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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 10, pp. 1--18.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2008 Texas State University - San Marcos.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2008/10\hfil Blowup and life span bounds]
{Blowup and life span bounds for a reaction-diffusion
equation with a time-dependent generator}

\author[E. T. Kolkovska, J. A. L\'opez-Mimbela,  A. P\'erez\hfil EJDE-2008/10\hfilneg]
{Ekaterina T. Kolkovska, Jos\'e Alfredo L\'opez-Mimbela,
 Aroldo P\'erez}  % in alphabetical order

\address{Ekaterina T. Kolkovska \newline
 Centro de Investigaci\'on en Matem\'{a}ticas\\
 Apartado Postal 402, 36000 Guanajuato, Mexico}
\email{todorova@cimat.mx}

\address{Jos\'{e} Alfredo L\'opez-Mimbela \newline
 Centro de Investigaci\'{o}n en Matem\'{a}ticas \\
Apartado Postal 402, 36000 Guanajuato, Mexico}
\email{jalfredo@cimat.mx}

\address{Aroldo P\'erez P\'erez \newline
 Universidad Ju\'{a}rez Aut\'onoma de Tabasco\\
Divisi\'on Acad\'emica de Ciencias B\'asicas\\
 Km. 1 Carretera Cunduac\'an-Jalpa de M\'endez\\
C.P. 86690 A.P. 24, Cunduac\'an, Tabasco, Mexico}
\email{aroldo.perez@dacb.ujat.mx}

\thanks{Submitted August 24, 2007. Published January 21, 2008.}
\subjclass[2000]{60H30, 35K55, 35K57, 35B35}
\keywords{Semilinear evolution equations; Feynman-Kac representation;
\hfill\break\indent
 critical exponent; finite time blowup; nonglobal solution; life span}

\begin{abstract}
 We consider the nonlinear equation
 $$
  \frac{\partial}{\partial t} u (t) = k (t) \Delta _{\alpha }u (t)
  + u^{1+\beta } (t),\quad u(0,x)=\lambda \varphi (x),\;
  x\in \mathbb{R} ^{d},
 $$
 where $\Delta _{\alpha }:=-(-\Delta)^{\alpha /2}$
 denotes the fractional power of the Laplacian;
  $0<\alpha \leq 2$,  $\lambda$, $\beta >0$ are
 constants;  $ \varphi$ is bounded, continuous, nonnegative function
 that does not vanish identically; and $k$ is a locally integrable function.
 We prove that  any combination of positive parameters
 $d,\alpha,\rho,\beta$, obeying  $0<d\rho\beta /\alpha<1$, yields
 finite time blow up of any nontrivial positive solution.
 Also we  obtain upper and lower bounds for the life span of the solution,
and show that the life span satisfies
 $T_{\lambda\varphi}\sim \lambda^{-\alpha \beta /(\alpha -d\rho \beta )}$
 near $\lambda=0$.
\end{abstract}

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

\section{Introduction}

We study positive solutions for the semilinear non-autonomous
Cauchy problem
\begin{equation} \label{cauchyec}
\begin{gathered}
\frac{\partial u ( t,x ) }{\partial t} =k ( t ) \Delta _{\alpha }u
( t,x ) +u^{1+\beta } ( t,x ), \\
 u ( 0,x ) = \varphi  ( x )\ge0 ,\quad x\in \mathbb{R} ^{d},
\end{gathered}
\end{equation}
where $\Delta _{\alpha }:=-(-\Delta )^{\alpha /2}$ denotes the
fractional power of the Laplacian, $0<\alpha \leq 2$ and $\beta
\in(0,\infty)$ are constants, the initial value $\varphi$ is bounded,
continuous and not identically zero, and $k:[0,\infty )\to
[ 0,\infty )$ is a locally integrable function satisfying
\begin{equation}  \label{condK}
\varepsilon _{1}t^{\rho }\leq \int_{0}^{t}k(
r)dr\leq \varepsilon _{2}t^{\rho }
\end{equation}
for all $t$ large enough, where $\varepsilon _{1}$, $\varepsilon
_{2}$ and $\rho$ are given positive constants. Solutions are
understood in the mild sense, so that \eqref{cauchyec} is
meaningful for any bounded measurable initial value.


  Recall (see e.g. \cite{Pazy}, Chapter 6) that there exists a number
$T_{\varphi }\in (0,\infty ]$ such that \eqref{cauchyec} has a unique
continuous solution $u$ on $[0,T_{\varphi })\times \mathbb{R}^{d}$,
which is given by
\begin{equation}
u(t,x)=U(t,0)\varphi (x)+\int_{0}^{t}(U
(t,s)u^{1+\beta }(s,\cdot ))(x)\,ds,
\label{MILD}
\end{equation}
and is bounded on $[0,T]\times \mathbb{R}^{d}$ for any
$0<T<T_{\varphi }$. Moreover, if $T_{\varphi }<\infty $, then
$\Vert u(t,\cdot )\Vert _{\infty }\to \infty $ as
$t\uparrow T_{\varphi }$. Here $\{U(t,s),0\leq s\leq t\}$ denotes
the evolution system corresponding to the family of generators
$\{k(t)\Delta _{\alpha }$, $t\geq 0\}$. When $T_{\varphi }=\infty $
we say that $u$ is a global solution, and when
$T_{\varphi}<\infty $ we say that $u$ blows up in finite time or that $u$ is
nonglobal.
The extended real number $T_{\varphi }$ is termed life span of
\eqref{cauchyec}.

Equations of the form \eqref{cauchyec} arise in fields like
statistical physics, hydrodynamics and molecular biology
\cite{Shlesinger}. Generators of the form $k(t)\Delta_{\alpha}$,
$\alpha\in(0,2]$, allow for non-local integro-differential or
pseudodifferential   terms that have been used in models of
anomalous growth of certain fractal interfaces \cite{Mann} and in
hydrodynamic models with modified diffusivity \cite{Bardos}.


This work can be considered as a continuation of our previous
article \cite{JALM-APP}, to which we refer for more background and
additional references. In \cite{JALM-APP} we proved that
$d>\frac{\alpha}{\rho\beta}$ implies existence of non-trivial
global solutions of \eqref{cauchyec} for all sufficiently small
initial values, and that, under the additional assumption
$\beta\in\{1,2,\dots,\}$, the condition
$d<\frac{\alpha}{\rho\beta}$ yields finite time blowup of any
positive solution. Moreover, the case $ \rho=0$, which under
condition (\ref{condK}) corresponds to an integrable $ k $, yields
finite time blow up of \eqref{cauchyec} for any non-trivial
initial value, regardless of the spatial dimension and the
stability exponent $\alpha$. Here we consider  the  case
$d<\frac{\alpha}{\rho\beta}$ with $\beta\in(0,\infty)$, and focus
on the asymptotic behavior of the life span % $T_{\lambda\varphi}$
of \eqref{cauchyec} when the initial value is of the form $\lambda
\varphi $, where $\lambda >0$ is a parameter.

The life span asymptotics  of semilinear parabolic Cauchy problems
give  insight on how the ``size" of the initial value affects the
blowup time of their positive solutions; see
\cite{GuiWa, Huang,Kobayashi,LiNi, MiYa, Pin2} and the
references therein. Given two functions
$f,g:[0,\infty)\to[0,\infty)$, let us say that
 $f\sim g$ near $c\in\{0,\infty\}$ if there exist two positive
constants $C_1$, $C_2$
such that $C_1f(r)\ge g(r)\ge C_2 f(r)$ for all $r$ which are
sufficiently close to $c$.
 In \cite{LiNi} it was proved,
initially for $k(t)\equiv1$ and $\alpha=2$, that
$T_{\lambda\varphi}\sim \lambda^{-\beta}$ near $\infty$ provided
$\varphi\ge0$ is bounded, continuous and does not vanish
identically. Later on, Gui and Wang \cite{GuiWa} showed that, in
fact,
$\lim_{\lambda\to\infty}T_{\lambda\varphi}\cdot\lambda^{\beta}=
\beta^{-1}\|\varphi\|^{-\beta}_ {L^{\infty}(\mathbb{R}^d)}$. The behavior
of $T_{\lambda \varphi}$ as $\lambda$ approaches $0$ was also
investigated by Lee and Ni in \cite{LiNi}. One of their results
addresses the case of initial values $\varphi\ge0$ obeying growth
conditions of the form $0<\liminf_{|x|\to\infty}|x|^{a}\varphi(x)$
 and $ \limsup_{|x|\to\infty}|x|^{a}\varphi(x)<\infty$,
where $a>0$ is a given constant different from $d$. They proved
that, in this case,
\begin{equation}\label{growth}
T_{\lambda\varphi} \sim
 (1/\lambda)^{(1/\beta\:-\:\frac{1}{2}\min\{a,d\})^{-1}}
 \quad\text{as } \lambda\to0.
\end{equation}

In the present paper we obtain upper and lower bounds for the life
span $T_{\lambda \varphi }$ of \eqref{cauchyec}, and provide in this
way a description of the behavior of $T_{\lambda \varphi }$ as
$\lambda \to \infty $ and $\lambda \to 0$. Here is
a brief outline.

First we prove that  any combination of positive parameters
$d,\alpha,\rho,\beta$, obeying  $0<d\rho\beta /\alpha<1$, yields
finite time blow up of any nontrivial positive solution of
\eqref{cauchyec}. This is carried out by bounding from below the
mild solution of \eqref{cauchyec} by a subsolution which locally
grows to $\infty$. Finite-time blowup of \eqref{cauchyec} is then
inferred from a classical comparison procedure that dates back to
\cite{KST} (see also \cite{BLMW}, Sect. 3). The construction of
our subsolution uses the  Feynman-Kac representation of
\eqref{cauchyec}, and requires to control the decay of certain conditional
probabilities of $W\equiv\{W(t),\ t\ge0\}$, where $W$ is the
$\mathbb{R}^d$-valued Markov process  corresponding to the evolution system
$\{U(t,s)$, $t\ge s\ge0\}$; see \cite{BLMW}, \cite{B-LM-W2} and
\cite{LM-P} for the time-homogeneous case.


