\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 116, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/116\hfil Instantaneous blow-up]
{Instantaneous blow-up of semilinear
non-autonomous equations with \\ fractional diffusion}

\author[J. Villa-Morales \hfil EJDE-2017/116\hfilneg]
{Jos\'e Villa-Morales}

\address{Jos\'e Villa-Morales \newline
Departamento de Matem\'aticas y F\'isica\\
Universidad Aut\'onoma de Aguascalientes\\
Av. Universidad No. 940, Cd. Universitaria\\
Aguascalientes, Ags., C.P. 20131, Mexico}
\email{jvilla@correo.uaa.mx}


\thanks{Submitted September 5, 2016. Published May 2, 2017.}
\subjclass[2010]{35K05, 45K05, 60G52, 34G20}
\keywords{Generalized Osgood's condition; semilinear equations;
\hfill\break\indent fractional diffusion; instantaneous blow-up}

\begin{abstract}
 We consider the Cauchy initial value problem
 \begin{gather*}
 \frac{\partial }{\partial t}u(t,x) =k(t)\Delta _{\alpha}u(t,x)+h(t)f(u(t,x)), \\
 u(0,x) = u_0(x),
 \end{gather*}
 where $\Delta _{\alpha }$ is the fractional Laplacian for $0<\alpha \leq 2$.
 We prove that if the initial condition $u_0$ is non-negative,
 bounded and measurable then the problem has a global integral solution
 when the source term $f$ is non-negative, locally Lipschitz and satisfies the
 generalized Osgood's condition
 \[
 \int_{\|u_0\|_{\infty }}^{\infty }\frac{ds}{f(s)}\geq \int_0^{\infty}h(s)ds.
 \]
 Also, we prove that if the initial data is unbounded then the
 generalized Osgood's condition does not guarantee the  existence of
 a global solution. It is important to point out that the proof of the
 existence hinges on the role of the function $h$. Analogously, the
 function $k$ plays a central role in the proof of the instantaneous blow-up.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}

We consider integral solutions of the semilinear
non-autonomous parabolic equation
\begin{equation}  \label{edif}
\begin{gathered}
\frac{\partial }{\partial t}u(t,x)
= k(t)\Delta _{\alpha }u(t,x)+h(t)f(u(t,x)),\quad t>0,\; x\in \mathbb{R}^{d},
 \\
u(0,x) = u_0(x),\quad  x\in \mathbb{R}^{d},  
\end{gathered}
\end{equation}
where the diffusion $\Delta _{\alpha }=-(-\Delta )^{\alpha /2}$ is the
fractional Laplacian (or $\alpha $-Laplacian), $0<\alpha \leq 2$, $u_0$ is
the initial data and $f$ is the source term. The diffusion and the source
terms are multiplicatively perturbated by continuous functions $k,h$.

Basic references for the study of the fractional Laplacian are the books
\cite{La} and \cite{St}. However, it is worth noticing that the systematic 
study of partial differential equations considering fractional diffusion 
is relatively new. This area of mathematics has been actively 
studied in the last decade by Caffarelli, V\'azquez and many others
(see for instance, \cite{Ca-Va,Va} and the references therein).

We  denote by $p$ the real-valued function determined by
\begin{equation}
\int_{\mathbb{R}^{d}}p(t,z)e^{iz\cdot \xi }dz
=e^{-t|\xi |^{\alpha }},\quad
\text{for all }t>0,\; \xi \in \mathbb{R}^{d}.  \label{defdenest}
\end{equation}
The space of all real-valued essentially bounded functions defined on
$\mathbb{R}^{d}$ will be denoted by $L^{\infty }(\mathbb{R}^{d})$. Let us
consider the family $\{T_{t},t\geq 0\}$ of bounded linear operators defined
on $L^{\infty }(\mathbb{R}^{d})$ as
\[
T_{t}g(x)=\int_{\mathbb{R}^{d}}p( t,x-y) g(y) dy,\quad
x\in \mathbb{R}^{d}.
\]
It is well known that $\{T_{t},t\geq 0\}$ is a strongly continuous semigroup
with infinitesimal generator $\Delta _{\alpha }$ (see \cite{Ap}).

