# On Blow-Up Structures for a Generalized Periodic Nonlinearly Dispersive Wave Equation

Authors: Ying Wang; Yunxi Guo; Fang Li
Journal: Mathematical Problems in Engineering
 (2014)

---

## Abstract

The local well-posedness for a generalized periodic nonlinearly dispersive wave equation is established. Under suitable assumptions on initial valueu
0, a precise blow-up scenario and several sufficient conditions about blow-up results to the equation are presented.


---

## Body

## 1. Introduction

Recently, Hu and Yin [1] and Yin [2] investigate the following equation:
(1)
u
t
-
u
t
x
x
+
2
ω
u
x
+
3
u
u
x
=
γ
(
2
u
x
u
x
x
+
u
u
x
x
x
)
,
where ω is nonnegative number and γ is arbitrary real number. It is shown from [1, 2] that (1) has solitary wave solutions and blow-up solutions for nonperiodic case and also solutions which blow up in finite time for periodic case.Ifγ
=
0, (1) becomes the famous BBM equation modelling the motion of internal gravity waves in shallow channel [3]. Some results related to the equation can be found in [4, 5]. It is worthwhile to mention that the equation does not have integrability and its solitary waves are not solitons [5].Ifγ
=
1 in (1), attention is attracted to the well-known Camassa-Holm equation, which models the unidirectional propagation of shallow water waves over a flat bottom. Here, u
(
t
,
x
) represents the free surface above a flat bottom and ω is a nonnegative parameter related to the critical shallow water speed [6]. As a model to describe the shallow water motion, the Camassa-Holm equation possesses a bi-Hamiltonian structure and infinite conservation laws [7–9] and is completely integrable [10]. It is regarded as a reexpression of geodesic flow on the diffeomorphism group of circle if ω
=
0 [11] and on the Bott-Virasoro group if ω
>
0 [12]. Recently, some significant results of dynamical behaviors have been obtained for the Cauchy problem of the Camassa-Holm equation. For example, the local well-posedness of corresponding solution for initial data u
0
∈
H
s
(
ℝ
) with s
>
3
/
2 was given by several authors (see [13–15]). Under certain assumptions on initial data u
0, the equation has global strong solutions and blow-up solutions for periodic and nonperiodic case (see [13, 16–22]). The existence and uniqueness of global weak solutions in H
1
(
ℝ
) for the equation were proved (see [23–25]). It is shown from [6] that the solitary waves of the equation are peakon solitons and are orbitally stable.Ifω
=
0 and γ
∈
ℝ, (1) changes into the rod equation derived by Dai [26] recently, which describes finite-length and small amplitude radial deformation waves in thin cylindrical compressible hyperelastic rods (see [26]), and u
(
t
,
x
) represents the radial stretch relative to a prestressed state in one-dimensional variable. The first investigation of the Cauchy problem of the rod equation on the line was done by Constantin and Strauss [27], the precise blow-up scenario, some blow-up results of strong solution, and the stability of a class of solitary waves to the rod equation are presented. In [28, 29], Zhou found the sufficient conditions to guarantee the finite blow-up of corresponding solution for periodic case. Moreover, Yin [30] discusses the rod equation on the circle and gives some interesting blow-up results.In this paper, we consider a generalized nonlinearly dispersive wave equation on the circle(2)
u
t
-
u
t
x
x
+
2
ω
u
x
+
a
u
u
x
+
β
(
u
-
u
x
x
)
=
γ
(
2
u
x
u
x
x
+
u
u
x
x
x
)
,
t
>
0
,
x
∈
ℝ
,
u
(
0
,
x
)
=
u
0
(
x
)
,
x
∈
ℝ
,
u
(
t
,
x
+
1
)
=
u
(
t
,
x
)
,
t
>
0
,
x
∈
ℝ
,
where ω and a are nonnegative fixed constants and β and γ are fixed arbitrary constants. Obviously, (2) reduces to (1) if we define a
=
3 and β
=
0. Actually, Wu and Yin [31] consider a nonlinearly dissipative Camassa-Holm equation which includes a nonlinearly dissipative term L
(
u
), where L is a differential operator. Thus, we can regard the term β
(
u
-
u
x
x
) as a dissipative term.Because of the termβ
(
u
-
u
x
x
), (2) does not admit conservation laws in previous works [1, 2]:
(3)
E
(
u
)
=
∫
𝕊
‍
(
u
2
+
u
x
2
)
d
x
.
Several estimates are established to prove several blow-up solutions. More precisely, we establish the local well-posedness of strong solutions for (2) subject to initial value u
0
∈
H
r
(
𝕊
)
,
r
>
3
/
2, with 𝕊
=
ℝ
/
ℤ (the circle of unit length) and give a precise blow-up scenario. Under suitable assumptions on the initial value u
0, relying on the classical mathematical techniques, the several sufficient conditions about blow-up solutions are found.


## 2. Local Well-Posedness

In this section, we establish the local well-posedness and blow-up scenario for the Cauchy problem (2) in H
r
(
𝕊
), r
>
3
/
2.We denote by* the convolution. Note that if G
(
x
)
:
=
cosh
⁡
(
x
-
[
x
]
-
1
/
2
)
/
2
sinh
⁡
(
1
/
2
), where [x] stands for the integer part of x
∈
R, then (
1
-
∂
x
2
)
-
1
f
=
G
*
f for all f
∈
L
2
(
ℝ
) and G
*
(
u
-
u
x
x
)
=
u. Using this identity, (2) becomes
(4)
u
t
+
γ
u
u
x
+
β
u
+
∂
x
G
*
[
a
-
γ
2
u
2
+
γ
2
(
u
x
)
2
+
2
ω
u
]
=
0
,
u
t
+
γ
u
u
x
+
β
u
+
∂
x
G
u
t
+
γ
u
u
x
+
β
u
+
∂
x
l
t
>
0
,
x
∈
ℝ
,
u
(
0
,
x
)
=
u
0
(
x
)
,
u
(
t
,
x
+
1
)
=
u
(
t
,
x
)
,
which is equivalent to
(5)
y
t
+
γ
u
y
x
+
2
γ
y
u
x
+
2
ω
u
x
+
β
y
+
(
a
-
3
γ
)
u
u
x
=
0
,
t
>
0
,
x
∈
ℝ
,
y
(
0
,
x
)
=
u
0
(
x
)
-
u
0
,
x
x
(
x
)
,
x
∈
ℝ
,
y
(
t
,
x
)
=
y
(
t
,
x
+
1
)
,
t
>
0
,
x
∈
ℝ
.Theorem 1.
Givenu
0
∈
H
r
(
𝕊
)
(
r
>
3
/
2
), there exist a maximal T
=
T
(
a
,
b
,
γ
,
ω
,
u
0
) and a unique solution u to problem (2), such that
(6)
u
=
u
(
·
,
u
0
)
∈
C
(
[
0
,
T
)
;
H
r
(
𝕊
)
)
⋂
‍
C
1
(
[
0
,
T
)
;
H
r
-
1
(
𝕊
)
)
.Proof.
