# A Note on the Generalized Camassa-Holm Equation

Authors: Yun Wu; Ping Zhao
Journal: Journal of Function Spaces
 (2014)

---

## Abstract

We study the generalized Camassa-Holm equation which contains the Camassa-Holm (CH) equation and Novikov equation as special cases with the periodic boundary condition. We get a blow-up scenario and obtain the global existence of strong and weak solutions under suitable assumptions, respectively. Then, we construct the periodic peaked solutions and apply them to prove the ill-posedness inH
s with s
<
3
/
2.


---

## Body

## 1. Introduction

In this paper we study the global strong and weak solutions to the generalized Camassa-Holm equation with periodic boundary condition:(1)
u
t
-
u
x
x
t
+
n
+
2
u
n
u
x
=
(
n
+
1
)
u
n
-
1
u
x
u
x
x
+
u
n
u
x
x
x
,
t
≥
0
,
x
∈
T
,
(2)
u
0
,
x
=
u
0
x
,
x
∈
T
,
where n
≥
1, n
∈
N, and T
=
R
/
2
π
Z.Whenn
=
1, (1) reduces to the the well-known CH equation:
(3)
u
t
-
u
x
x
t
+
3
u
u
x
=
2
u
x
u
x
x
+
u
u
x
x
x
.
The CH equation was derived independently by Fokas and Fuchssteiner in [1] and by Camassa and Holm in [2]. Fokas and Fuchssteiner derived (3) in studying completely integrable generalizations of the KdV equation with bi-Hamiltonian structures, while Camassa and Holm proposed (3) to describe the unidirectional propagation of shallow water waves over a flat bottom.As shown in [2], the CH equation is completely integrable and possesses an infinite number of conservation laws. Moreover, the CH equation is such an equation that exhibits both phenomena of soliton interaction (peaked soliton solutions) and wave breaking (the solution remains bounded while its slope becomes unbounded in finite time [3]), while the KdV equation does not model breaking waves [4]. In fact, wave breaking is one of the most intriguing long-standing problems of water wave theory [5]. The essential feature of CH should be pointed out: the fact that the traveling waves have a peak at their crest is exactly like for the waves of greatest height solutions of the governing equations for water waves (see [6–8] for the details).From a mathematical point of view the Camassa-Holm equation is well studied and a series of achievements had been made. Constantin [9] and Misiołek [10] investigated the Cauchy problem for the periodic Camassa-Holm equation. Constantin et al. [3, 11–14] studied the wave breaking of the Cauchy problem for the CH equation. Recently, Jiang et al. gave a new and direct proof for McKean’s theorem in [15]. Xin and Zhang [16] proved that (3) has global weak solutions for initial data in H
1
(
R
). Bressan and Constantin developed a new approach to the analysis of the CH equation and proved the existence of the global conservative and dissipative solutions in [17, 18]. Holden and Raynaud [19, 20] also obtained the global conservative and dissipative solutions. The large time behavior of the CH equation was firstly established in [21]. In [22], Himonas et al. studied the persistence properties and infinite propagation speed for the CH equation.In 2009, Novikov [23] found a new integrable equation:
(4)
u
t
-
u
x
x
t
+
4
u
2
u
x
=
3
u
u
x
u
x
x
+
u
2
u
x
x
x
.
It is derived that (4) possesses a bi-Hamiltonian structure and an infinite sequence of conserved quantities and admits exact peaked solutions u
(
t
,
x
)
=
±
c
e
-
|
x
-
c
t
| with c
>
0 [24, 25], as well as the explicit formulas for multipeakon solutions [25, 26].By using the Littlewood-Paley decomposition and Kato’s theory, the well-posedness of the Novikov equation has been studied in Besov spacesB
p
,
r
s
(
R
) and in the Sobolev space H
s
(
R
) (see [27, 28]). Wu and Yin [29] established some results on the existence and uniqueness of global weak solutions to the Novikov equation. Jiang and Ni [30] established some results about blow-up phenomena of the strong solution to the Cauchy problem for (4). For the periodic boundary condition case, Tıg
˘lay [31] proved that for s
>
5
/
2 the periodic Novikov equation is locally well-posed in H
s
(
T
). Later the range of regularity index of local well-posedness was extended to s
>
3
/
2 in [32]; furthermore, it is shown that the solution maps for both periodic boundary value problem and Cauchy problem of the Novikov equation are not uniformly continuous from any bounded subset in H
s into C
(
[
0
,
T
]
;
H
s
). When s
<
3
/
2, Grayshan [33] proved that the properties of the solution map for (4) are not (globally) uniformly continuous in Sobolev spaces H
s. For the nonuniform dependence and ill-posedness results in Besov spaces, we refer to [34–37].In this paper we consider the generalized CH equation (1) with n
≥
3. Let y
=
u
-
u
x
x, then (1) takes the form of a quasi-linear evolution equation of hyperbolic type:
(5)
y
t
+
(
n
+
1
)
u
n
-
1
u
x
y
+
u
n
y
x
=
0
,
x
∈
T
,
t
≥
0
.
Applying the operator 1
-
∂
x
2
-
1, (1) can be expressed as the following nonlocal form:
(6)
u
t
+
u
n
u
x
+
1
-
∂
x
2
-
1
∂
x
u
n
+
1
+
2
n
-
1
2
u
n
-
1
u
x
2
+
1
-
∂
x
2
-
1
n
-
1
2
u
n
-
2
u
x
3
=
0
.
From the above equation, we can view (1) as a nonlocal perturbation of the Burgers-type equation:
(7)
u
t
+
u
n
∂
x
u
=
0
.We recall the following results in [38] for (1) and (2).Proposition 1.
Ifs
>
3
/
2 and u
0
∈
H
s, then there exists a T
=
T
(
u
0
)
>
0 such that (1) and (2) have a unique solution u
∈
C
[
0
,
T
]
;
H
s
⋂
‍
C
1
(
[
0
,
T
]
;
H
s
-
1
) which depends continuously on the initial data u
0. Moreover, u satisfies the solution estimate
(8)
u
(
t
)
H
s
≤
2
u
0
H
s
,
0
≤
t
≤
T
=
1
2
n
c
s
u
0
H
s
n
,
where c
s
>
0 is a constant depending on s.In this paper we use the following notations. We use≲ to denote estimates that hold up to some universal constant which may change from line to line but whose meaning is clear from the context. f
≈
g stands for f
≲
g and f
≳
g. All function spaces are over T and we drop T in all function spaces if there is no ambiguity. For linear operators A and B, we denote [
A
,
B
]
=
A
B
-
B
A. The Fourier transform of the function f
(
x
) is defined by f
^
(
ξ
)
=
F
f
(
ξ
)
=
∫
T
‍
e
-
i
x
ξ
f
(
x
)
d
x, ξ
∈
Z. The inverse Fourier transform is given by f
(
x
)
=
F
-
1
f
^
(
x
)
=
(
1
/
2
π
)
∑
ξ
∈
Z
‍
f
^
(
ξ
)
e
i
x
ξ.The operatorD
s
=
1
-
∂
x
2
s
/
2 for any real number s is defined by D
s
f
^
(
ξ
)
=
1
+
ξ
2
s
/
2
f
^
(
ξ
).  H
s
(
T
) is the standard Soblev space on T whose norm is defined by
(9)
f
H
s
2
=
∑
ξ
∈
Z
‍
1
+
ξ
2
s
f
^
2
=
D
s
f
L
2
2
.For eachε
∈
(
0,1
], J
ε stands for the Friedrichs mollifier defined by
(10)
J
ε
f
x
=
j
ε
*
f
x
,
where * stands for the convolution. Here j
ε
(
x
)
=
F
-
1
j
^
(
ε
ξ
) and j
(
x
) is a Schwartz function satisfying 0
≤
j
^
(
ξ
)
≤
1 for all the ξ
∈
R and j
^
(
ξ
)
=
1 for any ξ
∈
[
-
1,1
].We will also use another mollifier. Define(11)
ρ
(
x
)
=
C
exp
⁡
1
x
2
-
1
,
|
x
|
<
1
,
0
,
|
x
|
≥
1
,
where the constant C
>
0 is chosen so that ∫
R
‍
ρ
(
x
)
d
x
=
1. For ε
∈
(
0,1
], we set ρ
ε
(
x
)
=
(
1
/
ε
)
ρ
(
(
1
/
ε
)
x
). To define the mollifier J
~
ε, we first let χ
Ω be the characteristic function on Ω
⊂
R and
(12)
f
~
ε
(
x
)
=
ρ
ε
/
2
*
χ
[
ε
,
2
π
-
ε
]
f
(
x
)
,
x
∈
[
0,2
π
]
.
It follows that supp
⁡
f
~
ε
⊂
[
ε
/
2
,
2
π
-
ε
/
2
] and when f is smooth, f
~
ε
-
χ
[
ε
,
2
π
-
ε
]
f
→
0 as ε
→
0 and x
∈
[
0,2
π
]. Then we can define J
~
ε by
(13)
J
~
ε
f
=
the
unique
periodic
extension
of
f
~
ε
with
period
2
π
.
Obviously, if f
∈
H
s
(
T
), s
>
0, then J
~
ε
f
∈
H
∞
(
T
) and f
-
J
~
ε
f
H
s
→
0 (as ε
→
0). Moreover, if f
(
x
)
≥
0, then J
~
ε
f
≥
0.Now we present our results.Theorem 2.
Letu
0
∈
H
s, s
>
3
/
2, and let T
u
0
⋆ be the maximal existence time of the corresponding solution u to (1) and (2). Then u blows up at T
u
0
⋆
<
∞ if and only if
(14)
limsup
⁡
t
→
T
u
0
⋆
u
x
(
t
)
L
∞
=
∞
.Theorem 3.
Letu
0
∈
H
s, s
>
3
/
2. If u
0
-
∂
x
2
u
0 does not change sign, then the corresponding solution to (1) and (2) exists globally.Theorem 4.
Supposeu
0
∈
H
s with s
∈
1
,
3
/
2 such that (
1
-
∂
x
2
)
u
0 does not change sign. Then (1) and (2) have a unique global weak solution u
∈
L
loc
⁡
∞
(
[
0
,
∞
)
;
H
s
) in the sense of distribution. Moreover, u
x
∈
L
loc
⁡
∞
(
[
0
,
∞
)
×
T
).Whens
<
3
/
2, the solution map is not uniformly continuous, and we establish the ill-posedness as follows. We refer to [33, 36, 39] for the ill-posedness results for the CH equation, Novikov equation, and the b-family equation.Theorem 5.
Ifs
<
3
/
2, then (1) and (2) are ill-posed in H
s in the sense that the solution map is not uniformly continuous from H
s into C
(
[
0
,
T
]
;
H
s
) for any T
>
0. More precisely, there exist two sequences of weak solutions u
m and v
m in H
s of (1) and (2) such that, for any t
∈
[
0
,
T
], there hold the following estimates:
(15)
u
m
(
0
,
·
)
-
v
m
(
0
,
·
)
H
s
≤
C
1
(
s
)
m
,
u
m
(
t
,
·
)
-
v
m
(
t
,
·
)
H
s
≥
C
2
(
s
)
t
3
/
2
-
s
m
s
+
2
|
s
|
+
1
/
2
,
where C
1
(
s
)
,
C
2
(
s
) only depend on s.We outline the rest of the paper. In the next section, we give some preliminaries. We deal with the blow-up criterion and prove Theorems2 and 3 in Section 3. In Section 4, we study the global weak solution and prove Theorem 4. We demonstrate Theorem 5 in Section 5.


