# Fujita Exponent for a Nonlinear Degenerate Parabolic Equation with Localized Source

Authors: Yulan Wang; Xiaojun Song; Chao Ye
Journal: Advances in Mathematical Physics
 (2014)

---

## Abstract

This paper is devoted to understand the blow-up properties of reaction-diffusion equations which combine a localized reaction term with nonlinear diffusion. In particular, we study the critical exponent of ap-Laplacian equation with a localized reaction. We obtain the Fujita exponent q
c of the equation.


---

## Body

## 1. Introduction

This short paper is devoted to the Fujita critical exponent of the followingp-Laplacian equation:
(1)
u
t
=
(
|
u
x
|
p
-
2
u
x
)
x
+
a
(
x
)
u
q
,
x
∈
R
,
t
>
0
,
u
(
x
,
0
)
=
u
0
(
x
)
,
x
∈
R
,
where p
>
2, q
>
0, and the initial data u
0
(
x
) is nonnegative continuous and compactly supported.We take the coefficienta
(
x
)
≥
0 which is a compactly supported function, which means that the reaction term acts onlylocally. Thus, the model may be used to describe a chemical reaction-diffusion process in which, due to the effect of the catalyst, the reaction takes place only at some local sites [1].As a representative example of quasilinear reaction-diffusion equation, classicalp-Laplacian equation with no source term (a
(
x
)
≡
0 in (1)) and its variants had been extensively studied in the past few decades (see [2–6] and references therein). It is well known that p-Laplacian equation in general does not allow for classical solvability due to the nonlinearity and degeneracy. Therefore, the “solution” to a p-Laplacian means the weak solution in the usual integral sense (see, e.g., [4]). Among other things, previous studies show that blow-up happens for the problem with homogeneous reaction; that is, a
(
x
)
≡
1:
(2)
u
t
=
(
|
u
x
|
p
-
2
u
x
)
x
+
u
q
,
x
∈
R
,
t
>
0
,
u
(
x
,
0
)
=
u
0
(
x
)
,
x
∈
R
.
It is shown that if 0
<
q
≤
1, every solution is global in time, if 1
<
q
≤
2
p
-
1, all solutions blow up, and if q
>
2
p
-
1, both global in time solutions and blowing up solutions exist. In this case, the numbers q
0
=
1 and q
c
=
2
p
-
1 are called theglobal existence exponent and theFujita exponent, respectively. On the other hand, we remark that there is a close connection of problem (1) with the problem of diffusion with flux conditions on the boundary:
(3)
u
t
=
(
|
u
x
|
p
-
2
u
x
)
x
,
x
>
0
,
t
>
0
,
-
|
u
x
|
p
-
2
u
x
=
u
q
,
x
=
0
,
t
≥
0
,
u
(
x
,
0
)
=
u
0
(
x
)
,
x
>
0
.
In fact, if we take a sequence of reaction coefficients converging to a Dirac delta at the origin (i.e., if a
n
(
x
)
→
δ
0
(
x
)), the corresponding solutions u
n of (1) should converge to a solution of problem (3). In [7], authors proved that the global existence exponent of (3) is q
0
=
2
(
p
-
1
)
/
p and the Fujita exponent is q
c
=
2
(
p
-
1
). After that, there are plenty of works that extended their study to doubly degenerate equation or fast diffusion case (see, e.g., [8–10] and references therein).The motivation of this short paper is the wish to understand the blow-up properties of reaction-diffusion equations which combine a nonlinear diffusion and localized reaction term, and this is the main difference with the existing studies of blow-up for similar reaction-diffusion equations. For the porous medium equation caseu
t
=
(
u
m
)
x
x
+
a
(
x
)
u
q, Ferreira et al. [11] proved that q
0
=
(
m
+
1
)
/
2, q
c
=
m
+
1.In our previous work, we has obtained the global existence exponent of (1) is q
0
=
2
(
p
-
1
)
/
p. In this note, we will prove thatthe Fujita exponent of(1) is q
c
=
2
(
p
-
1
). Thus, it is the same as that of (3), and this is coincident with our intuitive judgement. We will modify several techniques developed in [11] to prove our main result.


