# Nonexistence of Global Solutions to the Initial Boundary Value Problem for the Singularly Perturbed Sixth-Order

Authors: Changming Song; Jina Li; Ran Gao
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

We are concerned with the singularly perturbed Boussinesq-type equation including
the singularly perturbed sixth-order Boussinesq equation, which describes the bidirectional
propagation of small amplitude and long capillary-gravity waves on the surface of shallow water
for bond number (surface tension parameter) less than but very close to 1/3. The nonexistence
of global solution to the initial boundary value problem for the singularly perturbed Boussinesq-type
equation is discussed and two examples are given.


---

## Body

## 1. Introduction

In the numerical study of the ill-posed Boussinesq equation,(1)
u
t
t
=
u
x
x
+
(
u
2
)
x
x
+
u
x
x
x
x
.
Darapi and Hua [1] proposed the singularly perturbed Boussinesq equation
(2)
u
t
t
=
u
x
x
+
(
u
2
)
x
x
+
u
x
x
x
x
+
δ
u
x
x
x
x
x
x
as a dispersive regularization of the ill-posed classical Boussinesq equation (1), where δ
>
0 is a small parameter. The authors use both filtering and regularization techniques to control growth of the errors and to provide better approximate solutions of this equation. Dash and Daripa [2] presented a formal derivation of (2) from two-dimensional potential flow equations for water waves through an asymptotic series expansion for small amplitude and long wave length. The physical relevance of (2) in the context of water waves was also addressed in [2]; it was shown that (2) actually describes the bidirectional propagation of small amplitude and long capillary-gravity waves on the surface of shallow water for bond number (surface tension parameter) less than but very close to 1
/
3. On the basis of far-field analysis and heuristic arguments, Daripa and Dash [3] proved that the traveling wave solutions of (2) are weakly nonlocal solitary waves characterized by small amplitude fast oscillations in the far-field and obtained weakly nonlocal solitary wave solutions of (2). Feng [4] investigated the generalized Boussinesq equation including the singularly perturbed Boussinesq equation
(3)
u
t
t
=
[
Q
(
u
)
]
x
x
+
∑
i
=
1
n
‍
b
i
u
(
2
i
+
2
)
x
,
where Q
(
u
)
=
u
+
b
0
u
r, u
(
2
i
+
2
)
x
=
(
∂
2
i
+
2
u
)
/
(
∂
x
2
i
+
2
), r and b
i (i
=
1,2
,
…
,
n) are all real constants. It is easily seen that the choices b
0
=
1, r
=
2, n
=
2, b
1
=
1, and b
2
=
δ lead (3) to the singularly perturbed Boussinesq equation (2). By the means of two proper ansatzs, the author obtained explicit traveling solitary wave solutions of the generalized Boussinesq equation (3). To the best of our knowledge, however, there have not been any discussions on global solutions of the initial boundary value problem for (2) in the literature; recently, Song et al. [5] discussed the initial boundary value problem for the singularly perturbed Boussinesq-type equation
(4)
u
t
t
=
u
x
x
+
σ
(
u
)
x
x
+
α
u
x
4
+
β
u
x
6
,
x
∈
Ω
,
t
>
0
,
with the initial boundary value conditions
(5)
u
x
(
0
,
t
)
=
u
x
(
1
,
t
)
=
u
x
3
(
0
,
t
)
=
u
x
3
(
1
,
t
)
=
u
x
5
(
0
,
t
)
=
u
x
5
(
1
,
t
)
=
0
,
t
>
0
,
u
(
x
,
0
)
=
u
0
(
x
)
,
u
t
(
x
,
0
)
=
u
1
(
x
)
,
x
∈
Ω
¯
,
or with
(6)
u
(
0
,
t
)
=
u
(
1
,
t
)
=
u
x
x
(
0
,
t
)
=
u
x
x
(
1
,
t
)
=
u
x
4
(
0
,
t
)
=
u
x
4
(
1
,
t
)
=
0
,
t
>
0
,
u
(
x
,
0
)
=
u
0
(
x
)
,
u
t
(
x
,
0
)
=
u
1
(
x
)
,
x
∈
Ω
¯
,
where, and in the sequel u
x
i
=
∂
i
u
/
∂
x
i, σ
(
s
) is a given nonlinear function, α
>
0 and β
>
0 are real numbers, u
0
(
x
) and u
1
(
x
) are given initial value functions, and Ω
=
(
0,1
). By virtue of the Galerkin method and prior estimates, under the assumption “σ
′
(
s
) is bounded below and σ
(
s
) satisfies some smooth condition,” the existence and uniqueness of the global generalized solution and the global classical solution of the initial boundary value problem (4), (5) and (4), (6) are proved, respectively. But if σ
′
(
s
) is not bounded below, does the above-mentioned problem have any global solution? In this paper, we employ the energy method and the Jensen inequality to prove that the global solutions of the initial boundary value problem (4), (5) and (4), (6) cease to exist in a finite time, respectively. At last, we show that the global solution of the initial boundary value problem (2), (6) blows up in a finite time.The paper is organized as follows. In Section2, the main results are stated. The nonexistence of global solution of problem (4), (5) and (4), (6) is discussed in Section 3. In Section 4, we study the initial boundary problem (2), (6) and give two examples satisfying the theorems (Theorems 1–6).