As in \cite{L-R-S,Q-P}, we introduce the following concept.

\begin{definition} \rm
Let $u_0\geq 0$ be a measurable function. We say that \eqref{edif} has a
local integral solution on $[0,T)$ if there is a measurable function 
$u:[0,T)\times \mathbb{R}^{d}\to [ 0,\infty ]$ that is finite
almost everywhere and
\begin{equation}
u(t)=T_{K_0(t)}u_0+\int_0^{t}h(s)T_{K(s,t)}f(u(s))ds,\quad t\in [
0,T),  \label{EcIntg}
\end{equation}
holds almost everywhere (a.e.) in $[0,T)\times \mathbb{R}^{d}$, where
\[
K(s,t)=\int_{s}^{t}k(r)dr\quad \text{and}\quad  K_0(t)=K(0,t).
\]
We will say that \eqref{edif} has a global integral solution if \eqref{edif}
has a local integral solution for all $T>0$.
\end{definition}

A solution of the differential equation \eqref{edif} is called a classical
solution. It is clear that the non-existence of a local integral solution
for \eqref{edif} implies the non-existence of a classical solution 
(see \cite{Q-P}).

In this article we consider the following hypotheses:
\begin{itemize}
\item[(H1)] $u_0\geq 0$ and $0<\|u_0\|_{q}<\infty $, with $1\leq
q\leq \infty ;$

\item[(H2)] $f:[0,\infty )\to [ 0,\infty )$ is
non-decreasing, locally Lipschitz, $f(0)=0,$ and $f>0$ on $(0,\infty )$;

\item[(H3)] $h,k:(0,\infty )\to (0,\infty )$ are continuous
functions:
\begin{itemize}
\item[(a)] $\lim_{t\to 0}\int_0^{t}h(s)ds=0$,

\item[(b)] $\lim_{t\to 0}\int_0^{t}k(s)ds=0$.
\end{itemize}
\end{itemize}
In what follows we will use the notation
\[
H(t)=\int_0^{t}h(s)ds,\quad F_{x_0}(x)=\int_{x_0}^{x}\frac{ds}{f(s)},
\]
where $x_0\geq 0$.

It is not difficult to prove that under the hypotheses (H2) and (H3),
the initial value problem
\begin{equation} \label{edo}
\begin{gathered}
\frac{dy(t)}{dt} =h(t)f(y(t)),\quad t>0,   \\
y(0) =x_0\geq 0,
\end{gathered}
\end{equation}
has a unique solution if and only if 
$\operatorname{im}(H)\subseteq \operatorname{im}(F_{x_0})$.
Moreover, the solution is given by $y(t)=F_{x_0}^{-1}(H(t))$, $t\geq 0$.
Such criterion, of existence and uniqueness for \eqref{edo}, is called the
generalized Osgood's test (for a proof see for example 
\cite[Lemma 2.2]{C-M-V}).

Assume the hypotheses (H2), (H3) and (H3)(a). 
Theorem \ref{existenciaglobal} will establish that \eqref{edif} has a global 
integral solution if $0\leq u_0\in L^{\infty }(\mathbb{R}^{d})$ and
 $\operatorname{im}(H)\subseteq \operatorname{im}(F_{\|u_0\|_{\infty }})$. 
Thus the generalized Osgood's test is still valid in some sense.

The following question arouse naturally in \cite{L-R-S} for the case 
$h\equiv k\equiv 1$ and $\alpha =2$:
\begin{equation}
\parbox{10cm}{Does $\operatorname{im}(H)\subseteq
 \operatorname{im}(F_{\|u_0\|_{q}})$ (with $1\leq q<\infty$) guarantees 
 the existence
 of a global integral solution for \eqref{edif}?}
\label{question}
\end{equation}
To answer the above question we define the closed
ball with center at $x$ and radius $R$ by
$B_{R}(x)=\{y\in \mathbb{R}^{d}:|x-y|\leq R\}$. 

\begin{definition}\label{defexpinst} \rm
Let $u_0\geq 0$ be a measurable function. We say that a
measurable function $u\geq 0$ is a local $p$-integrable solution to 
\eqref{edif} if there are $x\in \mathbb{R}^{d}$, $p\geq 1$, $R>0$ and $T>0$,
such that $u(t,\cdot )\in L^{p}(B_{R}(x))$, for all $t<T$, with $u(0)=u_0$
and the equality $\eqref{EcIntg}$ is satisfied a.e. on $[0,T)\times B_{R}(x)$.
\end{definition}

Of course, the non-existence of $p$-integrable local solutions implies the
non-existence of local integral solutions. Therefore
we say that an equation \textit{blows-up instantaneously} if it does
not have a $p$-integrable local solution.

In \cite{L-R-S}, an initial condition $u_0$ satisfying (H1)
with $1\leq q<\infty $, and a source term $f$ satisfying (H2)
and $\operatorname{im}(H)\subseteq \operatorname{im}(F_{\|u_0\|_{q}})$ are assumed. 
Under these conditions it is readily proved that any local integral solution 
for \eqref{edif} does not belong to $L^{1}(B_{R}(0))$ for all $t<\tau $ and $R>1$.