## 2. Preliminaries

Rewrite (1) and (2) as follows:
(16)
∂
t
u
+
u
n
∂
x
u
+
F
(
u
)
=
0
,
x
∈
T
,
t
∈
R
+
,
u
(
0
,
x
)
=
u
0
(
x
)
∈
H
s
,
x
∈
T
,
where F
(
u
)
=
F
1
(
u
)
+
F
2
(
u
)
+
F
3
(
u
), with
(17)
F
1
(
u
)
=
D
-
2
∂
x
u
n
+
1
,
F
2
(
u
)
=
2
n
-
1
2
D
-
2
∂
x
u
n
-
1
∂
x
u
2
,
F
3
(
u
)
=
n
-
1
2
D
-
2
u
n
-
2
∂
x
u
3
.The following estimates are useful in our work.Lemma 6 (Kato-Ponce commutator estimate [40]).
Ifs
>
0, f
∈
H
s
⋂
‍
W
1
,
∞, and g
∈
H
s
-
1
⋂
‍
L
∞, then
(18)
D
s
,
f
g
L
2
≤
C
s
D
s
f
L
2
g
L
∞
+
∂
x
f
L
∞
D
s
-
1
g
L
2
.Lemma 7 (see [40] or see the Moser estimate in [41]).
Ifs
>
0, then H
s
⋂
‍
L
∞ is an algebra, and
(19)
f
g
H
s
≤
C
s
(
f
H
s
g
L
∞
+
f
L
∞
g
H
s
)
,
hhhhhhhhhhhhhhihhhhh
∀
f
,
g
∈
H
s
⋂
‍
L
∞
.From the construction of the mollifierj
ε, we know that j
^
ε
(
ξ
)
=
j
^
(
ε
ξ
) and, for any u
∈
H
s and r
≤
s,
(20)
u
-
J
ε
u
H
r
=
o
(
ε
s
-
r
)
.In addition, we have(21)
D
s
J
ε
=
J
ε
D
s
,
(22)
J
ε
f
,
g
L
2
=
f
,
J
ε
g
L
2
,
(23)
J
ε
u
H
s
≤
u
H
s
.Lemma 8 (see [42]).
Letw
(
x
) be a function such that ∂
x
w
L
∞
<
∞. Then, there is a c
>
0 such that, for any f
∈
L
2,
(24)
[
J
ε
,
w
]
∂
x
f
L
2
≤
c
∂
x
w
L
∞
f
L
2
.The following Calderon-Coifman-Meyer type commutator estimate is also useful (see Proposition 4.2, [43]).Lemma 9.
Ifρ
>
3
/
2 and 0
≤
σ
+
1
≤
ρ, then there is a c
>
0 such that
(25)
[
D
σ
∂
x
,
f
]
v
L
2
≤
c
f
H
ρ
v
H
σ
,
∀
f
∈
H
ρ
,
v
∈
H
σ
.


