Blow-up set for type I blowing up solutions for a semilinear heat equation

Ann. I. H. Poincaré – AN 31 (2014) 231–247

www.elsevier.com/locate/anihpc

Blow-up set for type I blowing up solutions for a semilinear heat equation

Yohei Fujishima

, Kazuhiro Ishige

aDivision of Mathematical Science, Department of Systems Innovation, Graduate School of Engineering Science, Osaka University, Toyonaka 560-8531, Japan

bMathematical Institute, Tohoku University, Aoba, Sendai 980-8578, Japan Received 3 August 2012; received in revised form 1 March 2013; accepted 8 March 2013

Available online 14 March 2013

Abstract

Letube a type I blowing up solution of the Cauchy–Dirichlet problem for a semilinear heat equation,

⎧⎪

⎨

⎪⎩

∂tu=u+u^p, x∈Ω, t >0, u(x, t)=0, x∈∂Ω, t >0, u(x,0)=ϕ(x), x∈Ω,

(P)

whereΩis a (possibly unbounded) domain inR^N,N1, andp >1. We prove that, ifϕ∈L^∞(Ω)∩L^q(Ω)for someq∈ [1,∞), then the blow-up set of the solutionuis bounded. Furthermore, we give a sufficient condition for type I blowing up solutions not to blow up on the boundary of the domainΩ. This enables us to prove that, ifΩis an annulus, then the radially symmetric solutions of(P )do not blow up on the boundary∂Ω.

©2013 Elsevier Masson SAS. All rights reserved.

1. Introduction

This paper concerns the blow-up problem for a semilinear heat equation,

⎧⎪

⎨

⎪⎩

∂_tu=u+u^p inΩ×(0, T ),

u(x, t )=0 on∂Ω×(0, T )if∂Ω= ∅, u(x,0)=ϕ(x)0 inΩ,

(1.1)

whereΩ is a (possibly unbounded) domain inR^N,N1,∂_t =∂/∂t,p >1,T >0, andϕ∈L^∞(Ω). LetT be the maximal existence time of the unique bounded solutionuof (1.1). IfT <∞, then

lim sup

t→T

u(t )

L^∞(Ω)= ∞,

* Corresponding author.

E-mail address:fujishima@sigmath.es.osaka-u.ac.jp(Y. Fujishima).

0294-1449/$ – see front matter ©2013 Elsevier Masson SAS. All rights reserved.

http://dx.doi.org/10.1016/j.anihpc.2013.03.001

and we callT the blow-up time of the solutionu. The blow-up ofuis said to be of type I if lim sup

t→T

(T−t )^p1−1u(t )

L^∞(Ω)<∞.

Furthermore, the blow-up ofuis said to be of O.D.E. type if lim sup

t→T

(T−t )^p−11 u(t )

L^∞(Ω)=κ withκ= 1

p−1

1/(p−1)

.

If the blow-up ofuis not of type I, then we say that the blow-up ofuis of type II. We denote byB(u)the blow-up set of the solutionu, that is,

B(u)= x∈Ω: there exists a sequence

(x_n, t_n)

⊂Ω×(0, T ) such that lim

n→∞(x_n, t_n)=(x, T ), lim

n→∞u(x_n, t_n)= +∞

.

We remark thatB(u)is a closed set inΩ.

The blow-up set for problem (1.1) has been studied intensively since the pioneering work due to Weissler[32].

See for example[1–3,6–27,31–35], and references therein. See also[30], which includes a good list of references in this topic. Among others, Friedman and McLeod[6]studied the blow-up set by using the comparison principle, and proved the following (see[6, Theorem 3.3]):

(a) IfΩis convex, then the boundary blow-up does not occur, that is,B(u)∩∂Ω= ∅.

In[14–16], Giga and Kohn studied blow-up problem (1.1), and established a blow-up criterion for the solutions in the case where(N−2)p < N+2. This criterion implies the following:

(b) IfΩ is a (possibly unbounded) convex domain and(N−2)p < N+2, then the blow-up set B(u)is bounded provided thatϕ∈H¹(Ω);

(c) IfΩis strictly star-shaped abouta∈∂Ω and(N−2)p < N+2, thena /∈B(u).

