ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
BLOW-UP FOR A SEMILINEAR HEAT EQUATION WITH MOVING NONLINEAR REACTION
RAUL FERREIRA
Communicated by Jesus Ildefonso Diaz
Abstract. We study the behavior of solutions of the semilinear problem ut=uxx+ (1 + (T−t)−αχ{|x|<(T−t)1/2})up, x∈R, t∈(0, T),
u(x,0) =u0(x)≥0, x∈R,
withα >0 and p >0. We describe, in terms of the parameters when the solution is bounded and when it blows up. For blowing up solutions we find the blow-up rate and the blow-up set.
1. Introduction
In this article we study the Cauchy problem, for the equation
ut=uxx+h(x, t)up, (1.1)
withα >0,p >0 and
h(x, t) = 1 + (T−t)−αχ{|x|<(T−t)1/2}.
Moreover, we assume thatu0 is a nonnegative, nontrivial regular function.
Existence of a solution can be easily achieved, and it is given by the representa- tion formula
u(x, t) = Z
R
Γ(x−y, t)u0(y)dy+ Z t
0
Z
R
Γ(x−y, t−s)up(y, s)dy ds +
Z t 0
Z
{|y|<(T−s)1/2}
Γ(x−y, t−s)up(y, s)(T −s)−αdy ds.
where Γ(x, t) = (4πt)−1/2e−x2/4t is the heat Kernel.
Uniqueness is standard forp≥1, but for 0< p <1 the reaction termf(x, t, u) = h(x, t)up is non-Lipschitz on u and the uniqueness fails, see [1]. Nevertheless, in this case we can we can construct a maximal solution just by taking the limit
u= lim
ε→0u(ε),
2010Mathematics Subject Classification. 35K58, 35B44, 35B40.
Key words and phrases. Semilinear parabolic equation; blow-up; asymptotic behaviour.
c
2018 Texas State University.
Submitted July 13, 2017. Published January 22, 2018.
1
whereu(ε)is the unique solution to our problem with initial condition u(ε)(x,0) = u0(x) +ε, and with the reactionf(x, t, u) replaced by
f(ε)(x, t, u) =h(x, t)
(up ifs≥ε, εp−1u ifs < ε,
see [10]. We note that the maximal solution satisfies the above representation formula. Moreover, a comparison principle among maximal solutions can be easily obtained.
Equation (1.1) can be considered in some way as a perturbation of the system ut=uxx+up1vq1
vt=vxx+up2vq2
Indeed, if we assume that thev component blows up in a finite timeT, typically the behaviour near the blow-up time is
v∼
((T −t)−q |x|<(T−t)1/2 1 |x|>(T−t)1/2
see [4], [13] (f ∼g means that there exist finite positive constantsc1,c2 such that c1g ≤f ≤c2g). Now including this behaviour on the equation ofu, we obtain an equation like (1.1).
Systems of this kind are common in population dynamics. In this context u andv represent two different species with a symbiotic behaviour. The cooperation between them is represented by the coupled reaction terms.
Probably the first study in blow-up for the semilinear heat equation is given by Fujita, [6], where the caseh= 1 is studied. He proves that for 1< p <1+2/nevery non-trivial positive solution blows up in finite time, while forp >1 + 2/nthere are both blow up and global solutions. The valuespg= 1 andpc = 1 + 2/n are called the critical global exponent and the critical blow-up exponent, respectively. In the border casep=pc every positive solution blow up, see [9, 15].
For the case of localized reaction (h(x, t) = h(x) with compact support) it is known thatpg = 1 andpc = 2 ifn= 1 while pg =pc = 1 forn≥2, see [11]. For further result on the blow-up phenomena in one dimensional case with localized reaction we refer to [5].
If the coefficient reaction is given byh(x, t) =ts|x|σ, Qi shows in [12] thatpg= 1 andpc= 1 + (2 + 2s+σ)/n. Since, in our problem, the reaction coefficient blows up ast→T, the critical exponents are independent ofp, see Theorem 1.1 below.