The proof of Theorem1 can be finished by using Kato's semigroup theory (see [1] or [2]). Here, we omit the detailed proof.


## 3. Blow-Up Solutions

Theorem 2.
Letu
0
∈
H
r
(
𝕊
)
,
r
>
3
/
2; the solution of u
(
·
,
u
0
) of problem (2) is uniformly bounded. Blow-up in finite time T
<
+
∞ occurs if and only if
(7)
liminf
⁡
t
↑
T
{
inf
⁡
x
∈
S
[
γ
u
x
(
t
,
x
)
]
}
=
-
∞
.Before proving the theorem, we give several useful lemmas.Lemma 3 (Kato and Ponce [32]).
Ifr
>
0, then H
r
∩
L
∞ is an algebra. Moreover
(8)
∥
u
v
∥
H
r
≤
c
(
∥
u
∥
L
∞
∥
v
∥
H
r
+
∥
u
∥
H
r
∥
v
∥
L
∞
)
,
where c is a constant depending only on r.Lemma 4 (Kato and Ponce [32]).
Letr
>
0. If u
∈
H
r
∩
W
1
,
∞ and v
∈
H
r
-
1
∩
L
∞, then
(9)
∥
[
Λ
r
,
u
]
v
∥
L
2
≤
c
(
∥
∂
x
u
∥
L
∞
∥
Λ
r
-
1
v
∥
L
2
+
∥
Λ
r
u
∥
L
2
∥
v
∥
L
∞
)
.Lemma 5.
Lets
≥
3
/
2 and u
(
t
,
x
) is the corresponding solution of (4) with initial data u
0
(
x
)
∈
H
r
(
𝕊
); it holds that if q
∈
(
0
,
r
-
1
], there is a constant c depending only on q such that
(10)
∫
𝕊
‍
(
Λ
q
+
1
u
)
2
d
x
≤
∫
𝕊
‍
(
Λ
q
+
1
u
0
)
2
d
x
+
c
∫
0
t
‍
(
∥
u
x
∥
L
∞
+
1
)
∥
u
∥
H
q
+
1
2
d
s
.Proof.
We rewrite (2) in the following equivalent form:
(11)
u
t
-
u
t
x
x
=
-
a
u
u
x
-
2
ω
u
x
+
γ
2
∂
x
3
u
2
-
γ
2
∂
x
(
u
x
2
)
-
β
u
+
β
∂
x
2
u
.
For q
∈
(
0
,
s
-
1
], applying (
Λ
q
u
)
Λ
q on both sides of (11) and integrating the new equation with respect to x by parts, we obtain
(12)
1
2
d
d
t
∫
𝕊
‍
(
(
Λ
q
u
)
2
+
(
Λ
q
u
x
)
2
)
d
x
=
-
a
∫
𝕊
‍
(
Λ
q
u
)
Λ
q
(
u
u
x
)
-
γ
∫
𝕊
‍
(
Λ
q
+
1
u
)
Λ
q
+
1
(
u
u
x
)
+
γ
∫
𝕊
‍
(
Λ
q
u
)
Λ
q
(
u
u
x
)
+
γ
2
∫
𝕊
‍
(
Λ
q
u
x
)
Λ
q
(
u
x
2
)
-
β
∫
𝕊
‍
(
Λ
q
+
1
u
)
2
d
x
,
where the Parseval equality
(13)
∫
𝕊
‍
(
Λ
q
u
)
Λ
q
∂
x
3
(
u
2
)
d
x
=
-
2
∫
S
‍
(
Λ
q
+
1
u
)
Λ
q
+
1
(
u
u
x
)
d
x
∫
𝕊
‍
(
Λ
q
u
)
Λ
q
∂
x
3
(
u
2
)
d
x
+
2
∫
𝕊
‍
(
Λ
q
u
)
Λ
q
(
u
u
x
)
d
x
is used. We will estimate each of the terms on the right-hand side of (11). For the first and the third terms, using integration by parts, the Cauchy-Schwartz inequality, and Lemmas 3 and 4, we have
(14)
∫
𝕊
‍
(
Λ
q
u
)
Λ
q
(
u
u
x
)
d
x
=
∫
𝕊
‍
(
Λ
q
u
)
[
Λ
q
(
u
u
x
)
-
u
Λ
q
u
x
]
d
x
+
∫
𝕊
‍
(
Λ
q
u
)
u
Λ
q
(
u
x
)
d
x
≤
c
∥
u
∥
H
q
2
∥
u
x
∥
L
∞
+
1
2
∥
u
x
∥
L
∞
∥
Λ
q
u
∥
L
2
2
≤
c
∥
u
∥
H
q
2
∥
u
x
∥
L
∞
,
where c only depends on q. Using the above estimate to the second term yields
(15)
∫
𝕊
‍
(
Λ
q
+
1
u
)
Λ
q
+
1
(
u
u
x
)
d
x
≤
c
∥
u
∥
H
q
+
1
2
∥
u
x
∥
L
∞
.
For the fourth term, using Lemma 3 gives rise to
(16)
∫
𝕊
‍
(
Λ
q
u
x
)
Λ
q
(
u
x
2
)
d
x
≤
∥
Λ
q
u
x
∥
L
2
∥
Λ
q
u
x
2
∥
L
2
≤
c
∥
u
∥
H
q
+
1
(
∥
u
x
∥
L
∞
∥
u
x
∥
H
q
)
≤
c
∥
u
∥
H
q
+
1
2
∥
u
x
∥
L
∞
.
It follows from (12)–(16) that
(17)
1
2
d
d
t
∫
𝕊
‍
(
(
Λ
q
u
)
2
+
(
Λ
q
u
x
)
2
)
d
x
≤
c
(
∥
γ
u
x
∥
L
∞
+
1
)
∥
u
∥
H
q
+
1
2
,
which leads to
(18)
∫
𝕊
‍
(
(
Λ
q
u
)
2
+
(
Λ
q
u
x
)
2
)
d
x
-
∫
𝕊
‍
(
(
Λ
q
u
0
)
2
+
(
Λ
q
u
0
x
)
2
)
d
x
≤
c
∫
0
t
‍
(
∥
γ
u
x
∥
L
∞
+
1
)
∥
u
∥
H
q
+
1
2
d
s
.
It completes the proof of the lemma.Proof of Theorem2.
Applying (10) with q
=
r
-
1, we have
(19)
∥
u
∥
H
r
(
𝕊
)
2
≤
∥
u
0
∥
H
r
(
𝕊
)
2
+
c
∫
0
t
‍
(
∥
γ
u
x
∥
L
∞
(
𝕊
)
+
1
)
∥
u
∥
H
r
(
𝕊
)
2
d
s
.
It follows from (19) and Gronwall’s inequality that
(20)
∥
u
∥
H
r
(
𝕊
)
2
≤
∥
u
0
∥
H
r
(
𝕊
)
2
exp
⁡
(
c
∫
0
t
‍
(
∥
γ
u
x
∥
L
∞
(
𝕊
)
+
1
)
d
s
)
.
