\documentstyle[twoside]{article}
\pagestyle{myheadings}
\markboth{\hfil On the Existence of Steady Flow \hfil}%
{\hfil Chunqing Lu \hfil}
\begin{document}
\setcounter{page}{159}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Differential Equations and Computational Simulations III}\newline
J. Graef, R. Shivaji, B. Soni J. \& Zhu (Editors)\newline
Electronic Journal of Differential Equations, Conference~01, 1997, pp. 149--160. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  147.26.103.110 or 129.120.3.113 (login: ftp)}
 \vspace{\bigskipamount} \\
 On the Existence of Steady Flow in a Channel with one Porous Wall
 or Two Accelerating Walls 
\thanks{ {\em 1991 Mathematics Subject Classifications:} 34B15, 76D05.
\hfil\break\indent
{\em Key words and phrases:} laminar flow, similarity solutions, 
Navier-Stokes equations.
\hfil\break\indent
\copyright 1998 Southwest Texas State University  and University of
North Texas. \hfil\break\indent
Published November 12, 1998.} }

\date{}
\author{Chunqing Lu}
\maketitle

\begin{abstract}
This paper presents a rigorous proof of the existence of steady flows in a
channel either with no-slip at one wall and constant uniform suction or
injection through another wall, or with two accelerating walls.
The flows are governed  by the fourth order nonlinear differential equation
$F^{iv}+R(FF'''-F'F'')=0$.
In the former case, the flow is subject to the boundary conditions
$F(-1)=F'(-1)=F'(1)=0$, $F(1)=-1$.
In the latter case, the boundary conditions are $F(-1)=F(1)=0$, $F'(-1)=-1$,
$F'(1) = 1$.
\end{abstract}

\section{Introduction}

A similarity solution of the Navier-Stokes equation that describes the
steady two-dimensional flow of a viscous incompressible fluid through a
channel with porous walls was first introduced by Berman in 1953. Since
the flows have numerous applications such as the control of boundary layer
separation with suction of injection, membrane separation processes,
filtration, and biological transport processes, etc., the Berman problem
has attracted many researchers' attention and Berman's results have been
extended since then. A special case that the channel has one porous wall
and one impermeable wall was studied by Cox in \cite{cox}, in which he
classified all possible types of solutions for the flows not only in a
channel with one porous and one impermeable, but also in a channel where
both walls are impermeable, but accelerating. This paper presents a
rigorous proof of existence of those solutions. It is very interesting
to study the special flow in a channel with one wall impermeable because
\cite{cox}
\begin{quote}
in laboratory experiment such a flow could be visualized easily
than in the case which has previously been considered where there is
suction through both walls, and experimental apparatus would simply
obscure the view.
\end{quote}

Assume that the fluid is viscous and incompressible, and that the channel
is sufficiently long, and therefore, can be described as $-\infty 
< x < \infty$, $-1<y<1$, where $y=-1$ denotes the lower wall
and $y=1$ the upper wall. Let $u$ be the velocity component along the
channel, and $v$ the velocity component across the channel. Then, a stream
function $\Phi (x,y)$ can be introduced which satisfies 
\[
u=\frac{\partial \Phi (x,y)}{\partial y}, \quad v=-\frac{\partial
\Phi (x,y)}{\partial x}.
\]
There is a similarity solution of the form $\Phi (x,y)=xF(y)$, and the
Navier-Stokes equations can be reduced to the Berman equation 
\begin{equation}
F^{iv}+R(FF'''-F'F'')=0
\label{eq:berman}
\end{equation}
where $R$ is the Reynolds number, and $R>0$ corresponds to suction and
$R<0$
to injection. If we assume the lower wall is impermeable and constant
suction is through the upper wall, then the boundary
conditions become 
\begin{eqnarray}
&F(-1) = 0, \quad F'(-1)=0\,,&  \label{bd:noslip-1} \\
&F(1) = -1, \quad F'(1)=0\,.&  \label{bd:noslip1}
\end{eqnarray}
If the channel has two accelerating walls with equal rates of
acceleration, the boundary conditions are 
\begin{eqnarray}
&F(-1) = 0, \quad F'(-1)=-1\,,&  \label{bd:2walls-1} \\
&F(1) = 0, \quad F'(1)=1\,. & \label{bd:2walls1}
\end{eqnarray}
When only the upper wall is accelerating. the boundary conditions are 
\begin{eqnarray}
&F(-1) =0, \quad F'(-1)=0\,,&  \label{bd:1wall-1} \\
&F(1)  = 0, \quad F'(1)=1\,.&  \label{bd:1wall1}
\end{eqnarray}
The paper proves the existence of solutions for these boundary
value problems. For brief, we will call the boundary value problem related
to only one porous wall (\ref{bd:noslip-1})-(\ref{bd:noslip1}) as the
first boundary value problem, or BVP (1), the
boundary value problem related to two accelerating walls
(\ref{bd:2walls-1})-(\ref{bd:2walls1}) as the second
boundary value problem, or BVP (2), and the boundary value problem related
to the upper accelerating wall (\ref{bd:1wall-1})-(\ref{bd:1wall1}) as the
third boundary value problem, or BVP (3).

