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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 302, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/302\hfil Analysis of stagnation point flow]
{Analysis of stagnation point flow of an upper-convected
Maxwell fluid}

\author[J. E. Paullet \hfil EJDE-2017/302\hfilneg]
{Joseph E. Paullet}

\address{Joseph E. Paullet \newline
School of Science, Penn State Behrend,
Erie, PA 16563-0203, USA}
\email{jep7@psu.edu}


\thanks{Submitted August 24, 2017. Published December 6, 2017.}
\subjclass[2010]{35B15, 76D10, 76A05}
\keywords{Stagnation point; boundary value problem;
\hfill\break\indent upper-convected Maxwell model}

\begin{abstract}
Several recent papers have investigated the two-dimensional
stagnation point flow of an upper-convected Maxwell fluid by
employing a similarity change of variable
to reduce the governing PDEs to a nonlinear third order ODE boundary
value problem (BVP). In these previous works, the
BVP was studied numerically and several conjectures regarding the
existence and behavior of the solutions were made.
The purpose of this article is to mathematically verify these conjectures.
We prove the existence of a solution to the BVP for all relevant
values of the elasticity parameter.
We also prove that this solution has monotonically increasing
first derivative, thus verifying the conjecture that no
``overshoot'' of the boundary condition occurs. Uniqueness
results are presented for a large range of parameter space and bounds on the
skin friction coefficient are calculated.
\end{abstract}

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

\section{Introduction}

This article analyzes the boundary value problem (BVP) governing
two-dimen\-sional stagnation-point flow of a fluid obeying
the upper-convected Maxwell model \cite{kn},
\cite{lip}, \cite{shs}.
Models such as this have been developed to describe the viscoelastic
properties that certain fluids exhibit, and which not
captured using standard Newtonian theory. Several previous
studies concerning stagnation-point flow of viscoelastic fluids
investigated a second-grade fluid model \cite{bw,rr}.
However, Sadeghy et al.\ \cite{shs} note that
the second-grade model is valid only for slow flows involving
small levels of elasticity. Also, overshoot of the
free stream velocity
inside the boundary layer has been observed for
the second-grade model
\cite{a1,bw}. A later study \cite{gr} suggested that the
second-grade model of \cite{bw} needed to be augmented and
in this augmented model no overshoot was observed
\cite{a2,s}.

Given these limitations and uncertainties
surrounding the behavior of stagnation-point
flow of a second-grade fluid, Sadeghy et al.\ \cite{shs}
conducted an investigation of
the same physical configuration using an
upper-convected Maxwell (UCM) model. They note that
the upper-convected Maxwell model is
valid for much larger values of elasticity than is the
second-grade model. They endeavored to determine whether
the behavior observed in the second-grade model regarding the
velocity profiles and the skin friction coefficient are present
in the UCM model.

Specifically, the problem derived in \cite{shs}
(see also \cite{kn,lip}) is
\begin{equation}
[1+kf(\eta)^2]f'''(\eta)+
[1-2kf'(\eta)]f(\eta)f''(\eta)+1-f'(\eta)^2=0,
\quad 0<\eta<\infty, \label{ode}
\end{equation}
subject to
\begin{equation}
f(0)=0,\quad f'(0)=0,\quad f'(\infty)=1.
\label{bcs}
\end{equation}
where $k>0$ is the elasticity parameter

Sadeghy  et al.\ \cite{shs} investigate this BVP numerically
and report a unique solution for each value of $k$.
They also detect no overshoot in the velocity profiles for
any value of $k$. (Overshoot
would involve a solution to the BVP with velocities
that exceed the free-stream velocity far from the plate and
would be exhibited by $f'(\eta)>1$ on some
interval(s) of $\eta$.) Sadeghy et al.\ also find
that the skin friction coefficient, which is proportional
to $f''(0)$, decreases as the elasticity parameter $k$ increases.

The objective of this paper is to determine whether these
numerical observations can be verified mathematically
from direct analysis of the BVP.
We note that since the BVP \eqref{ode}-\eqref{bcs}
is nonlinear,
there is no guarantee that a solution even exists, or that it
is unique. Thus our first goal
(Section 2) is to prove that for any $k>0$,
a solution to the BVP does indeed exist. This solution will
be shown to satisfy $0<f'(\eta)<1$ and $f''(\eta)>0$ for all
$\eta>0$. Thus a solution without overshoot does exist for
this model. In fact, since it can be seen from the ODE
\eqref{ode} that $f'$ cannot have a maximum above $f'=1$,
solutions with overshoot are not possible in this model.
Next, for $k\ge k_0\approx 3.5584$ (defined in Section 3) we prove
that there cannot be two solutions which both satisfy
$0<f'<1$ and $f''>0$ for all $\eta >0$.
We show that any second (or further) solutions
must have very specific behavior, including a minimum of
$f'$ below $-1$. Thus, any further solutions would have to
exhibit the physically dubious behavior of flow reversal ($f'<0$).
While such solutions are not unknown in some flow configurations
\cite{w}, we conjecture, based on our numerical investigations,
that they do not occur in the current model.
In Section 4, we discuss the behavior of the skin
friction coefficient as a function of the elasticity parameter
as well as other qualitative properties of the solution.
Finally, in Section 5, we discuss open problems for the
current model and how the analysis presented here might
be applied to other models involving the upper-convected
Maxwell fluid model. We note that techniques employed to other
stagnation point problems, such as those used in \cite{bar},
might fruitfully be applied to the current problem.


To investigate the BVP \eqref{ode}-\eqref{bcs} we
study a family of initial value
problems (IVPs) given by the ODE \eqref{ode} along with
the conditions at $\eta=0$, \eqref{bcs}a,b. To this we
add a third condition at $\eta=0$, namely,
\begin{equation}
f''(0)=\alpha.
\label{al}
\end{equation}
Equations \eqref{ode},\eqref{bcs}a,b and \eqref{al}
constitute a well-posed IVP.
By standard existence and uniqueness
theory, for each value of $\alpha$, this IVP will have a unique
local solution
on some open interval containing $\eta=0$.
Denote this solution by $f(\eta;\alpha)$. Occasionally the dependence
of $f$ on $\eta$ or $\alpha$ or both will be dropped for convenience
and ease of reading.
In the next section we show that
a value $\alpha^*$ exists such that $f(\eta;\alpha^*)$ exists
for all $\eta>0$ and also satisfies the boundary condition
at infinity, \eqref{bcs}c, giving a solution to the BVP.