## 2. Some Preliminary Lemmas

In this section, we will give some preliminary lemmas, whose proofs may be independent and interesting. We first have the following lemma from [3].Lemma 1.
There exists a positive constantL
s
<
∞ such that problem
(4)
(
|
w
′
|
p
-
2
w
′
)
′
-
1
p
-
2
w
+
w
p
-
1
=
0
,
0
<
x
<
L
s
,
w
′
(
0
)
=
0
,
w
′
(
L
s
)
=
w
(
L
s
)
=
0
has a unique nonnegative solution w
(
x
).Whenq
>
p
-
1, we need to introduce the energy functional of (1) as follows:
(5)
E
(
t
)
=
1
p
∫
R
|
u
x
|
p
d
x
-
1
q
+
1
∫
R
a
(
x
)
u
q
+
1
d
x
.Lemma 2.
Whenq
>
p
-
1, if there exists t
0
>
0 such that E
(
t
0
)
<
0, then all solutions to problem (1) blow up in finite time.Proof.
It is easy to verify that(6)
E
′
(
t
)
=
∫
R
u
x
t
(
|
u
x
|
p
-
2
u
x
)
d
x
-
∫
R
a
(
x
)
u
q
u
t
d
x
=
-
∫
R
|
u
t
|
2
≤
0
,
which shows that E
(
t
) is decreasing. Thus, E
(
t
)
≤
E
(
t
0
)
<
0 for any t
≥
t
0.
Define(7)
M
(
t
)
=
1
2
∫
0
t
∫
R
u
2
(
x
,
s
)
d
x
d
s
.
For any t
≥
t
0, we have M
(
t
)
>
0 and
(8)
M
′
(
t
)
=
1
2
∫
R
u
2
(
x
,
s
)
d
x
>
0
,
(9)
M
′′
(
t
)
=
∫
R
u
u
t
d
x
=
∫
R
u
(
(
|
u
x
|
p
-
2
u
x
)
x
+
a
(
x
)
u
q
)
d
x
=
-
∫
R
|
u
x
|
p
d
x
+
∫
R
a
(
x
)
u
q
+
1
d
x
=
(
q
+
1
)
(
-
1
p
∫
R
|
u
x
|
p
d
x
+
1
q
+
1
∫
R
a
(
x
)
u
q
+
1
d
x
)
-
(
1
-
q
+
1
p
)
∫
R
|
u
x
|
p
d
x
.
From q
>
p
-
1, it follows that -
(
1
-
(
q
+
1
)
/
p
)
>
0. Thus,
(10)
M
′′
(
t
)
≥
(
q
+
1
)
(
-
E
(
t
)
)
>
(
q
+
1
)
(
-
E
(
t
0
)
)
>
0
.
Namely, for any t
≥
t
0, we have M
′′
(
t
)
>
c
>
0 for some positive constant c.
SinceM
′
(
t
0
)
>
0 and M
(
t
0
)
>
0, we obtain that M
′
(
t
)
≥
C
t
+
C
1, where C
1 is a constant. Therefore, M
′
(
t
)
→
∞ as t
→
∞.
Next, we will prove that there existsT
<
∞ such that M
(
t
)
→
∞ as t
→
T.
It follows fromE
′
(
t
)
=
∫
R
‍
|
u
t
|
2
d
x that
(11)
E
(
t
)
-
E
(
t
0
)
=
∫
t
0
t
E
′
(
s
)
d
s
=
-
∫
t
0
t
∫
R
|
u
s
|
2
d
x
d
s
.
Since E
(
t
0
)
<
0, we have E
(
t
)
<
-
∫
t
0
t
‍
∫
R
‍
|
u
s
|
2
d
x
d
s. Thus, by (9),
(12)
M
′′
(
t
)
=
(
q
+
1
)
(
-
E
(
t
)
)
-
(
1
-
q
+
1
p
)
∫
R
|
u
x
|
p
d
x
≥
(
q
+
1
)
∫
t
0
t
∫
R
|
u
s
|
2
d
x
d
s
.
Therefore, by using Hölder inequality, we could obtain
(13)
M
M
′′
(
t
)
≥
q
+
1
2
(
∫
t
0
t
∫
R
|
u
s
|
2
d
x
d
s
)
(
∫
t
0
t
∫
R
u
2
d
x
d
s
)
>