## 3. Blow-Up Criterion and Global Existence of Strong Solutions

The aim of this section is to prove Theorems2 and 3 which show that the solution blows up only when the slope of the wave blows up and the solution exists globally if y
0
=
u
0
-
u
0
x
x does not change sign. Rewrite (1) as the following form:
(26)
u
t
-
u
x
x
t
=
-
(
n
+
2
)
(
u
n
u
x
)
+
∂
x
2
u
n
u
x
-
n
∂
x
(
u
n
-
1
u
x
2
)
+
u
n
-
1
u
x
u
x
x
.
First, we have the following lemma.Lemma 10.
Lets
>
3
/
2, 0
<
q
≤
s
-
1, and u
(
t
,
x
) is a solution to (1) and (2) with u
0
∈
H
s. Then there exists a constant C
s such that
(27)
u
(
t
)
H
q
+
1
2
-
u
(
0
)
H
q
+
1
2
≤
C
s
∫
0
t
‍
u
H
q
+
1
2
u
L
∞
n
-
1
u
x
L
∞
+
u
L
∞
n
-
2
u
x
L
∞
2
d
τ
.Proof.
Since the term-
n
∂
x
(
u
n
-
1
u
x
2
)
+
u
n
-
1
u
x
u
x
x is only in H
s
-
2, we cannot apply D
q to either side of (26) when q
>
s
-
2. So we apply the operator D
q
J
ε to (26), multiply both sides of the resulting equation by 2
D
q
J
ε
u, and integrate over T to obtain
(28)
d
d
t
∫
T
‍
D
q
J
ε
u
2
+
D
q
J
ε
u
x
2
d
x
=
-
2
(
n
+
1
)
∫
T
‍
D
q
J
ε
u
·
D
q
J
ε
u
n
u
x
d
x
-
2
∫
T
‍
D
q
+
1
J
ε
u
·
D
q
+
1
J
ε
u
n
u
x
d
x
+
2
n
∫
T
‍
D
q
J
ε
u
x
·
D
q
J
ε
u
n
-
1
u
x
2
d
x
+
2
∫
T
‍
D
q
J
ε
u
·
D
q
J
ε
u
n
-
1
u
x
u
x
x
d
x
=
N
1
+
N
2
+
N
3
+
N
4
,
where we used ∂
x
2
=
1
-
D
2 and
(29)
∫
T
‍
D
q
J
ε
u
·
D
q
∂
x
2
J
ε
u
n
u
x
d
x
=
∫
T
‍
D
q
J
ε
u
·
D
q
J
ε
u
n
u
x
d
x
-
∫
T
‍
D
q
+
1
J
ε
u
·
D
q
+
1
J
ε
u
n
u
x
d
x
.
We now estimate N
i, 1
≤
i
≤
4, respectively. For N
1, we first note that J
ε is self-adjoint, then commute the operator D
q with u
n, and use (21) and (22) to get
(30)
N
1
=
∫
T
‍
D
q
,
u
n
∂
x
u
D
q
J
ε
2
u
d
x
+
∫
T
‍
J
ε
,
u
n
∂
x
D
q
u
D
q
J
ε
u
d
x
-
1
2
∫
T
‍
u
n
x
D
q
J
ε
u
2
d
x
.
By using the Cauchy-Schwarz inequality, Lemmas6, 7, and 8, integration by parts, and (23), we have
(31)
∫
T
‍
D
q
,
u
n
∂
x
u
D
q
J
ε
2
u
d
x
≲
u
n
H
q
∂
x
u
L
∞
+
∂
x
u
n
L
∞
∂
x
u
H
q
-
1
u
H
q
≲
u
L
∞
n
-
1
u
x
L
∞
u
H
q
2
,
∫
T
‍
J
ε
,
u
n
∂
x
D
q
u
D
q
J
ε
u
d
x
≲
J
ε
,
u
n
D
q
∂
x
u
L
2
u
H
q
≲
u
L
∞
n
-
1
u
x
L
∞
u
H
q
2
,
1
2
∫
T
‍
∂
x
u
n
D
q
J
ε
u
2
d
x
≲
u
n
x
L
∞
u
H
q
2
≲
u
L
∞
n
-
1
u
x
L
∞
u
H
q
2
.
Therefore
(32)
N
1
≲
u
L
∞
n
-
1
u
x
L
∞
u
H
q
2
.
In the same way,N
2 can be estimated as
(33)
N
2
≲
u
L
∞
n
-
1
u
x
L
∞
u
H
q
+
1
2
.
ForN
3, we use the Cauchy-Schwarz inequality, Lemma 7, and (23) to obtain
(34)
N
3
≲
u
H
q
+
1
∂
x
u
n
u
x
H
q
≲
u
H
q
+
1
u
L
∞
n
-
1
u
x
L
∞
u
H
q
+
1
∂
x
u
n
H
q
u
x
L
∞
h
h
h
h
h
h
h
h
+
u
L
∞
n
-
1
u
x
L
∞
u
H
q
+
1
≲
u
L
∞
n
-
1
u
x
L
∞
u
H
q
+
1
2
.
For the estimate for N
4, we have
(35)
N
4
=
2
∫
T
‍
D
q
J
ε
u
·
D
q
J
ε
h
h
h
×
∂
x
(
u
n
-
1
u
x
2
)
-
(
n
-
1
)
u
n
-
2
u
x
3
d
x
∫
T
‍
≲
∫
T
‍
D
q
J
ε
u
x
·
D
q
J
ε
(
u
n
-
1
u
x
2
)
d
x
+
∫
T
‍
D
q
J
ε
u
·
D
q
J
ε
(
u
n
-
2
u
x
3
)
d
x
=
K
1
+
K
2
.
Similar to (34), K
1
≤
u
L
∞
n
-
1
u
x
L
∞
u
H
q
+
1
2, while
(36)
K
2
≲
u
H
q
u
n
-
2
u
x
L
∞
u
x
2
H
q
+
u
n
-
2
u
x
H
q
u
x
L
∞
2
≲
u
H
q
u
L
∞
n
-
2
u
x
L
∞
2
u
x
H
q
+
∂
x
u
n
-
1
H
q
u
x
L
∞
2
≲
u
H
q
u
L
∞
n
-
2
u
x
L
∞
2
u
x
H
q
+
u
H
q
+
1
u
L
∞
n
-
2
u
x
L
∞
2
≲
u
L
∞
n
-
2
u
x
L
∞
2
u
H
q
+
1
2
.
Hence we obtain that
(37)
N
4
≲
u
L
∞
n
-
2
u
x
L
∞
2
u
H
q
+
1
2
.
Combining (32), (33), (34), and (37) yields
(38)
d
d
t
∫
T
‍
D
q
J
ε
u
2
+
D
q
J
ε
u
x
2
d
x
≲
u
H
q
+
1
2
u
L
∞
n
-
1
u
x
L
∞
+
u
L
∞
n
-
2
u
x
L
∞
2
.
Integrating both sides with respect to t results in
(39)
J
ε
u
(
t
)
H
q
+
1
2
-
J
ε
u
(
0
)
H
q
+
1
2
≲
∫
0
t
‍
u
H
q
+
1
2
u
L
∞
n
-
1
u
x
L
∞
+
u
L
∞
n
-
2
u
x
L
∞
2
d
τ
.
Let ε tend to 0; we get (27) and therefore complete the proof of this lemma.Remark 11.
Takeq
′
=
q
+
1 in (27); then for q
′
∈
(
1
,
s
], we have
(40)
u
(
t
)
H
q
′
2
-
u
(
0
)
H
q
′
2
≤
C
s
∫
0
t
‍
u
H
q
′
2
u
L
∞
n
-
1
u
x
L
∞
+
u
L
∞
n
-
2
u
x
L
∞
2
d
τ
.
This estimate will be also used in the proofs of Lemma 20.
Letu be the solution to problem (1) and (2). We define T
u
0
⋆ to be its lifespan,
(41)
T
u
0
⋆
=
sup
⁡
T
>
0
:
u
∈
C
0
,
T
;
H
s
⋂
‍
C
1
0
,
T
;
H
s
-
1
.
Then the following alternative property holds:
(42)
i
T
u
0
⋆
=
∞
,
or
ii
T
u
0
⋆
<
∞
,
lim
⁡
t
→
T
u
0
⋆
u
(
t
,
·
)
H
s
=
∞
.Proof of Theorem2.
From (1) we can deduce that
(43)
d
d
t
∫
T
‍
u
2
+
u
x
2
d
x
=
0
,
which implies that
(44)
u
H
1
=
u
0
H
1
.
Taking q
=
s
-
1 in Lemma 10 and using u
L
∞
≲
u
H
1
=
u
0
H
1
≲
1, we obtain that
(45)
u
H
s
≤
u
0
H
s
exp
⁡
C
s
∫
0
t
‍
u
x
L
∞
+
u
x
L
∞
2
d
τ
.
Hence we know that if limsup
⁡
t
→
T
u
0
⋆
u
x
(
t
)
L
∞ is finite, then u
(
t
)
H
s is bounded and the case (ii) in (42) would not occur, which implies that u can be extended beyond T
u
0
⋆. On the other hand, if u does not blow up, then u
(
t
)
H
s is bounded on [
0
,
T
] for any T
>
0. Since H
s
-
1
↪
L
∞, ∥
u
x
∥
L
∞ is also bounded on [
0
,
T
u
0
⋆
]. Thus we complete the proof.Remark 12.
Actually, we can prove a more precise blow-up criterion for sufficiently regular solutions to (1); that is, if u
0
∈
H
s, with s
≥
2, then the solution u
(
t
,
x
) blows up if and only if
(46)
liminf
⁡
t
→
T
u
0
⋆
inf
⁡
x
∈
T
u
n
-
1
u
x
x
,
t
=
-
∞
.
In fact, multiplying both sides of (5) by y
=
u
-
u
x
x and integrating over T, we have
(47)
1
2
d
