satisfy suitable
 conditions.  The proof of our main result is based on the shooting method
 and the Sturm comparison theorem.
\end{abstract}

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

\section{Introduction}

In this paper, we consider the uniqueness of positive solutions for
the problem
\begin{gather}
u''+a(t)u'+f(t,u)=0 ,\quad t\in (0,b) \label{e1.1}\\
u(0)=0,\quad u(b)=0\,, \label{e1.2}
\end{gather}
where $0<b<\infty$, $a\in C^1[0,\infty)$ and
$f\in C^1([0,\infty)\times [0, \infty), [0, \infty))$ satisfy suitable
conditions.

By a {\it positive solution} of \eqref{e1.1}-\eqref{e1.2}, we understand a
function $u(t)$ which is positive in $t\in (0,b)$ and satisfies
the differential equation \eqref{e1.1} and the boundary conditions \eqref{e1.2}.

To study the uniqueness problem of \eqref{e1.1}-\eqref{e1.2}, we will use the
shooting method and the Sturm comparison theorem. Some of the
ideas used in this paper are motivated by Lynn Erbe and Moxun Tang \cite{e1,e2}
and Fu and Lin \cite{f1}.

This paper is organized as follows. In section 2, we
state the main result (see Theorem \ref{thm1} below) and give some examples
to illustrate the applicability of our results. In section 3, we
show some preliminary results. Finally we prove the main result in
Section 4.

 \section{Main Result}

To investigate the uniqueness of \eqref{e1.1}-\eqref{e1.2}, we
introduce the initial-value problem
\begin{gather}
u''+a(t)u'+f(t,u)=0 \label{e2.1} \\
u(0)=0,\quad u'(0)=\alpha>0. \label{e2.2}
\end{gather}
For a given number $\alpha>0$, we know from the assumptions
$a\in C^1[0,\infty)$ and $f\in C^1([0, \infty)\times [0, \infty), [0,
\infty))$ that \eqref{e2.1}-\eqref{e2.2} has a unique solution $u(t,\alpha)$
defined  on $(0,T_{\alpha})$, where $T_{\alpha}$ is either
$+\infty$ or a positive number such that $u$ can not be further
continued to the right of $T_{\alpha}$.
If $\alpha>0$, then $u(0,\alpha)=0,\ u'(0,\alpha)=\alpha>0$.
Therefore, there exists a positive number $\epsilon\in
(0,T_{\alpha}) $ such that
$$
u(t,\alpha)>0,\quad  t\in (0,\ \epsilon).
$$
When $u(t,\alpha)$ vanishes at some $t_0\in (0,T_{\alpha})$, we
define $b(\alpha)$ to be the first zero of $u(t,\alpha)$ in $(0,
T_{\alpha})$. More precisely, $b(\alpha)$ is a function of
$\alpha$ which has the properties
$$
u(t,\alpha)>0,\quad  t\in (0, b(\alpha));\quad u(b(\alpha),\alpha)=0.
$$
If $u(t,\alpha)$ is positive in $(0, T_{\alpha})$, then we define
$b(\alpha)=T_{\alpha}$.  Denote
$$
N:=\{\alpha:\alpha>0,\; b(\alpha)<T_{\alpha}\}.
$$
It is obviously that  problem \eqref{e1.1}-\eqref{e1.2} has no positive
solution if $N$ is an empty set. Hence, we suppose
$N\neq\emptyset$.

Now, for any given $b>0$, if we can prove there exists at most one
$\alpha \in N$ such that $b=b(\alpha)$, then we conclude the
uniqueness of positive solutions of the problem
\eqref{e1.1}-\eqref{e1.2}.

