# On the Study of Global Solutions for a Nonlinear Equation

Authors: Haibo Yan; Ls Yong
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

The well-posedness of global strong solutions for a nonlinear partial differential equation including the Novikov equation is established provided that its initial valuev
0
(
x
) satisfies a sign condition and v
0
(
x
)
∈
H
s
(
R
) with s
>
3
/
2. If the initial value v
0
(
x
)
∈
H
s
(
R
)
(
1
≤
s
≤
3
/
2
) and the mean function of (
1
-
∂
x
2
)
v
0
(
x
) satisfies the sign condition, it is proved that there exists at least one global weak solution to the equation in the space v
(
t
,
x
)
∈
L
2
(
[
0
,
+
∞
)
,
H
s
(
R
)
) in the sense of distribution and v
x
∈
L
∞
(
[
0
,
+
∞
)
×
R
).


---

## Body

## 1. Introduction

Recently, Wu [1] obtained the existence of local solutions in the space C
(
[
0
,
T
)
;
H
s
(
R
)
)
∩
C
1
(
[
0
,
T
)
;
H
s
-
1
(
R
)
) with s
>
3
/
2 for the following nonlinear equation:
(1)
v
t
-
v
t
x
x
+
k
v
m
v
x
+
(
m
+
3
)
v
m
+
1
v
x
=
(
m
+
2
)
v
m
v
x
v
x
x
+
v
m
+
1
v
x
x
x
+
λ
(
v
-
v
x
x
)
,
where m
≥
0 is a natural number, k
≥
0, and λ is a constant. Letting m
=
0 and λ
=
0, (1) becomes the Camassa-Holm equation [2]. If m
=
1
,
k
=
0, and λ
=
0, (1) reduces to the Novikov equation [3].A lot of works have been carried out to study various dynamic properties for the Camassa-Holm and the Novikov equations. Xin and Zhang [4] proved that there exists a global weak solution for the Camassa-Holm equation in the space H
1
(
R
) without the assumption of sign conditions on the initial value. Coclite et al. [5] investigated the global weak solutions for a generalized hyperelastic rod wave equation or a generalized Camassa-Holm equation. It is shown in Constantin and Escher [6] that the blowup occurs in the form of breaking waves; namely, the solution remains bounded but its slope becomes unbounded in finite time. After wave breaking, the solution can be continued uniquely either as a global conservative weak solution [7] or a global dissipative solution [8–10]. The periodic and the nonperiodic Cauchy problems for the Novikov equation were discussed by Grayshan [11] in the Sobolev space. Using the Galerkin-type approximation method, Himonas and Holliman [12] established the well-posedness for the Novikov model in the Sobolev space H
s
(
R
) with s
>
3
/
2 on both the line and the circle. The scattering theory was employed in Hone et al. [13] to find nonsmooth explicit soliton solutions with multiple peaks for the Novikov equation. Wu and Zhong [14] proved the existence of local strong and weak solutions for a generalized Novikov equation.The objective of this work is to study (1) with k
=
0. Namely, we investigate the problem
(2)
v
t
-
v
t
x
x
+
(
m
+
3
)
v
m
+
1
v
x
=
(
m
+
2
)
v
m
v
x
v
x
x
+
v
m
+
1
v
x
x
x
+
λ
(
v
-
v
x
x
)
,
v
(
0
,
x
)
=
v
0
(
x
)
,
where m
,
k, and λ are described in (1). Assuming that the initial value v
0
(
x
) satisfies a sign condition and v
0
(
x
)
∈
H
s
(
R
)
,
s
>
3
/
2, we will show that there exists a unique global strong solution in the Sobolev space C
(
[
0
,
∞
)
;
H
s
(
R
)
)
∩
C
1
(
[
0
,
∞
)
;
H
s
-
1
(
R
)
). If the initial value v
0
(
x
)
∈
H
s
(
R
)
(
1
≤
s
≤
3
/
2
) and the mean function of (
1
-
∂
x
2
)
v
0
(
x
) satisfies the sign condition, it is shown that there exists at least one global weak solution to the equation in the space v
(
t
,
x
)
∈
L
2
(
[
0
,
+
∞
)
,
H
s
(
R
)
) in the sense of distribution and v
x
∈
L
∞
(
[
0
,
+
∞
)
×
R
).The structure of this paper is as follows. The main results are given in Section2. Several lemmas are given in Section 3. Section 4 establishes the proof of the main results.


