# A nonlinear diffusion equation with reaction localized to the half-line

R. Ferreira and A. de Pablo  Raúl Ferreira Departamento de Matemáticas, U.
Complutense de Madrid, 28040 Madrid, Spain. e-mail<EMAIL_ADDRESS>Arturo de Pablo Departamento de Matemáticas, U. Carlos III de Madrid, 28911
Leganés, Spain. e-mail<EMAIL_ADDRESS>
###### Abstract.

We study the behaviour of the solutions to the quasilinear heat equation with
a reaction restricted to a half-line

$u_{t}=(u^{m})_{xx}+a(x)u^{p},$

$m,p>0$ and $a(x)=1$ for $x>0$, $a(x)=0$ for $x<0$. We first characterize the
global existence exponent $p_{0}=1$ and the Fujita exponent $p_{c}=m+2$. Then
we pass to study the grow-up rate in the case $p\leq 1$ and the blow-up rate
for $p>1$. In particular we show that the grow-up rate is different as for
global reaction if $p>m$ or $p=1\neq m$.

_In memoriam of our friend Ireneo Peral. Master of Mathematics_

##  1\. Introduction

We consider the following Cauchy problem

(1.1)
$\left\\{\begin{array}[]{ll}u_{t}=(u^{m})_{xx}+a(x)u^{p},&x\in\mathbb{R},\;t>0,\\\
u(x,0)=u_{0}(x).\end{array}\right.$

We take exponents $m,p>0$ and the coefficient is the characteristic function
of a half-line, $a(x)=\mathds{1}_{(0,\infty)}(x)$. The initial value $u_{0}\in
L^{1}(\mathbb{R})\cap L^{\infty}(\mathbb{R})$ is assumed to be continuous and
nonnegative, so that nonnegative solutions $u\geq 0$ are considered. We are
interested in characterizing and describing the phenomena of blow-up and grow-
up for the solutions to (1.1) in terms of the parameters of the problem, the
exponents $m$ and $p$ and the initial datum $u_{0}$. By a solution $u$ having
blow-up we mean that there exists a finite time $T$ such that $u$ is well
defined and finite for $t<T$ and

(1.2) $\lim_{t\to T^{-}}\|u(\cdot,t)\|_{\infty}=\infty.$

When $T=\infty$ we say that $u$ has grow-up.

The problem with global reaction $a(x)=1$ has been deeply studied in the last
years mainly concerning blow-up and $p>1$, see for instance the survey book
[14], but also in relation to grow-up, and thus $p\leq 1$, see [1, 12]. In
fact there can exist blow-up solutions only if $p>1$, and in that range small
initial data produce global solutions if and only if $p>m+2$. The global
solutions are unbounded if $p\leq 1$, i.e. they have grow-up, while they are
globally bounded if $p>m+2$. The exponents $p_{0}=1$ and $p_{c}=m+2$ are
called, respectively, global existence exponent and Fujita exponent. For the
related case in which the reaction coefficient is
$a(x)=\mathds{1}_{(-L,L)}(x)$, $0<L<\infty$, the exponents are
$p_{0}=\max\\{1,\frac{m+1}{2}\\}$ and $p_{c}=m+1$, see [2, 4, 13].

The first result in the paper establishes precisely for which exponents and
data we have such phenomena of blow-up or grow-up. We prove that the exponents
are the same as for the case $a(x)=1$.

######  Theorem 1.1.

  1. a)

If $0<p\leq p_{0}=1$ all the solutions to problem (1.1) are globally defined
and unbounded.

  2. b)

If $1<p\leq p_{c}=m+2$ all the solutions blow up in finite time.

  3. c)

If $p>m+2$ solutions may blow up in finite time or not depending on the
initial data. Global solutions are bounded.

The second question to deal with is the speed at which the unbounded solutions
tend to infinity, both in the grow-up and in the blow-up cases. For global
solutions we show that these rates are not the natural ones given by the
corresponding no diffusion ODE (2.7). This in fact gives an upper estimate of
the grow-up rate by comparison,

(1.3) $u(x,t)\leq\left\\{\begin{array}[]{ll}ct^{\frac{1}{1-p}},&p<1,\\\
ce^{t},&p=1.\end{array}\right.$

We remark that when $p<1$ the reaction function is not Lipschitz, and
uniqueness does not necessarily hold, neither comparison, see [1, 12]. In that
case we can use for comparison a maximal solution or a minimal solution, [12].

In the case of global reaction $a(x)=1$, it is proved in [1, 11] that the
above is indeed the grow-up rate when $0<p<1$, that is

(1.4) $u(x,t)\sim t^{\frac{1}{1-p}}$

uniformly in compact sets. By $f\sim g$ we mean $0<c_{1}\leq f/g\leq
c_{2}<\infty$.

However, for $p=1$ it is well known, through and easy change of variables that
eliminates the reaction, that $u(x,t)\sim t^{-1/2}e^{t}$ if $m=1$ and

$u(x,t)\sim e^{\gamma t},\qquad\gamma=\min\\{1,\frac{2}{m+1}\\},$

when $m\neq 1$, for $t$ large uniformly in compact sets of $\mathbb{R}$, see
[15] .

On the other hand, when $a(x)=\mathds{1}_{(-L,L)}$ it is proved in [3] that
estimate (1.3) is far from being sharp in most of the cases. In particular

$u(x,t)\sim\left\\{\begin{array}[]{ll}t^{\frac{1}{m+1-2p}},&\text{if }p\leq
1<m,\\\ t^{\frac{1}{1-p}},&\text{if }m\leq p<1,\\\ e^{t},&\text{if
}m<p=1,\end{array}\right.$

uniformly in compact sets in the first case, only for $|x|<L$ in the last two
cases. For $|x|>L$ the rate is different in the case $p>m$, namely

$u(x,t)\sim t^{\frac{1}{1-m}}.$

In the limit case of linear diffusion and linear reaction, $m=p=1$, it holds

$\lim_{t\to\infty}\frac{\log u(x,t)}{t}=\omega,$

where $\omega=\omega(L)\in(0,1)$, $\lim_{L\to\infty}\omega(L)=1$.

For our problem (1.1) we show that the rate is the same as for global reaction
only if $p\leq m$ with $p<1$ or $p=m=1$; it is the same as for
$a(x)=\mathds{1}_{(-L,L)}$ if $p>m$, and strictly in between of those two
problems if $p=1<m$. Again the rate is different for $p>m$ inside or outside
the support of the reaction coefficient $a(x)$.

######  Theorem 1.2.

Let $u$ be a solution to problem (1.1) with $p=1$.

  1. a)

If $m>1$ then $u(x,t)\sim e^{\alpha t}$ uniformly in compact sets of
$\mathbb{R}$, where $\alpha\in(1/m,2/(m+1))$ depends on the behaviour of
$u_{0}$ at infinity.

  2. b)

If $m=1$ then $u(x,t)\sim e^{t}$ uniformly in compact sets of $\mathbb{R}$.

  3. c)

If $m<1$ then $u(x,t)\sim e^{t}$ uniformly in compact sets of $\mathbb{R}^{+}$
and $u(x,t)\sim t^{\frac{1}{1-m}}$ uniformly in compact sets of
$\mathbb{R}^{-}$, provided
$u_{0}(x)\sim|x|^{\frac{-2}{1-m}}\left(\log|x|\right)^{\frac{1}{1-m}}$ for
$x\sim-\infty$.

######  Theorem 1.3.

Let $u$ be a solution to problem (1.1) with $p<1$.

  1. a)

If $m\geq p$ then $u(x,t)\sim t^{\frac{1}{1-p}}$ uniformly in compact sets of
$\mathbb{R}$.

  2. b)

If $m<p$ then $u(x,t)\sim t^{\frac{1}{1-p}}$ uniformly in compact sets of
$\mathbb{R}^{+}$ and $u(x,t)\sim t^{\frac{1}{1-m}}$ uniformly in compact sets
of $\mathbb{R}^{-}$, provided $u_{0}(x)\sim|x|^{\frac{-2}{1-m}}$ for
$x\sim-\infty$.

We compare in the next table the grow-up rates for the three problems, with
global recation $a(x)=1$, localized reaction $a(x)=\mathds{1}_{(-L,L)}$, and
reaction confined to the half-line $a(x)=\mathds{1}_{(0,\infty)}$.

| $\mathbf{p=1}$ | $\mathbf{p<1}$
---|---|---
$m>1$ | $m=1$ | $m<1$ | $m>p$ | $m=p$ | $m<p$
$\mathbb{R}$ | $e^{ct}$ | $e^{t}$ | $e^{t}$ | $t^{a}$ | $t^{a}$ | $t^{a}$
$(-L,L)$ | $t^{a}$ | $e^{\omega t}$ | $e^{t}\,/\,t^{b}$ | $t^{d}$ | $t^{d}$ | $t^{a}\,/\,t^{b}$
$(0,\infty)$ | $e^{\alpha t}$ | $e^{t}$ | $e^{t}\,/\,t^{b}$ | $t^{a}$ | $t^{a}$ | $t^{a}\,/\,t^{b}$

The exponents are

$\begin{array}[]{l}a=\dfrac{1}{1-p},\quad b=\dfrac{1}{1-m},\quad
c=\dfrac{2}{m+1},\quad d=\dfrac{1}{m+1-2p},\\\ \omega<1\;\text{ depends on
}L,\\\ \alpha\leq\alpha^{*}(m)<c\;\text{ depends on the behaviour of $u_{0}$
at infinity}.\end{array}$

In the case $p>m$ we have two different rates, inside or outside the support
of $a(x)$.