A further consequence of the Feynman-Kac representation of
\eqref{cauchyec} is the inequality $T_{\lambda\varphi}\le
\mathrm{Const.}\lambda^{-\frac{\alpha \beta }{\alpha -d\rho \beta
}}$, which holds for small positive $\lambda$ when
$0<d\rho\beta/\alpha < 1$. This, together with the lower bound of
$T_{\lambda\varphi}$ given in Section \ref{seccion6}, implies
(again under the condition $0<d\rho\beta/\alpha< 1$) that
\begin{equation}\label{lb}
T_{\lambda\varphi}\sim
\lambda^{-\frac{\alpha \beta }{\alpha -d\rho \beta }}
\end{equation} near $0$.
Note that (\ref{lb}) yields  (\ref{growth}) when $d<a$, $\alpha=2$
and $\rho=1$. We also provide an upper bound for
$T_{\lambda\varphi}$ which is valid for all $\lambda>0$, namely
\begin{equation}
\label{***} T_{\lambda\varphi} \leq \Big(C\lambda ^{-\beta }+[ (10
{\varepsilon _{2}}/{\varepsilon _{1}})^{1/\rho }\theta ]
^{\frac{\alpha -d\rho \beta }{\alpha }}\Big)^{\frac{\alpha }{
\alpha -d\rho \beta }}+\eta ,
\end{equation}
where $C,\theta$ and $\eta $ are suitable positive constants.

We remark that many of our arguments rely on the assumption
$d\rho\beta/\alpha<1$. Thus, the blowup behavior and life span
asymptotics of \eqref{cauchyec} in the ``critical''  case
$d\rho\beta/\alpha=1$ remain to be investigated.

As this paper is partly aimed at the multidisciplinary reader, in
the next section  we recall some basic facts regarding the
Feynman-Kac formula. In Section \ref{bridgebounds} we obtain
semigroup and bridge estimates that we shall need in the sequel.
Section \ref{seccion_explosion} is devoted to prove that
\eqref{cauchyec} does not admit nontrivial global solutions if
$d<\frac{\alpha}{\rho\beta}$. In the remaining sections
\ref{upper_estimate} and \ref{seccion6} we prove our bounds for
the life span of \eqref{cauchyec}.


\section{The Feynman-Kac representation and subsolutions}

For any $T>0$ let us consider the initial-value problem
\begin{equation} \label{1.4}
\begin{gathered}
\frac{\partial \varrho(t,x)}{\partial t}
=k(t)\Delta _{\alpha}\varrho(t,x)+\zeta
(t,x)\varrho(t,x),\quad 0<t\leq T,   \\
\varrho(0,x)=\varphi (x),\quad x\in \mathbb{R}^{d},
\end{gathered}
\end{equation}
where $k:[0,\infty )\to [ 0,\infty )$ is integrable on any bounded
interval, and $ \zeta $ and $\varphi $ are nonnegative bounded
continuous functions on $ [0,T]\times \mathbb{R}^{d}$ and
$\mathbb{R}^{d}$, respectively. It is well known that in the
classical setting $k\equiv 1$, $\alpha =2$, $\zeta (t,x)\equiv
\zeta (x)$, the solution of \eqref{1.4} can be expressed via the
Feynman-Kac formula, see e.g. \cite{Freidlin}. Theorem \ref{F-K}
below gives  the Feynman-Kac representation corresponding to \eqref{1.4}.

Let $W\equiv \{ W(t)\}_{t\geq 0}$ be the (time-inhomogeneous)
 c\`{a}dl\`{a}g
Feller process corresponding to the family of generators
$\{k (t)\Delta _{\alpha }\} _{t\geq 0}$. Note that $W$ can be
constructed by performing a deterministic time change of the
symmetric $\alpha$-stable process $Z\equiv\{Z(t)\}_{t\ge0}$.  We
designate $P_{x}$ the distribution of $\{W(t)
\}_{t\geq 0}$ such that $P_x[W(0)=x]=1$, and write $E_{x}$ for
the expectation with
respect to $P_{x}$,  $x\in\mathbb{R}^d$.

\begin{theorem}\label{F-K}
Let $k$, $\zeta$ and $\varphi$ be as
above. Then, the solution of \eqref{1.4} admits the Feynman-Kac
representation
\begin{equation}
\varrho(t,x)=E_{x}\Big[ \varphi (W(t)
)\exp \{\int_{0}^{t}\zeta(t-s,W(
s))\, ds\}\Big],\quad (t,x)\in
[0,T]\times\mathbb{R}^d.  \label{FeyKacR}
\end{equation}
\end{theorem}

\begin{proof}
The proof is a simple adaptation to our time-inhomogeneous context
of the approach used in \cite{Applebaum}.
We shall assume that $0<\alpha<2$; the case $\alpha=2$ is simpler and
can be handled in a similar fashion.

Recall \cite{Sato} that there exists a Poisson random measure
$N(dt,dx)$ on $[ 0,\infty )\times \mathbb{R} ^{d}\backslash
\{0\}$ having expectation $EN ( dt,dx)=dt\,\nu (dx)$,
with
\begin{equation*}
\nu(dx)=\frac{ \alpha
2^{\alpha-1}\Gamma((\alpha+d)/2)}{\pi^{d/2}\Gamma(1-
\alpha/2)\|x\|^{\alpha+d}}\,dx,
\end{equation*}
and such that the paths of $Z$ admit the L\'{e}vy-It\^{o} decomposition
\begin{equation}  \label{levy-ito}
Z(t)=\int_{| x| <1}x\widetilde{N}
(t,dx)+\int_{| x| \geq 1}xN (t,dx),\quad t\ge0,
\end{equation}
where $N (t,dx):=\int_{0}^{t}N (dt,dx)$, and $\widetilde{N}(t,dx)$
is the compensated Poisson random measure
\begin{equation*}
\widetilde{N} (t,B)=N (t,B)-t\nu (B), \quad t\geq 0,\;
B\in \mathcal{B}( \mathbb{R}^{d});
\end{equation*}
here $\mathcal{B}( \mathbb{R}^{d}) $ denotes the Borel
$\sigma$-algebra in $ \mathbb{R}^{d}$. The process $W$ also admits a
L\'{e}vy-It\^{o} decomposition, with corresponding Poisson random
measure $k(t)N( dt,dx)$.

Let us write $W(p^-)$ for the limit of $W$ from the left of $p$.
From the integration by parts formula we obtain
\begin{align*}
&d\big[ \varrho(t-s,W(s))\exp \big\{
\int_{0}^{s}\zeta (t-r,W(r))dr\big\} \big]  \\
&=\varrho(t-s,W(s^{-}))\zeta (t-s,W(s^{-}))\exp \big\{\int_{0}^{s}\zeta (
t-r,W(r^{-}))dr\big\}\,ds \\
&\quad +\exp \big\{\int_{0}^{s}\zeta (t-r,W(r^{-}))
\,dr\big\}\,d\varrho(t-s,W(s)).
\end{align*}
Using It\^{o}'s formula  \cite[Section 4.4]{Applebaum},  to
calculate $ d\varrho(t-s,W(s))$, we obtain
\begin{align*}
&d\big[ \varrho(t-s,W(s))\exp \big\{
\int_{0}^{s}\zeta (t-r,W(r))dr\big\}\big]  \\
&=\exp \big\{\int_{0}^{s}\zeta (t-r,W(r^{-}))\,dr\big\}\\
&\quad\times \Big\{\varrho(t-s,W(s^{-}))\zeta (t-s,W(s^{-}))\,ds
  -\frac{d}{ds}\varrho(t-s,W(s^{-})) \\
&\quad +k(s)\int_{| x| <1}\big[ \varrho(
t-s,W(s^{-})+x)-\varrho(t-s,W(s^{-}))\big] \widetilde{N}(ds,dx)\\
&\quad +k(s)\int_{| x| \geq 1}\big[ \varrho(
t-s,W(s^{-})+x)-\varrho(t-s,W(s^{-}))\big] N(ds,dx)\\
&\quad + k(s)\int_{| x| <1}\Big[ \varrho(
t-s,W(s^{-})+x)-\varrho(t-s,W(s^{-})) \\
&\quad -\sum_{i}x_{i}\frac{d}{dx_{i}}\varrho(t-s,W(s^{-})
)\Big] \nu (dx)ds\Big\}.
\end{align*}
Integrating from $0$ to $t$, and taking expectation with respect
to $P_{x}$, yields

\begin{align*}
\lefteqn{E_{x}\Big[ \varphi (W(t))\exp \big\{
\int_{0}^{t}\zeta (t-s,W(s))\,ds\big\}\Big]-\varrho(t,x)} \\
&=E_{x}\int_{0}^{t}\exp \big\{\int_{0}^{s}\zeta (
t-r,W(r^{-}))\,dr\big\}\\
&\quad\times\Big\{\varrho(t-s,W(s^{-}))\zeta (t-s,W(s^{-}))
-\frac{d}{ds}\varrho(t-s,W(s^{-})) \\
&\quad +k(s)\int_{| x| <1}\Big[ \varrho(
t-s,W(s^{-})+x)-\varrho(t-s,W(s^{-}))\phantom{\sum_i} \\
&\quad -\sum_{i}x_{i}\frac{d}{dx_{i}}\varrho(t-s,W(s^{-}))
\Big] \nu (dx)\\
&\quad + k(s)\int_{| x| \geq 1}\big[
\varrho(t-s,W(s^{-})+x)-\varrho(t-s,W(s^{-})
)\big] \,\nu (dx)\Big\}\,ds \\
&=0,
\end{align*}
where in the first equality we used the identity $\widetilde{N}(
ds,dx)=N(ds,dx)-ds\,\nu (dx)$, and the fact that the stochastic
integrals with respect to $\widetilde{N}(ds,dx)$ are martingales,
and therefore have expectation $0$.
\end{proof}


