# Global Existence and Blow-Up of Solutions for Nonlinear Klein-Gordon Equation with Damping Term and Nonnegative Potentials

Authors: Wen-Yi Huang; Wen-Li Chen
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

This paper is concerned with the nonlinear Klein-Gordon equation with damping term and
nonnegative potentials. We introduce a family of potential wells and discuss the invariant sets and vacuum
isolating behavior of solutions. Using the potential well argument, we obtain a new existence theorem of
global solutions and a blow-up result for solutions in finite time.


---

## Body

## 1. Introduction

In this paper, we consider the nonlinear Klein-Gordon equation with damping term and a real valued potentialT
(
x
)
(1)
u
t
t
-
Δ
u
+
T
(
x
)
u
+
u
+
|
u
t
|
m
-
2
u
t
=
|
u
|
p
-
2
u
,
t
>
0
,
x
∈
R
n
,
u
(
0
,
x
)
=
u
0
,
u
t
(
0
,
x
)
=
u
1
,
x
∈
R
n
,
where n
≥
2, m
≥
2, and
(2)
2
<
p
<
∞
if
n
≤
2
,
2
<
p
<
2
n
n
-
2
if
n
≥
3
.
u
=
u
(
t
,
x
) is a complex-valued function of (
t
,
x
)
∈
R
+
×
R
n, Δ is the Laplace operator on R
n, and u
t is called the damping term [1]. Ha and Nakagiri [1] studied the local existence for the Cauchy problem (1). Here we are interested in the sharp criteria for global existence and blow-up of solutions of the Cauchy problem (1).The Klein-Gordon equation is a relativistic version of the Schrödinger equation, which describes relativistic electrons. Levine [2], Ball [3], Payne and Sattinger [4], Zhang [5], and Gan and Zhang [6] applied the potential well theory and studied the blowing up properties of the nonlinear Klein-Gordon equations. In [7], Huang and Zhang studied the global existence and blow-up of solutions for the nonlinear Klein-Gordon equation with linear damping term (
m
=
2
). In [8, 9], the authors studied the existence of global solutions and decay for the energy of solution for the Klein-Gordon equation.The case of nonlinear damping and source terms(
m
>
2
,
p
>
2
) is considered by many authors. For instance, Georgiev and Todorova [10] prove that if m
≥
p, a global weak solution exists for any initial data; while 2
<
m
<
p the solution blows up in finite time when the initial energy is sufficiently negative. Ikehata [11] considers the solutions of (1) with small positive initial energy, using the so-called potential well theory introduced by Payne and Sattinger in [4]. Todorova and Vitillaro [12] prove that for any given numbers α
≥
0
,
λ
≥
0 there exist infinitely many data φ
(
x
)
,
ψ
(
x
) in the energy space such that the initial energy E
(
0
)
=
λ, the gradient norm ∥
∇
φ
∥
2
=
α, and the solution of (1) blows up in finite time.In this paper, we consider the interaction between the nonlinear damping and source terms(
m
>
2
,
p
>
2
) for the Cauchy problem (1). For the local well-posedness of the Cauchy problem (1), the readers may refer to [13, 14]. We have considered the global existence and the finite time blowing up. The potential well theory, which was introduced by Liu [15] and has been used for Schrödinger equations in [16–18], was applied to study the Cauchy problem (1). Based on the results, we show the sharp criteria for global existence and blowing up of its solutions. Applying the perturbed energy method we prove the uniform stabilization for p
≤
m. And using concavity arguments we prove that the blow-up solutions exist for p
>
m. The results can be extended to the case of more general nonlinearities under suitable assumptions. We extend parts of results in [7] and obtain several new results for system (1).This paper is organized as follows. In Section2, a family of potential wells are introduced and a series of properties are given. In Section 3, the invariant sets under the flow of problem (1) and the vacuum isolating behavior of solutions for 0
<
E
(
0
)
<
d and E
(
0
)
≤
0 are discussed. In Section 4, the global existence and blowing up of solutions for problem (1) are proved. In Section 5, the theorem on asymptotic behavior of solutions when m
=
2 is proved.


## 2. Potential Wells and Their Properties

For the Cauchy problem (1), we define the energy space as
(3)
H
:
=
{
u
∈
H
1
(
R
n
)
;
∫
R
n
‍
T
(
x
)
∥
u
∥
2
d
x
<
∞
}
.
H becomes a Hilbert space, continuously embedded in H
1
(
R
n
), which is endowed with the inner product
(4)
(
u
,
v
)
H
≔
∫
R
n
‍
[
∇
u
∇
v
+
T
(
x
)
u
v
+
u
v
]
d
x
,
whose associated norm is denoted by ∥
·
∥
H.Throughout this paper, we make the following assumptions onT
(
x
):
(5)
inf
⁡
x
∈
R
n
T
(
x
)
=
T
¯
(
x
)
>
0
,
T
(
x
)
is
positive
and
a
C
1
bounded
measurable
function