d
t
∫
T
‍
y
2
d
x
=
-
n
+
1
∫
T
‍
u
n
-
1
u
x
y
2
d
x
-
∫
T
‍
u
n
y
x
y
d
x
=
-
n
+
2
2
∫
T
‍
u
n
-
1
u
x
y
2
d
x
.
If u
n
-
1
u
x is bounded from below on [
0
,
T
u
0
⋆
)
×
T, that is, for some M
>
0, u
n
-
1
u
x
>
-
M for (
t
,
x
)
∈
[
0
,
T
u
0
⋆
)
×
T, then
(48)
∫
T
‍
y
2
d
x
≤
(
n
+
2
)
M
2
∫
0
t
‍
∫
T
‍
y
2
d
x
+
C
,
where C
>
0. By Gronwall’s inequality, we obtain that the L
2 norm of y is bounded on [
0
,
T
u
0
⋆
) which is equivalent to the boundedness of u
H
2 since u
H
2
≤
y
L
2
≤
2
u
H
2. On the other hand, since s
≥
2 and H
s
-
1 is an algebra, we know that u
n
-
1
u
x
L
∞
≲
u
n
-
1
H
s
u
x
H
s
-
1
≲
u
H
s
n.Remark 13.
The new blow-up criterion (46) is better than the one obtained in Theorem 2, which is quite common for nonlinear hyperbolic PDE (see [5, 44]). For the Camassa-Holm and related equations, the blow-up criterion is often written as
(49)
liminf
⁡
t
→
T
u
0
⋆
inf
⁡
x
∈
T
u
x
t
,
x
=
-
∞
which is different from (46). In fact, if u blows up and u
n
-
1
<
0, then u
x
→
+
∞.
Letq
(
t
,
x
) be the particle curve evolved by the solution; that is, it satisfies
(50)
d
q
(
t
,
x
)
d
t
=
u
n
t
,
q
t
,
x
,
t
∈
0
,
T
u
0
⋆
,
q
(
0
,
x
)
=
x
,
x
∈
R
.
Since u
∈
C
(
[
0
,
T
u
0
⋆
)
;
H
s
)
⋂
‍
C
1
(
[
0
,
T
u
0
⋆
)
;
H
s
-
1
), s
>
3
/
2, we see that u
(
t
,
x
) and u
x
(
t
,
x
) belong to C
1
[
0
,
T
]
×
T for any T
<
T
u
0
⋆. Therefore, for a fixed x, (50) has a unique solution q
(
t
,
x
)
∈
C
1
(
[
0
,
T
]
×
R
). Moreover, we have the following lemma.Lemma 14.
Letu
0
∈
H
s, s
>
3
/
2. The map q
(
t
,
·
) is an increasing diffeomorphism of R, and q
x
(
t
,
x
)
>
0 for (
t
,
x
)
∈
[
0
,
T
u
0
⋆
)
×
R.Proof.
Differentiating (50) with respect to x yields that
(51)
d
q
x
(
t
,
x
)
d
t
=
n
u
n
-
1
u
x
t
,
q
q
x
,
t
∈
0
,
T
u
0
⋆
,
q
x
(
0
,
x
)
=
1
,
x
∈
R
.
Solving the above equation, we obtain
(52)
q
x
(
t
,
x
)
=
exp
⁡
n
∫
0
t
‍
u
n
-
1
u
x
τ
,
q
τ
,
x
d
τ
.
Thus q
x is positive and q
(
t
,
x
) is an increasing diffeomorphism of R before the blow-up time.The following property for the strong solution is important in the proof of global existence.Lemma 15.
Letu
0
∈
H
s, s
≥
3, and u is the solution to (1) and (2). One has the following identity:
(53)
y
(
t
,
q
(
t
,
x
)
)
q
x
(
n
+
1
)
/
n
(
t
,
x
)
=
y
0
(
x
)
,
where (
t
,
x
)
∈
[
0
,
T
u
0
⋆
)
×
R, y
=
u
-
u
x
x, and y
0
=
u
0
-
u
0
x
x. Moreover, if y
0
n
/
(
n
+
1
)
∈
L
1, then
(54)
∫
T
‍
y
n
/
(
n
+
1
)
d
x
=
∫
T
‍
y
0
n
/
(
n
+
1
)
d
x
.Proof.
Differentiatingy
(
t
,
q
(
t
,
x
)
)
q
x
(
n
+
1
)
/
n
(
t
,
x
) with respect to t, we have
(55)
d
d
t
y
(
t
,
q
(
t
,
x
)
)
q
x
(
n
+
1
)
/
n
(
t
,
x
)
=
y
t
q
x
(
n
+
1
)
/
n
+
n
+
1
n
y
q
x
(
(
n
+
1
)
/
n
)
-
1
q
x
t
+
y
x
q
t
q
x
(
n
+
1
)
/
n
=
q
x
(
n
+
1
)
/
n
y
t
+
(
n
+
1
)
y
u
n
-
1
u
x
+
y
x
u
n
=
0
.
Solving the above equation with q
x
(
0
,
x
)
=
1, we obtain (53). Furthermore, under the condition in the lemma, we have
(56)
∫
T
‍
y
t
,
x
n
/
(
n
+
1
)
d
x
=
∫
T
‍
y
t
,
q
t
,
x
n
/
(
n
+
1
)
q
x
(
t
,
x
)
d
x
=
d
d
t
∫
T
‍
y
t
,
q
t
,
x
q
x
(
n
+
1
)
/
n
t
,
x
n
/
(
n
+
1
)
d
x
=
∫
T
‍
y
0
n
/
(
n
+
1
)
d
x
,
which implies (54).Remark 16.
Ifu
0
(
x
) has a compact support in an interval [
a
,
b
], so does y
0
(
x
)
=
(
1
-
∂
x
2
)
u
0
(
x
). Because of (53), we know y
(
t
,
x
) is compactly supported in [
q
(
t
,
a
)
,
q
(
t
,
b
)
] within its lifespan.
We note that the Green’s function of1
-
∂
x
2
-
1 is G
T
=
cosh
⁡
⁡
x
-
2
π
x
/
2
π
-
π
/
2
sinh
⁡
⁡
π for x
∈
T, where [
x
] stands for the integer part of x. For f
∈
L
2
(
T
), we have
(57)
1
-
∂
x
2
-
1
f
x
=
G
T
*
f
(
x
)
=
∫
T
‍
G
T
x
-
η
f
η
d
η
.
A direct computation gives rise to
(58)
u
t
,
x
=
e
x
-
π
4
sinh
⁡
(
π
)
∫
0
x
‍
e
-
ξ
y
(
t
,
ξ
)
d
ξ
+
e
-
x
+
π
4
sinh
⁡
(
π
)
∫
0
x
‍
e
ξ
y
(
t
,
ξ
)
d
ξ
+
e
x
+
π
4
sinh
⁡
(
π
)
∫
x
2
π
‍
e
-
ξ
y
(
t
,
ξ
)
d
ξ
+
e
-
x
-
π
4
sinh
⁡
(
π
)
∫
x
2
π
‍
e
ξ
y
(
t
,
ξ
)
d
ξ
,
u
x
t
,
x
=
e
x
-
π
4
sinh
⁡
(
π
)
∫
0
x
‍
e
-
ξ
y
(
t
,
ξ
)
d
ξ
-
e
-
x
+
π
4
sinh
⁡
(
π
)
∫
0
x
‍
e
ξ
y
(
t
,
ξ
)
d
ξ
+
e
x
+
π
4
sinh
⁡
(
π
)
∫
x
2
π
‍
e
-
ξ
y
(
t
,
ξ
)
d
ξ
-
e
-
x
-
π
4
sinh
⁡
(
π
)
∫
x
2
π
‍
e
ξ
y
(
t
,
ξ
)
d
ξ
.
Therefore, we have
(59)
u
+
u
x
(
t
,
x
)
=
1
2
sinh
⁡
(
π
)
∫
0
2
π
‍
e
(
x
-
ξ
-
2
π
(
x
-
ξ
)
/
2
π
-
π
)
y
(
t
,
ξ
)
d
ξ
,
u
-
u
x
(
t
,
x
)
=
1
2
sinh
⁡
(
π
)
∫
0
2
π
‍
e
(
-
x
+
ξ
+
2
π
(
x
-
ξ
)
/
2
π
+
π
)
y
(
t
,
ξ
)
d
ξ
.Lemma 17.
Ifu
0
∈
H
s, s
>
3
/
2, and y
0
=
(
1
-
∂
x
2
)
u
0
(
x
) does not change sign, then the solution u to (1) and (2) satisfies
(60)
u
x
L
∞
≤
u
L
∞
≲
u
H
1
=
u
0
H
1
,
x
∈
T
,
t
∈
[
0
,
T
u
0
⋆
)
.Proof.
We discuss the following results for the cases
≥
3; the lemma follows by using a simple density argument. By (53) and the positivity of G
T, we know y keeps the sign of y
0, and hence u
=
G
T
*
y keeps the sign of y
0. Therefore, employing (59), we obtain that, for x
∈
T, t
∈
[
0
,
T
u
0
⋆
),
(61)
-
u
≤
u
x
≤
u
,
if
y
0
=
(
1
-
∂
x
2
)
u
0
(
x
)
>
0
,
x
∈
T
,
u
≤
u
x
≤
-
u
,
if
y
0
=
(
1
-
∂
x
2
)
u
0
(
x
)
<
0
,
x
∈
T
.
That is, if y
0 does not change sign, then |
u
x
|
≤
|
u
|. By H
1
↪
L
∞, we obtain the desired estimate.Proof of Theorem3.
Combining Lemma17 and Theorem 2, we have Theorem 3.