\noindent\textbf{Nomenclature:}
\begin{itemize}
\item $f:$ dimensionless stream function

\item $\eta:$ similarity variable

\item $k:$ elasticity parameter

\item $\alpha:$ topological shooting parameter

\item $v:$ derivative of $f$ with respect to $\alpha$

\item $\eta_0, \eta_1, \eta_2, \eta_3, \eta_4, \eta_5, \overline{\eta}, \overline{\overline{\eta}}$:
various values of $\eta$ used in the proofs

\item $k_0, k_1, k_2, \hat k$:
various values of $k$ used in the proofs

\item $\alpha_1, \alpha_2, \overline \alpha, \alpha^*$:
various values of $\alpha$ used in the proofs
\end{itemize}

\section{Existence and qualitative properties}

\begin{theorem} \label{thm1} 
For every $k>0$ there exists a solution to the BVP \eqref{ode}-\eqref{bcs}.
Further, this solution satisfies
$0<f'(\eta)<1$ and $f''(\eta)>0$ for all $\eta>0$.
\end{theorem}

\begin{proof}
The proof of existence will involve the following
subsets of $(0, \infty)$:
\begin{gather*}
\mathcal{A}=\big\{ \alpha>0  : f''(\eta;\alpha)=0
\text{ before } f'(\eta;\alpha)=1\big\}, \\
\mathcal{B}=\big\{ \alpha>0 : f'(\eta;\alpha)=1 \text{ before }
 f''(\eta;\alpha)=0\big\}.
\end{gather*}

We will show that each of these sets is non-empty and open.
For $\mathcal{A}$ this is just a matter of continuity of
the solutions of the IVP
\eqref{ode}, \eqref{bcs}a,b and \eqref{al} in its initial
conditions.
We claim that for all $\alpha>0$ sufficiently small,
$\alpha \in \mathcal{A}$.
To see this, first note that
\begin{equation}
f'''(0;\alpha)=-1<0.
\end{equation}
If we take $\alpha=0$, then
for small $\eta >0$ we have $f'(\eta;0)<1$
and $f''(\eta;0)<0$; say on $(0,\varepsilon]$ for some $\varepsilon>0$.
By continuity of
the solutions of the initial value problem in its initial conditions,
for $\alpha>0$ sufficiently small, $f'(\eta;\alpha)$ will stay close to
$f'(\eta;0)$, i.e.\ we can arrange that
$f'(\eta;\alpha)<1$ on $(0,\varepsilon]$ with
$f''(\varepsilon;\alpha)<0$.
But as $f''(0;\alpha)>0$,
there must exist a first $\eta_0\in (0,\varepsilon)$
such that $f''(\eta_0;\alpha)=0$ with $f'(\eta;\alpha)<1$ on $[0,\eta_0]$.
Thus for $\alpha>0$ sufficiently small we have $\alpha \in \mathcal{A}$.
To show that $\mathcal{A}$ is open, consider
$\overline \alpha \in \mathcal{A}$. We will show that all $\alpha$
sufficiently close to $ \overline \alpha$ are also in $\mathcal{A}$. At
$\eta_0(\overline \alpha)$ we have
$0<f'(\eta_0;\overline \alpha)<1$ and $f''(\eta_0;\overline \alpha)=0$.
Evaluating \eqref{ode} at $\eta_0(\overline \alpha)$ implies that
\begin{equation}
f'''(\eta_0;\overline \alpha)=
\frac{f'(\eta_0;\overline \alpha)^2-1}{1+kf(\eta_0;\overline \alpha)^2}\ne 0.
\end{equation}
Thus, by continuity of the solutions of the IVP in its
initial conditions,
for $\alpha$ sufficiently close to
$\overline \alpha$,
$f''(\eta;\alpha)$ will also
have a root, $\eta_0(\alpha)$,
near $\eta_0(\overline \alpha)$ with $f'(\eta_0;\alpha)<1$.
Thus $\alpha \in \mathcal{A}$ and $\mathcal{A}$ is open.

The study of the set $\mathcal{B}$ will require bounds on $f$ and $f'$.
First note that integrating \eqref{ode}, (by parts where necessary),
from $0$ to $\eta$ gives
\begin{equation}
f''(\eta)[1+kf(\eta)^2]=\alpha-\eta
+(2kf'-1)ff'+2\int_0^{\eta}f'(t)^2[1-kf'(t)]\,dt
\label{int}
\end{equation}
We claim that for large positive $\alpha$,
$f'=1$ in the interval $(0,1]$
strictly before $f''=0$. (In fact in this case, if $f'=1$ before
$f''=0$, then $f''=0$ cannot occur at all.)
Suppose that the assertion is false.
Then one of the following must occur: (i) $f''=0$ at some first
point in $(0,1]$ with $f'<1$, (ii) $f''>0$ and $f'<1$ for all 
$\eta \in (0,1]$, or (iii) $f''=0$ and $f'=1$ simultaneously. We eliminate each
of these in turn.
To begin with (i), suppose that there exists a first $\eta_1 \in (0,1]$ with
\begin{equation}
f''(\eta_1)=0
\label{f''=0}
\end{equation}
with $0\le f' <1$ for $\eta \in [0,\eta_1]$.
Integrating $0\le f' <1$
from 0 to $\eta$ gives $0 \le f <\eta\le 1$
on $[0,\eta_1]\subset [0,1]$.
Using these bounds on $f$ and $f'$ and
ignoring the positive terms other than
$\alpha$ on the right hand side of
\eqref{int} we obtain:
\begin{equation}
f''(\eta)\ge \frac{\alpha -2(k+1)}{k+1} \quad \eta \in [0,\eta_1].
\end{equation}
Thus if we choose $\alpha>2(k+1)$ then $f''(\eta_1)>0$
contradicting \eqref{f''=0}. A similar argument shows that if
$\alpha>3(k+1)$ then we cannot have (ii) $f''>0$ and $f'<1$ on
all of $(0,1]$. (i.e.\ $f'(1)$ will be greater than $1$.)
This leaves only case (iii) $f'=1$ and $f''=0$ simultaneously;
however, substituting this information into \eqref{ode}
gives $f'''=0$ implying
 that $f'(\eta) \equiv 1$, contradicting the basic existence
and uniqueness theorem for initial value problems,
as $f'(0)=0\ne 1$. Thus if $\alpha >3(k+1)$ then
we must have $f'=1$ strictly before $f''=0$ and therefore
$\alpha \in \mathcal{B}$. An argument similar to that
for the set $\mathcal{A}$ shows that $\mathcal{B}$ is also open.


Thus, the sets $\mathcal{A}$ and $\mathcal{B}$ are non-empty
and open. They are also obviously disjoint.
But the interval $(0,\infty)$ is connected and thus
$\mathcal{A} \cup \mathcal{B}\ne (0,\infty)$. Therefore, there exists
some $\alpha^*$ such that $\alpha^* \notin \mathcal{A}$ and 
$\alpha^* \notin \mathcal{B}$. For such a value of $\alpha^*$ the only possibility
is for the solution $f(\eta;\alpha^*)$ to exist for all $\eta>0$ with
$0<f'(\eta;\alpha^*)<1$ and $f''(\eta;\alpha^*)>0$.

