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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 104, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/104\hfil Existence of scale invariant solutions]
{Existence of scale invariant solutions to horizontal flow with a
Fujita type diffusion coefficient}

\author[G. A. Braga, P. C. Carri\~ao, A. A. G. Ruas \hfil EJDE-2012/104\hfilneg]
{Gast\~ao A. Braga, Paulo C. Carri\~ao, Antonio A. G. Ruas}  % in alphabetical order

\address{Gast\~ao A. Braga \newline
Departamento de Matem\'atica - UFMG,
Caixa Postal 1621,
30161-970  Belo Horizonte, MG, Brazil}
\email{gbraga@mat.ufmg.br}

\address{Paulo C. Carri\~ao \newline
Departamento de Matem\'atica - UFMG,
Caixa Postal 1621, 
30161-970  Belo Horizonte,  MG, Brazil}
\email{carrion@mat.ufmg.br}

\address{Antonio A. G. Ruas \newline
Departamento de Matem\'atica - UFMG,
Caixa Postal 1621, 
30161-970  Belo Horizonte, MG, Brazil}
\email{gaspar@mat.ufmg.br}

\thanks{Submitted October 11, 2011. Published June 21, 2012.}
\subjclass[2000]{34E10, 34E13, 35C06, 35Q86}
\keywords{Water infiltration; nonlinear diffusion; self-similar solutions;
 \hfill\break\indent Fujita diffusion coefficient}

\begin{abstract}
 In this article, we study a boundary-initial value problem on the half-line
 for the diffusion equation with a Fujita type diffusion coefficient
 that carries a parameter $\alpha $. The equation models
 the flow of water in soil within an approximation where gravitational
 effects are not taken into account and, when $\alpha = 1$, an explicit
 self-similar solution $\psi(x/\sqrt t)$ can be found.
 We prove that if $\alpha > 1$ then the above problem, with
 uniform boundary conditions, posses self-similar solutions.
 This is the first step towards a multiscale (renormalization group)
 asymptotic analysis of solutions to more general equations than the ones
 studied here.
\end{abstract}

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

\section{Introduction}

The aim of this note is to study the following boundary-initial value problem (BIVP)
for the unknown $\theta = \theta(x,t)$
\begin{equation}
\label{eq:bivp}
\begin{gathered}
\frac{\partial \theta}{\partial t} = \frac{\partial}{\partial x}
\Big( \Big[ \frac{C \, (\theta - \theta_r)^{\alpha - 1}}{(\theta_s - \theta)^2}\Big]
\frac{ \partial \theta}{\partial x}\Big),
\quad  0<x<\infty, \;  t>0, \\
\theta(0, t) = \theta_0,\;  \forall t\geq 0; \quad
\theta(x,0) = \theta_i, \;  \theta_r < \theta_0,\;  \theta_i < \theta_s,
\end{gathered}\end{equation}
with $C>0$. When $\alpha = 1$,
the above problem describes the flow of water in soil
within an approximation where gravitational effects are not taken
into account, which is the case, for instance, of horizontal flow or
vertical flow at early times, see \cite{bib:smith}.
Here, we consider $\alpha$ as a parameter satisfying $\alpha\geq 1$.
Sticking to the hydrology's nomenclature,
$\theta(x,t)$ is the soil's \emph{water content} at height $x$, measured from the
soil's surface downwards, and at time $t$. $\theta_r$  and $\theta_s$ are
the  \emph{residual} and the \emph{saturated} values of the soil's water content,
respectively, and we assume that $0<\theta_r < \theta_s <1$.
The quantity between square brackets in \eqref{eq:bivp} will be
denoted by $D = D(\theta)$ and it is called the
\emph{hydraulic diffusion coefficient}. Observe that $D(\theta_r) = 0$
if $\alpha > 1$ and that $D(\theta)$ is convex around $\theta_r$ if
$\alpha > 2$.

The above BIVP is a natural extension of the $\alpha=1$ case, which
is within  Fujita's class defined in
\cite{bib:smith}, and it has been studied by several authors
\cite{bib:storm, bib:fujita, bib:knig-phil}.
Assuming that the soil is uniformly wet at the begining ($\theta_i$ is 
assumed to be constant)
and that the water content at the soil's surface is kept constant and equal to
$\theta_0$ at later times, the $\alpha=1$ case can be
explicitly solved under the additional condition
$\theta(x,t)\equiv  \varphi(x/\sqrt t)$.
Solutions of this form are said to be \emph{self-similar} or
\emph{scale invariant} because $\theta(x,t) = \theta(Lx,L^2t)$
for any $L>0$.
Existence and uniqueness of self-similar solutions is an important
issue within the context of asymptotic analysis as $t\to \infty$
of solutions to partial differential equations, see \cite{bib:barenblatt1}.
Nonlinearities could be added to the right hand side of equation
\eqref{eq:bivp} and, if so,  one would like to know
under which conditions they will be ``irrelevant",
``marginal" or ``relevant"
in the Renormalization Group (RG) sense, see
\cite{bib:bric-kupa,bib:gold92}.
The RG method is based upon a multiscale analysis that
provides the right asymptotic behavior of solutions to differential
equations and an essential step towards its rigorous
study is to prove the existence of the RG fixed points (the
self-similar solutions), see
\cite{bib:bric-kupa-lin,bib:braga-furt-mor-rolla-tp}.