As for blow-up, the rate at which the solutions approach infinity in a finite
time has been studied for the case of global reaction under different
conditions on the initial datum and exponents, with special care in the
multidimensional case, see [14] and the references therein. For dimension one,
as is our situation, any solution with blow-up at time $t=T$ satisfies, for
$t$ close to $T$,

$\|u(\cdot,t)\|_{\infty}\sim(T-t)^{-\frac{1}{p-1}}.$

For localized reaction $a(x)=\mathds{1}_{(-L,L)}$ the rates have been
established in [2, 4], giving a different rate depending on $p$ being bigger
or smaller than $m$,

$\|u(\cdot,t)\|_{\infty}\sim(T-t)^{-\gamma},\qquad\gamma=\max\\{\frac{1}{p-1},\frac{1}{2p-m-1}\\}.$

In addition the property $\partial_{t}u\geq 0$ is required in the proof of
this result.

We prove here for problem (1.1) that the rate is the same as for global
reaction, assuming again $\partial_{t}u\geq 0$, but this is required only
above the Fujita exponent, i.e. for $p>m+2$.

######  Theorem 1.4.

Let $u$ be a solution to problem (1.1) with $p>1$ such that becomes infinity
for $t\to T^{-}$, and assume further that $\partial_{t}u\geq 0$ if $p>m+2$.
Then

(1.5) $\|u(\cdot,t)\|_{\infty}\sim(T-t)^{-\frac{1}{p-1}}.$

We end the description of solutions of problem (1.1) by studying the set where
the solution tends to infinity, the blow-up set

$B(u)=\\{x\in\mathbb{R}\,:\,\exists\,x_{j}\to x,\;t_{j}\to
T,\;u(x_{j},t_{j})\to\infty\\}.$

In the global reaction case it has been proved the three possibilities
according to the reaction exponent: single point blow-up, $B(u)$ is a discrete
set, if $p>m$; regional blow-up, $B(u)$ is a compact set of positive measure,
if $p=m$; and global blow-up, $B(u)=\mathbb{R}$, if $p<m$. See again [14]. The
same happens for localized reaction $a(x)=\mathds{1}_{(-L,L)}$, at least for
$m>1$ and symmetric nondecreasing initial values, see [4]. In our case we
prove that the same happens, and we additionally show where this blow-up set
can lie in the case where the blow-up is not the whole line. To do that we
assume in the case $p\geq m$ that there exists some point $x_{0}$ for which
the blow-up rate (1.5) holds, i.e.

(1.6) $u(x_{0},t)\geq c(T-t)^{-\frac{1}{p-1}}.$

######  Theorem 1.5.

Let $u$ be a blow-up solution to problem (1.1), with compactly supported
initial datum. Assume also (1.6). We have for the blow-up set $B(u)$:

  1. a)

if $p>m$ then $B(u)\subset\mathbb{R}^{+}$. Moreover if $m>1$ it is bounded;

  2. b)

if $p=m$ then $B(u)$ is bounded with nontrivial measure;

  3. c)

if $p<m$ then $B(u)=\mathbb{R}$.

We remark that due to the lack of symmetry in the problem it is not clear the
existence of the point assumed in the statement. In general we can prove that
$B(u)=[x_{1},\infty)$ for some $-\infty\leq x_{1}<\infty$ if $p<m$, and $B(u)$
is bounded if $p=m$.

Organization of the paper: We characterize the critical exponents, Theorem
1.1, in Sections 2 and 3. The grow-up rates, Theorems 1.2 and 1.3 are proved
in Section 4, while the blow-up rates, Theorem 1.4 is proved in Section 5.
Finally we devote Section 6 to describe the blow-up sets, Theorem 1.5.

##  2\. Blow-up versus global existence

We prove in this section that the global existence exponent is $p_{0}=1$.
First it is obvious that if $0<p\leq 1$ every solution to problem (1.1) is
global. Just use comparison with the flat supersolution

(2.7) $U^{\prime}=U^{p},\qquad U(0)=\|u_{0}\|_{\infty}.$

######  Remark 2.1.

Though in the case $p<1$ there is in general no uniqueness, and therefore no
comparison (the reaction is not Lipschitz), we always can compare with a
supersolution which is a maximal solution of the equation, like the function
$U$ in (2.7) is, see [12].

In order to complete the proof of the first item in Theorem 1.1 we observe
that all the solutions have grow-up if $p\leq 1$.

######  Lemma 2.1.

Let $u$ be a solution of (1.1). If $p\leq 1$ then

$u(x,t)\to\infty$

uniformly in compact sets.

###### Proof.

We only note that this occurs for the solutions to the problem if the reaction
is localized in a bounded interval, $a(x)=\mathds{1}_{(-L,L)}$, see [3], and
any solution to that problem (translated) is a subsolution to our problem. ∎

We now show that for $p>1$ there exist solutions that blow up in finite time
provided the initial value is large in some sense.

######  Lemma 2.2.

If $p>\max\\{m,1\\}$ problem (1.1) has blow-up solutions.

###### Proof.

We observe that $u$ is a supersolution to the Dirichlet problem

$\left\\{\begin{array}[]{ll}w_{t}=(w^{m})_{xx}+w^{p},&x\in(A,B),\,t>0\\\
w(A,t)=w(B,t)=0,\\\ w(x,0)=w_{0}(x),\end{array}\right.$

for any interval $(A,B)\subset(0,\infty)$. Use then the results in [14]. ∎

######  Lemma 2.3.

If $1<p\leq m$ there exist blow-up solutions.

###### Proof.

We construct a self-similar subsolution

$\underline{u}(x,t)=(T-t)^{-\alpha}f(\xi)\qquad\xi=x(T-t)^{-\beta},$

satisfying $u(0,t)=0$. The self-similar exponents are given by

$\alpha=\frac{1}{p-1},\qquad\beta=\frac{p-m}{2}\alpha,$

and the self-similar profile satisfies

(2.8) $(f^{m})^{\prime\prime}-\beta\xi f^{\prime}+f^{p}-\alpha f=0,\qquad
f(0)=0.$

Using $(f^{m})^{\prime}(0)=\mu$ as shooting parameter we claim that there
exists some $\mu_{0}>0$ such that the corresponding profile $f_{0}$ satisfies

$f_{0}(\xi)>0\quad\mbox{in }(0,\xi_{0})\quad\mbox{and }\quad
f_{0}(\xi_{0})=0,$

for some $\xi_{0}>0$. This gives the desired blow-up subsolution with profile

$f(\xi)=\left\\{\begin{array}[]{ll}f_{0}(\xi),&\xi\in(0,\xi_{0})\\\
0,&\mbox{otherwise}\end{array}\right.$

Then, if $u_{0}(x)>\underline{u}(x,0)$ the solution of (1.1) blows up.

In order to prove the claim we argue by contradiction, assuming that for every
large $\mu$ the corresponding profiles $f_{\mu}$ are positive in $(0,\infty)$.
Given any of such profiles with $\mu>1$ we take $k=\mu^{\frac{p+m}{2}}$ and
consider the function

$g_{k}(\xi)=\frac{1}{k^{m}}f^{m}(k^{\frac{m-p}{2}}\xi).$

It satisfies the initial value problem

$\left\\{\begin{array}[]{l}g_{k}^{\prime\prime}+g_{k}^{p/m}-k^{1-p}\left(\beta\xi(g_{k}^{1/m})^{\prime}-\alpha
g_{k}^{1/m}\right)=0,\qquad\xi>0,\\\ g_{k}(0)=0,\\\
g_{k}^{\prime}(0)=1.\end{array}\right.$

We define the energy of the system at a point $\xi$ as

$E(\xi)=\frac{1}{2}(g_{k}^{\prime})^{2}+V(g_{k}),\qquad
V(s)=\frac{m}{p+m}s^{\frac{p+m}{m}}-\frac{\alpha
m}{1+m}k^{1-p}s^{\frac{m+1}{m}}.$

Multiplying the equation by $g_{k}^{\prime}$ we get that

$E^{\prime}(\xi)=\frac{\beta}{m}k^{1-p}\xi
g_{k}^{\frac{1-m}{m}}(g_{k}^{\prime})^{2}\leq 0,$

since $\beta\leq 0$. Thus,

$E(\xi)\leq E(0)=\frac{1}{2}.$

Also, calculating the minimum of the potential $V$ we have

$E(\xi)\geq-ck^{-(p+m)}\geq-c.$

Since $p>1$ this implies that there exists two constants $C_{1},C_{2}$
depending on $m$ and $p$ such that

$0\leq g_{k}\leq C_{1},\qquad|g_{k}^{\prime}|\leq C_{2}.$

Hence, letting $k\to\infty$ we have that $g_{k}$ converges uniformly in
compact sets to a non negative function $G$. It is clear that $G$ satisfies

$\left\\{\begin{array}[]{l}G^{\prime\prime}+G^{p/m}=0,\qquad\xi>0,\\\
G(0)=0,\\\ G^{\prime}(0)=1.\end{array}\right.$

However the solution of the above problem crosses the axis at some finite
point with non-zero slope. This is a contradiction and the claim is proved. ∎

##  3\. Fujita exponent

In this section we prove that the Fujita exponent is $p_{c}=m+2$, that is, all
solutions blow up if $1<p\leq m+2$, and if $p>m+2$ not all solutions do so. In
this last range $p>m+2$, it is easy to see that small initial data produce
global solutions, by comparison with the global supersolutions corresponding
to the case $a(x)=1$, see for instance the book [14]. In fact they tend to
zero for $t\to\infty$.

We divide the proof of blow-up below $p_{c}$ in three cases, $1<p\leq m$,
$m<p<m+2$ and $p=m+2$, the most difficult case being the last one.

######  Lemma 3.1.

If $1<p\leq m$ then all solutions blow up in finite time.

###### Proof.

We only have to check that the self-similar subsolution constructed in Lemma
2.3 can be put below any solution if we let pass enough time.