## 2. Main Theorems

Throughout this paper, we use the abbreviations∥
·
∥
=
∥
·
∥
L
2
(
Ω
). In the following we state the main results of this paper, where the existence of Theorems 1–4 has been proved in [5].Theorem 1 (see [5]).
Assume thatu
0
∈
H
6
(
Ω
), u
1
∈
H
3
(
Ω
), ∫
0
1
‍
u
0
(
x
)
d
x
=
∫
0
1
‍
u
1
(
x
)
d
x
=
0, u
0
x
2
k
+
1
(
0
,
t
)
=
u
0
x
2
k
+
1
(
1
,
t
)
=
u
1
x
2
k
+
1
(
0
,
t
)
=
u
1
x
2
k
+
1
(
1
,
t
)
=
0 (k
=
0,1
,
2), σ
∈
C
5
(
R
), and σ
′
(
s
) is bounded below; namely, there exists a constant C
0 such that σ
′
(
s
)
≥
C
0, for any s
∈
R. Then, for any T
>
0, the initial boundary value problem (4), (5) admits a unique global generalized solution u
(
x
,
t
) with
(7)
u
∈
C
(
[
0
,
T
]
;
H
6
(
Ω
)
)
∩
C
1
(
[
0
,
T
]
;
H
3
(
Ω
)
)
∩
C
2
(
[
0
,
T
]
;
L
2
(
Ω
)
)
.Theorem 2 (see [5]).
Assume that the assumptions of Theorem1 hold, u
0
∈
H
10
(
Ω
), u
1
∈
H
7
(
Ω
), and σ
∈
C
9
(
R
). Then, the initial boundary value problem (4), (5) admits a unique global classical solution u
(
x
,
t
).Theorem 3 (see [5]).
Assume thatu
0
∈
H
6
(
Ω
), u
1
∈
H
3
(
Ω
), u
0
x
2
k
(
0
,
t
)
=
u
0
x
2
k
(
1
,
t
)
=
u
1
x
2
k
(
0
,
t
)
=
u
1
x
2
k
(
1
,
t
)
=
0 (k
=
0,1
,
2), σ
∈
C
5
(
R
), σ
(
2
i
)
(
0
)
=
0 (i
=
1,2), and σ
′
(
s
) is bounded below. Then, for any T
>
0, the initial boundary value problem (4), (6) admits a unique global generalized solution u
(
x
,
t
) with
(8)
u
∈
C
(
[
0
,
T
]
;
H
6
(
Ω
)
)
∩
C
1
(
[
0
,
T
]
;
H
3
(
Ω
)
)
∩
C
2
(
[
0
,
T
]
;
L
2
(
Ω
)
)
.Theorem 4 (see [5]).
Assume that the assumptions of Theorem3 hold, u
0
∈
H
10
(
Ω
), u
1
∈
H
7
(
Ω
), σ
∈
C
9
(
R
), and σ
(
2
i
)
(
0
)
=
0 (i
=
3,4). Then, the initial boundary value problem (4), (6) admits a unique global classical solution u
(
x
,
t
).Theorem 5.
Assume that (1)σ
(
s
)
s
≤
μ
Γ
(
s
), Γ
(
s
)
≤
-
γ
|
s
|
m
+
1, where Γ
(
s
)
=
∫
0
s
‍
σ
(
τ
)
d
τ, μ
>
2, γ
>
0, and m
>
1 are constants, and (2) u
0
∈
H
2, u
1
∈
L
2, ∫
0
1
‍
u
0
(
x
)
d
x
=
∫
0
1
‍
u
1
(
x
)
d
x
=
0, and
(9)
E
0
=
∫
0
1
‍
(
∫
0
x
‍
u
1
(
ξ
)
d
ξ
)
2
d
x
+
∥
u
0
∥
2
-
α
∥
u
0
′
∥
2
+
β
∥
u
0
′′
∥
2
+
2
∫
0
1
‍
∫
0
u
0
‍
σ
(
s
)
d
s
d
x
≤
-
[
2
D
(
1
-
e
(
1
-
m
)
/
4
)
2
]
2
/
(
m
-
1
)
,
where D
=
γ
(
μ
-
2
)
/
[
2
(
m
-
7
)
/
2
(
m
+
3
)
]. Then the solution u
(
x
,
t
) of initial boundary value problem (4), (5) blows up in a finite time T
0; namely,
(10)
∥
u
(
t
)
∥
L
1
(
Ω
)
2
+
∫
0
t
‍
∥
u
(
τ
)
∥
2
d
τ
⟶
+
∞
a
s
t
⟶
T
0
-
,
where T
0 is defined in the proof.Theorem 6.
