# Blow-Up Criteria for the Modified Novikov Equation

Authors: Caochuan Ma; Wujun Lv
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We investigate the Cauchy problem for the modified Novikov equation.
We establish blow-up criteria on the initial data to guarantee the corresponding solution blowing
up in finite time.


---

## Body

## 1. Introduction

In this paper, we consider the following Cauchy problem of the modified Novikov equation:(1)
u
t
-
u
x
x
t
+
(
a
+
b
)
u
2
u
x
=
a
u
u
x
u
x
x
+
b
u
2
u
x
x
x
,
t
>
0
,
x
∈
ℝ
,
u
(
0
,
x
)
=
u
0
(
x
)
,
x
∈
ℝ
,
where the coefficients a and b are positive constants.In [1], Lai et al. presented the global existence of strong solutions and gave a blow-up scenario of strong solutions to the equation.By using Green’s functionG
(
x
)
=
(
1
/
2
)
e
-
x for the operator Λ
=
(
1
-
∂
x
2
)
1
/
2, (1) is equivalent to the nonlocal equation
(2)
u
t
+
b
u
2
u
x
=
G
*
[
-
a
u
2
u
x
+
a
-
6
b
2
(
u
u
x
2
)
x
+
2
b
-
a
2
u
x
3
]
=
G
*
[
-
a
u
2
+
a
-
6
b
2
(
u
u
x
2
)
x
]
t
>
0
,
x
∈
ℝ
,
u
(
0
,
x
)
=
u
0
(
x
)
,
x
∈
ℝ
,
where notation * denotes the spatial convolution.Lettinga
=
3
b and using the scaling translation u
~
=
b
u, (1) can be reformulated into the Novikov equation
(3)
u
~
t
-
u
~
x
x
t
+
4
u
~
2
u
~
x
=
3
u
~
u
~
x
u
~
x
x
+
u
~
2
u
x
x
x
,
which was derived by Novikov in a symmetry classification of nonlocal PDEs with quadratic or cubic nonlinearity [2]; subsequently, he found a scalar Lax pair for the Novikov equation (also see [3]) and proved that the Novikov equation is integrable. The equation has been investigated by many scholars. Hone and Wang gave a matrix Lax pair for the Novikov equation in [4] and showed how it was related by a reciprocal transformation to a negative flow in the Sawada-Kotera hierarchy. By using the matrix Lax pair, Hone et al. calculated the explicit formulas for multipeak on solutions of (1) in [3]. Ni and Zhou showed that the Novikov equation is well-posed in H
s
,
s
>
3
/
2 by applying Kato’s semigroup theory and the Novikov equation is locally well-posed in the Besov spaces B
2
,
r
s with the critical index s
=
3
/
2 and also considered the persistence properties of the solution. In [5], Jiang and Ni gave sufficient conditions on the initial data to guarantee the formulation of singularities in finite time and a global existence result was also established in [6]. It is worth pointing out recent many works have been done for the Novikov equation and the related equations, one can refer to [7–12] and the references therein.Now, we give some elementary results and a blow-up scenario of strong solutions which will be used in this paper.Theorem 1 (see [13]).
Givenu
(
x
,
t
=
0
)
=
u
0
∈
H
s
(
ℝ
) with s
>
3
/
2, then there exist a maximal T
=
T
(
u
0
)
>
0 and a unique solution u to (1) such that
(4)
u
=
u
(
·
,
u
0
)
∈
C
(
[
0
,
T
)
;
H
s
(
ℝ
)
)
∩
C
1
(
[
0
,
T
)
;
H
s
-
1
(
ℝ
)
)
.
Moreover, the solution depends continuously on the initial data; that is, the mapping u
0
→
u
(
·
,
u
0
)
:
H
s
(
ℝ
)
→
C
(
[
0
,
T
)
;
H
s
(
ℝ
)
)
∩
C
1
(
[
0
,
T
)
;
H
s
-
1
(
ℝ
)
) is continuous.Theorem 2 (see [1]).