For assertion (b), see[16, Theorem 5.1, Remarks 5.2 and 5.4]and for assertion (c), see[16, Theorem 5.3]. Assertion (b) was also proved in[12]and[13]for the one dimensional case, with the initial functionϕwhich deceases monotonically to 0 and which satisfies 0ϕ(x)C|x|^−2/(p−1) for some constantC. On the other hand, in[19], the second author of this paper and Mizoguchi proved that a blow-up criterion similar to that of[14–16]holds for type I blowing up solutions without the convexity of the domainΩ, and obtained the following:

(d) IfΩ is a bounded smooth domain inR^N and(N−2)pN+2, then type I blowing up solutions do not blow up on the boundary∂Ω.

Unfortunately, ifΩ is not convex, then there are few results, except assertion (d), identifying whether the boundary blow-up occurs or not, and the following problem is still open as far as we know:

(P ) LetΩbe an annulus inR^N. Then does the radially symmetric solution of (1.1) blow up on the boundary∂Ω?

We remark that there exists a solution blowing up on the boundary of the domain for the equation

∂_tu=u_xx+k u^m

x+u^2m−1, wherem >1 and large enoughk >2/√

m(see[4]).

In this paper we prove that the blow-up set of the solutionuof (1.1) is bounded if the blow-up of the solutionu is of type I and the initial function ϕ∈L^∞(Ω)∩L^q(Ω)for some q ∈ [1,∞). Furthermore, we give a sufficient condition for the solutionunot to blow up on the boundary of the domainΩ, and prove that, ifΩ is annulus, then the radially symmetric solution does not blow up on the boundary∂Ω. In addition, we prove that, ifΩ satisfies the

exterior sphere condition and the solutionuof (1.1) exhibits O.D.E. type blow-up, then the solution does not blow-up on the boundary∂Ω.

We introduce some notation. LetB(x, r)= {y∈R^N: |y−x|< r}forx∈R^Nandr >0. For any bounded contin- uous functionf onΩand any constantη, we put

M(f, η):=

x∈Ω: f (x)fL^∞(Ω)−η . For anyφ∈L^∞(R^N), let

e^t ϕ

(x):=(4π t)^−N2

R^N

e^−|x−y|

2

4t φ(y) dy.

For anyλ >0, letζλbe a solution ofζ=ζ^pwithζ (0)=λ, that is, ζ_λ(t):=κ(S_λ−t )^−p−11 withS_λ=λ^−(p−1)

p−1 . (1.2)

Now we are ready to state the main results of this paper. The first theorem concerns the boundedness of the blow-up set for problem (1.1).

Theorem 1.1.Letube a solution of (1.1)which exhibits type I blow-up att=T. Ifϕ∈L^∞(Ω)∩L^q(Ω)for some q∈ [1,∞), then

sup

x∈Ω\B(0,R), t∈(0,T )

u(x, t )<∞

for someR >0. In particular, the blow-up setB(u)is bounded.

In the second theorem we give a result on the relationship between the location of the blow-up set and the level sets of the solution just before the blow-up time. Theorem1.2also gives a sufficient condition for type I blowing up solutions of (1.1) not to blow up on the boundary∂Ω.

Theorem 1.2.Letube a solution of(1.1)which exhibits type I blow-up att=T. Assume

tlim→T(T−t )^p−11+12∇u(t )

L^∞(Ω)=0. (1.3)

Then the blow-up ofuis of O.D.E. type. Furthermore, for anyη∈(0, κ), there exists a constantT∈(0, T )such that

B(u)⊂

T<t <T

M

(T −t )^p−11u(t ), η

. (1.4)

In particular, the solutionudoes not blow up on the boundary∂Ω, that is,B(u)∩∂Ω= ∅.

Here we remark that, ifΩis a smooth bounded domain and(N−2)p < N+2, then the blow-up of the solution is of type I and (1.3) holds (see Theorem 1.1 in[26]).

