\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 307, pp. 1--17.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/307\hfil Existence of global and blowup]
{Existence of global and blowup solutions for a singular second-order ODE}

\author[R. Colucci \hfil EJDE-2015/307\hfilneg]
{Renato Colucci}

\address{Renato Colucci \newline
Department of Mathematical Sciences,
Xi'an Jiaotong-Liverpool University,
111 Ren'ai Road, SIP, Suzhou 215123, China}
\email{renatocolucci@hotmail.com}

\thanks{Submitted November 2, 2015. Published December 16, 2015.}
\subjclass[2010]{76M55, 35B40, 35B44}
\keywords{Self-similar solutions; blow up; asymptotic behavior; 
\hfill\break\indent forward-backward equations}

\begin{abstract}
 We study the existence of global solutions of a singular ordinary differential
 equation arising in the construction of self similar solution for a
 backward-forward parabolic equation. Also we present several numerical
 simulations and obtain an upper bound for the blowup time.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

The evolution equation
\begin{equation}\label{eq}
u_t=\frac{1}{2} [W'(u_x)]_x,
\end{equation}
arises in the study of nonlinear elasticity and
phase transitions \cite{MU:99,PE}). Here
$W(p)=\frac{1}{2}(p^2-1)^2$ is the so called double well potential
and  equation \eqref{eq} describes the $L^2-$gradient dynamics of the Energy
\begin{equation}\label{min}
\frac{1}{2}\int_IW(u_x)dx,
\end{equation}
where $I=(0,1)$. Both solving equation \eqref{eq} and minimizing \eqref{min}
are ill-posed problems. In particular, the equation \eqref{eq} has a forward
backward character depending on the sign of $W''$,
while the functional \eqref{min} is non convex and admits infinitely many
minimizers.
Another interesting choice of the potential function
is the following
\begin{equation*}
W(p)=\frac{1}{2}\ln(1+p^2).
\end{equation*}
The corresponding evolution equation
is the so called Perona-Malik equation \cite{PM:90} which has a similar 
(see \cite{KI:97})
ill-posed behavior as that of equation \eqref{eq}.

It is well known that the Perona-Malik equation admits global or not global
solutions depending on the initial datum. In particular if the
evolution is restricted in a bounded interval $I$ with
zero Neumann boundary conditions we have:
\begin{itemize}
\item[(1)] If $|u_{0x}(x)|<1$ for all $x\in I$
then there exists a unique global solution such that $|u_{x}(x,t)|<1$ for all
$t\geq0$ and $x\in I$ (see \cite{KK:98} ).

\item[(2)] If there exists a point $x_0\in I$ such that $|u_{x0}(x_0)|>1$
then then  there are no global solutions (see \cite{KK:98} and \cite{GO:07}).
\end{itemize}
In \cite{GO:07} the author also provided
a bound for the maximum interval of time of existence.

We would like to remark that there is a large body of literature
(see \cite{BG:2008,BFG:06}, \cite{CAC:14}-\cite{CO:12})
about the regularization of the ill-posed problems \eqref{eq} and \eqref{min}.
In general the idea is to add an high order term depending on a small
parameter $\epsilon$ which makes the equation well-posed (see for example
\cite{SL:91,BFG:06}).
The further step is to study the limit of the solutions
$u_\varepsilon$ of the regularized problem \cite{BFG:06} as
$\varepsilon\to0$.