We denote the {\it variation} of $u(t,\alpha)$ by
$$
\phi (t,\alpha)=\partial u(t,\alpha)/\partial \alpha,
 \quad t\in [0, T_{\alpha}).
$$
Then $\phi(t,\alpha)$ satisfies
\begin{gather}
\phi''+a(t)\phi'+f_u(t,u)\phi=0,\quad t\in [0, T_{\alpha}) \label{e2.3} \\
\phi(0)=0,\quad \phi'(0)=1,  \label{e2.4}
\end{gather}
where the notation $f_u(t,u)$ denotes $\partial f(t,u)/\partial u$.
Let $L$ be the linear operator
\begin{equation}
L(\phi)=\phi''+a(t)\phi'+f_u(t,u)\phi,\quad t\in [0, T_{\alpha})
\label{e2.5}
 \end{equation}
 and
 \begin{equation}
G_h(t)=u(t,\alpha)+\frac{h-1}{2}\frac{v(t)}{v'(t)}u'(t,\alpha)
,\quad t\in [0, T_{\alpha})
 \label{e2.6}
\end{equation}
where
$$
v(t)=\int_0^t\exp\big(-\int_0^\tau a(s)ds\big)d\tau %\label{e2.7}
$$
and accordingly
$$
v'(t)= \exp\big(-\int_0^t a(s)ds\big).
$$
It is easy to check that
$$
v''(t)+a(t)v'(t)=0, \quad t\in [0, T_{\alpha}).
$$
A different function $G_h(t)$ has been used by Erbe and Tang \cite{e1}.
However, the function $G_h(t)$ defined by \eqref{e2.6} is first
introduced here.

Differentiating $G_h(t)$ with respect to $t$,  we get
\begin{align*}
G_h'(t)&=u'(t,\alpha)+\frac{h-1}{2}(\frac{v(t)}{v'(t)})'u'(t,\alpha)+
\frac{h-1}{2}\frac{v(t)}{v'(t)}u''(t,\alpha)\\
&=u'(t,\alpha)+\frac{h-1}{2}(\frac{v(t)}{v'(t)})'u'(t,\alpha)\\
&\quad + \frac{h-1}{2}\frac{v(t)}{v'(t)}(-f(t,u(t,\alpha))-a(t)u'(t,\alpha))\\
&=\big(1+\frac{h-1}{2}(\frac{v(t)}{v'(t)})'-\frac{h-1}{2}a(t)\frac{v(t)}{v'(t)}\big)u'(t,\alpha)
-\frac{h-1}{2}\frac{v(t)}{v'(t)}f(t,u(t,\alpha))
\end{align*}
and
\begin{align*}
G_h''(t)
&=[\frac{h-1}{2}(\frac{v}{v'})''-\frac{h-1}{2}a'(t)\frac{v}{v'}
-\frac{h-1}{2}a(t)(\frac{v}{v'})']u'\\
&\quad +(1+\frac{h-1}{2}(\frac{v}{v'})'-\frac{h-1}{2}a(t)\frac{v}{v'})(-f(t,u)-a(t)u')\\
&\quad
-\frac{h-1}{2}(\frac{v}{v'})'f(t,u)-\frac{h-1}{2}\frac{v}{v'}f_u(t,u)u'
-\frac{h-1}{2}\frac{v}{v'}f_t(t,u)\\
&=[-a(t)-\frac{h-1}{2}a(t)(\frac{v}{v'})'+a^2(t)\frac{h-1}{2}\frac{v}{v'}]u'\\
&\quad -[1+(h-1)(\frac{v}{v'})'-\frac{h-1}{2}a(t)\frac{v}{v'}]f(t,u)
        -\frac{h-1}{2}\frac{v}{v'}f_u(t,u)u'\\
&\quad -\frac{h-1}{2}\frac{v}{v'}f_u(t,u).
\end{align*}
where $f_t(t,u)$ denotes $\partial f(t,u)/\partial t$.
So we have
\begin{align*}
L(G_h(t))&= G_h''(t)+a(t)G_h'(t)+f_u(t,u)G_h(t)\\
&= G_h''(t)+a(t)(1+\frac{h-1}{2}(\frac{v}{v'})'-\frac{h-1}{2}a(t)\frac{v}{v'})u'\\
&\quad -\frac{h-1}{2}\frac{v}{v'}a(t)f(t,u)
        +f_u(t,u)u+\frac{h-1}{2}\frac{v}{v'}f_u(t,u)u'\\
&=f_u(t,u)u-[1+(h-1)(\frac{v}{v'})']f(t,u)-\frac{h-1}{2}\frac{v}{v'}f_u(t,u).
\end{align*}
Denote $L_h(t)=L(t,u,\alpha)=L(G_h(t))$, then
$$
L_h(t)=f_u(t,u)u-[1+(h-1)(\frac{v}{v'})']f(t,u)-\frac{h-1}{2}\frac{v}{v'}f_u(t,u).
%\label{e2.8}
$$
Note that when $h=1$, we simply have $G_1(t)=u(t,\alpha),$ and
$L_1(t)=uf_u(t,u)-f(t,u)$.