It remains to show that $f'(\infty ;\alpha^*)=1$, satisfying the
boundary condition at infinity \eqref{bcs}c. Since $f'(\eta;\alpha^*)$
is positive, increasing, and bounded above by 1 we conclude that
$f'(\infty;\alpha^*)=L\le 1$ exists. Suppose for contradiction that
$0<L<1$.

To begin, we claim that $f'''\le 0$ for all $\eta >0$. Since $f'''(0)=-1$,
$f'''$ starts off negative. Suppose that $f'''$ were to
assume a positive value at some point, say $f'''(\eta_2)>0$ for
some $\eta_2>0$. We could not have $f'''>0$ for all
$\eta > \eta_2$ since two integrations from $\eta_2$ to
$\eta>\eta_2$
of the inequality $f'''>0$ would imply that $f'\to +\infty$
as $\eta \to \infty$ contradicting the boundedness of $f'$.
Thus $f'''$ would at some point have to decrease
back through zero.
Recalling the other properties of $f$ from above,
this would require a point at which $f>0$, $0<f'<1$, $f''>0$,
$f'''=0$ and $f^{(4)}\le 0$. Differentiating \eqref{ode} and
evaluating at such a point implies that
\begin{equation}
(1+kf^2)f^{(4)}-f'f''-2k(f')^2f''-2kf(f'')^2=0. \label{f'''=0}
\end{equation}
However, each term on the left side of \eqref{f'''=0} is non-positive
with the last three necessarily negative and we have a
contradiction. Thus $f'''$ cannot become positive and we have
$f'''\le 0$ for all $\eta >0$. This along with
$f''>0$ for all $\eta>0$, implies that
$f''(\infty)$ exists and since
$f'(\infty)$ also exists we must have $f''(\infty)=0$.

Next note that from the properties above on $f$ through $f'''$
we can conclude from the equation for the fourth derivative,
\begin{equation}
(1+kf^2)f^{(4)}+ff'''-(1+2kf')f'f''-2kf(f'')^2=0,
\label{f4}
\end{equation}
that $f^{(4)}>0$ for all $\eta>0$. This along with $f'''\le 0$
implies that $f'''(\infty)$ exists, and must be zero since
$f''(\infty)$ also exists.

Next, rewrite the ODE \eqref{ode} as
\begin{equation}
(1-2kf')ff''+kf^2f'''=(f')^2-1-f'''.
\label{ode2}
\end{equation}
On the right side of \eqref{ode2} we have that $(f')^2-1-f''' \to
L^2-1=-m<0$ as $\eta \to \infty$. Thus there exists
an $\eta_3>0$ such that
\begin{equation}
(1-2kf')ff''+kf^2f'''<-\frac m 2 \quad \forall  \eta >\eta_3. \label{ineq1}
\end{equation}
Dividing by $f>0$ we have
\begin{equation}
f''-2kf'f''+kff'''<-\frac m 2 \frac 1 f
\quad \forall \eta >\eta_3.
\label{ineq2}
\end{equation}

Since $0<f'< L$ for all $\eta>0$ (not just $\eta > \eta_3$)
we have on
integration from 0 to $\eta>0$ that
\begin{equation}
f <L\eta, \quad \forall \eta>0, \label{ineq3}
\end{equation}
or, after some rearrangement,
\begin{equation}
-\frac 1 f <-\frac{1}{L\eta}, \quad \forall \eta>0.
\label{ineq4}
\end{equation}

Using \eqref{ineq4} in \eqref{ineq2} we obtain
\begin{equation}
f''-2kf'f''+kff'''<-\frac{m}{2L\eta}, \quad \forall \eta>\eta_3.
\label{ineq5}
\end{equation}
Integrating (by parts where necessary) from $\eta_3$ to
$\eta>\eta_3$, collecting some terms
and aggregating all constants on the left into a quantity
called $C$ we obtain
\begin{equation}
f'-\frac{3k}{2}(f')^2+kff''+C<-\frac{m}{2L}\left(
 \ln \eta-\ln \eta_3\right),
\quad \forall \eta>\eta_3. \label{ineq6}
\end{equation}
Since $f'$ is bounded, letting $\eta \to \infty$ we conclude
that $ff'' \to -\infty$, giving a contradiction since
$f>0$ and $f''>0$.
Thus the assumption that
$f'\to L<1$ leads to a contradiction and we must
have $f'(\infty;\alpha^*)= 1$ giving a solution to the BVP.
\end{proof}

\section{Uniqueness results}

\begin{theorem} \label{thm2}
If $k\ge k_0 \approx 3.5584$, then there cannot be two solutions
$f(\eta)$ of
the BVP \eqref{ode}-\eqref{bcs} both of which satisfy
$0<f'(\eta)<1$ and $f''(\eta)>0$
for all $\eta >0$. The value $k_0$ will be defined precisely
in the proof.
\end{theorem}

\begin{proof} By contradiction, assume the the existence of two solutions,
$f(\eta;\alpha_1)$ and $f(\eta;\alpha_2)$, of the BVP
\eqref{ode}-\eqref{bcs} both of which satisfy
$0<f'(\eta;\alpha_i)<1$ and $f''(\eta;\alpha_i)>0$
for all $\eta >0$, $i=1,2$. Without loss of generality
assume that $\alpha_2>\alpha_1$.

To arrive at a contradiction we will
 use the quantity
$v=\partial f/\partial \alpha$. Differentiating the ODE
\eqref{ode} along with the initial conditions for
$f(\eta; \alpha)$ with respect to $\alpha$ we see that
$v(\eta;\alpha)$ satisfies the following IVP:
\begin{equation}
(1+kf^2)v'''+(1-2kf')fv''-2(kff''+f')v'+[(1-2kf')f''+2kff''']v=0,
\label{vode}
\end{equation}
subject to
\begin{equation}
v(0)=0, \quad v'(0)=0, \quad v''(0)=1.
\label{vics}
\end{equation}
Evaluating this at $\eta =0$ gives $v'''(0)=0$.
Differentiating \eqref{vode} with respect to $\eta$ gives
\begin{equation}
\begin{aligned}
&(1+kf^2)v^{(4)}+fv'''-[(1+2kf')f'+4kff'']v''
-(4kf'+1)f''v'\\
&+[2kff^{iv}+f'''-2kf''^2]v=0.
\end{aligned} \label{v4ode}
\end{equation}
Evaluating at $\eta=0$ we have $v^{(4)}(0)=0$.
Finally, differentiating
\eqref{v4ode} with respect to $\eta$ and evaluating at $\eta=0$
gives $v^{(5)}(0)=2\alpha > 0$.
Our main interest is in the behavior of $v(\eta;\alpha)$
and its derivatives
for $\alpha_1 \le \alpha \le \alpha_2$ and $\eta>0$.
Note that for small $\eta >0$ we have
$v>0$, $v'>0$, $v''>0$, $v'''>0$
and $v^{(4)}>0$. Ultimately, we wish to show that $v'$ cannot
have a maximum and thus $v'>0$ will be bounded away from zero
as $\eta \to \infty$. Also note, that before $v'$ can have
a maximum, $v''$ must first have a maximum.