Assumeu
0
(
x
)
∈
H
s, s
≥
3
/
2, and let T be the maximal existence time of the solution u
(
x
,
t
) to (1) with the initial data u
0
(
x
). If a
>
b, then the corresponding solution blows up in finite time if and only if
(5)
lim
⁡
t
↑
T
liminf
⁡
x
∈
ℝ
(
u
u
x
)
(
x
,
t
)
=
-
∞
.We also need to introduce the classical particle trajectory method. Supposeu
(
x
,
t
) is a solution of the Novikov equation; let q
(
x
,
t
) be the particle line evolved by the solution u:
(6)
d
q
(
x
,
t
)
d
t
=
b
u
2
(
q
(
x
,
t
)
,
t
)
,
q
(
x
,
t
=
0
)
=
x
.
Then
(7)
q
x
(
x
,
t
)
=
exp
⁡
(
2
∫
0
t
‍
u
u
x
(
q
,
s
)
d
s
)
,
q
x
(
x
,
0
)
=
1
,
which is always positive before the blow-up time. Therefore, the function q
(
x
,
t
) is an increasing diffeomorphism of the line before blow-up.Lety
=
Λ
2
u
=
(
1
-
∂
x
2
)
u; the following identity can be obtained:
(8)
y
(
q
(
x
,
t
)
,
t
)
q
x
2
(
x
,
t
)
=
y
0
(
x
)
e
(
4
b
-
a
)
∫
0
t
‍
u
u
x
d
s
.
In fact, direct computation yields
(9)
d
d
t
(
y
(
q
)
q
x
2
)
=
y
t
q
x
2
+
2
y
q
x
q
x
t
+
y
x
q
t
q
x
2
=
y
t
q
x
2
+
4
b
y
u
u
x
q
x
2
+
b
u
2
y
x
q
x
2
=
(
u
t
-
u
t
x
x
+
a
u
u
x
(
u
-
u
x
x
)
+
b
u
2
(
u
x
-
u
x
x
x
)
)
q
x
2
-
a
u
u
x
y
q
x
2
+
4
b
u
u
x
y
q
x
2
=
(
4
b
-
a
)
u
u
x
y
q
x
2
.Remark 3.
From (8), it follows that if y
0
(
x
)
=
Λ
2
u
0
(
x
)
≥
0 then y
(
t
,
x
)
=
Λ
2
u
(
t
,
x
)
≥
0. Since Λ
-
2
f
=
G
*
f, for f
∈
L
2
(
ℝ
), therefore, we obtain u
(
x
)
≥
0. If y
0
=
(
1
-
∂
x
2
)
u
0
(
x
)
≤
0, the result is similar.


## 2. Blow-Up Criteria

In this section, we present the following blow-up criteria on the initial data to guarantee that the corresponding strong solution of (1) blowing up. Our method is partially motivated by [14].Theorem 4.
Ifa
>
3
b, suppose that u
0
∈
H
s
(
ℝ
)
,
s
>
3
/
2, and there exists x
0
∈
ℝ such that u
0
(
x
0
)
≥
0 and y
0
(
x
0
)
=
(
1
-
∂
x
2
)
u
0
(
x
0
)
=
0,
(10)
y
0
(
x
)
≥
0
(
≢
0
)
f
o
r
x
∈
(
-
∞
,
x
0
)
,
y
0
(
x
)
≤
0
(
≢
0
)
f
o
r
x
∈
(
x
0
,
∞
)
.
Then the corresponding solution u
(
x
,
t
) to the modified Novikov equation (1) with u
0
(
x
) as the initial datum blows up in finite time.Proof.
By the local well-posedness theorem and a density argument, it suffices to consider the cases
≥
3; without loss of generality, we take s
=
3 for simplicity of notation. We also assume u
0
≢
0; otherwise, solutions are trivial.