## 2. Main Results

We define(3)
ϕ
(
x
)
=
{
e
1
/
(
x
2
-
1
)
,
|
x
|
<
1
,
0
,
|
x
|
≥
1
,
and let ϕ
ɛ
(
x
)
=
ɛ
-
1
/
4
ϕ
(
ɛ
-
1
/
4
x
) with 0
<
ɛ
<
1
/
4. For the convolution v
ɛ
0
=
ϕ
ɛ
⋆
v
0, we know that v
ɛ
0
∈
C
∞ for any v
0
∈
H
s with s
>
0. Notation (
1
-
∂
x
2
)
v
∈
N
+
(
R
) (or equivalently (
1
-
∂
x
2
)
v
∈
N
-
(
R
)) means that the mean function of (
1
-
∂
x
2
)
v is nonnegative; namely, (
1
-
∂
x
2
)
v
⋆
ϕ
ɛ
≥
0 (or equivalently (
1
-
∂
x
2
)
v
⋆
ϕ
ɛ
≤
0) for an arbitrary sufficiently small ɛ
>
0. For T
>
0 and nonnegative number s, we let C
(
[
0
,
T
)
;
H
s
(
R
)
) denote the Frechet space of all continuous H
s-valued functions on [
0
,
T
) and write Λ
=
(
1
-
∂
x
2
)
1
/
2.We state the result of global strong solutions for problem (2).Theorem 1.
Letv
0
(
x
)
∈
H
s
(
R
)
,
s
>
3
/
2, and (
1
-
∂
x
2
)
v
0
≥
0 for all x
∈
R or (
1
-
∂
x
2
)
v
0
≤
0 for all x
∈
R. Then problem (2) has a unique strong solution satisfying
(4)
v
(
t
,
x
)
∈
C
(
[
0
,
∞
)
;
H
s
(
R
)
)
∩
C
1
(
[
0
,
∞
)
;
H
s
-
1
(
R
)
)
.Definition 2.
A functionv
(
t
,
x
)
∈
L
2
(
[
0
,
+
∞
)
,
H
s
(
R
)
) is called a global weak solution to problem (2) if for every T
>
0 and all φ
(
t
,
x
)
∈
C
0
∞
(
[
0
,
T
]
×
R
), it holds that
(5)
∫
0
T
‍
∫
R
‍
[
v
t
-
v
t
x
x
+
(
m
+
3
)
v
m
+
1
v
x
-
(
m
+
2
)
v
m
v
x
v
x
x
-
v
m
+
1
v
x
x
x
-
λ
(
v
-
v
x
x
)
]
φ
(
t
,
x
)
d
x
d
t
=
0
with v
(
0
,
x
)
=
v
0
(
x
).Now we give the main result of global weak solution for problem (2).Theorem 3.
Letv
0
(
x
)
∈
H
s
(
R
)
,
1
≤
s
≤
3
/
2, (
1
-
∂
x
2
)
v
0
∈
N
+
(
R
) (or equivalently (
1
-
∂
x
2
)
v
0
∈
N
-
(
R
)). Then problem (2) has a unique global weak solution v
(
t
,
x
)
∈
L
2
(
[
0
,
+
∞
)
,
H
s
(
R
)
) in the sense of distribution and v
x
∈
L
∞
(
[
0
,
+
∞
)
×
R
).