The Feynman-Kac representation is suitable for
constructing subsolutions of reaction-diffusion equations of the type
\begin{equation}
\frac{\partial w(t,y)}{\partial t}=k(t)\Delta
_{\alpha }w(t,y)+ w^{1+\beta }(t,y)
,\qquad w(0,y)=\varphi (y),\quad y\in \mathbb{R}^{d},  \label{ee}
\end{equation}
where $\beta >0$ is a constant, and $k$, $\varphi $ are as in
(\ref {1.4}). From Theorem \ref{F-K} we know that
\begin{equation*}
w(t,y)=E_{y}\Big[ \varphi (W(t))\exp \Big(\int_{0}^{t}w^{\beta }(
t-s,W(s))\,ds\Big)\Big] ,\quad (t,y)\in [ 0,T]\times \mathbb{R}^{d},
\end{equation*}
for every positive $T< T_{\varphi}$. Hence, for every $y\in
\mathbb{R}^{d}$,
\begin{equation*}
w(t,y)\,\geq \,E_{y}[ \varphi (W(t))]
=:v_{0}(t,y),\quad t\geq 0,
\end{equation*}
so that $v_{0}$ is a subsolution of \eqref{ee}; i.e.,
$w( 0,\cdot)=v_{0}(0,\cdot )$ and $w(t,\cdot ) \geq v_{0}(t,\cdot )$ for
every $t>0$. The next lemma, which we will need in the following
section, is a direct consequence of the Feynman-Kac
representation.


\begin{lemma}\label{sub}
 Let $k$, $\varphi $ be as in \eqref{1.4}, and let
$\zeta (\cdot ,\cdot )$ be a nonnegative, bounded and continuous
subsolution of \eqref{ee}. Then, any solution of
\begin{equation*}
\frac{\partial \varrho(t,y)}{\partial t}=k(t)\Delta _{\alpha
}\varrho(t,y)+\zeta ^{\beta }(t,y)\varrho(t,y),\qquad \varrho(0,\cdot )
=\varphi ,
\end{equation*}
remains a subsolution of \eqref{ee}.
\end{lemma}

\section{Bridge and semigroup bounds\label{bridgebounds}}

Let us denote by $p(t,x)$, $t\geq 0$, $x\in\mathbb{R}^{d}$, the
transition densities of the $d$-dimensional symmetric
$ \alpha $-stable process $\{Z(t)\}_{t\geq 0}$. Recall that
$p(t,\cdot)$, $t>0$, are strictly positive, radially symmetric
continuous functions that satisfy the following properties.

\begin{lemma}\label{Lemma1}
For any $s,t>0$, and $x,y\in \mathbb{R}^{d}$, $p(t,x)$ satisfies
\begin{itemize}
\item[(i)]  $p(ts,x)=t^{-\frac{d}{\alpha}}p(s,t^{-\frac{1}{ \alpha }}x)$,
\item[(ii)]  $p(t,x)\leq p(t,y)$  when $| x| \geq |y| $,
\item[(iii)] $p(t,x)\geq (\frac{s}{t})^{d/\alpha}p(s,x)$
 for $t\geq s$,
\item[(iv)] $p( t,\frac{1}{\tau }(x-y))\geq p(t,x) p(t,y)$ if
$p(t,0)\leq 1$ and $\tau \geq 2$.
\end{itemize}
\end{lemma}


For a proof of the above lemma, see \cite[page 493]{GuKi}
or \cite[pages 46 and 47]{Sug}.

Let $\varphi:\mathbb{R}^d\to[0,\infty)$ be bounded
and measurable, and let $k:[0,\infty)\to[0,\infty)$ be locally
integrable. Notice that the transition probabilities of the Markov
process $\{W(t),\, t\ge0\}$ are given by
\begin{equation}\label{MICHA}
P(W(t)\in dy|W(s)=x)=
p(\textstyle\int_s^tk(r)\,dr,y-x)\,dy,\quad 0\le s\le
t, \ x\in\mathbb{R}^d.
\end{equation}
We define the function
\begin{equation}  \label{v0}
v_{0}(t,x)=E_{x}\left[ \varphi (W(t)) \right] = E_{x}\left[
\varphi (Z(K( t,0)) )\right]= \int
p(K(t,0),y-x)\varphi(y)\,dy,
\end{equation}
where $t\ge0$, $x\in\mathbb{R}^d$,
 $K(t,s):=\int_{s}^{t}k(r)dr$, $0\leq s\leq t$,
 and write $B(r)\equiv B_r\subset\mathbb{R}^d$ for the ball of
radius $r$, centered at the origin.

\begin{lemma} \label{lem4}
There exists a constant $c_0>0$ satisfying
\begin{equation}
v_{0}(t,x)\geq c_{0}K^{-d/\alpha}(
t,0)1_{B_{1}}\big(K^{-1/\alpha}(t,0)
x\big)\label{firstest}
\end{equation}
for all $x\in\mathbb{R}^d$, and all $t>0$ such that $K(t,0)\geq1$.
\end{lemma}


\begin{proof}
From Lemma \ref{Lemma1} (i), (ii) and the radial symmetry
of $p(t,\cdot )$ we have, for $K^{1/\alpha}(t,0)\geq 1$, $x\in B_{K^{1/\alpha}(
t,0)}$ and $z\in \partial B_{2}$, that
\begin{align*}
v_{0}(t,x)
&= E_{0}[ \varphi (Z(K(t,0))+x )] \\
&= E_{0}\big[ \varphi (K^{1/\alpha}(t,0)
 ( Z(1)+ K^{-1/\alpha}(t,0)x))\big] \\
&\geq \int_{B_{1}}\varphi (K^{1/\alpha} (
t,0)y)P\left[ Z(1)\in dy-K^{-1/\alpha }(t,0)x\right] \\
&= \int_{B_{1}}\varphi (K^{1/\alpha}(
t,0)y)p(1, y-K^{-1/\alpha}(t,0)x)dy \\
&\geq p(1,z)\int_{B_{1}}\varphi (K^{1/\alpha}(t,0)y)\, dy \\
&= p(1,z)K^{-d/\alpha}(t,0)
\int_{B_{K^{1/\alpha}(t,0)}}\varphi (y)dy \\
&\geq  p(1,z)K^{-d/\alpha}(t,0)
1_{B_{1}}(K^{-1/\alpha}(t,0)x)\int_{B_{1}}\varphi (y)dy.
\end{align*}
Letting $c_{0}=p(1,z)\int_{B_{1}}\varphi (y)dy $ yields
(\ref{firstest}).
\end{proof}


Fix $\theta >0$ such that (\ref{condK}) holds for all $t \geq
\theta$ and such that $K(\theta ,0)\geq 1$.
Define $ \delta _{0}=\min \{(\frac{ \varepsilon
_{1}}{2\varepsilon _{2}})^{1/\rho },1-(\frac{\varepsilon
_{1}}{2\varepsilon _{2}}) ^{1/\rho }\}$.


\begin{lemma}\label{Lemma2}
There exists $c>0$ such that for all $x$, $y\in B_{1}$ and
$t\ge \theta/\delta_0$,
\begin{equation*}
P_{x}\big[ W(s)\in B_{K^{1/\alpha}(t-s,0)}: W(t)=y \big] \geq c
\end{equation*}
for $s\in [ \theta ,\delta _{0}t] $.
\end{lemma}


\begin{proof}
Using (\ref{condK}) and Lemma \ref{Lemma1} (i), we obtain
\begin{equation}\label{ec1est}
\begin{aligned}
&P_{x}\big[ W(s)\in B_{K^{1/\alpha}(t-s,0)}:W(t)=y \big]    \\
&=\int_{B_{K^{1/\alpha}(t-s,0)}}\frac{p(K(s,0),x-z)p(K(t,s)
,z-y)}{p(K(t,0),x-y)}\,dz   \\
&\geq \int_{B_{rs^{\rho/\alpha}}}\frac{K^{-d/\alpha }(s,0)
}
{K^{- d/\alpha}(t,0)
}\\
&\quad\times \frac{p(1,K^{-1/\alpha}(s,0)(x-z))K^{-d/\alpha}( t,s)p( 1,K^{-1/\alpha}(t,s)(z-y))}
{p(1,K^{-1/\alpha}(t,0)(x-y))}
\,dz,
\end{aligned}
\end{equation}
with $r=\varepsilon _{1}^{1/\alpha}(\frac{1}{\delta _{0}}-1)^{\rho/\alpha}$.
It is straightforward to verify that $K^{-1/\alpha}(s,0)(x-z)\in B_{r_{1}}$,
where $r_{1}=2(\frac{ \varepsilon_{2}}{\varepsilon _{1}})^{1/\alpha}
\delta ^{ \frac{\rho }{\alpha }}$ with
$\delta =\max \{1,\frac{1}{\delta _{0}}-1\}$, hence
Lemma \ref{Lemma1} (ii) and radial symmetry of
$p(t,\cdot )$ imply
\begin{equation*}
p(1,K^{-1/\alpha}(s,0)(x-z))\geq p(1,\varsigma )\equiv c_{1}
\end{equation*}
for any $\varsigma \in \partial B_{r_{1}}$.