Instead of considering a possible regularization of  problem
\eqref{eq} or of \eqref{min}, we limit our analysis to the class of self similar
solutions (see \cite{BA:96}).
In general self similar solutions are useful to understand the
transient dynamics or the asymptotic behavior of solutions of partial
differential equations. The idea of this work is to describe in details
the global/non global existence for this class of solutions depending on the
initial data.
Consider the transformation
\begin{equation*}
(u,x,t) \quad\to\quad (\hat u, \hat x, \hat t),
\end{equation*}
with
\begin{equation*}
u=U \hat u, \quad x= L \hat x, \quad    t= T\hat t,
\end{equation*}
where $U$, $L$ and $T$ are abstract positive numbers 
(see \cite{BA:96} for a complete discussion).
Then, by dimension Analysis, we have that the equation is invariant under
the above transformation if and only if
\begin{equation*}
\frac{U^2T}{L^4}=1,\quad \frac{T}{L^2}=1,
\end{equation*}
this suggests that quantities $\frac{x}{t^{1/2}}$ and $x$ are dimensionless and
will play an important role in the study of the solutions.
The standard procedure (see \cite{BA:96}) consists in looking for general 
Self-Similar Solutions of the form
\begin{equation}\label{sss}
\begin{gathered}
   u(x,t)=t^\beta f(y),  \\
   y=\frac{x}{t^\alpha}. 
  \end{gathered}
\end{equation}
We observe that
\begin{gather*}
 y_t=-\frac{\alpha}{t}y,\quad
 y_x=t^{-\alpha},\\
 y_{xx}=0, \quad
 u_t = \beta t^{\beta-1}f(y) -\alpha t^{\beta-1} yf'(y),\\
 u_x= t^{\beta-\alpha}f'(y),\quad
 u_{xx}= t^{\beta-2\alpha}f''(y).
\end{gather*}
By putting the previous expression into \eqref{eq} we obtain
\begin{equation*}
\beta f(y)-\alpha y f'(y)=\frac{1}{2}W''(t^{\beta-\alpha}f'(y))f''(y)t^{1-2\alpha},
\end{equation*}
from which, $\alpha=\beta=1/2$.
Then we obtain the equation
\begin{equation*}
f(y)- y f'(y)=W''(f'(y))f''(y),
\end{equation*}
that is,
\begin{equation}\label{eq:ss}
f(y)-yf'(y)=2[3(f')^2-1]f''(y).
\end{equation}
Any linear function of the form  $f(y)=ay$ satisfies the previous equation,
giving rise to the well known class of affine solutions (see \cite{GO:07}):
\begin{equation*}
u(x,t)=t^{1/2} f(y)=ax.
\end{equation*}
Then the interesting case is when
\begin{equation*}
f(y)\not\equiv ay.
\end{equation*}
We first suppose that there exist $C^2$ solutions of \eqref{eq:ss}
with non constant first derivative and constant second derivative:
\begin{equation*}
f''(y)=b.
\end{equation*}
By substituting the expression $f(y)=f(0)+yf'(0) + \frac{1}{2}by^2$ 
into \eqref{eq:ss} we obtain the condition $b=f(0)=0$ and we obtain again 
the above class of affine solutions.
What it is expected is that the second derivative will play the important role for
the blow-up of solution when the first derivative $f'$ approaches the critical
values $\pm\frac{1}{\sqrt{3}}$.

In the next sections we will consider the  problem of finding
the interval of existence of solutions of \eqref{eq:ss} distinguishing two cases.
We will deal with solutions with infinite interval of existence $[0,+\infty)$
in section 2 with initial data satisfying
\begin{equation*}
f(0)>0 \; (\text{resp } <0),\quad f'(0)>\frac{1}{\sqrt{3}}
\;(\text{resp } <-\frac{1}{\sqrt{3}}).
\end{equation*}
The case of a finite interval $[0,y^*)$ of existence will be
analyzed in section 3. This case corresponds to the choice
$|f'(0)|<\frac{1}{\sqrt{3}}$, moreover,
in the same section, we will deal with the remaining possible choices of initial
data.
In section 4 we provide a bound for the length of the interval
$[0,y^*)$ in case of blow up of solutions. Moreover, using a method suggested
in \cite{GO:07} for the Perona-Malik equation, we find
a bound for the time interval of existence for non global solutions of
\eqref{eq}.
The results of sections 2-4 are illustrated by numerical experiments in
section 5.

\section{Global Solutions}

 In this section we study the case in which
equation \eqref{eq:ss} admits global solutions $f(y)$ with $y\in[0,+\infty)$.
The existence of such solutions depends on the values of $f'(0)$,
moreover we will see that the second derivative is not identically zero
and $f(y)$ approaches a linear solution of the form $Ky$ in finite
or infinite time.


\begin{theorem} \label{thm1}
Suppose that $f(0)>0$ and $f'(0)>1/\sqrt{3}$.
Then the $C^2$ solutions of \eqref{eq:ss} are global and satisfy
\begin{equation*}
\lim_{y\to\infty}f'(y)=K>\frac{1}{\sqrt{3}}.
\end{equation*}
Moreover
\begin{equation*}
\lim_{y\to\infty}[f(y)-yf'(y)]=0,
\end{equation*}
with
\begin{equation*}
f(y)-yf'(y)\geq0,\quad \forall\ y\geq0.
\end{equation*}
Moreover,
\begin{equation*}
\lim_{y\to\infty}|f(y)-Ky|=0, \quad
\lim_{y\to\infty}f''(y)=0,
\end{equation*}
with
$f''\not\equiv0$ and $f''(y)\geq 0$ for all $y\geq 0$.
\end{theorem}

\begin{proof}
Using the hypothesis on $f(0)$ and $f'(0)$ we have
\begin{equation*}
f(0)=2[3(f'(0))^2-1]f''(0)>0,
\end{equation*}
from which $f''(0)>0$. For $C^2$ solutions we have that there exists $\delta>0$:
\begin{equation*}
f(y)>0,\quad f'(y)>\frac{1}{\sqrt{3}},\quad f''(y)>0, \quad y\in[0,\delta),
\end{equation*}
from which $f$ and  $f'$ are increasing in $[0,\delta)$.
We have that
\begin{gather*}
f(y)-yf'(y)=2[3(f')^2-1]f''(y)>0,\quad y\in(0,\delta), \\
[f(y)-yf'(y)]'=-yf''(y)<0, \quad y\in(0,\delta),
\end{gather*}
from which $f(y)-yf'(y)$ is positive and decreasing in $(0,\delta)$ and
the same is true for $f''(y)$ (since $f'(y)$ is increasing).

