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\markboth{ A singular boundary-value problem }%
{ Robert M. Houck \& Stephen B. Robinson }
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\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
Fourth Mississippi State Conference on Differential Equations and\newline 
Computational Simulations, 
Electronic Journal of Differential Equations, \newline
Conference 03, 1999, pp 75--90. \newline
http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu or ejde.math.unt.edu (login: ftp)}
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  A singular nonlinear boundary-value problem 
\thanks{ {\em Mathematics Subject Classifications:} 34B15.
\hfil\break\indent
{\em Key words:} singular nonlinear boundary value problems, 
 existence and uniqueness, \hfil\break\indent 
asymptotic estimates.
\hfil\break\indent
\copyright 2000 Southwest Texas State University  and University of
North Texas. \hfil\break\indent
Published July 10, 2000. } }

\date{}
\author{ Robert M. Houck \& Stephen B. Robinson }
\maketitle

\begin{abstract} 
In this paper we prove an existence and uniqueness theorem for the 
singular nonlinear boundary-value problem 
$$\displaylines{
(|y'(t)|^py'(t))'+\frac{\phi}{y^{\lambda}(t)}=0 \mbox{ in } (0,1),\cr
y(0)=0=y(1),
}$$ 
where $p\geq 0$, $\lambda>0$ and $\phi$ is a positive function in
$L^1_{{\rm loc}}(0,1)$. 
Moreover, we derive asymptotic estimates describing the 
behavior of the solution and its derivative at the boundary. 
\end{abstract}

\newtheorem{lemma}{Lemma}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}{Corollary}

\section{Introduction}
In this paper we study the singular nonlinear boundary value 
problem 
\begin{equation}\begin{array}{c}
\displaystyle (|y'(t)|^py'(t))'+\frac{\phi}{y^{\lambda}}=0 
\quad \mbox{in } (0,1), \\[7pt] 
\displaystyle y(0)=y(1)=0\,, 
\end{array} \label{bvp}
\end{equation}
where we assume throughout that $p\geq 0$, $\lambda>0$ and 
$\phi$ is a positive function in $L^1_{{\rm loc}}(0,1)$. 
Boundary value problems such 
as (\ref{bvp}) occur in a wide variety of applications. For 
example, see \cite{CF}, \cite{CN} and \cite{S} for applications  
to fluid dynamics. 

The primary motivation for our work comes from \cite{T}, in which 
Taliaferro studied (\ref{bvp}) for the case where $p=0$ and 
$\phi\in C(0,1)$. Taliaferro showed that (\ref{bvp}) has a smooth 
positive solution iff 
\[ \int_0^1 t(1-t)\phi(t)\, dt<\infty,\]
and that the given solution is unique. Taliaferro also showed that 
$ \lim_{t\to 0^+}y'(t)$ exists and is finite 
iff 
\[ \int_0^{1/2}\frac{\phi(t)}{t^{\lambda}}\, dt<\infty, \]
where a similar condition applies at $t=1$. Moreover, Taliaferro 
derived asymptotic formulae describing  the behavior of the 
solution near the boundary for both the finite and infinite slope 
cases. Generalizations of Taliaferro's paper include Gatica, 
Oliker, and Waltman \cite{GOW}, Gatica, Hernandez, and Waltman 
\cite{GHW}, Baxley \cite{B}, and Baxley and Martin \cite{BM}. The 
papers \cite{GOW}, \cite{GHW} and \cite{B} study the case $p=0$, 
and the work in \cite{BM} is related to the case $-2\leq p<0$. 

In this paper we generalize Taliaferro's work to the case $p\geq 
0$. This generalization allows us to study fluids with velocity 
dependent diffusion, and it forces us to confront the inevitable 
difficulties that follow when a well understood linear 
differential operator is replaced by a nonlinear differential 
operator. 

One example of the differences between the cases $p=0$ and $p>0$ 
is that in the latter case the problem has an additional 
singularity where $y'(t)=0$. Moreover, this singularity cannot be 
avoided, because a positive solution of (\ref{bvp}) must be 
concave down  and must be $0$ at the boundaries, and therefore 
must have a unique critical point where it achieves a maximum. To 
get a different point of view on this singularity assume that $y$ 
is twice differentiable and rewrite (\ref{bvp}) as 
$$\displaylines{
y''(t)+\frac{\phi}{(p+1)|y'|^p y^{\lambda}}=0 
\quad \mbox{ in } (0,1), \cr 
y(0)=0=y(1), 
}$$ 
which is similar in form to the problem studied in \cite{BM}. From 
this representation of the problem it is clear that a solution 
cannot be twice differentiable at its critical point. 

In Section 2 we prove that (\ref{bvp}) has a positive solution iff 
$\phi\in L^1_{{\rm loc}}(0,1)$ such that 
\[
\int_0^1 \big(\int_{t}^{1/2}\phi \, ds \big)^{\frac{1}{p+1}}\,dt<\infty, 
\]
and that this solution is unique.  (This is equivalent to 
Taliaferro's condition when $p=0 $.) The argument begins by 
examining initial value problems that start in the interior of the 
interval and then ``shoot" towards the boundary. Of particular 
interest are the solutions whose initial slopes are $0$. Thus we 
shoot from the singularity mentioned above, and it is important to 
check that suitable existence, uniqueness, comparison, and 
continuous dependence theorems are still available. We prove these 
lemmas and then show that for each $T\in (0,1)$ there exist left 
and right {\it half solutions} $y_0$ and $y_1$ defined on $[0,T]$ 
and $[T,1]$, respectively, such that $y_0(0)=0=y_1(1)$ and 
$y_0'(T)=0=y_1'(T)$. Finally, we show that the parameter $T$ can 
be adjusted until $y_0(T)=y_1(T)$, thus bringing the half 
solutions together to create the unique solution to (\ref{bvp}). 


In Section 3 we examine the boundary behavior of the solution. We 
prove that the solution has finite slope at $t=0$ iff 
\[ \int_0^{1/2}\frac{\phi(t)}{t^{\lambda}}\, dt<\infty, \]
just as in \cite{T}, where a similar condition applies at $t=1$. 
For the finite slope case we quickly obtain the asymptotic 
formulae 
\[ 
y'(t) = \left ( \frac{1}{A}\right )^{\frac{\lambda}{p+1}}\Big( 
\int _t^T\frac{\phi}{s^{\lambda}}(1+o(1))\, ds \Big)^{\frac{1}{p+1}}, 
\]
and
\[
y(t)=\left ( \frac{1}{A}\right )^{\frac{\lambda}{p+1}}\int _0^t 
\Big( \int _t^T\frac{\phi}{s^{\lambda}}(1+o(1))\, ds \Big)^{\frac{1}{p+1}}, 
\]
which are similar to those in \cite{T}. For the remainder of 
Section 3 we concentrate on the infinite slope case. Using 
comparison arguments we prove that if $\psi$ is positive and 
locally integrable such that $ \lim_{t\to 
0^+}\frac{\psi (t)}{\phi (t)}=1$ and if $z(t)$ is any solution of 
\begin{equation}
\begin{array}{c}
\displaystyle (|z'(t)|^pz'(t))'+\frac{\psi (t)}{z^{\lambda}(t)}=0 
\mbox{ in } (0,\delta), \\[7pt] z(0)=0, 
\end{array}
\label{compbvp}
\end{equation}
for some $\delta>0$, then $ \lim_{t\to 
0^+}\frac{z'(t)}{y'(t)}=\lim_{t\to 
0^+}\frac{z(t)}{y(t)}=1$. We apply this general result to the 
special case where $\phi$ behaves like a power of $t$. In this 
case we see that if $\phi$ is asymptotically comparable to $ct^r$, 
then $y(t)$ is asymptotically comparable to $\gamma t^{\rho}$ for 
appropriate $\gamma$ and $\rho$ . These results complement those 
in \cite{T} and \cite{BM}. 