## 3. Several Lemmas

Lemma 4 (see [1]).
Letv
0
(
x
)
∈
H
s
(
R
) with s
>
3
/
2. Then the Cauchy problem (2) has a unique local solution
(6)
v
(
t
,
x
)
∈
C
(
[
0
,
T
)
;
H
s
(
R
)
)
∩
C
1
(
[
0
,
T
)
;
H
s
-
1
(
R
)
)
,
where T
>
0 depends on ∥
v
0
∥
H
s
(
R
).Using the first equation of system (2) derives
(7)
d
d
t
∫
R
‍
(
v
2
+
v
x
2
)
d
x
=
2
λ
∫
R
‍
(
v
2
+
v
x
2
)
d
x
,
which yields the conservation law
(8)
∫
R
‍
(
v
2
+
v
x
2
)
d
x
=
∫
R
‍
(
v
0
2
+
v
0
x
2
)
d
x
+
2
λ
∫
0
t
∫
R
(
v
2
+
v
x
2
)
d
x
d
t
.Lemma 5 (see [1]).
Lets
>
3
/
2 and the function v
(
t
,
x
) is a solution of problem (2) and the initial data v
0
(
x
)
∈
H
s. Then the following inequalities hold:
(9)
∥
v
∥
H
1
2
≤
∫
R
‍
(
v
2
+
v
x
2
)
d
x
≤
∫
R
‍
(
v
0
2
+
v
0
x
2
)
d
x
,
i
f
λ
≤
0
.
∥
v
∥
H
1
2
≤
∫
R
‍
(
v
2
+
v
x
2
)
d
x
≤
e
2
λ
t
∫
R
‍
(
v
0
2
+
v
0
x
2
)
d
x
,
i
f
λ
>
0
.
Forq
∈
(
0
,
s
-
1
], there is a constant c such that
(10)
∫
R
‍
(
Λ
q
+
1
v
)
2
d
x
≤
∫
R
‍
(
Λ
q
+
1
v
0
)
2
d
x
+
c
∫
0
t
‍
∥
v
∥
H
q
+
1
2
(
|
λ
|
+
(
∥
v
∥
L
∞
m
-
1
+
∥
v
∥
L
∞
m
)
∥
v
x
∥
L
∞
+
∥
v
∥
L
∞
m
-
1
∥
v
x
∥
L
∞
2
)
d
τ
.
Forq
∈
[
0
,
s
-
1
], there is a constant c such that
(11)
∥
v
t
∥
H
q
≤
c
∥
v
∥
H
q
+
1
(
|
λ
|
+
(
∥
v
∥
L
∞
m
-
1
+
∥
v
∥
L
∞
m
)
∥
v
∥
H
1
+
∥
v
∥
L
∞
m
∥
v
x
∥
L
∞
+
∥
v
∥
L
∞
m
-
1
∥
v
x
∥
L
∞
2
)
.Consider the differential equation(12)
p
t
=
v
m
+
1
(
t
,
p
)
,
t
∈
[
0
,
T
)
,
p
(
0
,
x
)
=
x
,
where v
(
t
,
x
) is the solution of problem (2) and T is the maximal existence time of the solution.Lemma 6.
Letv
0
∈
H
s
(
R
)
,
s
≥
3, and let T
>
0 be the maximal existence time of the solution to problem (2). Then system (12) has a unique solution p
(
t
,
x
)
∈
C
1
(
[
0
,
T
)
×
R
). Moreover, the map p
(
t
,
·
) is an increasing diffeomorphism of R with p
x
(
t
,
x
)
>
0 for (
t
,
x
)
∈
[
0
,
T
)
×
R.Proof.
From Lemma4, we know that there exists a unique solution
(13)
v
(
t
,
x
)
∈
C
(
[
0
,
T
)
;
H
s
(
R
)
)
∩
C
1
(
[
0
,
T
)
;
H
s
-
1
(
R
)
)
.
The Sobolev imbedding theorem derivesH
s
(
R
)
∈
C
1
(
R
). This means that two functions v
(
t
,
x
) and v
x
(
t
,
x