We will use the following assumptions:
\begin{itemize}
\item[(A1)] $f(t,0)\equiv 0$, and $uf_u(t,u)>f(t,u)>0$ for all $t>0$, $u>0$

\item[(A2)] $a(t)\le 0$, $a(t)v(t)+v'(t)\ge 0$, for all $t\ge 0$

\item[(A3)] If $\alpha>0$ and $h\ge 1$, and there
exists a $\widetilde{t}\in (0, b(\alpha))$, such that
$L_h(\widetilde{t}, \alpha)\ge 0$, then
$L_h(t,\alpha)\ge 0$ for all $t\in [\widetilde{t},\ b(\alpha))$

\item[(A4)] $(v(t)/v'(t))'f(t,u)+(v(t)/2v'(t))f_u(t,u)>0$, for all $t> 0$,
$u>0$.
\end{itemize}

The main result of this paper is as follows.

\begin{theorem} \label{thm1}
Assume (A1)-(A4) hold. Then \eqref{e1.1}-\eqref{e1.2} has at most
one positive solution.
\end{theorem}

\begin{example} \label{ex1} \rm
Let $f(t,u)=u^p$ $(p>1)$, $a(t)\equiv -1$. Then
$$
a(t)v(t)+v'(t)=1>0,\quad  L_h(t)=u^p[p-1-(h-1)e^{-t}].
$$
Obviously, (A1), (A2) are satisfied. Since $e^{-t}$ is strictly
decreasing for $t>0$, (A3) is satisfied. Meanwhile,
$$
(\frac{v}{v'})'f(t,u)+\frac{v}{2v'}f_u(t,u)=e^{-t}u^p>0,\quad
t>0,\; u>0.
$$
(A4) is also satisfied.
\end{example}

\begin{example} \label{ex2} \rm
Let $f(t,u)=t^lu^p$ $(l>0,\,p>1)$ and $a(t)\equiv 0$. Then
$$
a(t)v(t)+v'(t)=1>0,\quad L_h(t)=t^lu^p[p-h-(h-1)l/2]
$$
and
$$
(\frac{v}{v'})'f(t,u)+\frac{v}{2v'}f_u(t,u)=t^lu^p(1+\frac{l}{2t^2})>0,
\quad t>0,\ u>0.
$$
Obviously, (A1)-(A4) are satisfied.
\end{example}

\section{Preliminary Results}

\begin{lemma} \label{lem2}
 Suppose $f(t,0)\equiv 0$ for all $t\ge 0$, and
\begin{equation}
 \phi(b(\alpha),\alpha)\neq 0, \quad \alpha \in N.  \label{e3.1}
\end{equation}
Then one of the following cases must occur
\begin{itemize}
\item[(i)]  $N$ is an open interval