If there is a constant M
>
0 such that ∥
γ
u
x
∥
L
∞
≤
M on (
0
,
T
], then ∥
u
∥
H
r
(
𝕊
)
2 does not blow up. It completes the proof of Theorem 2.Lemma 6 (see [17]).
Letu
(
t
) be the solution to (2) on [
0
,
T
) with initial data u
0
∈
H
s
(
𝕊
), s
>
3
/
2. Then there exists at least one point ξ
(
t
)
∈
𝕊 with
(21)
m
(
t
)
=
inf
⁡
x
∈
𝕊
[
u
x
(
x
,
t
)
]
=
u
x
(
t
,
ξ
(
t
)
)
.
The function m
(
t
) is almost everywhere differentiable on (
0
,
t
) with
(22)
d
m
(
t
)
d
t
=
u
t
x
(
t
,
(
ξ
(
t
)
)
)
.Lemma 7 (see [21, 22]).
(i) For everyu
∈
H
1
(
𝕊
), one has
(23)
max
⁡
x
∈
[
0,1
]
u
2
(
x
)
≤
e
+
1
2
(
e
-
1
)
∥
u
∥
H
1
(
𝕊
)
2
,
where the constant (
e
+
1
)
/
2
(
e
-
1
) is sharp.
(ii) For everyu
∈
H
3
(
𝕊
), one has
(24)
max
⁡
x
∈
[
0,1
]
u
2
(
x
)
≤
c
∥
u
∥
H
1
(
𝕊
)
2
,
with the best possible constant c lying within the range (
1
,
13
/
12
]. Moreover, the best constant c is (
e
+
1
)
/
2
(
e
-
1
).Lemma 8 (see [2]).
Iff
∈
H
3
(
𝕊
) is such that ∫
𝕊
‍
f
(
x
)
d
x
=
a
0
/
2, then, for every ε
>
0, one has
(25)
max
⁡
x
∈
[
0,1
]
f
2
(
x
)
≤
ε
+
2
24
∫
𝕊
‍
f
x
2
d
x
+
ε
+
2
4
ε
a
0
2
.
Moreover,
(26)
max
⁡
x
∈
[
0,1
]
f
2
(
x
)
≤
ε
+
2
24
∥
f
∥
H
1
2
+
ε
+
2
4
ε
a
0
2
.Lemma 9.
Letu
∈
H
3
(
𝕊
) and T
>
0 be the maximal existence time of the solution u
(
t
,
x
) to problem (4). Then it holds that (i)
∫
𝕊
‍
u
(
t
,
x
)
d
x
=
∫
𝕊
‍
u
0
(
x
)
d
x
=
∫
𝕊
‍
y
(
t
,
x
)
d
x
=
∫
𝕊
‍
y
0
(
x
)
d
x,
(ii)
∥
u
∥
H
1
2
=
∥
u
∥
H
1
2
e
-
2
β
t.Proof.
The proof of (i) is similar to that of [2, lemma 3.6], so we omit it.
Multiplyingu to both sides of (2) and integrating by parts, we get
(27)
1
2
d
d
t
∫
𝕊
‍
(
u
2
+
u
x
2
)
d
x
=
-
β
∫
𝕊
‍
(
u
2
+
u
x
2
)
d
x
,
which yields
(28)
∥
u
∥
H
1
(
𝕊
)
2
=
∥
u
0
∥
H
1
(
𝕊
)
2
e
-
2
β
t
.
It finishes the proof.Lemma 1010 (see [29]).
Assume that a differential functiony
(
t
) satisfies
(29)
y
′
≤
-
C
y
2
(
t
)
+
K
,
with constants C
,
K
>
0. If the initial datum y
(
0
)
=
y
0
<
-
K
/
C, then the solutions to (29) go to -
∞ in finite time.We now present the first blow-up result.Theorem 11.
Letβ
<
0 and u
0
∈
H
s
(
𝕊
), s
≥
3.(i)
If0
<
γ
<
a
/
3 and there is a x
0
∈
𝕊 such that
(30)
γ
u
0
′
(
x
0
)
<
-
β
-
2
(
β
2
2
+
γ
(
a
-
2
γ
)
(
e
+
1
)
4
(
e
-
1
)
∥
u
0
∥
H
1
2
e
2
β
T
γ
u
′
0
(
x
0
)
v
v
+
4
γ
ω
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
e
β
T
)
1
/
2
,then the corresponding solution to (2) blows up in finite time.(ii)
Ifa
/
3
<
γ
<
a and there is a x
0
∈
𝕊 such that
(31)
γ
u
0
′
(
x
0
)
<
-
β
-
2
(
β
2
2
+
γ
(
a
-
γ
)
(
e
+
1
)
8
(
e
-
1
)
∥
u
0
∥
H
1
2
e
2
β
T
γ
u
′
0
(
x
0
)
v
v
+
4
γ
ω
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
e
β
T
)
1
/
2
,then the corresponding solution to (2) blows up in finite time.(iii)
Ifγ
<
0 or γ
>
a and there is a x
0
∈
𝕊 such that
(32)
γ
u
0
′
(
x
0
)
<
-
β
-
2
(
β
2
2
+
γ
(
γ
-
a
)
(
e
+
1
)
4
(
e
-
1
)
∥
u
0
∥
H
1
2
e
2
β
T
γ
u
′
0
(
x
0
)
v
v
+
4
|
γ
|
ω
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
e
β
T
)
1
/
2
,then the corresponding solution to (2) blows up in finite time.Proof.
LetT
>
0 be the maximal time of existence of the solution u to (2) with the initial data u
0. Note that ∂
x
2
G
*
f
=
G
*
f
-
f. Differentiating (4) with respect to x and multiplying the obtained equation by γ, we get
(33)
γ
u
t
x
=
-
γ
2
2
u
x
2
-
γ
2
u
u
x
x
+
γ
(
a
-
γ
)
2
u
2
+
2
γ
ω
u
-
γ
β
u
x
-
G
*
[
(
a
-
γ
)
γ
2
u
2
+
γ
2
2
u
x
2
+
2
γ
ω
u
]
.
Noting that u
x
x
(
t
,
ξ
(
t
)
)
=
0, for all t
∈
[
0
,
T
), and defining m
(
t
)
=
γ
u
x
(
t
,
ξ
(
t
)
)
+
β
=
inf
⁡
x
∈
𝕊
(
γ
u
x
(
t
,
x
)
+
β
), we obtain
(34)
m
′
(
t
)
=
-
1
2
m
2
(
t
)
+
β
2
2
+
γ
(
a
-
γ
)
2
u
2
+
2
γ
ω
u
-
G
*
[
(
a
-
γ
)
γ
2
u
2
+
γ
2
2
u
x
2
+
2
γ
ω
u
]
.
Next, we divide (34) into three cases to prove the theorem.