\section{Existence and Uniqueness}

In this section we prove
\begin{theorem}
The boundary-value problem (\ref{bvp}) has a positive solution iff
\[
\int_0^1 \Big(\int_{t}^{1/2}\phi \, ds \Big)^{\frac{1}{p+1}}\, dt<\infty\,. 
\]
Moreover, this solution is unique. \label{bvpex} 
\end{theorem}

 Let $T\in (0,1)$ and consider the initial value problem
\begin{equation}
\begin{array}{c}
\displaystyle (|y'(t)|^py'(t))'+\frac{\phi}{y^{\lambda}}=0 , \\[7pt]
y(T)=h,\, y'(T)=k,
\end{array}
\label{ivp}
\end{equation}
where $h>0$ and $k\geq 0$. We begin this section by investigating 
the positive solutions of (\ref{ivp}) on intervals whose right 
endpoint is $T$. 

Before proceeding it is helpful to rewrite (\ref{ivp}) as an 
equivalent integral equation. Assuming that $y$ is a smooth 
positive solution of (\ref{ivp}) on the interval $(a,T]$, we 
integrate once to get 
\[
|y'(t)|^py'(t)=k^{p+1}+\int_t^T\frac{\phi(s)}{y^{\lambda}(s)}\, 
ds,\, t\in (a,T]. 
\]
Clearly, $y'$ is nonnegative, so
\[
y'(t)=\Big( k^{p+1}+\int_t^T\frac{\phi}{y^{\lambda}}\, ds
\Big)^{\frac{1}{p+1}},\, t\in (a,T]. 
\]
A second integration yields the equation
\begin{equation}
y(t)=h-\int_t^T\Big(k^{p+1}+\int_{\tau}^T\frac{\phi}{y^{\lambda}}\,
 ds\Big)^{\frac{1}{p+1}}\, d\tau,\, t\in (a,T]. \label{inteq} 
\end{equation}
Our proofs will be based upon this representation of the problem. 
Several straightforward comments regarding (\ref{inteq}) are 
collected in the following lemma. 
\begin{lemma}
Suppose that $y\in C(a,T]$ is a positive function satisfying 
(\ref{inteq}). Then $y$ is increasing and concave down, $y\in 
C[a,T]\bigcap C^1(a,T]$, $|y'|^py'=(y')^{p+1}$ is differentiable 
a.e., and $y$ satisfies (\ref{ivp}) . \label{lem1} 
\end{lemma}
Thus for any positive $y\in C(a,T]$ we define
\[ Fy(t)=
h-\int_t^T\Big(k^{p+1}+\int_{\tau}^T\frac{\phi}{y^{\lambda}}\, 
ds\Big)^{\frac{1}{p+1}}\, d\tau, 
\]
and it is clear that positive solutions of (\ref{ivp}) correspond 
to fixed points of $F$. 


\begin{theorem} Suppose that $T>0$, $h>0$, and $k\geq 0$. Then there exists an $a \in [0,T)$ and a unique positive $y\in C[a,T]\bigcup C^1(a,T]$ such that $y$ is a solution of (\ref{ivp}) on $[a,T]$. Moreover we may assume that the interval $[a,T]$ is maximal in the sense that either $a=0$ or $y(a)=0$. 
\end{theorem}

\paragraph{Proof:}
We begin by proving existence and uniqueness over some interval 
$[a_0,T]$. Our primary tool will be the Contraction Mapping 
Theorem. The estimates below will be useful in later arguments, so 
they are proved in slightly greater generality than is needed for 
this theorem. Suppose that $y\in C[a_0,T]$ such that $0<m \leq 
y(t)$, where $a_0$ is to be determined. Then, for $t\in(a_0,T)$, 
we have 
\[
 0 \leq (Fy)'-k  =  \Big( k^{p+1}+\int_t^T\frac{\phi}{y^{\lambda}}\, ds
\Big)^{\frac{1}{p+1}}-\left ( k^{p+1}\right )^{\frac{1}{p+1}}
 \leq \Big( \int_t^T\frac{\phi}{y^{\lambda}}\, ds\Big)^\frac{1}{p+1},
\]
since $ 
|b^{\frac{1}{p+1}}-a^{\frac{1}{p+1}}|\leq|b-a|^{\frac{1}{p+1}}$ 
for any $a,b\geq 0$. Therefore 
\[
k\leq (Fy)' \leq k+\left (\frac{1}{m}\right 
)^{\frac{\lambda}{p+1}}\Big(\int_t^T\phi \, ds\Big)^{\frac{1}{p+1}}. 
\]
Integrating from $t$ to $T$ yields
\begin{equation}
k(T-t)\leq  h-Fy \leq k(T-t) +\left (\frac{1}{m}\right 
)^{\frac{\lambda}{p+1}}\int_t^T\Big(\int_{\tau}^T\phi \, 
ds\Big)^{\frac{1}{p+1}}\, d\tau. \label{est1} 
\end{equation}
 Recall that $\phi\in L^1_{{\rm loc}}(0,1)$.  Thus, given any $\epsilon\in (0,h)$, we can choose $a_0$ close enough to $T$ so that  
\[
k(T-t)\leq  h-Fy \leq k(T-t) +\frac{\epsilon}{2}.
\]
 Further restricting $a_0$ so that $|a_0-T|<\frac{\epsilon}{2k}$ we  have $|Fy(t)-h|\leq \epsilon$ for all $t\in [a_0,T]$. In particular, for $m=h-\epsilon$, and an appropriate choice of $a_0$, we get $F:{\overline B}_{\epsilon}(h)\to {\overline B}_{\epsilon}(h)$, where ${\overline B}_{\epsilon}(h):=\{y\in C[a_0,T]: |y(t)-h|\leq \epsilon\; \forall t\in [a_0,T]\}$.