\section{Main Results}

The main results in the paper are the following three theorems.
\newtheorem{thm}{Theorem}
\begin{thm}
\label{thm:noslip} For each real number $R$, the equation
(\ref{eq:berman})
with the boundary conditions (\ref{bd:noslip-1}) and (\ref{bd:noslip1})
has
at least one solution. If $R<0$, the solution is unique.
\end{thm}

\begin{thm}
\label{thm:2walls} For each real number $R$, the equation
(\ref{eq:berman})
with the boundary conditions (\ref{bd:2walls-1})-(\ref{bd:2walls1}) has at
least one solution. If $R$ is sufficiently large, then there are at least
three solutions.
\end{thm}

\begin{thm}
\label{thm:1wall} For each real number $R$, the equation (\ref{eq:berman})
with the boundary conditions (\ref{bd:1wall-1})-(\ref{bd:1wall1}) has at
least one solution.
\end{thm}

The proofs of the theorems rely on the following transformation and
lemmas.

\section{Transformation and lemmas}

We first consider BVP (1). Let 
\begin{equation}
RF(y)=-\frac{1}{2}bg(\xi ),\quad \xi =\frac{1}{2}b(y+1),
\label{change}
\end{equation}
where $b>0$ is a positive parameter to be determined. Substituting (\ref
{change}) into (\ref{eq:berman}), we obtain 
\begin{equation}
g^{iv}=gg'''-g'g''
\label{eq:g4}
\end{equation}
And the substitution converts the boundary conditions (\ref{bd:noslip-1})
and (\ref{bd:noslip1}) to the forms: 
\begin{eqnarray}
&g(0) =g'(0)=0\,,&  \label{bd:g0g'0} \\
&g(b) =\frac{2R}{b},\quad g'(b)=0\,.&  \label{bd:gbg'b}
\end{eqnarray}
This change of variables transforms the original boundary value problem
into an initial value problem that has two parameters $R$ and $b$. If we
consider the initial value problem (\ref{eq:g4}) subject to 
\begin{equation}
g(0)=g'(0)=0,g''(0)=\beta ,g'''(0)=\gamma ,  \label{ini:noslip}
\end{equation}
then any solution of the boundary value problem, (\ref{change}) with (\ref
{bd:noslip-1}) and (\ref{bd:noslip1}), must correspond to a pair $(\beta
,\gamma )$. If a solution $g(\xi ,\beta ,\gamma )$ of the initial value
problem, (\ref{eq:g4}) with (\ref{ini:noslip}), has a zero of its first
derivative at a positive point, then we say that the pair $(\beta ,\gamma
)$
provides a solution to the boundary value problem. The idea of the proof
of
the theorem \ref{thm:noslip} is to find all possible pairs $(\beta ,\gamma
)$
such that the corresponding solution of the initial value problem has a
positive zero $b$. It is clear that the values of $b$ and
$R=bg(b)/2$ are functions of $\beta $ and $\gamma $. Throughout the paper, we always
consider nonzero independent variable $\xi$.

For BVP (3), following the same change of variables (\ref{change}), the
boundary conditions
are converted to 
\begin{eqnarray}
&g(0) = g'(0)=0\,,&  \label{bd:bvp3-0} \\
&g(b) = 0,\quad g'(b)=-4R/b^2\,.&  \label{bd:bvp3-b}
\end{eqnarray}
As for BVP (2), we only consider the symmetric solutions, i.e., an odd
function $F(y)$ satisfying (\ref{eq:berman}) and (\ref{bd:2walls-1})-(\ref
{bd:2walls1}). Thus, $F(0)=F''(0)=0$, we may only consider
the half channel $[0,1]$. In this case, we set 
\[
\xi =by,\quad RF(y)=-bg(\xi ). 
\]
The resulting equation is exactly the same as (\ref{eq:g4}), but the
boundary conditions take the form: 
\begin{eqnarray}
&g(0) = g''(0)=0\,,&  \label{bd:bvp2-0} \\
&g(b) = 0,\quad g'(b)=-R/b^2 \,.&  \label{bd:bvp2-b}
\end{eqnarray}                       
The idea to find a solution of BVP (2) or BVP (3) is the same as that used
in solving BVP (1).

In order to cover all the three cases, we may consider more general cases
as
follows. Let $g(\xi ,\alpha ,\beta ,\gamma )$ denote the solution of (\ref
{eq:g4}) with the initial conditions 
\begin{equation}
g(0)=0,\quad g'(0)=\alpha ,\quad g''(0)=\beta ,\quad g'''(0)=\gamma \,.
\label{ini:general}
\end{equation}
Thus, in order to get a solution to a boundary value problem above, we
need only
find a suitable triple ($\alpha ,\beta ,\gamma )$ such that the solution
of (\ref{eq:g4}) with (\ref{ini:general}) has a zero point of $g'$ so
that it provides an $R=R(\alpha ,\beta ,\gamma )$. To find all solutions
of the boundary value problems, we must consider all possible such triples.
The following two lemmas hold for the initial value problem.

\newtheorem{lemma}{Lemma}
\begin{lemma}
\label{lm:g4<0} If $g(\xi ,\alpha ,\beta ,\gamma )$ is a nonlinear
solution
of (\ref{eq:g4}) - (\ref{ini:general}) solving the above three boundary
value problems, then $g^{iv}(\xi ,\alpha ,\beta ,\gamma )<0$ for $\xi >0$
and as long as the solution exists.
\end{lemma}