on
⁡
R
n
,
lim
⁡
x
→
∞
T
(
x
)
=
∞
.We define the energy functions(6)
E
(
t
)
=
1
2
∥
u
t
∥
2
2
+
1
2
∥
∇
u
∥
2
2
+
1
2
T
(
x
)
∥
u
∥
2
2
+
1
2
∥
u
∥
2
2
-
1
p
∥
u
∥
p
p
,
t
∈
[
0
,
T
)
,
(7)
E
(
t
)
+
∫
0
t
‍
∥
u
t
∥
m
m
d
τ
=
E
(
0
)
,
t
∈
[
0
,
T
)
,
and two functionals
(8)
S
(
u
)
=
1
2
∥
∇
u
∥
2
2
+
1
2
T
(
x
)
∥
u
∥
2
2
+
1
2
∥
u
∥
2
2
-
1
p
∥
u
∥
p
p
,
(9)
R
(
u
)
=
∥
∇
u
∥
2
2
+
T
(
x
)
∥
u
∥
2
2
+
∥
u
∥
2
2
-
∥
u
∥
p
p
.
Then we define the potential well W as follows:
(10)
W
≡
{
u
∈
H
(
R
n
)
∣
R
(
u
)
>
0
,
S
(
u
)
<
d
M
}
∪
{
0
}
,
and the outside set V of the corresponding potential well
(11)
V
≡
{
u
∈
H
(
R
n
)
∣
R
(
u
)
<
0
,
S
(
u
)
<
d
M
}
,
where
(12)
d
M
=
inf
⁡
M
S
(
u
)
,
M
=
{
u
∈
H
(
R
n
)
∖
{
0
}
∣
R
(
u
)
=
0
,
∥
u
∥
H
≠
0
}
.Forδ
>
0, we define
(13)
R
δ
(
u
)
=
δ
(
∥
∇
u
∥
2
2
+
T
(
x
)
∥
u
∥
2
2
+
∥
u
∥
2
2
)
-
∥
u
∥
p
p
,
d
M
(
δ
)
=
inf
⁡
M
δ
S
(
u
)
,
M
δ
=
{
u
∈
H
(
R
n
)
∖
{
0
}
∣
R
δ
(
u
)
=
0
,
∥
u
∥
H
≠
0
}
,
W
δ
≡
{
u
∈
H
(
R
n
)
∣
R
δ
(
u
)
>
0
,
S
(
u
)
<
d
M
(
δ
)
}
∪
{
0
}
,
0
<
δ
<
p
2
,
V
δ
≡
{
u
∈
H
(
R
n
)
∣
R
δ
(
u
)
<
0
,
S
(
ϕ
)
<
d
M
(
δ
)
}
,
0
<
δ
<
p
2
.Lemma 1.
If0
<
∥
u
∥
H
<
r
(
δ
), then R
δ
(
u
)
>
0. Particularly, if 0
<
∥
u
∥
H
<
r
(
1
), then R
(
u
)
>
0, where
(14)
r
(
δ
)
=
(
δ
C
*
p
)
1
/
(
p
-
2
)
,
C
* is an embedding constant of H
(
R
n
)
↪
L
p
(
R
n
).Proof.
If0
<
∥
u
∥
H
<
r
(
δ
), then
(15)
∥
u
∥
p
p
≤
C
*
p
∥
u
∥
H
p
=
C
*
p
∥
u
∥
H
p
-
2
∥
u
∥
H
2
<
δ
∥
u
∥
H
2
.
It follows that R
δ
(
u
)
>
0.Lemma 2.
IfR
δ
(
u
)
<
0, then ∥
u
∥
H
>
r
(
δ
). Particularly, if R
(
u
)
<
0, then ∥
u
∥
H
>
r
(
1
).Proof.
IfR
δ
(
u
)
<
0, then ∥
u
∥
H
≠
0. From
(16)
δ
∥
u
∥
H
2
<
∥
u
∥
p
p
≤
C
*
p
∥
u
∥
H
p
-
2
∥
u
∥
H
2
,
we obtain ∥
u
∥
H
>
r
(
δ
).Lemma 3.
Assume that (2) holds; then(1)
d
M
(
δ
)
≥
(
1
/
2
-
δ
/
p
)
r
(
δ
) for 0
<
δ
<
p. In particular, we have d
≥
2
/
α
C
⋆
α, α
=
p
/
(
p
-
2
).
(2)
d
M
(
δ
)
=
δ
2
/
(
p
-
2
)
(
1
/
2
-
δ
/
p
)
(
2
p
/
(
p
-
2
)
)
d.Proof.
(1) Foru
∈
M
δ, we get ∥
u
∥
H
2
≥
r
(
δ
) and
(17)
S
(
u
)
=
(
1
2
-
δ
p
)
∥
u
∥
H
2
+
1
p
R
δ
(
u
)
=
(
1
2
-
δ
p
)
∥
u
∥
H
2
≥
(
1
2
-
δ
p
)
r
(
δ
)
,
which yields d
M
(
δ
)
≥
(
1
/
2
-
δ
/
p
)
r
(
δ
) for 0
<
δ
<
p
/
2.
(2) Letu
∈
M be a minimizer; that is, d
M
=
S
(
u
). For any δ
>
0, define λ
=
λ
(
δ
) by(18)
δ
∥
λ
u
∥
H
2
=
∥
λ
u
∥
p
p
;
that is,
(19)
δ
∥
u
∥
H
2
=
λ
p
-
2
∥
u
∥
p
p
.
Then, for each δ
>
0, there exists a unique
(20)
λ
(
δ
)
=
(
δ
∥
u
∥
H
2
∥
u
∥
p
p
)
1
/
(
p
-
2
)
,
satisfying (18) which implies λ
(
δ
)
u
∈
M
δ. Since u
∈
M implies ∥
u
∥
H
2
=
∥
u
∥
p
p, we get λ
(
δ
)
=
δ
1
/
(
p
-
2
). Therefore,
(21)
d
M
(
δ
)
≤
S
(
λ
(
δ
)
u
)
=
1
2
δ
2
/
(
p
-
2
)
∥
u
∥
H
2
-
1
p
δ
p
/
(
p
-
2
)
∥
u
∥
p
p
=
(
1
2
-
δ
p
)
δ
2
/
(
p
-
2
)
∥
u
∥
H
2
.
Noting that
(22)
S
(
u
)
=
1
2
∥
u
∥
H
2
-
1
p
∥
u
∥
p
p
=
p
-
2
2
p
∥
u
∥
H
2
,
we have
(23)
d
M
(
δ
)
≤
δ
2
/
(
p
-
2
)
(
1
2
-
δ
p
)
p
-
2
2
p
S
(
u
)
=
δ
2
/
(
p
-
2
)
(
1
2
-
δ
p
)
p
-
2
2
p
d
.
On the other hand, letδ
>
0, u
∈
M
δ be a minimizer; that is, d
M
(
δ
)
=
S
(
u
). It follows that
(24)
d
M
(
δ
)
≥
δ
2
/
(
p
-
2
)
(
1
2
-
δ
p
)
p
-
2
2
p
d
.
Therefore, the conclusion follows from the above discussion.Lemma 4.
Assume that (2) holds; then(1)
lim
⁡
δ
→
0
⁡
d
M
(
δ
)
=
0, d
M
(
p
/
2
)
=
0. d
M
(
δ
) is continuous on 0
<
δ
≤
p
/
2.
(2)
d
M
(
δ
) is increasing on 0
<
δ
≤
1, is decreasing on 1
<
δ
≤
p
/
2, and takes the maximum d
=
d
M
(
1
) at δ
=
1.Proof.
From Lemma3, we obtain
(25)
d
M
′
(
δ
)
=
d
p
-
2
[
(
p
-
2
δ
)
δ
2
/
(
p
-
2
)
]
′
=
d
p
-
2
[
(
p
-
2
δ
)
2
p
-
2
δ
(
4
-
p
)
/
(
p
-
2
)
-
2
δ
2
/
(
p
-
2
)
]
=
2
d
p
(
p
-
2
)
2
δ
(
4
-
p
)
/
(
p
-
2
)
(
1
-
δ
)
.


