Computers and Mathematics with Applications 60 (2010) 670–679
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Computers and Mathematics with Applications
journal homepage: www.elsevier.com/locate/camwa
Blow-up and global existence for nonlinear parabolic equations with
Neumann boundary conditionsI
Juntang Ding a,∗ , Bao-Zhu Guo a,b,c
a
School of Mathematical Sciences, Shanxi University, Taiyuan 030006, PR China
b
Academy of Mathematics and Systems Science, Academia Sinica, Beijing 100190, PR China
c
School of Computational and Applied Mathematics, University of the Witwatersrand, Wits 2050, Johannesburg, South Africa
article
info
Article history:
Received 4 January 2010
Received in revised form 17 May 2010
Accepted 17 May 2010
Keywords:
Parabolic equation
Global solution
Blow-up solution
Neumann boundary condition
abstract
In this paper, we study the global and blow-up solutions of the following problem:

(h(u)) = ∇ · (a(u, t )b(x)∇ u) + g (t )f (u)

 ∂u t
=0

 ∂n
u(x, 0) = u0 (x) > 0
in D × (0, T ),
on ∂ D × (0, T ),
in D,
where D ⊂ RN is a bounded domain with smooth boundary ∂ D. By constructing auxiliary
functions and using maximum principles and comparison principles, the sufficient
conditions for the existence of global solution, an upper estimate of the global solution,
the sufficient conditions for the existence of the blow-up solution, an upper bound for the
‘‘blow-up time’’, and an upper estimate of the ‘‘blow-up rate’’ are specified under some
appropriate assumptions on the functions a, b, f , g , and h.
© 2010 Elsevier Ltd. All rights reserved.
1. Introduction
Blow-up and global existence for nonlinear parabolic equations are discussed by many authors (see e.g., [1–8]). In this
paper, we study the global and blow-up solutions of the following problem:

(h(u))t = ∇ · (a(u, t )b(x)∇ u) + g (t )f (u)


 ∂u
=0

∂n


u(x, 0) = u0 (x) > 0
in D × (0, T ),
on ∂ D × (0, T ),
(1.1)
in D,
where D ⊂ R is a bounded domain with smooth boundary ∂ D, ∂/∂ n represents the outward normal derivative on ∂ D,
u0 is the initial value, T is the maximal existence time of u, and D is the closure of D. Set R+ = (0, +∞). We assume,
throughout the paper, that a is a positive C 2 (R+ × R+ ) function, b is a positive C 1 (D) function, f is a positive C 2 (R+ )
function, g is a positive C (R+ ) function, h is a C 2 (R+ ) function, h0 (s) > 0 for any s ∈ R+ , and u0 is a positive C 2 (D) function.
Under these assumptions, the classical parabolic equation theory assures that there exists a unique classical solution u(x, t )
N
I This work was supported by the National Natural Science Foundation of China, the Natural Science Foundation of Shanxi Province, and the Foundation
of Shanxi Province for Overseas Returned Scholars.
∗ Corresponding author.
E-mail address: [email protected] (J. Ding).
0898-1221/$ – see front matter © 2010 Elsevier Ltd. All rights reserved.
doi:10.1016/j.camwa.2010.05.015
J. Ding, B.-Z. Guo / Computers and Mathematics with Applications 60 (2010) 670–679
671
for problem (1.1) with some T > 0 and the solution is positive over D × [0, T ). Moreover, by regularity theorem [9],
u ∈ C 3 (D × (0, T )) ∩ C 2 (D × [0, T )).
The problems of global and blow-up solution for nonlinear parabolic equations with Neumann boundary conditions have
been investigated extensively by many authors (see [10–16]). Some special cases of (1.1) have been treated already. Lair and
Oxley [17] studied the following problem:

ut = ∇ · (a(u)∇ u) + f (u)


 ∂u
=0

∂n


u(x, 0) = u0 (x) > 0
in D × (0, T ),
on ∂ D × (0, T ),
in D,
where D is a bounded domain of RN with smooth boundary. The necessary and sufficient conditions characterized by
functions a and f were given for the existence of global and blow-up solutions. Imai and Mochizuki [18] considered the
following problem:

(h(u))t = 1u + f (u)


 ∂u
=0

∂n


u(x, 0) = u0 (x) > 0
in D × (0, T ),
on ∂ D × (0, T ),
in D,
where D is a bounded domain of RN with smooth boundary. The sufficient conditions were developed for the existence of
global and blow-up solutions. Gao, Ding and Guo [19] discussed the following problem:

(h(u))t = ∇ · (a(u)∇ u) + f (u)


 ∂u
=0

∂n


u(x, 0) = u0 (x) > 0
in D × (0, T ),
on ∂ D × (0, T ),
in D,
N
where D is a bounded domain of R with smooth boundary. The sufficient conditions were obtained there for the existence
of global solution and blow-up solution. Meanwhile, the upper estimate of the global solution, the upper bound of the ‘‘blowup time’’ and upper estimate of ‘‘blow-up rate’’ were also given.
It seems that the method of [19] is not applicable for the problem (1.1). In this paper, by constructing completely different
auxiliary functions with those in [19] and technically using maximum principles and comparison principles, we investigate
the problem (1.1). We obtain some existence theorems of global solution, an upper estimate of the global solution, existence
theorems of blow-up solution, an upper bound of ‘‘blow-up time’’, and an upper estimates of ‘‘blow-up rate’’. Our results
extend and supplement those obtained in [19].
We proceed as follows. In Section 2 we study the global solution of the problem (1.1). Section 3 is devoted to the blow-up
solution of (1.1). A few examples are given in Section 4 to illustrate the applications of the abstract results.
2. Global solution
Our main result is the following Theorem 2.1.
Theorem 2.1. Let u be a solution of (1.1). Assume that
(i) For any (s, t ) ∈ R+ × R+ ,
1
a(s, t )f (s)
a(s, t )
+
h0 (s)
s
s
1
g (t )
as (s, t )
a(s, t )
≤ 0.
(2.1)
t
(ii)
Z
+∞
h0 (s)
f (s)
M0
+∞
Z
g (t )dt ,
ds ≥
0
M0 = max u0 (x).
(2.2)
D
Then the solution u of (1.1) must be a global solution, and
u(x, t ) ≤ φ
−1
t
Z
g (t )dt
,
∀(x, t ) ∈ D × R+ ,
(2.3)
0
where
φ(z ) =
Z
z
M0
h0 ( s )
f (s)
ds,
∀z ≥ M0 ,
and φ −1 is the inverse function of φ .
(2.4)
672
J. Ding, B.-Z. Guo / Computers and Mathematics with Applications 60 (2010) 670–679
Proof. Let u be a solution of the following problem:

(h(u))t = ∇ · (a(u, t )b(x)∇ u) + g (t )f (u)