As an application of Theorem1.2, we give the following result, which gives an affirmative answer to problem(P ).

Corollary 1.1.Let Ω=

x∈R^N: a <|x|< b

, 0< a < b <∞.

Then the radially symmetric solution of (1.1)does not blow up on the boundary∂Ω.

Furthermore, we give the following theorem with the aid of Corollary1.1.

Theorem 1.3.LetΩbe a bounded domain inR^Nsatisfying the exterior sphere condition. Letube a solution of (1.1) which exhibits O.D.E. type blow-up. Then the solutionudoes not blow up on the boundary∂Ω.

In this paper we improve the arguments in [8], and give a blow-up criterion for the semilinear heat equations with small diffusion (see Proposition2.1). This blow-up criterion enables us to study the location of the blow-up set for problem (1.1) by using the profile of the solution just before the blow-up time and to obtain Theorems1.1 and1.2. Furthermore, for the radially symmetric solutions of (1.1) in an annulus, we apply the arguments in[5]

and[28]with the aid of[26,27,29], and obtain the blow-up estimates of the solution and its gradient. Then we can prove Corollary1.1with the aid of Theorem1.2. In addition, we prove Theorem1.3by using Proposition2.1and Corollary1.1.

The rest of this paper is organized as follows: In Section 2 we give some preliminary results on the blow-up problem (1.1). Section 3 is devoted to the proofs of Theorems1.1,1.2, and Corollary1.1. In Section 4 we prove Theorem1.3.

2. Preliminaries

In this section we give preliminary results on the blow-up problem for the semilinear heat equations. We first give a lemma on O.D.E. type blowing up solutions.

Lemma 2.1.Assume the same conditions as in Theorem1.2. Then the blow-up of the solutionuis of O.D.E. type.

Proof. We denote byT the blow-up time of the solutionuof (1.1). Let >0 be a sufficiently small constant. Put w(x, t):=^p1−1u(x, T −+t ), ϕ(x):=^p1−1u(x, T −). (2.1) Thenwblows up att=1 and satisfies

⎧⎪

⎨

⎪⎩

∂_tw=w+w^p inΩ×(0,1), w(x, t)=0 on∂Ω×(0,1), w(x,0)=ϕ(x) inΩ.

(2.2)

By the comparison principle we see that w(t)

L^∞(Ω)ζλ(t) for 0< t < Sλ,

whereλ= ϕL^∞(Ω), and obtainS_λ1. This together with (1.2) implies

ϕL^∞(Ω)κ. (2.3)

Furthermore, since the blow-up ofuis of type I, by (2.1) we can find a positive constantCsuch that ϕL^∞(Ω)=^p−11u(T −)

L^∞(Ω)^p−11 ·C

T −(T−)_−p−¹₁

C. (2.4)

On the other hand, by (1.3) and (2.1) we have

lim→0¹²∇ϕL^∞(Ω)=lim

→0^p−11+12∇u(T−)

L^∞(Ω)

=lim

t→T(T−t )^p−11+12∇u(t )

L^∞(Ω)=0. (2.5)

Then, by (2.3)–(2.5) we apply[7, Proposition 1]to problem (2.2), and obtain

lim→0S_λ=1.

This together with (1.2) yields lim→0ϕL^∞(Ω)=κ, and we obtain

tlim→T(T −t )^p−11u(t )

L^∞(Ω)=lim

→0^p1−1u(T−)

L^∞(Ω)=lim

→0ϕL^∞(Ω)=κ.

Thus the blow-up of the solutionuis of O.D.E. type, and Lemma2.1follows. 2

Next we consider the blow-up problem for a semilinear heat equation with small diffusion. Letube a solution of

⎧⎪

⎨

⎪⎩

∂tu=u+u^p inΩ×(0, T),

u(x, t )=0 on∂Ω×(0, T)if∂Ω= ∅, u(x,0)=ϕ(x)0 inΩ,

(2.6)

whereN1, Ω is a domain inR^N,p >1, >0, andϕ∈L^∞(Ω). LetT andB be the blow-up time and the blow-up set of the solutionu of problem (2.6), respectively. The rest of this section is devoted to the proof of the following proposition, which is the main ingredient of this paper and a modification of[8, Proposition 4.1].