\paragraph{Proof}. For BVP (1) and BVP (3), the initial value is (0,$\beta
,\gamma )$, and BVP
(3) is related to the initial value $(\alpha ,0,\gamma )$. In either case,
we see from (\ref{eq:g4}) that $g^{iv}(0)=0$. Differentiating
(\ref{eq:g4})
once with respect to $\xi$, we obtain 
\begin{equation}
g^{v}=gg^{iv}-(g'')^{2},  \label{eq:g5}
\end{equation}
from which 
\begin{equation}
g^{iv}=-\int_{0}^{\xi }(g'')^{2}e^{\int_{s}^{\xi }g\,dt}\,ds.
\label{ex:g4}
\end{equation}
Since $g$ is not linear, $g''\neq 0$ at least for small
$\xi>0$, hence, $(g'')^{2}>0$ initially, which leads to the
lemma.  \hfill$\Box$

\begin{lemma}
\label{lm:homo} $g(\xi ,\beta ,\gamma )=\lambda g(\lambda \xi
,\alpha \lambda ^{-2},\beta \lambda ^{-3},\gamma\lambda ^{-4})$
for any positive $\lambda $ and as long as the solution exists.
\end{lemma}

\paragraph{Proof.}
Let $f(\xi )=\lambda g(\lambda \xi ,\alpha\lambda^{-2},\beta\lambda ^{-3},
\gamma\lambda ^{-4})$. A simple substitution shows
that $f$ solves the equation (\ref{eq:g4}) and satisfies the initial
condition $f(0)=0$, $f'(0)=\alpha $, $f''(0)=\beta $,
and $f'''(0)=\gamma $. By the uniqueness of the
solution to the initial value problem, the lemma follows. \hfill$\Box$

\paragraph{Remark.} This lemma was first introduced by Wang and Hwang in
\cite{wang}.

The following corollary is a consequence of lemma \ref{lm:homo}.

\newtheorem{coro}{Corollary}
\begin{coro}
\label{coro:homo}$R(\alpha ,\beta ,\gamma )=R(\alpha\lambda^{-2},
\beta\lambda ^{-3},\gamma \lambda ^{-4})$ for any $\lambda >0$ as long
 as one of them exists.
\end{coro}

\begin{lemma}
\label{lm:global} For any triple $(\alpha ,\beta ,\gamma )$ with $\alpha
=0$
or $\beta =0$, the solution $g(\xi ,\alpha ,\beta ,\gamma )$ of the
initial
value problem exists for all $\xi \in [0,\infty ).$
\end{lemma}

\paragraph{Proof.}
Suppose not. There would be a triple $(\alpha ,\beta ,\gamma )$ such that
the corresponding solution $g$ only exists on the maximal interval $[0,l)$
for a real number $l$.
By lemma \ref{lm:g4<0}, $g'''$ is decreasing, and
hence
must diverge to $-\infty $ as $\xi \rightarrow l^{-}$ for otherwise $%
g''',g'',g'$, and $g$ are all
bounded, hence $g^{iv}$ is bounded in $[0,l)$ and the solution can be
extend
to the point $l$. Since $g''$ is concave down and
$g''' \rightarrow -\infty$, it follows that $g'' \rightarrow -\infty $
as $\xi \rightarrow l^{-}$. Consecutive
integration of (\ref{eq:g4}) gives 
\begin{eqnarray}
&g''' = gg''-(g')^{2}+\alpha^{2}+\gamma \,,&  \label{eq:g3} \\
&g'' = gg'-2\int_{0}^{\xi }(g^{\prime})^{2}dt+(\alpha ^{2}+\gamma )\xi
+\beta\,. &\label{eq:g2}
\end{eqnarray}
From (\ref{eq:g2}), $g'$ must be unbounded as $\xi \rightarrow l^{-}$.
Since also $g'<\alpha +\beta \xi +\frac{1}{2}\gamma \xi ^{2}$,
$g'$ must diverge to $-\infty $ as $\xi \rightarrow l^{-}$. Then,
for sufficiently large number $N>0$, there is a number $k=k(N)$ such that
$g'<-N,g''<-N$ and $g'''<-N$, and $g$ keeps one sign in $(k,l)$.
Also, $|g'|$, $|g''|$, and $|g'''|$ are bounded by $N$ in $[0,k]$.

If $g<0$ in $(k,l)$, from (\ref{eq:g3}), $gg''-(g')^{2}<0$ in $(k,l)$
which implies
\begin{equation}
\frac{g''}{g'}<\frac{g'}{g}\,.
\label{ineq:lm3}
\end{equation}
Integrating (\ref{ineq:lm3}) once from $k$ to $\xi \in (k,l)$ shows 
\begin{equation}
g'(\xi )\geq cg(\xi ),  \label{ineq:lm3-}
\end{equation}
for all  $\xi \in (k,l)$ where $c=\frac{g'(k)}{g(k)}>0$ is a
constant , from which $g$ $\rightarrow -\infty $ as
$\xi \rightarrow l^{-}$.
An integration of (\ref{ineq:lm3-}) produces $g(\xi )\geq -|g(k)|e^{c(\xi
-k)}$,  and $\xi \in (k,l)$. This contradicts the fact  that $g$ is
unbounded at $l$.

If $g>0$ in $(k,l)$, from (\ref{eq:g2}), since $g'<-N$ for $\xi \in (k,l)$
 and $|g'|\leq N$ in $[0,k]$, there exists a number $M>0$
such that $g''(\xi )>-2\int_{0}^{\xi }(g')^{2}dt>-Mg'(\xi )^{2}$. Hence
\begin{equation}
\frac{g''}{g'}<-Mg'.  \label{ineq:lm3+}
\end{equation}
Integrating (\ref{ineq:lm3+}) from $k$ to any $\xi \in (k,l)$ leads to
$|g'(\xi )|\leq Ce^{-Mg(\xi)}$, where $C = |g'(k)|e^{Mg(k)}$, a
constant. Since $g(\xi) > 0$, this contradicts the assumption $%
g'\rightarrow -\infty $. The proof is completed. \hfill$\Box$


\paragraph{Remark.} Lemma \ref{lm:global} generalizes the similar lemma in
\cite{Lu}

Since the proofs of the three theorems are similar, for the sake of
briefness, we only provide the detailed proof of theorem 1, outline the
proof of  theorem 2 and omit the proof of theorem 3. 