) are bounded, Lipschitz in space and C
1 in time. Using the existence and uniqueness theorem of ordinary differential equations, we derive that problem (12) has a unique solution p
(
t
,
x
)
∈
C
1
(
[
0
,
T
)
×
R
).
Differentiating (12) with respect to x gives rise to
(14)
d
d
t
p
x
=
(
m
+
1
)
v
m
v
x
(
t
,
p
)
p
x
,
t
∈
[
0
,
T
)
,
p
x
(
0
,
x
)
=
1
,
from which we obtain
(15)
p
x
(
t
,
x
)
=
exp
⁡
(
∫
0
t
‍
(
m
+
1
)
v
m
v
x
(
τ
,
p
(
τ
,
x
)
)
d
τ
)
.
For every T
′
<
T, applying the Sobolev imbedding theorem results in
(16)
sup
⁡
(
τ
,
x
)
∈
[
0
,
T
′
)
×
R
|
v
x
(
τ
,
x
)
|
<
∞
.
Therefore, we know that there exists a constantM
>
0 such that p
x
(
t
,
x
)
≥
e
-
M
t for (
t
,
x
)
∈
[
0
,
T
)
×
R. The proof is completed.Lemma 7.
Letv
0
∈
H
s with s
≥
3, and let T
>
0 be the maximal existence time of the problem (2); it holds that
(17)
y
(
t
,
p
(
t
,
x
)
)
p
x
2
(
t
,
x
)
=
y
0
(
x
)
e
∫
0
t
‍
(
m
v
m
v
x
+
λ
)
d
τ
,
where (
t
,
x
)
∈
[
0
,
T
)
×
R and y
:
=
v
-
v
x
x.Proof.
We have(18)
d
d
t
[
y
(
t
,
p
(
t
,
x
)
)
p
x
2
(
t
,
x
)
]
=
y
t
p
x
2
+
2
y
p
x
p
x
t
+
y
x
p
t
p
x
2
=
y
t
p
x
2
+
2
y
(
m
+
1
)
v
m
v
x
p
x
2
+
v
m
+
1
y
x
p
x
2
=
[
y
t
+
(
m
+
2
)
v
m
v
x
y
+
y
x
v
m
+
1
]
p
x
2
+
m
v
m
v
x
y
p
x
2
=
[
v
m
+
1
v
t
-
v
t
x
x
+
(
m
+
2
)
v
m
v
x
(
v
-
v
x
x
)
+
v
m
+
1
(
v
x
-
v
x
x
x
)
-
λ
(
v
-
v
x
x
)
]
p
x
2
+
(
m
v
m
v
x
+
λ
)
y
p
x
2
=
[
v
t
-
v
t
x
x
+
(
m
+
3
)
v
m
+
1
v
x
-
(
m
+
2
)
v
m
v
x
v
x
x
-
v
m
+
1
v
x
x
x
-
λ
(
v
-
v
x
x
)
]
p
x
2
+
(
m
v
m
v
x
+
λ
)
y
p
x
2
=
(
m
v
m
v
x
+
λ
)
y
p
x
2
,
from which we have
(19)
y
(
t
,
p
(
t
,
x
)
)
p
x
2
(
t
,
x
)
=
p
x
(
0
,
x
)
y
0
(
x
)
e
∫
0
t
‍
(
m
v
m
v
x
+
λ
)
d
τ
.
Using p
x
(
0
,
x
)
=
1 completes the proof.Lemma 8.
Ifv
0
∈
H
s
(
R
)
,
s
≥
3
/
2, (
1
-
∂
x
2
)
v
0
≥
0 or (
1
-
∂
x
2
)
v
0
≤
0, then the solution of problem (2) satisfies
(20)
∥
v
x
∥
L
∞
≤
∥
v
∥
L
∞
.Proof.
We only need to prove this lemma for the casev
0
-
v
0
x
x
≥
0 since the proof of the other case (
1
-
∂
x
2
)
v
0
≤
0 is similar. It follows from Lemmas 6 and 7 that v
-
v
x
x
≥
0. Letting ξ
(
t
,
x
)
=
v