Proposition 2.1.Letube a solution of(2.6)withT=1such that sup

0<<₀

sup

0<t <1

(1−t )^p1−1u(t)

L^∞(Ω)C_∗ (2.7)

for some0>0andC_∗>0. LetΩbe a domain such thatΩ⊂Ω and{ ˜ϕ}0<<₀ a family of functions belonging toW^1,∞(Ω)such that

0ϕϕ˜ inΩfor all∈(0, 0), (2.8)

sup

0<<0

˜ϕL^∞(Ω)<∞. (2.9)

Assume that there exists a constantη >0such that

˜

ϕ(x) < κ−η on∂Ωif∂Ω= ∅. (2.10)

Then, for anyδ >0, there exist positive constantsσ and₁such that, if sup

0<<1

¹²∇ ˜ϕ_L∞({x∈Ω:κ−ηϕ˜(x)κ})σ, (2.11)

then there holds B⊂

x∈Ω: ϕ˜(x)κ−δ

, 0< < 1. (2.12)

Here the constantsσ and1are independent of the domainΩ.

Letδ >0. Letσ and₁ be sufficiently small positive constants to be chosen later, and assume (2.11). Letα∈ (0,min{κ, η}/10). For any∈(0, 1), put

ϕ^∗(x)=

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

κ−α ifx∈Ωandϕ˜(x)κ−α,

˜

ϕ(x) ifx∈Ωandκ−10αϕ˜(x)κ−α, κ−10α ifx∈Ωandϕ˜(x)κ−10α,

κ−10α ifx∈R^N\Ω.

(2.13)

By (2.10) and (2.13) we see thatϕ^∗∈W^1,∞(R^N). Letβ andγ be positive constants to be chosen later, and put z(x, t ):=

e^tϕ^∗

(x), (2.14)

w(t):=(κ−3α)^−(p−1)+βσ

1−(1−t )¹²

, (2.15)

and

fγ(t):=e^{γ t}

e^2(p−1)γ−e^{(p−1)γ t}_−p−¹₁ . Here the functionf_γ satisfies

f_γ(t)=γ

f_γ(t)+f_γ(t)^p

, 0< t <2, (2.16)

and there exists a positive constantc_γ, depending only onpandγ, such that c_γ inf

0<t <1f_γ(t) < sup

0<t <1

f_γ(t)c⁻_γ¹. (2.17)

Furthermore, we define the following three functionsv₁,v₂, andvby v₁(x, t):=

z(x, t )^−(p−1)−(p−1)t₋ ¹

p−1, (2.18)

v₂(x, t):=

z(x, t )^−(p−1)−w(t)_−p−1¹

, (2.19)

v(x, t ):=v1(x, t)+σ^p−21v2(x, t)²+fγ(t). (2.20)

Then we prove the following proposition.

Lemma 2.2. Assume the same conditions as in Proposition2.1. Then, for any α∈(0,min{κ, η}/10), there exist positive constantsβ₁,γ,σ, and₁such that, ifϕ˜satisfies(2.11), then the functionvdefined by(2.20)satisfies

∂_tvv+v^p inE (2.21)

for anyββ₁and∈(0, 1), where E:=

(x, t)∈R^N×(0,1): z(x, t )^−(p−1)−w(t)1 2C⁻

p−1

∗ 2 σ (1−t )¹²

. (2.22)

HereC_∗is the constant given in(2.7).