The concept of local $p$-integrable solution is weaker than that of local integral 
solution introduced in \cite{L-R-S}. In this new context we will have the 
following consequences:

(i) The answer given in \cite{L-R-S} for the question \eqref{question} uses 
strongly the fact that $R>1$ (this assumption is essential
in the study of the set $\{|x|\leq (R-1)/2\sqrt{t-s}\}$ introduced in the
proof of \cite[Theorem 4.1]{L-R-S}). We overcome this difficulty through
Lemma \ref{newpropaesta}: see the inequality \eqref{estpanue} below.

(ii) Some new phenomenon are found. \newline
(a) The existence of a global integral solution depends only of the
combined source term $h$ and $f$. More precisely, the existence of a global
integral solution depends of (H1) ($q=\infty $), (H2), (H3), (H3)(a) and
 $\operatorname{im}(H)\subseteq \operatorname{im}(F_{\|u_0\|_{q}})$.
This, at an intuitively level, implies that the diffusion term can be
perturbed by a very large term $k$ and we still have the existence of a
global integral solution. \\
(b) Symmetrically, given (H1) ($1\leq q<\infty $) and (H3),  to 
construct a convenient source term $\tilde{f}$ satisfying 
$\int_{\|u_0\|_{q}}^{\infty }\frac{ds}{\tilde{f}(s)}=\infty $
one requires  Hypotheses (H3)(b). In this case $h$ is
arbitrary, which means that $\int_0^{\infty }h(s)ds$ could be finite or 
infinite. In any case the generalized Osgood's condition is satisfied.


It is worth mention that Osgood-type conditions appears naturally in some
applied problems (see the references in \cite{L-R-S}). On the other hand,
from \cite{V} we have that if we require additionally that the source
term $f$ is convex then we have that the local integral solution of 
\eqref{edif} blows-up in finite time when 
$\int_{\|u_0\|_{\infty }}^{\infty } \frac{ds}{f(s)}<\infty $ and 
$\int_0^{\infty }h(s)ds=\infty $. The case 
$\int_{\|u_0\|_{\infty }}^{\infty }\frac{ds}{f(s)}<\int_0^{\infty
}h(s)ds<\infty $ is still open, but we believe that we also have blow-up in
finite time. Concerning the global existence of \eqref{edif}, we need additional 
conditions to guarantee a solution in a strong sense. For example, when 
$h\equiv k\equiv 1$ and $\alpha \in (1,2)$, the authors of \cite{D-I} proved that
 \eqref{edif} has a mild solution (or even such
solution is classical). Moreover, \eqref{edif} has a mild global solution
if we assume (see \cite{R-V})
\[
\sup \Big\{ \int_0^{t}\Big(\int_{s}^{t}k(r)dr\Big) ^{-1/\alpha}ds:
t\in [ 0,T]\Big\} <\infty ,\quad \text{for all }T>0.
\]

The importance of the study of equations like $\eqref{pdae}$ with
fractional diffusion is well known in applied mathematics. For example,
they arise in fields like molecular biology, hydrodynamics and statistical
physics \cite{S-Z-F}. Also, notice that generators of the form 
$k(t)\Delta_{\alpha }$ arise in models of anomalous growth of certain fractal
interfaces \cite{M-W}. The study of partial differential equations with
fractional diffusion is becoming more popular. In fact the number of
works both theoretical and practical are increasing: see for example 
\cite{Ca-Va,D-I,Va} and the references therein.

This article is organized as follows. In Section 2 we prove the existence of
global mild solutions of \eqref{pdae} and we present some basic properties
of $p(t,x)$. Section 3 provides a negative answer to the question 
\eqref{question}, that is, we give an initial condition 
$\phi _{\theta }\in L^{q}(\mathbb{R}^{d})$, $1\leq q<\infty $, 
$0<\theta <d/q$ and a source term $\tilde{f}$ that satisfy the generalized
 Osgood's condition for which there is instantaneous blow-up.

\section{Global existence and preliminary results}

The existence of local integral solutions for \eqref{EcIntg} follows form
the Banach contraction principle, as we shall see.
First, we state some well known properties of $p$.

\begin{lemma}\label{pdae} 
Let $t>0$ and $x,y\in \mathbb{R}^{d}$ then:
\begin{itemize}
\item[(a)] $p(t,x)\geq 0$ and $\int_{\mathbb{R}
^{d}}p(t,z)dz=1$ (density property).

\item[(b)] $p(t,x)=t^{-d/\alpha }p(1,t^{-1/\alpha }x)$ (scaling
property).