Assume that (1)σ
∈
C
2
(
R
), σ
(
0
)
=
0, and one of the following conditions holds: (i) σ
(
s
) is a convex and even function, σ
(
s
)
≥
a
s
m, where a
>
0 and m
>
1 are real numbers, (ii) σ
(
s
) is a convex function, σ
(
s
)
≥
a
s
m, where a
>
0 is a real number and m
≥
2 is an even number, and (2) -
(
π
/
2
)
∫
0
1
‍
u
0
(
x
)
sin
π
x
d
x
>
max
⁡
{
0
,
(
(
β
π
4
-
α
π
2
+
1
)
/
a
)
1
/
(
m
-
1
)
}, -
(
π
/
2
)
∫
0
1
‍
u
1
(
x
)
sin
π
x
d
x
>
0. Then the solution u
(
x
,
t
) of the initial boundary value problem (4), (6) blows up in a finite time T
1; namely,
(11)
∥
u
(
t
)
∥
⟶
+
∞
,
a
s
t
⟶
T
1
-
,
where T
1 is defined in the proof.


## 3. Nonexistence of Global Solutions of Problem (4), (5) and (4), (6)

We first quote the following lemmas.Lemma 7 (see Li [6]).
Assume thatu
˙
=
G
(
t
,
u
), v
˙
≥
G
(
t
,
v
), G
∈
C
(
[
0
,
+
∞
)
×
R
), t
0
≤
t
<
+
∞, and u
(
t
0
)
=
v
(
t
0
). Then u
(
t
)
≥
v
(
t
) as t
≥
t
0.Lemma 8 (Jensen inequality [7]).
Assume thatφ
(
u
)
:
u
∈
[
α
,
β
]
↦
R is a convex function, f
:
x
∈
Ω
↦
[
α
,
β
], and P
(
x
) is a continuous function, P
(
x
)
≥
0, P
(
x
)
≢
0. Then
(12)
φ
(
∫
Ω
‍
f
(
x
)
P
(
x
)
d
x
∫
Ω
‍
P
(
x
)
d
x
)
≤
∫
Ω
‍
φ
(
f
(
x
)
)
P
(
x
)
d
x
∫
Ω
‍
P
(
x
)
d
x
.
Integrating both sides of (4) over (
0,1
) and using (5) and the assumption of Theorem 1, we obtain ∫
0
1
‍
u
(
x
,
t
)
d
x
=
0, t
≥
0. Let v
(
x
,
t
)
=
∫
0
x
‍
u
(
ξ
,
t
)
d
ξ; then u
=
v
x and v satisfies
(13)
v
t
t
=
v
x
x
+
σ
(
v
x
)
x
+
α
v
x
4
+
β
v
x
6
,
x
∈
Ω
,
t
>
0
,
(14)
v
(
0
,
t
)
=
v
(
1
,
t
)
=
v
x
x
(
0
,
t
)
=
v
x
x
(
1
,
t
)
=
v
x
4
(
0
,
t
)
=
v
x
4
(
1
,
t
)
=
0
,
t
>
0
,
(15)
v
(
x
,
0
)
=
v
0
(
x
)
,
v
t
(
x
,
0
)
=
v
1
(
x
)
,
x
∈
Ω
¯
,
where v
0
(
x
)
=
∫
0
x
‍
u
0
(
ξ
)
d
ξ and v
1
(
x
)
=
∫
0
x
‍
u
1
(
ξ
)
d
ξ.Proof of Theorem5.
Multiplying both sides of (13) by 2
v
t, integrating by parts, and using condition (2) of Theorem 5, we have
(16)
E
˙
(
t
)
=
0
,
E
(
t
)
=
E
(
0
)
=
E
0
<
0
,
t
>
0
,
where
(17)
E
(
t
)
=
∥
v
t
(
t
)
∥
2
+
∥
v
x
(
t
)
∥
2
-
α
∥
v
x
x
(
t
)
∥
2
+
β
∥
v
x
3
(
t
)
∥
2
+
2
∫
0
1
‍
Γ
(
v
x
(
x
,
t
)
)
d
x
.
Let
(18)
F
(
t
)
=
∥
v
(
t
)
∥
2
+
∫
0
t
‍
∫
0
τ
‍
∥
v
x
(
s
)
∥
2
d
s
d
τ
.