Proof. Letσand1be positive constants to be chosen later, and assume (2.11). We first prove the following inequal- ities,

κ−10αz(x, t )κ−α inR^N×(0,∞), (2.23)

∇z(t )

L^∞(R^N)⁻¹²σ in(0,∞), (2.24)

v₁(x, t)C inR^N×(0,1), (2.25)

v^p−v₁^pC

σ^p−12 v²₂+σ

p−12p v₂^2p+f_γ+f_γ^p

inE, (2.26)

for all ∈(0, 1), where C is a positive constant, independent ofβ andγ. The inequality (2.23) easily follows from (2.13) and the comparison principle. By (2.11) and (2.13) we have

sup

t >0

∇z(t )

L^∞(R^N)∇ϕ^∗

L^∞(R^N)∇ ˜ϕ_L∞({x∈Ω:κ−10αϕ˜(x)κ−α})⁻¹²σ, and obtain the inequality (2.24). On the other hand, since

(κ−α)^−(p−1)−(p−1)=(p−1)

1−κ⁻¹α₋(p−1)

−1

>0, by (2.18) and (2.23) we have

v₁(x, t)

(κ−α)^−(p−1)−(p−1)₋ ¹

p−1 =κ

1−κ⁻¹α₋(p−1)

−1₋ ¹

p−1,

and obtain (2.25). The inequality (2.26) is obtained by the same argument as in (3.19) of[8], and we omit its details.

Next we prove (2.21) by using (2.23)–(2.26). Letβ andγ be positive constants to be chosen later. By (2.16) and (2.20) we obtain

∂_tv−

v+v^p

2

p−1σ^p2−1w(t)v^p₂⁺¹+γ

f_γ(t)+f_γ(t)^p

−pv^2p₁ ⁻¹z^−2p|∇z|²−2(p+1)σ^p−21v₂^2pz^−2p|∇z|²−

v^p−v^p₁ for all(x, t)∈E. Then, by (2.23)–(2.26) there exists a constantC₁, independent ofβandγ, such that

∂tv−

v+v^p

2

p−1σ^p2−1w(t)v^p₂⁺¹+γ

fγ(t)+fγ(t)^p

−C₁σ²−C₁σ^p−21+2v₂^2p−C₁

σ^p−21v²₂+σ

2p

p−1v^2p₂ +f_γ+f_γ^p

(2.27)

for all(x, t)∈E. Letγbe a positive constant such thatγ3C1. By (2.17), taking a sufficiently smallσif necessary, we have

(γ−C1)

fγ(t)+fγ(t)^p

−C1σ²2C1

cγ+c_γ^p

−C1σ²C1cγ. This together with (2.15) and (2.27) implies that

∂tv−

v+v^p

β

p−1σ

p+1

p−1(1−t )⁻¹²v^p₂⁺¹+C1cγ−C1

σ^p2−1v₂²+2σ

2p p−1v₂^2p

(2.28) for all(x, t)∈E.

Let

βmax

8(p−1)C1C

p−1

∗2 , c_γ^−(p−1),4C²₁(p−1)²

. (2.29)

By (2.19) and (2.22) we have v₂(x, t)^p−1=

z(x, t )^−(p−1)−w(t)₋1

2C

p−1

∗2 σ⁻¹(1−t )⁻¹², (x, t )∈E, and by (2.29) we obtain

2C1σ

p−12p v₂^2p=2C1σ v^p₂⁻¹·σ

p+1 p−1v^p₂⁺¹ 4C1C

p−1

∗2 (1−t )⁻¹²σ^p+1p−1v₂^p+1 β

2(p−1)(1−t )⁻¹²σ^p+1p−1v₂^p+1 (2.30) for all(x, t)∈E. Therefore, by (2.28) and (2.30), we obtain

∂_tv−

v+v^p

β

2(p−1)σ

p+1

p−1(1−t )⁻¹²v₂^p+1+C₁c_γ −C₁σ^p−21v²₂ (2.31) for all(x, t)∈E.