## 4. Global Weak Solutions

In this section, we prove that (1) and (2) have a unique global weak solution in lower-order Sobolev space H
s, 1
<
s
≤
3
/
2. First, we establish some estimates for the strong solutions to (1) with s
>
3
/
2.Lemma 18.
Lets
>
3
/
2, 0
<
q
≤
s
-
1, and u
∈
C
(
[
0
,
T
u
0
⋆
)
;
H
s
) is the solution to (1) and (2) with initial value u
0
∈
H
s. Then there is a constant C
s such that, for t
∈
[
0
,
T
u
0
⋆
), one has
(62)
∂
t
u
H
q
≤
C
s
u
H
q
+
1
×
u
L
∞
n
+
u
L
∞
n
-
1
u
x
L
∞
+
u
L
∞
n
-
2
u
x
L
∞
2
.Proof.
By using the operatorD
-
2
=
1
-
∂
x
2
-
1, we can rewrite (26) as
(63)
u
t
=
D
-
2
∂
x
-
n
+
2
n
+
1
(
u
n
+
1
)
-
n
(
u
n
-
1
u
x
2
)
+
D
-
2
∂
x
2
u
n
u
x
+
D
-
2
u
n
-
1
u
x
u
x
x
.
For q
∈
(
0
,
s
-
1
], taking the H
q norm of both sides of (63), it follows that
(64)
u
t
H
q
≤
D
-
2
∂
x
-
n
+
2
n
+
1
(
u
n
+
1
)
-
n
(
u
n
-
1
u
x
2
)
H
q
+
D
-
2
∂
x
2
u
n
u
x
H
q
+
D
-
2
u
n
-
1
u
x
u
x
x
H
q
=
N
1
′
+
N
2
′
+
N
3
′
.
The estimate for N
i
′
(
i
=
1,2
,
3
) is straightforward. By using Lemma 7, we have
(65)
N
1
′
≲
u
n
+
1
H
q
-
1
+
∂
x
(
u
n
)
u
x
H
q
-
1
≲
u
n
+
1
H
q
+
∂
x
(
u
n
)
u
x
H
q
≲
u
L
∞
n
u
H
q
+
1
+
u
L
∞
n
-
1
u
x
L
∞
u
H
q
+
1
.
Using ∂
x
2
=
1
-
D
2, we have
(66)
N
2
′
≲
∂
x
u
n
+
1
H
q
-
2
+
∂
x
u
n
+
1
H
q
≲
u
L
∞
n
u
H
q
+
1
.
Using Lemma 7, it follows that
(67)
N
3
′
≲
u
n
-
1
∂
x
(
u
x
2
)
H
q
-
2
≲
∂
x
u
n
-
1
u
x
2
-
u
x
2
∂
x
u
n
-
1
H
q
-
2
≲
u
n
-
1
u
x
2
H
q
+
u
x
2
∂
x
u
n
-
1
H
q
≲
u
H
q
+
1
u
L
∞
n
-
1
u
x
L
∞
+
u
L
∞
n
-
2
u
x
L
∞
2
.
Inserting (65), (66), and (67) into (64) yields the inequality (62).To show the existence of weak solution to (1) and (2) in lower-order Sobolev space H
s with 1
<
s
≤
3
/
2, we will consider the following problem first:
(68)
∂
t
u
+
u
n
∂
x
u
+
F
(
u
)
=
0
,
x
∈
T
,
t
∈
R
+
,
u
(
0
,
x
)
=
J
~
ε
u
0
(
x
)
∈
H
∞
,
where F
(
u
) is given in (17) and J
~
ε is the mollifier introduced in (11)–(13). It follows from Proposition 1 that, for each ε
>
0, there exists a T
ε
>
0 such that the above problem has a unique solution u
ε
(
t
,
x
)
∈
C
∞
(
[
0
,
T
ε
]
;
H
∞
).Lemma 19.
Letu
0
∈
H
s, 1
<
s
≤
3
/
2. If (
1
-
∂
x
2
)
u
0 does not change sign, then u
ε
(
t
,
x
)
∈
C
(
[
0
,
∞
)
;
H
∞
) and ∥
∂
x
u
ε
∥
L
∞
≤
∥
u
ε
∥
L
∞
≲
∥
u
0
∥
H
1.Proof.
We first note that, by the construction ofJ
~
ε, if f
≥
0 (or f
≤
0), then J
~
ε
f
≥
0 (or J
~
ε
f
≤
0). Thus, if (
1
-
∂
x
2
)
u
0 does not change sign, so does (
1
-
∂
x
2
)
J
~
ε
u
0
=
J
~
ε
(
1
-
∂
x
2
)
u
0. Using the notation u
ε
=
u in (61) implies
(69)
∂
x
u
ε
L
∞
≤
u
ε
L
∞
≲
u
0
H
1
.
Hence Theorem 3 yields that u
ε
(
t
,
x
) is a global solution.Lemma 20.
Ifs
∈
1
,
3
/
2, ε
∈
(
0,1
], and (
1
-
∂
x
2
)
u
0 does not change sign, then for any T
>
0, {
u
ε
}
⊂
L
∞
(
[
0
,
T
]
;
H
q
′
), {
∂
x
u
ε
}
⊂
L
∞
(
[
0
,
T
]
;
H
q
′
-
1
) with q
′
∈
(
1
,
s
], and {
∂
t
u
ε
}
⊂
L
∞
(
[
0
,
T
]
;
H
r
) with r
∈
(
0
,
s
-
1
].Proof.
By using (40), Lemma 19, and the Gronwall’s inequality, we obtain that
(70)
u
ε
H
q
′
≲
J
ε
u
0
H
q
′
×
exp
⁡
C
s
∫
0
t
‍
∂
x
u
ε
L
∞
+
∂
x
u
ε
L
∞
2
d
τ
≲
u
0
H
s
e
C
(
T
)
.
Therefore ∂
x
u
ε
H
q
′
-
1
≲
u
0
H
s
e
C
(
T
), where C
(
T
) only depends on T. Similarly, by using (62) and Lemma 19 we have
(71)
∂
t
u
ε
H
r
≲
u
ε
H
s
≲
u
0
H
s
e
C
(
T
)
.
Thus for q
′
∈
(
1
,
s
], we know {
u
ε
} is uniformly bounded in L
∞
(
[
0
,
T
]
;
H
q
′
), and {
∂
x
u
ε
} is uniformly bounded in L
∞
(
[
0
,
T
]
;
H
q
′
-
1
). Besides, {
∂
t
u
ε
} is uniformly bounded in L
∞
(
[
0
,
T
]
;
H
r
) with r
∈
(
0
,
s
-
1
].