\item[(ii)]  $N=(0, j_1)\cup(j_2, \infty)$ with $0<j_1< j_2<+\infty$.
Moreover, $b'(\alpha)>0$ for all $(0,j_1)$;\ $b'(\alpha)<0$ for
all $(j_2, \infty)$.
\end{itemize}
\end{lemma}

\begin{proof} From the definition of $b(\alpha)$, we have that
 $u'(b(\alpha),\alpha)\le 0$ and for all $\alpha\in N$,
\begin{equation}
u(b(\alpha),\alpha)=0.  \label{e3.2}
\end{equation}
 If $u'(b(\alpha),\alpha)=0$, then \eqref{e3.2} with the
assumption $f(t,0)\equiv 0$ for $t\geq 0$ imply
$$
u(t, \alpha)\equiv 0, \quad t>0
$$
However this contradicts the fact $u'(0,\alpha)=\alpha>0$.
Therefore, we must have
\begin{equation}
u'(b(\alpha),\alpha)<0.  \label{e3.3}
\end{equation}
By the Implicit Function Theorem,  $b(\alpha)$ is well-defined as a
function of $\alpha$ in $N$ and $b(\alpha)\in C^1(N)$.
Furthermore, it follows from \eqref{e3.3} that $N$ is an open set.

Differentiating both sides of the identity \eqref{e3.2} with respect to
$\alpha$, we obtain
\begin{equation}
u'(b(\alpha),\alpha)b'(\alpha) +\phi(b(\alpha),\alpha)=0. \label{e3.4}
\end{equation}
Combining this with \eqref{e3.1}, it follows that
$b'(\alpha)\neq 0$.

We note that if $\bar \alpha\in (0,\infty)\setminus N$ with
$\{\alpha_n\}\subset N$ and $\alpha_n\to\bar\alpha$ as
$n\to\infty$, then
$$
b(\alpha_n)\to+\infty.
$$
Otherwise, on the contrary , we may suppose that $b(\alpha_n)\to
t_1$ as $n\to\infty$. Let $n\to \infty$,  we have from
$u(b(\alpha_n),\alpha_n)=0$ that $u(t_1,\bar \alpha)=0$. However
this contradicts $\bar\alpha\notin  N$.

If $N$ be not an open interval, and let $J_1=(j_0,j_1)$ and
$J_2=(j_2,j_3)$ be two distinct components of $N$ with $0<j_1<
j_2<\infty$. Then
$$
\lim_{\alpha\to j_1^-}b(\alpha) =\lim_{\alpha\to
j_2^+}b(\alpha)=+\infty
$$
Since $b(\alpha)$ is strictly monotonic in each component of $N$,
we conclude that $b'(\alpha)>0$ in $J_1$, and $b'(\alpha)<0$ in
$J_2$. Meanwhile
$$
\lim_{\alpha\to j_0^+}b(\alpha)<+\infty,\quad
\lim_{\alpha\to j_3^-}b(\alpha)<+\infty
$$
It follows that $j_0=0$ and $j_3=+\infty$. Therefore
$N=(0,j_1)\cup(j_2,\infty)$ with $b'(\alpha)>0$ in $(0,j_1)$, and
$b'(\alpha)<0$ in $(j_2,\infty)$. The proof is completed.
\end{proof}

\begin{lemma} \label{lem3}
Let $\alpha\in N $, and let $f(t,u)$ satisfy (A1).
Then $\phi(t,\alpha)$ has at least one zero in $(0,b(\alpha))$.
\end{lemma}