Put E_,1=

(x, t)∈E: z(x, t )^−(p−1)−w(t)β¹²σ

, E_,2=E\E₁. By (2.19) and (2.29) we have

C1σ^p−21v₂²C1σ^p2−1

β¹²σ_−p−1²

=C1β^−p1−1 C1c_γ (2.32)

for all(x, t)∈E_,1. On the other hand, since (1−t )⁻¹² 1, σ v^p₂⁻¹σ

β¹²σ₋1

=β⁻¹², for all(x, t)∈E_,2, by (2.29) we have

β

2(p−1)σ^p+1p−1(1−t )⁻¹²v^p₂⁺¹= β

2(p−1)(1−t )⁻¹²σ v₂^p−1·σ^p2−1v₂²

β^1/2

2(p−1)σ^p−12 v₂²C₁σ^p−12 v²₂ for all(x, t)∈E_,2. This together with (2.32) implies

β 2(p−1)σ

p+1

p−1(1−t )⁻¹²v^p₂⁺¹+C₁c_γC₁σ^p−12 v₂² (2.33) for all(x, t)∈E. Therefore, by (2.31) and (2.33) we have (2.21) for all(x, t)∈E. Thus Lemma2.2follows. 2

Letβ₁be the constant given in Lemma2.2, and put β=max

β1,C_∗^−(p−1)/2 2

. (2.34)

Letχbe aC^∞smooth function inRsuch that

χ (z)=1/4 forz0, χ (z)=z forz1/2, 0χ(z)1 inR, and put

u(x, t)=v₁(x, t)+C_∗(1−t )^−p−11χ

z(x, t )^−(p−1)−w(t) C⁻_∗^(p−1)/2σ (1−t )^1/2

_−p²₋₁

+f_γ(t). (2.35)

This together with (2.20) and (2.22) implies that

u(x, t)=v(x, t ) inE. (2.36)

Here we prove the following lemma.

Lemma 2.3.Letube the function defined in(2.35). Then

u(x,0)ϕ(x), x∈Ω. (2.37)

Proof. For anyx∈Ω withϕ˜(x)κ−2α, by (2.8) and (2.13) we have

u(x,0)v₁(x,0)=ϕ^∗(x)ϕ˜(x)ϕ(x). (2.38)

On the other hand, for anyx∈Ωwithϕ˜(x) > κ−2α, we have z(x,0)=ϕ^∗(x) > κ−2α.

Then, by (2.15) and (2.23) we have

z(x,0)^−(p−1)−w(0) < (κ−2α)^−(p−1)−(κ−3α)^−(p−1)0.

This together with (2.7) and (2.35) implies u(x,0)C_∗χ

z(x,0)^−(p−1)−w(0) C_∗^−(p−1)/2σ

_−p²₋₁

=16^p−11C_∗

C_∗u(x,0)=ϕ(x). (2.39)

Therefore, by (2.38) and (2.39) we have the inequality (2.37), and Lemma2.3follows. 2 Now we are ready to complete the proof of Proposition2.1.

Proof of Proposition2.1. Leth∈C¹(R)be such that

h(z)= −1 forz1, h(z)=1 forz4,0h(z)1 inR.

By (2.7) we have h

u(x, t)^p−1 C_∗^p−1(1−t )⁻¹

= −1 inΩ×(0,1), and see thatusatisfies

∂tu=u+u^p +1 2

h

u^p−1 C^p_∗⁻¹(1−t )⁻¹

+1

G(x, t) inΩ×(0,1), (2.40)

where

G(x, t)=∂tu−

u+u^p .

On the other hand, by Lemma2.2and (2.36) we have

∂_tuu+u^p indE, that is, G0 inE. (2.41)

Furthermore, since χ

z(x, t)^−(p−1)−w(t) C_∗^−(p−1)/2σ (1−t )^1/2

1

2 inR^N× [0,1)\E, we have

u(x, t)4^1/(p−1)C_∗(1−t )^−1/(p−1), (x, t )∈R^N× [0,1)\E, and obtain

h

u(x, t)^p−1 C_∗^p−1(1−t )⁻¹

=1, (x, t )∈R^N× [0,1)\E. (2.42)

Sinceh1, by (2.41) and (2.42) we have

∂tu−

u+u^p +1 2

h

u^p−1 C_∗^p−1(1−t )⁻¹

+1

G(x, t)

=1 2

1−h

u^p−1 C_∗^p−1(1−t )⁻¹

G(x, t)0 inΩ×(0,1). (2.43)

Therefore, by (2.37), (2.40), and (2.43) we apply the comparison principle to obtain

u(x, t)u(x, t) inΩ× [0,1). (2.44)

Without loss of generality we can assume thatδ∈(0,min{κ, η}/2), and let α=δ/5∈

0,min{κ, η}/10 .