### 4.1. Existence of Global Weak Solution

Now we prove the existence of a global weak solution to problem (16).Proposition 21.
Supposeu
0
∈
H
s
(
R
) with s
∈
1
,
3
/
2 such that (
1
-
∂
x
2
)
u
0 does not change sign. Then (1) and (2) have a global weak solution u
∈
L
loc
⁡
∞
(
[
0
,
∞
)
;
H
s
) in the sense of distribution. Moreover, u
x
∈
L
loc
⁡
∞
(
[
0
,
∞
)
×
T
).Proof.
For an arbitrary fixedT
>
0, from Lemma 19, {
u
ε
k
} and {
∂
x
u
ε
k
} are bounded in L
∞
[
0
,
T
]
×
T. From Lemma 20, when q
′
∈
(
1
,
s
] and r
∈
(
0
,
s
-
1
], we have
(72)
u
ε
⊂
L
2
0
,
T
;
H
q
′
,
{
∂
x
u
ε
}
⊂
L
2
0
,
T
;
H
q
′
-
1
,
{
∂
t
u
ε
}
⊂
L
2
(
[
0
,
T
]
;
H
r
)
.
Let q
′
=
s, r
=
s
-
1. By the Aubin compactness theorem [45], there is a subsequence {
u
ε
k
}
⊂
{
u
ε
} such that u
ε
k converges to some u weakly in L
2
(
[
0
,
T
]
;
H
s
) and ∂
t
u
ε
k
→
∂
t
u, ∂
x
u
ε
k
→
∂
x
u weakly in L
2
(
[
0
,
T
]
;
H
s
-
1
) respectively. Moreover,
(73)
u
ε
k
⟶
u
in
L
2
0
,
T
;
H
q
strongly
for
q
∈
0
,
s
,
∂
t
u
ε
k
⟶
∂
t
u
,
∂
x
u
ε
k
⟶
∂
x
u
in
L
2
[
0
,
T
]
;
H
r
hhhhhhhhhhhhhhhihh
strongly
for
r
∈
(
0
,
s
-
1
)
.
Thus, the sequences {
u
ε
k
n
+
1
}, {
u
ε
k
n
-
1
∂
x
u
ε
k
2
}, {
u
ε
k
n
-
2
∂
x
u
ε
k
3
} converge to u
n
+
1, u
n
-
1
u
x
2, u
n
-
2
u
x
3 strongly in L
2
(
[
0
,
T
]
;
H
r
) for any r
∈
(
0
,
s
-
1
), respectively. Therefore, u
∈
C
(
[
0
,
T
]
;
H
r
) is a solution to (16) in the sense of distribution. By Lemmas 19 and 20, u
∈
L
loc
⁡
∞
(
[
0
,
∞
)
;
H
s
) and u
x
∈
L
loc
⁡
∞
(
[
0
,
∞
)
×
T
).


### 4.2. Uniqueness of the Global Weak Solution

Proposition 22.
Supposeu
0
∈
H
s with s
∈
1
,
3
/
2. If (
1
-
∂
x
2
)
u
0 does not change sign, then the weak solution to (1) and (2) is unique.Proof.
Letu
,
w be two solutions to (16) with the same initial data u
0; then v
=
u
-
w satisfies
(74)
v
t
=
-
u
n
v
x
-
f
v
∂
x
w
-
F
(
u
)
-
F
(
w
)
,
v
(
0
,
x
)
=
0
,
where f
=
∑
i
=
0
n
-
1
‍
w
n
-
1
-
i
u
i. Calculating the L
2 energy of v yields the equation
(75)
1
2
d
d
t
v
(
t
)
L
2
2
=
-
∫
T
‍
u
n
v
x
·
v
d
x
-
∫
T
‍
f
v
∂
x
w
·
v
d
x
-
∫
T
‍
F
(
u
)
-
F
(
w
)
·
v
d
x
.
Since (
1
-
∂
x
2
)
u
0 does not change sign, Lemma 19 implies that ∂
x
u
L
∞
≤
u
L
∞
≲
u
0
H
1 and ∂
x
w
L
∞
≤
w
L
∞
≲
u
0
H
1. Employing the Cauchy-Schwarz inequality and integrating by parts yield
(76)
∫
T
‍
u
n
v
x
·
v
d
x
≤
u
n
-
1
∂
x
u
L
∞
v
L
2
2
≲
v
L
2
2
,
∫
T
‍
f
v
∂
x
w
·
v
d
x
≤
f
L
∞
∂
x
w
L
∞
v
L
2
2
≲
v
L
2
2
.
For the estimate of the L
2 norm of F
(
u
ε
)
-
F
(
w
ε
), we have
(77)
F
1
(
u
)
-
F
1
(
w
)
L
2
≲
u
n
+
1
-
w
n
+
1
H
-
1
F
1
u
.
-
F
1
(
w
)
L
2
≲
∑
i
=
0
n
‍
u
n
-
i
w
i
L
∞
v
L
2
≲
v
L
2
,
F
2
(
u
)
-
F
2
(
w
)
L
2
≲
u
n
-
1
u
x
2
-
w
x
2
H
-
1
+
w
x
2
u
n
-
1
-
w
n
-
1
H
-
1
≲
u
n
-
1
(
u
x
+
w
x
)
L
∞
+
∑
i
=
0
n
-
2
‍
u
n
-
2
-
i
w
i
L
∞
w
x
2
L
∞
v
L
2
≲
v
L
2
,
F
3
(
u
)
-
F
3
(
w
)
L
2
≲
u
n
-
2
u
x
3
-
w
x
3
H
-
2
+
w
x
3
u
n
-
2
-
w
n
-
2
H
-
2
≲
u
n
-
2
·
P
L
∞
+
∑
i
=
0
n
-
3
‍
u
n
-
3
-
i
w
i
L
∞
w
x
3
L
∞
v
L
2
≲
v
L
2
,
where P
=
u
x
2
+
u
x
w
x
+
w
x
2. Hence
(78)
∫
T
‍
F
(
u
)
-
F
(
w
)
·
v
d
x
≲
v
L
2
2
.
Combining these inequalities, we obtain that
(79)
1
2
d
d
t
v
(
t
)
L
2
2
≲
v
L
2
2
.
Using v
(
0
,
x
)
=
0, we have v
=
0, which implies u
=
w.Proof of Theorem4.
With the aid of Propositions21 and 22, we complete the proof of Theorem 4.