  1. a)

It is clear when $p<m$ that we can do it since $\underline{u}(x,0)$ is small
taking $T$ large, as well as its support is small, due to the fact that
$\beta<0$.

  2. b)

For $p=m$ we note that $\underline{u}(x,0)$ is still small if $T$ is large but
it has a fixed support $[0,\xi_{0}]$ since $\beta=0$. Nevertheless, using the
penetration property of the solutions of the porous medium equation we obtain
that there exists $t_{0}>0$ such that the support of $u(\cdot,t_{0})$ contains
any interval.

∎

######  Lemma 3.2.

If $m<p<m+2$ then all solutions blow up in finite time.

###### Proof.

The proof is the same as for the global reaction and is an easy consequence of
the energy argument of [10], also called concavity argument. In fact, defining
the energy of a function $v$ as

(3.9)
$E_{v}(t)=\dfrac{1}{2}\int_{-\infty}^{\infty}|(v^{m})_{x}|^{2}-\frac{m}{p+m}\int_{0}^{\infty}v^{p+m},$

we have that if for a solution $u$ to (1.1) there exists some $t_{0}$ such
that $E_{u}(t_{0})<0$ then $u$ blows up in finite time. Now we consider the
Barenblatt function

(3.10)
$B(x,t;D)=t^{-\frac{1}{m+1}}\left(D-kx^{2}t^{-\frac{2}{m+1}}\right)_{+}^{\frac{1}{m-1}},$

where $k=\frac{m-1}{2m(m+1)}$, $D>0$. It is a subsolution to our equation and
it satisfies, for some constants $c_{1},c_{2}$ depending only on $m$, $p$ and
$D$,

$E_{B}(t)=c_{1}t^{-\frac{2m+1}{m+1}}-c_{2}t^{-\frac{p+m-1}{m+1}},$

which is negative for $t$ large provided $p<m+2$. The final step is a standard
comparison argument: we make $B(x,1;D)$ small by taking $D$ small, so that it
can be put below $u_{0}$; this implies $u(x,t)\geq B(x,t+1;D)$ for $t>0$; let
$t_{1}$ be such that $E_{B}(t_{1})<0$; let $v$ be the solution corresponding
to the initial value $B(\cdot,t_{1};D)$, which by the above energy argument
blows up in finite time; since $u\geq v$ so does $u$. In the case $m<1$ we
need the behaviour at infinity of every solution, see [9], since the function
(3.10) is positive, while for $m=1$ a Gaussian is used instead of a Barenblatt
function. ∎

We observe that the fact that the integral in the reaction term is performed
only in $(0,\infty)$ does not affect the original argument. In [4] we used the
fact that the integral in $(0,L)$ produces a different time power term if $L$
is finite, and so the Fujita exponent is different in that case.

######  Lemma 3.3.

If $p=m+2$ then all solutions blow up in finite time.

###### Proof.

We use the method introduced in [7] to prove blow-up for the critical exponent
in the case $a(x)=1$, but here the nonsymmetry of the problem makes things
more involved. The argument goes like this: assuming by contradiction that the
solution is global, we rescale and pass to the limit in time, thus obtaining a
solution to some problem for which we prove nonexistence.

Let $u$ be a global solution, and let $t_{0}\geq 1$ and $D$ be such that
$u(x,t_{0})\geq B(x,t_{0};D)$, where $B$ is given by (3.10) (if $m\neq 1$, for
$m=1$ we use instead a Gaussian like in the proof of Lemma 3.2). We define the
rescaled function

$v(\xi,\tau)=t^{\alpha}u(x,t),\qquad\xi=xt^{-\alpha},\quad\tau=\log
t,\quad\alpha=\frac{1}{m+1}.$

We have that $v$ is a solution, for $\tau>\tau_{0}=\log t_{0}$, of the
equation

(3.11) $v_{\tau}=(v^{m})_{\xi\xi}+\alpha(\xi v)_{\xi}+a(\xi)v^{m+2}.$

If $g$ is the solution to equation (3.11) with $g(\xi,\tau_{0})=B(\xi,1;D)$,
by comparison we have that $v\geq g$ for every $\tau>\tau_{0}$, and in
particular $g$ is globally defined in $\tau$. For the special form of the
initial value, it is easy to see that $g$ is nondecreasing in $\tau$, and
therefore there exists the limit

$\lim_{\tau\to\infty}g(\xi,\tau)=f(\xi)\in[0,\infty].$

We claim the following alternative:

$a)$ $f$ is locally bounded. Thus we can pass to the limit in (3.11), by means
of a Lyapunov functional, to get that $f$ is a positive solution of

(3.12) $(f^{m})^{\prime\prime}+\alpha(\xi
f)^{\prime}+\rho(\xi)f^{m+2}=0\qquad\xi\in\mathbb{R},$

see [7]. Now we observe that the function

$\mathcal{E}(\xi)=(f^{m})^{\prime}+\alpha\xi f$

satisfies $\mathcal{E}^{\prime}(\xi)=-\rho(\xi)f^{m+2}(\xi)$, so it is
constant for $\xi<0$ and decreasing for $\xi>0$. Then, if we assume
$\mathcal{E}(0)=\mathcal{E}_{0}>0$, we have that

$(f^{m})^{\prime}(\xi)\geq\mathcal{E}(\xi)\geq\mathcal{E}_{0},\qquad\xi<0.$

This implies that there exists a point $\xi_{1}<0$ such that $f(\xi_{1})=0$
and $(f^{m})^{\prime}(\xi_{1})\neq 0$. Therefore $\mathcal{E}(0)\leq 0$, and
there exists some $\xi_{2}>0$ such that
$\mathcal{E}(\xi_{2})=\mathcal{E}_{2}<0$. Exactly as before

$(f^{m})^{\prime}(\xi)\leq\mathcal{E}(\xi)\leq\mathcal{E}_{2},\qquad\xi>\xi_{2},$

so there exists a point $\xi_{3}>\xi_{2}$ such that $f(\xi_{3})=0$,
$(f^{m})^{\prime}(\xi_{3})\neq 0$. This gives a contradiction and $f$ cannot
exist.

$b)$ There exists $\xi_{0}$ such that $f(\xi_{0})=\infty$. Then $g$ is large
in a nontrivial interval and this would imply that it blows up in a finite
time. This is again a contradiction, and the theorem would be proved.

We have that $f$ satisfies equation (3.12) in any interval in which it is
bounded. It is clear that $f$ cannot have any minima since at such a point we
would have from the equation $(f^{m})^{\prime\prime}<0$. This implies

(3.13)
$\lim\limits_{\xi\to\xi_{0}^{-}}f(\xi)=\limsup\limits_{\xi\to\xi_{0}^{-}}(f^{m})^{\prime}(\xi)=\infty.$

Assume $\xi_{0}>0$. If $f$ is bounded in some interval
$(\xi_{0}-\delta,\xi_{0})$, $\delta\leq\xi_{0}$, then $f$ is increasing in
that interval with

$(f^{m})^{\prime}(\xi)\leq\mathcal{E}(\xi_{0}-\delta/2),\qquad\xi_{0}-\delta/2<\xi<\xi_{0}.$

This is a contradiction and thus $f(\xi)=\infty$ for every
$0\leq\xi\leq\xi_{0}$. Moreover, if $f$ is bounded in $\xi<0$, we have

$(f^{m})^{\prime}(\xi)+\alpha\xi f(\xi)=c<0,$

by the above. Thus by (3.13), there is a sequence $\xi_{j}\to 0^{-}$ such that
$|\xi_{j}|f(\xi_{j})\to\infty$. The same argument works from the left to the
right, assuming $\xi_{0}<0$. In conclusion $f$ is large in some interval
$|\xi|\leq\xi_{*}$, that could be small, but it satisfies that
$\xi_{*}f(\xi_{*})$ is large.

Let us now show that in this situation the function $g$ blows up in finite
time. By the monotonicity of $g$ in time we have that for any large constant
$A_{*}>0$ there exists $M>0$, $\xi_{M}>0$ and $\tau_{M}$ such that
$M\xi_{M}\geq A_{*}^{3/2}$ and $g(\xi,\tau)\geq M$ for every
$|\xi|\leq\xi_{M}$, $\tau\geq\tau_{M}$. Now we argue as in [4]. Let
$z(x,t)=e^{-\alpha\tau}g(\xi,\tau)$ be the function $g$ in the original
variables, and define $h(x,t+e^{\tau_{M}})$ the solution of (1.1) with initial
datum

$W(x)=\lambda^{-1}(A-\lambda^{-2}x^{2})_{+}$

where

$A=(\xi_{M}M)^{2/3},\qquad\lambda=e^{\alpha\tau_{M}}(\xi_{M}^{2}/M)^{1/3}.$

It is clear that $W(x)\leq z(x,e^{\tau_{M}})$, since

$\begin{array}[]{l}z(x,e^{\tau_{M}})=e^{-\alpha\tau_{M}}g(\xi,\tau_{M})\geq
e^{-\alpha\tau_{M}}M\qquad\text{for }|x|\leq
e^{-\alpha\tau_{M}}\xi_{M},\\\\[5.69054pt] W(x)\leq
W(0)=\lambda^{-1}A=e^{-\alpha\tau_{M}}M,\\\\[5.69054pt]
supp(W)=\\{|x|\leq\lambda A^{1/2}\\}=\\{|x|\leq
e^{-\alpha\tau_{M}}\xi_{M}\\}.\end{array}$

Moreover,

$E_{h}(0)=\lambda^{-(2m+1)}A^{2m+1/2}\Big{(}c_{1}-c_{2}A^{2}\Big{)},$

for some $c_{1},c_{2}$ depending only on $m$. This is negative for
$A>A_{*}=A_{*}(m)$. Thus $h$ blows up in finite time, and by comparison $z$,
or which is the same $g$, also blows up. This ends the proof. ∎

##  4\. Grow-up rates

The aim of this section is to study the speed at which the global unbounded
(grow-up) solutions to problem (1.1) tend to infinity. We therefore consider
the range $p\leq 1$. In order to avoid nonuniqueness issues when $p<1$ we
assume in that case that the initial value is positive for $x>0$, that is
where the non-Lipschitz reaction applies.