In this article, we take $\alpha \geq 1$ and we  prove that  \eqref{eq:bivp}
has a scale invariant solution for specific choices of
$\theta_0, \theta_i \in (0,1)$. We will show that


\begin{theorem} \label{teo:main}
Let  $\alpha \geq 1$ and  $\theta_0, \theta_i\in (\theta_r, \theta_s)$.
If $\theta_0 \leq \theta_i$ then  \eqref{eq:bivp} has a
unique classical solution
$\theta(x,t)$ of the form $\theta(x,t)= \psi(x/\sqrt t)$,
where $\psi : [0, \infty) \to (\theta_r, \theta_s)$ is a
$C^2([0, \infty))$ function satisfying $\psi(0) = \theta_0$
and $\psi(\infty) = \theta_i$.
Furthermore, given  $\theta_0\in (\theta_r, \theta_s)$,
there exists $\varepsilon > 0$ such that
 \eqref{eq:bivp} has a self-similar solution
for any choice of $\theta_i\in (\theta_0 -\varepsilon, \theta_s)$.
\end{theorem}



To prove Theorem \ref{teo:main}, we restate 
\eqref{eq:bivp} in terms of a boundary value problem as follows.
 Define
$$
\varphi= \frac{1}{ 1 - \sigma} -1,
$$
where $\varphi= \varphi(\zeta)$, with $\zeta$ equals the similarity variable 
$x/\sqrt t$, and  $\sigma = \sigma(x,t)\equiv (\theta(x,t) - \theta_r)/\Delta\theta$,
with $\Delta \theta \equiv \theta_s - \theta_r$. Then,  
$0< \sigma < 1$ and $0<\varphi <\infty$
so that  \eqref{eq:bivp} is rewritten as the  boundary value problem (BVP)
\begin{equation}\label{eq:bvp}
\begin{gathered}
\varphi '' + \Big( \frac{\alpha - 1}{\varphi(\varphi +1)}\Big) (\varphi ')^2 +
\frac{\zeta}{2K_1 }  \Big(\frac{(\varphi+ 1)^{\alpha - 3}}{\varphi^{\alpha - 1}}\Big)
\varphi ' = 0,\\
\varphi(0) = \varphi_0, \quad   \varphi(\infty) = \varphi_i,
\end{gathered}
\end{equation}
where  $K_1 \equiv C (\Delta\theta)^{\alpha - 3}$
and   $\varphi_0$ ($\varphi_i$)
corresponds to $\theta_0$ ($\theta_i$) through the relation
$$
\frac{\theta_k - \theta_r}{\theta_s - \theta_k} = \varphi_k, \quad k=0, i.
$$
It is straightforward to see that Theorem \ref{teo:main} is a corollary of
following result.

\begin{theorem} \label{teo:main1}
Consider the Boundary Value Problem \eqref{eq:bvp} with $\alpha \geq 1$,
and $\varphi_0$ and $\varphi_i$ in $(0, \infty)$.
\begin{enumerate}

\item If $\varphi_0 \leq \varphi_i$ then  \eqref{eq:bvp} has a unique classical
$C^2([0,\infty))$ solution
$\varphi: [0,\infty) \to [\varphi_0, \varphi_i]$;

\item if $\varphi_0$ is fixed then there exists $\varepsilon > 0$ such that
 \eqref{eq:bvp}, with $\varphi_i\in (\varphi_0 -\varepsilon, \varphi_0)$, 
has a classical $C^2([0,\infty))$ solution 
$\varphi: [0,\infty) \to [\varphi_0 - \varepsilon, \varphi_0]$;

\item if $\alpha = 2$ then $\varepsilon = \varphi_0$ in the above statements.

\end{enumerate}
\end{theorem}

Finally, instead of studying \eqref{eq:bvp}
directly, we replace it by the  initial value
problem 
\begin{equation}\label{eq:ivp}
\begin{gathered}
\varphi '' + \Big( \frac{\alpha - 1}{\varphi(\varphi +1)}\Big) (\varphi ')^2 +
\frac{t}{2K_1 }  \Big(\frac{(\varphi+ 1)^{\alpha - 3}}{\varphi^{\alpha - 1}}\Big)
\varphi ' = 0,\\
\varphi(0) = \varphi_0 > 0, \quad   \varphi '(0) = \varphi '_0\in\mathbb{R},
\end{gathered}
\end{equation}
where $t$ is the independent variable.
Let $\varphi(t):[0, \omega)\to [0,\infty)$
be the solution to  \eqref{eq:ivp} on its maximal interval $[0, \omega)$ and let
$\varphi '_0\equiv x$.
Observe that both $\varphi(t)$ and $\omega$ are functions of $x$ and,
when appropriate, we express this dependence as $\varphi(t,x)$ and $\omega(x)$,
respectively. It is  straightforward to see that Theorem \ref{teo:main}
is a corollary of the following theorem that will be proved in next section.

\begin{theorem} \label{teo:main2} 
Let $\varphi(t):[0, \omega)\to [0,\infty)$ be the unique classical
solution to \eqref{eq:ivp} on its maximal interval $[0, \omega)$, 
where $\varphi_0 > 0$ and $\varphi '_0 = x\in\mathbb{R}$.
\begin{enumerate}

\item Suppose $\alpha \geq 1$.
There exists a negative number $\bar x$ such that if $x\in (\bar x, \infty)$
then $\omega(x) = \infty$, i.e., $\varphi(t)$ is a global solution.
The limit $\lim_{t\to \infty} \varphi(t,x)$
exists and defines a continuous function $f(x)$ on $(\bar x, \infty)$ for which
$f(0) = \varphi_0$. Furthermore, $f(x)$
is a homeomorphism between $[0, \infty)$ and $[\varphi_0, \infty)$.
\item Suppose $1 < \alpha \leq 2$. There exists a negative number $\underbar x$
such that if $x\in (-\infty, \underbar x)$ then $\omega(x) < \infty$, i.e.,
$\varphi(t)$ ceases to be a classical solution at finite time. In particular,
for all $x< \underbar x$, $\lim_{t\to \omega^-} \varphi(t,x) = 0$ .


\item If $\alpha = 2$ then there exists a negative number $\lambda$ such that
the function $f(x)$  maps the interval
$(\lambda, 0]$ onto the interval $(0, \varphi_0]$.