\item[(c)] If $|x|\geq |y|$ then $p(t,x)\leq p(t,y)$ (radially
decreasing).

\item[(d)] The function $(t,x)\to p(t,x)$ is in $C^{\infty
}((0,\infty )\times \mathbb{R}^{d})$ (regularity).

\item[(e)] There exists $c_0=c_0(d,\alpha )>0$, such that
\[
p(t,x)\geq c_0\min \big\{ \frac{t}{|x|^{d+\alpha }},\frac{1}{t^{d/\alpha }}\big\} .
\]
\end{itemize}
\end{lemma}

\begin{proof}
For the proofs of (a)--(c) see \cite[Section 2]{Su}. The
proof of property (d) can be found in \cite{D-I}, and that of property (e)
can be found in \cite{B-S-S}.
\end{proof}

In what follows, $c$ will denote a positive constant whose
specific value is unimportant and can change from place to place. On
the other hand, if the constant $c_{\cdot }$ has a subindex then we refer to
a specific constant.

Let us recall that, by $L^{\infty }(\mathbb{R}^{d})$, we denote the space of
all measurable functions $\varphi :\mathbb{R}^{d}\to \mathbb{R}$
such that
\[
\| \varphi \| _{\infty }=\inf \{M\geq 0:|\varphi (x) |\leq M\text{ holds for 
almost all }x\}<\infty .
\]
Let $\tau >0$ be a real number that we will fix later. Define
\[
E_{\tau }=\{u:[ 0,\tau ] \to L^{\infty }(\mathbb{R}^{d})
\text{ and }\||u\||<\infty \}, 
\]
where
\[
\||u\||=\sup \{\| u( t) \| _{\infty }:0\leq
t\leq \tau \}.
\]
Then $E_{\tau }$ is a Banach space and the sets ($r>0$)
\[
P_{\tau }=\{u\in E_{\tau }:u\geq 0, \text{ a.e. }\}, \quad
 B_{\tau ,r}=\{u\in E_{\tau }:\||u\||\leq r\},
\]
are closed subsets of $E_{\tau }$.

\begin{theorem} \label{existenciaglobal}
Let us assume {\rm (H1)}, with $q=\infty $, {\rm(H2), (H3), (H3)(a)}. Then
\eqref{edif} has a global integral solution if 
$\operatorname{im}(H)\subseteq \operatorname{im}(F_{\|u_0\|_{\infty }})$.
\end{theorem}

\begin{proof}
Define the operator $\Psi :B_{\tau ,r}\cap P_{\tau }\to E_{\tau }$
by
\[
\Psi ( u) ( t)=T_{K_0(t)}u_0+\int_0^{t}h( s) T_{K(s,t) }f(u( s) )ds.
\]
Since $u\geq 0$ and $u_0\geq 0$, it is clear that 
$\Psi ( u)\geq 0$, so $\Psi ( u) \in P_{\tau }$. Using 
$\int_{\mathbb{R}^{d}}p(t,x)dx=1$, for almost all $x\in \mathbb{R}^{d}$ and
$u\in B_{\tau ,r}$, we have 
\begin{equation}
\begin{aligned}
\Psi ( u) ( t,x)  
&\leq \|u_0\|_{\infty}\int_{\mathbb{R}^{d}}p( K_0( t) ,y-x) dy \\
&\quad +\int_0^{t}h( s) \int_{\mathbb{R}^{d}}p( K( s,t) ,y-x) 
f(\|u( s) \|_{\infty})dy\,ds   \\
&\leq \|u_0\|_{\infty }+\int_0^{t}h( s)\int_{\mathbb{R}^{d}}p( K( s,t) ,y-x) 
f(r)dy\,ds\\
&\leq \|u_0\|_{\infty }+f(r)\int_0^{\tau }h( s) ds\,.
\end{aligned} \label{ejeusdens}
\end{equation}
Then
\[
\||\Psi ( u) \||\leq \|u_0\|_{\infty
}+f(r)\int_0^{\tau }h( s) ds.
\]
Let us take $r=1+\|u_0\|_{\infty }$. By  Hypothesis (H3)(a) we
can choose $\tau >0$ small enough such that
\[
f(r)\int_0^{\tau }h( s) ds<1.
\]
Then $\Psi ( u) \in B_{\tau ,r}\cap P_{\tau }$, therefore 
$\Psi(B_{\tau ,r}\cap P_{\tau })\subset B_{\tau ,r}\cap P_{\tau }$.