So for the purpose of contradiction suppose that $v''$ has
a first maximum at $\overline{\eta}$ and $v'$ has a first maximum
at $\overline{\overline{\eta}}>\overline{\eta}$.
At $\overline{\eta}$ we have $v(\overline{\eta};\alpha)>0$,
$v'(\overline{\eta};\alpha)>0$, $v''(\overline{\eta};\alpha)>0$, $v'''(\overline{\eta};\alpha)=0$,
and $v^{(4)}(\overline{\eta};\alpha)\le0$.
Note also that, up until this point, $v(\eta;\alpha)$ and
all its derivative through
$v'''(\eta;\alpha)$ are positive. Thus $f(\eta;\alpha)$ and all its
derivatives through $f'''(\eta;\alpha)$ are increasing functions of
$\alpha$. Thus we conclude that
\begin{gather}
0<f(\eta;\alpha_1)\le f(\eta;\alpha)\le f(\eta;\alpha_2), \label{f} \\
0<f'(\eta;\alpha_1)\le f'(\eta;\alpha)\le f'(\eta;\alpha_2)<1, \label{f'} \\
0<f''(\eta;\alpha_1)\le f''(\eta;\alpha)\le f''(\eta;\alpha_2), \label{f''} \\
f'''(\eta;\alpha_1)\le f'''(\eta;\alpha)\le f'''(\eta;\alpha_2)< 0,
\label{f'''}
\end{gather}
for all $0<\eta \le \overline{\eta}$. With these inequalities in place,
we see from \eqref{f4} that
\begin{equation}
f^{(4)}(\eta;\alpha)>0 \label{f4>0}
\end{equation}
for all $0<\eta \le \overline{\eta}$ and $\alpha_1 \le \alpha \le \alpha_2$.

Evaluating \eqref{vode} at $\overline{\eta}$ gives
\begin{equation}
(1-2kf')fv''-2(kff''+f')v'+[(1-2kf')f''+2kff''']v=0 \quad \text{at } \overline{\eta}.
\label{vodeatetabar}
\end{equation}
Given the conditions listed in the last paragraph, a necessary condition
for \eqref{vodeatetabar} to hold is that $1-2kf'(\overline{\eta})>0$. Thus
$f'(\overline{\eta})<1/2k$. But since $f'$ is strictly increasing we have
\begin{equation}
f'(\eta) < \frac{1}{2k} \text{on } [0,\overline{\eta}].
\label{2k}
\end{equation}

Next, evaluating the ODE for $v^{(4)}$ \eqref{v4ode} at $\overline{\eta}$ we have
\begin{equation}
\begin{aligned}
&(1+kf^2)v^{(4)}-[(1+2kf')f'+4kff'']v''
-(4kf'+1)f''v' \\
&+[2kff^{(4)}+f'''-2kf''^2]v=0 \quad \text{at } \overline{\eta}.
\end{aligned}\label{v4odeatetabar}
\end{equation}
The terns involving $v'$ and $v''$ are strictly negative and the term
involving $v^{(4)}$ is less than or equal to zero. Since $v>0$, the only
way for \eqref{v4odeatetabar} to hold is if
\begin{equation}
2kff^{(4)}+f'''-2kf''^2>0 \quad\text{at } \overline{\eta}.
\label{ineq6b}
\end{equation}

Using \eqref{f4}, the inequality \eqref{ineq6} can be rewritten as
\begin{equation}
\begin{aligned}
&\frac{kf''^2(4kf^2-1)}{1+kf^2}+\frac{f'''(1-kf^2)}{1+kf^2}
+\frac{kf''(2ff'+4kff'^2-f'')}{1+kf^2} \\
&-\frac{2k^2f^2f''^2}{1+kf^2}>0 \quad \text{at } \overline{\eta}.
\end{aligned}\label{ineq7}
\end{equation} 

The last term of \eqref{ineq7} is negative. Next, several technical lemmas
will derive bounds on $\alpha$, $\overline{\eta}$ and $\overline{\overline{\eta}}$ as well as
show that the first two terms of \eqref{ineq7} are nonpositive. We begin
with several bounds on $f$ and its derivatives.

Using \eqref{f} through \eqref{f4>0} along with \eqref{2k}
we conclude 
\begin{gather}
-1<f'''<0 \quad \text{on } (0,\overline{\eta}], \label{f'''b} \\
\alpha - \eta <f''<\alpha \quad \text{on } (0,\overline{\eta}], \label{f''b} \\
\alpha \eta - \frac{\eta^2}{2} <f'<\min \big\{\alpha \eta, \frac{1}{2k}\big\}
\quad \text{on } (0,\overline{\eta}], \label{f'b} \\
\frac{\alpha \eta^2}{2} - \frac{\eta^3}{6} <f<
\min \big\{\frac{\alpha \eta^2}{2},\frac{\eta}{2k}\big\}\quad 
\text{on } (0,\overline{\eta}].
\label{fb}
\end{gather} 
\end{proof}

\begin{lemma} \label{lem1} 
The quantity
\begin{equation}
\frac{kf''^2(4kf^2-1)}{1+kf^2} \quad \text{at } \overline{\eta}
\end{equation}
is nonpositive.
\end{lemma}

\begin{proof} For contradiction suppose that
\begin{equation}
\frac{kf''^2(4kf^2-1)}{1+kf^2}>0 \quad \text{at } \overline{\eta}
\label{ineq8}
\end{equation}
For \eqref{ineq8} to hold we must have $4kf(\overline{\eta})^2-1>0$, or
\begin{equation}
f(\overline{\eta})>\frac{1}{2\sqrt{k}}.
\label{ineq9}
\end{equation}
From
\eqref{fb} we have
\begin{equation}
f(\eta)<\frac{\eta}{2k} \quad \text{on } [0,\overline{\eta}].
\label{ineq10}
\end{equation}
Combining \eqref{ineq9} and \eqref{ineq10} we have
\begin{equation}
\frac{1}{2\sqrt{k}}<f(\overline{\eta})<\frac{\overline{\eta}}{2k},
\label{ineq11}
\end{equation}
which implies that
\begin{equation}
\overline{\eta}>\sqrt{k}. \label{ineq12}
\end{equation}