Let 0< < 1 andx ∈Ω be such thatϕ˜(x) < κ −δ. Then there exists a positive constantR, depending on andx, such that

˜

ϕ(x) < κ−δ=κ−5α, x∈B(x, R)∩Ω.

Then, by (2.13) we have

z(x,0)=ϕ^∗(x)κ−5α (2.45)

for allx∈B(x, R)∩Ω. Furthermore, by[7, Lemma 1], taking sufficiently smallσ and₁if necessary, we have sup

0<<₁

sup

0<t <1

z(t )−z(0)

L^∞(R^N)< α.

This together with (2.45) implies that z(x, t )κ−4α, (x, t )∈

B(x, R)∩Ω

× [0,1), (2.46)

for all∈(0, 1). On the other hand, letC₁be a positive constant such that

(κ−4α)^−(p−1)−(κ−3α)^−(p−1)C₁. (2.47)

Then, by (2.15), (2.34), (2.46), and (2.47), taking a sufficiently smallσ if necessary, we obtain z(x, t )^−(p−1)−w(t)(κ−4α)^−(p−1)−

(κ−3α)^−(p−1)+βσ

1−(1−t )¹² C₁−βσ +βσ (1−t )¹² C₁

2 +1 2C⁻

p−1

∗ 2 σ (1−t )¹²

max

1 2C1,1

2C⁻

p−1

∗ 2 σ (1−t )¹²

(2.48) for all (x, t)∈(B(x, R)∩Ω)× [0,1). This implies that (B(x, R)∩Ω)× [0,1)⊂E (see (2.22)). Therefore, by (2.17), (2.20), (2.25), (2.36), (2.44), and (2.48) we have

u(x, t)u(x, t)=v(x, t )v₁(x, t)+σ^p−12 (C₁/2)^−p−12 +c_γ⁻¹C₂

for all(x, t)∈(B(x, R)∩Ω)× [0,1), whereC2is a constant. This impliesx∈/B. Therefore, by the arbitrariness ofx, we have (2.12) for all∈(0, 1), and the proof of Proposition2.1is complete. 2

3. Proof of Theorems1.1and1.2

We prove Theorem1.1and Theorem1.2by using Proposition2.1.

Proof of Theorem1.1. Let₀be a sufficiently small positive constant. Put

u(x, τ ):=^p1−1u(x, T −+τ ), ϕ(x):=^p−11u(x, T −), M:= sup

0<t <T−

u(t )

L∞(Ω), for all∈(0, 0). Thenusatisfies

⎧⎪

⎨

⎪⎩

∂τu=u+u^p inΩ×(0,1), u(x, τ )=0 on∂Ω×(0,1), u(x,0)=ϕ(x) inΩ,

(3.1) andu blows up atτ=1. This implies thatϕL^∞(Ω)κ (see (2.3)). Furthermore, since the blow-up of the solu- tionuis of type I, we have

d_∗:= sup

0<<₀ϕL^∞(Ω)<∞. (3.2)

On the other hand, lettingϕ=0 outsideΩ, we apply the comparison principle to obtain 0u(x, t )e^Mp−1t

e^t ϕ

(x) inΩ×(0, T −). (3.3)

Furthermore, sinceϕ∈L^q(R^N), for anyδ >0, we take a sufficiently largeRso that

R^N\B(0,R)

ϕ(y)^qdyδ.

This together with the Hölder inequality implies that e^t ϕ

(x)^q(4π t)^−N2

R^N

e^−|x−y|

2

4t ϕ(y)^qdy

=(4π t)^−N2

B(0,R)

+

R^N\B(0,R)

e^−|x−y|

2

4t ϕ(y)^qdy (4π t)^−N2e^−(|x|−R)

2

4t ϕ^q_Lq(R^N)+(4π t)^−N2δ (3.4)

for allx∈R^N\B(0, R). Therefore, sinceδis arbitrary, by (3.3) and (3.4) we have

Llim→∞u(T −)

L^∞(Ω\B(0,L))=0.