 ∂u
=0

∂n


u(x, 0) = M0 > 0
in D × (0, T ),
on ∂ D × (0, T ),
(2.5)
in D.
It follows from the comparison principle [20] that u is an upper solution of (1.1). We show that u must be global, and so is
for the solution of (1.1).
Consider the auxiliary function
Φ (x, t ) = h0 (u)ut − g (t )f (u).
(2.6)
We find
∇ Φ = h00 ut ∇ u + h0 ∇ ut − gf 0 ∇ u,
(2.7)
1Φ = h |∇ u| ut + 2h ∇ u · ∇ ut + h ut 1u + h 1ut − gf |∇ u| − gf 1u,
000
00
2
00
0
00
0
2
(2.8)
and
Φt = h0 (u)ut − g (t )f (u) t = [(h(u))t − g (t )f (u)]t
= [∇ · (a(u, t )b(x)∇ u)]t = ab1u + aū b|∇ u|2 + a∇ b · ∇ u
t
= aū but 1u + at b1u + ab1ut + aūū b|∇ u|2 ut + aūt b|∇ u|2 + 2aū b∇ u · ∇ ut
+ aū ut ∇ b · ∇ u + at ∇ b · ∇ u + a∇ b · ∇ ut .
(2.9)
By (2.8) and (2.9), it follows that
ab
h
1Φ − Φt =
0
abh000
h0
−
2abh00
abh00
− aūū b |∇ u|2 ut +
−
2a
b
∇
u
·
∇
u
+
−
a
b
ut 1u
ū
t
ū
0
0
h
abgf
00
+ aūt b |∇ u|2 −
h0
h
0
abgf
+ at b 1u − aū ut ∇ b · ∇ u − at ∇ b · ∇ u − a∇ b · ∇ ut .
h0
(2.10)
With (2.5), we have
1u =
h0
ab
ut −
aū
a
1
gf
b
ab
|∇ u|2 − ∇ b · ∇ u −
.
(2.11)
Substitute (2.11) into (2.10) to obtain
ab
h
1Φ − Φt =
0
abh000
(aū )2 b
aū h0
00
−
2a
b
∇
u
·
∇
u
+
h
−
(ut )2
ū
t
h0
a
h0
h0
a
ah00
aū gf
gfh00
at h 0
aū bgf 0
aū at b
abgf 00
− 0 ut ∇ b · ∇ u +
− 0 − gf 0 −
ut +
+
−
−
a
b
|∇ u|2
ūt
0
0
− aūū b +
h
+
−
aū bh00
a
agf
0
h0
2
∇b · ∇u +
g ff
|∇ u|2 ut +
h
0
+
h0
at gf
a
2abh00
a
h
a
h
− a∇ b · ∇ ut .
(2.12)
By (2.7), we have
∇ ut =
1
h
∇Φ −
h00
h0
ut ∇ u +
gf 0
h0
∇ u.
(2.13)
Substitute (2.13) into (2.12), to get
ab
h0
1Φ + 2b
+
h0
2abh00 gf 0
( h0 ) 2
+
a
aū gf
a
−
−
gfh00
h0
∇u +
ū
aū bgf 0
h0
a
h0
−
− gf 0 −
∇ b · ∇ Φ − Φt =
abgf 00
h0
at h 0
a
+
aū at b
ut +
a
h0
− aūū b +
2
00
(aū )2 b
− aūt b |∇ u| + h −
g 2 ff 0
h0
abh000
+
at gf
a
.
a
aū h0
a
+
aū bh00
h0
−
2ab h00
( h0 ) 2
2 !
|∇ u|2 ut
(ut )2
(2.14)
J. Ding, B.-Z. Guo / Computers and Mathematics with Applications 60 (2010) 670–679
673
With (2.6), it has
1
ut =
h
Φ+
gf
h0
.
(2.15)
Substitution (2.15) of (2.14) gives
a
1 a
gf
at
a
1a
2
|∇
1Φ
+
2b
∇
∇
b
·
∇
Φ
+
ab
(
Φ
+
gf
)
+
+
Φ − Φt
u
+
u
|
+
0
0
0
0
0
0
ab
h
h
1
af
a
h0
= −abg
h
ū
+
ū ū
a
1 aū g
a
t
h
a
ū ū
h
ū
h
a
|∇ u|2 .
(2.16)
It is seen from (2.1) that the right-hand side of (2.16) is nonnegative, i.e.
a
a
1Φ
+
2b
∇
u
+
∇
b
· ∇Φ
0
0
0
ab
h
h
+ ab
h
ū
1 a
h0
a
|∇ u|2 +
ū ū
1a
h0
a
ū
(Φ + gf ) +
gf
h0
+
at
a
Φ − Φt ≥ 0.
(2.17)
Furthermore, by (2.5), it follows that
Φ (x, 0) = ∇ · (a(M0 , 0)b(x)∇ M0 ) = 0,
(2.18)
and
∂Φ
∂ ut
∂u
∂u
= h00 ut
+ h0
− gf 0
= h0
∂n
∂n
∂n
∂n
∂u
∂n
= 0.
(2.19)
t
Combining (2.17)–(2.19) and by virtue of the maximum principles [21], we know that the maximum of Φ in D × [0, T ) is
zero. Hence
Φ ≤ 0 in D × [0, T ),
and
h 0 ( u)
f (u)
ut ≤ g (t ).
(2.20)
For each fixed x ∈ D, integrate (2.20) over [0, t ] to produce
t
Z
h0 (u)
0
f (u)
u(x,t )
Z
ut dt =
h0 (s)
f (s)
M0
t
Z
g (t )dt ,
ds ≤
0
which implies that u must be a global solution of (2.5). In fact, suppose that u(x, t ) blows up at finite time T :
lim u(x, t ) = +∞.
t →T −
Passing to the limit as t → T − in (2.21) gives
Z
+∞
M0
h0 (s)
f ( s)
T
Z
g (t )dt <
ds ≤
+∞
Z
0
g (t )dt ,
0
which is a contradiction. This shows that u is global, and so is for u. Moreover, (2.21) implies that
φ (u(x, t )) =
u(x,t )
Z
h0 (s)
f (s)
M0
t
Z
g (t )dt .
ds ≤
0
Therefore,
u(x, t ) ≤ φ −1
t
Z
g (t )dt
.
0
Since u is an upper solution of (1.1), we finally get
u(x, t ) ≤ u(x, t ) ≤ φ −1
t
Z
g (t )dt
0
The proof is complete.
.
(2.21)
674
J. Ding, B.-Z. Guo / Computers and Mathematics with Applications 60 (2010) 670–679
Remark 2.1. When
+∞
Z
g (t )dt = +∞,