As we have said in the Introduction, the upper estimate of the grow-up rate is
given by comparison with the function in (2.7). In the case of global reaction
$a(x)=1$ this is sharp if $p<1$ or $m<1$. In fact we have for $t$ large

(4.14) $u(x,t)\sim\left\\{\begin{array}[]{ll}t^{\frac{1}{1-p}},&p<1,\\\
e^{t}&m<1=p,\\\ t^{-\frac{1}{2}}e^{t},&m=1=p,\\\
e^{\frac{2}{m+1}t},&m>1=p,\end{array}\right.$

see [1, 11, 15].

On the other hand, when $a(x)=\mathds{1}_{(-L,L)}$ the rates are proved in
[3]. Though in that situation the global existence exponent is different,
$p_{0}=\max\\{1,\frac{m+1}{2}\\}$, we quote the results proved in [3] in our
range $p\leq 1$:

$i)$ if $p\leq 1<m$ then

(4.15) $u(x,t)\sim t^{\frac{1}{m+1-2p}},$

in compact sets.

$ii)$ if $m<p<1$ then

(4.16) $u(x,t)\sim\left\\{\begin{array}[]{ll}t^{\frac{1}{1-p}},&\text{for
}|x|<L,\\\ t^{\frac{1}{1-m}},&\text{for }|x|>L,\end{array}\right.$

provided that the initial datum satisfies

(4.17) $|x|^{2}u_{0}^{1-m}(x)\sim 1.$

$iii)$ if $m<p=1$ then

(4.18) $u(x,t)\sim\left\\{\begin{array}[]{ll}e^{t},&\text{for }|x|<L,\\\
t^{\frac{1}{1-m}},&\text{for }|x|>L,\end{array}\right.$

provided that the initial datum satisfies

(4.19) $|x|^{2}u_{0}^{1-m}(x)\sim\log(x).$

$iv)$ if $p=m=1$ then

(4.20) $\lim_{t\to\infty}\frac{\log u(x,t)}{t}=\omega(L)\in(0,1).$

We prove in this paper that for problem (1.1) the rate can be that
corresponding to global reaction or to reaction localized in a bounded
interval, or none of them, depending on the sign of $p-m$. We can also have a
different rate inside or outside the region where the reaction applies when
$p>m$, like in the case $a(x)=\mathds{1}_{(-L,L)}$.

###  4.1. Case $p=1$

Though the reaction is linear this is the more involved case. We consider
separately the three cases according to $m$ being larger, equal or smaller
than 1.

The proof of the grow-up rate follows by comparison with special selfsimilar
subsolutions and supersolutions. We construct such functions in the form

(4.21) $w(x,t)=e^{\alpha t}f(xe^{-\beta t}),$

where necessarily

$\beta=\frac{m-1}{2}\alpha.$

Also, by (4.14) we consider only $\alpha\leq 2/(m+1)$.

The profile $f$ will be given by matching two functions,

(4.22) $f(\xi)=\left\\{\begin{array}[]{ll}\psi(\xi),&\xi\geq 0,\\\
\phi(-\xi),&\xi\leq 0,\end{array}\right.$

where $\psi$ and $\phi$ are the truncation by zero of the solutions of the
initial value problems, for some $\lambda\in\mathbb{R}$,

(4.23)
$\left\\{\begin{array}[]{l}(\psi^{m})^{\prime\prime}+\beta\xi\psi^{\prime}+(1-\alpha)\psi=0,\qquad\xi>0,\\\
\psi(0)=1,\\\ \psi^{\prime}(0)=\lambda,\end{array}\right.$ (4.24)
$\left\\{\begin{array}[]{l}(\phi^{m})^{\prime\prime}+\beta\xi\phi^{\prime}-\alpha\phi=0,\qquad\xi>0,\\\
\phi(0)=1,\\\ \phi^{\prime}(0)=-\lambda.\end{array}\right.$

We start with $m>1=p$.

####  4.1.1. Slow diffusion, $m>1$.

The existence of solutions with compact support for equations of the above
type has been studied in [8]. Let us consider, as in that paper, the problem
for some $\xi_{0}>0$ given,

(4.25) $\left\\{\begin{array}[]{ll}(g^{m})^{\prime\prime}+\beta\xi
g^{\prime}-qg=0,&\xi<\xi_{0},\\\
g(\xi_{0})=(g^{m})^{\prime}(\xi_{0})=0.\end{array}\right.$

It is proved in [8],

######  Theorem 4.1.

Let $\beta>0$. There exists a continuous solution $g$ to problem (4.25) such
that $g(0)>0$ for $2\beta+q>0$; $g(0)=0$ for $2\beta+q=0$; and if $2\beta+q<0$
there exists a point $\xi_{1}\in(0,\xi_{0})$ with $g(\xi_{1})=0$. Moreover, in
the first case, $g^{\prime}(0)<0$ if $\beta+q>0$; $g^{\prime}(0)=0$ if
$\beta+q=0$; and $g^{\prime}(0)>0$ if $\beta+q>0$. Finally

(4.26) $g(\xi)\sim(\xi_{0}-\xi)^{\frac{1}{m-1}}\qquad\text{for
}\;\xi\to\xi_{0}^{-}.$

Translating this result to our problems (4.23) and (4.24), where $q$ takes the
values, respectively, $q=\alpha-1<0$ and $q=\alpha>0$, we obtain the following
results.

######  Corollary 4.2.

  1. a)

For each $\alpha>0$ there exists a unique $\lambda_{-}(\alpha)>0$ such that
problem (4.24) with $\lambda=\lambda_{-}(\alpha)$ has a decreasing solution
with compact support.

  2. b)

Problem (4.23) has solutions with compact support for some $\lambda$ if and
only if $\alpha>1/m$, the solution being unique for each $\alpha$ given, and
thus $\lambda=\lambda_{+}(\alpha)$. Moreover, $\lambda_{+}(\alpha)>0$ if
$\alpha<2/(m+1)$; and $\lambda_{+}(2/(m+1))=0$.

If we find some $\alpha\in(1/m,2/(m+1))$ such that
$\lambda_{-}(\alpha)=\lambda_{+}(\alpha)$, we will obtain a solution $w$ with
frofile $f$ defined in $\mathbb{R}$ which has compact support. But we are also
interested in subsolutions, and these are obtained constructing profiles with
compact support $[-a,b]$ with a bad behaviour at the interfaces
$(f^{m})^{\prime}(-a)>0$, $(f^{m})^{\prime}(b)<0$. On the other hand, positive
profiles will serve as supersolutions.

Thus, in order to study in more detail the solutions to the equation in (4.25)
we introduce the variables

(4.27) $X=\frac{\xi g^{\prime}}{g},\qquad
Y=\frac{1}{m}\xi^{2}g^{1-m},\qquad\eta=\log\xi.$

We also fix the value $g(0)=1$ and consider the different values of
$g^{\prime}(0)$. We obtain the differential system,

$\left\\{\begin{array}[]{l}\dot{X}=X(1-mX)+Y(q-\beta X),\\\
\dot{Y}=Y(2-(m-1)X),\end{array}\right.$

defined in the half-plane $Y\geq 0$, where $\dot{X}=dX/d\eta$. As we have said
only the values $q=\alpha$ and $q=\alpha-1$ are of interest, with
$\alpha\in(1/m,2/(m+1))$. We have two finite critical points

$P_{1}=(0,0),\qquad P_{2}=(1/m,0)$

(if $q=\alpha-1$ there exists a third critical point but it lies in the lower
half-plane), and three critical points at infinity

$\Lambda_{1}=(-\infty,\infty),\qquad\Lambda_{2}=(-\frac{q}{\beta},\infty),\qquad\Lambda_{3}=(\infty,\infty).$

The point $P_{1}$ is an unstable node: we have a trajectory $\Gamma_{0}$
escaping this point from $\eta=-\infty$ along the vector $(q,1)$, and a family
of trajectories $\Gamma_{\kappa}$, $\kappa\neq 0$, behaving near the origin
like

(4.28) $X\sim\kappa\sqrt{Y}.$

The first one produces a profile $g$ with $g^{\prime}(0)=0$. The profile
corresponding to each $\Gamma_{\kappa}$ satisfies
$g^{\prime}(0)=\kappa/\sqrt{m}$. The point $P_{2}$ is a saddle and plays no
role at this stage.

Now fix $q=\alpha-1<0$. We first observe that defining the energy associated
to the problem

$E_{g}(\xi)=\frac{1}{2}((g^{m})^{\prime})^{2}+\frac{1-\alpha}{1+m}g^{1+m},$

it satisfies

$E_{g}^{\prime}(\xi)=-\beta m\xi g^{m-1}(g)^{\prime 2}\leq 0.$

Therefore $g$ is bounded, and all the trajectories starting at $P_{1}$ must go
to one the points at infinity $\Lambda_{1}$ or $\Lambda_{2}$. In fact
$\Lambda_{3}$ is unstable.