\section{Proof of Theorem \ref{thm:noslip}\label{sec:noslip}}

In this case, $g'(0)=\alpha =0$, therefore, to get all solutions
of BVP (1), we need to consider all possible pairs ($\beta ,\gamma $).
Denote $g(\xi ,0,\beta ,\gamma )=g(\xi ,\beta ,\gamma )$, and
$R(0,\beta ,\gamma)=R(\beta ,\gamma )$. It is immediately seen from lemma
\ref{lm:g4<0} that
$g'(\xi ,\beta ,\gamma )$ has no zero at all if both $\beta $ and
$\gamma $ are non-positive. Then, there are only three possible cases that
need to be investigated: (1) $\beta >0$ and $\gamma \leq 0$, (2) $\beta
\geq 0$ and $\gamma \geq 0$, and (3) $\beta \leq 0$ and $\gamma \geq 0$. Of
course, the case $\beta =\gamma =0$ is excluded.

\noindent{\bf Case (1).} If $\gamma <0$, then, since $g^{iv}<0$ for all $\xi >0$, we
have $g'(\xi )<\beta \xi +\frac{\gamma }{2}\xi ^{2}$ for all $\xi
>0$, by Taylor's theorem. It then follows that there is a point $\xi =r$
at which $g'=0$. If $\gamma =0$, then from (\ref{eq:g5}),
$g^{v}(0)=-\beta ^{2}<0$, hence $g'''(x_{0})\approx
-\beta ^{2}\frac{x_{0}^{2}}{2}<\gamma _{0}<0$ for sufficiently small
$x_{0}>0
$ where $\gamma _{0}$ is a constant. Then, $g'(\xi )<\beta (\xi
-x_{0})+\frac{\gamma _{0}}{2}(\xi -x_{0}^{2})$ for all $\xi >x_{0}$. A
similar argument can show the existence of $r$ at which $g'=0$.
And this is the only zero point of $g'$, namely, $g'<0$ for
all $\xi >r$. This shows that any pair $(\beta ,\gamma )$ where $\beta >0$
and $\gamma <0$ presents an $R>0$. Recall that
$R=R(\beta ,\gamma)=bg(b)/2$ which implies that $R$ is a continuous
function of $\beta $ and
$\gamma $ since the zero of $g'$, in this case, is a simple zero,
i.e., $g''<0$ whenever $g'=0$ from (\ref{eq:g3}). 

From corollary \ref{coro:homo}, $R(\beta ,\gamma )=R(1,\frac{\gamma
}{\beta^{4/3}})=R(\frac{\beta }{|\gamma |^{3/4}},-1)$.
As $\frac{\gamma }{\beta^{4/3}}\rightarrow 0^{-},R\rightarrow R(1,0)$; and
as $\frac{\gamma}{\beta ^{4/3}}\rightarrow -\infty ,R\rightarrow R(0,-1)=0$.
Since ($-\infty ,0]$ is
connected and the range of continuous functions with a connected domain is
also connected, all the possible $R$ that solutions of this kind
contribute is an
interval (0,$R_{1})=I_{1}$ where $R_{1}=\sup R(1,s)$ for all
$s\in(-\infty,0]$. The continuity of $R$ guarantees $R_{1}<\infty $.
In addition, $R(1,0)\leq R_{1}$.

\noindent{\bf Case (2).} In this case, $g''>0$ initially (i.e., for
sufficiently small $\xi >0$ ), which implies that $g,g'>0$
initially. From (\ref{eq:g5}) and lemma \ref{lm:g4<0}, $g^{v}<-(g'')^{2}$
as long as $g>0$. This means that the graph of $g'''$ is concave down and
decreasing as long as $g>0$. Thus,
either $g'''=0$ or $g=0$ eventually. If $g=0$, then
there must be a point at which $g'=0;$ if $g'''=0$ with $g'>0$, then $g''$
becomes concave down and decreases, thus, $g''$ must cross the $\xi $
axis, then
$g'$ is concave down and decreasing and must have a zero point.
Similar to above, by applying corollary \ref{coro:homo}, the maximal
interval for $R$, which admits a solution of this kind, is ($\inf
R(1,b),\sup R(1,b))=I_{2}$ for $b\in (0,\infty )$. Since $R(1,0)\in I_{1}$
and is a limit point of $I_{2}$, we see that $I_{1}\cup I_{2}$ is
connected.
In addition, there is only one zero of $g'$ in this case. To see
this, let $r$ be the first zero point of $g'$. Since $g''>0$ initially
and $r$ is the first zero of $g'$, it must be
that $g''(r)\leq 0$ and $g'>0$ in $(0,r)$. Then,
there is a point, say $s$, in $(0,r)$ such that $g'''(s)<0$.
By lemma \ref{lm:g4<0}, $g'''<0$ and hence
$g'',g'<0$ for all $\xi >r$. Note that $I_{2}$ must
be an open interval and $R(1,0)$ is either $\inf R(1,b)$ or $\sup R(1,b)$
for $b\in (0,\infty )$.