## 3. The Invariant Sets of Solutions

Lemma 5.
Assume thatu
0
∈
H
(
R
n
), u
1
∈
L
m
(
R
n
), 0
<
E
(
0
)
<
d
M, δ
1, and δ
2
(
δ
1
<
δ
2
) are the solutions of the function d
M
(
δ
)
=
E
(
0
).(1)
IfR
(
u
0
)
>
0 or ∥
u
0
∥
H
=
0, then u
(
t
)
∈
W
δ for any δ
∈
(
δ
1
,
δ
2
).
(2)
IfR
(
u
0
)
<
0, then u
(
t
)
∈
V
δ for any δ
∈
(
δ
1
,
δ
2
).Proof.
(1) Letu
(
t
) be any solution of the Cauchy problem (1) with
(26)
E
(
0
)
=
1
2
∥
u
1
∥
2
2
+
S
(
u
0
)
+
∫
0
t
‍
∥
u
1
∥
m
m
d
τ
=
d
M
(
δ
)
<
d
M
,
which gives S
(
u
0
)
<
d
M. If R
(
u
0
)
>
0, then from the definition of W
δ we obtain u
0
(
x
)
∈
W
δ. If ∥
u
0
∥
H
=
0, then u
0
(
x
)
∈
W
δ. Therefore u
0
(
x
)
∈
W
δ
,
∀
δ
∈
(
δ
1
,
δ
2
).
Next, we proveu
(
t
)
∈
W
δ
,
∀
δ
∈
(
δ
1
,
δ
2
)
,
t
≥
0. If it is not true, then there must exist a δ
~
∈
(
δ
1
,
δ
2
) and a t
~
>
0 such that u
(
t
~
)
∈
∂
W
δ
~; that is,
(27)
R
δ
~
(
u
(
t
~
)
)
=
0
,
∥
u
0
(
t
~
)
∥
H
≠
0
,
or
S
(
u
(
t
~
)
)
=
d
M
(
δ
~
)
.
From the energy inequality, we have
(28)
1
2
∥
u
t
∥
2
2
+
S
(
u
)
+
∫
0
t
‍
∥
u
t
∥
m
m
d
τ
=
E
(
0
)
=
d
M
(
δ
)
<
d
M
,
δ
1
<
δ
<
δ
2
,
t
≥
0
.
Then S
(
u
(
t
~
)
)
=
d
M is impossible. On the other hand, if R
δ
~
(
u
(
t
~
)
)
=
0, ∥
u
0
(
t
~
)
∥
H
≠
0, then we obtain u
(
t
~
)
∈
M. By the definition of M, we have S
(
u
(
t
~
)
)
≥
d
M, which contradicts (28). Hence u
(
t
)
∈
W
δ is true.
(2) First we proveu
0
∈
V
δ. From the energy inequality
(29)
1
2
∥
u
1
∥
2
2
+
S
(
u
0
)
+
∫
0
t
‍
∥
u
1
∥
m
m
d
τ
=
E
(
0
)
=
d
M
(
δ
)
<
d
M
,
we have
(30)
S
(
u
0
)
<
d
M
(
δ
)
,
∀
δ
1
<
δ
<
δ
2
.
Using R
(
u
0
)
<
0 yields R
δ
(
u
0
)
<
0 for δ
1
<
δ
<
δ
2. Therefore we obtain u
0
∈
V
δ.
Next, we show thatu
(
t
)
∈
V
δ for δ
1
<
δ
<
δ
2 and t
≥
0. If it is false, there exist a δ
~
∈
(
δ
1
,
δ
2
) and a t
~
>
0 such that u
(
t
~
)
∈
∂
V
δ
~; that is,
(31)
R
δ
~
(
u
(
t
~
)
)
=
0
,
∥
u
0
(
t
~
)
∥
H
≠
0
or
S
(
u
(
t
~
)
)
=
d
M
(
δ
~
)
.
However, from the conservation law we get that S
(
u
(
t
~
)
)
=
d
M
(
δ
~
) is impossible. If R
δ
~
(
u
(
t
~
)
)
=
0, then R
δ
~
(
u
(
t
)
)
<
0 for 0
≤
t
<
t
~. At the same time, Lemma 2 yields that ∥
u
(
t
)
∥
H
>
r
(
δ
~
)
>
0, 0
≤
t
<
t
~, and ∥
u
(
t
~
)
∥
H
≥
r
(
δ
~
). Hence by the definition of d
M
(
δ
), we have S
(
u
(
t
~
)
)
≥
d
M
(
δ
~
), which contradicts S
(
u
(
t
)
)
<
d
M
(
δ
). So we obtain
(32)
u
(
t
)
∈
V
δ
,
∀
δ
1
<
δ
<
δ
2
,
t
≥
0
.Lemma 6.
Assumeu
0
∈
H
(
R
n
), u
1
∈
L
m
(
R
n
), 0
<
E
(
0
)
<
d
M
(
δ
), δ
1, and δ
2
(
δ
1
<
δ
2
) are the solutions of the function d
M
(
δ
)
=
E
(
0
). Then W
δ and V
δ are invariant sets under the flow generated by (1), ∀
δ
1
<
δ
<
δ
2.Proof.
Letu
0
∈
V
δ and u
(
t
) satisfy (1). From (6), (7), and (8), we have
(33)
S
(
u
(
t
)
)
≤
E
(
t
)
=
E
(
0
)
-
∫
0
t
‍
∥
u
t
∥
m
m
d
τ
<
d
M
(
δ
)
,
t
∈
[
0
,
T
)
.
To check u
(
t