Now we show that, given a further restriction on $a_0$, $F$ is a 
contraction on ${\overline B}_{\epsilon}(h)$. Assume that 
$y_1,y_2\in C[a_0,T]$ such that $0<m\leq y_1(t),y_2(t)\leq M< 
\infty$ for all $t\in [a_0,T]$. An application of the Mean Value 
Theorem implies 
\begin{eqnarray*}
 \big|(Fy_1)'-(Fy_2)'\big| & = & 
\Big| \Big( k^{p+1}+\int_t^T\frac{\phi}{y_1^{\lambda}}\, ds 
\Big)^{\frac{1}{p+1}} - \Big( 
k^{p+1}+\int_t^T\frac{\phi}{y_2^{\lambda}}\, ds 
\Big)^{\frac{1}{p+1}} \Big|. \\ 
 & \leq &  \left (\frac{1}{p+1}\right )(c(t))^{\frac{1}{p+1}-1}
\Big(\int_t^T \phi \big|\frac{1}{y_1^{\lambda}}-\frac{1}{y_2^{\lambda}}
\big| \, ds \Big) ,
\end{eqnarray*}
where
\[ k^{p+1}+\int_t^T\frac{\phi}{y_i^{\lambda}}\, ds\leq c(t) \leq  k^{p+1}+\int_t^T\frac{\phi}{y_j^{\lambda}}\, ds,
\]
and either $i=1$ and $j=2$ or $i=2$ and $j=1$. In either case we 
see that 
\[
c(t)\geq \left (\frac{1}{M}\right )^{\lambda}\int_t^T\phi\, ds .
\]
Substituting the minimal value for $c(t)$ leads to
\begin{eqnarray*}
\lefteqn{\left |(Fy_1)'-(Fy_2)'\right |}\\
& \leq& \left(\frac{1}{p+1}\right 
)\left (\frac{1}{M}\right 
)^{\frac{\lambda}{p+1}-\lambda}\Big(\int_t^T \phi \, 
ds\Big)^{\frac{1}{p+1}-1}\Big(\int_t^T \phi \left 
|\frac{1}{y_1^{\lambda}}-\frac{1}{y_2^{\lambda}}\right | \, ds 
\Big) , 
\end{eqnarray*}
Apply the Mean Value Theorem a second time to get
\[
\left |\frac{1}{y_1^{\lambda}(t)}-\frac{1}{y_2^{\lambda}(t)}\right 
|\leq \lambda (c(t))^{-\lambda-1}|y_1(t)-y_2(t)|, 
\]
where $y_i(t)\leq c(t)\leq y_j(t)$ and either $i=1$ and $j=2$ or 
$i=2$ and $j=1$. In either case we have $c(t)\geq m$. Substituting 
the minimal value for $c(t)$ leads to 
\[
\left |(Fy_1)'-(Fy_2)'\right |  \leq \left 
(\frac{\lambda}{p+1}\right )\left (\frac{1}{M}\right 
)^{\frac{\lambda}{p+1}-\lambda}\left (\frac{1}{m}\right )^{\lambda 
+1}\Big(\int_t^T \phi \, ds\Big)^{\frac{1}{p+1}}\|y_1-y_2\| , 
\]
where $\|\cdot \|$ represents the sup-norm on $C[a_0,T]$. 
Integration leads to 
\begin{equation}
\left |Fy_1-Fy_2 \right |  \leq  \left (\frac{\lambda}{p+1}\right 
)\left ( \frac{1}{M}\right )^{\frac{\lambda}{p+1}-\lambda}\left 
(\frac{1}{m}\right )^{\lambda +1}\int_t^T\left(\int_{\tau}^T \phi 
\, ds\right)^{\frac{1}{p+1}}\, d\tau \|y_1-y_2\| . \label{est2} 
\end{equation}
For the purposes of this theorem we choose $m=h-\epsilon$ and 
$M=h+\epsilon$, and then refine our previous choice of $a_0$ to 
get 
\[
\|Fy_1-Fy_2\|  \leq  \frac{1}{2}\|y_1-y_2\| .
\]
Hence $F$ is a contraction on ${\overline B}_{\epsilon}(h)$, and 
we have established the existence of a unique local solution. 

We now extend the solution to a maximal subinterval of $[0,T]$ by 
a standard form of argument. Suppose that $[a,T]$ is the maximal 
subinterval of $[0,T]$ such that (\ref{ivp}) has a unique positive 
solution $y\in C[a,T]\bigcap C^1(a,T]$. If either $a=0$ or 
$y(a)=0$, we are done, so suppose that $a>0$ and $y(a)>0$. We know 
that $y'(t)$ is decreasing and nonnegative. Also 
\[
0\leq y'(t)=\Big( k^{p+1}+\int_t^T\frac{\phi}{y^{\lambda}}\, 
ds\Big)^{\frac{1}{p+1}}\leq \Big( k^{p+1}+\big(\frac{1}{y(a)}
\big)^\lambda \int_{a}^T\phi \, ds\Big)^{\frac{1}{p+1}} <\infty, 
\]
because $\phi\in L^1_{{\rm loc}}(0,1)$. Therefore $ 
\lim_{t\to a^+}y'(t)$ exists and $y\in C^1[a,T]$. Using 
$h=y(a)>0$ and $k=y'(a)\geq 0$, we apply the local existence 
result above to extend $y$ to a unique positive solution of 
(\ref{ivp}) over some interval $[a-\delta,T]$, contradicting the 
assumption that $[a,T]$ is maximal. The theorem is proved. 
\vspace{.1in}

Observe that for the case $k>0$ Theorem 1 follows from the 
standard existence and uniqueness theory for ordinary differential 
equations, because we can work on an interval where potential 
solutions are bounded away from the singularities. It was our 
preference to treat the case $k\geq 0$ as a whole. For the 
remainder of this section we focus primarily on the case $k=0$, 
and invite the interested reader to generalize. 


Before continuing we introduce some useful notation. We refer to 
the initial value problem (\ref{ivp}) with the additional 
restriction $k=0$ as (\ref{ivp})$_0$. Given $T\in(0,1)$, let $y_h$ 
denote the positive solution of (\ref{ivp})$_0$  satisfying 
$y(T)=h$. We define 
\[
\begin{array}{l}
\displaystyle {\mathcal H}_T^-:=\{h>0: y_h \mbox{ has maximal 
interval } [0,T]\}, \\ \displaystyle h_T^-:=\inf {\mathcal H}_T^- 
\end{array}
\]
The $-$ sign is a reminder that we are currently working to the 
left of $T$. 


\begin{lemma}
If $\int_0^T\big( \int_s^T \phi \, ds\big)^{\frac{1}{p+1}} \,dt =\infty$ 
then (\ref{ivp})$_0$ has no positive solution whose maximal interval 
is $[0,T]$, i.e. ${\mathcal H}_T^-$ is empty. \label{neccond} 
\end{lemma}

\paragraph{Proof:}
If $y\in C[0,T]$ is such a solution, then 
\begin{equation}
\begin{array}{lll}
\displaystyle y & = & \displaystyle h-\int_t^T \Big( 
\int_{\tau}^T \frac{\phi}{y^{\lambda}}\, ds\Big)^{\frac{1}{p+1}}\, d\tau \\ 
 \displaystyle &\leq & \displaystyle h-\left (\frac{1}{h}\right )
^{\frac{\lambda}{p+1}}\int_t^T\Big( \int_s^T \phi \, ds\Big)
^{\frac{1}{p+1}} \, d\tau,
\end{array}
\label{lem2ineq}
\end{equation}
because $y\leq h$ on $(0,T]$. It follows that $ 
\lim_{t\to 0^+}y(t)=-\infty$, a contradiction.  \bigskip