(i)The first is0
<
γ
<
a
/
3; note that  (
a
-
3
γ
)
γ
/
2
≥
0 and G
*
(
u
2
+
(
1
/
2
)
u
x
2
)
≥
(
1
/
2
)
u
2. Then, we have
(35)
G
*
[
(
a
-
γ
)
γ
2
u
2
+
γ
2
2
u
x
2
]
=
G
*
[
γ
2
u
2
+
γ
2
2
u
x
2
+
(
(
a
-
γ
)
γ
2
-
γ
2
)
u
2
]
≥
γ
2
2
u
2
+
G
*
(
(
a
-
3
γ
)
γ
2
u
2
)
.
Note that G
*
(
(
(
a
-
3
γ
)
γ
/
2
)
u
2
)
≥
0. Thus, we obtain
(36)
m
′
(
t
)
≤
-
1
2
m
2
(
t
)
+
β
2
2
+
γ
(
a
-
2
γ
)
2
u
2
+
2
γ
ω
u
-
G
*
(
2
γ
ω
u
)
.
From Lemmas 8 and 9, we have
(37)
max
⁡
x
∈
𝕊
{
u
2
}
≤
e
+
1
2
(
e
-
1
)
∥
u
∥
H
1
2
≤
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
2
e
-
2
β
T
.
Thus,
(38)
∥
u
∥
L
∞
≤
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
e
-
β
T
.
Note that ∥
G
∥
L
1
(
𝕊
)
=
1. Using Young’s inequality, we get
(39)
|
G
*
u
|
≤
∥
u
∥
L
∞
∥
G
∥
L
1
(
𝕊
)
≤
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
e
-
β
T
.
Therefore, we deduce(40)
m
′
(
t
)
≤
-
1
2
m
2
(
t
)
+
β
2
2
+
γ
(
a
-
2
γ
)
2
e
+
1
2
(
e
-
1
)
aaaaaaa
×
∥
u
0
∥
H
1
2
e
-
2
β
T
+
4
γ
ω
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
e
-
β
T
,
which results in
(41)
m
′
(
t
)
≤
-
1
2
m
2
(
t
)
+
K
,
with
(42)
K
=
β
2
2
+
γ
(
a
-
2
γ
)
2
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
2
e
-
2
β
T
+
4
γ
ω
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
e
-
β
T
.
Note that, from Lemma 10, if m
(
0
)
<
-
2
K, then there exists T, such that lim
⁡
t
↑
T
m
(
t
)
=
-
∞. Applying Theorem 2, the solution u to (2) does not exist globally in time.
(ii) The second isa
/
3
≤
γ
≤
a. Similar to case (i), we have γ
(
3
γ
-
a
)
/
4
≥
0 and
(43)
G
*
[
(
a
-
γ
)
γ
2
u
2
+
γ
2
2
u
x
2
]
≥
(
a
-
γ
)
γ
4
u
2
[
(
a
-
γ
)
γ
2
u
2
+
γ
2
2
u
x
2
]
+
G
*
(
γ
(
3
γ
-
a
)
4
u
x
2
)
.
Therefore, we obtain(44)
m
′
(
t
)
≤
-
1
2
m
2
(
t
)
+
β
2
2
+
γ
(
a
-
γ
)
4
u
2
+
2
γ
ω
u
-
G
*
(
2
γ
ω
u
)
.
Using (37)–(39), we get
(45)
m
′
(
t
)
≤
-
1
2
m
2
(
t
)
+
β
2
2
+
γ
(
a
-
γ
)
4
e
+
1
2
(
e
-
1
)
mmll
×
∥
u
0
∥
H
1
2
e
-
2
β
T
+
4
γ
ω
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
e
-
β
T
,
which results in
(46)
m
′
(
t
)
≤
-
1
2
m
2
(
t
)
+
K
,
with
(47)
K
=
β
2
2
+
γ
(
a
-
γ
)
4
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
2
e
-
2
β
T
+
4
γ
ω
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
e
-
β
T
.
Note that, from Lemma 10, if m
(
0
)
<
-
2
K, then there exists T, such that lim
⁡
t
↑
T
m
(
t
)
=
-
∞. Repeating the proof of (i), we conclude that the solution u blows up in finite time.
(iii) The third isγ
<
0 or γ
>
a; note that (
a
-
γ
)
γ
/
2
<
0. From Young’s inequality, we get for all t
∈
[
0
,
T
)
(48)
|
G
*
u
2
|
≤
∥
u
∥
L
∞
2
∥
G
∥
L
1
≤
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
2
e
-
2
β
T
.
Using (37)–(39), we get
(49)
m
′
(
t
)
≤
-
1
2
m
2
(
t
)
+
β
2
2
+
γ
(
γ
-
a
)
2
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
2
e
-
2
β
T
+
4
|
γ
|
ω
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
e
-
β
T
,
which results in
(50)
m
′
(
t
)
≤
-
1
2
m
2
(
t
)
+
K
,
with
(51)
K
=
β
2
2
+
γ
(
γ
-
a
)
2
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
2
e
-
2
β
T
+
4
|
γ
|
ω
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
e
-
β
T
.
Note that, from Lemma 10, if m
(
0
)
<
-
2
K, then there exists T, such that lim
⁡
t
↑
T
m
(
t
)
=
-
∞. Repeating the proof of (i), we deduce that the solution u blows up in finite time. It finishes the proof of the theorem.Next, we give the second blow-up result.Theorem 12.
Letβ
<
0 and u
0
∈
H
s
(
𝕊
), s
≥
3, and ∫
𝕊
‍
u
0
d
x
=
a
0
/
2.(i)
If0
<
γ
<
a
/
3 and for all ϵ
>
0 there is a x
0
∈
𝕊 such that
(52)
γ
u
0
′
(
x
0
)
<
-
β
-
2
(
β
2
2
+
γ
(
a
-
2
γ
)
2
f
f
f
f
f
f
f
f
f
f
f
f
×
(
ϵ
+
2
24
∥
u
0
∥
H
1
2
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
)
+
4
γ
ω
f
f
f
f
f
f
f
f
f
f
f
f
×
ϵ
+
2
24
∥
u
0
∥
H
1
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
)
1
/
2
,then the corresponding solution to (2) blows up in finite time.(ii)
Ifa
/
3
<
γ
<
a and for all ϵ
>
0 there is a x
0
∈
𝕊 such that
(53)
γ
u
0
′
(
x
0
)
<
-
β
-
2
(
β
2
2
+
γ
(
a
-
γ
)
4
(
ϵ
+
2
24
∥
u
0
∥
H
1
2
e
-
2
β
T
mmmmmmmm
+
ϵ
+
2
4
ϵ
a
0
2
)
+
4
γ
ω
×
ϵ
+
2
24
∥
u
0
∥
H
1
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
)
1
/
2
,then the corresponding solution to (2) blows up in finite time. (iii)
Ifγ
<
0 or γ
>
a and for all ϵ
>
0 there is a x
0
∈
𝕊 such that
(54)
γ
u
0
′
(
x
0
)
<
-
β
-
2
(
β
2
2
+
γ
(
γ
-
a
)
2
(
ϵ
+
2
24
∥
u
0
∥
H
1
2
e
-
2
β
T
mmmmmmmm
+
ϵ
+
2
4
ϵ
a
0
2
)
+
4
|
γ
|
ω
×
ϵ
+
2
24
∥
u
0
∥
H
1
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
)
1
/
2
,then the corresponding solution to (2) blows up in finite time.Proof.