)
∈
V
δ, we need to prove
(34)
R
(
u
(
t
)
)
<
0
,
t
∈
[
0
,
T
)
.
If (34) is not true, by continuity, there would exist a t
¯
>
0 such that R
(
u
(
t
¯
)
)
=
0 because of R
(
u
0
)
<
0. It follows that u
(
t
¯
)
∈
M
δ. This is impossible for S
(
u
(
t
¯
)
)
<
d
M
(
δ
) and d
M
(
δ
)
=
inf
⁡
u
∈
M
δ
⁡
S
(
u
). Thus (34) is true. So V
δ is invariant under the flow generated by (1).
Similarly, we show thatW
δ is also invariant under the flow generated by (1). This completes the proof of Lemma 6.Lemma 7.
Let the initial data(
u
0
,
u
1
)
∈
H
(
R
n
)
×
L
m
(
R
n
), and u
(
t
,
x
) be a local solution of the Cauchy problem (1) on [
0
,
T
). If there exists a u
0
∈
V
δ and a E
(
0
)
<
d
M
(
δ
), then the inequality
(35)
∥
∇
u
∥
2
2
+
T
(
x
)
∥
u
∥
2
2
+
∥
u
∥
2
2
>
2
p
p
-
2
d
M
(
δ
)
is fulfilled for t
∈
[
0
,
T
).Proof.
By the definition ofd
M
(
δ
), we have
(36)
d
M
(
δ
)
=
inf
⁡
{
p
-
2
2
p
(
∥
∇
u
∥
2
2
+
T
(
x
)
∥
u
∥
2
2
+
∥
u
∥
2
2
)
}
.
According to Lemma 6, we have ∥
u
(
t
,
·
)
∥
p
p
>
∥
∇
u
(
t
,
·
)
∥
2
2 for t
∈
[
0
,
T
). From (36) and the identity (8), we get
(37)
d
M
(
δ
)
<
p
-
2
2
p
(
∥
∇
u
∥
2
2
+
T
(
x
)
∥
u
∥
2
2
+
∥
u
∥
2
2
)
,
which completes the proof of Lemma 7.In order to extend the caseE
(
0
)
>
0 to E
(
0
)
≤
0 we give the following lemma.Lemma 8.
Letu
0
∈
H
(
R
n
), u
1
∈
L
m
(
R
n
), and (2) hold. Assume E
(
0
)
=
0; then the solutions of problem (1) satisfy
(38)
∥
u
∥
H
2
≥
r
0
=
2
(
p
C
*
p
)
1
/
(
p
-
2
)
.Proof.
Letu be any solution of problem (1) with E
(
0
)
=
0 and ∥
u
0
∥
H
≠
0, T
max
⁡ the existence time. From (6), (7), and (9), we have
(39)
E
(
t
)
=
1
2
∥
u
t
∥
2
2
+
S
(
u
)
≤
E
(
0
)
=
0
,
and get S
(
u
)
≤
0 for 0
≤
t
<
T
max
⁡. Hence, using(40)
1
2
∥
u
∥
H
2
≤
1
p
∥
u
∥
p
p
≤
C
*
p
∥
u
∥
H
2
∥
u
∥
H
p
-
2
,
we see that either ∥
u
∥
H
2
=
0 or (38) hold. If ∥
u
∥
H
2
=
0, then ∥
u
∥
H
2
≡
0 for 0
≤
t
<
T
max
⁡ (otherwise there exists a t
0
∈
[
0
,
t
max
⁡
) such that 0
<
∥
u
(
t
0
)
∥
H
2
<
r
0), which contradicts the condition ∥
u
0
∥
H
≠
0.Theorem 9.
Letu
0
∈
H
(
R
n
), u
1
∈
L
m
(
R
n
), 2
<
m
<
p, and (2) hold. Assume E
(
0
)
<
0 or E
(
0
)
=
0 and ∥
u
0
∥
H
≠
0. Then the solutions of problem (1) belong to V
δ for 0
<
δ
<
p
/
2.Proof.
Letu be any solution of problem (1) with E
(
0
)
<
0 or E
(
0
)
=
0 and ∥
u
0
∥
H
≠
0, T
max
⁡ the existence time. From (6), we have
(41)
E
(
t
)
=
1
2
∥
u
t
∥
2
2
+
(
1
2
-
δ
p
)
∥
u
∥
H
2
+
1
p
R
δ
(
u
)
≤
E
(
0
)
,
0
<
δ
<
p
2
,
for 0
≤
t
<
T
max
⁡. From (41) we see that if E
(
0
)
<
0, then S
(
u
)
≤
E
(
t
)
<
0
<
d
M
(
δ
) for 0
≤
t
<
T
max
⁡. If E
(
0
)
=
0 and ∥
u
0
∥
H
≠
0, then by Lemma 8 we obtain ∥
u
∥
H
2
≥
r
0
>
0. Therefore, by (41) we have S
(
u
)
≤
E
(
t
)
≤
0
<
d
M
(
δ
) for 0
≤
t
<
T
max
⁡. Hence, for the above two cases, we have u
∈
V
δ for 0
<
δ
<
p
/
2.