 If $\delta=+\infty$ then
\begin{equation}
0\leq\lim_{y\to\infty}[f(y)-yf'(y)]=\lim_{y\to\infty}2[3(f'(y))^2-1]f''(y).
\end{equation}
From the continuity of $f'(y)$ we have that
\begin{equation*}
\lim_{y\to\infty}f'(y)>\frac{1}{\sqrt{3}},
\end{equation*}
and as a consequence
\begin{equation*}
\lim_{y\to\infty}f''(y)\geq0.
\end{equation*}
Suppose that
\begin{equation*}
\lim_{y\to\infty}f''(y)=K>0,
\end{equation*}
then $f'$ is increasing in $(0,\infty)$ and we have
\begin{equation*}
\lim_{y\to\infty}2[3(f'(y))^2-1]=+\infty.
\end{equation*}
On the other hand, since the term $f(y)-yf'(y)$ is positive and decreasing,
then its limit is finite,
\begin{equation*}
\lim_{y\to\infty}[f(y)-yf'(y)]=c\geq0,
\end{equation*}
but this is a contradiction. Then
\begin{equation*}
\lim_{y\to\infty}f''(y)=0,
\end{equation*}
and, as a consequence,
$\lim_{y\to\infty}[f(y)-yf'(y)]=0$.

 If $\delta\neq+\infty$ then
 $f(\delta)>0$ and
$f'(\delta)>1/\sqrt{3}$ since both $f$ and $f'$ are increasing in
$[0,\delta)$.
The we must have $f''(\delta)=0$, which means
\begin{equation*}
f(\delta)-\delta f'(\delta)=0.
\end{equation*}
Moreover,
\begin{equation*}
([f(y)-yf'(y)]')_{y=\delta}=-\delta f''(\delta)=0.
\end{equation*}
We claim that $f''(y)=0$ in the interval $[\delta,+\infty)$. Then
\begin{gather*}
f(y)-yf'(y)=f(\delta)-\delta f'(\delta)=0,\quad y\in[\delta,+\infty), \\
f'(y)=f'(\delta),\quad y\in[\delta,+\infty),
\end{gather*}
from which
\begin{equation*}
f(y)=\frac{f(\delta)}{\delta}y=f'(\delta)y,\quad y\in[\delta,+\infty).
\end{equation*}

To see that $f''(y)=0$ in $[\delta,+\infty)$, suppose that
there exists $\nu>0$ such that
\begin{equation*}
f''(y)>0,\quad f'(y)>\frac{1}{\sqrt{3}},\quad y\in(\delta,\nu).
\end{equation*}
As a consequence,
the term $f(y)-yf'(y)$ is decreasing on it and negative. Then, by \eqref{eq:ss}, 
we have
\begin{equation*}
3[f'(y)]^2-1<0,\quad y\in(\delta,\nu),
\end{equation*}
but this is a contradiction of the fact that $f'(y)>\frac{1}{\sqrt{3}}$ on the
same interval. The case $f''<0$ can be analyzed using  a similar argument.
\end{proof}

\begin{remark} \label{rmk1} \rm
The above proof also works for the case
\begin{equation*}
f(0)<0,\quad f'(0)< -\frac{1}{\sqrt{3}}.
\end{equation*}
Since the equation is invariant for the transformation $f\to(-f)$.
\end{remark}

\section{Blow-up Solutions}

We start by considering the case in which the solutions of 
\eqref{eq:ss} are not global. We prove that this is always the case in which
$|f'(0)|<1/\sqrt{3}$ and this strictly depends on the blow up of the
second derivative of $f$.

\begin{theorem} \label{thm2}
Suppose that
\begin{equation*}
f(0)>0, \quad 0<f'(0)<\frac{1}{\sqrt{3}},
\end{equation*}
then  the non trivial $C^2$ solutions of equation \eqref{eq:ss} are not global.
In particular there exists a point $y_*>0$ such that
\begin{equation*}
\lim_{y\to y_*}f''(y)=-\infty.
\end{equation*}
\end{theorem}