We next show that $\alpha \ge \sqrt{k}$.
For contradiction suppose that $\alpha < \sqrt{k} < \overline{\eta}$. The last part of this
inequality using \eqref{ineq12}. Then from \eqref{f'b} we have
\begin{equation}
\frac{\alpha^2}{2} < f'(\alpha) < f'(\overline{\eta}) < \frac{1}{2k}, \label{ineq13}
\end{equation}
from which we conclude that
\begin{equation}
\alpha < \frac{1}{\sqrt{k}}. \label{ineq14}
\end{equation}

Recall that
\begin{equation}
f''(\eta)[1+kf(\eta)^2]=\alpha-\eta
+(2kf'-1)ff'+2\int_0^{\eta}f'(t)^2[1-kf'(t)]\,dt.
\end{equation}
Using \eqref{2k} and \eqref{ineq14} we obtain
\begin{equation}
f''(\eta)[1+kf(\eta)^2]< \frac{1}{\sqrt{k}} + \Big(
\frac{1-2k^2}{2k^2}\Big)\eta \quad \text{on } [0,\overline{\eta}],
\label{ineq15}
\end{equation}
which is less than or equal to zero if $\eta \ge
2k^2/(\sqrt{k}(2k^2-1))$. Since $f''>0$ we must therefore have
\begin{equation}
\overline{\eta} < \frac{2k^2}{\sqrt{k}(2k^2-1)}. \label{ineq16}
\end{equation}
Combining \eqref{ineq16} with \eqref{ineq12} we have
\begin{equation}
\sqrt{k}<\overline{\eta} < \frac{2k^2}{\sqrt{k}(2k^2-1)},
\end{equation}
which implies
\begin{equation}
h(k)=2k^3-2k^2-k<0.
\end{equation}
The positive root of
$h(k)$ is $\hat k=(1+\sqrt{3})/2\approx 1.366$ with $h(k)<0$ if $k<
\hat k$. Since
$k\ge k_0 >\hat k$ we arrive at a contradiction.

Thus if $k>k_0$, we must have
\begin{equation}
\alpha \ge \sqrt{k}
\label{ineq17}
\end{equation}
Using \eqref{ineq17} in \eqref{f'b} we obtain
\begin{equation}
f'>\alpha \eta -\frac{\eta^2}{2} > \sqrt{k} \eta-\frac{\eta^2}{2}.
\end{equation}
Since $f'$ is increasing and $\overline{\eta} >\sqrt{k}$ we have
\begin{equation}
f'(\eta)> f'(\sqrt{k})>\frac{k}{2} \text{ on } [\sqrt{k},\overline{\eta}].
\label{ineq18}
\end{equation}

Combining \eqref{ineq18} with \eqref{2k} we have
\begin{equation}
\frac{k}{2} < f'(\overline{\eta}) <\frac{1}{2k},
\label{ineq20}
\end{equation}
which implies that $k<1$ giving a contradiction since
$k\ge k_0>1$, thus the proof is complete.
\end{proof}

Thus if $k \ge k_0$, we have
\begin{equation}
\frac{kf''^2(4kf^2-1)}{1+kf^2} \le 0 \quad \text{at } \overline{\eta}.
\label{ineq21}
\end{equation}
Note that this implies that $f(\overline{\eta})\le 1/2\sqrt{k}$, and since
$f$ is increasing we have
\begin{equation}
f(\eta) \le \frac{1}{2\sqrt{k}} \quad \text{on } [0,\overline{\eta}].
\label{ineq22}
\end{equation}

\begin{lemma} \label{lem2} The quantity
\begin{equation}
\frac{f'''(1-kf^2)}{1+kf^2} \quad \text{at } \overline{\eta}
\end{equation}
is nonpositive.
\end{lemma}

\begin{proof}
Note that $f'''(\overline{\eta})<0$. Also, from
\eqref{ineq22} we have $f(\overline{\eta})<1/2\sqrt{k}$. Thus
$1-kf(\overline{\eta})^2>1-k(1/2\sqrt{k})^2=3/4>0$.
Thus
\begin{equation}
\frac{f'''(1-kf^2)}{1+kf^2}<0 \quad\quad\text{at } \overline{\eta}
\end{equation}
which completes the proof.
\end{proof}

\begin{lemma} \label{lem3}
\begin{equation}
f''(\overline{\overline{\eta}})<\frac{1}{k^{3/2}} \quad \text{and}\quad
\overline{\overline{\eta}} > \alpha - \frac{1}{k^{3/2}} \label{ineq23}
\end{equation}
\end{lemma}

\begin{proof}
By Lemmas \ref{lem1} and \ref{lem2}, in order for \eqref{ineq7} to hold
we must have
\begin{equation}
\frac{kf''(2ff'(1+2kf')-f'')}{1+kf^2}>0 \quad \text{at } \overline{\eta}.
\label{ineq24}
\end{equation}
Since $f''(\overline{\eta})>0$, \eqref{ineq24} implies that
\begin{equation}
2ff'(1+2kf')-f''>0 \quad \text{at } \overline{\eta}. \label{ineq25}
\end{equation}

Using \eqref{2k} and \eqref{ineq22} in the quantity on the
left-hand side of \eqref{ineq25} we have
\begin{equation}
\begin{aligned}
&2f(\overline{\eta})f'(\overline{\eta})(1+2kf'(\overline{\eta}))-f''(\overline{\eta}) \\
&< 2\left(\frac{1}{2\sqrt{k}}\right)\left(\frac{1}{2k}\right)
\left(1+2k\left(\frac{1}{2k}\right)\right)-f''(\overline{\eta})
=\frac{1}{k^{3/2}}-f''(\overline{\eta}).
\end{aligned} \label{ineq26}
\end{equation}
We obtain a contradiction of \eqref{ineq24}
if the right most term of \eqref{ineq26}
is less than or equal to zero. Thus we must have
 $f''(\overline{\eta}) < 1/k^{3/2}$. Since $f''$ is decreasing and $\overline{\overline{\eta}} > \overline{\eta}$
we obtain
\begin{equation}
f''(\overline{\overline{\eta}})<\frac{1}{k^{3/2}}. \label{ineq27}
\end{equation}