The profiles satisfying (4.26) correspond to trajectories entering the point
$\Lambda_{1}$ linearly, since they satisfy

$\lim_{\xi\to\xi_{0}}X(\xi)=-\infty,\quad\lim_{\xi\to\xi_{0}}Y(\xi)=\infty,\quad\lim_{\xi\to\xi_{0}}\frac{Y(\xi)}{X(\xi)}=-D<0.$

Using the equation

(4.29) $\frac{dY}{dX}=\frac{Y(2-(m-1)X)}{X(1-mX)+Y(q-\beta X)},$

we get that the only possible behaviours near $\Lambda_{1}$ are

$Y\sim-X,\qquad Y\sim|X|^{\frac{m-1}{m}}.$

Thus from Corollary 4.2 we get that the trajectory $\Gamma_{\kappa^{+}}$ with
$\kappa^{+}=\lambda_{+}(\alpha)\sqrt{m}$ joins $P_{1}$ with $\Lambda_{1}$
satisfying $Y\sim-DX$ near $\Lambda_{1}$, and it is the unique trajectory with
that behaviour at infinity. All the other trajectories joining these two
points enter $\Lambda_{1}$ below $\Gamma_{\kappa^{+}}$, and above this
trajectory near the origin. This implies that the corresponding profiles have
slopes at the origin $g^{\prime}(0)<\lambda_{+}(\alpha)$. Observe that this
implies that $g$ vanishes at some point $b<\infty$ with
$g(\xi)\sim(b-\xi)^{\frac{1}{m}}$, so $(g^{m})^{\prime}(b)<0$.

On the other hand, the trajectories with $\kappa>\kappa^{+}$ must go to
$\Lambda_{2}$. The corresponding profiles are positive with

$g(\xi)\sim\xi^{-\frac{1-\alpha}{\beta}}$

for $\xi$ large.

Figure 1. Trajectories in the phase-plane for $q=\alpha-1$ (on the right a
zoom at the origin). The black line enters $\Lambda_{1}$ linearly; the blue
lines go to $\Lambda_{1}$ like $Y\sim|X|^{\frac{m-1}{m}}$; the red lines go to
$\Lambda_{2}$.

In summary we have proved the following result.

######  Lemma 4.3.

Let $\psi$ be a solution of (4.23) with some $\alpha\in(1/m,2/(m+1))$. There
exists some $\lambda_{+}=\lambda_{+}(\alpha)>0$ such that

  1. a)

If $\lambda<\lambda_{+}$ there exists $\xi_{0}<\infty$ such that
$\psi(\xi_{0})=0>(\psi^{m})^{\prime}(\xi_{0})$.

  2. b)

If $\lambda=\lambda_{+}$ there exists $\xi_{0}<\infty$ such that
$\psi(\xi_{0})=(\psi^{m})^{\prime}(\xi_{0})=0$.

  3. c)

If $\lambda>\lambda_{+}$ the solution $\psi$ is positive and
$\psi\sim\xi^{-\frac{1-\alpha}{\beta}}$ for $\xi$ large.

We comment by passing what is the behaviour in the case $\alpha=1/m$. If we
trace back the unique trajectory entering $\Lambda_{1}$ linearly, we see that
it goes to $P_{2}$. In fact integrating the equation between $\xi$ and $a$ we
get

$-(g^{m})^{\prime}(\xi)=\beta\xi g(\xi)-\beta\int_{\xi}^{a}g(s)ds.$

A second integration gives

$g^{m}(\xi)=\beta\xi\int_{\xi}^{a}g(s)ds.$

Then $\lim_{\xi\to 0}Y(\xi)=0$ trivially, while

$\lim_{\xi\to 0}X(\xi)=\lim_{\xi\to
0}\frac{\xi(g^{m})^{\prime}(\xi)}{mg^{m}(\xi)}=\frac{1}{m}\lim_{\xi\to
0}\frac{\xi\left(-\beta\xi
g(\xi)+\beta\int_{\xi}^{a}g(s)ds\right)}{\beta\xi\int_{\xi}^{a}g(s)ds}=\frac{1}{m}.$

Therefore all the trajectories starting at $P_{1}$ must go to $\Lambda_{1}$
like $Y\sim|X|^{\frac{m-1}{m}}$. We obtain profiles with a bad interface
behaviour for every value of $\lambda$.

The phase-space (6.37) is studied in the same way in the case $q=\alpha>0$.
Following the same argument for problem (4.24) we obtain:

######  Lemma 4.4.

Let $\phi$ be a solution of (4.24) with some $\alpha\in(0,2/(m+1))$. There
exist $\lambda_{-}=\lambda_{-}(\alpha)>0$ such that:

  1. a)

If $\lambda>\lambda_{-}$ there exists $\xi_{1}<\infty$ such that
$\phi(\xi_{1})=0>(\phi^{m})^{\prime}(\xi_{1})$.

  2. b)

If $\lambda=\lambda_{-}$ there exists $\xi_{1}<\infty$ such that
$\phi(\xi_{1})=(\phi^{m})^{\prime}(\xi_{1})=0$.

  3. c)

If $\lambda<\lambda_{-}$ the solution $\phi$ is positive and unbounded.

Moreover $\lambda_{-}(\alpha)=K/\sqrt{\alpha}$.

We now study the matching.

######  Lemma 4.5.

There exists a unique $\alpha_{*}\in(1/m,2/(m+1))$ such that
$\lambda_{-}(\alpha_{*})=\lambda_{+}(\alpha_{*})$.

###### Proof.

Define the continuous function
$h(\alpha)=\lambda_{+}(\alpha)-\lambda_{-}(\alpha)$. It is clear that
$h(2/(m+1))<0$. On the other hand, taking $\lambda=\lambda_{-}(1/m)$ in (4.24)
and (4.23) we obtain a profile which crosses the axis at some positive point
with bad interface behaviour. Therefore by continuous dependence of the
profile with respect to the parameter $\alpha$ we have the same behaviour for
$\alpha=\varepsilon+1/m$. This implies $h(\varepsilon+1/m)>0$. Then there
exists $\alpha_{*}\in(1/m,2/(m+1))$ with $h(\alpha_{*})=0$. The uniqueness
follows by comparison. Indeed, if we assume that
$h(\alpha_{1})=0=h(\alpha_{2})$ with $\alpha_{1}<\alpha_{2}$ we have that the
solutions $w_{1}$ with profile $f_{1},f_{2}$ given in (4.22) with
$\alpha=\alpha_{1},\alpha_{2}$ satisfy

$w_{1}(x,0)=f_{1}(x)>e^{-\alpha_{2}t}f_{2}(xe^{\beta_{2}t})=w_{2}(x,-t_{1})$

for some $t_{1}>0$. This implies $w_{1}(x,t)\geq w_{2}(x,t-t_{1})$ for any
$t>0$. In particular at $x=0$ this means $f_{1}(0)e^{\alpha_{1}t}\geq
f_{2}(0)e^{\alpha_{2}(t-t_{1})}$, which is impossible if $t$ is large. ∎

######  Theorem 4.6.

Let $u$ be the solution to problem (1.1) with $p=1<m$. Let $\alpha_{*}$ be
given in Lemma 4.5 and define
$\gamma_{*}=\frac{2(1-\alpha_{*})}{(m-1)\alpha_{*}}$.

  1. a)

If there exists some $1<\gamma<\gamma_{*}$ such that $u_{0}(x)\sim
x^{-\gamma}$ as $x\to\infty$, then

$C_{1}e^{\alpha(\gamma)t}\leq u(x,t)\leq C_{2}e^{\alpha(\gamma)t},$

where

$\alpha(\gamma)=\frac{2}{2+\gamma(m-1)}.$

  2. b)

If $\limsup_{x\to\infty}x^{-\gamma_{*}}u_{0}(x)<\infty$, then for all
$\varepsilon>0$

$C_{1}e^{\alpha_{*}t}\leq u(x,t)\leq
C(\varepsilon)e^{(\alpha_{*}+\varepsilon)t}.$

  3. c)

If $u_{0}$ has support bounded from the right then

$C_{1}e^{\alpha_{*}t}\leq u(x,t)\leq C_{2}e^{\alpha_{*}t}.$

The above estimates are uniform in compact subsets of $\mathbb{R}$ for $t$
large.

###### Proof.

The proof follows by comparison with the self-similar functions constructed
before. We define:

$\bullet$ $w_{*}$ the self-similar function given in (4.21) with
$\alpha=\alpha_{*}$. It is a solution to (1.1) with compact support

$\mbox{supp}(w(\cdot,t))=[-K_{-}e^{\beta t},K_{+}e^{\beta t}].$

$\bullet$ For $\alpha\in(\alpha_{*},2/(m+1))$, which implies
$\lambda_{+}(\alpha)<\lambda_{-}(\alpha)$, we consider $w_{-}$ the self-
similar function given in (4.21) with $\lambda=\lambda_{-}(\alpha)$. It is a
solution to (1.1) with support bounded from the left

$\mbox{supp}(w(\cdot,t))=[-K_{1}e^{\beta
t},\infty),\qquad\lim_{x\to\infty}x^{\frac{2(1-\alpha)}{(m-1)\alpha}}w(x,t)=K_{2}e^{t}.$

In this case, we also consider $w_{\lambda}$ the self-similar function given
in (4.21) with $\lambda_{+}(\alpha)<\lambda<\lambda_{-}(\alpha)$. It is a
positive solution to (1.1) such that

$\lim_{x\to\infty}x^{\frac{2(1-\alpha)}{(m-1)\alpha}}w_{\lambda}(x,t)=K_{2}e^{t},\qquad\lim_{x\to-\infty}w_{\lambda}(x,t)=\infty.$

We now consider the different cases in the statement of the theorem.