## 4. Sharp Condition for Global Existence and Blow-Up

Definition 10 (weak solution).
The functionu
(
t
,
x
)
∈
C
(
[
0
,
T
)
;
H
(
R
n
)
) with u
t
(
t
,
x
)
∈
C
(
[
0
,
T
)
;
L
m
(
R
n
)
) is called a weak solution of problem (1), such that u
(
0
,
x
)
=
u
0
(
x
) in H
(
R
n
), u
t
(
0
,
x
)
=
u
1
(
x
) in L
m
(
R
n
) and
(42)
〈
u
t
t
,
v
〉
+
∫
R
n
‍
∇
u
·
∇
v
d
x
+
∫
R
n
‍
T
(
x
)
u
v
d
x
+
∫
R
n
‍
u
v
d
x
+
∫
R
n
‍
|
u
t
|
m
-
2
u
t
v
d
x
=
∫
R
n
‍
|
u
|
p
-
2
u
v
d
x
,
for all v
∈
H
(
R
n
) and t
∈
[
0
,
T
).Theorem 11.
Let(
u
0
,
u
1
)
∈
H
(
R
n
)
×
L
m
(
R
n
), 2
<
p
≤
m, and (2) hold. Suppose that 0
<
E
(
0
)
<
d
M, R
(
u
0
)
>
0, or ∥
u
0
∥
H
=
0; then Cauchy problem (1) has a global weak solution u
(
t
,
x
)
∈
C
(
[
0
,
T
)
;
H
(
R
n
)
)
(
u
t
(
t
,
x
)
∈
C
(
[
0
,
T
)
;
L
m
(
R
n
)
) for some T
∈
(
0
,
∞
) with u
(
t
)
∈
W.Proof.
Let{
w
j
(
x
)
} be a system of base functions in H
(
R
n
). Construct the approximate solution u
m
(
t
,
x
) of problem (1)
(43)
u
m
(
t
,
x
)
=
∑
j
=
1
m
‍
g
j
m
(
t
)
w
j
(
x
)
,
m
=
1,2
,
…
,
satisfying
(44)
(
u
m
t
t
,
w
s
)
-
(
Δ
u
m
,
w
s
)
+
T
(
x
)
(
u
m
,
w
s
)
+
(
u
m
,
w
s
)
+
(
|
u
m
t
|
m
-
2
u
m
t
,
w
s
)
=
(
|
u
m
|
p
-
2
u
m
,
w
s
)
,
(45)
u
m
(
x
,
0
)
=
∑
j
=
1
m
‍
a
j
m
w
j
(
x
)
⟶
u
0
(
x
)
in
H
1
(
R
n
)
,
(46)
u
m
t
(
x
,
0
)
=
∑
j
=
1
m
‍
b
j
m
w
j
(
x
)
⟶
u
1
(
x
)
in
L
m
(
R
n
)
.
Multiplying (44) by g
j
m
′
(
t
) and summing for s, we have
(47)
d
d
t
[
1
2
(
∥
u
m
t
∥
2
2
+
∥
u
m
∥
H
2
)
-
1
p
∥
u
m
∥
p
p
]
+
∥
u
m
t
∥
m
m
=
0
.
Integrating with respect to t, we get
(48)
1
2
(
∥
u
m
t
∥
2
2
+
∥
u
m
∥
H
2
)
-
1
p
∥
u
m
∥
p
p
+
∫
0
t
∥
u
m
t
∥
m
m
d
τ
=
1
2
(
∥
u
m
t
(
0
)
∥
2
2
+
∥
u
m
(
0
)
∥
H
2
)
-
1
p
∥
u
m
(
0
)
∥
p
p
.
For E
(
0
)
<
d
M and R
(
u
0
)
>
0 or ∥
u
∥
H
2
=
0, we have
(49)
1
2
∥
u
m
t
∥
2
2
+
S
(
u
m
)
+
∫
0
t
∥
u
m
t
∥
m
m
d
τ
=
E
m
(
0
)
<
d
M
,
0
≤
t
<
∞
.
From
(50)
1
2
∥
u
1
∥
2
2
+
S
(
u
0
)
+
∫
0
t
∥
u
1
∥
m
m
d
τ
=
E
(
0
)
<
d
M
,
0
≤
t
<
∞
,
we have S
(
u
0
)
<
d
M. Hence from R
(
u
0
)
>
0, we obtain u
0
∈
W.
From (45) and (46), for sufficiently large m, we obtain
(51)
1
2
∥
u
m
t
∥
2
2
+
S
(
u
m
)
+
∫
0
t
∥
u
m
t
∥
m
m
d
τ
=
E
m
(
0
)
<
d
M
,
0
≤
t
<
∞
,
and u
m
(
0
)
∈
W. Similar with the proof of Lemma 5, from (51), for sufficiently large m and 0
≤
t
<
∞, we can prove u
m
(
t
)
∈
W and
(52)
S
(
u
m
)
=
1
2
∥
u
m
∥
H
2
-
1
p
∥
u
m
∥
p
p
=
(
1
2
-
1
p
)
∥
u
m
∥
H
2
+
1
p
R
(
u
m
)
≥
p
-
2
2
p
∥
u
m
∥
H
2
.
Thus we obtain(53)
1
2
∥
u
m
t
∥
2
2
+
p
-
2
2
p
∥
u
m
∥
H
2
<
d
M
,
0
≤
t
<
∞
;
then
(54)
∥
u
m
t
∥
2
2
≤
2
d
M
,
0
≤
t
<
∞
,
∥
u
m
∥
H
2
≤
2
p
p
-
2
d
M
,
0
≤
t
<
∞
,
∥
u
m
∥
p
2
≤
C
*
2
∥
u
m
∥
H
2
≤
C
*
2
2
p
p
-
2
d
M
,
0
≤
t
<
∞
,
∥
|
u
m
|
p
-
2
u
m
∥
q
q
=
∥
u
m
∥
p
p
≤
C
*
p
(
2
p
p
-
2
d
M
)
p
/
2
,
q
=
p
p
-
1
,
0
≤
t
<
∞
.
Using (54) and the method of compact, we obtain that u
(
t
,
x
)
∈
C
(
[
0
,
T
)
;
H
(
R
n
)
) is a global weak solution of problem (1). From Lemma 5, we have u
(
t
)
∈
W for 0
≤
t
<
∞.Theorem 12.
Let(
u