\noindent{\bf Case (3).} Unlike the two cases above, some $\beta $ and
$\gamma $ may not provide an $R$ at all. The discussion of this case
consists of the following three lemmas.

\begin{lemma}
\label{lm:nozero} If $\beta <0$ and $\gamma >0$ with $\beta \leq
-\sqrt{2}\gamma ^{3/4}$, then $g'(\xi ,\beta ,\gamma )$ has no
zero for all $\xi >0$.
\end{lemma}

\paragraph{Proof.}
By corollary \ref{coro:homo}, we may only consider $g(\xi ,l,1)$ where
$l=\frac{\beta }{\gamma ^{3/4}}$. Then, equation (\ref{eq:g3}) takes the form 
\begin{equation}
g'''=gg''-(g')^{2}+1
\label{eq:g3:1}
\end{equation}
subject to 
\begin{equation}
g(0)=g'(0)=0,\quad g''(0)=l.  \label{ini:noslip;3}
\end{equation}
From the condition of the lemma, $l\leq -\sqrt{2}$. Since $g^{iv}<0$ for
all 
$\xi >0$, we see $g'''<1$, then $g''\leq \xi +l$ and $g'\leq l\xi +
\frac{\xi ^{2}}{2}$ for all $\xi >0$.
Thus, $g''<0$ for $\xi <|l|$ and $g'(|l|)\leq -1$.
Hence, $g'<0$ for $0<\xi \leq |l|$. Next, we show that $g'<0$
for all $\xi >|l|$. To see this, it suffices to show that $g''<0$ for all
$\xi >0$. Suppose, for contradiction, that there is a first zero
of $g''$, say $x_{0}$. Then, $x_{0}>|l|$, $g'(x_{0})<-1$, and
$g'''(x_{0})\geq 0$ from (\ref{eq:g4}). However, from (\ref{eq:g3:1}),
$g'''(x_{0})<0$, a contradiction.  \hfill $\Box$

\begin{lemma}
\label{lm:2zeros} There exists a real number $l_{0}<0$ such that if
$0>\beta
>l_{0}\gamma ^{\frac{3}{4}}$ then, $g'(\xi ,\beta ,\gamma )$ has
exactly two zeros $r_{1}$ $<$ $r_{2}$ with $g(r_{1})<0$ and $g(r_{2})>0.$
Also, $g'',g'''>0$ at $r_{1}$ and $%
g'',g'''<0$ at $r_{2}$.
\end{lemma}

\paragraph{Proof.}
Consider the initial value problem (\ref{eq:g3:1})-(\ref{ini:noslip;3})
with $l=0$. Then, $g'''(0)=1$, which implies $g',g''>0$ initially.
Similar to the case (2) above, there is a unique zero point $r_{0}$ of
$g'(\xi ,1,0)$. Consider a compact set 
$[0,2r_{0}]$. From the proof of case (2), $g'(2r_{0}),g''(2r_{0})<0$.
It follows that for given $\epsilon >0$, there is a
$\delta _{0}>0$ such that for any $\delta \in (0,\delta _{0}),
|g^{(k)}(\xi, 0, 1)-g^{(k)}(\xi, -\delta ,1)|<\epsilon $ on $[0,2r_{0}]$
for $k=0,1,2,3$. Choosing $\epsilon $ small enough, we can make $g(\frac{%
r_{0}}{2},-\delta ,1)>0$, $g'(\frac{r_{0}}{2},-\delta ,1)>0$, $%
g'(2r_{0},-\delta ,1)<0$ and $g''(2r_{0},-\delta
,1)<0$. Noting that for this function $g$, the initial condition
$g''(0)<0$, hence $g$, $g',g''<0$ initially.
Thus, $g'$ must have a first zero point $r_{1}\in (0,\frac{r_{0}}{2})$
with $g(r_{1})<0$ and a second zero point $r_{2}\in (\frac{r_{0}}{2}
,2r_{0})$. Since the first zero point of $g'$ is the first relative
minimum and the second zero point of $g'$ the relative maximum,
$g''(r_{1})\geq 0$ and $g''(r_{2})\leq 0$.
Following the argument similar to above, there is no more zero point of
$g'$ for $\xi >r_{2}$. It is obvious that $g(r_{1})<0$. Since
$g'''$ decreases and changes signs in $(r_{1},r_{2})$,
we see $g'''<0$ for all $\dot{\xi}\geq r_{2}$. It can
be seen that $g(r_{2})>0$, for otherwise, $g'''(r_{2})\geq 1$
from (\ref{eq:g3:1}), a contradiction. \hfill$\Box$\smallskip

Now we show, by contradiction, that for any $\delta >0$, $g'(\xi
,-\delta ,1)$ has either no zero or two zeros. If not, then let $r_{1}$
denote the first zero of $g'$. Thus, $g(r_{1})<0$. Since
$g,g',g''<0$ initially, we first see $g''(r_{1})\geq 0$. Next we claim the
equal sign cannot hold, otherwise,
from (\ref{eq:g3:1}), that $g'=g''=0$ implies $g'''(r_{1})=1$, which is
impossible because of lemma \ref{lm:g4<0}. With $g(r_{1})<0,g'(r_{1})=0$,
and $g''(r_{1})>0$, one can see that after $r_{1}$ the function $g$ begins
increasing. Now the question is: can the function $g$ keep increasing and
negative for all $\xi $? If so, then there would be a zero point of $%
g''$ with $g<0$ and $g'''\leq 0$, which implies $g^{iv} \geq 0$. But,
this is impossible by lemma \ref{lm:g4<0}.