Blow-up phenomena has attracted an increasing interest among researchers in the last years, not only for its wide variety of applications, but also motivated by the mathematical analysis behind those kind of equations, see for instance the books [7, 14] and the surveys [3, 8].
Our first objective is to identify when either the solution is bounded or it blows up.
Theorem 1.1. (i) If p≤1 and α <1 all the solutions to problem (1.1) are bounded;
(ii) If p≤1andα≥1 all the solutions to problem (1.1)blow up a time t=T; (iii) Ifp >1andα≥1all the solutions to problem(1.1)blow up a timeT∗< T;
(iv) Ifp >1andα <1solutions may blow up at timeT∗≤T or not depending on the size of initial data.
Once we have characterized the exponents giving rise to either blow-up or bounded solutions, we want to study the way the blowing up solutions behave as approaching the blow-up time. This means that we must investigate the speed at which they blow up, the blow-up rate and where the solutions blow up, the blow-up set.
We note that if the blow-up time T∗ < T, the reaction coefficient h(x, t) is bounded. Then, in the study of the blow up does not play any role and the solution behaves like the solution of
wt=wxx+wp,
see [2]. Thus, we only study the rangep≤1≤αwhere the maximal solution blows up at timeT independently of initial data.
Theorem 1.2. Letube the maximal solution of (1.1). Ast approachesT we have that
(i) Forp <1 andα >1,
ku(·, t)k∞∼ (T −t)−α−11−p. (ii) Forp <1 andα= 1,
ku(·, t)k∞∼
log T T−t
(1−p)1 . (iii) Forp= 1andα >1,
t→Tlim(T−t)1−αlogku(·, t)k∞= 1 1−α.
(iv) Forp= 1 =α, there existsγ∗∈(0,1) such that for all ε >0 Cε(T−t)−γ∗+ε≤ ku(·, t)k∞≤C2(T−t)−γ∗. Finally, we study the blow-up set. Which it is defined by
B(u) ={x∈R:∃xn →x, tn %T withu(xn, tn)→ ∞}.
Theorem 1.3. Letua blow-up solution of (1.1). Then,B(u) ={0}forp <1≤α, while forp= 1we have
B(u) =
(−∞,∞) forα >2 [−2,2] forα= 2 {0} forα∈[1,2)
This paper is organized as follows. In the next Section, we study in terms of the parameterspandα, when the solution is bounded and it blows up, namely we prove Theorem 1.1. In Section 3 we find the blow rate for the range p≤1 ≤α, Theorem 1.2. Finally, in Section 4 Theorem 1.3 are proved.
2. Blow-up versus boundedness Lemma 2.1. If α <1 andp≤1then the solution uis bounded.
Proof. It follows by comparison with the flat supersolution w0(t) = (1 + (T−t)−α)wp, t >0,
w(0) =ku0k∞. (2.1)
Which is given by v(t) =
v01−p+ (1−p) t−1−α1 (T−t)1−α−T1−α1/(1−p)
, p <1, v0exp
t−1−α1 (T−t)1−α−T1−α
, p= 1.
(2.2)
Lemma 2.2. If α < 1 and p > 1, then there are both: bounded and blow-up solutions.
Proof. The fact that for small initial data the solution is bounded follows by com- parison with the flat supersolution (2.1).
On the other hand, the blow-up, is given by comparison with the problem vt=vxx+vp (x, t)∈(−L, L)×(0, T)
v(±L, t) = 0 t∈(0, T) v(x,0) =v0(x) x∈(−L, L)
Applying Kapplan’s method it is well known that we can takev0 large enough to
ensure thatv blows up before timeT.
Lemma 2.3. Let α ≥ 1 then the solution blows up at a finite time T∗ ≤ T.
Moreover, if T∗=T, thenu(0, t)→ ∞ ast→T.