\begin{proof}  Note that $L(\phi)=0$, i.e.,
\begin{equation}
\phi''+a(t)\phi'+f_u(t,u)\phi=0 \label{e3.5}
\end{equation}
Meanwhile,
\begin{equation}
G_h''(t)+a(t)G_h'(t)+f_u(t,u)G_h(t)=L_h(t) \label{e3.6}
\end{equation}
Multiply both sides of \eqref{e3.6} by $\exp\big(\int_0^t
a(s)ds\big)\phi(t,\alpha)$, and multiply both sides of
\eqref{e3.5} by $\exp\big(\int_0^t a(s)ds\big)G_h(t)$, then
subtract the resulting identities, we have
\begin{equation}
\big[\exp\big(\int_0^t a(s)ds\big)(G_h'\phi-G_h\phi')\big]' =
\exp\big(\int_0^t a(s)ds\big)\phi(t,\alpha)L_h(t). \label{e3.7}
\end{equation}
Set $h=1$ in \eqref{e3.7}, we get
\begin{equation}
\big[\exp\big(\int_0^t a(s)ds\big)(u'\phi-u\phi')\big]'=\exp\big(\int_0^t
a(s)ds\big)\phi(t,\alpha)(f_u(t,u)u-f(t,u))
 \label{e3.8}
\end{equation}
Suppose on the contrary that $\phi(t,\alpha)$ does not vanish in
$(0,b(\alpha))$. Then we know from \eqref{e2.4} that
$$
\phi(t,\alpha)>0, \quad t\in (0, b(\alpha)).
$$
This implies that the right hand side of \eqref{e3.8} is positive in
$(0,b(\alpha))$, and accordingly
\begin{equation}
\big[\exp\big(\int_0^t a(s)ds\big)(u'\phi-u\phi')\big]'>0,
 \quad t\in (0,b(\alpha)). \label{e3.9}
\end{equation}
Since
$$
\exp\big(\int_0^t a(s)ds\big)(u'\phi-u\phi')\Big|_{t=0}=0
$$
we have form \eqref{e3.9} that
\begin{equation}
\exp\big(\int_0^t a(s)ds\big)(u'\phi-u\phi')\Big|_{t=b(\alpha)}>0.
\label{e3.10}
\end{equation}
On the other hand
$$
\exp\big(\int_0^t a(s)ds\big)(u'\phi-u\phi')\big|\,_{t=b(\alpha)}
=\exp\big(\int_0^{b(\alpha)}
a(s)ds\big)u'(b(\alpha),\alpha)\phi(b(\alpha),\alpha)\le 0
$$
which contradicts \eqref{e3.10}.  Therefore, $\phi(t,\alpha)$ has
at least one zero in $(0, b(\alpha))$.
\end{proof}

\section{Proof of Theorem \ref{thm1}}

We note that for any $\alpha \in N$, \eqref{e3.4} holds. If we can show
that
\begin{equation}
\phi(b(\alpha),\alpha) <0,\quad \alpha \in N \label{e4.1}
\end{equation}
then from \eqref{e4.1}, \eqref{e3.3} and \eqref{e3.4}, it follows that
$b'(\alpha)<0$.
Combining this with Lemma \ref{lem2}, we conclude that $N$ is an open
interval and $ b'(\alpha)<0$ for all $\alpha\in N$. So $b(\alpha)$
is a strictly decreasing function in $N$. Thus, for any given
$b>0$, there is at most one $\alpha \in N$ such that
$b(\alpha)=b$. If such $\alpha $ exists exactly, then
$u(t,\alpha)$, which is the unique solution of the initial value
problem \eqref{e2.1}-\eqref{e2.2},  must be the positive solution of boundary
value problem \eqref{e1.1}-\eqref{e1.2}. Combining this with the uniqueness of
solution of the initial value problem, we know that the positive
solution of \eqref{e1.1}-\eqref{e1.2} is unique.

\begin{proof}[Proof of Theorem \ref{thm1}]
We need only to prove \eqref{e4.1}. To this end, we divide the
proof into six steps.


\noindent{\it Step 1.}\  We show that there exists unique
$c(\alpha)\in (0,b(\alpha))$ such that
\begin{equation}
u'(c(\alpha),\alpha)=0 \label{e4.2}
\end{equation}
and
\begin{equation}
u'(t,\alpha)>0  \mbox{ on } [0, c(\alpha));\quad  u'(t,\alpha)<0
 \mbox{ on } (c(\alpha),b(\alpha)] \label{e4.3}
\end{equation}
  In fact, if $\tau\in (0, b(\alpha))$ such that
$u'(\tau,\alpha)=0$. Then from \eqref{e2.1} and (A1), we have that
$$
u''(\tau)=f(\tau, u(\tau, \alpha))<0
$$
which means that $\tau$ is a local maximum of $u(t,\alpha)$.
Combining this with the fact $u(0, \alpha)=u(b(\alpha),\alpha)=0$,
it concludes  that there exists unique $c(\alpha)\in (0,
b(\alpha))$ such that \eqref{e4.2} and \eqref{e4.3} hold.