By virtue of condition (1) of Theorem 5 and noting that
(19)
μ
∫
0
1
‍
Γ
(
v
x
(
x
,
t
)
)
d
x
=
E
0
-
∥
v
t
(
t
)
∥
2
-
∥
v
x
(
t
)
∥
2
+
α
∥
v
x
x
(
t
)
∥
2
-
β
∥
v
x
3
(
t
)
∥
2
+
(
μ
-
2
)
∫
0
1
‍
Γ
(
v
x
)
d
x
,
we obtain
(20)
F
¨
(
t
)
=
2
∫
0
1
‍
v
v
t
t
d
x
+
2
∥
v
t
(
t
)
∥
2
+
∥
v
x
(
t
)
∥
2
=
-
2
∫
0
1
‍
(
v
x
2
-
α
v
x
x
2
+
β
v
x
3
2
+
σ
(
v
x
)
v
x
)
d
x
+
2
∥
v
t
(
t
)
∥
2
+
∥
v
x
(
t
)
∥
2
≥
-
2
E
0
+
∥
v
x
(
t
)
∥
2
+
4
∥
v
t
(
t
)
∥
2
-
2
(
μ
-
2
)
∫
0
1
‍
Γ
(
v
x
)
d
x
≥
-
2
E
0
+
2
γ
(
μ
-
2
)
∫
0
1
‍
|
v
x
|
m
+
1
d
x
.
It follows from (20) that
(21)
F
˙
(
t
)
≥
-
2
E
0
t
+
2
γ
(
μ
-
2
)
∫
0
t
‍
∫
0
1
‍
|
v
x
(
x
,
τ
)
|
m
+
1
d
x
d
τ
+
F
˙
(
0
)
,
(22)
F
(
t
)
≥
-
E
0
t
2
+
2
γ
(
μ
-
2
)
∫
0
t
‍
∫
0
τ
‍
∫
0
1
‍
|
v
x
(
x
,
s
)
|
m
+
1
d
x
d
s
d
τ
+
F
˙
(
0
)
t
+
F
(
0
)
,
where F
˙
(
0
)
=
2
∫
0
1
‍
(
∫
0
x
‍
u
0
(
ξ
)
d
ξ
∫
0
x
‍
u
1
(
ξ
)
d
ξ
)
d
x and F
(
0
)
=
∥
∫
0
x
‍
u
0
(
ξ
)
d
ξ
∥
2. Combining (20) with (22) leads to
(23)
F
¨
(
t
)
+
F
(
t
)
≥
-
E
0
t
2
+
F
˙
(
0
)
t
+
F
(
0
)
-
2
E
0
+
2
γ
(
μ
-
2
)
(
∫
0
1
‍
|
v
x
|
m
+
1
d
x
m
m
m
m
m
m
m
m
m
m
+
∫
0
t
‍
∫
0
τ
‍
∫
0
1
‍
|
v
x
(
x
,
s
)
|
m
+
1
d
x
d
s
d
τ
)
.
Making use of the Hölder inequality, we get
(24)
∫
0
1
‍
|
v
x
|
m
+
1
d
x
≥
∥
v
x
(
t
)
∥
m
+
1
,
∫
0
t
‍
∫
0
τ
‍
∫
0
1
‍
|
v
x
(
x
,
s
)
|
m
+
1
d
x
d
s
d
τ
≥
(
t
2
2
)
(
1
-
m
)
/
2
(
∫
0
t
‍
∫
0
τ
‍
∫
0
1
‍
|
v
x
(
x
,
s
)
|
2
d
x
d
s
d
τ
)
(
1
+
m
)
/
2
.
Substituting (24) into (23) and using the Poincaré inequality (
∥
v
x
(
t
)
∥
≥
∥
v
(
t
)
∥
) and the inequality a
n
+
b
n
≥
2
1
-
n
(
a
+
b
)
n (a
,
b
≥
0
,
n
≥
1), we conclude that when t
≥
1,
(25)
F
¨
(
t
)
+
F
(
t
)
≥
2
γ
(
μ
-
2
)
t
1
-
m
×
[
(
∫
0
t
‍
∫
0
τ
‍
∫
0
1
‍
|
v
x
(
x
,
s
)
|
2
d
x
d
s
d
τ
)
(
m
+
1
)
/
2
∫
0
t
‍
(
∥
v
x
(
t
)
∥
2
)
(
m
+
1
)
/
2
mmmm
+
(
∫
0
t
‍
∫
0
τ
‍
∫
0
1
‍
|
v
x
(
x
,
s
)
|
2
d
x
d
s
d
τ
)
(
m
+
1
)
/
2
]
-
E
0
t
2
+
F
˙
(
0
)
t
+
F
(
0
)
-
2
E
0
≥
2
(
3
-
m
)
/
2
γ
(
μ
-
2
)
t
1
-
m
F
(
1
+
m
)
/
2
(
t
)
-
E
0
t
2
+
F
˙
(
0
)
t
+
F
(
0
)
-
2
E
0
.