Similar to Theorem11, we divide (34) into three cases to prove the theorem.
(i) The first is0
<
γ
<
a
/
3; note that (
a
-
3
γ
)
γ
/
2
≥
0 and G
*
(
u
2
+
(
1
/
2
)
u
x
2
)
≥
(
1
/
2
)
u
2. Then, we have G
*
(
(
a
-
3
γ
)
γ
/
2
)
u
2
≥
0.
From Lemmas8 and 9, we have
(55)
max
⁡
x
∈
S
{
u
2
}
≤
ϵ
+
2
24
∥
u
∥
H
1
2
+
ϵ
+
2
4
ϵ
a
0
2
≤
ϵ
+
2
24
∥
u
0
∥
H
1
2
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
.
Thus,
(56)
∥
u
∥
L
∞
≤
ϵ
+
2
24
∥
u
0
∥
H
1
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
.
Note that ∥
G
∥
L
1
(
𝕊
)
=
1. Using Young’s inequality, we get
(57)
|
G
*
u
|
≤
∥
u
∥
L
∞
∥
G
∥
L
1
(
𝕊
)
mml
≤
ϵ
+
2
24
∥
u
0
∥
H
1
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
.
Therefore, we deduce(58)
m
′
(
t
)
≤
-
1
2
m
2
(
t
)
+
β
2
2
+
γ
(
a
-
2
γ
)
2
vvvv
×
(
ϵ
+
2
24
∥
u
0
∥
H
1
2
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
)
vvvv
+
4
γ
ω
ϵ
+
2
24
∥
u
0
∥
H
1
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
,
which results in
(59)
m
′
(
t
)
≤
-
1
2
m
2
(
t
)
+
K
with
(60)
K
=
β
2
2
+
γ
(
a
-
2
γ
)
2
(
ϵ
+
2
24
∥
u
0
∥
H
1
2
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
)
+
4
γ
ω
ϵ
+
2
24
∥
u
0
∥
H
1
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
.
Note that, from the lemma, if m
(
0
)
<
-
2
K, then there exists T, such that lim
⁡
t
↑
T
m
(
t
)
=
-
∞. Applying Theorem 2, the solution u blows up in finite time.
(ii) The second isa
/
3
≤
γ
≤
a. Similar to case (i), we have γ
(
3
γ
-
a
)
/
4
≥
0 and
(61)
G
*
[
(
a
-
γ
)
γ
2
u
2
+
γ
2
2
u
x
2
]
≥
(
a
-
γ
)
γ
4
u
2
+
G
*
(
γ
(
3
γ
-
a
)
4
u
x
2
)
.
Using (55)–(57), we get
(62)
m
′
(
t
)
≤
-
1
2
m
2
(
t
)
+
β
2
2
+
γ
(
a
-
γ
)
4
vvvv
×
(
ϵ
+
2
24
∥
u
0
∥
H
1
2
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
)
vvvv
+
4
γ
ω
ϵ
+
2
24
∥
u
0
∥
H
1
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
,
which results in
(63)
m
′
(
t
)
≤
-
1
2
m
2
(
t
)
+
K
with
(64)
K
=
β
2
2
+
γ
(
a
-
γ
)
4
ϵ
+
2
24
∥
u
0
∥
H
1
2
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
+
4
γ
ω
ϵ
+
2
24
∥
u
0
∥
H
1
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
.
Following the proof of (i) in Theorem 11, we derive that the solution u blows up in finite time.
(iii) The third isγ
<
0 or γ
>
a; note that (
a
-
γ
)
γ
/
2
<
0. From Young’s inequality, we get for all t
∈
[
0
,
T
)
(65)
|
G
*
u
2
|
≤
∥
u
∥
L
∞
2
∥
G
∥
L
1
≤
ϵ
+
2
24
∥
u
0
∥
H
1
2
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
.
Using (55)–(57) and (65), we get
(66)
m
′
(
t
)
≤
-
1
2
m
2
(
t
)
+
β
2
2
+
γ
(
γ
-
a
)
2
vvvv
×
(
ϵ
+
2
24
∥
u
0
∥
H
1
2
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
)
vvvv
+
4
|
γ
|
ω
ϵ
+
2
24
∥
u
0
∥
H
1
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
,
which results in
(67)
m
′
(
t
)
≤
-
1
2
m
2
(
t
)
+
K
with
(68)
K
=
β
2
2
+
γ
(
γ
-
a
)
2
ϵ
+
2
24
∥
u
0
∥
H
1
2
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
+
4
|
γ
|
ω
ϵ
+
2
24
∥
u
0
∥
H
1
e
-
2
β
T
+
ϵ
+
2
4
ϵ
a
0
2
.
Following the proof of (i) in Theorem 11, we obtain that the solution u blows up in finite time. It finishes the proof of the theorem.Next, we give the third blow-up result.Theorem 13.
Assume thatβ
<
0 and u
0
∈
H
s
(
𝕊
), s
≥
3, and ∥
u
0
∥
H
1
≠
0.(i)
If0
<
γ
<
a
/
3 is such that
(69)
∫
𝕊
‍
γ
u
0
x
3
d
x
<
-
2
β
∥
u
0
∥
H
1
2
e
-
2
β
T
-
2
3
(
β
2
2
∥
u
0
∥
H
1
4
e
-
4
β
T
+
3
γ
(
a
-
2
γ
)
(
e
+
1
)
4
(
e
-
1
)
mmmm
×
∥
u
0
∥
H
1
6
e
-
6
β
T
+
12
ω
γ
m
m
m
m
×
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
5
e
-
5
β
T
)
1
/
2
,then the corresponding solution to (2) blows up in finite time.(ii)
Ifa
/
3
<
γ
<
a is such that
(70)
∫
𝕊
‍
γ
u
0
x
3
d
x
<
-
2
β
∥
u
0
∥
H
1
2
e
-
2
β
T
-
2
3
(
β
2
2
∥
u
0
∥
H
1
4
e
-
4
β
T
+
3
γ
(
a
-
γ
)
(
e
+
1
)
8
(
e
-
1
)
m
m
m
n
×
∥
u
0
∥
H
1
6
e
-
6
β
T
+
12
ω
γ
m
m
m
l
n
×
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
5
e
-
5
β
T
)
1
/
2
,then the corresponding solution to (2) blows up in finite time.(iii)
Ifγ
>
a or γ
<
0 is such that
(71)
∫
𝕊
‍
γ
u
0
x
3
d
x
<
-
2
β
∥
u
0
∥
H
1
2
e
-
2
β
T
-
2
3
(
β
2
2
∥
u
0
∥
H
1
4
e
-
4
β
T
+
3
γ
(
γ
-
a
)
(
e
+
1
)
8
(
e
-
1
)
mmmm
×
∥
u
0
∥
H
1
6
e
-
6
β
T
+
12
ω
|
γ
|
m
m
m
m
×
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
5
e
-
5
β
T
)
1
/
2
,then the corresponding solution to (2) blows up in finite time.Proof.