Suppose that the solution exists globally. Due to (8) and the initial condition (10), we have
(11)
y
(
q
(
x
0
,
t
)
,
t
)
=
0
,
y
(
q
(
x
,
t
)
,
t
)
≥
0
(
≢
0
)
,
for
x
∈
(
-
∞
,
x
0
)
,
y
(
q
(
x
,
t
)
,
t
)
≤
0
(
≢
0
)
,
for
x
∈
(
x
0
,
∞
)
,
for all t
≥
0. Since u
(
x
,
t
)
=
G
*
y
(
x
,
t
), x
∈
ℝ, t
>
0, we can write u
(
x
,
t
) and u
x
(
x
,
t
) as follows:
(12)
u
(
x
,
t
)
=
1
2
e
-
x
∫
-
∞
x
‍
e
ξ
y
(
ξ
,
t
)
d
ξ
+
1
2
e
x
∫
x
∞
‍
e
-
ξ
y
(
ξ
,
t
)
d
ξ
,
u
x
(
x
,
t
)
=
-
1
2
e
-
x
∫
-
∞
x
‍
e
ξ
y
(
ξ
,
t
)
d
ξ
+
1
2
e
x
∫
x
∞
‍
e
-
ξ
y
(
ξ
,
t
)
d
ξ
.
As a result of (12)
(13)
(
u
+
u
x
)
(
x
,
t
)
=
e
x
∫
x
∞
‍
e
-
ξ
y
(
ξ
,
t
)
d
ξ
,
(14)
(
u
-
u
x
)
(
x
,
t
)
=
e
-
x
∫
-
∞
x
‍
e
ξ
y
(
ξ
,
t
)
d
ξ
,
for all t
≥
0.
From (12) and differentiating u
u
x
(
q
(
x
0
,
t
)
,
t
) with respect to t, we have
(15)
d
d
t
(
2
u
u
x
)
(
q
(
x
0
,
t
)
,
t
)
=
d
d
t
{
-
1
2
e
-
2
q
(
x
0
,
t
)
(
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
y
(
ξ
,
t
)
d
ξ
)
2
+
1
2
e
2
q
(
x
0
,
t
)
(
∫
q
(
x
0
,
t
)
∞
‍
e
-
ξ
y
(
ξ
,
t
)
d
ξ
)
2
}
=
b
u
2
(
q
(
x
0
,
t
)
,
t
)
e
-
2
q
(
x
0
,
t
)
(
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
y
(
ξ
,
t
)
d
ξ
)
2
-
e
-
2
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
y
(
ξ
,
t
)
d
ξ
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
y
t
(
ξ
,
t
)
d
ξ
+
b
u
2
(
q
(
x
0
,
t
)
,
t
)
e
2
q
(
x
0
,
t
)
(
∫
q
(
x
0
,
t
)
∞
‍
e
-
ξ
y
(
ξ
,
t
)
d
ξ
)
2
+
e
2
q
(
x
0
,
t
)
∫
q
(
x
0
,
t
)
∞
‍
e
-
ξ
y
(
ξ
,
t
)
d
ξ
∫
q
(
x
0
,
t
)
∞
‍
e
-
ξ
y
t
(
ξ
,
t
)
d
ξ
=
b
u
2
(
u
-
u
x
)
2
(
q
(
x
0
,
t
)
,
t
)
-
(
u
-
u
x
)
(
q
(
x
0
,
t
)
,
t
)
e
-
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
y
t
(
ξ
,
t
)
d
ξ
+
b
u
2
(
u
+
u
x
)
2
(
q
(
x
0
,
t
)
,
t
)
+
(
u
+
u
x
)
(
q
(
x
0
,
t
)
,
t
)
e
q
(
x
0
,
t
)
∫
q
(
x
0
,
t
)
∞
‍
e
-
ξ
y
t
(
ξ
,
t
)
d
ξ
.
Equation (1) can be rewritten as
(16)
y
t
=
-
b
(
y
u
2
)
x
+
(
2
b
-
a
)
y
u
u
x
.