Choose t
0
≥
1 such that
(26)
-
2
E
0
t
0
+
F
˙
(
0
)
≥
0
,
-
E
0
t
0
2
+
F
˙
(
0
)
t
0
+
F
(
0
)
-
2
E
0
≥
0
,
and thus (21) and (22) imply that F
˙
(
t
)
≥
0 and F
(
t
)
≥
0, as t
≥
t
0. Multiplying both sides of (25) by 2
F
˙
(
t
), we have
(27)
d
d
t
[
F
˙
2
(
t
)
+
F
2
(
t
)
]
≥
D
t
1
-
m
d
d
t
F
(
m
+
3
)
/
2
(
t
)
+
H
(
t
)
,
t
≥
t
0
,
where
(28)
D
=
γ
(
μ
-
2
)
2
(
m
-
7
)
/
2
(
m
+
3
)
,
H
(
t
)
=
2
F
˙
(
t
)
(
-
E
0
t
2
+
F
˙
(
0
)
t
+
F
(
0
)
-
2
E
0
)
.
Equation (27) implies that
(29)
d
d
t
[
t
m
-
1
(
F
˙
2
(
t
)
+
F
2
(
t
)
)
-
D
F
(
m
+
3
)
/
2
(
t
)
]
≥
t
m
-
1
H
(
t
)
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
≥
t
0
.
Since
(30)
∫
t
0
t
‍
τ
m
-
1
H
(
τ
)
d
τ
≥
∫
t
0
t
‍
2
(
-
2
E
0
τ
+
F
˙
(
0
)
)
(
-
E
0
τ
2
+
F
˙
(
0
)
τ
+
F
(
0
)
mmmmmmmmmmmmc
-
2
E
0
-
E
0
τ
2
)
d
τ
⟶
+
∞
,
t
⟶
+
∞
,
there exists a t
1
>
t
0 such that
(31)
∫
t
0
t
‍
τ
m
-
1
H
(
τ
)
d
τ
+
t
0
m
-
1
(
F
˙
2
(
t
0
)
+
F
2
(
t
0
)
)
-
D
F
(
m
+
3
)
/
2
(
t
0
)
≥
0
,
t
>
t
1
.
By (31), integrating both sides of (29) over (
t
0
,
t
), we obtain
(32)
t
m
-
1
[
F
˙
2
(
t
)
+
F
2
(
t
)
]
≥
D
F
(
m
+
3
)
/
2
(
t
)
,
t
≥
t
1
.
Namely,
(33)
F
˙
(
t
)
+
F
(
t
)
≥
2
D
t
(
1
-
m
)
/
2
F
(
m
+
3
)
/
4
(
t
)
,
t
≥
t
1
.
In order to use Lemma 7, we consider the following initial value problem of the Bernoulli equation:
(34)
X
˙
+
X
=
2
D
t
(
1
-
m
)
/
2
X
(
m
+
3
)
/
4
,
t
>
t
1
,
X
(
t
1
)
=
F
(
t
1
)
.
We can obtain the solution of the initial value problem (34) as follows:
(35)
X
(
t
)
=
e
-
(
t
-
t
1
)
[
∫
t
1
t
‍
m
-
1
4
F
(
1
-
m
)
/
4
(
t
1
)
-
m
-
1
4
2
D
m
m
m
m
m
c
-
m
-
1
4
2
D
×
∫
t
1
t
‍
τ
(
1
-
m
)
/
2
e
(
(
1
-
m
)
/
4
)
(
τ
-
t
1
)
d
τ
]
4
/
(
1
-
m
)
=
e
-
(
t
-
t
1
)
F
(
t
1
)
I
4
/
(
1
-
m
)
(
t
)
,
t
≥
t
1
,
where
(36)
I
(
t
)
=
1
-
m
-
1
4
2
D
F
(
m
-
1
)
/
4
(
t
1
)
×
∫
t
1
t
‍
τ
(
1
-
m
)
/
2
e
(
(
1
-
m
)
/
4
)
(
τ
-
t
1
)
d
τ
.
By (36), we know that I
(
t
1
)
=
1
>
0 and
(37)
J
(
t
)
=
m
-
1
4
2
D
F
(
m
-
1
)
/
4
(
t
1
)
∫
t
1
t
‍
τ
(
1
-
m
)
/
2
e
(
(
1
-
m
)
/
4
)
(
τ
-
t
1
)
d
τ
≥
m
-
1
4
2
D
F
(
m
-
1
)
/
4
(
t
1
)
(
1
+
t
1
)
(
1
-
m
)
/
2
×
∫
t
1
t
1
+
1
‍
e
(
(
1
-
m
)
/
4
)
(
τ
-
t
1
)
d
τ
=
2
D
F
(
m
-
1
)
/
4
(
t
1
)
(
1
+
t
1
)
(
1
-
m
)
/
2
×
(
1
-
e
-
(
m
-
1
)
/
4
)
,
t
≥
t
1
+
1
.