Define 
\[
A=\{ b<0 \ | \ g'(\xi ,\beta ,1) \ \mbox{has two zeros for
all }\ \beta \in (b,0)\}. 
\]
It is seen from lemma \ref{lm:2zeros} that the set $A$ is open and not
empty. It follows from lemma \ref{lm:nozero} that the set $A$ is bounded
below. Let 
\[
\beta _{1}=\inf A. 
\]
Then, $\beta _{1}<0$, $\beta _{1}\notin A$, and so $g'(\xi ,\beta
_{1},1)$ has no zero. Also, $g''(\xi ,\beta _{1},1)$ $<0$
for
all $\xi >0$, otherwise, if $g''=0$ at an $x$, then
$g'''(x)=g^{iv}(x)/g(x)>0$. It then follows that
$g'',g'''>0$ as long as $g'<0$
for otherwise, at the first zero point of $g''$ after $x$,
if any, $g'''<0$, which made $g^{iv}>0$, a contradiction. It
turns out that $g'$ becomes zero because of the concavity. From
the above argument, $g'$ would have a second zero. This contradicts
the definition of $\beta _{1}$.

\begin{lemma}
\label{lm:2infinity} Suppose that $\beta \in A$ and that $\beta \rightarrow \beta_{1}$.
Then $\lim g(r_{1})=-\infty $, and $\liminf r_{2}g(r_{2})=\infty $.
\end{lemma}

\paragraph{Proof. }
Since $g''(\xi ,\beta _{1},1)<0$ for all $\xi >0$, we see
$g(\xi ,\beta _{1},1)\rightarrow -\infty $ as $\xi \rightarrow \infty $ and
$g'(\xi ,\beta _{1},1)<-\sigma $ for some $\sigma >0$ and for all
$\xi >x_{0}$ where $x_{0}$ is a constant. Then, for any $M_{0}>0$, there
is an interval $(\beta _{1},b_{0})\subset A$ such that for $\beta \in (\beta
_{1},b_{0})$, $g'(M_{0},\beta ,1)<0$ and
$g(M_{0},\beta,1)<-\frac{\sigma }{2}M_{0}$. This implies that for
$\beta \in (\beta _{1},b_{0})$, $r_{1}>M_{0}$ where $r_{1}$ is the first
 zero of $g'$, and $g(r_{1})<-$ $\frac{\sigma }{2}M_{0}$.
This proves the first conclusion of
the lemma and $r_{1}\rightarrow \infty $ as
$\beta \rightarrow \beta_{1}^{+}$. To prove the second conclusion, we only
need to prove that
$\lim\inf g(r_{2})>0$. Suppose not. Then, for any small $\epsilon >0$ there
would be infinitely many $\beta \in A$ such that the corresponding
$g(r_{2})<\epsilon $. Consider the second zero point $x_{2}$ of
$g''$ at which $g'''<0$ and $g>0$ (it is impossible to have $g''=0,g'''<0$,
and $g\leq 0$ simultaneously). From (\ref{eq:g3:1}), $g'(x_{2})>1$.
Let $x_{2}$ be the point at which $g'=1$. Then, $\epsilon >g(r_{2})>g(x_{1})$.
Since $g''<0$ after $x_{2}$ and $g''(x_{1})=1$
,we have $g(\xi )>\xi -x_{2}+g(x_{2})$ for $\xi \in (x_{2},x_{1})$ since
$g'\geq 1$ over that interval. This implies $x_{1}-x_{2}<\epsilon$.
Hence, by the mean value theorem and lemma \ref{lm:g4<0},
 $g'''(x_{1})\leq \frac{0-g''(x_{1})}{x_{2}-x_{1}}<
\frac{g''(x_{1})}{\epsilon }$. On the other hand, from (\ref{eq:g4}),
$g'''=gg''$ at $x_{1}$, and so $g(x_{1})g''(x_{1})<
\frac{g''(x_{1})}{\epsilon }$, which implies $g(x_{1})>\frac{1}{\epsilon }$,
a contradiction. \hfill$\Box$\smallskip

Lemma \ref{lm:2infinity}, in fact, implies that the solutions of BVP (1)
exist for sufficiently large $|R|$. Furthermore, we can show that as
$\beta \rightarrow 0^{-}$, $g(r_{1})\rightarrow 0$ and $g(r_{2})\rightarrow
y_{0}=g(r,0,1)$. It then follows, from the fact that continuous functions
map connected set into connected sets, that the solution of BVP (1) exists
for all $R\leq 0$, and for all $R>R_{3}$ where $R_{3}=\inf R(\beta ,1)\leq
R(0,1)$.

Now the picture is clear, the solutions exist for all $R\leq 0$. As $R$
increases, the solutions of case (1) appear till $R_{1}$ (defined as
above)
while the solutions of case (2) appear for $R\in I$ where $I$ is given in
the proof of case (2). Then, the solutions of case (3) appear for all $%
R>R_{3.}$ It can be seen that $R_{3},R_{1}\in \overline{I_{2}}$, the
closure of $I_{2}$, since 
\[
R_{1}=\sup_{s\leq 0}R(1,s)\geq R(1,0)\geq \inf_{s>0}R(1,s) 
\]
which implies $R_{1}\in \overline{I_{2}}$. By the corollary, 
\[
\sup_{s\in (0,\infty )}R(1,s)=\sup_{t\in (0,\infty )}R(t,1) 
\]
which implies $R_{3\in \overline{I_{2}}}$ since $\inf R(\beta ,1)<\sup
R(\beta ,1)$. This shows that there exists at least one solution for each
$R\in (-\infty ,\infty )$.