0
,
u
1
)
∈
H
(
R
n
)
×
L
m
(
R
n
), 2
<
p
≤
m, and (2) hold. Assume that E
(
0
)
<
d, R
δ
(
u
0
)
>
0, or ∥
u
0
∥
H
=
0, where δ
1 and δ
2
(
δ
1
<
δ
2
) are two roots of equation d
M
(
δ
)
=
e. Then the problem (1) admits a unique global solution u
∈
C
(
[
0
,
T
)
;
H
(
R
n
)
) and u
∈
W
δ for δ
∈
(
δ
1
,
δ
2
) and 0
≤
t
<
∞.Proof.
From Theorem11, we see that to prove Theorem 12 we only need to prove R
δ
(
u
0
)
>
0. Indeed, if it is not true, then there exists a δ
¯
∈
[
1
,
δ
2
) such that R
δ
¯
(
u
0
)
=
0. Since R
δ
¯
(
u
0
)
>
0 implies ∥
u
0
∥
H
≠
0, we obtain S
(
u
0
)
≥
d
M
(
δ
¯
), which contradicts S
(
u
0
)
≤
E
(
0
)
<
d
M
(
δ
) for δ
∈
(
δ
1
,
δ
2
).Theorem 13.
Let(
u
0
,
u
1
)
∈
H
(
R
n
)
×
L
m
(
R
n
) and (2) hold. Assume E
(
0
)
<
d
M.(1)
If2
<
m
<
p, there exists t
0
∈
[
0
,
T
) such that u
(
t
0
)
∈
V; then the solution u
(
x
,
t
) of Cauchy problem (1) blows up in a finite time.
(2)
If2
<
p
≤
m, there exists t
0
∈
[
0
,
T
) such that u
(
t
0
)
∈
W; then the solution u
(
x
,
t
) of problem (1) globally exists on [
0
,
∞
). Moreover, for t
∈
[
0
,
∞
), u
(
x
,
t
) satisfies
(55)
∥
u
t
∥
2
2
+
p
-
2
p
×
(
∫
|
∇
u
0
|
2
d
x
+
∫
T
(
x
)
|
u
0
|
2
d
x
+
∫
|
u
0
|
2
d
x
)
<
2
d
M
.Proof.
ByE
(
0
)
<
d
M, we have S
(
u
0
)
≤
E
(
0
)
<
d
M.
Firstly, we prove (1) of Theorem13. From the energy identity we have
(56)
∫
0
t
‍
∥
u
t
∥
m
m
d
s
=
E
(
0
)
-
E
(
t
)
≤
d
M
,
for all t
≥
0.
DenotingJ
(
t
)
=
∥
u
(
t
,
·
)
∥
2
2, we have
(57)
J
′′
(
t
)
=
2
∥
u
t
∥
2
2
-
2
R
(
u
)
-
2
∫
‍
u
u
t
|
u
t
|
m
-
2
d
x
.
Using the Hölder inequality and the interpolation inequality, we obtain
(58)
|
∫
‍
u
u
t
|
u
t
|
m
-
2
d
x
|
≤
∥
u
∥
m
∥
u
t
∥
m
m
-
1
≤
∥
u
∥
2
δ
∥
u
∥
p
1
-
δ
∥
u
t
∥
m
m
-
1
,
with δ
=
(
1
/
m
-
1
/
p
)
/
(
1
/
2
-
1
/
p
). From R
(
u
)
<
0, we have
(59)
∫
‍
T
(
x
)
|
u
|
2
d
x
<
∥
u
∥
p
p
,
which together with Lemma 6 give
(60)
∥
u
∥
2
δ
∥
u
∥
p
1
-
δ
∥
u
t
∥
m
m
-
1
≤
C
∥
u
t
∥
m
m
-
1
∥
u
∥
p
1
-
p
/
m
-
δ
+
p
δ
/
2
∥
u
∥
p
p
/
m
.
Using the Young inequality and 1
-
p
/
m
-
δ
+
p
δ
/
2
=
0, we have
(61)
∥
u
∥
2
δ
∥
u
∥
p
1
-
δ
∥
u
t
∥
m
m
-
1
≤
C
(
ɛ
)
∥
u
t
∥
m
m
+
ɛ
∥
u
∥
p
p
,
since
(62)
-
R
(
u
)
≥
-
R
(
u
)
+
δ
(
E
(
t
)
-
E
(
0
)
)
≥
(
1
-
δ
p
)
∥
u
∥
p
p
+
δ
2
∥
u
t
∥
2
2
+
(
δ
2
-
1
)
×
(
∥
∇
u
∥
2
2
+
∫
‍
T
(
x
)
|
u
|
2
d
x
+
∥
u
∥
2
2
)
-
δ
E
(
0
)
,
then
(63)
1
2
J
′′
(
t
)
+
C
(
ɛ
)
∥
u
t
∥
m
m
≥
(
1
+
δ
2
)
∥
u
t
∥
2
2
+
(
1
-
δ
p
-
ɛ
)
∥
u
∥
p
p
+
(
δ
2
-
1
)
(
∥
∇
u
∥
2
2
+
∫
‍
T
(
x
)
|
u
|
2
d
x
+
∥
u
∥
2
2
)
-
δ
E
(
0
)
,
where the constant δ
>
2 is chosen as follows.
SinceE
(
0
)
<
d
M, we choose the constant δ so that
(64)
2
p
d
M
p
d
M
-
(
p
-
2
)
E
(
0
)
<
δ
<
p
.
This guarantees δ
>
2. Then, using this choice and Lemma 7 we get
(65)
(
δ
2
-
1
)
(
∥
∇
u
∥
2
2
+
∫
‍
T
(
x
)
|
u
|
2
d
x
+
∥
u
∥
2
2
)
-
δ
E
(
0
)
≥
(
δ
2
-
1
)
2
p
p
-
2
d
M
-
δ
E
(
0
)
≥
0
.
If the constant δ is fixed, we choose the constant ɛ such that
(66)
C
1
=
1
-
δ
p
-
ɛ
>
0
.
Finally, using the inequality (60), (63) and Lemma 7 we have