## 4.1. Existence of Global Weak Solution

Now we prove the existence of a global weak solution to problem (16).Proposition 21.
Supposeu
0
∈
H
s
(
R
) with s
∈
1
,
3
/
2 such that (
1
-
∂
x
2
)
u
0 does not change sign. Then (1) and (2) have a global weak solution u
∈
L
loc
⁡
∞
(
[
0
,
∞
)
;
H
s
) in the sense of distribution. Moreover, u
x
∈
L
loc
⁡
∞
(
[
0
,
∞
)
×
T
).Proof.
For an arbitrary fixedT
>
0, from Lemma 19, {
u
ε
k
} and {
∂
x
u
ε
k
} are bounded in L
∞
[
0
,
T
]
×
T. From Lemma 20, when q
′
∈
(
1
,
s
] and r
∈
(
0
,
s
-
1
], we have
(72)
u
ε
⊂
L
2
0
,
T
;
H
q
′
,
{
∂
x
u
ε
}
⊂
L
2
0
,
T
;
H
q
′
-
1
,
{
∂
t
u
ε
}
⊂
L
2
(
[
0
,
T
]
;
H
r
)
.
Let q
′
=
s, r
=
s
-
1. By the Aubin compactness theorem [45], there is a subsequence {
u
ε
k
}
⊂
{
u
ε
} such that u
ε
k converges to some u weakly in L
2
(
[
0
,
T
]
;
H
s
) and ∂
t
u
ε
k
→
∂
t
u, ∂
x
u
ε
k
→
∂
x
u weakly in L
2
(
[
0
,
T
]
;
H
s
-
1
) respectively. Moreover,
(73)
u
ε
k
⟶
u
in
L
2
0
,
T
;
H
q
strongly
for
q
∈
0
,
s
,
∂
t
u
ε
k
⟶
∂
t
u
,
∂
x
u
ε
k
⟶
∂
x
u
in
L
2
[
0
,
T
]
;
H
r
hhhhhhhhhhhhhhhihh
strongly
for
r
∈
(
0
,
s
-
1
)
.
Thus, the sequences {
u
ε
k
n
+
1
}, {
u
ε
k
n
-
1
∂
x
u
ε
k
2
}, {
u
ε
k
n
-
2
∂
x
u
ε
k
3
} converge to u
n
+
1, u
n
-
1
u
x
2, u
n
-
2
u
x
3 strongly in L
2
(
[
0
,
T
]
;
H
r
) for any r
∈
(
0
,
s
-
1
), respectively. Therefore, u
∈
C
(
[
0
,
T
]
;
H
r
) is a solution to (16) in the sense of distribution. By Lemmas 19 and 20, u
∈
L
loc
⁡
∞
(
[
0
,
∞
)
;
H
s
) and u
x
∈
L
loc
⁡
∞
(
[
0
,
∞
)
×
T
).


## 4.2. Uniqueness of the Global Weak Solution

Proposition 22.
Supposeu
0
∈
H
s with s
∈
1
,
3
/
2. If (
1
-
∂
x
2
)
u
0 does not change sign, then the weak solution to (1) and (2) is unique.Proof.
Letu
,
w be two solutions to (16) with the same initial data u
0; then v
=
u
-
w satisfies
(74)
v
t
=
-
u
n
v
x
-
f
v
∂
x
w
-
F
(
u
)
-
F
(
w
)
,
v
(
0
,
x
)
=
0
,
where f
=
∑
i
=
0
n
-
1
‍
w
n
-
1
-
i
u
i. Calculating the L
2 energy of v yields the equation
(75)
1
2
d
d
t
v
(
t
)
L
2
2
=
-
∫
T
‍
u
n
v
x
·
v
d
x
-
∫
T
‍
f
v
∂
x
w
·
v
d
x
-
∫
T
‍
F
(
u
)
-
F
(
w
)
·
v
d
x
.
Since (
1
-
∂
x
2
)
u
0 does not change sign, Lemma 19 implies that ∂
x
u
L
∞
≤
u
L
∞
≲
u
0
H
1 and ∂
x
w
L
∞
≤
w
L
∞
≲
u
0
H
1. Employing the Cauchy-Schwarz inequality and integrating by parts yield
(76)
∫
T
‍
u
n
v
x
·
v
d
x
≤
u
n
-
1
∂
x
u
L
∞
v
L
2
2
≲
v
L
2
2
,
∫
T
‍
f
v
∂
x
w
·
v
d
x
≤
f
L
∞
∂
x
w
L
∞
v
L
2
2
≲
v
L
2
2
.
For the estimate of the L
2 norm of F
(
u
ε
)
-
F
(
w
ε
), we have
(77)
F
1
(
u
)
-
F
1
(
w
)
L
2
≲
u
n
+
1
-
w
n
+
1
H
-
1
F
1
u
.
-
F
1
(
w
)
L
2
≲
∑
i
=
0
n
‍
u
n
-
i
w
i
L
∞
v
L
2
≲
v
L
2
,
F
2
(
u
)
-
F
2
(
w
)
L
2
≲
u
n
-
1
u
x
2
-
w
x
2
H
-
1
+
w
x
2
u
n
-
1
-
w
n
-
1
H
-
1
≲
u
n
-
1
(
u
x
+
w
x
)
L
∞
+
∑
i
=
0
n
-
2
‍
u
n
-
2
-
i
w
i
L
∞
w
x
2
L
∞
v
L
2
≲
v
L
2
,
F
3
(
u
)
-
F
3
(
w
)
L
2
≲
u
n
-
2
u
x
3
-
w
x
3
H
-
2
+
w
x
3
u
n
-
2
-
w
n
-
2
H
-
2
≲
u
n
-
2
·
P
L
∞
+
∑
i
=
0
n
-
3
‍
u
n
-
3
-
i
w
i
L
∞
w
x
3
L
∞
v
L
2
≲
v
L
2
,
where P
=
u
x
2
+
u
x
w
x
+
w
x
2. Hence
(78)
∫
T
‍
F
(
u
)
-
F
(
w
)
·
v
d
x
≲
v
L
2
2
.
Combining these inequalities, we obtain that
(79)
1
2
d
d
t
v
(
t
)
L
2
2
≲
v
L
2
2
.
Using v
(
0
,
x
)
=
0, we have v
=
0, which implies u
=
w.Proof of Theorem4.
With the aid of Propositions21 and 22, we complete the proof of Theorem 4.


## 5. Ill-Posedness

In this section we establish the ill-posedness of (1) and (2) in the sense that the solution map is not uniformly continuous from H
s into C
(
[
0
,
T
]
;
H
s
) with s
<
3
/
2 either. Firstly, we show that (1) and (2) possess periodic peaked solutions.