Proof. First we observe that by comparison with the heat equation we get that there existsm >0 such that
u≥m (x, t)∈(−T1/2, T1/2)×(0, T]. (2.3) Using the representation formula, we get
u(x, t) = Z
R
Γ(x−y, t)u0(y)dy+ Z t
0
Z
R
Γ(x−y, t−s)up(y, s)dy ds +
Z t 0
Z
R
Γ(x−y, t−s)up(y, s)(T−s)−αχ{|y|<(T−s)1/2}dy ds
≥ Z t
0
Z
R
Γ(x−y, t−s)mp(T−s)−αχ{|y|<(T−s)1/2}dy ds.
Now, we observe that forx= 0 and t=T the above integral becomes C
Z T 0
(T −t)−α Z
|y|<(T−s)1/2
1
(T −s)1/2e−y2/(4(T−s))dy ds
=C Z T
0
(T−t)−α Z 1/2
−1/2
e−r2dr ds.
Which is divergent forα≥1.
Lemma 2.4. Letα≥1. Ifp≤1the maximal solutionublows up at timeT∗=T, while forp >1the blow up time satisfies thatT∗< T.
Proof. The fact that for p≤1 the solution blows up at timeT, is given by com- parison with the flat supersolution given in (2.1).
Now, forp >1 we assume that ublows up at time T and we define v(ξ, τ) =u(x, t) ξ=x(T−t)−1/2, τ =−log T−t
T ,
which satisfies the equation vτ =vξξ−1
2ξvξ+
T e−τ+T1−αe(α−1)τχ{|ξ|≤1}
vp. Notice that
• Asv is symmetric and decreasing forξ >0 the term−ξvξ is non-negative.
• From (2.3) we have thatv > min [−1,1]×[0,∞].
• α≥1 implies thateα−1τ≥1.
Therefore, for all 1< q < p the functionv is a supersolution of the problem wτ=wξξ+T1−αχ{|ξ|≤1}mp−qwq R×(τ0,∞)
w(ξ, τ0) =v(ξ, τ0)
For this problem it is well know that for all non-negative initial data the solution blows up at finite time if 1 < q ≤2, see [5, 11]. Then, the function u blows up
before timeT.
3. Blow-up rates
Since forα≥1 andp≤1 the solution of Problem (1.1) blows up at timeT, the comparison with the flat supersolution defined in (2.1) gives us the upper blow-up rate. However, as we see below, this is not the correct blow-up rate in the linear casep=α= 1. More precisely,
Lemma 3.1. Let ube a solution of (1.1). Then, there existsC >0 such that
ku(·, t)k∞≤C
(T−t)−α−11−p p <1< α e(T−t)1
−α
α−1 p= 1< α log T1−t1−p1
p <1 =α To study the lower blow-up rate we consider different cases.
Lemma 3.2. Let p <1 andα >1. Then
ku(·, t)k∞≥(T−t)−α−11−p.
Proof. Let φ1 be the first eigenfunction of the Laplacian in (−1,1) normalized according tokφ1k∞= 1, and consider the function
u(x, t) =A(T−t)−α−11−pφ1(x(T−t)−1/2).
To see thatuis subsolution we need:
• Comparison of the initial data. Notice that u is a supersolution of the heat equation, then u(x, t) ≥ m > 0 in (−T1/2, T1/2)×(0, T). Thus taking A small enough,u(x,0)≤u0(x).
•An inequality for the equation: substituting uin the equation we need that Aα−1
1−pφ(ξ) +A
2ξφ01(ξ)≤ −λ1Aφ1(ξ) + ((T−t)α+ 1)Apφp1(ξ)
where ξ = x(T −t)−1/2. Observe that A2ξφ01(ξ) < 0 and (T −t)αApφp1(ξ) > 0, therefore the above inequality holds provided that
A1−pφ1−p1 (ξ)α−1 1−p +λ1
≤1.
Thus, takingAsmall enough we are done.
This implies thatu(x, t)≥u(x, t) and lower blow-up rate follows.