\begin{proof}
By hypotheses we obtain
\begin{equation*}
f(0)=2[3(f'(0))^2-1]f''(0),
\end{equation*}
from which $f''(0)<0$. By continuity there exists $\delta>0$ such that
\begin{gather*}
f(y)>0, \quad f''(y)<0,\quad y\in(0,\delta), \\
f'(y)<\frac{1}{\sqrt{3}}, \quad y\in(0,\delta).
\end{gather*}
Then both $f$ and $f(y)-yf'(y)$ are positive and increasing
and they cannot change sign at $y=\delta$. The only possibility is that
$f'(\delta)=0$,
from which
\begin{equation*}
f(\delta)=-2 f''(\delta)>0.
\end{equation*}
Then there exist $\nu>\delta$ such that
\begin{equation*}
f(y)>0, \quad f''(y)<0, \quad y\in(\delta,\nu),
\end{equation*}
and, as a consequence (using that $W''(f')$ must remain negative),
\begin{equation*}
f'(y)\in\big(-\frac{1}{\sqrt{3}},0\big), \quad y\in(\delta,\nu).
\end{equation*}
Then both $f$ and $f'$ are decreasing and the term $f(y)-yf'(y)$
is positive and  increasing.
We need to study these functions at $y=\nu$.
We observe that $W''(f'(\nu))\in(-2,0]$  then $f''(\nu)\neq0$
and remains negative at $y=\nu$. Then we have only three cases:
\begin{itemize}
  \item[(a)] $f(\nu)=0$,  $f'(\nu)=-\frac{1}{\sqrt{3}}$,
  \item[(b)] $f(\nu)=0$,  $f'(\nu)>-\frac{1}{\sqrt{3}}$,
  \item[(c)] $f(\nu)>0$,  $f'(\nu)=-\frac{1}{\sqrt{3}}$.
\end{itemize}

Cases (a) and (c) are compatible if
\begin{equation*}
\lim_{y\to\nu}f''(y)=-\infty.
\end{equation*}

Case (b): By hypotheses there exists $\eta>\nu$ such that
\begin{equation*}
f(y)<0,\quad f'(y)<-\frac{1}{\sqrt{3}},\quad f''(y)<0,\quad y\in(\nu,\eta).
\end{equation*}
Moreover, the term $f(y)-yf'(y)$ is positive and increasing and as a consequence
$f''$ is negative and decreasing. Then, since $f$ and $f'$ are negative and
decreasing we must have
$f'(\eta)=-1/\sqrt{3}$,
and, as a consequence,
\begin{equation}\label{bu}
\lim_{y\to\eta}f''(y)=-\infty.
\end{equation}
If $\eta=+\infty$, since $f$, $f'$, and $f''$ are negative and decreasing
we have that
\begin{equation*}
\lim_{y\to\infty}f''(y)\in[-\infty,0),
\end{equation*}
and then there exists a point $y_0>\nu$ such that
$f'(y_0)=-\frac{1}{\sqrt{3}}$,
and we obtain again \eqref{bu}.

The case $\nu=+\infty$ can be analyzed using the same idea of the case
$\eta=\infty$, observing that $f'$ and $f''$ are negative and decreasing.
\end{proof}

\begin{remark} \label{rmk2} \rm 
From the previous proof and the arguments
used in  Case 1, we can extend the statement of the previous theorem for 
the hypotheses
\begin{equation*}
f(0)<0,\quad f'(0)\in(-1/\sqrt{3},0).
\end{equation*}
Also, it is easy to see that the same proof works for
\begin{equation*}
f(0)<0,\quad f'(0)\in(0,1/\sqrt{3}).
\end{equation*}
In fact is it sufficient to invert the signs of $f''$ and of the 
term $f(y)-yf'(y)$. This means
that Theorem \ref{thm2} is true for the more general hypotheses
$|f'(0)|<1/\sqrt{3}$.
\end{remark}

 We note that the following cases are outside the hypotheses
of Theorems \ref{thm1} and \ref{thm2}:
\begin{gather}\label{n1}
f(0)<0, \quad f'(0)>\frac{1}{\sqrt{3}}, \\
\label{n2}
f(0)>0, \quad f'(0)<-\frac{1}{\sqrt{3}}.
\end{gather}
We analyze the case \eqref{n1} since the other case can be studied in 
a similar way.
What we expect is both global and blow up behavior:
\begin{enumerate}
\item The term $f(y)-yf'(y)$ remains negative while
$f'$ approaches $\frac{1}{\sqrt{3}}$ and $f''\to-\infty$ as $y$ 
approaches some critical point $y^*$.

\item The term $f(y)-yf'(y)$ approaches zero at some point $y_0$ while $f'$
remains away from the critical value $\frac{1}{\sqrt{3}}$. The derivative
of the solution will remain constant, that is $f'(y)=\frac{f(y_0)}{y_0}$, 
for $y\geq y_0$.
\end{enumerate}
The two situations depend on the value of
$f'$ at zero and that of the term $f(y)-yf'(y)$ near zero. In particular
near zero the derivative of the term $f(y)-yf'(y)$ satisfies
\begin{equation*}
|[f(y)-yf'(y)]'| \approx  y|f''(0)|=y\frac{|f(0)|}{W''(0)}.
\end{equation*}
Then, if $f'(0)$ is very close to $1/\sqrt{3}$,
we expect that the term $f(y)-yf'(y)$ converge to zero faster than  
$f'$  to $1/\sqrt{3}$.