(67)
J
′′
(
t
)
+
C
(
ɛ
)
∥
u
t
∥
m
m
≥
C
1
∥
u
∥
p
p
≥
C
1
(
∥
∇
u
∥
2
2
+
∫
‍
T
(
x
)
|
u
|
2
d
x
+
∥
u
∥
2
2
)
≥
C
1
2
p
p
-
2
d
M
,
where C
1
>
0. Since (56), integrating (67) over [
0
,
t
] we have
(68)
J
′
(
t
)
≥
C
1
p
d
M
p
-
2
t
-
C
(
ɛ
)
d
M
+
J
′
(
0
)
,
which concludes that there exists a t
1 such that J
′
(
t
)
|
t
=
t
1
>
0. Hence, J
(
t
) is increasing for t
>
t
1 (which is the interval of existence). Since R
(
u
)
<
0, there exists a t
2 such that ∥
u
(
t
,
x
)
∥
p
p is increasing for t
>
t
2. When t is large enough, the quantities ∥
u
t
(
t
,
x
)
∥
m
m and ∥
∇
u
(
t
,
x
)
∥
2
2 are small enough. Otherwise, assume that there is t
* such that ∥
u
t
(
t
,
x
)
∥
m
m
>
∥
u
t
(
t
*
,
x
)
∥
m
m for all t
>
t
*. By integrating the inequality, we obtain a contradiction with (56) and E
(
t
)
≥
0.
Thus in these cases, the quantity(69)
(
1
-
δ
p
-
ɛ
)
∥
u
∥
p
p
+
(
δ
2
-
1
)
×
(
∥
∇
u
∥
2
2
+
∫
‍
T
(
x
)
|
u
|
2
d
x
+
∥
u
∥
2
2
)
-
δ
E
(
0
)
-
C
(
ɛ
)
∥
u
t
∥
m
m
will eventually become positive. Therefore for t large enough, from (63) and (65) we have
(70)
J
′′
(
t
)
≥
(
1
+
δ
2
)
∥
u
t
∥
2
2
.
Using the Hölder inequality, we get
(71)
J
(
t
)
J
′′
(
t
)
≥
2
+
δ
8
[
J
′
(
t
)
]
2
.
Since
(72)
[
J
-
(
δ
-
6
)
/
8
(
t
)
]
′′
=
-
δ
-
6
8
J
-
(
δ
+
10
)
/
8
(
t
)
[
J
(
t
)
J
′′
(
t
)
-
2
+
δ
8
[
J
′
(
t
)
]
2
]
,
from (71) we have [
J
-
(
δ
-
6
)
/
8
(
t
)
]
′′
≤
0. Therefore J
-
(
δ
-
6
)
/
8
(
t
) is concave for sufficiently large t, and there exists a finite time T
* such that
(73)
lim
⁡
t
→
T
*
J
-
(
δ
-
6
)
/
8
(
t
)
=
0
.
From assumption on T
(
x
), we obtain(74)
∫
‍
T
(
x
)
|
u
|
2
d
x
≥
T
¯
∫
‍
|
u
|
2
d
x
.
Thus one gets T
<
∞ and
(75)
lim
⁡
t
→
T
-
∥
u
∥
H
=
∞
.
We complete the proof of (1) of Theorem 13.
Next, we prove (2) of Theorem13.
From (6), (7), and (55), we obtain E
(
0
)
<
d
M. It follows that u
0 satisfies
(76)
R
(
u
0
)
=
∫
‍
|
∇
u
0
|
2
d
x
+
∫
‍
T
(
x
)
|
u
0
|
2
d
x
+
∫
‍
|
u
0
|
2
d
x
-
∫
‍
|
u
0
|
p
d
x
>
0
,
which will be proved by contradiction. If (76) is not true, then we have R
(
u
0
)
≤
0. Thus there exists 0
<
μ
≤
1 such that u
0
≠
0 and
(77)
R
(
μ
u
0
)
=
μ
2
(
∫
‍
|
∇
u
0
|
2
d
x
+
∫
‍
T
(
x
)
|
u
0
|
2
d
x
+
∫
‍
|
u
0
|
2
d
x
)
-
μ
p
∫
‍
|
u
0
|
p
d
x
=
0
,
which implies μ
u
0
∈
M.
On the other hand, for0
<
μ
≤
1, u
0
∈
W and (55) yield
(78)
S
(
μ
u
0
)
<
μ
2
×
(
∫
‍
|
∇
u
0
|
2
d
x
+
∫
‍
T
(
x
)
|
u
0
|
2
d
x
+
∫
‍
|
u
0
|
2
d
x
)
≤
d
M
,
which is contradictory to Lemma 4.
Therefore, byR
(
u
0
)
>
0 and Lemma 6, we have R
(
u
)
>
0 and E
(
t
)
≤
E
(
0
)
<
d
M. Thus
(79)
E
(
t
)
-
1
p
S
(
u
)
≤
E
(
0
)
;
namely,
(80)
1
2
∫
‍
|
u
t
|
2
d
x
+
p
-
2
2
p
×
(
∫
|
∇
u
|
2
d
x
+
∫
T
(
x
)
|
u
|
2
d
x
+
∫
|
u
|
2
d
x
)
≤
E
(
0
)
.
Therefore we have established the bound of u
(
x
,
t
) in H for t
∈
[
0
,
T
) and thus the solution u
(
x
,
t
) of (1) exists globally on t
∈
[
0
,
∞
).
From (76), E
(
0
)
<
d
M, and Lemma 4, we have the estimate (55).
Thus, we complete the proof of Theorem13.