Lemma 3.3. Let p= 1andα >1. Then, forA >0 small enough ku(·, t)k∞≥A(T−t)π
2 4 e(T−t)1
−α α−1 .
Proof. As in the previous Lemma, we use a comparison argument. In this case we consider the subsolution
u=A(T−t)λ1e(T−t)1
−α
α−1 φ1(x(T−t)−1/2), γ >0,
where λ1 =π2/4 and φ1 are the first eigenvalue and the first eigenfunction of the
Laplacian in (−1,1).
We remark that in this case the extra term, (T−t)π2/4appears. We conjecture that this extra term is technical and it can be avoided. Using the comparison with the functionwgiven in (2.1) it is easy to obtain the following blow-up rate.
Corollary 3.4. Let p= 1 andα >1. Then
t→Tlim(T−t)1−αlog(u(0, t)) = 1 α−1. Lemma 3.5. Let α= 1andp <1. Then
ku(·, t)k∞≥
log T T−t
(1−p)1
Proof. Using the representation formula u(x, t) =
Z
R
Γ(x−y, t)u0(y)dy+ Z t
0
Z
R
Γ(x−y, t−s)up(y, s)dy ds +
Z t 0
Z
{|y|<(T−s)1/2}
Γ(x−y, t−s)up(y, s)(T−s)−1dy ds.
Observe that the first two integrals are positive, then u(x, t)≥
Z t 0
Z
{|y|<(T−s)1/2}
Γ(x−y, t−s)up(y, s)(T−s)−1dy ds
= 1
√π Z t
0
Z x+(T−s)1
/2 2(t−s)1/2
x−(T−s)1/2 2(t−s)1/2
up(x−2(t−s)1/2z, s) 1
T −se−z2dz ds.
As 0< s < t < T, we note that for 0≤x≤(T−t)1/2, x−(T−s)1/2
2(t−s)1/2 ≤0 and x+ (T −s)1/2
2(t−s)1/2 ≥ (T−s)1/2 2(t−s)1/2 ≥ 1
2, Therefore,
u(x, t)≥ 1
√π Z t
0
Z 1/2 0
up(x−2(t−s)1/2z, s) 1
T−se−z2dz ds. (3.1) On the other hand, by comparison with the heat equation, there exists C0 >0 such that
u(x, t)≥C0, x∈(−T1/2, T1/2), t∈(0, T). (3.2)
Since −T1/2 ≤ x−t1/2 ≤ x−2(t−s)1/2z ≤ x ≤ T1/2, we can use this lower estimate in (3.1) to improve it as follows
u(x, t)≥ C0p
√π Z t
0
Z 1/2 0
1
T−se−z2dz ds=C1log T T−t
, where
C1=AC0p and A= 1 T√
π Z 1/2
0
e−z2dz.
Which is a better lower estimate. Using this new lower estimate in (3.1) u(x, t)≥ C1p
√π Z t
0
Z 1/2 0
log T T−s
p 1
T−se−z2dz ds
=C1pA 2
log T T−t
p+1
. Iterating this procedure we get that
u(x, t)≥Ck
log T T −t
γk
, withγ0= 0,C0given in (3.2) and
γk+1=pγk+ 1, Ck+1 =Ckp A γk+1
. It is easy to see that ask→ ∞,
γk=
k−1
X
j=0
pj → 1
1−p and Ck →
(1−p)A1−p1 .
Therefore, for 0≤x≤(T −t)1/2, u(x, t)≥
(1−p)A1−p1
log T T−t
1−p1
and the proof is complete.
For the linear casep=α= 1 we perform the self-similar change of variables u(x, t) =Aet(T−t)−γv(ξ, τ) ξ=|x|(T−t)−1/2, τ = log 1
T−t . It is easy to see that the rescaled function satisfies
vτ=vξξ−1
2ξvξ+ (χ1−γ)v. (3.3)
Lemma 3.6. There exits a unique(γ∗, F∗)such thatγ∗∈(0,1) andF∗is an even, positive non-increasing (forξ >0) stationary solution of (3.3).