Now let us see that $\Psi $ is a contraction. 
Take $u,\tilde{u}\in B_{\tau,r}\cap P_{\tau }$,
\begin{align*}
&|\Psi ( u) ( t,x) -\Psi ( \tilde{u}) ( t,x) | \\
&=\big| \int_0^{t}h( s) \int_{\mathbb{R}^{d}}p( K( s,t) ,y-x) 
[ f(u( s,y) )-f(\tilde{u}( s,y) )] dy\,ds \big| \\
&\leq \sup_{t\in [ 0,\tau ] }\Big\{\int_0^{t}h( s) 
\int_{\mathbb{R}^{d}}p(K( s,t) ,y-x) |f(u( s,y) )-f(\tilde{u}(s,y) )|dy\,ds\Big\} .
\end{align*}
Since $\mathbb{R}$ is locally compact, then $f$ is Lipschitz on each compact
subset of $\mathbb{R}$. In particular, for $[0,r]$ there exists a constant 
$c>0$ such that
\[
|f(s)-f(t)|\leq c|s-t|,\quad \text{for all }s,t\in [ 0,r].
\]
From the above inequality we easily deduce
\[
|f(u( s,y) )-f(\tilde{u}( s,y) )| 
\leq c| u( s,x) -\tilde{u}( s,x) |
\leq c\|u(s)-\tilde{u}(s)\|_{\infty }.
\]
Consequently,
\begin{align*}
\||\Psi ( u) -\Psi ( \tilde{u}) \|| 
&\leq \sup_{t\in [ 0,\tau ] }\big\{ \int_0^{t}h( s)
c\| u(s)-\tilde{u}( s) \| _{\infty }ds \Big\} \\
&\leq ( c\int_0^{\tau }h( s) ds) \||u-\tilde{u}\||.
\end{align*}
Using again  Hypothesis (H3)(a), we can choose $\tau >0$ small
enough such that $\Psi $ is a contraction. Therefore $\Psi $ has a unique
fixed point (on $B_{\tau ,r}\cap P_{\tau }$): the local mild solution to the
equation \eqref{EcIntg}.

Let $[0,\tau _{\rm max})$ be the maximal interval for which the local integral
solution $u$ of \eqref{EcIntg} exists. Clearly, $\tau _{\rm max}\geq \tau$. 
Let us suppose that $\tau _{\rm max}<\infty $. Since $f$ is non decreasing,
we deduce
\begin{align*}
u( t,x)  
&\leq T_{K_0( t) }\|u_0\|_{\infty}( x) +\int_0^{t}h( s) T_{K(
s,t) }f(\|u( s) \|_{\infty })ds \\
&=\|u_0\|_{\infty }+\int_0^{t}h( s) f(\|u(s) \|_{\infty })ds,\quad
 0\leq t<\tau _{\rm max},\ x\in \mathbb{R}^{d}.
\end{align*}
Then
\[
\|u( t) \|_{\infty }\leq \|u_0\|_{\infty}+\int_0^{t}h( s) f(\|u( s) \|_{\infty })ds.
\]
Now let us consider the integral equation
\[
y( t) =\|u_0\|_{\infty }+\int_0^{t}h( s) f(y( s) )ds,
\]
whose solution $y$ is given by
\[
y(t)=F_{\|u_0\|_{\infty }}^{-1}(H(t)),\quad 0\leq t<\infty ,
\]
where we have used that
 $\operatorname{im}(H)\subseteq \operatorname{im}(F_{\|u_0\|_{\infty }})$. By
the Comparison Theorem (see \cite{C-M-V}) we have
\[
\|u( t) \|_{\infty }\leq F_{\|u_0\|_{\infty }}^{-1}(H(t)),\quad
0\leq t<\tau _{\rm max}.
\]
The continuity of $p$ (property (d) in Lemma \ref{pdae}) and the
Bounded Convergence Theorem (notice that 
$p(s)  \leq  2^{d/\alpha }p(\tau _{\rm max})$,  
$\tau _{\rm max}/2  \leq  s  \leq   \tau _{\rm max}$, this is
consequence of (b) and (c) in Lemma \ref{pdae}) allow us
to take the limit $t\uparrow \tau _{\rm max}$ in \eqref{EcIntg},
\begin{align*}
u(\tau _{\rm max},x) &:= \lim_{t\uparrow \tau _{\rm max}}u(t,x) \\
&= T_{K_0(\tau _{\rm max})}u_0(x)+\int_0^{\tau _{\rm max}}h(s)T_{K(s,\tau
_{\rm max})}f(u(s,\cdot ))ds \\
&\leq  F_{\|u_0\|_{\infty }}^{-1}(H(\tau _{\rm max})).
\end{align*}
From this, the measurability of $u(\tau _{\rm max},\cdot )$ follows easily.
Also that
\[
0<\|u_0\|_{\infty }\leq \|u(\tau _{\rm max})\|_{\infty }
\leq F_{\|u_0\|_{\infty }}^{-1}(H(\tau _{\rm max})). 
\]
Accordingly we can consider the equation \eqref{EcIntg} with initial 
condition $u(\tau _{\rm max},\cdot)(\in L^{\infty }(\mathbb{R}^{d}))$. 
By the first part of the proof, the
solution $u$ can be extended beyond $\tau _{\rm max}$, contradicting the
definition of $\tau _{\rm max}$.
\end{proof}