\end{enumerate}
\end{theorem}


\section{Proof of Theorem \ref{teo:main2}}
\label{sec:theo3}
%\label{sec:prel}

In this section we consider
 \eqref{eq:ivp} with $\alpha$ and $\varphi '_0$ in $\mathbb{R}$.   We
will show below that the derivative  $\varphi '(t)$
keeps the sign of $\varphi '(0) = \varphi '_0$ for all positive
times, i.e., the product $\varphi '_0\cdot \varphi '(t)$ is non negative for 
all $t\geq 0$
and it is zero if and only if $\varphi '_0 = 0$.

\begin{lemma} \label{lema:sign-deri}
Let $\varphi(t):[0, \omega) \to \mathbb{R}$ be the solution
to  \eqref{eq:ivp} with $\varphi_0 >0$ and $\alpha, \varphi '_0 \in \mathbb{R} $,
 where $[0,\omega)$ is the solution's maximal interval of existence. Then the
product
$\varphi '_0 \cdot \varphi '(t)$ is non negative for all $t\in [0, \omega)$
 and it is zero if and only if $\varphi '_0 = 0$.
\end{lemma}

\begin{proof}
 For any $\alpha, \varphi '_0\in\mathbb{R}$, the IVP \eqref{eq:ivp} has a unique local
positive solution as long as $\varphi_0>0$, see \cite{bib:codd-levi}, and it will
keep itself positive as long as it exists as a classical solution.
Of course, if $\varphi '_0 = 0$ then, by uniqueness, $\varphi(t) = \varphi_0$ 
for all $t$.
If $\varphi '_0\not= 0$ then  $\varphi '(t)\not= 0$ for $t$ close to $0$ and, 
of course, $\varphi '_0$ and $\varphi '(t)$ will have the same sign.
Let $[0, \omega')\subset [0, \omega)$ be the maximal interval on which
$\varphi '_0$ and $\varphi '(t)$ will have the same sign.
We will prove below that $\omega' = \omega$. Suppose, by contradiction, 
that $\omega'<\omega$.
Then, by continuity, $\varphi '(\omega') = 0$.
Divide \eqref{eq:ivp} through by $\varphi '(t)$, $t<\omega'$, and integrate out to
get that
\begin{equation} \label{sol}
\varphi'(t)= \varphi '_0 \Big(\frac{\varphi(t) +1}{\varphi(t)}\Big)^{\alpha - 1}\,
\Big(\frac{\varphi_0}{\varphi_0+1}\Big)^{\alpha - 1}\, h(t),
\end{equation}
where
\begin{equation}
\label{eq:h}
h(t) = \exp \Big( - \int_0^t \frac{t (\varphi(t) + 1)^{\alpha - 3}}{2 K
(\varphi(t))^{\alpha - 1}}\,\, {\rm d} t \Big).
\end{equation}
We conclude from \eqref{sol} and \eqref{eq:h}
that $\lim_{t\to\omega'^{-}} \varphi '(t) $ exists and is different from 0,
a contradiction.
\end{proof}


\subsection{Proof of Part I when $\varphi '_0 > 0$}
\label{subsec:parti}

\begin{lemma} \label{lema:posi-deri}
Let $\varphi(t):[0, \omega) \to \mathbb{R}$ be the solution
to \eqref{eq:ivp} with $\alpha \geq 1$ and $\varphi '_0 >0$, where
$[0,\omega)$ is the solution's maximal interval of existence. Then
$\varphi '(t)$ is a positive, monotonically decreasing, function on $[0,\omega)$.
\end{lemma}

\begin{proof}
It follows from Lemma \ref{lema:sign-deri} that $\varphi '(t)>0$ for all
$t\in [0,\omega)$ if $\varphi '_0>0$. That $\varphi '(t)$ is monotonically decreasing
comes from $\alpha\geq 1$ and from the ODE, rewritten as
\begin{equation} \label{eq:deri-segu}
\varphi ''(t)   =   - \Big( \frac{\alpha - 1}{\varphi(t)(\varphi(t)+1)}\Big)
(\varphi '(t))^2
 -\frac{t}{2K_1 }
\Big(\frac{(\varphi(t) + 1)^{\alpha - 3}}{\varphi^{\alpha - 1}(t)}\Big)\varphi '(t).
\end{equation}
\end{proof}

\begin{lemma} \label{positivex} 
Under the hypothesis of Lemma \ref{lema:posi-deri}, the
solution to  \eqref{eq:ivp} is well defined for all $t\geq 0$.
\end{lemma}

\begin{proof}
Since $\alpha\geq 1$ and $\varphi '_0>0$, it follows
from Lemma \ref{lema:posi-deri} that $\varphi'(t)$ is a positive, monotonically
decreasing, function on $[0,\omega)$.
In particular, $\varphi '(t) \leq \varphi '_0$ for all $t\in[0,\omega)$ which implies
that $\varphi(t) \leq \varphi_0 + t\,\varphi '_0$. But since $\varphi_0 \leq \varphi(t)$
for all $t$, it follows that $\varphi(t)$ is bounded above and below for all
$t\in [0, \omega)$ an it follows from the theorem of existence of solutions
to differential equations, see \cite{bib:codd-levi}, that  the solution
can be extended up to $\omega = \infty$.
\end{proof}

For the rest of this article, we replace $\varphi '_0$ by $x$ in \eqref{eq:ivp}.
To make the $x$ dependence explicit we sometimes
write $\varphi(t,x)$, $\varphi '(t,x)$ and $h(t,x)$ instead of $\varphi(t)$,
$\varphi '(t)$ and $h(t)$,
respectively. If $x \geq 0$ and $\alpha\geq 1$ then it follows from lemmas
\eqref{lema:posi-deri} and \eqref{positivex} that $\varphi(t,x)$ is a monotonically increasing
function of $t$ for all $t\geq 0$, implying that the limit
\begin{equation} \label{eq:f}
\lim_{t\to\infty}\varphi(t,x) \equiv f(x)
\end{equation}
is  well defined although it could be infinity. $f(x)$ is the boundary value of
$\varphi(t,x)$ at $t=\infty$.
The next two lemmas show
that if $x\geq 0$ and $\alpha\geq 1$ then $f(x)$ is a continuous
homeomorphism between $[0,\infty)$ and $[\varphi_0,\infty)$.
It follows from this result that  \eqref{eq:bvp} has a unique classical
solution if $\varphi_0 < \varphi_i$.