LetT
>
0 be the maximal time of existence of the solution u to (2) with the initial data u
0. Applying γ
u
x
2
∂
x to both sides of (2) and integrating by parts, we get
(72)
d
d
t
∫
𝕊
‍
γ
u
x
3
d
x
=
-
3
γ
2
2
∫
𝕊
‍
u
x
4
-
3
β
∫
𝕊
‍
γ
u
x
3
d
x
+
3
(
a
-
γ
)
γ
2
vvvvvvv
c
×
∫
𝕊
‍
u
x
2
u
2
d
x
+
6
ω
γ
∫
𝕊
‍
u
x
2
u
d
x
+
3
vvvvvvv
c
×
∫
𝕊
‍
u
x
2
G
*
(
(
a
-
γ
)
γ
2
u
2
+
γ
2
2
u
x
2
+
2
ω
γ
u
)
.
Since
(73)
|
∫
𝕊
‍
u
x
3
d
x
|
≤
(
∫
𝕊
‍
u
x
4
d
x
)
1
/
2
(
∫
𝕊
‍
u
x
2
d
x
)
1
/
2
,
thus
(74)
∫
𝕊
‍
u
x
4
d
x
≥
(
∫
𝕊
‍
u
x
3
d
x
)
2
∫
𝕊
‍
u
x
2
d
x
≥
(
∫
𝕊
‍
u
x
3
d
x
)
2
∥
u
∥
H
1
2
.
Therefore, we obtain(75)
d
d
t
∫
𝕊
‍
γ
u
x
3
d
x
≤
-
3
2
∥
u
0
∥
H
1
2
e
-
2
β
T
(
∫
𝕊
‍
γ
u
x
3
d
x
)
2
-
3
β
∫
𝕊
‍
γ
u
x
3
d
x
wwwwwv
+
3
(
a
-
γ
)
γ
2
∫
𝕊
‍
u
x
2
u
2
d
x
+
6
ω
γ
∫
𝕊
‍
u
x
2
u
d
x
wwwwwv
+
3
∫
𝕊
‍
u
x
2
G
*
(
(
a
-
γ
)
γ
2
u
2
mmmmmmmmmmmmmm
m
+
γ
2
2
u
x
2
+
2
ω
γ
u
)
d
x
.
Next, we divide (75) into three cases to prove the theorem.
(i) The first is0
<
γ
≤
a
/
3; from case (i) of Theorem 11, we know that γ
(
a
-
3
γ
)
≥
0 and G
*
(
(
(
a
-
3
γ
)
γ
/
2
)
u
2
)
≥
0.
From Holder's inequality and Yong's inequality, we have(76)
|
∫
𝕊
‍
u
x
2
u
2
d
x
|
≤
∥
u
∥
L
∞
2
∫
𝕊
‍
u
x
2
d
x
≤
e
+
1
2
(
e
-
1
)
∥
u
∥
H
1
4
≤
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
4
e
-
4
β
T
,
|
∫
𝕊
‍
u
x
2
u
d
x
|
≤
∥
u
∥
L
∞
∫
𝕊
‍
u
x
2
d
x
≤
e
+
1
2
(
e
-
1
)
∥
u
∥
H
1
3
≤
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
3
e
-
3
β
T
,
|
∫
𝕊
‍
u
x
2
G
*
u
d
x
|
≤
∥
G
*
u
∥
L
∞
∫
𝕊
‍
u
x
2
d
x
≤
e
+
1
2
(
e
-
1
)
∥
u
∥
H
1
3
≤
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
3
e
-
3
β
T
.
Using (76) and (75), it yields
(77)
d
d
t
∫
𝕊
‍
γ
u
x
3
d
x
≤
-
3
2
∥
u
0
∥
H
1
2
e
-
2
β
T
vvvvvvll
×
(
∫
𝕊
‍
γ
u
x
3
d
x
+
β
∥
u
0
∥
H
1
2
e
-
2
β
T
)
2
+
3
β
2
2
vvvvvvll
×
∥
u
0
∥
H
1
2
e
-
2
β
T
+
3
(
a
-
2
γ
)
γ
2
e
+
1
2
(
e
-
1
)
vvvvvvll
×
∥
u
0
∥
H
1
4
e
-
4
β
T
+
12
ω
γ
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
3
e
-
3
β
T
.
Setting(78)
m
(
t
)
=
∫
𝕊
‍
γ
u
x
3
d
x
+
β
∥
u
0
∥
H
1
2
e
-
2
β
T
,
(79)
K
=
3
β
2
2
∥
u
0
∥
H
1
2
e
-
2
β
T
+
3
(
a
-
2
γ
)
γ
2
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
4
e
-
4
β
T
+
12
ω
γ
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
3
e
-
3
β
T
,
we have
(80)
d
d
t
m
(
t
)
≤
-
3
2
∥
u
0
∥
H
1
2
e
-
2
β
T
m
2
(
t
)
+
K
.
Note that, from the lemma, ifm
(
0
)
<
-
2
K
/
3
∥
u
0
∥
H
1
e
-
β
T, similar to the proof in case (i) of Theorem 11, we conclude that the corresponding solution will blow up in finite time.
(ii) The second isa
/
3
<
γ
≤
a; from case (ii) of the theorem, we have γ
(
3
γ
-
a
)
≥
0 and G
*
(
(
γ
(
3
γ
-
a
)
/
4
)
u
x
2
)
≥
0.
Using (76) and (75), it yields
(81)
d
d
t
∫
𝕊
‍
γ
u
x
3
d
x
≤
-
3
2
∥
u
0
∥
H
1
2
e
-
2
β
T
mmmmm
×
(
∫
𝕊
‍
γ
u
x
3
d
x
+
β
∥
u
0
∥
H
1
2
e
-
2
β
T
)
2
+
3
β
2
2
mmmmm
×
∥
u
0
∥
H
1
2
e
-
2
β
T
+
3
(
a
-
γ
)
γ
4
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
4
mmmmm
×
e
-
4
β
T
+
12
ω
γ
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
3
e
-
3
β
T
.
Setting(82)
m
(
t
)
=
∫
𝕊
‍
γ
u
x
3
d
x
+
β
∥
u
0
∥
H
1
2
e
-
2
β
T
,
K
=
3
β
2
2
∥
u
0
∥
H
1
2
e
-
2
β
T
+
3
(
a
-
γ
)
γ
4
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
4
e
-
4
β
T
+
12
ω
γ
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
3
e
-
3
β
T
,
we have
(83)
d
d
t
m
(
t
)
≤
-
3
2
∥
u
0
∥
H
1
2
e
-
2
β
T
m
2
(
t
)
+
K
.