From the initial data we have that $f''(0)<0$, then by continuity we can find
$\delta$ such that
\begin{equation}\label{v1}
f(y)<0,\quad f''(y)<0,\quad y\in[0,\delta).
\end{equation}
The left hand side of  \eqref{eq:ss} is negative at zero and it is increasing
in $[0,\delta)$:
\begin{equation}\label{v2}
f(y)-yf'(y)<0,\quad y\in[0,\delta),
\end{equation}
and
\begin{equation}\label{v3}
[f(y)-yf'(y)]'=-yf''(y)>0,\quad y\in[0,\delta).
\end{equation}
As a consequence we have
\begin{equation}\label{v4}
W''(f'(y))>0,\quad y\in[0,\delta), \quad\text{that is}\quad
f'(y)>\frac{1}{\sqrt{3}},\quad y\in[0,\delta).
\end{equation}
Since $f'$ is decreasing, $W''(f')$ will be decreasing and $f''$  increasing
in the same interval.
We need to analyze the behavior of the solutions at $y=\delta$, 
there are several cases:
\begin{itemize}
\item[(A)] $f(\delta)=0$, $f''(\delta)<0$, $f(\delta)-\delta f'(\delta)<0$,
\item[(B)] $f(\delta)<0$, $f''(\delta)=0$, $f(\delta)-\delta f'(\delta)<0$,
\item[(C)] $f(\delta)<0$, $f''(\delta)<0$, $f(\delta)-\delta f'(\delta)=0$,
\item[(D)] $f(\delta)=0$, $f''(\delta)=0$, $f(\delta)-\delta f'(\delta)<0$,
\item[(E)] $f(\delta)=0$, $f''(\delta)<0$, $f(\delta)-\delta f'(\delta)=0$,
\item[(F)] $f(\delta)<0$, $f''(\delta)=0$, $f(\delta)-\delta f'(\delta)=0$,
\item[(G)] $f(\delta)=0$, $f''(\delta)=0$, $f(\delta)-\delta f'(\delta)=0$.
\end{itemize}

Case (G) has already been studied in Theorem \ref{thm1}, the solution
is $f(y)=\frac{f(\delta)}{\delta}y$ for $y\geq\delta$ while
the hypotheses of cases (B) and (D) are not consistent since the term $W''(f')$ is
bounded then if $f''$ is zero also the term $f(y)-yf'(y)$ must be zero. 
Moreover (E) and (F) are inconsistent  since they would imply that
$f'(\delta)=0$ and $f'(\delta)<0$ respectively.
We pass to analyze the others cases.
\smallskip

\noindent\textbf{Case (C)}
As a consequence of the hypotheses we have
$3(f'(\delta))^2-1=0$, that is $f'(\delta)=1/\sqrt{3}$,
from which
\begin{equation*}
f(\delta)-\frac{\delta}{\sqrt{3}}=0,
\end{equation*}
but this is not possible since $f(\delta)$ is negative.
\smallskip

\noindent\textbf{Case (A)}
We have that
\begin{equation*}
-\delta f'(\delta)=2[3(f'(\delta'))^2-1]f''(\delta),
\end{equation*}
from which we have $f'(\delta)>1/\sqrt{3}$, that is, $f$ is increasing.
Then there exists $\nu>\delta$ such that
\begin{equation*}
f(y)>0,\quad f''(y)<0,\quad   f(y)-yf'(y)<0,\quad y\in(\delta,\nu)
\end{equation*}
from which we obtain that $f$ is increasing and
\begin{equation*}
f(y)>0,\quad f'(y)>\frac{1}{\sqrt{3}},\quad y\in(\delta,\nu).
\end{equation*}
Then at the point $\nu$ we have the following cases:
the hypotheses
\begin{equation*}
f''(\nu)=0, \quad f(\nu)-\nu f'(\nu)<0,
\end{equation*}
are not compatible with $f'(\nu)\geq\frac{1}{\sqrt{3}}$ and $f'$ decreasing;
while the hypotheses
\begin{equation*}
f''(\nu)<0, \quad f(\nu)-\nu f'(\nu)=0,
\end{equation*}
are only compatible with $f'(\nu)=1/\sqrt{3}$.
From which
\begin{equation*}
f(y)=\frac{y}{\sqrt{3}},\quad y\geq\nu.
\end{equation*}
The other possibility is that
\begin{equation*}
f''(\nu)<0, \quad f(\nu)-\nu f'(\nu)<0, \quad  f'(\nu)=\frac{1}{\sqrt{3}}.
\end{equation*}
This means that
\begin{equation*}
\lim_{y\to\nu}f''(\nu)=-\infty.
\end{equation*}
By these arguments we expect that for initial data satisfying \eqref{n1} and
\eqref{n2} there are both global and no global solutions.
We will illustrate these two cases in section 5 below.