q
+
1
2
(
∫
t
0
t
∫
R
u
u
s
d
x
d
s
)
2
=
q
+
1
2
(
1
2
∫
R
u
2
(
x
,
t
)
d
x
-
1
2
∫
R
u
2
(
x
,
t
0
)
d
x
)
=
q
+
1
2
(
M
′
(
t
)
-
M
′
(
t
0
)
)
2
.
Due to the fact that M
′
(
t
)
→
∞ as t
→
∞ and since (
q
+
1
)
/
2
>
1, there exists a constant α
>
0 such that
(14)
M
(
t
)
M
′′
(
t
)
≥
(
1
+
α
)
(
M
′
(
t
)
)
2
,
when t is large enough. This is equivalent to the proposition that M
-
α is a concave function.
SinceM
(
t
)
≥
0 and M
(
t
) is increasing, there exists 0
<
T
<
∞ such that M
-
α
(
T
)
=
0. Therefore, M
(
t
)
→
∞ as t
→
T.
Furthermore, we claim thatM
′′
(
t
)
→
∞ as t
→
T. Otherwise, M
′′
(
t
) is bounded in [
0
,
T
]. Assume that M
′′
(
t
)
≤
2
C
2 for t
∈
[
0
,
T
], where C
2 is a positive constant.
Integrating the inequality twice over the interval[
0
,
T
], we obtain that
(15)
M
(
t
)
≤
C
2
t
2
+
C
3
t
+
C
4
,
t
∈
[
0
,
T
]
.
The right side is bounded in [
0
,
T
], and so is M
(
t
), which contradicts the fact that M
(
t
)
→
∞ as t
→
T. Hence,
(16)
M
′′
(
t
)
⟶
∞
in the limit t
→
T.
Next, we prove thatu
(
x
,
t
) blows up in finite time. By (1),
(17)
M
′′
(
t
)
=
-
∫
R
|
u
x
|
p
d
x
+
∫
R
a
(
x
)
u
q
+
1
d
x
≤
∫
R
a
(
x
)
u
q
+
1
d
x
≤
meas
⁡
(
supp
⁡
a
)
∥
u
(
·
,
t
)
∥
∞
q
+
1
.
Therefore, u
(
x
,
t
) blows up in the sense of L
∞ norm from (16).Lemma 3.
Ifq
>
2
(
p
-
1
) and σ
<
-
1, Cauchy problem
(18)
u
t
=
(
|
u
x
|
p
-
2
u
x
)
x
+
|
x
|
σ
u
q
,
(19)
u
(
x
,
0
)
=
u
0
(
x
)
has a global supersolution.Proof.
Consider the following Cauchy problem:(20)
u
t
=
(
|
u
x
|
p
-
2
u
x
)
x
,
x
∈
R
,
t
>
0
,
u
(
x
,
0
)
=
L
|
x
|
-
1
,
x
∈
R
.
The existence and uniqueness of the solution to this problem was established in [5, 6, 12]. The solution has the form
(21)
U
(
x
,
t
)
=
t
-
β
g
L
(
|
x
|
t
-
β
)
,
β
=
1
2
(
p
-
1
)
,
where g
L is a positive, bounded, and decreasing function and satisfies the following boundary value problem:
(22)
(
|
g
L
′
|
p
-
2
g
L
′
)
′
+
β
r
g
L
′
+
β
g
L
=
0
,
r
>
0
,
g
L
′
(
0
)
=
0
,
lim
⁡
r
→
∞
r
g
L
(
r
)
=
L
.
Setz
(
x
,
t
)
=
λ
U
(
x
,
λ
p
-
2
t
+
t
0
), where λ
>
0 is to be determined later; t
0 is a constant between 0 and 1.
One can see thatlim
⁡
r
→
∞
r
g
L
(
r
)
=
L and σ
<
-
1 show that lim
⁡
r
→
∞
r
σ
/
(
q
-
1
)
g
L
(
r
)
=