\begin{lemma} \label{lema:inje} 
If $x \geq 0$ and $\alpha\geq 1$ then $f(x)$, defined by \eqref{eq:f},
is a monotonically increasing function that satisfies
$f(x)\to\infty$ as $x\to\infty$.
\end{lemma}

\begin{proof} We first show that $f(x)\to\infty$ as $x\to\infty$.
It follows from Lemma \ref{lema:posi-deri} that, under the above hypothesis
on $\alpha$ and $x$, $\varphi(t,x)$ is an increasing function of $t$. In
particular, $\varphi(t,x) \geq \varphi_0$ for all $t\geq 0$ which, together with
\eqref{sol} and with $\alpha\geq 1$, leads to
$$
\varphi '(t,x) \geq x \Big(\frac{\varphi_0}{\varphi_0+1}\Big)^{\alpha - 1}
{\rm e}^{-[\frac{(1+\varphi_0)^{\alpha -3}}{K\varphi_0^{\alpha -1}}] \frac{t^2}{4}}.
$$
Integration on both sides of the above inequality gives the result.

To prove that
$f$ is increasing in the interval $[0,\infty)$, we take
$ \bar{x} > x \geq 0$ and define $\delta (t) \equiv \bar\varphi(t)-\varphi(t)$
where $\varphi(t) = \varphi(t,x)$ and $\bar\varphi(t) = \varphi(t,\bar x)$ are the
solutions to \eqref{eq:ivp} with initial derivatives
$x$ and $\bar x$, respectively.
In the sequel we will prove that  $\delta '(t) > 0$ for all $t \geq 0$.
It then follows that our result is proven because
$0< \bar{x} - x = \delta (0) \leq \delta (t) \leq f(\bar{x}) - f(x)$, where
the last inequality is obtained by taking the limit of $\delta (t)$ as $t\to \infty$
and using its monotonicity.


Let $[0,\bar t)$ be the maximal interval
where $\delta '(t)>0$. Of course, $[0,\bar t)$ is not empty because $\delta (t)$ is
continuous and  $\delta '(0) = \bar x - x >0$.
We will show  that $\bar t = \infty$. Suppose by contradiction that $\bar{t}<\infty$;
i.e., $\delta '(\bar{t}) = 0$.
It follows from \eqref{eq:ivp} that
\begin{align*}
\delta ''(t) 
&=-  \frac{(\alpha - 1)\bar\varphi'^{\,2}}{\bar\varphi^2 + \bar\varphi} -
\frac{t}{2K}\Big[\frac{(\bar \varphi+ 1)}{\bar\varphi}\Big]^{\alpha - 1}\frac{1}{(\bar\varphi + 1)^2}
\bar\varphi' \\
&\quad + \frac{(\alpha - 1)\varphi'^{\,2}}{\varphi^2 + \varphi} + \frac{t}{2K}
\Big[\frac{(\varphi+ 1)}{ \varphi}\Big]^{\alpha - 1}\frac{1}{( \varphi + 1)^2} \varphi'.
\end{align*}
Now, since $\delta '(t) > 0$  for $t \in [0,\bar t)$, then $\delta (t)$ is increasing
and $\delta (t) = \bar\varphi(t) -\varphi(t) \geq  0$ for $t \in [0,\bar t)$.
It then follows from the above identity that
$$
\delta ''(t) \geq
-  \frac{\alpha - 1}{\varphi^2 + \varphi}(\bar\varphi'^{\,2} - \varphi'^{\,2}) -
\frac{t}{2K}\,
[\frac{(\bar \varphi+ 1)}{\bar\varphi}]^{\alpha - 1}\frac{1}{(\bar\varphi + 1)^2}(\bar\varphi' - \varphi').
$$
For $t \in [0,\bar t)$, we divide the last inequality by 
$\bar \varphi '(t) - \varphi '(t) = \delta '(t)$
and integrate out from $0$ to $\bar t$ to obtain
\begin{gather*}
0 = \delta '(\bar t) \geq (\bar x - x)\,\exp (U),\\
U = \Big[- \int_0^{\bar t}
\Big( \frac{\alpha - 1}{\varphi^2 + \varphi}(\bar\varphi' + \varphi') +
\frac{t}{2K}\,
\big[\frac{(\bar \varphi+ 1)}{\bar\varphi}\big]^{\alpha - 1}
\frac{1}{(\bar\varphi + 1)^2} \Big)\, {\rm d} t \Big],
\end{gather*}
which is a contradiction because the right hand side of the last inequality is
positive.
\end{proof}

\begin{lemma} \label{lema:fini}
If $x \geq 0$ and $\alpha\geq 1$ then $f(x)$, defined by \eqref{eq:f},
is a continuous function.
\end{lemma}

\begin{proof}
The continuity of $f(x)$ comes from the following claim which we will prove
below:
If $x \geq 0$ and $\alpha\geq 1$ then there exist positive constants
$C$ and $t_0$ such that
\begin{equation}
\label{eq:phi'-bound}
\varphi '(t,x) \leq Cx/t^2
\end{equation}
for all $t>t_0$. It follows from the above upper bound that
$\varphi(t,x)$  converges, as $t\to\infty$,   to $f(x)$, the convergence
being uniform on compact sets  because:
$$
|\varphi(t',x) - \varphi(t,x)| 
= \Big| \int_t^{t'}\varphi '(t,x)\,\, {\rm d} t \Big|
\leq Cx \int_t^{t'}\frac{1}{t^2}\,\, {\rm d} t
\leq \frac{Cx}{t}
$$
for all $t'\geq t > t_0$.