-
v
x
x, we have
(21)
v
=
1
2
e
-
x
∫
-
∞
x
‍
e
η
ξ
(
t
,
η
)
d
η
+
1
2
e
x
∫
x
∞
‍
e
-
η
ξ
(
t
,
η
)
d
η
,
which derives
(22)
∂
x
v
(
t
,
x
)
=
-
1
2
(
e
-
x
∫
-
∞
x
‍
e
η
ξ
(
t
,
η
)
d
η
+
e
x
∫
x
∞
‍
e
-
η
ξ
(
t
,
η
)
d
η
)
+
e
x
∫
x
∞
‍
e
-
η
ξ
(
t
,
η
)
d
η
=
-
v
(
t
,
x
)
+
e
x
∫
x
∞
‍
e
-
η
ξ
(
t
,
η
)
d
η
≥
-
v
(
t
,
x
)
.
On the other hand, we have
(23)
∂
x
v
(
t
,
x
)
=
1
2
(
e
-
x
∫
-
∞
x
‍
e
η
ξ
(
t
,
η
)
d
η
+
e
x
∫
x
∞
‍
e
-
η
ξ
(
t
,
η
)
d
η
)
-
e
-
x
∫
-
∞
x
‍
e
η
ξ
(
t
,
η
)
d
η
=
v
(
t
,
x
)
-
e
-
x
∫
-
∞
x
‍
e
η
ξ
(
t
,
η
)
d
η
≤
v
(
t
,
x
)
.
The inequalities (22) and (23) derive that inequality (20) is valid.Lemma 9.
Fors
>
0
,
u
∈
H
s
(
R
), and u
ɛ
=
ϕ
ɛ
⋆
u, it holds that
(24)
∥
u
ɛ
x
∥
L
∞
≤
c
∥
u
x
∥
L
∞
,
∥
u
ɛ
∥
H
q
≤
c
,
i
f
q
≤
s
,
∥
u
ɛ
∥
H
q
≤
c
ɛ
(
s
-
q
)
/
4
,
i
f
q
>
s
,
∥
u
ɛ
-
u
∥
H
q
≤
c
ɛ
(
s
-
q
)
/
4
,
i
f
q
≤
s
,
∥
u
ɛ
-
u
∥
H
s
=
o
(
1
)
,
where c is a constant independent of ɛ.The proof of this lemma can be found in [15, 16].From Lemma4, it derives that the Cauchy problem
(25)
v
t
-
v
t
x
x
=
-
(
m
+
3
)
v
m
+
1
v
x
+
(
m
+
2
)
v
m
v
x
v
x
x
+
v
m
+
1
v
x
x
x
+
λ
(
v
-
v
x
x
)
=
-
m
+
3
m
+
2
(
v
m
+
2
)
x
+
1
m
+
2
∂
x
3
(
v
m
+
2
)
-
(
m
+
1
)
∂
x
(
v
m
v
x
2
)
+
v
m
v
x
v
x
x
+
λ
(
v
-
v
x
x
)
,
v
(
0
,
x
)
=
v
ε
0
(
x
)
,
has a unique solution v depending on the parameter ɛ. We write v
ɛ
(
t
,
x
) to represent the solution of problem (25). Using Lemma 4 derives that v
ɛ
(
t
,
x
)
∈
C
∞
(
[
0
,
T
)
,
H
∞
(
R
)
) since v
ε
0
(
x
)
∈
C
0
∞
(
R
).Lemma 10.
Provided thatv
0
∈
H
s
(
R
)
,
1
≤
s
≤
3
/
2, and (
1
-
∂
x
2
)
v
0
∈
N
+
(
R
) (or equivalently (
1
-
∂
x
2
)
v
0
∈
N
-
(
R
)), then there exists a constant c
>
0 independent of ɛ and t such that the solution of problem (25) satisfies
(26)
∥
v
ɛ
x
∥
L
∞
≤
c
e
c
t
.Proof.
Using Lemmas5 and 9, if v
0
∈
H
s
(
R
) with 1
≤
s
≤
3
/
2, we have
(27)
∥
v
ɛ
∥
L
∞
(
R
)
≤
c
∥
v
ɛ
∥
H
1
(
R
)
≤
c
e
c
t
∥
v
ɛ
0
∥
H
1
(
R
)
≤
c
e
c
t
,
where c is independent of ɛ and t.
From Lemma8, we have
(28)
∥
v
ɛ
x
∥
L
∞
(
R
)
≤
∥
v
ɛ
∥
L
∞
(
R
)
,
which completes the proof.