Firstly, we can estimate the first term as(17)
e
-
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
y
t
(
ξ
,
t
)
d
ξ
=
-
b
e
-
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
(
y
(
ξ
,
t
)
u
2
(
ξ
,
t
)
)
ξ
d
ξ
+
(
2
b
-
a
)
e
-
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
y
(
ξ
,
t
)
u
(
ξ
,
t
)
u
ξ
(
ξ
,
t
)
d
ξ
=
b
e
-
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
y
(
ξ
,
t
)
u
2
(
ξ
,
t
)
d
ξ
+
(
2
b
-
a
)
e
-
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
y
(
ξ
,
t
)
u
(
ξ
,
t
)
u
ξ
(
ξ
,
t
)
d
ξ
=
b
e
-
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
(
u
3
-
u
2
u
x
x
)
d
ξ
+
(
2
b
-
a
)
e
-
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
(
u
2
u
x
-
u
u
x
u
x
x
)
d
ξ
=
3
b
-
a
3
u
3
-
b
u
2
u
x
-
2
b
-
a
2
u
u
x
2
+
a
3
e
-
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
u
3
d
ξ
+
6
b
-
a
2
e
-
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
u
u
x
2
d
ξ
+
2
b
-
a
2
e
-
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
u
x
3
d
ξ
.
We also apply the following inequality in [6]:
(18)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
(
2
u
3
+
3
u
u
ξ
2
-
u
ξ
3
)
(
ξ
,
t
)
d
ξ
≥
e
q
(
x
0
,
t
)
u
3
(
q
(
x
0
,
t
)
,
t
)
.
So we can derive
(19)
a
3
e
-
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
u
3
d
ξ
+
6
b
-
a
2
e
-
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
u
u
x
2
d
ξ
+
2
b
-
a
2
e
-
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
u
x
3
d
ξ
=
1
6
e
-
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
[
(
a
-
3
b
)
(
2
u
3
-
3
u
u
x
2
-
3
u
x
3
)
+
3
b
(
2
u
3
+
3
u
u
x
2
-
3
u
x
3
)
]
d
ξ
≥
a
-
3
b
6
u
3
+
3
b
6
u
3
=
a
6
u
3
.
Putting (19) into (17), we have
(20)
e
-
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
y
t
(
ξ
,
t
)
d
ξ
≥
-
b
u
2
(
q
(
x
0
,
t
)
,
t
)
u
x
(
q
(
x
0
,
t
)
,
t
)
-
2
b
-
a
2
u
(
q
(
x
0
,
t
)
,
t
)
u
x
2
(
q
(
x
0
,
t
)
,
t
)
+
6
b
-
a
6
u
3
(
q
(
x
0
,
t
)
,
t
)
.
Similarly, we have(21)
e
q
(
x
0
,
t
)
∫
q
(
x
0
,
t
)
∞
‍
e
-
ξ
y
t
(
ξ
,
t
)
d
ξ
≥
-
b
u
2
(
q
(
x
0
,
t
)
,
t
)
u
x
(
q
(
x
0
,
t
)
,
t
)
+
2
b
-
a
2
u
(
q
(
x
0
,
t
)
,
t
)
u
x
2
(
q
(
x
0
,
t
)
,
t
)
+
a
-
6
b
6
u
3
(
q
(
x
0
,
t
)
,
t
)
.
Putting (21) and (22) into (15), we obtain
(22)
d
d
t
2
u
u
x
(
q
(
x
0
,
t
)
,
t
)
≤
2
b
(
u
4
+
u
2
u
x
2
)
(
q
(
x
0
,
t
)
,
t
)
-
(
u
-
u
x
)
(
6
b
-
a
6
u
3
-
b
u
2
u
x
-
2
b
-
a
2
u
u
x
2
)
+
(
u
+
u
x
)
(
-
6
b
-
a
6
u
3
-
b
u
2
u
x
+
2
b
-
a
2
u
u
x
2
)
=
a
3
u
2
(
u
2
-
u
x
2
)
(
q
(
x
0
,
t
)
,
t
)
.
Here we use the facts that (
u
-
u
x
)
(
q
(
x
,
t
)
,
t
)
≥
0, x
∈
(
-
∞
,
q
(
x
0
,
t
)
), from (10) and (14), and (
u
+
u
x
)
(
q
(
x
,
t
)
,
t
)
≤
0, x
∈
(
q
(
x
0
,
t
)
,
∞
), from (10) and (13).