## 5. Asymptotic Behaviour of Solutions

We now state and prove the following theorem on asymptotic behavior of solutions whenm
=
2.Theorem 14.
Letm
=
2 in problem (1). Assume 0
<
E
(
0
)
<
d
m, R
(
u
0
)
>
0, or ∥
u
0
∥
H
=
0. For the global solution of the problem (1) given in Theorem 13, we have
(81)
E
(
t
)
≤
C
e
-
k
t
,
0
≤
t
<
∞
,
for some positive constants C and k.Proof.
Letu be a global solution of the problem (1); then by Theorem 13, we obtain u
∈
W
δ for δ
1
<
δ
<
δ
2 and 0
≤
t
<
∞, where δ
1 and δ
2 (δ
1
<
δ
2) are two roots of equation d
M
(
δ
)
=
E
(
0
). Differentiating (7) with respect to t and multiplying the obtained equality by e
γ
t
(
γ
>
0
), we have
(82)
d
d
t
(
e
γ
t
E
(
t
)
)
+
e
γ
t
∥
u
t
∥
2
2
=
γ
e
γ
t
E
(
t
)
,
0
≤
t
<
∞
.
Integrating (82) with respect to t, we get
(83)
e
γ
t
E
(
t
)
+
∫
0
t
‍
e
γ
τ
∥
u
τ
∥
2
2
d
τ
=
E
(
0
)
+
γ
∫
0
t
‍
e
γ
τ
E
(
τ
)
d
τ
.
It follows from u
(
t
)
∈
W and
(84)
E
(
t
)
=
1
2
∥
u
t
∥
2
2
+
p
-
2
2
p
∥
u
∥
p
p
+
1
2
R
(
u
)
;
then
(85)
E
(
t
)
≤
1
2
∥
u
τ
∥
2
2
+
1
2
R
(
u
)
+
p
-
2
2
p
∥
u
∥
H
2
.
Moreover, taking v
=
u in (42), we obtain
(86)
d
d
t
(
u
,
u
t
)
-
∥
u
t
∥
2
2
+
∥
u
∥
H
2
+
1
2
d
d
t
∥
u
∥
2
2
=
∥
u
∥
p
p
,
which implies
(87)
R
(
u
)
=
∥
u
t
∥
2
2
-
d
d
t
(
u
,
u
t
)
-
1
2
d
d
t
∥
u
∥
2
2
.
From (83), (85), and (87), we get
(88)
e
γ
τ
E
(
τ
)
d
τ
+
∫
0
t
‍
e
γ
τ
∥
u
τ
∥
2
2
d
τ
≤
E
(
0
)
+
γ
∫
0
t
‍
e
γ
τ
(
1
2
∥
u
τ
∥
2
2
d
τ
+
1
2
R
(
u
)
+
p
-
2
2
p
∥
u
∥
H
2
)
d
τ
≤
E
(
0
)
+
γ
∫
0
t
‍
e
γ
τ
×
(
1
2
∥
u
τ
∥
2
2
d
τ
+
p
-
2
2
p
∥
u
∥
H
2
)
d
τ
-
γ
4
∫
0
t
‍
e
γ
τ
d
d
τ
(
(
u
,
u
τ
)
+
∥
u
∥
2
2
)
d
τ
,
(89)
∫
0
t
‍
e
γ
τ
d
d
τ
(
(
u
,
u
t
)
+
∥
u
∥
2
2
)
d
τ
=
2
(
u
0
,
u
1
)
+
∥
u
0
∥
2
2
-
e
γ
t
(
2
(
u
,
u
t
)
+
∥
u
∥
2
2
)
+
γ
∫
0
t
‍
e
γ
τ
(
2
(
u
,
u
τ
)
+
∥
u
∥
2
2
)
d
τ
≤
2
∥
u
0
∥
2
2
+
∥
u
1
∥
2
2
+
e
γ
t
(
2
∥
u
∥
2
2
+
∥
u
t
∥
2
2
)
+
γ
∫
0
t
‍
e
γ
τ
(
2
∥
u
∥
2
2
+
∥
u
τ
∥
2
2
)
d
τ
.
From (88) and (89), it follows that
(90)
e
γ
τ
E
(
τ
)
d
τ
+
∫
0
t
‍
e
γ
τ
∥
u
τ
∥
2
2
d
τ
≤
E
(
0
)
+
γ
∫
0
t
‍
e
γ
τ
×
(
1
2
∥
u
τ
∥
2
2
d
τ
+
p
-
2
2
p
∥
u
∥
H
2
)
d
τ
+
γ
4
(
2
∥
u
0
∥
2
2
+
∥
u
1
∥
2
2
)
+
γ
4
e
γ
t
(
2
∥
u
∥
2
2
+
∥
u
t
∥
2
2
)
+
γ
2
4
∫
0
t
‍
e
γ
τ
(
2
∥
u
∥
2
2
+
∥
u
τ
∥
2
2
)
d
τ
≤
E
(
0
)
+
C
0
+
γ
4
e
γ
t
(
2
∥
u
∥
2
2
+
∥
u
t
∥
2
2
)
+
γ
2
4
∫
0
t
‍
e
γ
τ
{
(
2
+
2
(
p
-
2
)
γ
p
)
∥
u
∥
2
2
+
(
1
+
γ
4
)
∥
u
τ
∥
2
2
}
d
τ
,
where C
0
=
(
γ
/
4
)
(
2
∥
u
0
∥
2
2
+
∥
u
1
∥
2
2
); then
(91)
e
γ
t
E
(
t
)
≤
E
(
0
)
+
C
0
+
γ
4
e
γ
t
(
2
∥
u
∥
2
2
+
∥
u
t
∥
2
2
)
+
γ
2
4
∫
0
t
‍
e
γ
τ
{
(
2
+
2
(
p
-
2
)
γ
p
)
∥
u
∥
2
2
+
(
1
+
4
(
γ
-
1
)
γ
2
)
∥
u
τ
∥
2
2
}
d
τ
.
Furthermore, from
(92)
E
(
t
)
=
1
2
∥
u
t
∥
2
2
+
p
-
2
2
p
∥
u
∥
H
2
+
1
p
R
(
u
)
>
1
2
∥
u
t
∥
2
2
+
p
-
2
2
p
∥
u
∥
H
2
>
p
-
2
2
p
(
∥
u
t
∥
2
2
+
∥
u
∥
H
2
)
,
we obtain
(93)
∥
u
t
∥
2
2
+
∥
u
∥
H
2
<
2
p
p
-
2
E
(
t
)
.
Let δ
1
=
max
⁡
{
2
+
(
2
(
p
-
2
)
)
/
γ
p
,
1
+
(
4
(
γ
-
1
)
)
/
γ
2
}; then
(94)
e
γ
t
E
(
t
)
≤
E
(
0
)
+
C
0
+
2
γ
4
e
γ
t
(
2
∥
u
∥
H
2
+
∥
u
t
∥
2
2
)
+
γ
2
δ
1
4
∫
0
t
‍
e
γ
τ
(
2
∥
u
∥
H
2
+
∥
u
t
∥
2
2
)
d
τ
≤
E
(
0
)
+
C
0
+
2
γ
p
2
(
p
-
2
)
e
γ
t
E
(
t
)
+
γ
2
δ
1
p
2
(
p
-
2
)
∫
0
t
‍
e
γ
τ
E
(
τ
)
d
τ
.
From (93), we obtain(95)
e
γ
t
E
(
t
)
≤
C
1
+
C
2
∫
0
t
‍
e
γ
τ
E
(
τ
)
d
τ
,
where C
1
=
(
E
(
0
)
+
C
0
)
/
(
1
-
2
γ
p
/
2
(
p
-
2
)
), C
2
=
γ
2
δ
1
p
/
2
(
p
-
2
)
(
1
-
2
γ
p
/
2
(
p
-
2
)
). Choosing k sufficiently small and together with Gronwall inequality, we have
(96)
E
(
t
)
≤
C
1
e
-
k
t
,
where k
=
γ
(
1
-
γ
δ
1
p
/
2
(
p
-
2
)
(
1
-
2
γ
p
/
2
(
p
-
2
)
)
)
>
0.



---

*Source: 142892-2014-10-14.xml*