Proof. We look for solutions of the problem Fγ00−1
2ξFγ0 + (χ1−γ)Fγ = 0, ξ >0, Fγ(0) = 1, Fγ0(0) = 0.
(3.4) To study this problem, we use a shooting method. We define
Λ+=
γ∈(0,1) :Fγ >0 and there existsxγ such thatFγ(xγ)>1 Λ−=
γ∈(0,1) : there existsxγ such that Fγ(xγ)<0
Λ∗=
γ∈(0,1) :Fγ≥0 andFγ0 ≤0 .
Notice that these sets are disjoint. Moreover, by continuous dependence of the solution with respect to the parameter γ both sets, Λ+ and Λ− are open sets.
Then, if we prove that both sets are non-empty, we get that Λ∗ is a non-empty closed set and the result follows.
To prove that we consider the boundary casesγ= 1 andγ= 0.
(i) For γ = 1 it is trivial to see that F1(ξ) = 1 for 0 ≤ ξ ≤ 1. Moreover F100(1+) = 1, then the profileF1(ξ) >1 for ξ >1. Now, applying the continuous dependence of the solution with respect to the parameter γ, we get that γ ∈ Λ+
forγ∼1.
(ii) Forγ= 0 we rewrite the equation (3.4) as (e−ξ2/4F00)0=−χ1F0e−ξ2/4,
to get that for 0≤ξ≤1, the profile satisfies thatF00 <0 in the positivity set ofF0. Therefore if there exists ξ0 ∈(0,1] such that F0(ξ0) = 0, the profile is negative in (ξ0, ξ0+ε). On the other hand, ifF0(ξ)>0 in 0≤ξ≤1, we have thatF00(1)<0 and
F00(ξ) =F00(1)eξ
2−1
4 <0, ξ >1.
Therefore, the profile crosses the axis at some pointξ0. The continuous dependence of the solution with respect to the parameterγ implies thatγ∈Λ− forγ∼0.
The uniqueness follows from the fact thatGγ =e−ξ2/4Fγ is a solution of G00γ+1
2ξG0γ+ (χ1+1
2 −γ)Gγ = 0.
Let us suppose that there existsγ16=γ2 in Λ∗. Then γ1
Z
R
Fγ1Gγ2dξ= Z
R
(Fγ001−1
2ξFγ01+χ1Fγ1)Gγ2dξ
= Z
R
F1(G00γ2+1
2ξG0γ2+ (χ1+1
2)Gγ2)dξ
=γ2 Z
R
Fγ1Gγ2dξ,
which is a contradiction.
Remark 3.7. The parameterγ∗can be seen as the first eigenvalue of the operator L(w) =w00−12ξw0+χ1w. Then
−γ∗= inf
w∈X
R
R|w0|2e−ξ2/4dξ−R
Rχ1w2e−ξ2/4dξ R
Rw2e−ξ2/4dξ , whereX is the weightedHρ1(R) space with weightρ(ξ) =e−ξ2/4.
Lemma 3.8. Let v be a solution of (3.3)withγ=γ∗. Then,v is bounded.
Proof. Let us define H(v) =
Z
R
|vξ|2e−ξ2/4dξ− Z
R
(χ1−γ∗)v2e−ξ2/4dξ.
Notice that by the definition ofγ∗ in Remark 3.7, H(v)≥0. On the other hand, multiplying the equation (3.3) byeξ2/4v we get
∂
∂τ2 Z
R
v2e−ξ2/4dξ=−H(v)≤0.
This monotonicity implies thatvis bounded almost everywhere. Therefore, there existsξ0∈R\[−1,1] such thatv(ξ0, τ)≤C. Now we observe that forξ > ξ0, the functionvis a subsolution of
wτ =wξξ−1
2ξwξ−γ∗w ξ > ξ0, τ >0 w(ξ0, τ) =C
w(ξ,0) =v0(ξ).