Thus, the term in the right-hand side of
(\ref{ec1est}) is bounded from below by
\begin{equation*}
\int_{B_{rs^{\rho/\alpha}}}\frac{c_{1}K^{-d/\alpha}(s,0)K^{-d/\alpha}(t,s)p(1,K^{- \frac{1}{\alpha }}(t,s)(z-y))}
{K^{-d/\alpha}(t,0)p(1,K^{-1/\alpha}( t,0)(x-y))}\,dz.
\end{equation*}
Using (\ref{condK}), and the facts that $K(t,0)\geq K(t,s)$ and
$p(t,x)\leq p( t,0)$ for all $t>0$
and $x\in \mathbb{R}^{d}$, it follows that
\begin{equation}
P_{x}\big[ W(s)\in B_{K^{1/\alpha}( t-s,0)}:W(t) =y \big] \geq
\int_{B_{rs^{\rho/\alpha}}}c_{2}s^{-d\rho/\alpha}p(1,K^{-1/\alpha}
(t,s)( z-y))dz,  \label{ec2est}
\end{equation}
where $c_{2}=\frac{c_{1}\varepsilon _{2}^{-d/\alpha}}{p(1,0)}$.
Since $\theta \leq s\leq \delta _{0}t$, we have from
(\ref{condK}) and the definition of $\delta _{0}$ that
\begin{equation*}
K^{-1/\alpha}(t,s)=[ K(t,0) -K(s,0)]
^{-1/\alpha}\leq ( \varepsilon _{1}t^{\rho }-\varepsilon
_{2}\delta _{0}^{\rho }t^{\rho })^{-1/\alpha} \leq
c_{3}t^{-\rho/\alpha},
\end{equation*}
where $c_{3}=(\frac{2}{\varepsilon _{1}}) ^{1/\alpha}$.
Since $y\in B_{1}$, $z\in B_{rs^{\rho/\alpha}}$ and
$\theta \leq s\leq \delta _{0}t$, we deduce that, for $t\geq 1$,
$y\in B_{t^{\rho/\alpha}}$ and $z\in B_{r\delta
_{0}^{\rho/\alpha}t^{\rho/\alpha}}$. Letting
$\gamma =\max \{1,\;r\delta _{0}^{\rho/\alpha}\}$, it follows that
$z-y\in B_{2\gamma t^{\rho/\alpha}}$, and thus $K^{-1/\alpha}(t,s)(z-y)\in
B_{2\gamma c_{3}}$. Therefore,
\begin{equation*}
p(1,K^{-1/\alpha}(t,s)(
z-y))\geq p(1,\varsigma )\equiv c_{4}
\end{equation*}
for any $\varsigma \in \partial B_{2\gamma c_{3}}$. From
(\ref{ec2est}) we conclude that
\begin{equation*}
P_{x}\big[ W(s)\in B_{K^{1/\alpha}( t-s,0)\text{
}}:W(t) =y \big] \geq
\int_{B_{rs^{\rho/\alpha}}}c_{5}s^{-d\rho /\alpha}dz\equiv c.
\end{equation*}
\end{proof}


\section{Nonexistence of positive global solutions \label{seccion_explosion}}


In this section we shall use the Feynman-Kac representation to
construct a subsolution of \eqref{cauchyec} which grows to
infinity uniformly on the unit ball. As we are going to prove
afterward, this guarantees nonexistence of nontrivial positive
solutions of \eqref{cauchyec}.


Let $v$ solve the semilinear nonautonomous equation
\begin{equation} \label{ec1iter}
\begin{gathered}
\frac{\partial v(t,x)}{\partial t}
=k(t)\Delta _{\alpha }v(t,x)+v_{0}^{\beta }(t,x)v(t,x),   \\
v(0,x)=\varphi (x),\quad x\in \mathbb{R}^{d},
\end{gathered}
\end{equation}
where $k$ and $\varphi$ are as in \eqref{cauchyec}, and $v_0$ is
defined in (\ref{v0}). Since $v_{0}\leq u$, where $u$ is the
solution of (\ref{cauchyec}), it follows from Lemma \ref{sub}
that $v\leq u$ as well. Without loss of generality we shall assume
that $\varphi$ does not a.e. vanish on the unit ball.


\begin{proposition}\label{Prop3}
There exist $c^{\prime }$, $c''>0$
such that, for all $x\in B_{1}$ and all $t>0$ large enough,
\begin{equation*}
v(t,x)\geq c^{\prime }t^{-d\rho/\alpha}\exp
(c''t^{1-\frac{d\rho \beta }{\alpha }}).
\end{equation*}
\end{proposition}

\begin{proof}
In the sequel, $c_{0},c_{1},\dots,c_8 $\ denote suitable
positive constants, $c_0,\dots,c_5$ being defined in Lemma \ref{Lemma2}.
From Theorem \ref{F-K} we know that
\begin{equation*}
v(t,x)=\int_{\mathbb{R}^{d}}\varphi ( y)p(K(t,0),x-y)
E_{x}\Big[\exp \int_{0}^{t}v_{0}^{\beta }(t-s,W(s))ds \:\Big|\:W(t)=y \Big] dy.
\end{equation*}
Let $\theta $ and $\delta _{0}$ be as in Lemma \ref{Lemma2}. For
any $\theta \leq s\leq \delta _{0}t$, we have $t-s\geq t-\delta
_{0}t=(1-\delta _{0})t\geq \delta _{0}t\geq \theta $, and
therefore $K^{\frac{1}{ \alpha }}(t-s,0)\geq 1$. From here, using
(\ref{firstest}) and Jensen's inequality, we get
\begin{align*}
v(t,x) &\geq \int_{\mathbb{R}^{d}}\varphi
(y)p(K(t,0),x-y)\\
&\quad\times  E_{x}\Big[ \exp \int_{\theta }^{\delta
_{0}t}c_{0}^{\beta }K^{-d\beta/\alpha}(t-s,0)1_{B_{K^{
\frac{1}{\alpha }}(t-s,0)}}(W(s))\,ds\:\Big|\:
W(t)=y \Big] dy \\
&\geq \int_{B_{1}}\varphi (y)p(K(t,0),x-y)\\
&\quad\times \exp \Big\{\int_{\theta }^{\delta _{0}t}c_{0}^{\beta
}K^{-\frac{ d\beta }{\alpha }}(t-s,0)P_{x}\Big[ W(s)\in B_{K^{
\frac{1}{\alpha }}(t-s,0)\text{ }}\:\Big|\:W(t)=y\Big] \,ds\Big\}\,dy.
\end{align*}
It follows from Lemma \ref{Lemma1} and Lemma \ref{Lemma2}  that
\begin{align*}
v(t,x)&\geq \int_{B_{1}}\varphi (y)p(K(t,0),x-y)\exp
\int_{\theta }^{\delta
_{0}t}c_{6}K^{-d\beta/\alpha}(t-s,0)\,ds\,dy \\
&=\int_{B_{1}}\varphi (y) K^{-d/\alpha}(t,0)p(
1,K^{-1/\alpha}(t,0)(x-y))dy\\
&\quad\times \exp \int_{\theta }^{\delta_{0}t}c_{6}
K^{- d\beta/\alpha}(t-s,0)\,ds.
\end{align*}
Let $x$, $y\in B_{1}$. Then $K^{-1/\alpha}( t,0)(x-y)\in
B_{2}$. Radial symmetry of $p(t,\cdot )$ implies
\begin{equation*}
p\big(1,K^{-1/\alpha}(t,0)(x-y)\big)\geq p(1,\varsigma )\equiv c_{7}
\end{equation*}
for any $\varsigma \in \partial B_{2}$. Therefore,
\begin{equation}
v(t,x)\geq \int_{B_{1}}c_{7}\varphi (
y)K^{-d/\alpha}(t,0)dy\exp
\int_{\theta }^{\delta _{0}t}c_{6}K^{-d\beta/\alpha}(t-s,0)ds. \label{ec3est}
\end{equation}
Let $c_{8}=c_{7}\int_{B_{1}}\varphi (y)dy$. Using (\ref {condK})
and the fact that $K(t,0)\geq K( t-s,0)$,
the term in the right of (\ref{ec3est}) is bounded below by
\begin{align*}
&c_{8}K^{-d/\alpha}(t,0)\exp \Big( c_{6}\int_{\theta }^{\delta
_{0}t}K^{-\frac{d\beta }{\alpha } }(t,0)ds\Big)\\
&\geq c_{8}\varepsilon_{2}^{-d/\alpha} t^{-d\rho/\alpha}
\exp \big[c_{6}\varepsilon _{2}^{-d\beta/\alpha}(\delta
_{0}t^{1-\frac{d\beta \rho }{\alpha }}-\theta t^{- \frac{d\beta
\rho }{\alpha }})\big]
\end{align*}
if $t>0$ is large. It follows that
\begin{equation}
\label{cprima} v(t,x)\geq c^{\prime }t^{- \frac{d\rho }{\alpha
}}\exp (c''t^{1-\frac{d\beta \rho }{\alpha }})
\end{equation}
 for sufficiently large $t$, where
 $c'=c_{8}\varepsilon _{2}^{-d/\alpha}\exp \big(
-c_{6}\theta \varepsilon _{2}^{-d\beta/\alpha}\big)$
and $c''=c_{6}\delta _{0}\varepsilon_{2}^{-d\beta/\alpha}$.
\end{proof}

As a consequence of Proposition \ref{Prop3}, if
$0<\frac{d\beta \rho }{\alpha }<1$, then $\inf_{x\in
B_{1}}v(t,x)\to \infty $ when $t\to \infty $. As $v$ is subsolution of
Equation \eqref{cauchyec}, this implies that
\begin{equation}  \label{Mathias}
C(t):=\inf_{x\in B_{1}}u(t,x)
\to \infty \quad \text{as }t\to \infty .
\end{equation}