To obtain the upper bound \eqref{eq:phi'-bound} we first prove that
$\varphi '(t,x) \to 0$ as $t\to \infty$. It follows from \eqref{sol} that
$$
\varphi '(t,x) \leq xh(t,x)
$$
if $\alpha\geq 1$ and $x\geq 0$ because
the product $((\varphi +1)/\varphi)^{\alpha -1}((\varphi_0+1)/\varphi_0)^{\alpha -1}$ 
will be at most $1$.
Then, it is sufficient to prove
that  $h(t,x)\to 0$ as $t\to \infty$.
But, since $h(t,x)\leq 1$ for all $t, x \geq 0$, the above inequality
implies that $\varphi(t,x) \leq xt + \varphi_0$
for $t\in [0,\infty)$ which gives rise to the following
lower bound for the integrand in \eqref{eq:h}:
\begin{equation}
\label{eq:lowe-boun-inte}
\frac{t}{2K}
\frac{ (\varphi+ 1)^{\alpha - 3}}{ (\varphi)^{\alpha - 1}}
 =  \frac{t}{2K}(\frac{\varphi+ 1}{\varphi})^{\alpha -1}\frac{1}{(\varphi+ 1)^{2}}
\geq  \frac{t}{2K} \frac{1}{(xt + \varphi_0 + 1)^{2}}
\end{equation}
and implying that $h(t,x)\to 0$ as $t\to \infty$.

Now, since $\varphi '(t,x) \to 0$ as $t\to\infty$ and since $\varphi '(t,x)$ is
monotonically decreasing, see the proof of Lemma \ref{positivex},
it follows that given $r>0$ and small, there exist $s>0$ and
$t_0>0$ such that $\varphi(t,x) \leq r t + s$ for all $t\geq t_0$ and,
as in \eqref{eq:lowe-boun-inte}, we get
$$
\frac{t}{2K}
\frac{ (\varphi+ 1)^{\alpha - 3}}{ (\varphi)^{\alpha - 1}}
\geq \frac{1}{4r^2 K}\,\, \frac{1}{t}
$$
for $t>t_0$. Then, \eqref{eq:phi'-bound} is proven
once we choose $r$ such that $1/(4r^2K)\geq 2$.
\end{proof}


\subsection{Proof of Part I when $\varphi '_0 < 0$}
\label{subsec:exis-none}

In this section and in Section \ref{subsec:partii}
we consider  \eqref{eq:ivp} with $\varphi '_0 <0$.
Let $\varphi(t):[0, \omega) \to \mathbb{R}$ be its solution on the maximal
interval $[0, \omega)$. It follows from Lemma
\ref{lema:sign-deri} that $\varphi '(t)<0$ for all $t\in[0,\omega)$ and therefore
that $\varphi(t)$ is a monotonically decreasing function on  $[0,\omega)$.
It also follows from \eqref{eq:deri-segu} that
if $\alpha \geq 1$ then $\varphi ''(t) < 0$, i.e. $\varphi(t)$ is concave, at least
for small values of $t$.
For larger values of $t$ there will be a competition between the two
parcels on the right hand side of \eqref{eq:deri-segu}, one keeping
itself positive while the other one keeping negative, and it
is a matter to decide who is going to win as $t$ gets larger.
The question whether $\varphi(t)$ changes concavity or not as
$t$ increases is related to how negatively large or small is $\varphi '_0$
and we deal with this problem in sections \ref{subsec:partii}
and \ref{subsec:partiii}.
In this section we show that there exists a negative real number
$\bar x$ such that $\varphi(t,x)$ is well defined for all $t$ and for
$x> \bar x$. In particular, $\varphi(t,x)$ changes concavity at some
point $t_0$ and, similarly to Lemma \ref{lema:fini},
we prove that the limit  \eqref{eq:f} is a well defined continuous
function on $(\overline{x}, 0)$.


\begin{lemma} \label{lema:glob-exis}
Let $\varphi(t):[0, \omega) \to \mathbb{R}$ be the solution
to  \eqref{eq:ivp} with $\alpha \geq 1$ and $x$ satisfying
\begin{equation}
\label{eq:x-boun}
-\frac{2}{\sqrt \pi}
\frac{1}{\alpha } \Big[ \frac{(\varphi_0 + 1)^{\alpha - 3}}{4 K (\varphi_0)
^{\alpha - 3}}\Big]^{1/2}<    x \leq 0,
\end{equation}
where $[0,\omega)$ is the solution's maximal interval of existence. Then
$\omega=\infty$.
\end{lemma}

\begin{proof}
We will show that if Condition \eqref{eq:x-boun} is satified then the limit
$\lim_{t\to \omega^-}\varphi(t)$ is positive. This is enough to conclude that
$\omega=\infty$ because it follows from this limit and  from \eqref{sol} that
the limit $\lim_{t\to \omega^-}\varphi '(t)$
exists and is also positive, implying that the solution can always be
extended to the right of $\omega$, for any positive $\omega$.