The following estimates will be essentials in the proof of
Theorem \ref{TeoPrincipal}.

\begin{lemma} \label{Estpoabdprob}
Let $R>0$. There exists a $\delta =\delta (\alpha ,R)>0$
such that
\begin{equation}
\int_{|y|\leq R}p( t,y-x) dy\geq c_1,\quad \text{for all }
0<t\leq \delta ,\; |x|\leq R,  \label{estdens}
\end{equation}
where $c_1=c_1(d,\alpha )$ is a positive constant.
\end{lemma}

\begin{proof}
The change of variable theorem and the scaling property of $p$ imply
\begin{align*}
\int_{|y|\leq R}p( t,y-x) dy 
&= \int_{|y+x|\leq R}p(t,y) dy \\
&= t^{-d/\alpha }\int_{|y+x|\leq R}p(1,t^{-1/\alpha }y)dy \\
&= \int_{|t^{1/\alpha }y+x|\leq R}p( 1,y) dy \\
&= \int_{|y+t^{-1/\alpha }x|\leq t^{-1/\alpha }R}p( 1,y) dy.
\end{align*}
We are going to set $\delta =R^{\alpha }$ and
\[
\tilde{x}=\begin{cases}
0, & x=0, \\
-\frac{1}{2|x|}x, & x\neq 0.
\end{cases}
\]
If $0<t\leq \delta $ and $|x|\leq R$, then
\[
B_{1/2}(\tilde{x})\subset B_1(0)\cap B_{t^{-1/\alpha }R}(-t^{-1/\alpha }x).
\]
Therefore
\begin{align*}
\int_{|y|\leq R}p( t,y-x) dy 
&\geq \int_{B_{1/2}(\tilde{x})}p( 1,y) dy \\
&\geq \inf_{z\in B_1(0)}p( 1,z) \int_{B_{1/2}(\tilde{x})}dy \\
&= \operatorname{Vol}(B_{1/2}(0))\inf_{z\in B_1(0)}p( 1,z) ,
\end{align*}
where $\operatorname{Vol}(B_{1/2}(0))$ is the volume of the ball 
$B_{1/2}(0)$.
\end{proof}


Let $1\leq q<\infty $. We will consider the real-valued function
\begin{equation}
\phi _{\theta }(x)=\frac{1}{|x|^{\theta }}1_{B_1(0)\backslash \{0\}}(x),\quad
x\in \mathbb{R}^{d}.  \label{dfteta}
\end{equation}
Observe that
\begin{equation}
\phi _{\theta }\in L^{q}(\mathbb{R}^{d}), \quad \text{if }0<\theta <\frac{d}{q}.
\label{cdqto}
\end{equation}

\begin{lemma} \label{newpropaesta}
Let $0<t\leq 1$ and $|x|\leq t^{1/\alpha }/2$, then
\begin{equation}
T_{t}\phi _{\theta }(x)\geq c_2t^{-\theta /\alpha },  \label{EsxabSmg}
\end{equation}
where $c_2=c_2(d,\alpha )$ is a positive constant.
\end{lemma}

\begin{proof}
By property (e) in Lemma \ref{pdae} we obtain
\[
\int_{|y|\leq 1}\frac{p( t,y-x) }{|y|^{\theta }}dy
\geq
c_0\int_{|y|\leq 1,|x-y|\leq t^{1/\alpha }}\frac{\min \{
t|y-x|^{-d-\alpha },t^{-d/\alpha }\} }{|y|^{\theta }}dy.
\]
Since $|x-y|\leq t^{1/\alpha }$ is equivalent to
\[
\frac{t}{|y-x|^{d+\alpha }}\geq \frac{1}{t^{d/\alpha }},
\]
we have
\[
T_{t}\phi _{\theta }(x)\geq c_0t^{-d/\alpha }\int_{|y|\leq 1,|x-y|\leq
t^{1/\alpha }}\frac{dy}{|y|^{\theta }}.
\]
Moreover
\begin{align*}
\int_{|y|\leq 1,|x-y|\leq t^{1/\alpha }}\frac{dy}{|y|^{\theta }} 
&\geq\int_{|y|\leq t^{1/\alpha }-|x|}\frac{dy}{|y|^{\theta }} \\
&= c\int_0^{t^{1/\alpha }-|x|}r^{d-1-\theta }dr
\geq ct^{d/\alpha -\theta /\alpha }.
\end{align*}
In the last inequality we have used \eqref{cdqto} and 
$|x|\leq t^{1/\alpha}/2$.
\end{proof}