Claim
1. u
u
x
(
q
(
x
0
,
t
)
,
t
)
<
0 is decreasing and u
2
(
q
(
x
0
,
t
)
,
t
)
<
u
x
2
(
q
(
x
0
,
t
)
,
t
) for all t
≥
0.
Suppose that there exists at
0 such that u
2
(
q
(
x
0
,
t
)
,
t
)
<
u
x
2
(
q
(
x
0
,
t
)
,
t
) on [
0
,
t
0
) and u
2
(
q
(
x
0
,
t
0
)
,
t
0
)
≥
u
x
2
(
q
(
x
0
,
t
0
)
,
t
0
). Now, let
(23)
I
(
t
)
:
=
u
(
u
-
u
x
)
(
q
(
x
0
,
t
)
,
t
)
,
I
I
(
t
)
:
=
u
(
u
+
u
x
)
(
q
(
x
0
,
t
)
,
t
)
.
Firstly, differentiating I
(
t
), we get
(24)
d
I
(
t
)
d
t
=
-
b
u
2
(
q
(
x
0
,
t
)
,
t
)
×
(
e
-
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
y
(
ξ
,
t
)
d
ξ
)
2
+
(
e
-
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
y
(
ξ
,
t
)
d
ξ
)
×
(
e
-
q
(
x
0
,
t
)
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
y
t
(
ξ
,
t
)
d
ξ
)
+
1
2
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
y
t
(
ξ
,
t
)
d
ξ
∫
q
(
x
0
,
t
)
∞
‍
e
-
ξ
y
(
ξ
,
t
)
d
ξ
+
1
2
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
y
(
ξ
,
t
)
d
ξ
∫
q
(
x
0
,
t
)
∞
‍
e
-
ξ
y
t
(
ξ
,
t
)
d
ξ
≥
-
b
u
2
(
u
-
u
x
)
2
(
q
(
x
0
,
t
)
,
t
)
+
(
6
b
-
a
6
u
3
-
b
u
2
u
x
-
2
b
-
a
2
u
u
x
2
)
u
(
q
(
x
0
,
t
)
,
t
)
+
1
2
(
6
b
-
a
6
u
3
-
b
u
2
u
x
-
2
b
-
a
2
u
u
x
2
)
×
(
u
-
u
x
)
(
q
(
x
0
,
t
)
,
t
)
-
1
2
(
6
b
-
a
6
u
3
+
b
u
2
u
x
-
2
b
-
a
2
u
u
x
2
)
×
(
u
-
u
x
)
(
q
(
x
0
,
t
)
,
t
)
≥
a
6
u
2
(
u
x
2
-
u
2
)
,
on
⁡
[
0
,
t
0
)
.
Secondly, differentiating I
I
(
t
), we get
(25)
d
I
I
(
t
)
d
t
=
b
u
2
(
q
(
x
0
,
t
)
,
t
)
×
(
e
q
(
x
0
,
t
)
∫
q
(
x
0
,
t
)
∞
‍
e
-
ξ
y
(
ξ
,
t
)
d
ξ
)
2
+
(
e
q
(
x
0
,
t
)
∫
q
(
x
0
,
t
)
∞
‍
e
-
ξ
y
(
ξ
,
t
)
d
ξ
)
×
(
e
q
(
x
0
,
t
)
∫
q
(
x
0
,
t
)
∞
‍
e
-
ξ
y
t
(
ξ
,
t
)
d
ξ
)
+
1
2
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
y
t
(
ξ
,
t
)
d
ξ
∫
q
(
x
0
,
t
)
∞
‍
e
-
ξ
y
(
ξ
,
t
)
d
ξ
+
1
2
∫
-
∞
q
(
x
0
,
t
)
‍
e
ξ
y
(
ξ
,
t
)
d
ξ
∫
q
(
x
0
,
t
)
∞
‍
e
-
ξ
y
t
(
ξ
,
t
)
d
ξ
≤
b
u
2
(
u
+
u
x
)
2
(
q
(
x
0
,
t
)
,
t
)
-
(
6
b
-
a
6
u
3
+
b
u
2
u
x
-
2
b
-
a
2
u
u
x
2
)
×
u
x
(
q
(
x
0
,
t
)
,
t
)
+
1
2
(
6
b
-
a
6
u
3
-
b
u
2
u
x
-
2
b
-
a
2
u
u
x
2
)
×
(
u
+
u
x
)
(
q
(
x
0
,
t
)
,
t
)
-
1
2
(
6
b
-
a
6
u
3
+
b
u
2
u
x
-
2
b
-
a
2
u
u
x
2
)
×
(
u
+
u
x
)
(
q
(
x
0
,
t
)
,
t
)
=
-
a
6
u
2
(
u
x
2
-
u
2
)
,
on
⁡
[
0
,
t
0
)
.