If $\alpha \geq 1$ and $x  <0$ then it follows from
Lemma \ref{lema:sign-deri} that $\varphi '(t) < 0$ for all $t\in [0,\omega)$ implying,
together with \eqref{sol}, \eqref{eq:h} and that $\varphi(t) \leq \varphi_0$ for all $t\in [0, \omega)$,
that
$$
\varphi^{\alpha -1}(t)\varphi'(t)\geq  x \varphi_0 ^{\alpha - 1}\,
\exp \Big( -  \frac{t^2 (\varphi_0 + 1)^{\alpha - 3}}{4 K (\varphi_0)^{\alpha - 1}} 
\Big)
$$
for any $t\in[0,\omega)$. Define
$$
\gamma^2 \equiv \frac{(\varphi_0 + 1)^{\alpha - 3}}{4 K (\varphi_0)^{\alpha - 1}}
$$
and integrate the above inequality from $0$ to $t$ to obtain
$$
\varphi^\alpha(t) \geq \varphi^\alpha_0 
+ \alpha x \varphi_0^{\alpha -1} \int_0^t {\rm e}^{-\gamma^2 t^2} {\rm d} t 
= \varphi^\alpha_0 + \frac{\alpha x \varphi_0^{\alpha -1}}{\gamma}
 \int_0^{\gamma t} {\rm e}^{-u^2} {\rm d} u \equiv (\varphi_{m1}(t))^\alpha.
$$
It follows from the above inequality that $\varphi(t)$ is positive whenever 
$\varphi_{m1}(t)$ is positive; i.e., whenever $x$ is such that
$$
\frac{-1}{\alpha x} \Big[ \frac{(\varphi_0 + 1)^{\alpha - 3}}{4 K (\varphi_0)
^{\alpha - 3}}\Big]^{1/2}
>   \int_0^{\gamma t} {\rm e}^{-u^2} {\rm d} u,
$$
which is fulfilled if Condition \eqref{eq:x-boun} is satisfied because
$$
\frac{\sqrt \pi}{2} = \int_0^{\infty} {\rm e}^{-u^2} {\rm d} u 
> \int_0^{\gamma t} {\rm e}^{-u^2} {\rm d} u
$$
for all positive $t$.
\end{proof}

Define
$$
\overline{x} \equiv -\frac{2}{\sqrt \pi}
\frac{1}{\alpha } \Big[ \frac{(\varphi_0 + 1)^{\alpha - 3}}{4 K 
(\varphi_0)^{\alpha - 3}}\Big]^{1/2}.
$$
It follows from Lemma \ref{lema:glob-exis} that
if $\alpha \geq 1$ then $f(x)$, given by the limit  \eqref{eq:f}, is well defined
on the interval $(\overline{x}, 0)$. In next lemma we prove that
$f(x)$ is continuous on $(\overline{x}, 0)$ and this result
implies that $f(x)$ is onto the interval $f(\overline{x}, 0)$.


\begin{lemma} \label{lema:cont}
If $\alpha\geq 1$ and $x \in (\bar{x}, 0)$ then $f(x)$, defined by \eqref{eq:f},
is a continuous function.
\end{lemma}

\begin{proof}
Let $I\equiv [x_1, x_2] \subset (\bar{x}, 0)$. The proof of this lemma
is similar to the proof of Lemma \ref{lema:fini} and it is enough to show
that $|\varphi '(t)|$ is bounded by an integrable function of $t$, uniformly
in  $x\in I$.
To do so, we observe that it follows from
the proof of Lemma \ref{lema:glob-exis} that there exists $m>0$ such that
$m \leq f(x) \leq \varphi_0$ for any $x\in I$. Therefore, from \eqref{sol},
we obtain
$$
|\varphi '(t)| \leq |x| \Big(\frac{m+1}{m}\Big)^{\alpha - 1}\,
\Big(\frac{\varphi_0}{\varphi_0+1}\Big)^{\alpha - 1}
\exp \Big( -  \frac{t^2 (\varphi_0 + 1)^{\alpha - 3}}{4 K (\varphi_0)^{\alpha - 1}}
 \Big)
$$
which completes the proof.
\end{proof}


\subsection{Proof of Part II}
\label{subsec:partii}

In Lemma \ref{lema:omeg-infi} below we prove that if $\alpha \geq 2$
and if $\varphi(t, x)$ changes
concavity at some point $t_0$   then $\varphi(t, x)$ exists for all $t$.
We use this result to prove, in Lemma \ref{finiteomega},
that classical solutions will not be defined  on $[0, \infty)$.


\begin{lemma} \label{lema:omeg-infi} 
Let $\alpha\geq 2$ and suppose that there exists
$t_0>0$ such that $\varphi ''(t_0) = 0$. Then, $t_0$ is unique and
the solution $\varphi(t)$ to  \eqref{eq:ivp} is well defined for all $t\geq 0$.
\end{lemma}

\begin{proof}
 Since $\{t: \varphi ''(t) = 0\}$ is nonempty,  
$t_0\equiv \inf\{t:\varphi ''(t)=0\}$ is well defined.
We first prove that if $\alpha\geq 2$ then $\{t: \varphi ''(t) = 0\} = \{t_0\}$.
 From \eqref{eq:ivp}, $\varphi ''(t)$ can be
written as
\begin{equation} \label{gdef}
\varphi''(t) =  g(t) \Big(\frac{ - \varphi'(t) }{\varphi^2(t) +\varphi(t) }\Big),
\end{equation}
where $g(t)$ is given by
\begin{equation} \label{eq:g}
g(t) \equiv (\alpha - 1)\,\varphi'(t)
+ \frac{t}{2\,K} \Big(\frac{\varphi(t)+1}{\varphi(t)}\Big)^{\alpha - 2}.
\end{equation}
It follows from \eqref{gdef} that both $\varphi''(t)$ and $g(t)$ have the same sign
because $\varphi(t)$ and $-\varphi '(t)$ are both positive for $t\in[0,\omega)$.
 From \eqref{eq:g}, we obtain
\begin{equation}
\label{gderiv}
g'(t) =  (\alpha - 1)\,\varphi''(t)
+ \frac{1}{2\,K} \Big(\frac{\varphi(t)+1}{\varphi(t)}\Big)^{\alpha - 2}
+ \frac{(\alpha-2)}{2\,K} \Big(\frac{\varphi(t)+1}{\varphi(t)}\Big)^{\alpha - 3}
\Big(\frac{-\varphi'(t)}{\varphi^2(t)}\Big).
\end{equation}

It follows from \eqref{gderiv} that $g'(t_0)>0$ if $\alpha \geq 2$.
We claim that $g'(t)>0$ for all $t \geq t_0$.
If not, there would exist $t_2>t_0$ such that $g'(t_2)=0$.
Of course, $g(t_2)>0$ because $g(t)$ is increasing up to $t_2$
and because, from \eqref{gdef},  $g(t_0)=0$. Then we conclude,
from \eqref{gderiv}, that
$\varphi''(t_2) < 0$.  Therefore, again from \eqref{gdef}, we conclude
that $g(t_2)<0$, a contradiction. In particular, $g(t)$ is an increasing
function for $t> t_0$.