\section{Instantaneous blow-up}

Let us consider the function $\phi _{\theta }$ defined in \eqref{dfteta},
with $d>q\theta >0$. We choose
\[
r_1:=\|\phi _{\theta }\|_{q}+1>\|\phi _{\theta }\|_{q},
\]
and define recursively the sequence $(r_i)$ as
\[
r_{i+1}:=2\Big\{ r_i+\Big( \int_0^{K_0^{-1}( (
r_i/c_2) ^{-\alpha /\theta }) }h(s)K_0(s)^{d/\alpha
}ds\Big) ^{-1}\Big( K_0( \frac{1}{r_i}) \Big)^{-2}\Big\} ,
\]
for $i=2,3,\dots$. Inasmuch as $r_{i+1}>2r_i$, then $(r_i)$ is strictly
increasing and $r_i\uparrow \infty $. We define
\[
\tilde{f}(r_i)=\big( \frac{r_{i+1}}{2}-r_i\big) K_0( \frac{1}{r_i}) ,
\]
for $i=1,2,\dots$. Notice that
\begin{align*}
\tilde{f}(r_i) 
&= \Big( \int_0^{K_0^{-1}( (
r_i/c_2) ^{-\alpha /\theta }) }h(s)K_0(s)^{d/\alpha
}ds\Big) ^{-1}\Big( K_0( \frac{1}{r_i}) \Big) ^{-1} \\
&\leq \Big( \int_0^{K_0^{-1}( ( r_{i+1}/c_2)
^{-\alpha /\theta }) }h(s)K_0(s)^{d/\alpha }ds\Big) ^{-1}\Big(
K_0( \frac{1}{r_{i+1}}) \Big) ^{-1}
=\tilde{f}(r_{i+1}),
\end{align*}
then we can define $\tilde{f}:[0,\infty )\to [ 0,\infty )$ as
\begin{equation}
\tilde{f}(x)=\begin{cases}
\frac{\tilde{f}(r_1)}{r_1}x, & 0\leq x\leq r_1, \\
\tilde{f}(r_i), & r_i\leq x\leq \frac{1}{2}r_{i+1}, \\
\text{linear interpolation}, & \frac{1}{2}r_{i+1}\leq x\leq r_{i+1}.
\end{cases}   \label{deffam}
\end{equation}
In this way $\tilde{f}$ is non-decreasing. Also, from the definition of 
$\tilde{f}$ follows that it is locally Lipschitz. Moreover,
\begin{align*}
\int_{\|\phi _{\theta }\|_{q}}^{\infty }\frac{ds}{\tilde{f}(s)} 
&\geq \sum_{i=1}^{\infty }\int_{r_i}^{\frac{1}{2}r_{i+1}}\frac{1}{\tilde{
f}(r_i)}ds \\
&= \sum_{i=1}^{\infty }\Big( K_0( \frac{1}{r_i})\Big) ^{-1}=\infty ,
\end{align*}
because of Hypothesis (H3)(b). In this way, the function $\tilde{f}$ 
satisfies  Hypothesis (H2). Also, the generalized
Osgood's condition is satisfied for all continuous functions 
$h:(0,\infty)\to (0,\infty )$ because 
$\int_{\|\phi _{\theta }\|_{q}}^{\infty }\frac{ds}{\tilde{f}(s)}=\infty $ 
($1\leq q<\infty $).

Our main result reads as follows.

\begin{theorem} \label{TeoPrincipal}
Let us suppose $q\in [1,\infty )$, $\tilde{f}$ is
as in \eqref{deffam} and  Hypotheses {\rm (H3), (H3)(b)}. 
If $0<\theta <d/q$, and $u_0=\phi _{\theta }$, then 
equation \eqref{edif} possesses no local $p$-integrable solution for all
 $p\geq 1$.
\end{theorem}

\begin{proof}
We proceed by contradiction. Suppose that there are $p\geq 1,$ $R>0$ and 
$\tau >0$, such that \eqref{EcIntg} has an $p$-integrable solution 
$u(t,\cdot)$ on $B_{R}(0),$ for all $t<\tau $, with $u_0=\phi _{\theta }$. 
Let us take a $t<\min \{ \tau ,K_0^{-1}(1),K_0{}^{-1}(\delta
),K_0{}^{-1}((2R)^{\alpha })\} $, where $\delta $ is given in Lemma
\ref{Estpoabdprob}.