But for this problemw(ξ) =C is a supersolution. Then,v is uniformly bounded forξ≥ξ0. The same argument provides thatv is uniformly bounded for ξ≤ −ξ0. Finally, forξ∈[−ξ0, ξ0],v is a subsolution of the problem
wτ=wξξ−1
2ξwξ+ (χ1−γ∗)w ξ∈(−ξ0, ξ0), τ >0 w(±ξ0, τ) =C
w(ξ,0) =v0(ξ).
Observe that the function w(ξ) = AF∗(ξ) is a supersolution for A large enough.
Thus,v is uniformly bounded.
From these results we obtain the following blow-up rates.
Lemma 3.9. Let p=α= 1 and(γ∗, F∗)given in Lemma 3.6. Then, there exists a positive constantC1 such that
ku(·, t)k∞≤C1(T−t)−γ∗. Moreover:
(1) If for some 0 ≤ t0 < T, u(x, t0) > F∗(x) for x large, then there exists C2>0 such that
ku(·, t)k∞≥C2(T−t)−γ∗.
(2) In the general case, we have that for allε >0there exists Cε>0such that ku(·, t)k∞≥Cε(T −t)−γ∗+ε.
Proof. The upper blow-up rate follows by the fact thatv is bounded.
If for some timet0 andxlarge enoughu(x, t0)> F∗(x), we only note that forA small enough the functionu=A(T−t)−γ∗F∗(x(T−t)−1/2) is a subsolution of the problem (1.1). Then, the lower blow-up rate follows.
For the general case we can use the profiles given in Lemma 3.6 withγ∈Λ− to obtain the subsolutionu=A(T−t)−γmax{F(x(T−t)1/2),0}.
4. Blow-up set
Using the blow-up rate we can construct subsolutions and supersolutions which determine the blow-up set. We consider the problem
wt=wxx+λw x > x0, 0< t < T w(x0, t) =K(T−t)νe(T−t)
−n
α−1 0< t < T w(x,0) =w0(x) x > x0
(4.1)
It is well know that, see for instance [7],
B(w) =
[x0,∞) forn >1 x0, x0+ 2 α−11 1/2
forn= 1 {x0} forn∈(0,1) {x0} forn= 0 and ν <0 Notice that for fixedn >0 the blow-up set is independent ofν andK.
Lemma 4.1. Let p < 1 ≤ α. Then the blow-up set of a solution of (1.1) is the origin, that is,B(u) ={0}.
Proof. We first note thatuis a supersolution of the equation vt=vxx+vp.
Sincep <1 we get thatv≥((1−p)t)1/(1−p), see [1]. Then
u(x, t)≥v(x, t)≥((1−p)t)1/(1−p). (4.2) Now, we assume thatx1 >0 is a blow-up point, that is x1 ∈B(u), and define 0< x0< x1 andt0 such thatx0= (T−t0)1/2. Using the estimate (4.2),
up(x, t)≤λu, λ=p−1 t0
−1 , fort≥t0. On the other hand, from the upper blow-up rate is
u(x0, t)≤C(T−t)−(α−1)/(1−p) forα <1, while
u(x0, t)≤C
log T T−t
1−p1
≤C(T−t)−1−p1 , forα= 1.
Summing up we obtain that forα≤1 the solutionuis a subsolution of the problem vt=vxx+λv, (x0,∞)×(t0, T)
v(x0, t) =K(T −t)−γ v(x, t0) =u(x, t0)
Note that this is problem (4.1) with λ= ((p−1)t0)−1, ν =γ and n= 0. Then, B(v) ={x0}. Hence, by comparisonu(x1, t) is bounded. A contradiction.
Lemma 4.2. Let p= 1andua solution of (1.1). Then, the blow-up set is
B(u) =
(−∞,∞) forα >2 [−2,2] forα= 2 {0} forα∈[1,2)
Proof. Note that from Lemmas 3.9 and 3.1,uis a subsolution of (4.1) withλ= 1, ν =γ∗, n=α−1>0 andklarge enough. Arguing as in the previous Lemma we obtainB(u) ={0} forα∈[1,2). While for α= 2 we get that ifx0 ∈B(u) then u(x, t) is bounded forx > x0+ 2.