0. From (21), we know that
(23)
sup
⁡
⁡
(
|
x
|
σ
·
∥
z
(
x
,
t
)
∥
∞
)
≤
C
λ
(
λ
p
-
2
t
+
t
0
)
-
β
(
1
-
σ
/
(
q
-
1
)
)
for some C
>
0. Therefore, z
(
x
,
t
) is the solution of the following problem:
(24)
z
t
=
(
|
z
x
|
p
-
2
z
x
)
x
,
x
∈
R
,
t
>
0
,
z
(
x
,
0
)
=
λ
U
(
x
,
t
0
)
,
x
∈
R
.
Next, we introduce the functionv
(
x
,
t
)
=
A
(
t
)
z
(
x
,
B
(
t
)
), where A
(
t
) and B
(
t
) are solutions of the following ordinary differential system:
(25)
A
′
(
t
)
=
C
q
-
1
λ
q
-
1
(
λ
p
-
2
B
(
t
)
+
t
0
)
-
β
(
q
-
1
-
σ
)
A
q
(
t
)
,
t
>
0
,
B
′
(
t
)
=
A
p
-
2
(
t
)
,
t
>
0
,
A
(
0
)
=
1
,
B
(
0
)
=
0
.
Then, it is easy to verify that v
(
x
,
t
) satisfies
(26)
v
t
≥
(
|
v
x
|
p
-
2
v
x
)
x
+
|
x
|
σ
v
q
,
x
∈
R
,
t
>
0
,
v
(
x
,
0
)
=
λ
U
(
x
,
t
0
)
,
x
∈
R
.
Therefore, v
(
x
,
t
) is a supersolution of (18) if u
(
x
,
0
)
≤
λ
U
(
x
,
t
0
). If q
>
2
(
p
-
1
), it is a global supersolution from the following Lemma 4.Lemma 4.
Ifq
>
2
(
p
-
1
) and σ
<
-
1, then there exists a positive constant λ
0 such that problem (25) has a global solution (
A
(
t
)
,
B
(
t
)
) with A
(
t
) bounded in [
0
,
+
∞
) if 0
≤
λ
<
λ
0.Proof.
The local existence and uniqueness of solution(
A
(
t
)
,
B
(
t
)
) of (25) follows from the standard theory of initial value problem on ordinary differential equation. From (25), for t
>
0, we have A
′
(
t
)
>
0, A
(
t
)
>
1 and the solution continues as long as the solution exists and A
(
t
) is finite.
By (25), B
(
t
) is uniquely defined by B
(
t
)
=
∫
0
t
‍
A
p
-
2
(
s
)
d
s. Moreover,
(27)
B
(
s
)
=
∫
0
s
A
p
-
2
(
s
)
d
τ
≥
A
p
-
2
(
0
)
s
=
s
for all s
∈
[
0
,
t
]. From σ
<
-
1 and q
>
2
(
p
-
1
), we have β
(
q
-
1
-
σ
)
-
1
>
0. So
(28)
1
-
A
1
-
q
(
t
)
=
(
q
-
1
)
C
q
-
1
λ
q
-
1
∫
0
t
(
λ
p
-
2
B
(
s
)
+
t
0
)
-
β
(
q
-
1
-
σ
)
d
s
≤
q
-
1
β
(
q
-
1
-
σ
)
-
1
C
q
-
1
λ
q
-
1
+
p
t
0
-
β
(
q
-
1
-
σ
)
+
1
.
Letλ
=
λ
0 be a positive constant defined by
(29)
q
-
1
β
(
q
-
1
-
σ
)
-
1
C
q
-
1
λ
q
-
1
+
p
t
0
-
β
(
q
-
1
-
σ
)
+
1
=
1
2
and, given a fixed λ
∈
(
0
,
λ
0
), by (28), we have 1
≤
A
(
t
)
≤
2
1
/
(
q
-
1
) as long as A
(
t
) exists globally. Meanwhile, it is easy to see that t
≤
B
(
t
)
≤
2
(
p
-
2
)
/
(
q
-
1
) for all t
>
0; thus, B
(
t
) is global.