An immediate consequence of this lemma is that  
$\int_0^T\left ( \int_s^T \phi \, ds\right )^{\frac{1}{p+1}} \, dt 
<\infty$ is a necessary condition for the existence of a solution 
to (\ref{bvp}). Similarly, $\int_T^1\left ( \int_s^T 
\phi \, ds\right )^{\frac{1}{p+1}} \, dt <\infty$ is necessary. 
Hence 
\[
\int_0^1\Big( \int_s^T \phi \, ds\Big)^{\frac{1}{p+1}} \, dt 
<\infty 
\]
is necessary. Since $\phi$ is locally integrable is suffices to 
write this condition using $T=1/2$. 


\begin{lemma}
Let $K:=\int_0^T\left ( \int_s^T \phi \, ds\right 
)^{\frac{1}{p+1}} \, dt$. If $K<\infty$ then ${\mathcal H}_T^-$ is 
nonempty, and is bounded below by $K^{\frac{p+1}{p+1+\lambda}}$ 
\end{lemma}

\paragraph{Proof:}
Let $y$ be a solution of (\ref{ivp})$_0$. Apply inequality 
(\ref{est1}) with $m=\frac{h}{2}$, $k=0$, and $Fy=y$ to get 
\[
0\leq  h-y \leq \left (\frac{2}{h}\right )^{\frac{\lambda}{p+1}}K.
\]
For large $h$ we have $0\leq h-y\leq \frac{h}{4}$, so $y\geq 
\frac{3h}{4}>0$. Thus if $y\geq\frac{h}{2}$ on an interval, then 
$y\geq\frac{3h}{4}$ on that interval. Hence, for large $h$, the 
solution remains above $\frac{3h}{4}$ no matter how far it is 
extended. Therefore the solution must have maximal interval 
$[0,T]$. 

Substitute $t=0$ into inequality (\ref{lem2ineq}) to get
\[ 0\leq y(0)\leq h-\left (\frac{1}{h}\right )^{\frac{\lambda}{p+1}}K, \]
for $h\in{\mathcal H}_T^-$. This simplifies to 
\[ h\geq K^{\frac{p+1}{p+1+\lambda}} .\]
\vspace{.1in}


Before characterizing ${\mathcal H}_T^-$ and $h_T^-$ further we 
need a comparison lemma. 

\begin{lemma}
Let $T\in (0,1)$ and any let $y_1,y_2\in C[a_0,T]$ be solutions of 
(\ref{ivp}) with $0<y_1(T)\leq y_2(T)$ and $0\leq y_2'(T)\leq 
y_1'(T)$. Moreover, assume that at least one of these inequalities 
is strict. Then $y_2-y_1$ is strictly decreasing in $ [a_0,T]$. 
\label{comparison}
\end{lemma}

\paragraph{Proof:}
Since at least one of the inequalities is strict, we know that 
$y_1(t)<y_2(t)$ in some interval $(T-\delta,T]$. Suppose that 
there exists a $t_0\in (a_0,T)$ such that $y_1(t_0)=y_2(t_0)$ and 
$y_1(t)<y_2(t)$ for $t\in(t_0,T)$. Therefore, for $t\in (t_0,T)$, 
\[
\begin{array}{ll}
y_2'(t)& \displaystyle =\Big( (y_2'(T))^{p+1}+\int_t^T 
\frac{\phi}{y_2^{\lambda}}\, ds\Big)^{\frac{1}{p+1}}\\ 
 & \displaystyle < \Big( (y_1'(T))^{p+1}+\int_t^T 
\frac{\phi}{y_1^{\lambda}}\, ds\Big)^{\frac{1}{p+1}}=y_1'(t).
\end{array}
\]
Thus $y_2-y_1$ is strictly decreasing in $[t_0,T]$ with 
$y_2(T)-y_1(T)\leq 0$, which contradicts $y_1(t_0)=y_2(t_0)$. 
Hence $y_1(t)<y_2(t)$ in $[a_0,T)$, and it follows, by the same 
inequality, that $y_2'(t)<y_1'(t)$ in $[a_0,T)$. 

\begin{corollary}
Let $T\in (0,1)$  and assume that $\int_0^T\left ( 
\int_s^T \phi \, ds\right )^{\frac{1}{p+1}}\, dt<\infty$. If 
$h>h_T^-$ then $h\in {\mathcal H}_T^-$, and if $h<h_T^-$ then 
$h\not\in {\mathcal H}_T^-$. \label{cor1} 
\end{corollary}

\paragraph{Proof:}
We have already shown that ${\mathcal H}_T^-$ is nonempty and 
bounded below by $K^{\frac{p+1}{p+1+\lambda}}$, so $h_T^->0$. 
Suppose that $h_1\in {\mathcal H}_T^-$ and $h_2>h_1$. Let 
$y_{h_1}$ and $y_{h_2}$ represent the corresponding solutions of 
(\ref{ivp})$_0$. Then Lemma \ref{comparison} shows that 
$y_{h_2}>y_{h_1}$ on any common interval of definition, and 
therefore $h_2\in {\mathcal H}_T^-$. The lemma follows. 

\vspace{.1in}

Given Lemma \ref{comparison} and its consequences, it is 
reasonable to guess that the positive solution of (\ref{ivp})$_0$ 
satisfying $y(T)=h_T^-$ will have maximal interval $[0,T]$ and 
will satisfy $y(0)=0$. This is indeed the case as the following 
lemmas will show. 

\begin{lemma}
Let $T\in (0,1)$  and assume that $\int_0^T\left ( 
\int_s^T \phi \, ds\right )^{\frac{1}{p+1}} \, dt<\infty$. Let 
$y_1,y_2\in C[a_0,T]$ be solutions of (\ref{ivp})$_0$ and $0<m\leq 
y_1(t)\leq y_2(t)\leq M$ for all $t\in [a_0,T]$. Then given any 
$\epsilon >0$ there is a $\delta >0$ such that if 
$|y_1(T)-y_2(T)|<\delta$, then $|y_1(t)-y_2(t)|<\epsilon$ for all 
$t\in [a_0,T]$. \label{condep} 
\end{lemma} 