We need the following lemma.

\begin{lemma}\label{lemma_nuevo}
Let
$$
\xi_1:=\min_{x\in B_1}\min_{0\le r\le1}P_x[Z(r)\in B_1],
$$
where $\{Z(t),\, t\ge0\}$ denotes the symmetric $\alpha$-stable process.
Then
\begin{itemize}
\item[(i)] $\xi_1>0$

\item[(ii)] For any $0\le s\le t$,
$$
\xi:=\int_{B_1}p(K(t_0+t,t_0),y-x)\,dy\ge\int_{B_1}p(K(t+t_0,s+t_0),y-x)\,dv
\ge \xi_1^{\lfloor K(t+t_0,s+t_0)\rfloor},
$$
where $\lfloor x\rfloor$ denotes the least integer no smaller
than $x\in [0,\infty)$.
\end{itemize}
\end{lemma}

\begin{proof}
Let $\epsilon>0$, and let $f_{\epsilon}:\mathbb{R}^d\to\mathbb{R}_+$
be a continuous function bounded by 1 such that $f_{\epsilon}=1$ on $B_1$,
and $f_{\epsilon}=0$ on the complement of the ball $B_{1+\epsilon}$.
Because of strong continuity of the semigroup $\{S(t)\}_{t\ge0}$
corresponding to  $\{Z(t)$, $t\ge0\}$,
$\lim_{t\to0}S(t)f_{\epsilon}=f_{\epsilon}$ uniformly on $\mathbb{R}^d$.
Moreover, uniformly on $B_1$,
$$
1_{B_1} = \lim_{\epsilon\downarrow0}f_{\epsilon} =
\lim_{\epsilon\downarrow0}\lim_{t\to0} S(t)f_{\epsilon} =
\lim_{t\to0}\lim_{\epsilon\downarrow0} S(t)f_{\epsilon}
= \lim_{t\to0}S(t)1_{B_1},
$$
where the last equality follows from the bounded convergence
theorem. Hence, there exists $\epsilon\equiv\epsilon_{1/2}>0$ such
that $\sup_{x\in B_1}|S(t)1_{B_1}(x)-1_{B_1}(x)| < \frac{1}{2}$ for all
$t<\epsilon $, which implies
$$
P_x[Z(r)\in B_1] = S(r)1_{B_1}(x)\ge \frac{1}{2}
$$
for all $r<\epsilon$ and $x\in B_1$. Therefore, in order to prove (i)
it suffices to show that
\begin{equation}\label{xi}
\inf_{\epsilon \le r\le 1}P_x[Z(r)\in B_1]>\mathrm{A}
\end{equation}
for all $x\in B_1$, where the constant $\mathrm{A}>0$ does not depend on $x$.
Using Lemma \ref{Lemma1} (i) we get
\begin{align*}
P_x[Z(r)\in B_1]&=\int_{B_1}p(r,z-x)\,dz\\
&= r^{-d/\alpha}\int_{B_1}p(1,r^{-1/\alpha}(z-x))\,dz\\
&\ge \inf_{y\in B_{2\epsilon^{-1/\alpha}}}p(1,y)\int_{B_1}dz,
\end{align*}
which yields (\ref{xi}).

The assertion in part (ii) is deduced from the Chapman-Kolmogorov
equation as follows. If $l:=K(t_0+t, t_0+s)\le1$, the statement follows
directly from part (i). If $l>1$, then
\begin{align*}
P_x[Z(l)\in B_1]
& = \int_{B_1}\int_{\mathbb{R}^d}p(l-1,y-x)p(1,z-y)\,dy\,dz \\
&\ge \int_{B_1}\int_{B_1}p(l-1,y-x)p(1,z-y)\,dy\,dz\\
&= S(l-1)((S(1)1_{B_1})1_{B_1})\\
&\ge \xi_1 P_x[Z(l-1)\in B_1].
\end{align*}
Thus, applying the above procedure  $\lfloor l-1\rfloor$ times we
obtain the assertion.
\end{proof}

Now we are ready to prove that (\ref{Mathias}) is sufficient to
guarantee finite-time blow up of \eqref{cauchyec}.


\begin{theorem}\label{Theo4}
If  $0<\frac{d\rho \beta }{\alpha}<1$, then all nontrivial positive
solutions of \eqref{cauchyec}
are nonglobal.
\end{theorem}

\begin{proof}
Let $u$ be the solution of \eqref{cauchyec}, and let
$t_{0}>0$ be such that $\Vert u(t_{0},\cdot )\Vert _{\infty}<\infty $.
Then
\begin{align*}
u(t+t_{0},x)&=\int_{\mathbb{R}^{d}}p(K(t+t_{0},t_{0}),y-x)u(t_{0},y)dy \\
&\quad+\int_{0}^{t}\int_{\mathbb{R}^{d}}p(
K(t+t_{0},s+t_{0}),y-x)u^{1+\beta }(s+t_{0},y)\,dy\,ds \\
&\geq \int_{B_{1}}p(K(t+t_{0},t_{0}),y-x)u(t_{0},y)dy \\
&\quad +\int_{0}^{t}\int_{B_{1}}p(K(t+t_{0},s+t_{0}),y-x)u^{1+\beta }(
s+t_{0},y)\,dy\,ds.
\end{align*}
Therefore, $w(t,\cdot ):=u(t_{0}+t,\cdot ) $ satisfies
\begin{align*}
w(t,x)&\geq C(t_{0})
\int_{B_{1}}p(K(t+t_{0},t_{0}),y-x)\, dy\\
&\quad +\int_{0}^{t}\int_{B_{1}}p(K(
t+t_{0},s+t_{0}),y-x)\big(\min_{z\in
B_{1}}w(s,z)\big)^{1+\beta }\,dy\,ds.
\end{align*}
It follows from Lemma \ref{lemma_nuevo} that for all
$t\in [ 0,1] $,
\begin{equation*}
\min_{x\in B_{1}}w(t,x)\geq \xi C(t_{0})
+\xi \int_{0}^{t}\big(\min_{z\in B_{1}}w(s,z)\big)^{1+\beta }ds.
\end{equation*}
We put $\mathrm{w}(t)\equiv \min_{z\in B_{1}}w(t,z)$,
$t\geq 0$, and consider the integral equation
\begin{equation*}
v(t)=\xi C(t_{0})+\xi \int_{0}^{t}v^{1+\beta }(s)\,ds,
\end{equation*}
whose solution satisfies
\begin{equation}\label{chi(t)}
v^{\beta }(t)=\frac{\left[ \xi C(t_{0})
\right] ^{\beta }}{1-\beta \xi ^{1+\beta }C^{\beta }(t_{0})t}.
\end{equation}
Choosing $t_{0}$ large enough  that the blow up time of $v$ is smaller
than one, renders
\begin{equation*}
\mathrm{w}(1)=\min_{x\in B_{1}}w(1,x)\geq v(1)=\infty ,
\end{equation*}
which proves blow up of $u$ at time $t_0+1$.
\end{proof}


\section{Upper estimates for the life span \label{upper_estimate}}


  In this section we obtain two upper bounds for the life span of
the solution to \eqref{cauchyec} with initial value
$u(0,\cdot )=\lambda \varphi (\cdot )$, where $\lambda $ is a
positive parameter. We first consider the case of small and positive
$\lambda$.


\begin{proposition} \label{prop9}
If $0<d\rho \beta /\alpha \leq n/(n+1)$ with $ n\in \mathbb{N}$,
then there exists a constant $C_n>0$ such that for all
sufficiently small $\lambda>0$,
$$
T_{\lambda\varphi} \leq C_n\lambda^{-\frac{\alpha\beta}{\alpha-d\beta\rho}}.
$$
\end{proposition}


\begin{proof}
From \eqref{cprima} and (\ref{Mathias})
it follows that
$$
C(t) \geq \lambda c^{\prime }t^{-d\rho/\alpha}
\exp \big(c''t^{1-\frac{d\beta \rho }{\alpha }}\big)
$$
for all $t\geq \frac{\theta }{ \delta _{0}}$;
here we require $t\ge \theta/\delta_0$  in order to have an
interval $[\theta,\delta_0t]$ so that $K^{-1/\alpha}(t,s)\le {\rm
Const.} t^{-\rho/\alpha}$ for $s\in [\theta,\delta_0t]$, and then
to use (\ref{ec2est}).