  1. a)

$u_{0}(x)\sim x^{-\gamma}$ with $1<\gamma<\gamma_{*}$. Taking
$\alpha=\alpha(\gamma)\in(\alpha_{*},2/(m+1))$ we have
$\lambda_{+}(\alpha)<\lambda_{-}(\alpha)$, and the functions $u_{0}$, $w_{-}$
and $w_{\lambda}$ have the same behaviour at infinity. Then there exists
$t_{1}$ large enough such that

$w_{-}(x,-t_{1})\leq u_{0}(x)\leq w_{\lambda}(x,t_{1}),$

and by comparison

$w_{-}(x,t-t_{1})\leq u(x,t)\leq w_{\lambda}(x,t+t_{1}).$

The grow-up rate follows.

  2. b)

$u_{0}(x)\leq x^{-\gamma_{*}}$. The lower bound follows by comparison with
$w_{*}(x,t-t_{1})$. For the upper bound we compare with
$w_{\lambda}(x,t+t_{1})$ with $\alpha=\alpha_{*}+\delta$, $\delta>0$ small.

  3. c)

$u_{0}$ with compact support. We compare from below as in the previous case,
and from above with $w_{*}(x,t+t_{1})$.

∎

We observe that for any initial value $u_{0}$ the grow-up rate is always
exponential, like for global reaction $a(x)=1$, but with an exponent strictly
smaller $\alpha\leq\alpha_{*}<2/(m+1)$. In the case of a localized reaction,
$a(x)=\mathds{1}_{(-L,L)}$, the grow-up was polynomial.

The second case to consider when $p=1$ is $m=1$, where things are more or less
explicit.

####  4.1.2. Linear diffusion $m=1$

######  Lemma 4.7.

Let $u$ be the solution to (1.1) with $p=m=1$. Then,

$C_{\varepsilon}e^{(1-\varepsilon)t}\leq u(x,t)\leq C_{2}e^{t},$

uniformly in compact subsets of $\mathbb{R}$.

###### Proof.

The upper estimate is given by comparison with the function in (2.7). For the
lower bound we use again comparison, this time with an exponential selfsimilar
function, see (4.23), (4.24). Since here $\beta=0$, we look for a function in
separated variables

$w(x,t)=e^{\alpha t}f(x),\qquad 0<\alpha<1,$

where the profile $f$ satisfies

$\left\\{\begin{array}[]{ll}f^{\prime\prime}+(1-\alpha)f=0,&x>0,\\\
f^{\prime\prime}-\alpha f=0,&x<0,\\\ f(0)=1.\end{array}\right.$

This gives

$f(x)=\left\\{\begin{array}[]{ll}C_{1}e^{\sqrt{\alpha}x}+C_{2}e^{-\sqrt{\alpha}x},&x<0,\\\
C_{3}\sin(\sqrt{1-\alpha}\ x)+\cos(\sqrt{1-\alpha}\
x),&x>0.\end{array}\right.$

The matching condition at $x=0$ means

$C_{1}+C_{2}=1,\qquad C_{3}=\sqrt{\frac{\alpha}{1-\alpha}}(C_{1}-C_{2}).$

Notice that for any $x_{-}<0$ given we can take

$C_{2}=\frac{e^{\sqrt{\alpha}x_{-}}}{e^{\sqrt{\alpha}x_{-}}-e^{-\sqrt{\alpha}x_{-}}},$

so that $f(x_{-})=0$. Moreover, for

$x_{+}=\frac{1}{\sqrt{1-\alpha}}\arctan\left(\frac{-1}{C_{3}}\right)\in\left(\frac{\pi}{2\sqrt{1-\alpha}},\frac{\pi}{\sqrt{1-\alpha}}\right),$

we have $f(x)>0$ in $(0,x_{+})$ and $f(x_{+})=0$. This profile gives us a
subsolution by the procedure of truncation by zero. We denote by $f_{x_{-}}$
this truncated profile.

Let now $u$ be a solution of (1.1). Since the heat equation has infinite speed
of propagation, we can assume without loss of generality that $u_{0}(x)>0$.
Then there exists $t_{1}>0$ such that

$u_{0}(x)\geq e^{-\alpha t_{1}}f_{x_{-}}(x).$

By comparison we deduce $u(x,t)\geq w_{x_{-}}(x,t-t_{1})$. We obtain the lower
grow-up rate for compact subsets of $(-\infty,\frac{\pi}{2\sqrt{1-\alpha}})$
and for every $\alpha<1$.

Finally we observe that for $A>0$ the function $w_{x_{-}}(x-A,t-t_{1})$ is
also a subsolution to (1.1), so we obtain the lower grow-up rate for any
compact subset of $\mathbb{R}$. ∎

We end by considering the case $m<1=p$.

####  4.1.3. Fast diffusion $m<1$.

Here the rate is different for $x>0$ and for $x<0$, as in the case of a
localized reaction, $a(x)=\mathds{1}_{(-L,L)}$.

We first show that the grow-up rate given in (4.14) is sharp for compact
subset of $\mathbb{R}^{+}$ by proving the lower bound. To do that we compare
with a subsolution in separated variables with compact support in
$\mathbb{R}^{+}$,

$\underline{u}(x,t)=f(x)g(t).$

Notice that since $u$ has global grow-up, see Lemma 2.1, we have
$u(x,t_{0})\geq\underline{u}(x,0)$ for $t_{0}$ large enough, so then the
comparison of the initial data is granted by a time shift.

######  Lemma 4.8.

Let $u$ be a solution of (1.1) with $m<1=p$. Then,

$u(x,t)\geq ce^{t}$

uniformly in compact subset of $\mathbb{R}^{+}$.

###### Proof.

Let $\phi$ be the solution to the problem

$\left\\{\begin{array}[]{l}(\phi^{m})^{\prime\prime}+\phi=0,\quad\xi>0,\\\
\phi(0)=0,\\\ \phi^{\prime}(0)=1.\end{array}\right.$

Since $\phi^{m}$ is concave $\phi$ must vanish at some point $x_{0}<\infty$.
Now we consider the rescaled function

$f(x)=A\phi(A^{\frac{1-m}{2}}x),$

which satisfies the same equation and vanishes at $x=x_{0}A^{-\frac{1-m}{2}}$.
This is the spatial part of our subsolution. The time part $g$ is defined as
the solution to

$\left\\{\begin{array}[]{l}g^{\prime}=g-g^{m},\quad\xi>0,\\\
g(0)>1.\end{array}\right.$

We have $u(x,t+t_{0})\geq f(x)g(t)$ for any $x>0$ and $t>0$. Since
$g^{\prime}\sim g$ as $t\to\infty$, the comparison gives the desired lower
bound. ∎

In order to obtain the grow-up rate for $\mathbb{R}^{-}$, we note that by
(4.14) $u$ is a subsolution of the problem

$\left\\{\begin{array}[]{ll}w_{t}=(w^{m})_{xx},&x<0,t>0,\\\
w(0,t)=C_{1}e^{t},&t>0,\\\ w(x,0)=w_{0}(x),&x<0.\end{array}\right.$

It is proved in [3] that there exists a unique self-similar solution of
exponencial type

$W(x,t)=e^{t}f(xe^{\frac{1-m}{2}t}),$

which is increasing in both variables $x$ and $t$. Moreover, for $|\xi|$ large

$f(\xi)\sim|\xi|^{\frac{-2}{1-m}}\left(\log|\xi|\right)^{\frac{1}{1-m}}.$

Then, if the initial datum satisfies

(4.30)
$u_{0}(x)\sim|x|^{\frac{-2}{1-m}}\left(\log|x|\right)^{\frac{1}{1-m}},\qquad
x\sim-\infty,$

we can take as a supersolution $\overline{w}(x,t)=AW(x,t)$. Notice that from
the property $W_{t}\geq 0$ we have

$\overline{w}_{t}-\overline{w}_{xx}=(A-A^{m})w_{t}\geq 0,$

provided $A>1$. Moreover, taking $A$ large enough we get
$\overline{w}(x,0)\geq u_{0}(x)$.

On the other hand, by Lemma 4.8 we have that $u$ is a supersolution to the
problem

$\left\\{\begin{array}[]{ll}w_{t}=(w^{m})_{xx},&x<1,t>0,\\\
w(1,t)=C_{2}e^{t},&t>0,\\\ w(x,0)=w_{0}(x),&x<1.\end{array}\right.$

and $\underline{w}(x,t)=AW(x,t)$ with $A$ small enough to have
$\underline{w}(x,0)\leq u_{0}(x)$ is a subsolution.

As a conclusion,

######  Lemma 4.9.

Let $u$ be a solution of (1.1) with $m<1=p$, such that the initial datum
$u_{0}$ satisfies the condition (4.30). Then, for $x<0$

$u(x,t)\sim t^{\frac{1}{1-m}}.$

###  4.2. Case $p<1$

Here we distinguish between $m<p$ and $m\geq p$.

####  4.2.1. Case $m<p$.

######  Lemma 4.10.

Let $m<p<1$. If $u_{0}$ satisfies (4.17) for $x\sim-\infty$, then

$u(x,t)\sim\left\\{\begin{array}[]{ll}t^{\frac{1}{1-p}},&x>0,\\\
t^{\frac{1}{1-m}},&x<0,\end{array}\right.$

uniformly in compact sets.

###### Proof.

The proof follows in the same way as in the case $p=1$, using here the
selfsimilar profile

$W(x,t)=t^{\frac{1}{1-p}}f(xt^{\frac{p-m}{2(1-p)}})$

constructed in [3], which is again increasing in both variables $x$ and $t$,
and that satisfies, for $|\xi|$ large,

$f(\xi)\sim|\xi|^{\frac{-2}{1-m}}.$

∎

####  4.2.2. Case $m\geq p$.

######  Lemma 4.11.

Let $p<1\leq m$. then

$u(x,t)\geq ct^{\frac{1}{1-p}}.$

uniformly in compact sets of $\mathbb{R}$.