\paragraph{Proof:}
The proof will follow from an extension of estimate (\ref{est2}) 
to this special case. Notice that we can substitute $y'$ for 
$(Fy)'$. Also, Lemma \ref{comparison} shows that $y_2(t)-y_1(t)$ 
is positive and decreasing, so 
$\sup_{[t,T]}|y_2-y_1|=(y_2(t)-y_1(t))$ on any $[t,T]\subset 
[a_0,T]$. Thus 
\[
0\leq y_1'(t)-y_2'(t)  \leq \big(\frac{\lambda}{p+1}\big)
\big(\frac{1}{M}\big)^{\frac{\lambda}{p+1}-\lambda}
\big(\frac{1}{m}\big)^{\lambda +1}\Big(\int_t^T \phi \, 
ds\Big)^{\frac{1}{p+1}}(y_2(t)-y_1(t)) . 
\]
Let $C:=\left (\frac{\lambda}{p+1}\right )\left 
(\frac{1}{M}\right )^{\frac{\lambda}{p+1}-\lambda}\left 
(\frac{1}{m}\right )^{\lambda +1}$, $P(t):=C\left(\int_t^T \phi \, 
ds\right)^{\frac{1}{p+1}}$ and $w:=y_1-y_2$. We restate the 
inequality above as 
\[
w'(t)+P(t)w(t)\leq 0.
\]
Thus
\[
\left ( e^{P(t)}w(t)\right )'\leq 0,
\]
and so
\[ 
e^{P(T)}w(T)-e^{P(t)}w(t)\leq 0.
\]
Notice that $P(T)=0$ and $w(T)=y_1(T)-y_2(T)$, so
\[
e^{-P(t)}(y_1(T)-y_2(T))\leq y_1(t)-y_2(t)\leq 0.
\]
The assumption $\int_0^T\left ( \int_s^T \phi \, 
ds\right )^{\frac{1}{p+1}} \, dt<\infty$ guarantees that $P(t)$ is 
bounded even in the case $a_0=0$. Therefore the lemma follows. 
\vspace{.1in}


It is important to observe that the choice of $\delta$ in the 
previous proof depends upon $\epsilon,m,M$ and $K$, but does not 
depend upon the interval $[a_0,T]$. 

\begin{lemma}
Let $T\in (0,1)$ and assume that $\int_0^T\left ( 
\int_s^T \phi \, ds\right )^{\frac{1}{p+1}} \, dt<\infty$. Then 
$h_T^-\in {\mathcal H}_T^-$, and the solution, $y$, of 
(\ref{ivp})$_0$ with $y(T)=h_T^-$ has maximal interval $[0,T]$ and 
satisfies $y(0)=0$. 
\end{lemma}

\paragraph{Proof:}
Suppose that $y$ has a maximal interval $[a,T]$ with $a>0$. Recall 
that $y(a)=0$ in this case. Let $h_a^-$ be defined as before, and 
choose $a_1\in (a,T)$ such that $0<y(a_1)<h_a^-$. Let $m=y(a_1), 
M=h^-_T+1$ and $\epsilon< h_a^--y(a_1)$, and choose $\delta$ as in 
Lemma \ref{condep}. Let $y_{\delta}$ represent the solution of 
(\ref{ivp})$_0$ satisfying $y_{\delta}(T)=h_T^-+\delta$. Since 
$y_{\delta}(T)>h_T^-$ we know that $y_{\delta}$ has maximal 
interval $[0,T]$, and by Lemma \ref{condep}, we know that 
$y(a_1)<y_{\delta}(a_1)<h_a^-$. Since $y_{\delta}$ is increasing 
it follows that $y_{\delta}(a)<h_a^-$. Now compare $y_{\delta}$ to 
the solution, ${\overline y}$, of (\ref{ivp})$_0$ satisfying 
${\overline y}(a)=y_{\delta}(a)$ . It is clear that 
$y_{\delta}'(a)>0$, so, by Lemma \ref{comparison}, we know that 
$y_{\delta}<{\overline y}$ on any common interval of definition. 
However, since ${\overline y}(a)<h_a^-$ we know that ${\overline 
y}$ has maximal interval $[{\overline a},a]$ for some ${\overline 
a}\in (0,a)$ and we know ${\overline y}({\overline a})=0$. Thus 
$y_{\delta}({\overline a})<0$, a contradiction. Therefore $a=0$, 
and $y\in C[0,T]\bigcap C^1(0,T]$. 

Suppose that $y(0)=\alpha >0$. Let $m=\frac{\alpha}{2}, M=h_T^-$ 
and $\epsilon =\frac{\alpha}{4}$, and choose $\delta$ as in Lemma 
\ref{condep}. Let $y_{\delta}$ represent the solution of 
(\ref{ivp})$_0$ with $y_{\delta}(T)=h_T^--\delta$. By Lemma 
\ref{condep} we know that if $\frac{\alpha}{2}\leq y_{\delta}\leq 
y\leq h_T^-$ on some interval $[a_0,T]$, then $|y_{\delta}-y|\leq 
\frac{\alpha}{4}$ and thus $\frac{3\alpha}{4}\leq y_{\delta}\leq 
y\leq h_T^-$. It follows that $y_{\delta}$ is bounded below by 
$\frac{3\alpha}{4}$ on any subinterval of $[0,T]$, and so its 
maximal interval is $[0,T]$. However, this contradicts the fact 
that $y_{\delta}(T)=h_T^--\delta\not\in {\mathcal H}_T^-$. The 
theorem is proved. 

\vspace{.1in}

It is easy to show, by Lemma \ref{comparison}, that $h_T^-$ is the 
unique $h\in{\mathcal H}_T^-$ such that the associated solution 
satisfies $y(0)=0$. We refer to the solution, $y$, of 
(\ref{ivp})$_0$ such that $y(T)=h_T^-$ as the {\it left half 
solution} on $[0,T]$. By identical arguments, which we omit, we 
introduce the quantity $h_T^+$ and the {\it right half solution} 
on the interval $[T,1]$. 

The remaining lemmas in this section will prove that $h_T^-=h_T^+$ 
for exactly one $T$. This will allow us to join the two half 
solutions to create the unique positive solution of (\ref{bvp}). 


\begin{lemma} $h_T^-$ is a monotone increasing function in $T$.
\end{lemma}

\paragraph{Proof:}
Let $T_1,T_2\in (0,1)$ such that $T_2<T_1$. Let $y_1, y_2$ 
represent the left half solutions on $[0,T_1]$ and $[0,T_2]$, 
respectively. Suppose that $y_2(T_2)=h_{T_2}^-\geq h_{T_1}^- 
=y_1(T_1)$. Since $y_1$ is increasing and $y_{1}'$ is decreasing, 
it is clear that $y_1(T_2)<y_2(T_2)$ and 
$y_{1}'(T_2)>0=y_{2}'(T_2)$. By Lemma \ref{comparison} we know 
that $y_2-y_1$ is strictly decreasing, which implies that 
$y_1(0)<y_2(0)=0$, a contradiction. Hence $h_{T_2}^-<h_{T_1}^-$. 


\begin{lemma}
$h_T^-$ is a continuous function of $T$.
\end{lemma}

\paragraph{Proof:}
Let $\{T_n\}\subset (0,1)$ be an increasing sequence that 
converges to $T\in (0,1)$, and let $\{y_n\}$ and $y$ represent the 
corresponding left half solutions. From the previous lemma we know 
that $\{h_{T_n}^-\}$ is monotone increasing and bounded above by 
$h_T^-$, and thus converges to some $h\leq h_T^-$. 

 Observe that for $m<n$ we can compare $y_m$ and $y_n$ on the interval $[0,T_m]$. We know that $y_n'(T_m)< 0$. If $y_n(T_m)\leq h_{T_n}^-$, then Lemma \ref{comparison} implies that $y_n<y_m$ in $[0,T_m)$. Thus $y_n(0)<0$, a contradiction. Hence $y_n(T_m)>h_{T_m}^-$, A similar contradiction arises if $y_n(t)=y_m(t)$ for any $t\in (0,T_m]$, so $y_m(t)<y_n(t)$ for $t\in (0,T_m]$. 