0
(2.2) implies
h0 (s)
+∞
Z
f (s)
M0
M0 = max u0 (x).
ds = +∞,
D
3. Blow-up solution
Theorem 3.1. Let u be a solution of (1.1). Assume that
(i) For any (s, t ) ∈ R+ × R+ ,
1
a(s, t )f (s)
a(s, t )
+
h0 (s)
s
1
as (s, t )
g (t )
s
a( s , t )
≥ 0.
(3.1)
t
(ii)
h0 (s)
+∞
Z
f ( s)
M0
+∞
Z
ds <
M0 = max u0 (x).
g (t )dt ,
(3.2)
D
0
(iii)
min ∇ · (a(u0 , 0)b(x)∇ u0 (x)) = 0.
(3.3)
D
Then the solution u must blow up in finite T , and
T ≤η
−1
Z
+∞
h0 (s)
f (s)
M0
ds ,
T
Z
u(x, t ) ≤ ψ −1
g (t )dt
(3.4)
,
(3.5)
t
where
η(z ) =
z
Z
g (t )dt ,
∀z ≥ 0,
ψ(z ) =
0
+∞
Z
z
h0 ( s )
f (s)
ds,
∀z ≥ 0,
(3.6)
η−1 and ψ −1 are the inverse functions of η and ψ respectively.
Proof. Construct an auxiliary function
Ψ (x, t ) = h0 (u)ut − g (t )f (u).
(3.7)
Replacing Φ and u with Ψ and u in (2.16), respectively, gives
a
1a
gf
a
1 a
at
2
1Ψ
+
2b
∇
u
+
∇
b
·
∇
Ψ
+
ab
|∇
u
|
+
(
Ψ
+
gf
)
+
+
Ψ − Ψt
0
0
0
0
0
0
ab
h
h
1
af
a
h0
= −abg
h
u
+
u u
a
1 au g
a
t
h
a
u u
h
u
|∇ u|2 .
h
a
(3.8)
From assumption (3.1), the right-hand side of (3.8) is nonpositive, i.e.
a
a
1Ψ
+
2b
∇
u
+
∇
b
· ∇Ψ
0
0
0
ab
h
h
+ ab
u
1 a
a
h0
u u
h
|∇ u|2 +
1a
a
h0
u
(Ψ + gf ) +
gf
h0
+
at
a
Ψ − Ψt ≤ 0 .
(3.9)
Now, by (3.3), it has
min Ψ (x, 0) = min ∇ · (a(u0 , 0)b(x)∇ u0 (x)) = 0.
D
D
(3.10)
J. Ding, B.-Z. Guo / Computers and Mathematics with Applications 60 (2010) 670–679
675
By (1.1), it follows that
∂Ψ
∂u
∂ ut
∂u
= h00 ut
+ h0
− gf 0
= h0
∂n
∂n
∂n
∂n
∂u
∂n
= 0.
(3.11)
t
Combining (3.9)–(3.11) and applying the maximum principles, we know that the minimum of Ψ in D × [0, T ) is zero. Hence
Ψ ≥ 0 in D × [0, T ),
i.e.
h 0 ( u)
f (u)
ut ≥ g (t ).
(3.12)
At the point x0 ∈ D where u0 (x0 ) = M0 , integrate (3.12) over [0, t ] to get
t
Z
h0 (u)
f (u)
0
u(x0 ,t )
Z
ut dt =
h0 (s)
f ( s)
M0
t
Z
g (t )dt ,
ds ≥
(3.13)
0
which implies that u must blow up in finite time. Actually, if u is a global solution of (1.1), then for any t > 0, it follows from
(3.13) that
+∞
Z
h0 (s)
f ( s)
M0
u(x0 ,t )
Z
ds ≥
h0 (s)
f (s)
M0
t
Z
g (t )dt .
ds ≥
(3.14)
0
Letting t → +∞ in (3.14) yields
+∞
Z
h0 (s)
f ( s)
M0
+∞
Z
g (t )dt ,
ds ≥
0
which contradicts with the assumption (3.2). This shows that u must blow up in a finite time t = T . Furthermore, letting
t → T in (3.13), we have
u(x0 ,t )
Z
h0 (s)
lim
t →T
f ( s)
M0
t
Z
g (t )dt ,
ds ≥ lim
t →T
0
i.e.
+∞
Z
M0
h0 (s)
f ( s)
T
Z
g (t )dt = η(T ),
ds ≥
0
which implies that
T ≤ η −1
+∞
Z
h0 (s)
f (s)
M0
ds .
By integrating the inequality (3.12) over [t , s](0 < t < s < T ), one has, for each fixed x, that
ψ (u(x, t )) ≥ ψ (u(x, t )) − ψ (u(x, s)) =
Letting s → T yields
ψ (u(x, t )) ≥
T
Z
g (t )dt ,
t
which implies that
u(x, t ) ≤ ψ
−1
T
Z
g (t )dt
t
The proof is complete.
Remark 3.1. When
+∞
Z
g (t )dt = +∞,
0
.
Z
u(x,s)
u(x,t )
h0 (s)
f ( s)
Z
ds ≥
t
s
g (t )dt .
676
J. Ding, B.-Z. Guo / Computers and Mathematics with Applications 60 (2010) 670–679
(3.2) implies that
h0 (s)
+∞
Z
f (s)
M0
ds < +∞,
M0 = max u0 (x).
D
In Theorem 3.1, the assumption (3.3) is not essential for the existence of blow-up solution. We state the following
Corollary 3.1 by removing condition (3.3).
Corollary 3.1. Let u be a solution of (1.1). Assume that
(i) For any (s, t ) ∈ R+ × R+ ,
1
a(s, t )f (s)
a(s, t )
+
h0 (s)
s
s
1
g (t )
as ( s , t )
a(s, t )
≥ 0.
t
(ii)
h0 (s)
+∞
Z
f (s)
m0
ds <
+∞
Z
g (t )dt ,
m0 = min u0 (x).
0
(3.15)
D
Then the solution u must blow up in finite T and
T ≤η
−1
+∞
Z
h0 (s)
f ( s)
m0
ds ,
where
η(z ) =
z
Z
g (t )dt ,
∀z ≥ 0,
0
η−1 is the inverse function of η.
Proof. Let u be a solution of the following problem:

(h(u))t = ∇ · (a(u, t )b(x)∇ u) + g (t )f (u)


 ∂u
=0

∂n


u(x, 0) = m0
in D × (0, T ),
on ∂ D × (0, T ),
in D.
By the comparison principle [20], u is a lower solution of (1.1). Therefore, Theorem 3.1 can be applied to get that u blows up
in a finite time T ∗ , and
T ∗ ≤ η −1
+∞
Z
m0
h0 ( s )
f (s)
ds .
Hence, u blows up in a finite time T and
T ≤T ≤η
∗
−1
Z
+∞
m0
The proof is complete.
h0 (s)
f (s)
ds .
Remark 3.2. When
+∞
Z
g (t )dt = +∞,
0
(3.15) implies
Z
+∞
m0
h0 (s)
f (s)
ds < +∞,
m0 = min u0 (x).
D
4. Applications
When a(u, t ) ≡ a(u), b(x) ≡ 1, and g (t ) ≡ 1, the conclusions of Theorems 2.1 and 3.1, and Corollary 3.1 still hold true.
In this sense, our results extend and supplement the results of [19].
In what follows, we present several examples to demonstrate the applications of the obtained abstract results.
J. Ding, B.-Z. Guo / Computers and Mathematics with Applications 60 (2010) 670–679
677
Example 4.1. Let u be a solution of the following problem:
 m
u t = ∇ · (un ∇ u) + uq



∂u
=0


 ∂n
u(x, 0) = u0 (x) > 0
in D × (0, T ),
on ∂ D × (0, T ),
in D,
where D ⊂ RN is a bounded domain with smooth boundary ∂ D, m > 0, −∞ < n < +∞, −∞ < q < +∞. Here
h(u) = um ,
a(u, t ) = un ,
b(x) = 1,
g (t ) = 1,
f ( u) = uq .
By Theorem 2.1, if 0 < m − q ≤ n + 1 or m = q, then u must be a global solution, and
u(x, t ) ≤


 m−q
m


m−q
m1−q
t + M0
,
t
M0 e m ,
0 < m − q ≤ n + 1,
m = q,
where M0 = maxD u0 (x). By Corollary 3.1, if m − q ≤ n + 1 < 0 or m − q < 0 ≤ n + 1, u blows up in a finite time T , and
T ≤
m
m−q
q−m
m0
,
where m0 = minD u0 (x).
Example 4.2. Let u be a solution of the following problem:
in D × (0, T ),
 mu e
= ∇ · enu ∇ u + equ