###### Proof.

We consider a subsolution in selfsimilar form

$w(x,t)=t^{\alpha}f(\xi),\qquad\xi=xt^{\beta},$

where

$\alpha=\frac{1}{1-p},\qquad\beta=-\frac{m-p}{2}\alpha,$

and the selfsimilar profile satisfies

$\mathfrak{L}(f):=(f^{m})^{\prime\prime}-\beta\xi
f^{\prime}+a(\xi)f^{p}-\alpha f\geq 0.$

We construct the profile gluing four functions. Let $A>0$ be a constant to be
fixed and put $\xi_{0}=-\sqrt{2/\alpha}A^{\frac{m-1}{2m}}$.

  1. a)

For $\xi\leq\xi_{0}$ we put $f_{1}(\xi)=0$.

  2. b)

For $\xi_{0}\leq\xi\leq 0$ we define

$f_{2}^{m}(\xi)=A+\sqrt{2\alpha}A^{\frac{1+m}{2m}}\xi+\alpha
A^{1/m}\frac{\xi^{2}}{2}.$

Notice that $f^{m}_{2}(\xi_{0})=(f_{2}^{m})^{\prime}(\xi_{0})=0$. Moreover
since $\beta\leq 0$ and $f_{2}$ is non-decreasing

$\mathfrak{L}(f_{2})\geq(f_{2}^{m})^{\prime\prime}-\alpha f_{2}\geq\alpha
A^{\frac{1}{m}}-\alpha f_{2}(0)=0.$

  3. c)

For $0\leq\xi\leq\xi_{1}=\sqrt{2\alpha}A^{\frac{1+m-2p}{2m}}$ we define

$f_{3}^{m}(\xi)=A+\sqrt{2\alpha}A^{\frac{1+m}{2m}}\xi-(A^{p/m}-\alpha
A^{1/m})\frac{\xi^{2}}{2}.$

We have $(f_{3}^{m})(0)=(f_{2}^{m})(0)=A$,
$(f_{3}^{m})^{\prime}(0)=(f_{2}^{m})^{\prime}(0)$, so this function $f_{3}$
matches well with $f_{2}$. Also $f_{3}$ is increasing in $0<\xi<\xi_{1}$, with

$f_{3}(\xi_{1})=A^{\frac{1}{m}}\left(1+\alpha\frac{A^{\frac{1-p}{m}}}{1-\alpha
A^{\frac{1-p}{m}}}\right)^{\frac{1}{m}},\qquad(f_{3})^{\prime}(\xi_{1})=0.$

Since $p<1$ we get that for $A$ small enough, both $\xi_{1}$ and
$f_{3}(\xi_{1})$ are small. Hence, the function $f_{3}^{p}(\xi)-\alpha
f_{3}(\xi)$ is increasing. Then

$\mathfrak{L}(f_{3})\geq(f_{3}^{m})^{\prime\prime}+f_{3}^{p}-\alpha
f_{3}\geq(f_{3}^{m})^{\prime\prime}+f_{3}^{p}(0)-\alpha f_{3}(0)=0.$

  4. d)

For $\xi>\xi_{1}$ we consider $f_{4}=g_{+}$, where $g$ is the solution to the
initial value problem

$\left\\{\begin{array}[]{ll}(g^{m})^{\prime\prime}-\beta\xi
g^{\prime}+g^{p}-\alpha g=0,&\xi>\xi_{1},\\\ g(\xi_{1})=f_{3}(\xi_{1}),\\\
g^{\prime}(\xi_{1})=0.\end{array}\right.$

It is clear that if $f_{3}(\xi_{1})<(1/\alpha)^{\alpha}$ then $g$ is
nonincresing and positive for $\xi_{1}\leq\xi<\xi_{2}\leq\infty$.

The final function putting together $f_{i}$, $i=1,\cdots,4$ is a subsolution
to our problem with zero initial value. This gives the desired lower bound of
the grow-up rate. ∎

##  5\. Blow-up rates

###### Proof of Theorem 1.4.

The lower blow-up rate is obtained easily by (strict) comparison with the
supersolution

$U(t)=C_{p}(T-t)^{-\frac{1}{p-1}}.$

Indeed, if we assume that there exists $t_{0}\in(0,T)$ such that

$\|u(\cdot,t_{0})\|_{\infty}<U(t_{0}),$

it also holds

$\|u(\cdot,t_{0})\|_{\infty}<U(t_{0}-\varepsilon)$

for some $\varepsilon>0$, which is a contradiction with the fact that $u$
blows up at time $T$.

In order to prove the upper blow-up rate we use a rescaling technique inspired
in the work [6].

Let us define

(5.31) $M(t)=\max_{\mathbb{R}\times[0,t]}u(x,\tau),$

and consider, for any fixed $t_{0}\in(0,T)$, the increasing sequence of times

$t_{j+1}=\sup\\{t\in(t_{j},T)\,:\,M(t)=2M(t_{j})\\}.$

Observe that for this sequence we have $\|u(\cdot,t_{j})\|_{\infty}=M(t_{j})$.
We also observe that since the reaction only takes place for $x>0$, we get
that near the blow-up time the maximum of $u$ is achieved in $\mathbb{R}^{+}$.
Therefore we can take $x_{j}\geq 0$ such that

$u(x_{j},t_{j})=M(t_{j}).$

We consider the sequence

$z_{j}=(t_{j+1}-t_{j}){M^{p-1}(t_{j})}.$

Let us observe that if $z_{j}$ is bounded, we get that

$t_{j+1}-t_{j}\leq cM^{1-p}(t_{j})=c2^{j(1-p)}M^{1-p}(t_{0}).$

Performing the sum,

$T-t_{0}\leq
cM^{1-p}(t_{0})\sum_{j=0}^{\infty}2^{j(1-p)}=c^{\prime}M^{1-p}(t_{0})\leq
c^{\prime}\|u(\cdot,t_{0})\|_{\infty}^{1-p},$

that is, the desired upper blow-up rate. Therefore, in order to arrive at a
contradiction, we assume that there exists a subsequence of times, still
denoted $t_{j}$, such that

(5.32) $\lim_{j\to\infty}z_{j}=\infty.$

Now we define the functions

$\varphi_{j}(y,s)=\frac{1}{M_{j}}u(M_{j}^{\frac{m-p}{2}}y+x_{j},M_{j}^{1-p}s+t_{j}),$

for

$y\in\mathbb{R},\qquad s\in I_{j}=(-t_{j}M_{j}^{p-1},(T-t_{j})M_{j}^{p-1}),$

where $M_{j}=M(t_{j})$. Notice that $I_{j}\to\mathbb{R}$ as $j\to\infty$ and
$\varphi_{j}$ is a solution to the equation

$(\varphi_{j})_{s}=(\varphi_{j}^{m})_{xx}+a(y-x_{j}M_{j}^{\frac{p-m}{2}})\varphi_{j}^{p},\qquad(y,s)\in\mathbb{R}\times
I_{j}.$

It also satisfies

$\varphi_{j}(0,0)=1\quad\text{and}\quad\varphi_{j}(y,s)\leq 2\quad\mbox{in
}\mathbb{R}\times I_{j}.$

The uniform bounds for $\varphi_{j}$ impliy that $\varphi_{j}$ is Hölder
continuous with uniform coeffcient. Since $\varphi_{j}(0,0)=1$ we have a
uniform nontrivial lower bound for every $\varphi_{j}$, that is,

(5.33) $\varphi_{j}(y,0)\geq g(y)\geq 0,$

for some nontrivial function $g$.

We claim that, under the assumption of Theorem 1.4, each function
$\varphi_{j}$ blows up at a finite time $S_{j}$ which is uniformly bounded,
that is $S_{j}<S$. This is a contradiction with the fact that
$\varphi_{j}(y,s)\leq 2$ for $s\in(0,z_{j})$. Indeed, since $z_{j}\to\infty$
we can take $j$ large such that $z_{j}>S$. Therefore (5.32) can not be true
and the blow-up rate follows.

In order to prove the claim we first observe that since $x_{j}\geq 0$, we have
that $\varphi_{j}$ is a supersolution of the equation

(5.34) $h_{s}=(h^{m})_{yy}+a(y)h^{p},\qquad(y,s)\in\mathbb{R}\times(0,z_{j}).$

Therefore, for $p\leq m+2$, we can apply Theorem 1.1 to get that the solution
of the above equation with initial datum $h(x,0)=g(y)$ blows up at some time
$S$. Then, by comparison, $\varphi_{j}$ blows up at time $S_{j}<S$.

For the case $p>m+2$ we need the extra hypothesis $\partial_{t}u\geq 0$, which
implies $(\varphi_{j})_{s}\geq 0$. Therefore $\varphi_{j}$ is a supersolution
of the problem

(5.35)
$\left\\{\begin{array}[]{ll}w_{s}=(w^{m})_{yy},&(y,s)\in\mathbb{R}^{+}\times(0,z_{j}),\\\
w(0,s)=1,\\\ w(y,0)=0.\end{array}\right.$

Since $w\leq 1$ we can pass to the limit, by means of a Lyapunov functional,
to get that $w(x,s)\to 1$ as $s\to\infty$ uniformly in compact sets of
$\mathbb{R}^{+}$. Actually in the linear case $m=1$ the solution to problem
(5.35) is explicit, while if $m\neq 1$ the solution is the so-called
Polubarinova-Kochina solution. Notice that this behaviour is also true if we
consider the problem in $\mathbb{R}^{-}$. Therefore, by comparison
$\varphi_{j}(y,s)\geq 1/2$ in $|y|<K$ and $s>s_{K}$, and then

$\varphi_{j}(y,s_{K})\geq h_{0}(y)=\frac{1}{2}(1-x^{2}/K^{2})_{+}^{1/m}.$

Observe that the energy of $h_{0}$ given in (3.9) satisfies

$E_{h_{0}}=\dfrac{1}{2}\int_{-\infty}^{\infty}|(h_{0}^{m})_{x}|^{2}-\frac{m}{p+m}\int_{0}^{\infty}h_{0}^{p+m}=C_{1}K^{-1}-C_{2}K<0$

for $K$ large. Then, applying the concavity argument the solution of (5.34)
with initial datum $h_{0}$ blows up at finite time $S$ and by comparison
$\varphi_{j}$ also blows up at a time $S_{j}<S$. ∎

##  6\. Blow-up sets

We prove here Theorem 1.5. We first consider the case $p\geq m$.