For the lower bound on $\overline{\overline{\eta}}$, we first consider $\overline{\eta}$.
Suppose that $\overline{\eta} \le \alpha$.
Then, since $\alpha - \eta < f''(\eta)$ on $[0,\overline{\eta}]$
we have
\begin{equation}
2f(\overline{\eta})f'(\overline{\eta})(1+2kf'(\overline{\eta}))-f''(\overline{\eta})<
\frac{1}{k^{3/2}}-f''(\overline{\eta})<
\frac{1}{k^{3/2}} +\overline{\eta} - \alpha.
\label{ineq28}
\end{equation}
Again, we have a contradiction if the right most term of
\eqref{ineq28} is less than or equal to zero. Thus
\begin{equation}
\overline{\overline{\eta}} > \overline{\eta} > \alpha -\frac{1}{k^{3/2}}.
\label{ienq29}
\end{equation}
\end{proof}

\begin{lemma} \label{lem4} 
For $\alpha_1\le \alpha \le \alpha_2$,
\begin{equation}
\alpha > \frac{\sqrt{54k^2-2}}{9k^{3/2}}.
\label{ineq30}
\end{equation}
\end{lemma}

\begin{proof}
Multiplying \eqref{ode} by $f''$ and integrating
(by parts where necessary) from 0 to $\eta$ we have
\begin{equation}
\frac{1}{2}(1+kf^2)f''^2-\frac{1}{2}\alpha^2+\int_0^{\eta}
(1-3kf')ff''^2\,dt+f'-\frac{1}{3}f'^3=0.
\label{int2}
\end{equation}
Let $\eta_4$ be the point where $f(\eta;\alpha_1)$
increases through $1/3k$.
(Note that $k\ge k_0 > 1/3$ so that $1/3k < 1$).
Then
\begin{equation}
\frac{1}{2}\alpha_1^2=\frac{27k^2-1}{81k^3}+
\frac{1}{2}(1+kf(\eta_4)^2)f''(\eta_4)^2+\int_0^{\eta_4}
(1-3kf')ff''^2\,dt.
\label{al2}
\end{equation}
Since all three terms on the right side of \eqref{al2} are positive
we have
\begin{equation}
\frac{1}{2}\alpha_1^2>\frac{27k^2-1}{81k^3},
\end{equation}
or, for $\alpha_1\le \alpha \le \alpha_2$,
\begin{equation}
\alpha \ge \alpha_1 > \frac{\sqrt{54k^2-2}}{9k^{3/2}}.
\end{equation} 

Recall that at a first maximum of $v'$ at $\overline{\overline{\eta}}$, we have
$v>0$, $v'>0$, $v''=0$ and $v'''\le 0$. Evaluating \eqref{vode}
at $\overline{\overline{\eta}}$ we have
\begin{equation}
(1+kf^2)v'''-2(kff''+f')v'+[(1-2kf')f''+2kff''']v=0
\quad \text{at } \overline{\overline{\eta}}.
\label{vodeetabb}
\end{equation}
The first term of \eqref{vodeetabb} is nonpositive and the
second is strictly negative. Since $v>0$, we must therefore
have
\begin{equation}
(1-2kf')f''+2kff'''>0 \quad \text{at } \overline{\overline{\eta}}
\label{ineq31}
\end{equation}
in order for \eqref{vodeetabb} to hold. Recall that
$f>0$, $f''>0$ and $f'''<0$. Thus a necessary condition for
\eqref{vodeetabb} to hold is that $f'(\overline{\overline{\eta}})<1/2k$.
Since $f'$ is increasing, we thus have
\begin{equation}
f'(\eta) < \frac{1}{2k} \quad \text{on } [0,\overline{\overline{\eta}}].
\label{ineq32}
\end{equation}

Next we use \eqref{ode} to rewrite \eqref{ineq31} as
\begin{equation}
(1-2kf')f''+2kf\frac{f'^2-1+(2kf'-1)ff''}{1+kf^2}>0\quad\text{at } \overline{\overline{\eta}} ,
\end{equation}
or
\begin{equation}
\frac{1}{1+kf^2}\left[(1-2kf')(1-kf^2)f''+2kf(f'^2-1)\right]
>0 \quad\text{at } \overline{\overline{\eta}}.
\label{ineq33}
\end{equation}

We will show that \eqref{ineq33} cannot
hold, giving us our final contradiction to the assumption
that $v'$ can have a positive maximum.
To this end, first
note that since $f(\overline{\overline{\eta}})>0$, $f'(\overline{\overline{\eta}})<1/2k<1$ and
$f''(\overline{\overline{\eta}})>0$, a necessary condition for \eqref{ineq33} to
hold is that
\begin{equation}
f(\overline{\overline{\eta}}) < 1/\sqrt{k}. \label{fetabb}
\end{equation}
The argument will take two
paths depending on whether $\overline{\overline{\eta}} > \alpha$ or $\overline{\overline{\eta}} \le \alpha$.

If $\overline{\overline{\eta}} > \alpha$, then since $f$ is increasing we have from \eqref{fb}
\begin{equation}
\frac{\alpha^3}{3}<f(\alpha)<f(\overline{\overline{\eta}}). \label{ineq34}
\end{equation}

Using the facts that $\alpha^3/3<f(\overline{\overline{\eta}})<1/\sqrt{k}$, $f'(\overline{\overline{\eta}})<1/2k<1$ and
$f''(\overline{\overline{\eta}}) <1/k^{3/2}$ in the quantity in brackets in \eqref{ineq33}
we have
\begin{equation}
\begin{aligned}
&(1-2kf'(\overline{\overline{\eta}}))(1-kf(\overline{\overline{\eta}})^2)f''(\overline{\overline{\eta}})+2kf(\overline{\overline{\eta}})(f'(\overline{\overline{\eta}})^2-1)\\
&<\frac{1}{k^{3/2}}+2k\frac{\alpha^3}{3}\Big(\frac{1}{4k^2}-1\Big).
\end{aligned} \label{ineq35}
\end{equation}
Next, using \eqref{ineq30} we have
\begin{equation}
\begin{aligned}
&(1-2kf'(\overline{\overline{\eta}}))(1-kf(\overline{\overline{\eta}})^2)f''(\overline{\overline{\eta}})+2kf(\overline{\overline{\eta}})(f'(\overline{\overline{\eta}})^2-1)\\
&<\frac{1}{k^{3/2}}+\frac{(54k^2-2)^{3/2}(1-4k^2)}{6\cdot9^3\cdot
k^{11/2}}.
\end{aligned} \label{ineq36}
\end{equation}
We obtain a contradiction of \eqref{ineq33} if the right side of
\eqref{ineq36} is less than or equal to zero. This occurs if
$k\ge k_1 >\approx 2.8618$ where $k_1$ is the root of
\begin{equation}
\frac{1}{k^{3/2}}+\frac{(54k^2-2)^{3/2}(1-4k^2)}{6\cdot9^3\cdot
k^{11/2}}=0.
\end{equation}