## 3. Main Results and Their Proofs

We first deal with the case ofq
>
2
(
p
-
1
), which is easier than the other cases.Theorem 5.
Ifq
>
2
(
p
-
1
), then there exist both global solutions and blowing-up solutions to problem (1).Proof.
Lemma3 states that when q
>
2
(
p
-
1
) and σ
<
-
1, equation
(30)
u
t
=
(
|
u
x
|
p
-
2
u
x
)
x
+
|
x
|
σ
u
q
has a global supersolution.
Note that the reaction coefficient of this equation is bigger thana
(
x
), multiplied by a constant if necessary. So, we conclude that problem (1) has global solutions if q
>
2
(
p
-
1
) and σ
<
-
1.
On the other hand, it is well known (see, e.g., [13]) that problem
(31)
u
t
=
(
|
u
x
|
p
-
2
u
x
)
x
+
u
q
,
x
∈
[
-
L
,
L
]
,
t
>
0
,
u
(
±
L
,
t
)
=
0
,
t
>
0
,
u
(
x
,
0
)
=
u
0
(
x
)
,
x
∈
[
-
L
,
L
]
,
admits a blowing-up solution, which is a subsolution to our problem for suitable L. We thus complete our proof.Our remainder objective is to prove that if2
(
p
-
1
)
/
p
<
q
≤
2
(
p
-
1
), then all solutions blow up in finite time. And this result is composed of several theorems.To simplify the exposition, we may takea
(
x
), a characteristic function; a
(
x
)
=
χ
[
-
L
,
L
] for some 0
<
L
<
+
∞ in the remainder of this paper. Note that such assumption imposing on a
(
x
) is not essential since it can be modified to a more general compactly function.Theorem 6.
If2
(
p
-
1
)
/
p
<
q
<
p
-
1, then all the solutions to problem (1) blow up in finite time.Proof.
We will prove this result by the similar idea of Lemma  10 in [11]. We firstly construct here a blow-up subsolution matching of a self-similar function with a blowing-up parabola.
Fix a point0
<
x
0
≤
L and consider the even function obtained by the reflection of
(32)
u
_
(
x
,
t
)
=
{
(
A
(
t
)
-
B
(
t
)
x
p
/
(
p
-
1
)
)
+
,
0
≤
x
≤
x
0
,
V
(
x
-
x
0
,
t
)
,
x
≥
x
0
,
where A
(
t
) and B
(
t
) are to be determined later and V is a self-similar solution of the problem
(33)
V
t
=
(
|
V
x
|
p
-
2
V
x
)
x
,
x
>
0
,
t
>
0
,
-
|
V
x
|
p
-
2
V
x
=
V
s
,
x
=
0
,
t
>
0
.
It is known from [7] that problem (33) admits blowing-up self-similar solution if s
>
2
(
p
-
1
)
/
p, and they have the form
(34)
V
(
x
,
t
)
=
(
T
-
t
)
-
γ
Θ
(
x
(
T
-
t
)
(
s
-
(
p
-
1
)
)
/
(
p
-
1
)
·
γ
)
,
where γ
=
(
p
-
1
)
/
(
s
·
p
-
2
(
p
-
1
)
).
In order to have aC
1 function u
_
(
x
,
t
), we need
(35)
A
(
t
)
-
B
(
t
)
x
0
p
/
(
p
-
1
)
=
V
(
0
,
t
)
,
-
p
p
-
1
B
(
t
)
x
0
1
/
(
p
-
1
)
=
V
′
(
0
)
.
To this end, we take
(36)
A
(
t
)
=
V
(
0
,
t
)
+
p
-
1
p
x
0
V
s
/
(
p
-
1
)
(
0
,
t
)
,
B
(
t
)
=
p
-
1
p
x
0
-
1
/
(
p
-
1
)
V
s
/
(
p
-
1
)
(
0
,
t
)
.
Henceforth, for 0
≤
x
≤
x
0, the function u
_
(
x
,
t
) has the following form:
(37)
u
_
(
x
,
t
)
=
(
T
-
t
)
-
γ
I
(
x
,
t
)
,
I
(
x
,
t
)
=
Θ
(
0
)
+
p
-
1
p
x
0
1
/
(
1
-
p
)
(
x
0
p
/
(
p
-
1
)
-
x
p
/
(
p
-
1
)
)
×
Θ
s
/
(
p
-
1
)
(
0
)
(
T
-
t
)
γ
(
1
-
s
/
(
p
-
1
)
)
.
Notice that if s
≤
p
-
1, I
(
x
,
t
)
~
const. In order to see that u
_ is a subsolution to problem (1), we only have to look at the interval (
0
,
x
0
). Direct calculations show