\section{An estimate for the blowup point $y_*$}

 Using a slight modification of a method proposed by Gobbino
in \cite{GO:07} for the Perona-Malik equation, we derive an
estimate for the maximum interval of time of existence for the solutions
of \eqref{eq}, in the spatial interval I=$(0,1)$, whose initial 
datum $u(x,0)=v(x)$ satisfies
\begin{equation*}
 v_x(I)\cap \Big(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\Big)\neq\emptyset.
\end{equation*}
We observe that this is related to the finite time blowup
of the the self similar solutions. In fact
\begin{equation*}
u_{x}(x,t)=\left[t^{1/2}f\left(\frac{x}{t^{1/2}}\right)\right]_x=f'(y),
\end{equation*}
and we have observed that if for
$|f'(0)|<1/\sqrt{3}$
then there exists a point $y^*\neq+\infty$ such that
$\lim_{y\to y^*}f''(y)=-\infty$.

Let $\psi$ be a $C^2$ function, then (see \cite{GO:07}):
\begin{align*}
\frac{d}{dt}\int_I \psi(u_x)dx
&=\int_I\psi'(u_x)u_{xt}dx
 =-\int_I \psi''(u_x)u_{xx}u_tdx +
\left(\psi'(u_x)u_t\right)_{|_{\partial I}}\\
&=-\int_{I}\psi''(u_x)W''(u_x)[u_{xx}]^2dx,
\end{align*}
where we have used Neumann boundary conditions.
Integrating in the time interval $(t_1,t_2)$ we obtain
\begin{equation}\label{ineq}
\int_I \psi(u_x(x,t_2))dx + \int_{t_1}^{t_2}\int_{I}[\psi''(u_x)W''(u_x)][u_{xx}]^2
\,dx\,dt
=\int_I \psi(u_x(x,t_1))dx.
\end{equation}
Let $x_0\in I$ such that
$v(x_0)\in(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$,
and chose 
\begin{equation*}
\psi(p)= \begin{cases}
\big(|p|-\frac{1}{\sqrt{3}}\big)^4, 
 & p\in (-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}), \\
0, & \text{elsewhere}.
\end{cases}
\end{equation*}
Then we observe that $\psi(p)\geq0$ and $\psi''(p)W''(p)\leq0$ for all $p$.
Using inequality \eqref{ineq} with $t_1=0$ and $t_2=t$ we observe that
the term on the right is positive. Since $\psi''(p)W''(p)\leq0$
then also the first term of the left-hand side must be positive which means that for
every $t$ there exists at least one point
$x_0(t)\in(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$, and if $u$ is of class $C^2$
then by continuity for any $t\geq0$ there exists $x_1(t)\in I$ such that
$|u_x(x_1(t),t)|=1/\sqrt{3}$.
Then by Neumann boundary conditions:
\begin{align*}
&|W'(u_x(x_1(t),t))-W'(u_x(0,t))|\\
&= W'(\frac{1}{\sqrt{3}})\leq\int_0^{x_1(t)}|[W'(u_x)]_x|dx\\
&\leq \{x_1(t)\}^{1/2}\big\{ \int_0^{x_1(t)} \{[W'(u_x)]_x\}^2dx
\big\}^{1/2}
\leq \Big\{\int_I [W''(u_x)u_{xx}]^2dx\Big\}^{1/2}.
\end{align*}
Then by putting $\psi(p)=W(p)$ in the inequality \eqref{ineq} we obtain:
\begin{equation*}
t[W'(1/\sqrt{3})]^2\leq
\int_0^t\int_{I}[W''(u_x)]^2[u_{xx}]^2\,dx\,dt\leq\int_I W(v_x)dx,
\end{equation*}
from which we have the bound of the maximum interval of existence
\begin{equation*}
T\leq\frac{1}{[W'(\frac{1}{\sqrt{3}})]^2}\int_I W(v_x)dx.
\end{equation*}
Again using an argument from \cite{GO:07},
 if $u$ is only of class $C^1$ we obtain
the same result using the inequality
\begin{equation*}
\int_I \psi(u_x(x,t))dx\geq\int_I \psi(u_x(x,0))dx.
\end{equation*}
For the solution of equation \eqref{eq:ss} we have a similar result:


\begin{theorem} \label{thm3}
Suppose that $f(0)>0$ and $f'(0)=0$. Then 
the point $y_*$ of blowup for $f''$ satisfies
\begin{equation*}
y_*\leq \frac{4}{3\sqrt{3}f(0)}.
\end{equation*}
\end{theorem}

\begin{remark} \label{rmk3} \rm 
We observe that the hypotheses $f'(0)=0$ is not restrictive since it well
describe the case in which
\begin{equation*}
f(0)>0, \; f'(0)\in(0,1/\sqrt{3}), \quad\text{or}\quad
f(0)<0, \; f'(0)\in(-1/\sqrt{3},0).
\end{equation*}
In fact, for both cases, by the proof of Theorem \ref{thm2}, 
there exists a point at which
the derivative pass trough zero.
\end{remark}