\noindent{\it Step 2.} We show that
\begin{equation}
\phi(t,\alpha)> 0,\quad t \in (0, c(\alpha)]. \label{e4.4}
\end{equation}
From the facts that $u(t,\alpha)>0$ and $u'(t,\alpha)>0$ on $(0,
c(\alpha))$, we have that
$$
G_h(t)>0,\quad t\in (0, c(\alpha)] % \label{e4.5}
$$
whenever  $h\ge 1$. Since
\begin{align*}
L_h(c(\alpha))
&=\big[f_u(t,u)u-\big(1+({h}-1)(\frac{v}{v'})'\big)f(t,u)
-\frac{h-1}{2}\frac{v}{v'}f_u(t,u)\big]_{t=c(\alpha)}\\
&=\big[f_u(t,u)u-\big(1-(\frac{v}{v'})'\big)f(t,u)+\frac{v}{2v'}f_u(t,u)\\
&\quad -h\ \big((\frac{v}{v'})'f(t,u)
+\frac{v}{2v'}f_u(t,u)\big)\big]_{t=c(\alpha)}\\
\end{align*}
Combing it with (A4), we can choose $\bar{h}\in (1, \infty)$ so
large such that
$$
L_{\bar{h}}(c(\alpha))<0 %\label{e4.6}
$$
which together with (A3) implies that
\begin{equation}
L_{\bar{h}}(t)<0,\quad t\in (0, c(\alpha)]. \label{e4.7}
\end{equation}
Suppose on the contrary that \eqref{e4.4} is not true, and let $t_2$ be
the first zero of $\phi(t, \alpha)$ in $(0,c(\alpha)]$. Then
\begin{equation}
\phi(t, \alpha)>0\ \mbox{on}\ t\in (0, t_2), \ \ \phi(t_2,
\alpha)=0. \label{e4.8}
\end{equation}
Note that  $\phi(t,\alpha)$ and $G_{\bar h}(t)$ satisfy
$$
\big[\exp\big(\int_0^t a(s)ds\big)(G'_{\bar h}\phi-G_{\bar h}\phi')\big]'
=\exp\big(\int_0^t a(s)ds\big)\phi(t,\alpha)L_{\bar h}(t) %\eqno\eqref{e4.9}
$$
which together with \eqref{e4.7}, \eqref{e4.8} imply
\begin{equation}
\big[\exp\big(\int_0^t a(s)ds\big)(G'_{\bar h}\phi-G_{\bar
h}\phi')\big]'<0, \quad t\in (0, t_2) \label{e4.10}
\end{equation}
Since $G_{\bar h}(0)=u(0,\alpha)=0$, it follows that
$$
\big[\exp\big(\int_0^t a(s)ds\big)(G'_{\bar h}\phi-G_{\bar
h}\phi')\big]_{t=0}=0 %\label{e4.11}
$$
which and \eqref{e4.10} yield
\begin{equation}
\big[\exp\big(\int_0^t a(s)ds\big)(G'_{\bar h}\phi-G_{\bar
h}\phi')\big]_{t=t_2}<0. \label{e4.12}
\end{equation}
On the other hand, since $\phi'(t_2)\leq 0$ and $G_{\bar
h}(t_2)>0$, we have
\begin{align*}
 &\big[\exp\big(\int_0^t a(s)ds\big)(G'_{\bar h}\phi-G_{\bar
   h}\phi')\big]_{t=t_2}\\
&= \exp\big(\int_0^{t_2} a(s)ds\big)(-G_{\bar h}(t_2)\phi'(t_2))
 \geq 0
\end{align*}% \label{e4.13}
This contradicts \eqref{e4.12}. Therefore \eqref{e4.4} holds.