Finally, if $\overline{\overline{\eta}} \le \alpha$, then from Lemma \ref{lem3} we have
$\overline{\overline{\eta}} > \alpha -1/k^{3/2}$. Using this in \eqref{fb}
and the fact that $f$ is increasing we have
\begin{equation}
\frac{\alpha}{3}\Big(\alpha -\frac{1}{k^{3/2}}\Big)^2 < f(\overline{\overline{\eta}}).
\label{ineq37}
\end{equation}
Using the facts that $\alpha(\alpha-1/k^{3/2})^2/3<f(\overline{\overline{\eta}})<1/\sqrt{k}$,
$f'(\overline{\overline{\eta}})<1/2k<1$ and
$f''(\overline{\overline{\eta}}) <1/k^{3/2}$ in the quantity in brackets in \eqref{ineq33}
we have
\begin{equation}
\begin{aligned}
&(1-2kf'(\overline{\overline{\eta}}))(1-kf(\overline{\overline{\eta}})^2)f''(\overline{\overline{\eta}})+2kf(\overline{\overline{\eta}})(f'(\overline{\overline{\eta}})^2-1)\\
&<\frac{1}{k^{3/2}}+2k\frac{\alpha}{3}\Big(\alpha -\frac{1}{k^{3/2}}\Big)^2
\Big(\frac{1}{4k^2}-1\Big).
\end{aligned}\label{ineq38}
\end{equation}
Using the result of Lemma \ref{lem4} in \eqref{ineq38},  at $\overline{\overline{\eta}}$ we have
\begin{equation}
\begin{aligned}
&(1-2kf')(1-kf^2)f''+2kf(f'^2-1) \\
&<\frac{1}{k^{3/2}}+\frac{2k}{3}\frac{\sqrt{54k^2-2}}{9k^{3/2}}
\Big(\frac{\sqrt{54k^2-2}}{9k^{3/2}} -\frac{1}{k^{3/2}}\Big)^2
\Big(\frac{1}{4k^2}-1\Big).
\end{aligned} \label{ineq39}
\end{equation}
We obtain a contradiction of \eqref{ineq33} if the right side of
\eqref{ineq39} is less than or equal to zero. This occurs if
$k\ge k_2 >\approx 4.9377$ where $k_2$ is the root of
\begin{equation}
\frac{1}{k^{3/2}}+\frac{2k}{3}\frac{\sqrt{54k^2-2}}{9k^{3/2}}
\Big(\frac{\sqrt{54k^2-2}}{9k^{3/2}} -\frac{1}{k^{3/2}}\Big)^2
\Big(\frac{1}{4k^2}-1\Big)=0.
\label{ineq40}
\end{equation}

The analysis of the last paragraph
used the rather crude bounds
$1-kf(\overline{\overline{\eta}})^2<1$ and $1-2kf'(\overline{\overline{\eta}}) <1$.
An improved value for $k$ can be obtained
by employing the lower bounds in \eqref{fb} and \eqref{f'b}.
We thus obtain a contradiction of \eqref{ineq33} if $k$ is larger
than the root of
\begin{align*}
&\Big(1-\frac{\sqrt{54k^2-2}(\sqrt{54k^2-2}-9)}{81k^2}\Big)
\Big(1-\frac{\sqrt{54k^2-2}(\sqrt{54k^2-2}-9)^2}
{3\cdot 9^3\cdot k^{7/2}}\Big)
\big(\frac{1}{k^{3/2}} \big) \\
&+\frac{2k}{3}\frac{\sqrt{54k^2-2}}{9k^{3/2}}
\Big(\frac{\sqrt{54k^2-2}}{9k^{3/2}} -\frac{1}{k^{3/2}}\Big)^2
\Big(\frac{1}{4k^2}-1\Big)=0.
\end{align*}
 The root of this equation defines
the value $k_0 \approx 3.5584$
mentioned in the statement of
the theorem.

This final contradiction proves that $v'$ cannot have a maximum.
Thus $v''>0$ for all $\eta >0$, and since $v'(0)=0$
we conclude that $v'>0$ is bounded away from zero for $\eta$ large.

Next, if two solutions with
the properties given in the statement of Theorem \ref{thm2} were to
exist, then by the Mean Value Theorem we would have
\begin{equation}
f'(\eta;\alpha_2)-f'(\eta;\alpha_1) =
 \Big(\frac{\partial f'(\eta;\alpha)}{\partial \alpha}
 \Big)_{\alpha=\hat \alpha} (\alpha_2-\alpha_1)
 =v'(\eta;\hat \alpha)(\alpha_2-\alpha_1),
\label{mvt}
\end{equation}
for $\eta>0$ where $\hat \alpha \in (\alpha_1,\alpha_2)$.
Since $v'(\eta;\hat \alpha)$ is bounded
away from zero for $\eta$ large, there exists a constant
$M>0$ such that
\begin{equation}
0=f'(\infty;\alpha_2)-f'(\infty;\alpha_1)
 =\lim_{\eta \to \infty} v'(\eta;\hat \alpha)(\alpha_2-\alpha_1)
 >M(\alpha_2-\alpha_1)>0.
\end{equation}
This contradiction proves Theorem \ref{thm2}.
\end{proof}

Our numerical investigations indicate solutions that
violate the inequalities $0<f'<1$ and $f''>0$ do not exist,
however we have not been able to prove this. We end this
section by discussing the properties that any second (or
further) solutions must have.
First note that from the ODE \eqref{ode},
$f'$ can only have a maximum in the range $-1<f'<1$ and
can only have a minimum if either $f'>1$ or $f'<-1$. Thus,
as mentioned in the introduction, there can be no solutions
exhibiting overshoot (values of $f'$ above $1$).
Further, any second solution would have to have at least one
minimum below $-1$, and thus would exhibit flow reversal
($f' < 0$), which is unlikely in this physical configuration.

\section{Bounds on the skin friction coefficient}

A quantity of much physical interest is the
skin friction coefficient which is proportional to $f''(0)=\alpha$.
The following result gives bounds on $\alpha$. (In this section,
$\alpha$ corresponds to the value $\alpha(k)$
that gives the solution to the
BVP \eqref{ode}-\eqref{bcs} with the properties $0<f'(\eta, \alpha)<1$
and $f''(\eta, \alpha)>0$ for all $\eta >0$.)

\begin{theorem} \label{thm3} 
For $0<k\le 1/3$ we have
\begin{equation}
f''(0)=\alpha>\frac{2}{\sqrt{3}}. \label{41}
\end{equation}
For $k > 1/3$ we have
\begin{equation}
f''(0)=\alpha>\frac{\sqrt{54k^2-2}}{9k^{3/2}}
\label{42}
\end{equation}
For $k\ge \hat k \approx 2.2825$ we have
\begin{equation}
f''(0)=\alpha<\frac{2}{\sqrt{3}}. \label{43}
\end{equation}
The exact value of $\hat k$ will be defined in the proof.
\end{theorem}