Note that, from the lemma, ifm
(
0
)
<
-
2
K
/
3
∥
u
0
∥
H
1
e
-
β
T, similar to the proof in case (i) of Theorem 11, we derive that the corresponding solution will blow up in finite time.
(iii) The third isγ
>
a or γ
<
0; note that (
a
-
γ
)
γ
/
2
<
0.
Using (76) and (75), it yields
(84)
d
d
t
∫
𝕊
‍
γ
u
x
3
d
x
≤
-
3
2
∥
u
0
∥
H
1
2
e
-
2
β
T
d
d
t
∫
𝕊
‍
γ
u
x
3
d
×
(
∫
𝕊
‍
γ
u
x
3
d
x
+
β
∥
u
0
∥
H
1
2
e
-
2
β
T
)
2
+
3
β
2
2
d
d
t
∫
𝕊
‍
γ
u
x
3
d
×
∥
u
0
∥
H
1
2
e
-
2
β
T
+
3
(
γ
-
a
)
γ
4
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
4
d
d
t
∫
𝕊
‍
γ
u
x
3
d
×
e
-
4
β
T
+
12
ω
|
γ
|
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
3
e
-
3
β
T
.
Setting(85)
m
(
t
)
=
∫
𝕊
‍
γ
u
x
3
d
x
+
β
∥
u
0
∥
H
1
2
e
-
2
β
T
,
K
=
3
β
2
2
∥
u
0
∥
H
1
2
e
-
2
β
T
+
3
(
γ
-
a
)
γ
4
e
+
1
2
(
e
-
1
)
×
∥
u
0
∥
H
1
4
e
-
4
β
T
+
12
ω
|
γ
|
×
e
+
1
2
(
e
-
1
)
∥
u
0
∥
H
1
3
e
-
3
β
T
,
we have
(86)
d
d
t
m
(
t
)
≤
-
3
2
∥
u
0
∥
H
1
2
e
-
2
β
T
m
2
(
t
)
+
K
.
Note that, from the lemma, ifm
(
0
)
<
-
2
K
/
3
∥
u
0
∥
H
1
e
-
β
T, similar to the proof in case (i) of Theorem 11, we deduce that the corresponding solution will blow up in finite time. It completes the proof of Theorem 13.Finally, we give the fourth blow-up result.Theorem 14.
Letβ
≥
0 and u
0
∈
H
s, s
≥
3. Assume that u
0
≠
0 and ∫
𝕊
‍
u
0
d
x
=
0. If γ and ω satisfy one of the following conditions:(i)
a
sinh
⁡
(
1
/
2
)
/
(
6
+
sinh
⁡
(
1
/
2
)
)
<
γ
≤
a and ω
<
(
6
3
/
169
)
(
3
γ
/
sinh
⁡
(
1
/
2
)
-
(
a
-
γ
)
/
2
)
∥
u
0
∥
H
1
e
-
4
β
T,
(ii)
γ
>
a and ω
<
(
6
3
/
169
)
(
3
γ
/
sinh
⁡
(
1
/
2
)
+
(
a
-
γ
)
/
2
)
∥
u
0
∥
H
1
e
-
4
β
T,
(iii)
γ
<
a
sinh
⁡
(
1
/
2
)
/
(
sinh
⁡
(
1
/
2
)
-
6
) and ω
<
(
6
3
/
169
|
γ
|
)
(
3
γ
2
/
sinh
⁡
(
1
/
2
)
+
γ
(
a
-
γ
)
/
2
)
∥
u
0
∥
H
1
e
-
4
β
T,then the corresponding solution of (2) blows up in finite time.Proof.
Supposing that the statement is not correct, then the solution of (2) exists globally in time.
Thus, applyingγ
u
x
2
∂
x to both sides of (4) and integrating by parts, we get
(87)
d
d
t
∫
𝕊
‍
γ
u
x
3
d
x
=
-
3
γ
2
2
∫
𝕊
‍
u
x
4
-
3
β
∫
𝕊
‍
γ
u
x
3
d
x
+
3
(
a
-
γ
)
γ
2
d
d
t
∫
𝕊
‍
γ
m
v
×
∫
𝕊
‍
u
x
2
u
2
d
x
+
6
ω
γ
∫
𝕊
‍
u
x
2
u
d
x
d
d
t
∫
𝕊
‍
γ
m
v
-
3
∫
𝕊
‍
u
x
2
G
*
(
a
-
γ
)
γ
2
u
2
d
x
d
d
t
∫
𝕊
‍
γ
m
v
-
3
∫
𝕊
‍
u
x
2
G
*
γ
2
2
u
x
2
d
x
-
6
ω
γ
∫
𝕊
‍
u
x
2
G
*
u
d
x
.
Due to∥
u
∥
L
∞
2
≤
(
1
/
12
)
∫
S
‍
u
x
2
d
x
≤
(
1
/
12
)
∥
u
∥
H
1
2 and ∫
S
‍
u
x
2
d
x
≥
(
12
/
13
)
∥
u
∥
H
1
2 (see [29]), we get
(88)
∫
𝕊
‍
u
2
u
x
2
d
x
≤
max
⁡
{
u
2
}
∫
𝕊
‍
u
x
2
d
x
≤
1
12
(
∫
𝕊
‍
u
x
2
d
x
)
2
,
∫
𝕊
‍
u
u
x
2
d
x
≤
max
⁡
{
|
u
|
}
∫
𝕊
‍
u
x
2
d
x
≤
3
6
(
∫
𝕊
‍
u
x
2
d
x
)
3
/
2
,
∫
𝕊
‍
u
x
2
G
*
(
u
2
)
d
x
≤
∥
G
∥
L
1
max
⁡
{
u
2
}
m
m
m
m
m
m
m
m
×
∫
𝕊
‍
u
x
2
d
x
≤
1
12
(
∫
𝕊
‍
u
x
2
d
x
)
2
,
∫
𝕊
‍
u
x
2
G
*
u
d
x
≤
∥
G
∥
L
1
max
⁡
{
u
}
∫
𝕊
‍
u
x
2
d
x
≤
3
6
(
∫
𝕊
‍
u
x
2
d
x
)
3
/
2
.
Since 1
/
2
sinh
⁡
(
1
/
2
)
≤
G
(
x
)
≤
cosh
⁡
(
1
/
2
)
/
2
sinh
⁡
(
1
/
2
), we have
(89)
∫
𝕊
‍
u
x
2
G
*
(
u
x
2
)
d
x
≥
1
2
sinh
⁡
(
1
/
2
)
(
∫
𝕊
‍
u
x
2
d
x
)
2
≥
72
169
sinh
⁡
⁡
(
1
/
2
)
∥
u
∥
H
1
4
.