(38)
u
_
t
=
(
T
-
t
)
-
γ
-
1
(
γ
I
-
γ
(
p
-
1
-
s
)
p
h
h
h
h
h
h
h
h
h
h
h
×
x
0
1
/
(
1
-
p
)
(
x
0
p
/
(
p
-
1
)
-
x
p
/
(
p
-
1
)
)
h
h
h
h
h
h
h
h
h
h
h
×
Θ
s
/
(
p
-
1
)
(
0
)
(
T
-
t
)
γ
(
1
-
s
/
(
p
-
1
)
)
γ
(
p
-
1
-
s
)
p
)
,
(
|
u
_
x
|
p
-
2
u
_
x
)
x
=
-
Θ
s
(
0
)
x
0
(
T
-
t
)
-
γ
s
,
u
_
q
=
(
T
-
t
)
-
q
γ
I
q
.
Therefore, when T is small, u
_ is a subsolution to problem (1) if the inequality
(39)
C
1
T
-
γ
-
1
≤
-
C
2
T
-
γ
s
+
C
3
T
-
q
γ
is valid.
This inequality could be achieved if we takes
<
min
⁡
{
q
,
(
p
-
1
)
(
q
+
1
)
/
p
}. Since 2
(
p
-
1
)
/
p
<
q
<
(
p
-
1
), q
<
(
p
-
1
)
(
q
+
1
)
/
p. So, we only need to choose 2
(
p
-
1
)
/
p
<
s
<
q to get the desired blow-up subsolution.
Now, we claim that all solutions to (1) blow up in finite time. Suppose on the contrary that u
(
x
,
t
) exist globally. From Lemma 6 of [11], we know that there exists some t
0
>
0 such that u
(
x
,
t
0
)
≥
u
_
(
x
,
0
). Henceforth, u
(
x
,
t
0
+
t
)
≥
u
_
(
x
,
t
) by comparison theorem. And this implies that u
(
x
,
t
) blows up. This contradicts the hypothesis that u
(
x
,
t
) exist globally.Theorem 7.
Ifq
=
p
-
1, then every solution to problem (1) blows up in finite time.Proof.
We first prove that problem (1) admits, for any length L
>
0, a unique symmetric blow-up solution in the self-similar form
(40)
U
(
x
)
=
(
T
-
t
)
-
1
/
(
p
-
2
)
F
(
x
)
.
Substituting U into the first equation of (1), we have that the profile F satisfies the equation
(41)
(
|
F
′
|
p
-
2
F
′
)
′
-
1
p
-
2
F
+
a
(
x
)
F
p
-
1
=
0
,
x
>
0
plus the boundary condition F
′
(
0
)
=
0.
IfL
≥
L
s, that is, [
-
L
s
,
L
s
]
⊂
supp
⁡
a
(
x
), (
T
-
t
)
-
1
/
(
p
-
2
)
w
(
x
) is the desired blow-up solution, where w
(
x
) is defined in Lemma 1.
IfL
<
L
s, we construct F by putting together two pieces as in [11]. For x
≥
L, since a
(
x
)
=
0, it is easy to verify that
(42)
F
(
x
)
=
C
(
L
0
-
x
)
+
p
/
(
p
-
2
)
satisfies the above equation for any L
0
>
L, where C
p
-
2
=
(
1
/
2
(
p
-
1
)
)
(
(
p
-
2
)
/
p
)
p
-
1. If 0
<
x
<
L, we consider the problem
(43)
(
|
ν
′
|
p
-
2
ν
′
)
′
-
1
p
-
2
ν
+
ν
p
-
1
=
0
,
0
<
x
<
L
,
ν
(
0
)
=
A
>
(
1
p
-
2
)
1
/
(
p
-
2
)
,
ν
′
(
0
)
=
0
,
ν
′
(
L
)
=
C
(
L
0
-
L
)
+
p
/
(
p
-
2
)
,
ν
′
(
L
)
=
-
p
p
-
2
C
(
L
0
-
L
)
+
p
/
(
p
-
2
)
.
Taking m
=
1 in Lemma 3.1 of [14], we know that, for every length L
<
L
s, there exists a positive solution ν to this problem. Therefore, (
T
-
t
)
-
1
/
(
p
-
12
)
ν
(
x
) is the desired self-similar blow-up solution.
We now check that the above self-similar solution constructed can be put below any solution if we let pass enough time. In fact, the self-similar solution has small initial value ifT
>
0 is large, but its support is not small, since the length L
0 is not small. We then use the penetration property of the solutions to the p-Laplacian equation to guarantee that there exists t
0
>
0 such that the support of u
(
·
,
t
0
) contains the interval [
-
L
0
,
L
0
]. Therefore, we could get u
(
x
,
t
0
)
≥
U
(
x
,
0
) by taking T which is large enough.