\begin{proof} 
Using the fact that $f'$ approaches one from below
in the ODE \eqref{ode}, we conclude that $ff''$ tends to zero.
Letting $\eta \to \infty$ in \eqref{int2} we conclude that
\begin{equation}
f''(0)=\alpha^* > \frac{2}{\sqrt{3}} \quad \text{for } 0<k\le \frac 1 3,
\label{44}
\end{equation}
whereas the result of Lemma \ref{lem4} gives
\begin{equation}
f''(0)=\alpha^* > \frac{\sqrt{54k^2-2}}{9k^{3/2}} \quad \text{for } k> \frac 1 3.
\label{45}
\end{equation}
To obtain the upper bound on $\alpha$
for $k> \hat k$ we again let $\eta \to \infty$
in \eqref{int2} to obtain
\begin{equation}
\int_0^{\infty} (1-3kf')ff''^2\, dt = \frac 1 2 \Big(\alpha^2-\frac 4 3 \Big).
\label{46}
\end{equation}
If the integral on the left hand side of \eqref{46} is negative
the result follows. Suppose for contradiction that the integral
is non-negative. Then $\alpha \ge 2/\sqrt{3}$. From \S 2 we have
that for a solution that satisfies $0<f'<1$ and $f''>0$ for all
$\eta>0$, we also have $f^{(4)}>0$ for all $\eta >0$. This again
leads to the bounds
\begin{gather}
-1<f'''(\eta)<0, \label{47} \\
\alpha-\eta<f''(\eta)<\alpha, \label{48} \\
\alpha \eta - \frac{\eta^2}{2}<f'(\eta)<\alpha \eta, \label{49} \\
\frac{\alpha \eta^2}{2} - \frac{\eta^3}{6}<f(\eta)<\frac{\alpha \eta^2}{2}.
\label{50}
\end{gather}

Using $\alpha \ge 2/\sqrt{3}$ and \eqref{49} we have
$f'(2/\sqrt{3})>2/3$.
Since $f'''<0$ we have
\begin{equation}
\frac{\eta}{\sqrt{3}} = l(\eta) < f'(\eta) \quad \text{on } [0,2/\sqrt{3}],
\label{51}
\end{equation}
where $l(\eta)=\eta/\sqrt{3}$ is the line
through $(0,0)$ and $(2/\sqrt{3},2/3)$.
Let $\eta_5$ be the point where $f'$ increases through $1/3k$.
Using \eqref{51} we have that $\eta_5<1/\sqrt{3}k$. Using
\eqref{48} through \eqref{51} we have
\begin{equation}
\begin{aligned}
\int_0^{\eta_5} (1-3kf')ff''^2\, dt
&< \int_0^{1/\sqrt{3}k} \Big(1-3k\big(\frac{t}{\sqrt{3}}\big) \Big)
\Big(\frac{\alpha t^2}{2}\Big)(\alpha)^2\, dt \\
&=\frac{\alpha^3}{72\sqrt{3}k^3}.
\end{aligned} \label{52}
\end{equation}
Next, again employing the bounds \eqref{48} through \eqref{51} we have
\begin{equation}
\begin{aligned}
\int_{\eta_5}^{1/\sqrt{3}} (1-3kf')ff''^2\, dt
&< \int_{1/\sqrt{3}k}^{1/\sqrt{3}}
(1-3k(\alpha t))
\Big(\frac{6k-1}{18\sqrt{3}k^3}\Big)
\Big(\frac{1}{\sqrt{3}}\Big)^2\, dt \\
&=\frac{(6k-1)(1+k-k^2)}{162k^4}.
\end{aligned} \label{53}
\end{equation}

Combining \eqref{52} and \eqref{53} we have
\begin{equation}
\int_0^{1/\sqrt{3}} (1-3kf')ff''^2\, dt<
\frac{\alpha^3}{72\sqrt{3}k^3}+
\frac{(6k-1)(1+k-k^2)}{162k^4}.
\label{54}
\end{equation}
We arrive at a contradiction (of the assumption that
$\int_0^{\infty}(1-3kf')ff''^2\,dt \ge 0$)
if the right hand side of
\eqref{54} is negative and thus we must have
\begin{equation}
\alpha > \Big(\frac{4\sqrt{3}(6k-1)(k^2-k-1)}{9k}\Big)^{1/3}.
\label{55}
\end{equation}
Using \eqref{55} in \eqref{49} we have
\begin{equation}
f'(1/\sqrt{3})>\Big(\frac{4\sqrt{3}(6k-1)(k^2-k-1)}{9k}\Big)^{1/3}
\frac{1}{\sqrt{3}}-\frac 1 6.
\label{56}
\end{equation}
We obtain a contradiction (of $0<f'<1$ for all $\eta >0$)
if the right side of \eqref{56} is
greater or equal to 1. This occurs if $k \ge \hat k
\approx 2.2825$ where $\hat k$ is the positive root of
\begin{equation}
\Big(\frac{4\sqrt{3}(6k-1)(k^2-k-1)}{9k}\Big)^{1/3}
\frac{1}{\sqrt{3}}-\frac 7 6 =0.
\end{equation}
Thus for $k \ge \hat k$ we have
$\alpha < 2/\sqrt{3}$.
\end{proof}

\section{Discussion and open problems}

Through direct analysis of the BVP \eqref{ode}-\eqref{bcs}
we have proven the existence of a solution for stagnation
point flow of an upper-convected Maxwell fluid. This
solution is shown to satisfy $0<f'<1$ and $f''>0$ for all $\eta >0$.
For $k \ge k_0$ we have also shown that a solution \emph{with these
properties} is unique. Any further solutions must exhibit the
physically unrealistic property of flow reversal in the boundary
layer.

The analysis presented here should prove useful in the study of
generalizations of the stagnation point problem posed
in Sadeghy et al.\ \cite{shs}. Kumari and Nath \cite{kn}
extended this model by considering the effects of heat transfer
and an induced magnetic field on the flow. (See equations (8) and (11)
in \cite{kn}.) More recently, Lok et al.\ \cite{lip}
considered a generalization that incorporated the effects of a
shrinking sheet with suction. (See equations (7) and (9) in
\cite{lip}.) Straightforward extensions of the arguments given
here should yield existence results for both of these problems.
The uniqueness results will no doubt prove more problematic.
One reason is that the already technical nature of Theorem \ref{thm2}
will no doubt become more involved in these more complicated
problems. Another
reason is that the numerical results of Lok et al.\
\cite{lip} indicate that no solution to their boundary value problem
exists for values of the
shrinking parameter $\lambda$ less than a critical value.
There are really no ``standard'' methods for proving nonexistence
of solutions to BVPs. Each problem generally has to be approached
on an ad-hoc basis. However, the techniques of \cite{m} may prove
applicable to the problem posed in \cite{lip}. Finally, we mention
that the techniques employed here involved a third order ODE with two
boundary conditions at $\eta=0$ and one boundary condition at
infinity. The method does generalize to higher dimensional ODEs
where the dimension of the topological shooting space may also
increase depending on the number of boundary conditions at each
boundary. Thus, the methods could be applied to the higher
dimensional problems considered in \cite{bar}, \cite{br}, \cite{kn} and
\cite{lip}.


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