It follows from (22) that
(38)
F
(
m
-
1
)
/
4
(
t
)
(
1
+
t
)
(
1
-
m
)
/
2
≥
[
-
E
0
t
2
+
F
˙
(
0
)
t
+
F
(
0
)
(
t
+
1
)
2
]
(
m
-
1
)
/
4
⟶
(
-
E
0
)
(
m
-
1
)
/
4
,
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
c
m
m
t
⟶
+
∞
.
Choose t
1 sufficiently large such that F
(
m
-
1
)
/
4
(
t
1
)
(
1
+
t
1
)
(
1
-
m
)
/
2
≥
(
-
E
0
)
(
m
-
1
)
/
4
/
2. Combining (37) with (9), we obtain
(39)
J
(
t
)
≥
1
2
2
D
(
-
E
0
)
(
m
-
1
)
/
4
(
1
-
e
(
1
-
m
)
/
4
)
≥
1
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
t
≥
t
1
+
1
.
Hence
(40)
I
(
t
)
=
1
-
J
(
t
)
≤
0
,
t
≥
t
1
+
1
.
By using the continuity of I
(
t
), there exists a finite time T
0, t
1
<
T
0
≤
t
1
+
1 such that I
(
T
0
)
=
0. Therefore, X
(
t
)
→
+
∞ as t
→
T
0
-. By virtue of Lemma 7, we deduce that F
(
t
)
≥
X
(
t
), t
≥
t
1. Hence
(41)
F
(
t
)
=
∫
0
1
‍
(
∫
0
x
‍
u
(
ξ
,
t
)
d
ξ
)
2
d
x
+
∫
0
t
‍
∫
0
τ
‍
∫
0
1
‍
u
2
(
ξ
,
s
)
d
ξ
d
s
d
τ
⟶
+
∞
as t
→
T
0
-. It follows from (41) that
(42)
F
(
t
)
≤
(
∫
0
1
‍
|
u
(
ξ
,
t
)
|
d
ξ
)
2
+
t
∫
0
t
‍
∥
u
(
τ
)
∥
2
d
τ
.
Therefore,
(43)
∥
u
(
t
)
∥
L
1
(
Ω
)
2
+
∫
0
t
‍
∥
u
(
τ
)
∥
2
d
τ
⟶
+
∞
as
t
⟶
T
0
-
.
Theorem 5 is proved.Proof of Theorem6.
Let(44)
y
(
t
)
=
-
π
2
∫
0
1
‍
u
(
x
,
t
)
sin
π
x
d
x
.
Multiplying both sides of (4) by (
π
/
2
)
sin
π
x, integrating by parts over [
0,1
], and making use of the Jensen inequality and condition (1) of Theorem 6, we have
(45)
y
¨
+
(
π
2
-
α
π
4
+
β
π
6
)
y
=
π
3
2
∫
0
1
‍
σ
(
u
)
sin
π
x
d
x
≥
π
2
σ
(
π
2
∫
0
1
‍
u
(
x
,
t
)
sin
π
x
d
x
)
≥
a
π
2
y
m
,
t
>
0
,
and, from (6) and condition (2) of Theorem 6, we get
(46)
y
(
0
)
=
-
π
2
∫
0
1
‍
u
0
(
x
)
sin
π
x
d
x
=
y
0
>
0
,
y
˙
(
0
)
=
-
π
2
∫
0
1
‍
u
1
(
x
)
sin
π
x
d
x
=
y
1
>
0
.
Thus, we claim that
(47)
y
(
t
)
>
0
,
y
˙
(
t
)
>
0
,
t
>
0
.
In fact, if it is not true, then there exists at
* such that y
˙
(
t
)
>
0, t
∈
[
0
,
t
*
) and y
˙
(
t
*
)
=
0. Then y
(
t
) is monotonically increasing on [
0
,
t
*
]; that is, y
(
t
)
≥
y
0, t
∈
[
0
,
t
*
]. By using (45) and condition (2) of Theorem 6, we obtain
(48)
y
¨
(
t
)
≥
π
2
y
(
a
y
m
-
1
-
β
π
4
+
α
π
2
-
1
)
>
π
2
y
0
(
a
y
0
m
-
1
-
β
π
4
+
α
π
2
-
1
)
>
0
,
(
0
,
t
*
]
,
and hence y
˙
(
t
) is monotonically increasing on [
0
,
t
*
], which contradicts the assumption y
(
t
*
)
=
0. So claim (47) is valid.
Multiplying both sides of (45) by 2
y
˙ and integrating the product over [
0
,
t
] lead to
(49)
y
˙
2
≥
2
a
π
2
m
+
1
(
y
m
+
1
-
y
0
m
+
1
)
-
(
π
2
-
α
π
4
+
β
π
6
)
(
y
2
-
y
0
2
)
+
y
1
2
=
G
(
y
)
.