In the sequel, we will prove that $\varphi(t)$ remains bounded away from zero
if it changes concavity at some point $t_0$. Since $\varphi(t)$ is also bounded above
by $\varphi_0$, the solution will exist for all $t\geq 0$, see \cite{bib:codd-levi}.
Fix $t_i > t_0$ and consider  $t\in [t_i, \omega)$.
Since $\varphi(t)$ is decreases and $g(t)$  increases as $t$ increases above
$t_0$, we get
$$
\frac{g(t)}{1+\varphi(t)} \geq \frac{g(t_i)}{1+\varphi(t_i)}
\equiv K_i > 0,
$$
which implies from \eqref{gdef} that
$$
K_i \, \Big( \frac{-\varphi'(t)}{\varphi(t)} \Big) \leq \varphi''(t)
$$
for all $t \geq t_i$.
Integrating both sides of the above inequality from $t_i$ to $t$, we get that
$$
K_i \ln \Big( \frac{\varphi(t_i)}{\varphi(t)} \Big) \leq
\varphi'(t)-\varphi'(t_i) \leq -\varphi'(t_i),
$$
where we have used that $\varphi'(t) < 0$ for all $t\in [0,\omega)$ to get the last inequality.
After exponentiating both sides of the above inequality we get
$$
\varphi(t) \geq \varphi(t_i)\,\exp{\frac{\varphi'(t_i)}{K_i}} > 0
$$
for all $t> t_i$ and we are done.
\end{proof}

\begin{lemma} \label{finiteomega}
Let $\varphi(t):[0, \omega) \to \mathbb{R}$ be the solution
to  \eqref{eq:ivp} with $1 < \alpha \leq 2$ and
\begin{equation}
\label{eq:x-upper-bound}
x < - \sqrt{\frac{\varphi_0}{2K (\alpha - 1)}},
\end{equation}
where $[0,\omega)$ is the solution's maximal interval of existence. Then
$\omega < \infty$.
\end{lemma}

\begin{proof}
 Let $L(t) \equiv \varphi_0 + t \, x$ and observe that
$L(-\varphi_0/x)=0$. It follows from the arguments given at the beginning of this
section that $\varphi(t) < L(t)$ holds, at least
for small positive values of $t$, because $\varphi''(t) < 0$ if $t \geq 0$ and
small and if $\alpha > 1$. We claim that $\omega < - \varphi_0/x$.
If not, there would exist a positive time $t_1$, with $t_1 < - \varphi_0/x$,
such that $\varphi(t_1) = L(t_1)$. It then follows that $\varphi(t)$ changes concavity
at some time  $t_0 < t_1$, i.e.,  $\varphi''(t_0) = 0$.
 From \eqref{eq:deri-segu} we read that
$$
\varphi '(t_0) = - \frac{t_0}{2 K (\alpha - 1)}\Big( \frac{\varphi(t_0) + 1}{\varphi(t_0)}\Big)^{\alpha -2}.
$$
Since $x > \varphi'(t_0)$ and  $t_0 < -\varphi_0/x$, we take
$\alpha \leq 2$ to obtain, from the above inequality, that
$$
x > \frac{\varphi_0}{x}\frac{1}{2 K (\alpha - 1)},
$$
which is in contradiction with  \eqref{eq:x-upper-bound}.
\end{proof}


\subsection{Proof of Part III}
\label{subsec:partiii}

For $\alpha = 2$, we show that
the set of solutions, parametrized by $\varphi '_0<0$, is organized
increasingly as  $\varphi '_0$ varies from $-\infty$ to some negative
number $\lambda$ and we use this result to prove that
$f(x)$ maps the interval $(\lambda, 0]$ onto the interval $(0, \varphi_0]$.
As in the proof of Lemma \ref{lema:inje},  we take
$ 0 >  \bar{x} > x $ and write $\delta (t) = \bar\varphi(t)-\varphi(t)$,
where $\varphi(t) = \varphi(t,x)$ and $\bar\varphi(t) = \varphi(t,\bar x)$ are the
solutions to \eqref{eq:ivp} with initial derivatives
$x$ and $\bar x$, respectively.

\begin{lemma} \label{lema:t_star}
Let $\alpha = 2$ and suppose that $\omega(\bar  x) < \infty$.  Then
$\omega(  x) < \omega(\bar  x)$ for all $x < \bar x$.
\end{lemma}