 Recall from \eqref{chi(t)} that $v(1)=\infty$
provided $\beta \xi ^{1+\beta }C^{\beta }(t_{0})=1$.
 Note that $ t_{0}\leq t_{1}$, where $t_{1}$ is such that $t_{1}\geq \frac{\theta }{
\delta _{0}}$ and
\[
\beta\xi^{1+\beta}\lambda^{\beta}(c')^{\beta}
t_{1}^{-d\beta\rho/\alpha}\exp \big(\beta
c''t_{1}^{1-\frac{d\beta \rho }{\alpha }}\big)=1.
\]
Choosing $\theta\ge\delta_0$,  from the inequality $e^{x}\geq
\frac{x^{n+1}}{(n+1)!}$ and the fact that the condition
$0<\frac{d\rho \beta }{\alpha }\leq \frac{n}{n+1}$ implies $
\frac{d\rho\beta}{n\alpha}\leq 1-\frac{d\rho\beta}{\alpha}$, we
have that $t_{1}\leq t_{2}$, where $t_{2}$ is such that
\[
 t_2\ge\frac{\theta}{\delta_0} \mbox{ \ \ and \ \ }
\frac{1}{(n+1)!}\beta^{n+2} \xi ^{1+\beta
}(c')^{\beta}(c'')^{n+1}\lambda ^{\beta
}t_{2}^{1-\frac{d\beta \rho }{\alpha }}=1,
\]
which is the same as
\[
t_{2}=\big[\frac{(n+1)!}{\beta^{n+2}\xi^{1+\beta}(c')^{\beta}
(c'')^{n+1}}\big]^{\frac{\alpha}{\alpha-d\beta\rho}}
\lambda^{-\frac{\alpha\beta}{\alpha-d\beta\rho}}.
\]
Choosing
\[
C_n=\big[ \frac{(n+1)!}{\beta^{n+2}
\xi^{1+\beta}(c')^{\beta}(c'')^{n+1}}\big]^{\frac{\alpha}{\alpha-d\beta\rho}}
\]
renders $ t_{0}\leq t_{1}\leq t_{2}=C_n\lambda ^{-\frac{\alpha
\beta }{\alpha -d\beta \rho }}. $ Hence $ T_{\lambda\varphi} \leq
C_n\lambda ^{-\frac{\alpha \beta }{\alpha -d\beta \rho }} $ for
all $\lambda>0$ such that $C_n\lambda ^{-\frac{\alpha \beta
}{\alpha -d\beta \rho }} \ge \frac{\theta}{\delta_0}$.
\end{proof}

Let us define
\begin{equation*}
\upsilon (t)=\int_{\mathbb{R}^{d}}p(
K(t,0),x)u(t,x)dx,
\end{equation*}
where $u$ is the solution of \eqref{cauchyec}, and let $\theta >0$
be such that (\ref{condK}) holds for all $t\geq \theta $.


\begin{lemma}\label{Lemma5}
If there exist $\tau_{0}\geq \theta $ such that $\upsilon (t)=\infty
$ for $t\geq \tau_{0}$, then the solution to \eqref{cauchyec}
blows up in finite time.
\end{lemma}

\begin{proof}
Since $p(K(t,0),0)=K(t,0)^{-d/\alpha}p(1,0)$ and
$K(t,0)\ge \varepsilon_1 t^{\rho}$ for all $t\ge\theta$, we can assume,
 by taking $\tau_0$ bigger if necessary,
that $p(K(t,0),0)\leq 1$  for all $t\geq \tau_0$.


If $\tau_{0}\leq \varepsilon _{1}^{1/\rho }t$  and
$\varepsilon _{1}^{1/\rho }t\leq r\leq (2\varepsilon
_{1})^{1/\rho }t$, we have, from the conditions on
$k(t)$, that
\begin{align*}
\tau &\equiv \big[ \frac{K((10\varepsilon _{2})^{1/\rho}t,r)}{K(r,0)}\big]
 ^{1/\alpha}
=\big[\frac{K((10\varepsilon _{2})^{1/\rho }t,0)-K(r,0)}{K(r,0)}
 \big] ^{1/\alpha} \\
&\geq
\big[ \frac{K((10\varepsilon _{2})^{1/\rho }t,0)}{K((2\varepsilon _{1})
^{1/\rho }t,0)}-1\big] ^{1/\alpha} \\
&\geq \big[ \frac{\varepsilon _{1}(10\varepsilon _{2})
t^{\rho}}{\varepsilon _{2}(2\varepsilon _{1})t^{\rho }}-1\big]
^{1/\alpha}=4^{1/\alpha}\geq 2.
\end{align*}
Using properties (i) and (iv) in Lemma \ref{Lemma1}, with
$\tau=\big[ \frac{ K((10\varepsilon _{2})^{1/\rho }t,r)}{
K(r,0)}\big] ^{1/\alpha}$, yields
\begin{align*}
&p\Big(K\big((10\varepsilon _{2})^{1/\rho }t,r\big),x-y\Big) \\
&= p\Big(K(r,0)\big[ \frac{K((
10\varepsilon _{2})^{1/\rho }t,r)}{K(r,0)}\big],x-y\Big)\\
&=\big[ \frac{K(r,0)}{K((10\varepsilon _{2})^{1/\rho }t,r)}\big]
^{d/\alpha}p(K(r,0),\big[ \frac{K(r,0)
}{K((10\varepsilon _{2})^{1/\rho }t,r)}\big] ^{1/\alpha}(x-y))\\
&\geq \big[ \frac{K(r,0)}{K(( 10\varepsilon _{2})^{1/\rho }t,r)}\big]
^{d/\alpha}p(K(r,0),x)p(K(r,0) ,y).
\end{align*}
Since $\upsilon (t)=\infty $ for all $t\geq \tau_{0}$, it follows
that
\begin{align*}
&\int_{\mathbb{R}^{d}}p\big(K((10\varepsilon _{2})^{1/\rho }t,r),x-y\big)
u( r,y)dy\\
&\geq \big[ \frac{K(r,0)}{K(( 10\varepsilon
_{2})^{1/\rho }t,r)}\big] ^{d/\alpha}p(K(r,0),x)\upsilon (r) =\infty .
\end{align*}
The solution $u(t,x)$ of \eqref{cauchyec} satisfies
\begin{align*}
u(t,x)
&=\lambda \int_{\mathbb{R}^{d}}p(K(t,0),x-y)\varphi (y)dy
+\int_{0}^{t}\Big(\int_{\mathbb{R}^{d}}p(K(t,r),x-y)u^{1+\beta }(r,y)dy\Big)dr
\\
&\geq \int_{0}^{t}\Big(\int_{\mathbb{R}^{d}}p(K(t,r),x-y)u^{1+\beta }(r,y)dy
\Big)dr.
\end{align*}
Thus,
\begin{equation*}
u\big((10\varepsilon _{2})^{1/\rho }t,x\big)
\geq \int_{0}^{(10\varepsilon _{2})^{1/\rho}t}
\Big(\int_{\mathbb{R}^{d}}p(K(( 10\varepsilon _{2})^{1/\rho }t,r),x-y)
 u^{1+\beta }(r,y)dy\Big)\, dr.
\end{equation*}
Jensen's inequality renders
\begin{equation*}
u\big((10\varepsilon _{2})^{1/\rho }t,x\big)
\geq \int_{\varepsilon _{1}^{1/\rho }t}^{(2\varepsilon _{1})^{1/\rho }t}
\Big(\int_{\mathbb{R}^{d}}p(K((10\varepsilon
_{2})^{1/\rho }t,r),x-y)u(r,y)dy\Big)^{1+\beta }dr=\infty ,
\end{equation*}
so that $u(t,x)=\infty $ for any
$t\geq (10\frac{ \varepsilon_{2}}{\varepsilon _{1}})^{1/\rho }\tau_{0}$
and $ x\in \mathbb{R}^{d}$.
\end{proof}


\begin{proposition}\label{Prop6}
Let $0<\frac{d\rho \beta }{\alpha}<1$. There exists a constant $C>0$ depending on $\alpha $, $\beta
$, $d$, $\varepsilon _{1}$, $\varepsilon _{2}$, $\theta $, $\rho $
and $\varphi $, such that
\begin{equation}
T_{\lambda\varphi} \leq \big\{C\lambda ^{-\beta }+[ (10
\frac{\varepsilon _{2}}{\varepsilon _{1}})^{1/\rho }\theta
] ^{\frac{\alpha -d\rho \beta }{\alpha }}\big\}
^{\frac{\alpha }{ \alpha -d\rho \beta }}+\eta ,\quad \lambda
>0, \label{upperest}
\end{equation}
where $\eta $ is any positive real number satisfying
$p( K(\eta,0),0)\leq 1$.
\end{proposition}