Since G
(
y
0
)
=
y
1
2
>
0 and
(50)
G
′
(
y
)
=
2
π
2
y
[
a
y
m
-
1
-
(
1
-
α
π
2
+
β
π
4
)
]
≥
2
π
2
y
0
[
a
y
0
m
-
1
-
(
1
-
α
π
2
+
β
π
4
)
]
≥
0
,
G
(
y
)
>
G
(
y
0
)
>
0, t
>
0. It follows from (49) that
(51)
y
˙
≥
[
2
a
π
2
m
+
1
(
y
m
+
1
-
y
0
m
+
1
)
m
c
c
-
(
π
2
-
α
π
4
+
β
π
6
)
(
y
2
-
y
0
2
)
+
y
1
2
2
a
π
2
m
+
1
]
1
/
2
,
t
>
0
,
and (51) implies that the interval [
0
,
T
1
) of the existence of y
(
t
) is finite; namely,
(52)
T
1
≤
∫
y
0
+
∞
‍
[
2
a
π
2
m
+
1
(
y
m
+
1
-
y
0
m
+
1
)
m
m
c
m
m
-
(
π
2
-
α
π
4
+
β
π
6
)
(
y
2
-
y
0
2
)
+
y
1
2
2
a
π
2
m
+
1
]
-
1
/
2
d
y
<
+
∞
,
and y
(
t
)
→
+
∞ as t
→
T
1
-; that is, ∫
0
1
‍
u
(
x
,
t
)
sin
π
x
d
x
→
-
∞ as t
→
T
1
-. By the Hölder inequality, we have ∥
u
(
t
)
∥
→
+
∞, as t
→
T
1
-. Theorem 6 is proved.


## 4. Initial Boundary Value Problem (2), (6) and Some Examples

By virtue of the Galerkin method [8] we can prove that initial boundary value problem (2), (6) admits a unique local generalized solution and a unique local classical solution. Moreover, by using Theorem 6, we obtain the following theorem.Theorem 9.
Assume thatu
(
x
,
t
) is the generalized solution of initial boundary value problem (2), (6) and the following condition holds:
(53)
y
0
=
-
π
2
∫
0
1
‍
u
0
(
x
)
sin
π
x
d
x
>
max
⁡
{
0
,
(
δ
π
4
-
π
2
+
1
)
}
,
y
1
=
-
π
2
∫
0
1
‍
u
1
(
x
)
sin
π
x
d
x
>
0
.
Then
(54)
∥
u
(
t
)
∥
⟶
+
∞
,
a
s
t
⟶
T
2
-
,
where
(55)
T
2
≤
∫
y
0
+
∞
‍
[
2
π
2
3
(
y
3
-
y
0
3
)
m
m
m
m
m
-
(
π
2
-
π
4
+
δ
π
6
)
(
y
2
-
y
0
2
)
+
y
1
2
2
π
2
3
]
-
1
/
2
d
y
<
+
∞
.Proof.
A simple verification shows that all conditions of Theorem6 are satisfied and thus Theorem 9 is proved immediately.Example 1.
We consider the following equation:(56)
u
t
t
=
u
x
x
+
a
(
|
u
|
m
-
1
u
)
x
x
+
α
u
x
4
+
β
u
x
6
,
x
∈
Ω
,
t
>
0
,
with the initial boundary value conditions
(57)
u
x
(
0
,
t
)
=
u
x
(
1
,
t
)
=
u
x
3
(
0
,
t
)
=
u
x
3
(
1
,
t
)
=
u
x
5
(
0
,
t
)
=
u
x
5
(
1
,
t
)
=
0
,
t
>
0
,
u
(
x
,
0
)
=
u
0
(
x
)
,
u
t
(
x
,
0
)
=
u
1
(
x
)
,
x
∈
Ω
¯
,
or with
(58)
u
(
0
,
t
)
=
u
(
1
,
t
)
=
u
x
x
(
0
,
t
)
=
u
x
x
(
1
,
t
)
=
u
x
4
(
0
,
t
)
=
u
x
4
(
1
,
t
)
=
0
,
t
>
0
,
u
(
x
,
0
)
=
u
0
(
x
)
,
u
t
(
x
,
0
)
=
u
1
(
x
)
,
x
∈
Ω
¯
,
where a
≠
0 and m
>
1 are all real numbers, u
0
(
x
)
=
u
1
(
x
)
=
K
0
cos
⁡
π
x, and K
0
>
0 is a constant.