\begin{proof}
By the hypotheses we have
\begin{equation*}
W'(f'(y_*))-W'(f'(0))
=W'(f'(y_*))
=\int_0^{y^*}[W'(f'(y))]_ydy
=\int_0^{y_*}[f(y)-yf'(y)]dy.
\end{equation*}
Since the term $f(y)-yf'(y)$ is positive and increasing in $[0,y_*)$ 
then we obtain
$y_*f(0)\leq W'(f'(y_*))$.
In details:
\begin{equation*}
y_*\leq \frac{4}{3\sqrt{3}f(0)}.
\end{equation*}
As expected the value of $y_*$ is proportional to
$f(0)^{-1}$. We give an illustration of this result in Section 5 below.
\end{proof}

\section{Numerical Experiments}

 In this section we give a brief description of the results from the
previous sections. In particular we consider two cases in which $f''(y)$ remains
bounded and $f''(y)\to\pm\infty$ respectively. We end the article by testing the
bound of the point $y_*$ of blow up of the second derivative of the solution.

\subsection*{Experiment 1}
We consider a first example in which
$f(0)=0.1$ and $f'(0)=0.6>1/\sqrt{3}$.
We represent the solution  $f(y)$, together with $f'(y)$ and $f''(y)$,
in figure \ref{fig:1exp1} and compare it with the straight line 
$f'(0)y + f(0)$
in figure \ref{fig:2exp1}. We observe that $f\to ay+b$ with $a>f'(0)$.

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.5\textwidth]{fig1} % f1e1
\end{center}
\caption{Experiment 1: the functions $f(y)$, $f'(y)$,
and $f''(y)$.}
\label{fig:1exp1}
\end{figure}


\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.5\textwidth]{fig2} % f2e1
\end{center}
 \caption{Experiment 1: the solution $f(y)$ and the straight line
$f'(0)y+ f(0)$.}\label{fig:2exp1}
\end{figure}
By Theorem \ref{thm1} we have that the left-hand side of  \eqref{eq:ss}
is positive and decreasing and this is represented in figure \ref{fig:3exp1}.

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.5\textwidth]{fig3} % f3e1}
\end{center}
\caption{Experiment 1: the term  $f(y)-yf'(y)$.}\label{fig:3exp1}
\end{figure}


\subsection*{Experiment 2}
A second experiment is proposed to describe the case of blow up of the second
derivative. Here the initial data are:
\begin{equation*}
f(0)=0.1,\quad f'(0)=0.5<\frac{1}{\sqrt{3}}.
\end{equation*}
%
The solution $f(y)$ together with its first and
second derivative is represented in
figure \ref{fig:1exp2} while we compare it with the straight line
$f'(0)y + f(0)$ in figure \ref{fig:2exp2}.
%

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.5\textwidth]{fig4} % f1e2
\end{center} \caption{Experiment 2: The functions $f(y)$, $f'(y)$ and $f''(y)$.}\label{fig:1exp2}
\end{figure}

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.5\textwidth]{fig5} % f2e2
\end{center} \caption{Experiment 2:  the solution $f(y)$ and the straight line
$f'(0)y+ f(0)$.}\label{fig:2exp2}
\end{figure}

 In figure \ref{fig:3exp2} below we represent the term  $f(y)-yf'(y)$
that is positive and increasing as suggested by Theorem \ref{thm2}.

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.5\textwidth]{fig6} % f3e2
\end{center} 
\caption{Experiment 2: The left-hand side. of the equation is positive and increasing}\label{fig:3exp2}
\end{figure}

 Finally in figure \ref{fig:4exp2}-\ref{fig:5exp2}
we describe the blow up of
the second derivative: the first derivative tends to the value
$-\frac{1}{\sqrt{3}}$ while the second derivative diverges to $-\infty$.

\begin{figure}[htb]\label{fig:4exp2}
\begin{center}
\includegraphics[width=0.5\textwidth]{fig7} % f4e2
\end{center} 
\caption{Experiment 2: The first derivative of the solutions tends to a zero of $W''(\cdot)$.}
\end{figure}

\begin{figure}[htb]\label{fig:5exp2}
\begin{center}
\includegraphics[width=0.5\textwidth]{fig8} % f5e2}
\end{center} 
\caption{Experiment 2: The second derivative diverges to $-\infty$.}
\end{figure}