In order to have a common interval for comparison we define
\[
{\overline y}_n(t):=\left \{ \begin{array}{l} y_n(t), 0\leq T_n \\ 
h_{T_n}, T_n<t\leq T \end{array}\right. 
\]
It is clear that ${\overline y}_n\in C[0,T]\bigcap C^1(0,T]$ and 
that for $m<n$ we have ${\overline y}_m(t)< {\overline y}_n(t)\leq 
y(t)$ in $(0,T]$. Thus $\{{\overline y}_n(t)\}$ is bounded and 
increasing for any $t\in (0,T]$ and we can define $ 
f(t)=\lim_{n\to\infty}{\overline y_n(t)}$. Moreover, if 
$t<T$, then $t<T_n$ for all $n$ large enough and we have 
$f(t)=\lim_{n\to\infty} y_n(t)$. 

Next we argue that the convergence is better than pointwise. For 
all $t\geq t_0 >0$ we have 
\[
0\leq {\overline y}_n'(t)\leq y_n'(t_0)=
\Big(\int_{t_0}^T\frac{\phi}{y_{n}^{\lambda}}\, ds 
\Big)^{\frac{1}{p+1}} \leq \left ( \frac{1}{y_1(t_0)}\right 
)^{\frac{\lambda}{p+1}}\Big(\int_{t_0}^T \phi \, ds 
\Big)^{\frac{1}{p+1}}<\infty 
\]
Thus, by the Arzela-Ascoli theorem, me may assume, without loss of 
generality, that $\{{\overline y}_n\}$ converges uniformly to $f$ 
on compact subsets of $(0,T]$. 

Thus $f\in C(0,T]$ and for $t\in (0,T)$
\begin{eqnarray*}
\displaystyle f(t) &=& \lim_{n\to\infty}y_n(t)\\ 
 &=&\lim_{n\to\infty}\Big( h_{T_n}^- -\int_{t}^{T_n}\Big( 
\int_{\tau}^{T_n}\frac{\phi}{y_n^{\lambda}}\, ds 
\Big)^{\frac{1}{p+1}} \, d\tau\Big)\\ 
&=&h-\int_{t}^{T}\Big( \int_{\tau}^{T}\frac{\phi}{f^{\lambda}}\, 
ds \Big)^{\frac{1}{p+1}}\, d\tau. 
\end{eqnarray*}
Hence $f$ is a solution of (\ref{ivp})$_0$ with maximal interval 
$[0,T]$. Thus $h\in{\mathcal H}_{T}^-$ and $h\geq h_T^-$. But we 
already proved that $h\leq h_T^-$ so $h=h_T^-$, and thus 
$ \lim_{n\to\infty} h_{T_n}^-=h_T^-$. 
Moreover, $f(t)\equiv y(t)$. 

If $\{T_n\}$ is a sequence converging to $T$ from the right, then 
an analagous argument holds. In fact the argument is simpler, 
because there is no need to extend the functions to a common 
interval. We omit this part of the proof. The lemma is proved. 

\begin{lemma}
$ \lim_{T\to 0^+}h_T^- =0$.
\end{lemma}

\paragraph{Proof:}
Let $y_1$ represent the left half solution of (\ref{ivp})$_0$ over 
$[0,\frac{1}{2}]$, and let $y_2$ represent the left half solution 
of (\ref{ivp})$_0$ over $[0,T]$ where $T\in (0,\frac{1}{2})$. 
Clearly, $y_1'(T)>0=y_2'(T)$. If $y_1(T)\leq h_T^-=y_2(T)$, then 
Lemma \ref{comparison} implies that $y_1(0)<y_2(0)=0$, a 
contradiction. Thus $0\leq h_T^- < y_1(T)$.  Hence $ 
\lim_{T\to 0^+}h_T^-=0$. 

\vspace{.1in}

Identical arguments show that $h_T^+$ is continuous and decreasing 
in $[0,1)$ with $ \lim_{T\to 1^-}h_T^+=0$. 
Therefore there is a unique $T \in (0,1)$ such that $h_T^-=h_T^+$. 
For this $T$ let $y_0$ represent the left half solution over 
$[0,T]$ and let $y_1$ represent the right half solution over 
$[T,1]$, and define 
\[
y(t):=\left \{ \begin{array}{ll} y_0(t),& t\in [0,T] \\ y_1(t), & 
t\in (T,1] \end{array}\right . 
\]
This $y$ is the unique solution of (\ref{bvp}). Thus Theorem 
\ref{bvpex} has been proved. 


\section{ Boundary Behavior }

In this section we assume throughout that  
$\int_0^{1/2}\big( \int_s^{1/2} \phi \, ds\big)^{\frac{1}{p+1}} \, dt
<\infty$ is satisfied, and we investigate 
the boundary behavior of the unique positive solution of 
(\ref{bvp}). More specifically, we concentrate on behavior near 
$0$ for left half solutions of (\ref{ivp})$_0$. Similar results 
apply for the behavior of right half solutions near $1$. 

We begin with the question of whether the slope at the boundary is 
finite or infinite. Recall that the solution, $y$, is concave 
down, so the quantity $ y'(0)=\lim_{t\to 
0^+}y'(t)\in (0,\infty]$ is well-defined. 
\begin{theorem}
Let $y$ be the left half solution of (\ref{ivp})$_0$ on $[0,T]$. 
Then $y^{\prime}(0)$ is finite if and only if 
$\int_0^{1/2} \frac{\phi}{t^{\lambda}}\, dt < \infty$. 
\end{theorem}

\paragraph{Proof:}
Assume that $y'(0)=A<\infty$. We know that 
\[
(y'(t))^{p+1}=\int_t^T\frac{\phi}{y^\lambda}\, ds,
\]
and that $0\leq y'(t)\leq A$ and thus $0\leq y(t)\leq At$ in 
$[0,T]$. Therefore 
\[
(y'(t))^{p+1}\geq \left (\frac{1}{A^{\lambda}}\right 
)\int_t^T\frac{\phi}{s^\lambda}\, ds\geq 0, 
\]
and it follows that 
\[
A^{p+1+\lambda}\geq \int_0^T\frac{\phi}{t^\lambda}\, ds.
\]
Hence $\int_0^{1/2}\frac{\phi}{y^\lambda}\, dt<\infty$. 