\begin{proof}
Observe that $\delta (0) = 0$ and $\delta '(0) > 0$. Then $\delta '(t) > 0$ 
for small values of $t$.
We will show below that $\delta '(t) > 0$ for all values of $t\in (0,\omega_m)$, 
where $\omega_m = \min\{\omega(x), \omega(\bar x)\}$; i.e., 
$\delta (t)$ is an increasing and strictly
positive function on $(0,\omega_m)$. It follows from
this result that $\omega(  x) < \omega(\bar  x)$. In fact, if
it happened that $\omega(  x) \geq \omega(\bar  x)$ then $\delta (t)=0$
at some point $t< \omega_m$ but this is not possible because $\delta (t) > 0$ 
for all $t\in (0,\omega_m)$.
So, suppose, by contradiction,  that there exists a first point $\tilde t$ such that
$\delta '(\tilde t) = 0$, i.e., $\bar \varphi'(\tilde t) = \varphi'(\tilde t)$.
It then follows from \eqref{eq:ivp} with $\alpha = 2$ that
$$
[\bar\varphi(\tilde t)^2+\bar\varphi(\tilde t)]\bar\varphi''(\tilde t)
= [\varphi(\tilde t)^2+\varphi(\tilde t)]\varphi''(\tilde t),
$$
and we conclude that $\bar\varphi''(\tilde t)$ and $\varphi''(\tilde t)$ are both
positive or negative. But, by hypothesis, $\bar \omega < \infty$ which implies
from Lemma~\ref{lema:omeg-infi} that $\bar\varphi''(t) < 0$ for all
$t \in [0,\bar \omega)$. Therefore, $\bar\varphi''(\tilde t)$ and
$\bar\varphi''(\tilde t)$ are both negative.
 From the above identity we get
$$
0 < \frac{\bar\varphi''(\tilde t)}{\varphi''(\tilde t)}
= \frac{\varphi(\tilde t)^2+\varphi(\tilde t)}{\bar\varphi(\tilde t)^2+
\bar\varphi(\tilde t)} < 1,
$$
leading that  $\bar\varphi''(\tilde t) - \varphi''(\tilde t) = \delta ''(\tilde t)  > 0$.
On the other hand, since $\delta (t)$ is increasing
in the interval $[0,\tilde t]$ and since $\delta '(\tilde t ) = 0$ then $\delta ''(\tilde t) \leq 0$,
a  contradiction with $\delta ''(\tilde t) > 0$.
\end{proof}

Once $\varphi_0$ is fixed and $1<\alpha\leq 2$, it  follows from 
 Lemma \ref{finiteomega} that
$\lambda \equiv \sup \{x \in (-\infty, 0) : \omega(x) < \infty\}$ is
well defined and it follows from Lemma \ref{lema:glob-exis} that $\lambda < 0$.
Next result states that the set $\{x \in (-\infty, 0) : \omega(x) < \infty\}$
is an interval.

\begin{corollary}\label{cor:sobre}
If $\alpha = 2$, then 
$(-\infty, \lambda] = \{x \in (-\infty, 0) : \omega(x) < \infty\}$.
\end{corollary}

\begin{proof} It is sufficient to show that
$(-\infty, \lambda] \subset \{x \in (-\infty, 0) : \omega(x) < \infty\}$.
We first observe that $\lambda \in \{x \in (-\infty, 0) : \omega(x) < \infty\}$
because the set $\{x \in (-\infty, 0) : \omega(x) = \infty\}$ is open
(this is so because of Lemma \ref{lema:omeg-infi} and the smoothly continuous
dependence of solutions on the initial conditions, see \cite{bib:codd-levi}).
 Then $\omega(\lambda) < \infty$ and it follows from
Lemma \ref{lema:t_star} that $\omega(x) < \omega(\lambda)$ if $x< \lambda$
and we are done.
\end{proof}

It follows from the corollary that $f(x)$ is a well defined function on 
$(\lambda, 0]$. In next lemma we characterize the set $f((\lambda,0])$.

\begin{lemma} \label{lema:fonto}
If $\alpha = 2$ then $f((\lambda,0]) = (0, \varphi_0]$.
\end{lemma}

\begin{proof} It follows from Lemma \ref{lema:cont} that
$f(x)$ is continuous on $(\lambda,0]$. Since $f(0) = \varphi_0$ then
or $f((\lambda,0]) = (0, \varphi_0]$
or there exists $a > 0$ such that $f(\lambda,0] \subset [a, \varphi_0]$.
In what follows we discard the second option.
Since $\omega(\lambda) < \infty$ then $\varphi(t, \lambda) \to 0$ as $t\to \omega(\lambda)$.
In particular, given $a > 0$ there exists $t_a > 0$
such that $\varphi(t_a,\lambda) < a/2$. By
the continuous dependence on initial conditions, $\varphi(t_a,x) < a/2$
if $x$ is close enough of $\lambda$, with $x > \lambda$.
But since $\varphi(t,x)$ is decreasing as a function of $t$, then
$\varphi(t,x) < a/2$ for all  $t>t_a$ and then $f(x) \leq a/2 < a$.
It follows from here that
$f((\lambda,0]) = (0, \varphi_0]$.
\end{proof}

We remark that
Lemma \ref{lema:omeg-infi} is proven only for $\alpha \geq 2$ and
Lemma \ref{finiteomega}
is proven only for $1< \alpha \leq 2$, but numerical simulations indicate that
both hold for all $\alpha \geq 1$.
 If so then Lemma \ref{lema:fonto}, equivalently
the third part of Theorem  \ref{teo:main2}, would
hold for any $\alpha \geq 1$. Numerical simulations also suggest
that the solutions to  \eqref{eq:ivp} are organized increasingly
as a function of $x$ and we have proved that this is the case if $x\geq 0$
and $\alpha \geq 1$ or if $x < \lambda$ and $\alpha = 2$  but it is an open 
problem to prove it for $x\in (\lambda, 0)$. If so then the result would imply that
$f(x): (\lambda, 0]\to (0,\varphi_0]$ is one-to-one and, together with 
lemmas \eqref{lema:cont} and \eqref{lema:fonto}, a homeomorphism.
Therefore, based upon numerical experiments, we conjecture that Theorem
\ref{teo:main} holds also when $\theta_i < \theta_0$, for any
$\theta_0, \theta_i\in
(\theta_r, \theta_s)$.


\subsection*{Acknowledgements}
 Gast\~{a}o A Braga thanks Aleksey Telyakovskii for
many helpful discussions related to this problem.  He also acknowledges 
the financial support of the brazilian agencies CNPq and FAPEMIG.

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\end{document}