We observe that \eqref{EcIntg} implies
\[
u(s,x)\geq T_{K_0(s)}\phi _{\theta }(x),\quad \text{for all }s\geq 0,\;
 x\in \mathbb{R}^{d}.
\]
From ($\ref{EcIntg}$), Jensen's inequality, and the above estimation we obtain
\begin{align*}
& \int_{|x|\leq R}u(t,x)^{p}dx \\
& \geq \int_{|x|\leq R}( \int_0^{t}h(s)\int_{\mathbb{R}
^{d}}p(K(s,t),y-x)\tilde{f}( T_{K_0(s)}\phi _{\theta }(y))
dy\,ds) ^{p}dx \\
& \geq cR^{d(1-p)}( \int_{|x|\leq R}\int_0^{t}h(s)\int_{\mathbb{R}
^{d}}p(K(s,t),y-x)\tilde{f}( T_{K_0(s)}\phi _{\theta }(y))
dy\,ds\,dx) ^{p}.
\end{align*}

Denoting the set $\{y:T_{K_0(s)}\phi _{\theta }(y)\geq r_i\}$ by 
$A_i$, and using that $\tilde{f}$ is non-decreasing we obtain
\begin{align*}
& \int_{|x|\leq R}u(t,x)^{p}dx \\
& \geq c( \int_{|x|\leq R}\int_0^{t}h(s)\int_{A_i}p(K(s,t),y-x)
\tilde{f}( T_{K_0(s)}\phi _{\theta }(y)) dy\,dsdx) ^{p} \\
& \geq c( \int_0^{t}h(s)\int_{A_i}\int_{|x|\leq R}p(K(s,t),y-x)dx
\tilde{f}(r_i)dy\,ds) ^{p}.
\end{align*}
Since $r_i\uparrow \infty $, there exists $i_0\in \mathbb{N}$ such that
\begin{equation}
t_i:=K_0^{-1}\Big( ( \frac{r_i}{c_2}) ^{-\alpha /\theta}\Big) 
\leq t,\quad  \text{for all } i\geq i_0.  \label{consobti}
\end{equation}
If $0\leq s\leq t_i$, the inequality \eqref{consobti} yields
\[
r_i<c_2K_0(s)^{-\theta /\alpha }.
\]
Moreover, if $|y|\leq K_0(s)^{1/\alpha }/2$, then \eqref{EsxabSmg} implies
\[
r_i<T_{K_0(s)}\phi _{\theta }(y).
\]
Hence,
\begin{equation}
A_i=\{ y:T_{K_0(s)}\phi _{\theta }(y)\geq r_i\} \supset
\{ y:|y|\leq 2^{-1}K_0(s)^{1/\alpha }\} .  \label{contecondemp}
\end{equation}
On the other hand, for $s<t$, we have $K(s,t)\leq K_0(t)\leq \delta $ and 
$2^{-1}K_0(s)^{1/\alpha }\leq 2^{-1}K_0(t)^{1/\alpha }<R$ which indicate that
we are able to use \eqref{estdens}. Then
\begin{equation}
\int_{|x|\leq R}p(K(s,t),y-x)dx\geq c_1,\quad \text{for all }
|y|\leq 2^{-1}K_0(s)^{1/\alpha }.  \label{estpanue}
\end{equation}
Using \eqref{contecondemp} and \eqref{estpanue} we obtain
\begin{align*}
\Big( \int_{|x|\leq R}u(t,x)^{p}dx\Big) ^{1/p} 
&\geq c\tilde{f} (r_i)\int_0^{t_i}h(s)
 \int_{\{ |y|\leq 2^{-1}K_0(s)^{1/\alpha }\} }dy\,ds \\
&= c\tilde{f}(r_i)\int_0^{t_i}h(s)K_0(s)^{d/\alpha }ds \\
&= c\Big( \int_0^{1/r_i}k(s)ds\Big) ^{-1}.
\end{align*}
Letting $i\to \infty $, $r_i\uparrow \infty $, therefore 
(H3)(b) implies $\int_{|x|\leq R}u(t,x)^{p}dx=\infty $, $t<\tau $. The
contradiction obtained proves the result.
\end{proof}

\subsection*{Acknowledgment}
This work was partially supported by the grant PIM16-4 of Universidad Aut\'onoma 
de Aguascalientes.
The author wants to thank the anonymous referee for his/her very careful review
of the original manuscript and  for the helpful comments.

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\end{document}