\begin{proof}
From Lemma \ref{Lemma1} we obtain
\begin{align*}
p(K(\eta ,0),x-y)
&=p\big(K(\eta,0),\frac{1}{2}(2x-2y)\big)\\
&\geq p\big(K(\eta ,0),2x\big)p(K(\eta ,0),2y)\\
&= 2^{-d}p(2^{-\alpha }K(\eta ,0),x)p(K(\eta ,0),2y).
\end{align*}
Therefore
\begin{align*}
u(\eta ,x)&\geq \lambda\int_{\mathbb{R}^{d}}p(
K(\eta ,0),x-y)\varphi (y)dy \\
&\geq 2^{-d}\lambda p(2^{-\alpha }K(\eta ,0)
,x)\int_{\mathbb{R}^{d}}p(K(\eta,0),2y) \varphi (y)dy \\
&=\lambda N_{0}p(2^{-\alpha }K(\eta ,0) ,x),
\end{align*}
where $N_{0}=2^{-d}\int_{\mathbb{R}^{d}}p(K( \eta ,0),2y)\varphi
(y)dy$. Thus, for any $\lambda >0$ , $t\geq 0$ and $x\in
\mathbb{R}^{d}$,
\begin{align*}
u(t+\eta ,x)
&=\int_{\mathbb{R}^{d}}p(K(t+\eta,\eta),x-y)u(\eta ,y)\,dy \\
&\quad+\int_{0}^{t}\Big(\int_{\mathbb{R}^{d}}p(K(t +\eta,r
+\eta),x-y)u^{1+\beta }(r+\eta ,y)\,dy\Big)\,dr
\\
&\geq \lambda N_{0}\int_{\mathbb{R}^{d}}p(K( t+\eta,\eta)
,x-y)p(2^{-\alpha }K(\eta ,0),y)dy \\
&\quad +\int_{0}^{t}\Big(\int_{\mathbb{R}^{d}}p(K(t +\eta,r
+\eta),x-y)u^{1+\beta }(r+\eta ,y)\,dy\Big)\,dr
\\
&\geq \lambda N_{0}p(K(t+\eta,\eta)+2^{-\alpha}K(\eta,0),x)\\
&\quad +\int_{0}^{t}\Big(\int_{\mathbb{R}^{d}}p(K(t +\eta,r
+\eta),x-y)u^{1+\beta }(r+\eta ,y)\,dy\Big)\,dr
\\
&\geq w(t,x),
\end{align*}
where $w$ solves the equation
\begin{equation} \label{intqw}
\begin{aligned}
w(t,x)&=\lambda N_{0}p(K( t+\eta,\eta)+2^{-\alpha}K(\eta ,0),x)  \\
&\quad +\int_{0}^{t}\Big(\int_{\mathbb{R}^{d}}p(K(t +\eta,r
+\eta),x-y)w^{1+\beta }(r,y)dy\Big)dr,
\quad t\geq 0,\; x\in \mathbb{R}^{d}.
\end{aligned}
\end{equation}
Hence, it is sufficient to prove that $w$ is non-global, and, because
of Lemma \ref{Lemma5}, it suffices to show finite time blowup of
\begin{equation*}
\upsilon (t)=\int_{\mathbb{R}^{d}}p(
K(t,0),x)w(t,x)\,dx,\quad t\geq
0.
\end{equation*}
Multiplying both sides of (\ref{intqw}) by $p(K( t,0),x)$ and
integrating, we obtain
\begin{align*}
v(t) &=\int_{\mathbb{R}^{d}}p(K(t,0),x)w(t,x)\,dx \\
&=\lambda N_{0}\int_{\mathbb{R}^{d}}p(K(t+\eta,\eta)
+2^{-\alpha }K(\eta ,0),x) p(K(t,0),x)\,dx \\
&\quad +\int_{\mathbb{R}^{d}}\int_{0}^{t}
 \int_{\mathbb{R}^{d}}p(K(t+\eta,r+\eta),x-y)p( K(t,0)
,x)w^{1+\beta }(r,y)\,dy\,dr\,dx \\
&=\lambda N_{0}p(K(t,0)+K(t+\eta,\eta)+2^{-\alpha }K(\eta,0),0)\\
&\quad +\int_{0}^{t}\int_{\mathbb{R}^{d}}p(K(t+\eta,r+\eta)+K(t,0),y)
w^{1+\beta }(r,y)\,dy\,dr,\quad t\geq 0.
\end{align*}
 From Lemma \ref{Lemma1} (i), we have $p(t,0)\leq p(s,0)$ for all
$0<s\leq t$. Hence
\begin{align*}
\upsilon (t)
&\geq \lambda N_{0}p(2K(t+\eta,0)+2^{-\alpha }K(\eta ,0),0)\\
&\quad +\int_{0}^{t}\int_{\mathbb{R}^{d}}p(
K(t+\eta,r+\eta)+K(t,0),y)w^{1+\beta }(r,y)\,dy\,dr.
\end{align*}
Using now Lemma \ref{Lemma1} (iii) we obtain,
\begin{align*}
\upsilon (t)&\geq \lambda N_{0}p(2K(t+\eta,0)
+2^{-\alpha }K(\eta ,0),0)\\
&\quad +\int_{0}^{t}\Big(\frac{K(r,0) }{K( t+\eta,r+\eta)+K(t,0)}
 \Big)^{d/\alpha}\int_{\mathbb{R}^{d}}p(K(r,0),y)w^{1+\beta }(r,y)\,dy\,dr.
\end{align*}
Jensen's inequality together with Lemma \ref{Lemma1} (i) gives
\begin{align*}
\upsilon (t)
&\geq \lambda N_{0}p(2K(t+\eta,0)+2^{-\alpha }K(\eta ,0),0)
 +\int_{0}^{t}\big(\frac{K(r,0)}{2K(t+\eta,0)}\big)^{d/\alpha}
 \upsilon ^{1+\beta }(r)\,dr \\
&=\lambda N_{0}[ 2K(t+\eta,0)+2^{-\alpha }K(\eta ,0)]^{-d/\alpha}p( 1,0)\\
&\quad +\int_{0}^{t}\big(\frac{K(r,0)}{2K(t+\eta,0)} \big)^{d/\alpha}
\upsilon ^{1+\beta }(r) \,dr.
\end{align*}
Let $f_{1}(t)=K^{d/\alpha}( t+\eta,0)\upsilon (t)$ and
$t\geq \theta $. We have
\begin{align*}
f_{1}(t)&\geq \lambda p(1,0)N_{0}[ \frac{ K(\theta+\eta
,0)}{2K(\theta+\eta ,0) +2^{-\alpha }K( \eta ,0)}]^{d/\alpha}\\
&\quad +2^{-d/\alpha} \int_{\theta}^{t}
 K^{-d\beta/\alpha}( r,0)f_{1}^{1+\beta }(r)\,dr,
\end{align*}
and if $N:=p(1,0)N_{0}[ \frac{K(\theta+\eta ,0)}{
2K(\theta+\eta ,0)+2^{-\alpha }K(\eta ,0)}] ^{d/\alpha}$, then
\begin{equation*}
f_{1}(t)\geq \lambda N+2^{-d/\alpha}\int_{ \theta
}^{t}K^{^{-d\beta/\alpha}}(r,0) f_{1}^{1+\beta}(r)dr,\quad t\geq \theta .
\end{equation*}
Let $f_{2}$ be the solution of the integral equation
\begin{equation*}
f_{2}(t)=\lambda N+2^{-d/\alpha}\int_{\theta }^{t}
K^{-d\beta/\alpha}(r,0)f_{2}^{1+\beta }(r)dr,\quad t\geq \theta ,
\end{equation*}
which satisfies
\begin{equation}
f_{2}^{\beta }(t)=\frac{(\lambda N)^{\beta }}{ 1-\beta (\lambda
N)^{\beta }(\frac{1}{2})^{d/\alpha}H(t)}  \label{solf2}
\end{equation}
with
$H(t)\equiv \int_{\theta}^{t}K^{-d\beta/\alpha}(r,0)dr$.
 From (\ref{condK}) and the assumption
$0<\frac{d\rho \beta }{\alpha }<1$, we get
\begin{equation*}
H(t)\geq {\varepsilon _{2}^{-d\beta/\alpha}}
\int_{\theta }^{t}r^{-\frac{d\rho \beta }{\alpha }}dr=\frac{\alpha
}{\alpha -d\rho \beta }\varepsilon _{2}^{-d\beta/\alpha}[ t^{ \frac{\alpha -d\rho \beta }{\alpha }}-\theta
^{\frac{\alpha -d\rho \beta }{ \alpha }}] \to \infty \quad
\text{ as }\quad t\to \infty .
\end{equation*}
Hence, there exists $\tau_{0}\geq \theta $ such that
$\beta(\frac{1}{2} )^{d/\alpha}(\lambda N)^{\beta}H(\tau_{0})=1$, and
therefore,
\[
 \int_{\theta}^{\tau_{0}}K^{^{-d\beta/\alpha}}(r,0)dr
=\frac{2^{d/\alpha}}{\beta }N^{-\beta }\lambda
^{-\beta },
\]
which together with (\ref{condK}) gives
$\int_{\theta }^{\tau_{0}}(\varepsilon _{2}r^{\rho })^{-
\frac{d\beta }{\alpha }}dr\leq \frac{2^{d/\alpha}}{\beta
}N^{-\beta}\lambda ^{-\beta }$.
Hence
\begin{equation*}
\tau_{0}^{\frac{\alpha -d\rho \beta }{\alpha }}\leq
\frac{2^{d/\alpha} [ \alpha -d\rho \beta ]
}{\alpha \beta }N^{-\beta }\varepsilon _{2}^{\frac{d\beta }{\alpha
}}\lambda ^{-\beta }+\theta ^{\frac{\alpha -d\rho \beta }{\alpha}},
\end{equation*}
or, equivalently,
\begin{equation}
\tau_{0}\leq \big\{\frac{2^{d/\alpha}\left[ \alpha
-d\rho \beta \right] }{\alpha \beta }N^{-\beta }\varepsilon
_{2}^{\frac{d\beta }{\alpha } }\lambda ^{-\beta }+\theta
^{\frac{\alpha -d\rho \beta }{\alpha }}\big\}
^{ \frac{\alpha }{\alpha -d\rho \beta }}.  \label{estTo}
\end{equation}
 From (\ref{solf2}), we deduce that $f_{2}(\tau_{0}) =\infty $. It
follows that
\begin{equation*}
K^{d/\alpha}(\tau_{0}+\eta,0)\upsilon (
\tau_{0})=f_{1}(\tau_{0})\geq f_{2}(\tau_{0})=\infty ,
\end{equation*}
which implies (as in the proof of Lemma \ref{Lemma5}) that
$w(t,x)=\infty $  if $ t\geq (10\frac{\varepsilon _{2}}{
\varepsilon _{1}})^{1/\rho }\tau_{0}$, and thus,
$u(t,x)=\infty $  provided
$ t \geq (10\frac{\varepsilon _{2}}{\varepsilon _{1}})^{1/\rho }\tau_{0}+\eta$.
 Therefore
\begin{equation*}
T_{\lambda\varphi}\leq \big(10\frac{\varepsilon
_{2}}{\varepsilon _{1}}\big)^{1/\rho }\tau_{0}+\eta .
\end{equation*}
 From (\ref{estTo}) we conclude that there is a positive
constant $ C=C(\alpha ,\beta ,d,\varepsilon _{1},\varepsilon
_{2},\theta ,\rho ,\varphi )$ satisfying (\ref{upperest}).
\end{proof}


\section{A lower estimate for the life span\label{seccion6}}

To bound from below the life span
$T_{\lambda\varphi}$ of the initial-value problem
\eqref{cauchyec}, we need to assume that (\ref{condK}) holds for
any $t\geq 0$, and that $\varphi$ is integrable.