######  Lemma 6.1.

Let $u$ be a blow-up solution to (1.1) with compactly supported initial datum.
Then $B(u)$ is bounded from the left. In fact,

i) $B(u)\subset\mathbb{R}^{+}$ if $p>m$;

ii) $B(u)\subset[-K,\infty)$ if $p=m$.

###### Proof.

Notice that by the upper blow-up rate, $u$ is a subsolution to the problem on
the left half-line

(6.36) $\left\\{\begin{array}[]{ll}w_{t}=(w^{m})_{xx},&x<0,\,0<t<T,\\\
w(0,t)=C(T-t)^{-1/(p-1)},\\\ w(x,0)=w_{0}(x),\end{array}\right.$

provided that $C$ is large. For $m=1$ we have and explicit formula for $w$,
and it is easy to see that $w$ is bounded for $x<0$, see for instance [14].
For $m\neq 1$ we use comparison with a selfsimilar solution in the form

$W(x,t)=(T-t)^{-\alpha}F(x(T-t)^{-\beta}),\qquad\alpha=\frac{1}{p-1},\ \
\beta=\frac{p-m}{2}\alpha,$

where the profile $F$ satisfies the equation

$(F^{m})^{\prime\prime}-\beta\xi F-\alpha F=0.$

Observe that for $p=m$, i.e. $\beta=0$, this equation is the same as for
problem (4.24), so by Lemma 4.4 there exists a profile $F_{1}$ with compact
support and satisfying $F_{1}(0)=1$. By scaling
$F(\xi)=CF(C^{\frac{1-m}{2}}\xi)$ is also a solution, with large support if
$C$ is large. We then take $C$ large so as to have that the corresponding
solution $W$ satisfies $W(x,0)\geq u_{0}(x)$ and by comparison we obtain the
bound of $B(u)$.

For the case $p>m$ ($\beta>0$), we introduce as in Seccion 4.1.1 the variables

(6.37) $X=\frac{|\xi|g^{\prime}}{g},\qquad
Y=\frac{1}{m}\xi^{2}g^{1-m},\qquad\eta=\log|\xi|,$

to obtain the differential system,

$\left\\{\begin{array}[]{l}\dot{X}=X(1-mX)+Y(\alpha+\beta X),\\\
\dot{Y}=Y(2-(m-1)X).\end{array}\right.$

It is easy to see that all the orbits in the second quadrant start at the
origin and have three posible behaviours: they cross the vertical axis; or the
horizontal variable goes to $-\infty$; or $(X,Y)\to(-\alpha/\beta,\infty)$.
The existence of a unique orbit joining the origin with
$(-\alpha/\beta,\infty)$ is given in [5] for $m<1$, but the argument works as
well for $m>1$. From this orbit we obtain a positive, increasing profile
$F_{1}$ such that $F_{1}(0)=1$ and $F_{1}(\xi)\sim|\xi|^{-\alpha/\beta}$ for
$\xi\sim-\infty$. Notice that, for $x<0$ and $t$ near $T$, we have

$W_{1}(x,t)=(T-t)^{-\alpha}F_{1}(x(T-t)^{-\beta})\sim|x|^{-\alpha/\beta},$

that is $W$ is bounded. Rescaling and comparison as before implies the same
property for our solution, $u$ is bounded for $x<0$. ∎

######  Remark 6.1.

Notice that for $p<m$ we have again an equation like (4.24) for the profile.
This give us a family of selfsimilar solutions

$W_{C}(x,t)=C(T-t)^{-\frac{1}{p-1}}F_{1}(C^{\frac{1-m}{2}}x(T-t)^{\frac{m-p}{2(p-1)}})$

with global blow-up.

######  Lemma 6.2.

Let $u$ be a blow-up solution to (1.1) with compactly supported initial datum.
Then $B(u)$ is bounded from the right provided $p\geq m>1$.

###### Proof.

We only have to notice that the support of the blow-up solutions to the
equation with global reaction, $a(x)=1$, are bounded if $p\geq m>1$, see [14].
The proof of this result uses the intersection comparison technique with self-
similar profiles in a neighborhood of the free boundary. Then the same result
holds for our equation near the right-hand free boundary. ∎

######  Lemma 6.3.

Let $u$ be a blow-up solution to (1.1) with $p<m$ and a compactly supported
initial datum. Assume also that there exists $x_{0}\in\mathbb{R}$ satisfying
(1.6). Then $B(u)=\mathbb{R}$.

###### Proof.

Thanks to the hypothesis at $x_{0}$ we have that $u$ is a supersolution to the
problem defined on the left of $x_{0}$,

$\left\\{\begin{array}[]{ll}w_{t}=(w^{m})_{xx},&x<x_{0},\,0<t<T,\\\
w(x_{0},t)=C_{1}(T-t)^{-1/(p-1)},\\\ w(x,0)=w_{0}(x),\end{array}\right.$

as well as to the problem on the right, $x>x_{0}$. Moreover, the self-similar
solution given in Remark 6.1 is a subsolution if we choose $C$ small enough
such that $W_{C}(x,0)<u_{0}(x)$. Comparison ends the proof. ∎

The same argument allows to prove that if $p=m$ the blow-up set contains some
nontrivial interval, thus concluding the proof of Theorem 1.5.

To finish this section we remark that, in the range $p<m$, but without the
hypothesis of the existence of $x_{0}$, it is easy to see that $B(u)$ is
unbounded at least from the right. In fact if we assume that at some point
$u(x_{1},t)<M$, then $u$ is a subsolution to

$\left\\{\begin{array}[]{ll}z_{t}=(z^{m})_{xx}+a(x)z^{p},&x<x_{1},\,t>0,\\\
z(x_{1},t)=M,\\\ z(x,0)=z_{0}(x).\end{array}\right.$

On the other hand, the stationary solution of the equation with $z(0)=K$ and
$z^{\prime}(0)=0$, is supersolution of the problem if $K$ is large enough.
Then by comparison $u$ must be bounded in $(-\infty,x_{1})$. Therefore if $u$
is bounded in some interval $(x_{1},\infty)$ then it cannot blow up.

## References

  * [1] J. Aguirre and M. Escobedo. A Cauchy problem for $u_{t}-\Delta u=u^{p}$ with $0<p<1$. Asymptotic behaviour of solutions. Ann. Fac. Sci. Toulouse Math. 8 (1986), 175–203.
  * [2] X. Bai, S. Zhou and S. Zheng. Cauchy problem for fast diffusion equation with localized reaction. Nonlinear Anal. 74 (2011), 2508–2514.
  * [3] R. Ferreira and A. de Pablo. Grow-up for a quasilinear heat equation with a localized reaction. J. Differential Equations 268 (2020), 6211–6229.
  * [4] R. Ferreira, A. de Pablo and J.L. Vázquez. Blow-up for the porous medium equation with a localized reaction. J. Differential Equations 231 (2006), 195–211
  * [5] R. Ferreira, A. de Pablo, F. Quirós and J. D. Rossi.The blow-up profile for a fast diffusion equation with a nonlinear boundary condition. Rocky Mountain J. Math. 33 (2003), 123–146
  * [6] M. Fila and P. Souplet.The blow-up rate for semilinear parbolic problems on general domains. Nonlinear differ. equ. appl. 8 (2001), 473–480.
  * [7] V.A. Galaktionov. Blow-up for quasilinear heat equations with critical Fujita’s exponents. Proc. Roy. Soc. Edinburgh Sect. A 124 (1994) 517–525.
  * [8] B. H. Gilding and L. A. Peletier. On a class of similarity solutions of the porous media equation. J. Math. Anal. Appl. 55 (1976), 351–364
  * [9] M.A. Herrero and M. Pierre. The Cauchy Problem for $u_{t}=\Delta u^{m}$ when $0<m<1$. Trans. Amer. Math. Soc. 291 (1985), 145–158.
  * [10] H.A. Levine and P. Sacks. Some existence and nonexistence theorems for solutions of degenerate parabolic equations. J. Differential Equations 52 (1984) 135–161.
  * [11] A. de Pablo and J. L. Vázquez. The balance between strong reaction and slow diffusion. Comm. Partial Differential Equations 15 (1990), 159–183.
  * [12] A. de Pablo and J. L. Vázquez. Travelling waves and finite propagation in a reaction-diffusion equation. J. Differential Equations 93 (1991), no. 1, 19–61.
  * [13] R.G. Pinsky. Existence and nonexistence of global solutions for $u_{t}=\Delta u+a(x)u^{p}$ in $R^{d}$. J. Differential Equations 133 (1997), 152–177
  * [14] A.A. Samarskii, V.A. Galaktionov, S.P. Kurdyumov and A.P. Mikhailov. Blow-up in Problems for Quasilinear Parabolic Equations. Nauka, Moscow, 1987 (in Russian); English translation: Walter de Gruyter, Berlin, 1995.
  * [15] J. L. Vázquez. “The porous medium equation. Mathematical theory”. Oxford Mathematical Monographs. The Clarendon Press, Oxford University Press, Oxford, 2007. ISBN: 978-0-19-856903-9.