(1) Ifa
>
0 and m
≥
9, a simple calculation shows that σ
(
=
a
|
s
|
m
-
1
s
)
∈
C
9
(
R
), σ
(
2
i
)
(
0
)
=
0 (i
=
1,2
,
3,4), and σ
′
(
s
)
=
a
m
|
s
|
m
-
1
≥
0; that is, σ
′
(
s
) is bounded below. And u
0
(
x
) and u
1
(
x
) satisfy the conditions of Theorems 2 and 4, respectively; then by Theorems 2 and 4 we know that the initial boundary value problem (56), (57) and (56), (58) admits a unique global classical solution, respectively.
(2) Ifa
<
0 and m
>
1, we have σ
(
s
)
s
=
a
|
s
|
m
-
1, Γ
(
s
)
=
a
|
s
|
m
-
1
/
(
m
+
1
); taking μ
=
m
+
1
>
2 and γ
=
-
a
/
(
m
+
1
), then σ
(
s
)
s
=
μ
Γ
(
s
), Γ
(
s
)
=
-
γ
|
s
|
m
-
1; obviously, u
0
∈
H
2, u
1
∈
L
2, ∫
0
1
‍
u
0
(
x
)
d
x
=
∫
0
1
‍
u
1
(
x
)
d
x
=
0, and
(59)
E
0
=
∫
0
1
‍
(
∫
0
x
‍
u
1
(
ξ
)
d
ξ
)
2
d
x
+
∥
u
0
∥
2
-
α
∥
u
0
′
∥
2
+
β
∥
u
0
′′
∥
2
+
2
a
m
+
1
∥
u
0
∥
L
m
+
1
(
Ω
)
m
+
1
=
(
1
π
2
+
1
-
α
π
2
+
β
π
4
)
K
0
2
2
+
2
a
K
0
m
+
1
m
+
1
∫
0
1
‍
|
cos
⁡
π
x
|
m
+
1
d
x
.
We can take K
0 suitable large such that
(60)
E
0
≤
-
[
2
D
(
1
-
e
(
1
-
m
)
/
4
)
2
]
2
/
(
m
-
1
)
,
where D
=
-
a
(
m
-
1
)
/
[
2
(
m
-
7
)
/
2
(
m
+
1
)
(
m
+
3
)
]. Thus, all assumptions of Theorem 5 are satisfied; then by Theorem 5 we conclude that the solution of initial boundary value problem (56), (57) must blow up in a finite time T
0; namely,
(61)
∥
u
(
t
)
∥
L
1
(
Ω
)
+
∫
0
t
‍
∥
u
(
τ
)
∥
2
d
τ
⟶
+
∞
as
t
⟶
T
0
-
.Example 2.
We consider the following equation:(62)
u
t
t
=
u
x
x
+
a
(
u
m
)
x
x
+
α
u
x
4
+
β
u
x
6
,
x
∈
Ω
,
t
>
0
,
with the initial boundary value conditions
(63)
u
x
(
0
,
t
)
=
u
x
(
1
,
t
)
=
u
x
3
(
0
,
t
)
=
u
x
3
(
1
,
t
)
=
u
x
5
(
0
,
t
)
=
u
x
5
(
1
,
t
)
=
0
,
t
>
0
,
u
(
x
,
0
)
=
u
0
(
x
)
,
u
t
(
x
,
0
)
=
u
1
(
x
)
,
x
∈
Ω
¯
,
or with
(64)
u
(
0
,
t
)
=
u
(
1
,
t
)
=
u
x
x
(
0
,
t
)
=
u
x
x
(
1
,
t
)
=
u
x
4
(
0
,
t
)
=
u
x
4
(
1
,
t
)
=
0
,
t
>
0
,
u
(
x
,
0
)
=
u
0
(
x
)
,
u
t
(
x
,
0
)
=
u
1
(
x
)
,
x
∈
Ω
¯
,
where a
>
0 is a real number and m
>
1 is a positive integer, u
0
(
x
)
=
u
1
(
x
)
=
-
K
1, and K
1
>
0 is a constant.
(1) Ifa
>
0 and m is an odd number, a simple verification shows that all conditions of Theorems 2 and 4 are satisfied; then by Theorems 2 and 4 we know that the initial boundary value problem (62), (63) and (62), (64) admits a unique global classical solution, respectively.
(2) Ifa
>
0 and m is an even number, then σ
(
s
) (=
a
s
m) is a convex and even function, and we can take K
1 suitable large such that
(65)
-
π
2
∫
0
1
‍
u
0
(
x
)
sin
π
x
d
x
=
K
1
>
max
⁡
{
0
,
(
β
π
4
-
α
π
2
+
1
a
)
1
/
(
m
-
1
)
}
,
-
π
2
∫
0
1
‍
u
1
(
x
)
sin
π
x
d
x
=
K
1
>
0
.
Thus, by Theorem 6, we deduce that the solution of initial boundary value problem (62), (64) must blow up in a finite time T
1; namely,
(66)
∥
u
(
t
)
∥
⟶
+
∞
,
as
t
⟶
T
1
-
.



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*Source: 928148-2014-05-26.xml*