\noindent{\it Step 3.}
 We show that if $h>1$ then $G_h(t)$ has exactly
one zero $\tau_h$ in $(c(\alpha), b(\alpha))$.
  If $h>1$, we have from the definition of $G_h(t)$ that
$$
G_h(c(\alpha))=u(c(\alpha))>0,\quad G_h(b(\alpha))<0 %\label{e4.14}
$$
which implies that  $G_h(t)$ with $h>1$ must have zeros in
$(c(\alpha), b(\alpha))$.

 Next we show that $G_h(t)$ with $h>1$ has at most one zero in $(c(\alpha),
b(\alpha))$.
For any given $h>1$, let $G_h(t)=0$ for some $t\in (c(\alpha),
b(\alpha))$. Then
$$
u(t,\alpha)+\frac{h-1}{2}\frac{v(t)}{v'(t)}u'(t,\alpha)=0
$$
and consequently
$$
\frac{u'(t)}{u(t)}= \frac{2}{1-h}\frac{v'(t)}{v(t)}. % \label{e4.15}
$$
Set
$$
w_1(s)=\frac{u'(s)}{u(s)},\quad s\in (c(\alpha),b(\alpha))
$$
and
$$
w_2(s)=\frac{2}{1-h}\frac{v'(s)}{v(s)}, \quad  s\in (c(\alpha),
b(\alpha))
$$
By (A2), we have
\begin{gather*}
w_1'(s)=\frac{-a(s)u'u-f(s,u)u-u'^2}{u^2}<0,\quad  s\in (c(\alpha),\ b(\alpha)) \\
w_2'(s)=\frac{2}{1-h}\frac{-v'(a(s)v+v')}{v^2}\ge 0,\quad
 s\in (c(\alpha),\ b(\alpha)).
\end{gather*}
Hence, $w_1(s)$ and $w_2(s)$ intersect at most once in
$(c(\alpha), b(\alpha))$, and accordingly $G_h(t)$ $(h>1)$ has at
most one zero in $(c(\alpha), b(\alpha))$.

\noindent{\it Step 4.}\
 Let $\theta(\alpha)$ be the first zero of
$\phi(t,\alpha)$ in $(c(\alpha), b(\alpha))$. We show that there
exists an unique $p\in (1, \infty)$ such that the unique zero,
$\tau_p$, of $G_p(t)$ in $(c(\alpha), b(\alpha)]$ satisfying
\begin{gather}
\tau_p=\theta(\alpha) \label{e4.16} \\
G_p(t)>0 \mbox{ on } (0, \theta(\alpha)),\quad
G_p(\theta(\alpha))=0,\quad   G_p(t)<0 \mbox{ on } (\theta(\alpha),
b(\alpha)], \label{e4.17} \\
\phi(t,\alpha)>0\mbox{ on }(0, \theta(\alpha)), \quad
\phi(\theta(\alpha),\alpha)=0. \label{e4.18}
\end{gather}
 Note that for $t\in (c(\alpha), b(\alpha))$, $u'(t,\alpha)<0$.
 From the definition of $G_h(t)$, we
have that for any fixed $\tau \in (c(\alpha), b(\alpha))$,
$G_h(\tau)$ is continuous and strictly decreasing with respect to
$h$. Since $G_1(\tau)=u(\tau,\alpha)>0$ and $\lim_{h\to
+\infty}G_h(\tau)=-\infty$, there must be a unique $h>1$ such that
$$
G_{h}(\tau)=0.
$$
In particular, for $\theta(\alpha)\in (c(\alpha), b(\alpha))$, there exists a unique
number $p\in (1, \infty)$ such that
$$
G_p(\theta(\alpha))=0
$$
i.e.
$\tau_p=\theta(\alpha)$.

Equations \eqref{e4.17} and \eqref{e4.18} can be deduced from the fact that
both $G_p$ has
unique zero in $(c(\alpha), b(\alpha))$ and $\theta(\alpha)$ is
the first zero of $\phi(t, \alpha)$ in $(c(\alpha), b(\alpha))$.