Next we prove the uniqueness of the solution of
BVP (1) for $R<0$, which is a consequence of the following lemma.

\begin{lemma}
 \label{lm:unique} Suppose that $r_{1}=r_{1}(\beta )$ denotes the
first zero point of $g'(\xi ,\beta ,1)$ for $\beta <0$. Then both
$r_{1}$ and $g(r_{1})$ are increasing function of $\beta$.
\end{lemma}

\paragraph{Proof.}
Let $\beta _{1}<\beta _{2}<0$ be given such that the corresponding
solutions 
$g_{1}$ and $g_{2}$ have zero points of their first derivatives. Since
$g_{1}'''(0)=g_{2}'''(0)=1$, ($g_{2}-g_{1})''>0$, and $(g_{2}-g_{1})'>0$
for sufficiently small $\xi >0$. This shows that $g_{2}^{2}<g_{1}^{2}$ ,
and $g_{2}>g_{1}$ initially. Using the integral form of $g^{iv}$ given by
(\ref{ex:g4}), we obtain that $g_{2}^{iv}>g_{1}^{iv}$ initially. Hence,
$g_{2}'''>g_{1}'''$ initially. We
claim that $g_{2}^{(k)}>g_{1}^{(k)}$ for $k=0,1,2,3$ as long as
$g_{1}''\leq 0$ and $g_{2}'\leq 0$, which implies
that $g_{2}''$ becomes zero before $g_{1}''$
does. To see this, we subtract the equation $g_{1}^{iv}=g_{1}g_{1}'''
-g_{1}'g_{1}''$ from $g_{2}^{iv}=g_{2}g_{2}'''-g_{2}'g_{2}''$ and obtain 
\[
(g_{2}-g_{1})^{iv}=g_{2}(g_{2}-g_{1})'''+(g_{2}-g_{1})g_{1}'''-
g_{2}'(g_{2}-g_{1})''-g_{1}''(g_{2}-g_{1})'\,. 
\]
Let $\psi =g_{2}-g_{1}$. The last equation can be written in the form 
\begin{equation}
\psi ^{iv}=g_{2}\psi '''-g_{2}'\psi ''-g_{1}''\psi '
+g_{1}'''\psi\,.  \label{eq:ff4}
\end{equation}
Also, all of $\psi ,\psi '$, $\psi '',\psi'''$, and
$\psi ^{iv}$ are positive in $(0,\delta )$ where $\delta $ is sufficiently
small. Note that $g_{1}'''>0$
for $\xi <r_{1}(\beta _{1})$. We then see from (\ref{eq:ff4}) that $\psi
'''>0$ as long as $g_{1}''\leq 0$ and
$g_{2}'\leq 0$ since at the first zero $x$ of $\psi '''$,
 if any, $\psi ^{iv}(x)>0$, a contradiction. This shows that $g_{2}''=0$
before $g_{1}''$ does.

Now there are two possibilities: (1). $g_{2}'=0$ before or on $g_{1}''=0$,
which implies the lemma, and (2). $g_{2}'=0$
at a point with $g_{1}''>0$. In the latter case, we begin
with the point $x_{1}$ where $g_{2}''=0$. Of course, $g_{1}''(x_{1})<0$,
and $\psi ^{(k)}(x_{1})>0$ for $k=0,1,2,3$.
Since $g_{2}''>0$ for $\xi >x_{1}$, $\psi ''$,
hence $\psi '$ and $\psi $ are positive as long as $g_{1}''\leq 0$.
Let $x_{2}$ be the first zero point of $g_{1}''$. Then, at
$\xi =x_{2}$, $\psi ,\psi '$, and $\psi''$ are positive, and
$g_{2}''(x_{2})>0$, $g_{1}'(x_{2})<0$, and $g_{1}'(x_{2})<0$. Consider the
third order equation of (\ref{eq:ff4}) 
\[
\psi '''=g_{2}\psi ''-(g_{2}'+g_{1}')\psi '+g_{1}''\psi\,,
\]
with the initial condition $\psi ^{(k)}(x_{2})>0$ for $k=0,1,2$. At the
first zero $x_{3}$ of $\psi $, if any, $\psi '''>0$
from the last equation.. This shows that after $x_{2}$, $0>g_{2}'>g_{1}'$
before one of them reaches the zero point. Consequently, $g_{2}'$ becomes
zero before $g_{1}'$ does. Therefore, $r_{2}<r_{1}$, and
$0>g_{2}(r_{1}(\beta _{2})\dot{)}>g_{1}(r_{1}(\beta ))$. 
The proof of theorem \ref{thm:noslip} is completed. \hfill$\Box$

\paragraph{Remark.} Lemma \ref{lm:unique} can be generalized  and then
applied to prove the uniqueness of the solution for the original Berman
problem with injection. This will be presented in another paper.

\section{Proof of theorem \ref{thm:2walls}}

BVP (2) is related the initial value problem with the initial conditions 
\[
g(0)=g''(0)=0,g'(0)=\alpha ,g'''(0)=\gamma\,.
\]
Any symmetric solution of BVP (2) corresponds to a solution $g(\xi ,\alpha
,\gamma )$ of the initial value problem which satisfies
\[
g(b)=0,g'(b)=-\frac{R}{b^{2}}\,.
\]

Then, there are only three possible cases to be considered:
(I) $\alpha>0$ and $\gamma \leq 0$, (II) $\alpha \geq 0$ and $\gamma \geq 0$
with ($\alpha ,\gamma )\neq (0,0)$, and (III) $\alpha <0$ and $\gamma >0$.

\paragraph{Case (I)} Applying lemma \ref{lm:g4<0}, one sees that $g'=0$ at
an $r>0$ because the concavity and monotonicity of $g'$, and then
$g$ must cross the $\xi $-axis at a $b>0$ with $g'(b)<0$ and $g''(b)<0$.
This is the only zero of $g$ in (0,$\infty )$.