Let $\{U(t,s)\}_{t\geq s\geq 0}$ be the evolution family on
$C_{b}(\mathbb{R}^{d})$ generated by the family of operators
$\{k(t)\Delta _{\alpha }\}_{t\geq 0}$,
which is given by
\begin{equation*}
U(t,s)\varphi (x)=\int_{\mathbb{R} ^{d}}\varphi (y)p(K(t,s)
,x-y)dy=S(K(t,s))\varphi ( x),
\end{equation*}
where $\{S(t)\}_{t\geq 0}$ is the semigroup with the
infinitesimal generator $\Delta _{\alpha }$.

\begin{proposition}\label{Prop7}
Let $0<\frac{d\rho \beta }{\alpha }<1$.
There exists a constant $c>0$, depending on $\alpha $, $\beta $,
$d$, $\varepsilon _{1}$, $\rho $ and $\varphi $, such that
\begin{equation}
T_{\lambda\varphi} \geq c\lambda ^{-\frac{\alpha \beta }{\alpha
-d\rho \beta }},\quad\lambda >0.  \label{lowest}
\end{equation}
\end{proposition}


\begin{proof}
The function
\begin{equation*}
\overline{u}(t,x): =\Big[ \lambda ^{-\beta }-\beta
\int_{0}^{t}\| U(r,0)\varphi \| _{\infty }^{\beta }dr\Big]
^{-1/\beta}U(t,0)\varphi (x),\qquad t\geq 0,\quad
x\in \mathbb{R}^{d}.
\end{equation*}
is a supersolution of \eqref{cauchyec}. Indeed,
$\overline{u}(0,\cdot)=\lambda \varphi ( \cdot)$ and
\begin{align*}
\frac{\partial \overline{u}(t,x)}{\partial t}
&=-\frac{1}{ \beta}\Big[ \lambda ^{-\beta }-\beta \int_{0}^{t}\|
 U(r,0)\varphi \| _{\infty }^{\beta }dr\Big]
^{-\frac{1}{\beta }-1}\Big[ -\beta \| U(t,0)\varphi \|
_{\infty }^{\beta }\Big] U(t,0)\varphi (x)\\
&\quad +\Big[ \lambda ^{-\beta }-\beta \int_{0}^{t}\| U(
r,0)\varphi \| _{\infty }^{\beta }dr\Big] ^{-1/\beta}
k(t)\Delta _{\alpha }U(t,0) \varphi (x).
\end{align*}
Since $-\frac{1}{\beta }-1=-\frac{\beta +1}{\beta }$, we get
\begin{align*}
\frac{\partial \overline{u}(t,x)}{\partial t}
&=\Big\{\Big[\lambda ^{-\beta }-\beta \int_{0}^{t}\| U(r,0) \varphi
 \| _{\infty }^{\beta }dr\Big] ^{-1/\beta}\Big\}^{\beta +1}
 \| U(t,0)\varphi \|_{\infty }^{\beta }U(t,0)\varphi (x)\\
&\quad +k(t)\Delta _{\alpha }\Big[ \lambda ^{-\beta
}-\beta \int_{0}^{t}\| U(r,0)\varphi
\| _{\infty }^{\beta }dr\Big] ^{-1/\beta}U(t,0)\varphi (x).
\end{align*}
Using the inequality
\begin{equation*}
\| U(t,0)\varphi \| _{\infty}^{\beta }U(t,0)\varphi (x)\geq [
U(t,0)\varphi (x)] ^{1+\beta }
\end{equation*}
it follows that
\begin{equation*}
\frac{\partial \overline{u}(t,x)}{\partial t}\geq k( t)\Delta
_{\alpha }\overline{u}(t,x)+\overline{u} ^{1+\beta }(t,x),
\end{equation*}
showing that $\overline{u}$ is a supersolution of
\eqref{cauchyec}. Writing $ \overline{L}(\lambda)$ for the life
span of $\overline{u}$, it follows that
\begin{equation*}
\overline{L}(\lambda)\leq T_{\lambda\varphi},\quad\lambda\ge0.
\end{equation*}
Now,
\begin{equation*}
\overline{u}(t,x)=\Big[ \lambda ^{-\beta }-\beta
\int_{0}^{t}\| U(r,0)\varphi
\| _{\infty }^{\beta }dr\Big] ^{-1/\beta}U(t,0)\varphi (x)=\infty
\end{equation*}
when $\lambda ^{-\beta }=\beta \int_{0}^{t}\| U(r,0)
\varphi \| _{\infty }^{\beta }\,dr$.
By definition of $\overline{L}(\lambda)$,
\begin{equation}
\beta ^{-1}\lambda ^{-\beta }=\int_{0}^{\overline{L}
(\lambda)}\| U(r,0)\varphi \| _{\infty }^{\beta}dr.  \label{lowest2}
\end{equation}
Note that, by Lemma \ref{Lemma1} (i), (ii),
\begin{align*}
U(t,0)\varphi (x)
&=S(K(t,0))\varphi (x)\\
&=\int_{\mathbb{R}^{d}}\varphi (y)p(K(t,0),x-y)dy \\
&\leq p(1,0)K^{-d/\alpha}(t,0)\| \varphi \| _{1},\quad t>0,\; x\in
\mathbb{R}^{d}.
\end{align*}
Since, by assumption, (\ref{condK}) holds for any $t\geq 0$, we
obtain
\begin{equation*}
\| U(t,0)\varphi \| _{\infty }\leq p(
1,0)(\varepsilon _{1}t^{\rho })^{-d/\alpha} \|
\varphi \| _{1}.
\end{equation*}
Inserting this inequality in (\ref{lowest2}) and using that
$0<\frac{d\rho \beta }{\alpha }<1$, we get
\begin{align*}
\beta ^{-1}\lambda ^{-\beta }
&\leq (p(1,0)\|\varphi \| _{1})^{\beta }
\varepsilon _{1}^{-d\beta/\alpha}\int_{0}^{\overline{L}(\lambda)}
r^{-d\rho \beta/\alpha}\,dr \\
&= \frac{\alpha }{\alpha -d\rho \beta }(p(1,0) \| \varphi
\| _{1})^{\beta }\varepsilon _{1}^{-d\beta/\alpha}
\overline{L}(\lambda)^{\frac{\alpha -d\rho \beta }{\alpha}},
\end{align*}
which gives
\begin{equation*}
\overline{L}(\lambda)^{\frac{\alpha -d\rho \beta }{\alpha }}\geq
\frac{ \alpha -d\rho \beta }{\alpha \beta }(p(1,0) \|
\varphi \| _{1})^{-\beta }\varepsilon _{1}^{\frac{d\beta
}{ \alpha }}\lambda ^{-\beta }.
\end{equation*}
In this way we obtain the inequality
\begin{equation*}
T_{\lambda\varphi} \geq \Big[ \frac{\alpha -d\rho \beta }{\alpha
\beta }\Big] ^{\frac{\alpha }{\alpha -d\rho \beta }}(
p(1,0)\| \varphi \| _{1}) ^{-\frac{\alpha \beta
}{\alpha -d\rho \beta }}\varepsilon _{1}^{\frac{d\beta }{\alpha
-d\rho \beta } }\lambda ^{-\frac{\alpha \beta }{\alpha -d\rho
\beta }},
\end{equation*}
which proves the existence of a constant
$c\equiv c(\alpha ,\beta,d,\varepsilon _{1},\rho ,\varphi )>0$
that satisfies (\ref{lowest}).
\end{proof}


Summarizing both, upper and lower bounds for the life span of (
\ref{cauchyec}), we get the following statement.


\begin{theorem}\label{Theo8}
 Let  $0<\frac{d\rho \beta }{\alpha}<1$, and let
$T_{\lambda\varphi}$ be the life span of the
nonautonomous semilinear equation
\begin{gather*}
\frac{\partial u(t,x)}{\partial t}
 =k(t)\Delta_{\alpha }u(t,x)+u^{1+\beta }(t,x)\\
u(0,x)= \lambda \varphi (x)\ge0 ,\quad x\in \mathbb{R}^{d},
\end{gather*}
where $\lambda>0$. Then
\begin{equation}
\lim_{\lambda \to 0}T_{\lambda\varphi}
=\infty ,\quad
\lim_{\lambda \to \infty }T_{\lambda\varphi}\in \big[
0,(10\frac{\varepsilon _{2}}{\varepsilon _{1}})^{1/\rho}
\theta +\eta \big] ,  \label{asybeha}
\end{equation}
where $\theta $ and $\eta $ are any positive numbers such that
$\varepsilon _{1}\theta ^{\rho }\leq K(\theta ,0)\leq \varepsilon
_{2}\theta ^{\rho }$ and $p(K(\eta ,0),0)\leq 1$, respectively.
\end{theorem}

\begin{proof}
Due to (\ref{upperest}) and (\ref{lowest}),
\begin{equation*}
c\lambda ^{-\frac{\alpha \beta }{\alpha -d\rho \beta }}
\leq T_{\lambda\varphi}
\leq \big\{C\lambda ^{-\beta }
+\big[ (10\frac{\varepsilon _{2}}{\varepsilon _{1}})^{1/\rho }\theta
\big] ^{\frac{ \alpha -d\rho \beta }{\alpha }}\big\}
^{\frac{\alpha }{\alpha -d\rho \beta }}+\eta ,
\end{equation*}
from which (\ref{asybeha}) follows directly using the fact that
$0<\frac{ d\rho \beta }{\alpha }<1$.
\end{proof}

\subsection*{Acknowledgements}
The authors are grateful to the anonymous referees for their careful
reading of the original manuscript, and for their useful suggestions.


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\end{document}