\subsection{Experiment 3}

Here we want to illustrate the cases not covered by 
Theorems \ref{thm1} and \ref{thm2}; that is,
\begin{equation*}
f(0)<0, \quad f'(0)>\frac{1}{\sqrt{3}}.
\end{equation*}
We have seen in Section 3 that both global solutions and blow up are possible.
If $f'(0)$ is big enough then the solution reaches a zero
and the solution is global, while if not, the solution remains negative and
there is the blow up of the second derivative.
For a first case we have chosen
\begin{equation*}
f(0)=-0.1,\quad f'(0)=\frac{1}{\sqrt{3}}+0.1
\end{equation*}
The initial datum $f'(0)$ is very close to the critical point,
then the term $f(y)-yf'(y)$ remains negative (figure \ref{Exp3fig2})
while $f'$ reaches the critical point (see figure \ref{Exp3fig3}).
This makes the second derivative diverge (figure \ref{Exp3fig4})

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.5\textwidth]{fig9} % NTBlow1
\end{center} 
\caption{Experiment 3: the solution and its derivatives 
when $f'(0)=\frac{1}{\sqrt{3}}+0.1$.}\label{Exp3fig1}
\end{figure}

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.5\textwidth]{fig10} % NTBlow2
\end{center} 
\caption{Experiment 3: the term $f(y)-yf'(y)$ remains negative 
in the interval of existence of the solution.}\label{Exp3fig2}
\end{figure}

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.5\textwidth]{fig11} % NTBlow3
\end{center} 
\caption{Experiment 3: the first derivative reaches the critical value.}
\label{Exp3fig3}
\end{figure}

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.5\textwidth]{fig12} % NTBlow4
\end{center} 
\caption{Experiment 3: the second derivative of
the solution diverges to $-\infty$.}\label{Exp3fig4}
\end{figure}

 The other possibility discussed in section 3 is that
if $f'(0)$ is big enough then the term $f(y)-yf'(y)$ approaches a zero
at the same time that $f'$ approaches the critical value $\frac{1}{\sqrt{3}}$.
Initial data have been chosen as follows

\begin{equation*}
f(0)=-0.1,\quad f'(0)=\frac{1}{\sqrt{3}}+0.15
\end{equation*}

\begin{figure}[htb]
\begin{center}
 \includegraphics[width=0.5\textwidth]{fig13} % NTGlobal1
\end{center} 
\caption{Experiment 3: the solution and its derivatives for
$f'(0)=\frac{1}{\sqrt{3}}+0.15$.}\label{Exp3fig5}
\end{figure}

\begin{figure}[htb]
\begin{center}
 \includegraphics[width=0.5\textwidth]{fig14} % NTGlobal2
\end{center} 
\caption{Experiment 3: the term $f(y)-yf'(y)$ is zero in the interval
$[y_0,+\infty)$.}\label{Exp3fig6}
\end{figure}

\begin{figure}[htb]
\begin{center}
 \includegraphics[width=0.5\textwidth]{fig15} %NTGlobal3
\end{center} 
\caption{Experiment 3: the  first derivative is
$\frac{1}{\sqrt{3}}$ in the interval $[y_0,+\infty)$.}\label{Exp3fig7}
\end{figure}

\begin{figure}[htb]
\begin{center}
 \includegraphics[width=0.5\textwidth]{fig16} NTGlobal4
\end{center} 
\caption{Experiment 3: the second derivative is
zero in the interval $[y_0,+\infty)$.}\label{Exp3fig8}
\end{figure}

\subsection{Experiment 4}

We consider a numerical experiment to illustrate the bound on the
point $y_*$ of the blow up of the second derivative of the solution.
The initial
data are $f(0)=0.1$, $f'(0)=0$.
In this case the point of blow up satisfies the above bound:
\begin{equation*}
y_*\approx 1.29 \leq \frac{4}{3\sqrt{3}f(0)}\approx 1.54
\end{equation*}
In figure \ref{Exp4:fig1} we represent the solution
while in figure \ref{Exp4:fig2} we represent the first derivative. We observe that
it converges to $-\frac{1}{\sqrt{3}}$  as it approaches $y_*$.

\begin{figure}[htb]
\begin{center}
 \includegraphics[width=0.5\textwidth]{fig17} % ExpBound1
\end{center} 
\caption{Experiment 4: the Solution f and its first and second derivative for
$f(0)=0.5$ and $f'(0)=0$.}\label{Exp4:fig1}
\end{figure}

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.5\textwidth]{fig18} ExpBound2
\end{center} 
\caption{Experiment 4: the first derivative of the solution
approaches $-\frac{1}{\sqrt{3}}$ as y tends to $y_*$.}\label{Exp4:fig2}
\end{figure}

Finally, in figure \ref{Exp4:fig3} below, we represent the second
derivative of the solution close to $y_*$.

\begin{figure}[htb]
\begin{center}
 \includegraphics[width=0.5\textwidth]{fig19} % ExpBound3
\end{center} 
\caption{Experiment 4: the second derivative of the solution
diverges to $-\infty$ as $y$ tends to $y_*\approx1.29$.}\label{Exp4:fig3}
\end{figure}

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\end{document}