Hence, from (24), (25), and the continuity property of ODEs, we can draw
(26)
u
2
(
u
x
2
-
u
2
)
(
q
(
x
0
,
t
)
,
t
)
=
-
I
(
t
)
I
I
(
t
)
>
-
I
(
0
)
I
I
(
0
)
>
0
,
for all t
>
0. This means t
0 can be extended to infinity. This is a contradiction, so the claim is true.
Moreover, using (24) and (25) again, we have the following inequality for u
2
(
u
x
2
-
u
2
)
(
q
(
x
0
,
t
)
,
t
):
(27)
d
d
t
u
2
(
u
x
2
-
u
2
)
(
q
(
x
0
,
t
)
,
t
)
≥
-
a
3
u
u
x
(
q
(
x
0
,
t
)
,
t
)
u
2
(
u
x
2
-
u
2
)
(
q
(
x
0
,
t
)
,
t
)
.
Putting (22) into (27) yields
(28)
d
d
t
u
2
(
u
x
2
-
u
2
)
(
q
(
x
0
,
t
)
,
t
)
≥
a
3
u
2
(
u
x
2
-
u
2
)
(
q
(
x
0
,
t
)
,
t
)
×
(
a
6
∫
0
t
‍
u
2
(
u
x
2
-
u
2
)
(
q
(
x
0
,
s
)
,
s
)
d
s
-
a
6
u
0
u
0
x
(
x
0
)
)
.
Before completing the proof, we want the following technical lemma.Lemma 5 (see [15]).
Suppose thatΦ is a twice continuous differential satisfying
(29)
Φ
′′
(
t
)
≥
C
0
Φ
′
(
t
)
Φ
(
t
)
,
t
>
0
,
C
0
>
0
,
Φ
(
0
)
>
0
,
Φ
′
(
0
)
>
0
.
ThenΦ
(
t
) blows up in finite time. Moreover the blow-up time T can be estimated in terms of the initial datum as
(30)
T
≤
max
⁡
{
2
C
0
Φ
(
0
)
,
Φ
(
0
)
Φ
′
(
0
)
}
.LetΦ
(
t
)
=
∫
0
t
‍
u
2
(
u
x
2
-
u
2
)
(
q
(
x
0
,
s
)
,
s
)
d
s
-
u
0
u
0
x
(
x
0
); then (28) is an equation of type (29) with C
0
=
a
2
/
18. The proof is complete by applying Lemma 5.When we change the signs ofu
0
(
x
0
) and y
0
(
x
) in Theorem 4, similarly, we have the following blow-up criterion.Theorem 6.
Ifa
>
3
b, suppose that u
0
∈
H
s
(
ℝ
)
,
s
>
(
3
/
2
), and there exists a x
0
∈
ℝ such that u
0
(
x
0
)
≤
0 and y
0
(
x
0
)
=
(
1
-
∂
x
2
)
u
0
(
x
0
)
=
0,
(31)
y
0
(
x
)
≤
0
(
≢
0
)
f
o
r
x
∈
(
-
∞
,
x
0
)
,
y
0
(
x
)
≥
0
(
≢
0
)
f
o
r
x
∈
(
x
0
,
∞
)
.
Then the corresponding solution u
(
x
,
t
) to the modified Novikov equation (1) with u
0
(
x
) as the initial datum blows up in finite time.



---

*Source: 243616-2014-04-30.xml*