Then we can take a positive constantLsatisfying

0ϕ(x)κ/2 (3.5)

for allx∈Ω with|x|L. For anyx∈R^N, we put

˜ ϕ(x)=

⎧⎪

⎨

⎪⎩

ϕL^∞(Ω) if|x|L,

−(|x| −L)+ ϕL^∞(Ω) ifL<|x|L+ ϕL^∞(Ω)−κ/2, κ/2 if|x|> L+ ϕL^∞(Ω)−κ/2.

Then we have

˜

ϕ∈W^1,∞ R^N

, ϕL^∞(Ω)= ˜ϕL^∞(R^N), ∇ ˜ϕL^∞(R^N)1, (3.6) and by (3.5) we obtain

ϕ(x)ϕ˜(x) inΩ. (3.7)

Therefore, by (3.2), (3.6), and (3.7) we apply Proposition2.1withδ=κ/4, and obtain B(u)=B(u)⊂

x∈Ω: ϕ˜(x)3κ/4

⊂B

0, L+ ϕL^∞(Ω)−κ/2

for all sufficiently small >0. This means thatB(u)is bounded, and the proof of Theorem1.1is complete. 2 Proof of Theorem1.2. We use the same notation as in the proof of Theorem1.1. By (1.3) we have

lim→0¹²∇ϕL^∞(Ω)=0.

Then, for anyη >0, we apply Proposition2.1withϕ˜=ϕ andΩ=Ω tou, and have

B(u)⊂M(ϕ, η), 0< < ₀, (3.8)

for some₀>0. Therefore, sinceB(u)=B(u), by (3.8) we have

B(u)⊂

0<<₀

M

^p−11 u(T −), η .

This implies (1.4). Furthermore, by Lemma2.1, we see that the blow-up of the solution u is of O.D.E. type, and Theorem1.2follows. 2

Next we prove Corollary1.1by using Theorem1.2with the aid of blow-up estimates of the solutions.

Proof of Corollary1.1. LetΩ= {a <|x|< b}with 0< a < b <∞. Letube a radially symmetric solution of (1.1) blowing up att=T. Then, due to Theorem1.2, it suffices to prove

sup

0<t <T

(T−t )^p−11u(t )

L^∞(Ω)<∞, (3.9)

tlim→T(T−t )^{p−11 +12}∇u(t )

L^∞(Ω)=0. (3.10)

We first prove (3.9) by the same argument as in the proof of[5, Theorem 2.1]. For anyt∈(0, T ), we put M(t ):= uL^∞(Ω×(0,t )), λ(t ):=M(t )^−p−21.

SinceM(t )is a positive, continuous, and nondecreasing function on(0, T )such thatM(t )→ ∞ast→T, we can defineτ (t )by

τ (t ):=max

τ∈(0, T ): M(τ )=2M(t )

, 0< t < T .

Then, similarly to[5], it suffices to prove that there exists a constantKsuch that λ(t )⁻²

τ (t )−t

K, t∈(T /2, T ). (3.11)

We prove (3.11) by contradiction. Assume that there exists a sequence{t_j}such that

jlim→∞λ(t_j)⁻²

τ (t_j)−t_j

= ∞.

For anyj=1,2, . . ., we take a sequence{(r_j,tˆ_j)} ⊂ [a, b] ×(0, tj]satisfying u(rj,tˆj)1

2M(tj).

Putλj=λ(tj)and vj(τ, s):=λ

2 p−1

j u

λjτ +rj, λ²_js+ ˆtj

for(τ, s)∈Ij×

−λ⁻_j²tˆj, λ⁻_j²(T− ˆtj) , whereI_j:= {τ∈R: λ_jτ+r_j∈(a, b)}. Thenv_j satisfies

∂svj=∂_τ²vj+λj

N−1

r_j+λ_jτ∂τvj+v^p_j inIj×

−λ⁻_j²tˆj, λ⁻_j²(T− ˆtj) .