On the other hand, Lemma 3.3 provides a lower bound of u(0, t), then uis a supersolution of (4.1) withλ= 1, ν=π2/4,n=α−1 andk small enough. Then, R=B(w)⊂B(u) forα >2, while forα= 2 we have that [−2,2] =B(w)⊂B(u).
Finally, for the critical case α = 2, we observe that for all ε > 0, the point x0=ε∈B(u), thenu(x, t) is bounded forx∈(ε+ 2,∞). Passing to the limit as
→0 we obtain that B(u) = [−2,2].
Acknowledgements. This works was supported by project MTM2014-53037-P, Spain.
References
[1] J. Aguirre, M. Escobedo;A Cauchy problem forut= ∆u+upwith0< p <1. Asymptotic behaviour of solutions.Ann. Fac. Sci. Toulouse Math., (5) 8 (1986/87), no. 2, 175–203.
[2] C. Br¨andle, F. Quir´os, J. D. Rossi; The role of non-linear diffusion in non-simultaneous blow-up.J. Math. Anal. Appl., 308 (2005), no. 1, 92–104.
[3] K. Deng, H. A. Levine; The role of critical exponents in blow-up theorems: the sequel. J.
Math. Anal. Appl.,243(2000), 85–126.
[4] M. Escobedo, H. A. Levine;Critical blowup and global existence numbers for a weakly coupled system of reaction-diffusion equations.Arch. Rational Mech. Anal., 129 (1995), no. 1, 47–100.
[5] R. Ferreira, A. de Pablo, J. L. V´azquez;Classification of blow-up with nonlinear diffusion and localized reaction.J. Differential Equations, 231 (2006), no. 1, 195–211.
[6] H. Fujita;On the blowing-up of solutions of the Cauchy problem forut= ∆u+u1+α, J. Fac.
Sci. Univ. Tokyo Sec. IA Math.,16(1966), 105–113.
[7] V. A. Galaktionov, S. P. Kurdyumov, A. P. Mikhailov, A. A. Samarski˘ı;Unbounded solutions of semilinear parabolic equations, Keldysh Ins. Appl. Math. Acad. Sci. USSR, Preprint No.
161 (1979).
[8] V. A. Galaktionov, J. L. V´azquez;The problem of blow-up in nonlinear parabolic equations, Discrete Contin. Dynam. Systems A, 8, 399–433, 2002.
[9] K. Hayakawa; On nonexistence of global solutions of some semilinear parabolic equations, Proc. Japan Acad., 49 (1973), 503–525.
[10] A. de Pablo, J. L. Vazquez;Travelling waves and finite propagation in a reaction-diffusion equation. J. Differential Equations, 93 (1991), 19–61.
[11] R. Pinsky; Existence and nonexistence of global solutions forut = ∆u+a(x)upin Rd, J.
Differential Equations, 133 (1997), no. 1, 152–177.
[12] Y. W. Qi;The critical exponents of parabolic equations and blow-up inRn. Proc. Roy. Soc.
Edinburgh Sect. A128(1998), 123–136.
[13] F. Quiros, J. D. Rossi;Non simultaneous blow-up in a semilinear parabolic system., Z. Angew.
Math. Phys., 55 (2004), no. 2, 357–362.
[14] P. Quittner, P. Souplet;Superlinear parabolic problems. blow-up, global existence and steady states, Birkhaeuser Verlag Basel - Boston - Berlin, 2007.
[15] F. B. Weissler;Existence and nonexistence of global solutions for a semilinear heat equation, Israel J. Math., 38 (1981), 29–40.
Raul Ferreira
Departamento de Matem´atica Aplicada, Univ. Complutense de Madrid, 28040 Madrid, Spain
E-mail address:raul [email protected]