## 4. Proof of Main Results

Proof of Theorem1.
Since∥
v
∥
L
∞
(
R
)
≤
c
∥
v
∥
H
1
(
R
)
≤
c
e
c
t and taking q
+
1
=
s in inequality (10), we have
(29)
∥
v
∥
H
s
2
≤
∥
v
0
∥
H
s
2
+
c
∫
0
t
‍
e
c
τ
∥
v
∥
H
s
2
(
∥
v
x
∥
L
∞
+
∥
v
x
∥
L
∞
2
)
d
τ
,
from which we obtain
(30)
∥
v
∥
H
s
≤
∥
v
0
∥
H
s
e
c
∫
0
t
‍
e
c
τ
(
∥
v
x
∥
L
∞
+
∥
v
x
∥
L
∞
2
)
d
τ
.
Applying Lemma 8 yields
(31)
∥
v
∥
H
s
≤
∥
v
0
∥
H
s
c
e
e
c
t
,
from which we complete the proof of Theorem 1.Provided that1
≤
s
≤
3
/
2, for problem (25), applying Lemmas 5, 8, and 10, and the Gronwall’s inequality, we obtain the inequalities
(32)
∥
v
ɛ
∥
H
1
≤
∥
v
ɛ
0
∥
H
1
≤
c
e
c
t
,
∥
v
ɛ
∥
H
q
≤
c
∥
v
ɛ
0
∥
H
q
exp
⁡
[
∫
0
t
‍
(
∥
v
ɛ
x
∥
+
∥
v
ɛ
x
∥
L
∞
2
)
d
τ
]
≤
c
e
e
c
t
,
∥
u
ɛ
t
∥
H
r
≤
c
∥
u
ɛ
∥
H
r
+
1
(
1
+
e
c
t
)
≤
c
(
1
+
e
c
t
)
,
where q
∈
(
0
,
s
]
,
r
∈
[
0
,
s
-
1
], and c is a constant independent of t and ɛ. Using the Aubin compactness theorem, we know that that there is a subsequence {
v
ε
n
} of {
v
ɛ
} such that {
v
ε
n
} and their temporal derivatives {
v
ε
n
t
} converge weakly to a function v
(
t
,
x
) and its derivative v
t in the space L
2
(
[
0
,
T
]
,
H
s
(
R
)
) and L
2
(
[
0
,
T
]
,
H
s
-
1
(
R
)
), respectively, where T is an arbitrary fixed positive number. In addition, for any real number M
1
>
0, {
v
ε
n
} converges strongly to the function v in the space L
2
(
[
0
,
T
]
,
H
q
(
-
M
1
,
M
1
)
) for q
∈
(
0
,
s
] and {
v
ε
n
t
} converges strongly to v
t in the space L
2
(
[
0
,
T
]
,
H
r
(
-
M
1
,
M
1
)
) for r
∈
[
0
,
s
-
1
].Proof of Theorem3.
For an arbitrary fixedT
>
0, using Lemma 10, we know that {
v
ε
n
x
}
(
ε
n
→
0
) is bounded in the space L
∞. Therefore, we derive that the sequences {
v
ε
n
}, {
v
ε
n
x
}, {
v
ε
n
x
2
}, and {
v
ε
n
x
3
} converge weakly to v, v
x, v
x
2, and v
x
3 in L
2
(
[
0
,
T
]
,
H
r
(
-
R
1
,
R
1
)
) for any r
∈
[
0
,
s
-
1
), separately. Applying the identity v
m
(
v
x
2
)
x
=
(
v
m
v
x
2
)
x
-
(
v
m
)
x
v
x
2, we conclude that v satisfies the equation
(33)
-
∫
0
T
‍
∫
R
‍
v
(
φ
t
-
φ
x
x
t
)
d
x
d
t
=
∫
0
T
‍
∫
R
‍
[
(
m
+
3
m
+
2
v
m
+
2
+
(
m
+
1
)
v
m
v
x
2
)
φ
x
-
1
m
+
2
v
m
+
2
φ
x
x
x
-
1
2
v
m
v
x
2
φ
x
-
m
2
v
m
-
1
v
x
3
φ
+
λ
v
(
φ
-
φ
x
x
)
]
d
x
d
t
,
where φ
(
t
,
x
)
∈
C
0
∞
(
[
0
,
T
]
×
R
). We know that Y
=
L
1
(
[
0
,
T
]
×
R
) is a separable Banach space and {
v
ε
n
x
} is a bounded sequence in the dual space Y
*
=
L
∞
(
[
0
,
T
]
×
R
) of Y. Thus, there exists a subsequence of {
v
ε
n
x
}, still denoted by {
v
ε
n
x
}, weakly star convergent to a function u in L
∞
(
[
0
,
T
]
×
R
). Since {
v
ε
n
x
} weakly converges to v
x in L
2
(
[
0
,
T
]
×
R
), it derives that v
x
=
u almost everywhere. Therefore, we obtain v
x
∈
L
∞
(
[
0
,
T
]
×
R
). Since T
>
0 is an arbitrary number, we complete the proof of existence of global weak solutions to problem (2).



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*Source: 808214-2014-04-14.xml*