Therefore, we deduce
(90)
d
d
t
∫
𝕊
‍
γ
u
x
3
d
x
=
-
3
γ
2
2
∫
𝕊
‍
u
x
4
-
3
β
∫
𝕊
‍
γ
u
x
3
d
x
+
2
3
ω
|
γ
|
×
(
∫
𝕊
‍
u
x
2
d
x
)
3
/
2
+
(
(
a
-
γ
)
γ
8
-
3
γ
2
4
sinh
⁡
(
1
/
2
)
)
(
∫
𝕊
‍
u
x
2
d
x
)
2
.
Next, we divide (90) three cases to prove the theorem.
(i) The first isa
sinh
⁡
(
1
/
2
)
/
(
6
+
sinh
⁡
(
1
/
2
)
)
<
γ
≤
a. Note that γ
(
a
-
γ
)
/
8
>
0 and γ
(
a
-
γ
)
/
8
-
3
γ
2
/
4
sinh
⁡
(
1
/
2
)
<
0. From the condition of Theorem 14, we have
(91)
d
d
t
∫
𝕊
‍
γ
u
x
3
d
x
≤
-
3
γ
2
2
∫
𝕊
‍
u
x
4
-
3
β
∫
𝕊
‍
γ
u
x
3
d
x
+
2
3
ω
γ
d
d
t
∫
𝕊
‍
γ
u
x
3
v
×
∥
u
0
∥
H
1
3
+
36
169
(
(
a
-
γ
)
γ
2
-
3
γ
2
sinh
⁡
(
1
/
2
)
)
d
d
t
∫
𝕊
‍
γ
u
x
3
v
×
∥
u
0
∥
H
1
4
e
-
4
β
T
d
d
t
∫
𝕊
‍
γ
u
x
3
v
≤
-
3
γ
2
2
∫
𝕊
‍
u
x
4
-
3
β
∫
𝕊
‍
γ
u
x
3
d
x
-
δ
,
where δ
=
2
3
ω
γ
∥
u
0
∥
H
1
3
+
(
36
/
169
)
(
(
a
-
γ
)
γ
/
2
-
3
γ
2
/
sinh
⁡
(
1
/
2
)
)
∥
u
0
∥
H
1
4
e
-
4
β
T
>
0. From Holder's inequality, we obtain
(92)
∫
𝕊
‍
γ
u
x
3
d
x
≤
γ
(
∫
𝕊
‍
u
x
3
d
x
)
3
/
4
.
Thus, we have
(93)
d
d
t
∫
𝕊
‍
γ
u
x
3
d
x
≤
-
3
γ
2
/
3
2
(
∫
𝕊
‍
γ
u
x
3
d
x
)
4
/
3
-
3
β
∫
𝕊
‍
γ
u
x
3
d
x
-
δ
.
From (93), we have
(94)
d
d
t
∫
𝕊
‍
γ
u
x
3
d
x
≤
-
3
β
∫
𝕊
‍
γ
u
x
3
d
x
-
δ
.
Solving (94), we obtain
(95)
∫
𝕊
‍
γ
u
x
3
d
x
≤
-
δ
3
β
+
C
e
-
3
β
t
,
where C
<
0 is constant which implies that there is a t
0
≥
0 such that
(96)
∫
𝕊
‍
γ
u
x
3
d
x
<
0
,
t
≥
t
0
.
On the other hand,
(97)
d
d
t
∫
𝕊
‍
γ
u
x
3
d
x
≤
-
3
γ
2
/
3
2
(
∫
𝕊
‍
γ
u
x
3
d
x
)
4
/
3
-
3
β
∫
𝕊
‍
γ
u
x
3
d
x
.
Solving (97), we obtain
(98)
0
>
(
∫
𝕊
‍
γ
u
x
3
d
x
)
-
1
≥
(
-
γ
2
/
3
2
β
+
C
1
e
β
t
)
3
,
where C
1
≥
0 is constant, which leads to a contradiction as t becomes large enough. Therefore, case (i) is proved.
(ii) The second isγ
≥
a. Note that γ
(
a
-
γ
)
/
8
<
0 and -
γ
(
a
-
γ
)
/
8
-
3
γ
2
/
4
sinh
⁡
(
1
/
2
)
<
0. From the condition of the theorem, we have
(99)
d
d
t
∫
𝕊
‍
γ
u
x
3
d
x
≤
-
3
γ
2
2
∫
𝕊
‍
u
x
4
-
3
β
∫
𝕊
‍
γ
u
x
3
d
x
+
2
3
ω
γ
d
d
t
∫
𝕊
‍
γ
m
v
×
∥
u
0
∥
H
1
3
+
36
169
(
-
(
a
-
γ
)
γ
2
-
3
γ
2
sinh
⁡
(
1
/
2
)
)
d
d
t
∫
𝕊
‍
γ
m
v
×
∥
u
0
∥
H
1
4
e
-
4
β
T
d
d
t
∫
𝕊
‍
γ
m
≤
-
3
γ
2
2
∫
𝕊
‍
u
x
4
-
3
β
∫
𝕊
‍
γ
u
x
3
d
x
-
δ
,
where δ
=
2
3
ω
γ
∥
u
0
∥
H
1
3
+
(
36
/
169
)
(
-
(
a
-
γ
)
γ
/
2
-
3
γ
2
/
sinh
⁡
(
1
/
2
)
)
∥
u
0
∥
H
1
4
e
-
4
β
T
>
0. Repeating the arguments in the proof of case (i), we deduce that the corresponding solution blows up.
(iii) The third isγ
<
a
sinh
⁡
(
1
/
2
)
/
(
sinh
⁡
(
1
/
2
)
-
6
). Note that γ
(
a
-
γ
)
/
8
<
0 and -
γ
(
a
-
γ
)
/
8
-
3
γ
2
/
4
sinh
⁡
(
1
/
2
)
<
0. From the condition of the theorem, we have
(100)
d
d
t
∫
𝕊
‍
γ
u
x
3
d
x
≤
-
3
γ
2
2
∫
S
‍
u
x
4
-
3
β
∫
𝕊
‍
γ
u
x
3
d
x
+
2
3
ω
|
γ
|
d
d
t
∫
𝕊
‍
γ
m
v
×
∥
u
0
∥
H
1
3
+
36
169
(
-
(
a
-
γ
)
γ
2
-
3
γ
2
sinh
⁡
(
1
/
2
)
)
d
d
t
∫
𝕊
‍
γ
m
v
×
∥
u
0
∥
H
1
4
e
-
4
β
T
d
d
t
∫
𝕊
‍
γ
m
≤
-
3
γ
2
2
∫
𝕊
‍
u
x
4
-
3
β
∫
𝕊
‍
γ
u
x
3
d
x
-
δ
,
where δ
=
2
3
ω
|
γ
|
∥
u
0
∥
H
1
3
+
(
36
/
169
)
(
-
(
a
-
γ
)
γ
/
2
-
3
γ
2
/
sinh
⁡
(
1
/
2
)
)
∥
u
0
∥
H
1
4
e
-
4
β
T
>
0. Repeating the arguments in the proof of case (i), we derive that the corresponding solution blows up. This finishes the proof of the theorem.



---

*Source: 751949-2014-04-07.xml*