By comparison,u must blow up in finite time.
Now, we turn to the case ofq
>
p
-
1.Theorem 8.
Ifp
-
1
<
q
<
2
(
p
-
1
), every solution to problem (1) blows up in finite time.Proof.
From Lemma2, we know that if there exists t
0
>
0, such that energy functional E
u
(
t
0
)
<
0, then all solutions to (1) blow up in finite time.
We now check that the so-called Barenblatt function(44)
u
B
(
x
,
t
;
d
)
=
t
-
1
/
2
(
p
-
1
)
D
(
d
p
/
(
p
-
1
)
-
x
p
/
(
p
-
1
)
t
-
p
/
2
(
p
-
1
)
2
)
+
(
p
-
1
)
/
(
p
-
2
)
satisfies the above requirement provided that t is large. In fact,
(45)
E
u
B
(
t
)
=
1
p
(
D
p
p
-
2
)
p
×
∫
-
∞
+
∞
t
-
p
(
2
p
-
1
)
/
2
(
p
-
1
)
2
h
h
h
h
h
h
h
×
(
d
p
/
(
p
-
1
)
-
x
p
/
(
p
-
1
)
t
-
p
/
2
(
p
-
1
)
2
)
+
p
/
(
p
-
2
)
h
h
h
h
h
h
h
×
x
p
/
(
p
-
1
)
d
x
-
D
q
+
1
×
∫
-
L
L
t
-
(
q
+
1
)
/
2
(
p
-
1
)
h
h
h
h
h
h
×
(
t
-
p
/
(
2
(
p
-
1
)
2
)
d
p
/
(
p
-
1
)
h
h
h
h
h
h
h
h
h
h
-
x
p
/
(
p
-
1
)
t
-
p
/
2
(
p
-
1
)
2
)
+
(
q
+
1
)
(
p
-
1
)
/
(
p
-
2
)
d
x
≤
C
1
(
∫
d
1
t
1
/
2
(
p
-
1
)
d
2
t
1
/
2
(
p
-
1
)
x
p
/
(
p
-
1
)
d
x
)
×
t
-
p
(
2
p
-
1
)
/
2
(
p
-
1
)
2
-
C
2
t
(
q
+
1
)
/
2
(
p
-
1
)
×
(
∫
0
c
L
t
-
1
/
2
(
p
-
1
)
(
1
-
z
p
/
(
p
-
1
)
)
+
(
q
+
1
)
(
p
-
1
)
/
(
p
-
2
)
d
x
)
×
t
1
/
2
(
p
-
1
)
≤
C
1
t
-
(
2
p
2
-
3
p
+
1
)
/
2
(
p
-
1
)
2
-
C
3
t
-
(
q
+
1
)
/
2
(
p
-
1
)
.
From q
<
2
(
p
-
1
), we know -
(
2
p
2
-
3
p
+
1
)
/
2
(
p
-
1
)
2
<
-
(
q
+
1
)
/
2
(
p
-
1
). Therefore, E
u
B
(
t
)
<
0 provided that t is large enough.
Finally, we may choosed
>
0 which is small in order to have u
B
(
x
,
1
;
d
)
≤
u
0
(
x
). Therefore,
(46)
u
B
(
x
,
t
+
1
;
d
)
≤
u
(
x
,
t
)
.
Since the solution to problem (1) with initial value u
B
(
·
,
t
1
;
d
) with t
1 which is large blows up, so does u.Now, we deal with the case ofq
=
2
(
p
-
1
).Theorem 9.
Ifq
=
2
(
p
-
1
), every solution to (1) blows up in finite time.Proof.
Here, we argue as in [15] or [7]. The proof consists of several steps.
Step
1. Assume by contradiction that there exists a global nonnegative nontrival solutions u
(
x
,
t
) to problem (1). According to the symmetry properties of equation with q
=
2
(
p
-
1
), we introduce the following rescaled function:
(47)
θ
(
ξ
,
τ
)
=
(
1
+
t
)
1
/
2
(
p
-
1
)
u
(
x
,
t
)
,
ξ
=
x
(
1
+
t
)
-
1
/
2
(
p
-
1
)
,
τ
=
ln
⁡
(
t
+
1
)
.
Then, θ solves the equation
(48)
θ
τ
=
(
|
θ
ξ
|
p
-
2
θ
ξ
)
ξ
+
1
2
(
p
-
1
)
(
ξ
θ
)
ξ
+
ψ
(
ξ
)
e
τ
/
2
(
p
-
1
)
θ
2
(
p
-
1
)
,
where ψ
(
ξ
)
=
χ
{
|
ξ
|
≤
L
e
-
τ
/
2
(
p
-
1
)
}.
Denote byg
(
ξ
,
τ
) the solution of (48) with initial data
(49)
g
(
ξ
,
0
)
=
D
(
d
p
/
(
p
-
1
)
-
ξ
p
/
(
p
-
1
)
)
+
(
p
-
1
)
/
(
p
-
2
)
=
u
B
(
ξ
,
1
;
d
)
.
Assuming that without loss of generality u
0
(
0
)
>
0, we have that there exists a small d
>
0 such that u
0
(
x
)
≥
u
B
(
x
,
1
;
d
). This implies that g
(
ξ
,
τ
)
≤
θ
(
ξ
,
τ
), and therefore g is also a global solution.
Step
2. We now discuss some important properties of g
(
ξ
,
τ
).
First, by applying the maximum principle to the linear parabolic equation for the derivativez