Assume $y'(0)=\infty$. Let $A>0$ such that $y(t)\geq At$ on 
$[0,T]$. Then 
\[
(y'(t))^{p+1} \leq  
\frac{1}{A^{\lambda}}\int_t^{T}\frac{\phi}{s^{\lambda}}\, ds. 
\]
Thus
\[
\lim_{t\to 0^+}\int_t^{T}\frac{\phi}{s^{\lambda}}\, ds\geq 
A^{\lambda}\lim_{t\to 0^+}(y'(t))^{p+1}=\infty. 
\]

For the finite slope case we derive asymptotic formulae for $y$ 
that are similar to those in \cite{T}. 

\begin{theorem}
Let $y$ be a left half solution of (\ref{ivp})$_0$ on $[0,T]$ with 
$y'(0)=A<\infty$. Then 
\[ 
y'(t) = \left ( \frac{1}{A}\right )^{\frac{\lambda}{p+1}}\Big( 
\int _t^T\frac{\phi}{s^{\lambda}}(1+o(1))\, ds \Big)^{\frac{1}{p+1}}, 
\]
and
\[
y(t)=\left ( \frac{1}{A}\right )^{\frac{\lambda}{p+1}}\int _0^t 
\Big( \int _{\tau}^T\frac{\phi}{s^{\lambda}}(1+o(1))\, ds 
\Big)^{\frac{1}{p+1}}\, d\tau\,. 
\]
\end{theorem}

\paragraph{Proof:} Substitute $\frac{1}{y^\lambda 
(s)}=\frac{1}{(As)^{\lambda}}(1+o(1))$. 
\vspace{.1in}

For the remainder of this section we concentrate on the infinite 
slope case and assume  $ \int_0^{1/2} 
\frac{\phi}{t^{\lambda}}\, dt =\infty$. The following theorem 
provides a general tool for comparing the boundary behavior of 
solutions. 

\begin{theorem}
Let $\psi\in L^1_{{\rm loc}}(0,\delta)$ be a positive function 
satisfying $ \int_0^{\delta} 
\frac{\psi}{t^{\lambda}}\, dt =\infty$, $ 
\int_0^{\delta}\left ( \int_t^{\delta} \psi \, ds\right 
)^{\frac{1}{p+1}} \, dt<\infty$, and $ 
\lim_{t\to 0^+}\frac{\psi}{\phi}=1$. Let $y$ be the left 
half solution of (\ref{ivp})$_0$ on $[0,T]$, and let $z$ be a 
particular solution of 
\[
\begin{array}{l}
\displaystyle (|z'|^pz')' + \frac{\psi}{z^\lambda}=0 \mbox{ in } 
(0,\delta),\\ z(0)=0. 
\end{array}
\]
Then $\displaystyle \lim_{t\to 
0^+}\frac{z'}{y'}=\lim_{t\to 0^+}\frac{z}{y}=1$. 
\label{bdry}
\end{theorem}
\vspace{.1in}

Before proving Theorem \ref{bdry} we provide an interesting 
application. 

\begin{corollary}
Assume that $ \lim_{t\to 0^+}\frac{c 
t^{r}}{\phi}=1$ where $c > 0$ and $-p-2 < r \leq \lambda -1$. Let 
$y$ be the left half solution of (\ref{ivp})$_0$ on $[0,T]$, and 
let $z(t)=\gamma t^\rho$ such that $ 
\rho=\frac{r+p+2}{\lambda +p+1}$ and $ \gamma=\left 
(\frac{c}{\rho^{p+1}(1-\rho)}\right )^{\frac{1}{\lambda+p+1}}$. 
Then $ \lim_{t\to 
0^+}\frac{z'}{y'}=\lim_{t\to 0^+}\frac{z}{y}=1$. 
\end{corollary}

\paragraph{Proof:} The restrictions on $r$ guarantee that 
$\psi = ct^r$ satisfies 
$$ \int_0^1 \big(\int_{t}^{1/2}\psi \, ds
 \big)^{\frac{1}{p+1}}\, dt<\infty$$ and $ \int_0^{1/2} 
\frac{\psi}{\tau^{\lambda}} =\infty$.
 It is straight forward to 
check that $z(t)$ is a solution of 
$$\displaylines{
 (|z'|^pz')' + \frac{\psi}{z^\lambda}=0 \quad \mbox{in }(0,\infty),\cr
 z(0)=0\,. 
}$$ 
Hence the result follows from Theorem \ref{bdry}. \vspace{.1in} 

The proof of Theorem \ref{bdry} depends upon several lemmas.

\begin{lemma}
Let $\psi\in L^1_{{\rm loc}}(0,\delta)$ be a positive function 
satisfying $ \int_0^{\delta} 
\frac{\psi}{t^{\lambda}}\, dt =\infty$, and $ 
\int_0^{\delta}\big( \int_t^{\delta} \psi \, ds 
\big)^{\frac{1}{p+1}} \, dt<\infty$. Let let $z_1$ and $z_2$ be 
solutions of 
$$\displaylines{
 (|z'|^pz')' + \frac{\psi}{z^\lambda}=0\quad \mbox{in } 
(0,\delta),\cr z(0)=0. 
}$$
 Then $\displaystyle \lim_{t\to 0^+}\frac{z_1'}{z_2'}=\lim_{t\to 0^+}\frac{z_1}{z_2}=1$.
\end{lemma}

\paragraph{Proof:}
Assume that $z_1$ and $z_2$ are distinct solutions with 
$z_1(t)<z_2(t)$ at some point. If $z_1'(t)\geq z_2'(t)$ at this 
same point, then, by Lemma \ref{comparison}, $z_2-z_1$ must be 
decreasing on $(0,t)$. But this leads to a contradiction of 
$z_1(0)=z_2(0)=0$. Thus $z_1'(t)<z_2'(t)$. It follows that $0\leq 
z_1'(t) < z_2'(t)$ and $0< z_1(t) < z_2(t)$ in $[0,\delta]$. 
Moreover, 
\begin{eqnarray*}
z_2'(t) &=& \Big( (z_2'(\delta))^{p+1} + 
\int_t^\delta \frac{\psi}{z_2^{\lambda}}\, ds\Big)^{\frac{1}{p+1}}\\ 
&<& \Big( (z_2'(\delta))^{p+1} +\int_t^\delta \frac{\psi}{z_1^{\lambda}}
\, ds\Big)^{\frac{1}{p+1}} \\
      &=& \Big( (z_2'(\delta))^{p+1} + (z_1'(t))^{p+1}
-(z_1'(\delta))^{p+1}\, ds \Big)^{\frac{1}{p+1}}.
\end{eqnarray*}
Thus
\[
1 < \frac{z_2'(t)}{z_1'(t)} < \Big( \big(\frac{z_2'(\delta)}{z_1'(t)}
\big)^{p+1} + 1 -\big(\frac{z_1'(\delta)}{z_1'(t)}\big)^{p+1}
\Big)^{\frac{1}{p+1}}. 
\]
Since $ \int_0^{\delta} \frac{\psi}{\tau^{\lambda}}\, 
d\tau =\infty$, we know that $ \lim_{t\to 
0^+}z_1'(t)=\infty$. Hence $ \lim_{t\to 
0^+}\frac{z_2'(t)}{z_1'(t)} =1$. By L'Hospital's rule we have 
$ \lim_{t\to 0^+}\frac{z_2(t)}{z_1(t)} =1$. 
The proof is done. 


\begin{lemma}
Let $\psi\in L^1_{{\rm loc}}(0,\delta)$ be a positive function 
satisfying $ \int_0^{\delta} 
\frac{\psi}{t^{\lambda}}\, dt =\infty$, $ 
\int_0^{\delta}\left ( \int_s^{\delta} \psi \, ds\right 
)^{\frac{1}{p+1}} \, dt<\infty$, and $\psi \geq \phi$ ($\psi \leq 
\phi$). Let $y$ be the left half solution of (\ref{ivp})$_0$ on 
$[0,T]$, and let $z$ be a particular solution of 
$$\displaylines{
(|z'|^pz')' + \frac{\psi}{z^\lambda}=0 \quad \mbox{in } 
(0,\delta),\cr z(0)=0. 
}$$
Then $ \liminf_{t\to 0^+}\frac{z}{y}\geq 1$  
$(\displaystyle \limsup_{t\to 0^+}\frac{z}{y}\leq 1)$. 
\end{lemma}

\paragraph{Proof:}
By the previous lemma we know that all solutions of 
$$\displaylines{
(|z'|^pz')' + \frac{\psi}{z^\lambda}=0 \quad\mbox{ in } 
(0,\delta),\cr z(0)=0, 
}$$
have asymptotically identical boundary behavior. Thus we may 
compare $y$ to the particular solution where 
$z'(\delta)=y'(\delta)$. 