\paragraph{Case (II)} We apply the equation (\ref{eq:g5}) and see that
 $g^{v}<0$ as long as $g>0$, which implies $g'''$ is concave down and
decreasing in $(0,\infty )$. Thus, $g'''$ must crass
$\xi $-axis, hence $g''$ becomes concave down and  decreasing.
Continuing the similar argument shows that there exists a unique $b$ at
which $g(b)=0$ with $g'(b)$, $g''(b)<0$.

\paragraph{\bf Case (III)} We can prove that $g$ has either none or two positive
zero points. Suppose that there is a zero point of $g$. Let $b_{1}$ be the
first such point. Then, $g'$ and $g''$ must be of the same sign
by lemma \ref{lm:g4<0}. It turns out that $g'(b_{1}),g''(b_{1})>0$.
Since the equation is autonomous, an argument similar to
that of case (II) shows the existence of the second positive zero point of
$g$. We now define 
\[
B=\{\alpha _{0}<0 | \ g(\xi ,\alpha ,1)\ \mbox{ has two
positive zeros for all }\alpha \in (\alpha _{0},0) \}. 
\]
It can be easily seen that the set $B$ nonempty sine $B$ consists of at
least some $\alpha <0$ with small absolute values. $B$ is also an open set
because of the transversality of the function when it crosses the $\xi $%
-axis. By the definition, the set $B$ is connected. Let $\inf
B=\tilde{\alpha}$. Then, for any
$\alpha \in (\tilde{\alpha},0)$, $g(\xi ,\alpha ,1)$ has
exact two zero pints in $(0$,$\infty )$. Note that the first zero $b_{1}$ of 
$g$ provides a negative $R$ while the second zero $b_{2}$ of $g$ gives a
positive $R$.  \smallskip

Next, we determine the feasible set for $R$ considering all possible pairs
$(\alpha ,\gamma )$.
In the case (I), the solution is concave down. As $\alpha \rightarrow 0^{+}$,
$R(\alpha ,-1)\rightarrow 0$ since both $b$ and $g(b)$ approach zero. If
$\alpha \rightarrow \infty $, we apply corollary \ref{coro:homo},
$R(\alpha ,-1)=R(1,-\frac{1}{\alpha ^{2}})$. Since $g(\xi ,1,0)$ is the
linear function $g=\xi $, we can choose $\alpha $ sufficiently large such
that the positive zero point $b>$ $M$ for any give $M>0$. To show that
$R\rightarrow \infty $, as above, we need to prove $g'(b)$ is bounded above
by a nonzero constant. An integration of (\ref{eq:g4}), with the initial
condition $g'(0)=\alpha $, $g''(0)=0$, $g'''(0)=-1$
where $\alpha \gg 1$, produces
\begin{equation}
g'''=gg''-(g')^{2}+\alpha^{2}-1.  \label{eq:g3:2}
\end{equation}
At $\xi =b$, $g'''<0$. This implies $g'(b)>\sqrt{\alpha ^{2}-1}$,
which shows that $g'(b)\rightarrow \infty.$
This shows that the type I solutions are valid for all $R\in (0,\infty )$.

In the case (II), similar to the case (I), $R\rightarrow \infty $ as
$\alpha \rightarrow \infty $. However, as $\alpha \rightarrow 0^{+}$,
$R\rightarrow R(0,1)>0$. let $R_{2}=\inf R(\alpha ,1)$ for
$\alpha \in (0,\infty )$. The type II solutions exist for all $R>R_{2}$.

In the case (III), we consider the processes $\alpha \rightarrow 0$ and
$\alpha \rightarrow \tilde{\alpha}$ which is give above. For positive $R$,
we see the feasible set is ($\inf b_{2}^{2}|g'(b_{2})|,\sup
b_{2}^{2}|g'(b_{2})|)=(R_{3},\infty )$. Here, we applied
corollary \ref{coro:homo} again, $R(\alpha ,1)=R(-1,\frac{1}{\alpha^{2}})$.
The argument for having $\sup b_{2}^{2}|g'(b_{2})|=\infty $ is
$b_{2}$ can be sufficiently large and $g'(b)$ is
bonded below as $\alpha \rightarrow \tilde{\alpha}$. Why is $b_{2}$ can
be sufficiently large? Since either $\tilde{\alpha}=-\infty $ or
$g(\xi,\tilde{\alpha},1)$ has no zero point at all. In the former case,
we consider $g(\xi,-1,0)=-\xi$; in the latter case,
$g(\xi ,\tilde{\alpha},1)$ is a decreasing function defined on
$(0,\infty )$. In either case, we can choose and $\alpha$ such that the
corresponding $b_{2}$ is sufficiently large (so is $b_{1})$. For the first
zero $b_{1}$, we see that as $\alpha\rightarrow 0^{-}$, $g(\xi ,\alpha ,1)$
and $g(\xi ,0,1)$ can be arbitrarily close on a closed interval $[0,T]$
where $T$ is a positive constant. This shows as $\alpha \rightarrow 0^{-}$,
$R\rightarrow 0^{-}$.  As $\alpha\rightarrow \tilde{\alpha}$,
$b_{1}\rightarrow \infty $, and $g'(b)$
must be bounded below, otherwise, if $g'(b_{1})\rightarrow 0$ (or
a sequence), then $g'''(b_{1})\rightarrow \tilde{\alpha}^{2}+1>1$,
which is impossible, since $g'''$ is a decreasing function. The proof of
theorem 2 is complete.

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\end{thebibliography}
\bigskip

{\sc Chunqing Lu}\\
 Department of Mathematics and Statistics\\
 Southern Illinois University at Edwardsville \\
 Edwardsville, Illinois 62026, USA\\
Email address: clu\@siue.edu 


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