\noindent{\it Step 5.}\
 We show that there exists $t_3\in (0,\theta(\alpha)]$ such that
\begin{equation}
L_p(t_3)\ge 0. \label{e4.19}
\end{equation}
Suppose on the contrary that $L_p(t)< 0$ on $(0, \theta(\alpha)]$.
Note that $\phi(t,\alpha)$ and $G_p(t)$ satisfy
\begin{equation}
\big[\exp\big(\int_0^t a(s)ds\big)(G_p'\phi-G_p\phi')\big]'
=\phi(t,\alpha)L_p(t)\exp\big(\int_0^t a(s)ds\big). \label{e4.20}
\end{equation}
Since the right-hand side of \eqref{e4.20} is negative on $(0,\theta(\alpha))$,
$$
\big[\exp\big(\int_0^t
    a(s)ds\big)(G_p'\phi-G_p\phi')\big]'<0, \quad t\in (0, \theta(\alpha))
%\label{e4.21}
$$
This and
$$
\exp\big(\int_0^t a(s)ds\big)(G_p'\phi-G_p\phi')\Big|_{t=0}=0
$$
imply that
\begin{equation}
\exp\big(\int_0^t
a(s)ds\big)(G_p'\phi-G_p\phi')\Big|_{t=\theta(\alpha)}<0
\label{e4.22}
\end{equation}
On the other hand, we have from \eqref{e4.16} and the definitions of $p$
and $\theta(\alpha)$ that
$$
\exp\big(\int_0^t a(s)ds\big)(G_p'\phi-G_p\phi')\Big|_{t=\theta(\alpha)}= 0
$$
This contradicts \eqref{e4.22}. Therefore, \eqref{e4.19} holds for some
$t_3\in (0, \theta(\alpha)]$.

\noindent{\it Step 6.}\  We show that $\theta(\alpha)$ is the unique zero
of $\phi(t, \alpha)$ in $(c(\alpha), b(\alpha)]$.
Suppose on the contrary that there exists $\tau_1\in
(\theta(\alpha), b(\alpha)]$ such that
$$
\phi(\tau_1, \alpha)=0, \quad \phi(t, \alpha)<0\quad
\mbox{on } (\theta(\alpha), \tau_1)
$$
and
\begin{equation}
\phi'(\tau_1, \alpha)>0. \label{e4.23}
\end{equation}
(Note that if $\phi'(\tau_1, \alpha)=0$, then $\phi\equiv 0$. This
contradicts \eqref{e2.4})
We have from (A3) and \eqref{e4.19} that
$$
L_p(t)\ge  0,\quad t\in [\theta(\alpha), b(\alpha)). %\label{e4.24}
$$
Integrating both sides of \eqref{e4.20} from $\theta(\alpha)$ to
$\tau_1$, we get
\begin{equation}
 -\exp\big(\int_0^{\tau_1} a(s)ds\big)G_p(\tau_1)\phi'(\tau_1,\alpha)
=\int_{\theta(\alpha)}^{\tau_1}\exp\big(\int_0^t
a(s)ds\big)\phi(t,\alpha)L_p(t)dt. %\label{e4.25}
\end{equation}
This together with the fact $\phi(t,\alpha)<0$ on
$(\theta(\alpha), \tau_1) $ implies that
\begin{equation}
-\exp\big(\int_0^{\tau_1}
a(s)ds\big)G_p(\tau_1(\alpha))\phi'(\tau_1,\alpha)\leq 0. \label{e4.26}
\end{equation}
On the other hand, we have from the fact $G_p(\tau_1)<0$ and
\eqref{e4.23} that
$$
-\exp\big(\int_0^{\tau_1} a(s)ds\big)G_p(\tau_1)\phi'(\tau_1,\alpha)> 0.
$$
This contradicts \eqref{e4.26}.
 Therefore, $\phi(t,\alpha)$ has not zero point in
$(\theta(\alpha), b(\alpha)]$, and consequently
$\phi(b(\alpha), \alpha)<0$.
\end{proof}

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\end{document}