Suppose that $y(t)>z(t)$ at some point in $(0,\delta)$. Then 
$y(t)>z(t)$ in some interval $(a_0,b_0)\subset (0,\delta)$ such 
that $z'(b_0)\geq y'(b_0)$. For $t\in (a_0,b_0)$ we have 
\begin{eqnarray*}{ll}
y'(t) &=& \Big( \left ( y'(b_0)\right )^{p+1} + 
\int_t^{b_0}\frac{\phi}{y^{\lambda}} \, ds 
\Big)^{\frac{1}{p+1}} \\ 
& <& \Big( \left 
(z'(t_0)\right )^{p+1} + \int_t^{b_0}\frac{\psi}{z^{\lambda}}\, ds 
\Big)^{\frac{1}{p+1}}\\
&=& z'(t)\,. 
\end{eqnarray*}
Thus $y-z$ is decreasing in $(a_0,b_0)$. Thus the maximal interval 
where $y>z$ must be $(0,b_0)$, and, by the same argument, $y-z$ is 
decreasing on $(0,b_0)$. But this implies $0=y(0)-z(0)< 
y(b_0)-z(b_0)\leq 0$, a contradiction. Hence $z(t)\geq y(t)$ in 
$[0,\delta]$. The case $\psi\leq\phi$ can be argued similarly so 
the lemma is proved. \vspace{.1in} 


\paragraph{\bf Proof of Theorem 5:}
Let $\epsilon > 0$ be given. Observe that for $c>0$ we have that 
$w=cy$ is a left half solution on $[0,T]$ of the problem 
\begin{equation}
\begin{array}{c}
\displaystyle (|w'|^pw')'+\frac{c^{\lambda + p + 
1}\phi_1}{w^{\lambda}}=0 , \\[7pt] w(T)=h,\, w'(T)=0, 
\end{array}
\label{scaledivp}
\end{equation}
Also, without loss of generality, we may assume that 
$(1-\epsilon)^{\lambda +p+1}\phi\leq\psi\leq(1+\epsilon)^{\lambda 
+p+1}\phi$ in $(0,\delta)$. Therefore, the previous lemma and the 
observation about (\ref{scaledivp}) show that $ 
\liminf_{t\to 0^+} \frac{z}{(1-\epsilon)y}\geq 1$ and 
$ \limsup_{t\to 0^+} 
\frac{z}{(1+\epsilon)y}\leq 1$. Hence $ 
\lim_{t\to 0^+} \frac{z}{y}= 1$. 

By further restricting the size of the interval $(0,\delta)$ we 
may now assume that $ (1-\epsilon)\leq \frac{\phi 
z^{\lambda}}{\psi y^{\lambda}}\leq (1+\epsilon)$. We have 
\begin{eqnarray*}
y'(t) &=&\Big( (y'(\delta))^{p+1}+\int_t^{\delta}\frac{\phi}{y^{\lambda}}\, 
ds\Big)^{\frac{1}{p+1}}\\ 
&=& \Big( 
(y'(\delta))^{p+1}+\int_t^{\delta}\frac{\psi}{z^{\lambda}}\frac{\phi 
z^{\lambda}}{\psi y^{\lambda}}\, ds \Big)^{\frac{1}{p+1}}. 
\end{eqnarray*}
Thus 
\begin{eqnarray*}
 \big( (y'(\delta))^{p+1}+\frac{1}{(1+\epsilon)}\int_t^{\delta}
\frac{\psi}{z^{\lambda}}\, ds\big)^{\frac{1}{p+1}}
&\leq& y'(t)\\
&\leq&\big( (y'(\delta))^{p+1}+\frac{1}{(1-\epsilon)}
\int_t^{\delta}\frac{\psi}{z^{\lambda}}\, ds \big)^{\frac{1}{p+1}}. 
\end{eqnarray*}
Divide this inequality through by $ z'(t)=\big( 
(z'(\delta))^{p+1}+\int_t^{\delta}\frac{\psi}{z^{\lambda}}\, ds 
\big)^{\frac{1}{p+1}}$ and let $t\to 0^+$ to get 
\[
\left ( \frac{1}{1+\epsilon}\right)^{\frac{1}{p+1}}\leq
\liminf_{t\to 0^+} \frac{y'(t)}{z'(t)} \leq 
\limsup_{t\to 0^+}  \frac{y'(t)}{z'(t)}\leq \left ( 
\frac{1}{1-\epsilon}\right)^{\frac{1}{p+1}}. 
\]
Hence $ \lim_{t\to 
0^+}\frac{z'(t)}{y'(t)}=1$, and Theorem \ref{bdry} is proved. 
\vspace{.1in}

In \cite{T} Taliaferro applies a result similar to Theorem \ref{bdry}
 to obtain a precise description of boundary behavior for a more general 
collection of functions $\phi$ than those described in Corollary 2.
 Taliaferro assumes that $ \lim_{t\to 0^+}\frac{tf''(t)}{f'(t)}=R$,
 where $ f(t):=\int_t^T \psi/s^\lambda \, ds$ and where $\psi$ 
is a smooth function such that $ \lim_{t\to 0^+} \psi/\phi=1$. 
We note that if $tf''(t)/f'(t)\equiv R$ then one can show 
that $\psi$ is of the form $ct^r$, so Corollary 2 is applicable. 
Taliaferro's more general condition implies that for any $\epsilon>0$ 
$\psi$ is bounded between some $ct^{r-\epsilon}$ and $ct^{r+\epsilon}$ 
in some neighborhood of $0$. Thus $\psi$ still behaves much like $ct^r$. 
Our methods can be used to find corresponding estimates on the boundary
 behavior of the solution, but we have not generalized Taliaferro's 
argument and results to this case. 




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\end{thebibliography}
\bigskip

\noindent{\sc Robert M. Houck} (e-mail: houck@mthcsc.wfu.edu)\newline 
{\sc Stephen B. Robinson} (e-mail: robinson@mthcsc.wfu.edu)\newline 
Department of Mathematics and Computer Science \newline 
Wake Forest University \newline 
Winston-Salem, NC  27109, USA

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