t


∂u
=0
 ∂n


u(x, 0) = u0 (x) > 0
on ∂ D × (0, T ),
in D,
where D ⊂ RN is a bounded domain with smooth boundary ∂ D, m > 0, −∞ < n < +∞, −∞ < q < +∞. Now
h(u) = emu ,
a(u, t ) = enu ,
b(x) = 1,
g (t ) = 1,
f (u) = equ .
By Theorem 2.1, if 0 < m − q ≤ n or m = q, then u must be a global solution, and
u(x, t ) ≤



1
ln
m−q

m + M ,
0
t
m−q
m
t +e
(m−q)M0
,
0 < m − q ≤ n,
m = q,
where M0 = maxD u0 (x). By Corollary 3.1, if m − q < 0 ≤ n or m − q ≤ n < 0, u blows up in a finite time T , and
T ≤
m
q−m
e(m−q)m0 ,
where m0 = minD u0 (x).
Example 4.3. Let u be a solution of the following problem:
 m
ln (1 + u) t = ∇ · (lnn (1 + u)∇ u) + lnq (1 + u)



∂u
=0

∂

 n
u(x, 0) = u0 (x) > 0
in D × (0, T ),
on ∂ D × (0, T ),
in D,
where D ⊂ RN is a bounded domain with smooth boundary ∂ D, m > 0, −∞ < n < +∞, −∞ < q < +∞. We now have
h(u) = lnm (1 + u),
a(u, t ) = lnn (1 + u),
b(x) = 1,
g (t ) = 1,
By Theorem 2.1, if 1 ≤ m − q ≤ n + 1, then u must be a global solution, and
u(x, t ) ≤ exp
"
m−q
m
t + ln
m−q
(1 + M0 )
m1−q #
− 1,
f (u) = lnq (1 + u).
678
J. Ding, B.-Z. Guo / Computers and Mathematics with Applications 60 (2010) 670–679
where M0 = maxD u0 (x). By Corollary 3.1, if m − q ≤ n + 1 < 0 or m − q < 0 ≤ n + 1, u blows up in a finite time T , and
T ≤
m
lnm−q (1 + m0 ) ,
q−m
where m0 = minD u0 (x).
Example 4.4. Let u be a solution of the following problem:
! !

3
X

−3 t
2
3 u

xi ∇ u + u4 (3 + u)eu−3t
u e t =∇ · u e 1+




i
=
1


 ∂u
=0

∂n


!2

3

X


2

xi
u(x, 0) = 1 + 1 −
in D × (0, T ),
on ∂ D × (0, T ),
in D,
i =1
n
where D = x = (x1 , x2 , x3 ) |
h(u) = u3 eu ,
o
P3
i=1
x2i < 1 is the unit ball of R3 . Now
a( u, t ) = u− 3 e t ,
b(x) = 1 +
3
X
x2i ,
g (t ) = e−3t ,
f (u) = u4 (3 + u)eu .
i=1
It is easy to check that (2.1) and (2.2) hold. It follows from Theorem 2.1 that u must be a global solution and
u(x, t ) ≤ φ
−1
t
Z
g (t )dt
=
0
6
1 + 2e−3t
.
Example 4.5. Let u be a solution of the following problem:
! !

3
X

3 u
3 t
2


u e t =∇ · u e 1+
xi ∇ u + u4 (3 + u)eu−t in D × (0, T ),



i=1
∂u


=0
on ∂ D × (0, T ),



 ∂n
u(x, 0) = 2
in D,
n
o
P3 2
3
where D = x = (x1 , x2 , x3 ) |
i=1 xi < 1 is the unit ball of R . Now we have
h(u) = u3 eu ,
a( u, t ) = u3 e t ,
b(x) = 1 +
3
X
x2i ,
g ( t ) = e− t ,
f (u) = u4 (3 + u)eu .
i =1
It is easy to check that (3.1)–(3.3) hold. It then follows from Theorem 3.1 that u blows up in a finite time T , and
T ≤ η −1
Z
+∞
M0
u(x, t ) ≤ ψ
−1
h0 (s)
f ( s)
ds
T
Z
g (t )dt
t
= ln 2,
=
1
e− t − e− T
.
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