=
g
ξ and by using a standard regularization argument, we have that g
(
ξ
,
τ
) is nonincreasing in ξ
>
0. Next, for the special form of the initial data, it is easy to verify by the same method of [7, 15] that g is increasing in τ. Therefore, there exist
(50)
f
(
ξ
)
=
lim
⁡
τ
→
+
∞
⁡
g
(
ξ
,
τ
)
.
We claim that f
(
ξ
)
<
+
∞.
In fact, this fails for someξ
0
>
0 and hence lim
⁡
τ
→
+
∞
g
(
ξ
0
,
τ
)
=
+
∞. Since g is nonincreasing in ξ, we conclude that given M
>
0 there exists τ
M which is large such that
(51)
g
(
ξ
,
τ
M
)
>
M
,
|
ξ
|
≤
ξ
0
.
Set w
(
x
,
t
)
=
e
-
τ
/
2
(
p
-
1
)
g
(
ξ
,
τ
). We will show that by taking M which is large enough w blows up in finite time, which contradicts the fact that g is global.
To this end we consider a special functionv
(
x
)
=
u
B
(
x
,
t
;
d
). Then v
(
x
) has negative energy, supp
⁡
(
v
)
⊂
(
-
ξ
0
e
τ
M
/
2
(
p
-
1
)
,
ξ
0
e
τ
M
/
2
(
p
-
1
)
) and v
(
x
)
≤
M
e
-
τ
M
/
2
(
p
-
1
). The first condition is achieved if d is large enough. The last two requirements are possible for some t
=
t
* provided that M is chosen as large enough, depending on ξ
0 and τ
M. This implies that the solution to (48) with initial value v blows up in finite time. w is larger than this solution, and this contradicts the fact that g is global. The contradiction shows that f
(
ξ
) defined as above is finite for every ξ
≠
0.
Step
3. Using the monotonicity property, we can introduce a Lyapunov functional as in [15]. By means of Lyapunov functional, we could obtain as in [15] that f is a weak symmetric solution to the equation
(52)
(
|
θ
′
|
p
-
2
θ
′
)
′
+
1
2
(
p
-
1
)
(
ξ
θ
)
′
=
0
.
for every ξ
≠
0.
In order to finish the blow-up argument in this case, we observe that the only self-similar solution to this equation is (see [7])
(53)
f
(
ξ
)
=
D
(
d
p
/
(
p
-
1
)
-
ξ
p
/
(
p
-
1
)
)
+
(
p
-
1
)
/
(
p
-
2
)
.
This is followed by uniqueness and a symmetrization argument. In particular, f
(
0
) is finite. Then, by passing to the limit in the weak formulation of (48), we could get the boundary condition at ξ
=
0
+ as follows:
(54)
lim
⁡
ξ
→
0
+
(
|
θ
′
|
p
-
2
θ
′
)
(
ξ
)
=
-
2
L
f
2
(
p
-
1
)
(
0
)
.
However, the profiles obtained do not satisfy this condition. This completes the proof in the critical case.


## 4. Conclusion and Further Discussion

Theorems6–9 show that if 2
(
p
-
1
)
/
p
<
q
≤
2
(
p
-
1
), then all solutions to problem (1) blow up. On the other hand, from Theorem 5, we know both blow-up solutions and global solutions exist in the case of q
>
2
(
p
-
1
). Thus, the Fujita exponent of problem (1) is q
c
=
2
(
p
-
1
).Comparing the proof of the main results in this paper and that of the corresponding part in [11], we find that we could deal with the doubly degenerate problem
(55)
u
t
=
(
|
(
u
m
)
x
|
p
-
2
(
u
m
)
x
)
x
+
a
(
x
)
u
q
,
x
∈
R
,
t
>
0
,
u
(
x
,
0
)
=
u
0
(
x
)
,
x
∈
R
,
by similar a technique to the present paper. One can conjecture that the Fujita exponent of this Cauchy problem is q
=
(
m
+
1
)
(
p
-
1
).



---

*Source: 301747-2014-08-24.xml*