### 5.1. Existence of Periodic Peaked Solutions

Proposition 23.
Letc
>
0 and θ
=
x
-
c
t
-
2
π
(
x
-
c
t
)
/
2
π
-
π. Then u
(
t
,
x
)
=
γ
cosh
⁡
θ is a periodic peaked solution to (1) if and only if γ
=
c
1
/
n
sech
⁡
π. Moreover, u
(
t
,
·
)
∈
H
s with s
<
3
/
2.Proof.
LetH
1
=
∂
t
(
1
-
∂
x
2
)
u, H
2
=
(
1
/
n
+
1
)
∂
x
(
1
-
∂
x
2
)
u
n
+
1, H
3
=
∂
x
u
n
+
1
+
(
(
2
n
-
1
)
/
2
)
u
n
-
1
u
x
2, and H
4
=
(
(
n
-
1
)
/
2
)
u
n
-
2
u
x
3. Then (1) can be expressed as ∑
i
=
1
4
‍
H
i
=
0. Assuming that u
(
x
,
t
)
=
γ
cosh
⁡
θ is the periodic solution to (1), we have
(80)
u
x
=
γ
sinh
⁡
θ
,
(81)
u
x
x
=
γ
cosh
⁡
θ
-
2
γ
sinh
⁡
π
δ
c
t
,
where δ
c
t is the periodic Dirac delta function at x
=
c
t (mod 2
π). Thus
(82)
H
1
=
-
2
c
γ
sinh
⁡
π
δ
c
t
′
.
Direct computation shows that
(83)
u
n
+
1
x
x
=
(
n
+
1
)
γ
n
+
1
×
cosh
⁡
n
+
1
(
θ
)
+
n
cosh
⁡
n
-
1
(
θ
)
sinh
⁡
2
(
θ
)
h
h
h
h
h
-
2
cosh
⁡
n
(
π
)
sinh
⁡
π
δ
c
t
′
cosh
⁡
n
+
1
(
θ
)
+
n
cosh
⁡
n
-
1
(
θ
)
sinh
⁡
2
(
θ
)
=
(
n
+
1
)
γ
n
+
1
×
(
n
+
1
)
cosh
⁡
n
+
1
(
θ
)
-
n
cosh
⁡
n
-
1
(
θ
)
h
h
h
h
h
-
2
cosh
⁡
n
(
π
)
sinh
⁡
π
δ
c
t
′
(
n
+
1
)
cosh
⁡
n
+
1
(
θ
)
-
n
cosh
⁡
n
-
1
(
θ
)
.
Therefore, we have
(84)
H
2
=
1
n
+
1
∂
x
u
n
+
1
-
u
n
+
1
x
x
=
γ
n
+
1
(
-
n
2
-
2
n
)
cosh
⁡
n
(
θ
)
sinh
⁡
θ
h
h
h
h
h
i
+
(
n
2
-
n
)
cosh
⁡
n
-
2
(
θ
)
sinh
⁡
θ
+
2
γ
n
+
1
cosh
⁡
n
-
1
(
π
)
sinh
⁡
π
δ
c
t
′
.
Using (80), we can compute that
(85)
H
3
=
γ
n
+
1
(
n
-
1
)
(
2
n
-
1
)
2
3
n
cosh
⁡
n
(
θ
)
sinh
⁡
θ
h
h
h
h
i
i
+
(
n
-
1
)
(
2
n
-
1
)
2
cosh
⁡
n
-
2
(
θ
)
sinh
⁡
3
(
θ
)
=
γ
n
+
1
2
n
2
+
3
n
-
1
2
cosh
⁡
n
(
θ
)
sinh
⁡
θ
h
h
h
h
i
i
-
(
n
-
1
)
(
2
n
-
1
)
2
cosh
⁡
n
-
2
(
θ
)
sinh
⁡
θ
2
n
2
+
3
n
-
1
2
cosh
⁡
n
(
θ
)
sinh
⁡
θ
.
Similarly,
(86)
H
4
=
n
-
1
2
γ
n
+
1
cosh
⁡
n
-
2
(
θ
)
sinh
⁡
3
(
θ
)
=
n
-
1
2
γ
n
+
1
cosh
⁡
n
(
θ
)
sinh
⁡
θ
-
cosh
⁡
n
-
2
(
θ
)
sinh
⁡
θ
.
Putting these results together, we see that
(87)
∑
i
=
1
4
‍
H
i
=
2
γ
n
+
1
cosh
⁡
n
(
π
)
sinh
⁡
π
δ
c
t
′
-
2
c
γ
sinh
⁡
π
δ
c
t
′
=
0
,
which implies γ
=
c
1
/
n
sech
⁡
π.
Obviously, ifu
(
t
,
x
)
=
c
1
/
n
sech
⁡
π
cosh
⁡
(
x
-
c
t
-
2
π
(
x
-
c
t
)
/
2
π
-
π
), we have
(88)
u
^
t
,
ξ
=
c
1
/
n
sech
⁡
π
×
∫
T
‍
e
-
i
x
ξ
cosh
⁡
(
x
-
c
t
-
2
π
x
-
c
t
2
π
-
π
)
d
x
=
2
c
1
/
n
sech
⁡
π
·
e
-
i
c
t
ξ
×
∫
0
2
π
‍
e
(
1
-
i
ξ
)
x
-
π
+
e
-
(
1
+
i
ξ
)
x
+
π
d
x
=
2
c
1
/
n
tanh
π
·
e
-
i
c
t
ξ
1
+
ξ
2
.
Therefore, when s
<
3
/
2, we have
(89)
u
H
s
2
=
∑
ξ
=
-
∞
+
∞
‍
1
+
ξ
2
s
u
^
2
=
4
tanh
2
(
π
)
c
2
/
n
∑
ξ
=
∞
+
∞
‍
1
+
ξ
2
s
-
2
<
∞
.
Hence u
(
t
,
x
)
=
c
1
/
n
sech
⁡
π
cosh
⁡
θ
∈
H
s with s
<
3
/
2.


### 5.2. Proof of Theorem5

By Proposition23, we know that (1) has two sequences of periodic peakon (weak) solutions:
(90)
u
m
(
t
,
x
)
=
a
m
1
/
n
sech
⁡
π
cosh
⁡
x
-
a
m
t
-
2
π
x
-
a
m
t
2
π
-
π
,
v
m
(
t
,
x
)
=
c
m
1
/
n
sech
⁡
π
cosh
⁡
x
-
c
m
t
-
2
π
x
-
c
m
t
2
π
-
π
,
where a
m
,
c
m
∈
R are constants velocity which will be specified later. By (88), we know
(91)
u
m
^
(
t
,
ξ
)
=
2
a
m
1
/
n
tanh
(
π
)
·
e
-
i
a
m
t
ξ
1
+
ξ
2
,
v
m
^
(
t
,
ξ
)
=
2
c
m
1
/
n
tanh
(
π
)
·
e
-
i
c
m
t
ξ
1
+
ξ
2
.Notes
<
3
/
2; when t
=
0, we have
(92)
u
m
(
0
,
·
)
-
v
m
(
0
,
·
)
H
s
2
(93)
=
∑
ξ
=
-
∞
+
∞
‍
1
+
ξ
2
s
u
m
^
(
0
,
ξ
)
-
v
m
^
(
0
,
ξ
)
2
=
4
tanh
2
(
π
)
a
m
1
/
n
-
c
m
1
/
n
2
∑
ξ
=
-
∞
+
∞
‍
1
+
ξ
2
s
-
2
=
C
1
s
2
a
m
1
/
n
-
c
m
1
/
n
2
,
where C
1
(
s
)
=
2
tanh
(
π
)
∑
ξ
=
-
∞
+
∞
‍
1
+
ξ
2
s
-
2
1
/
2
<
+
∞.Whent
>
0, it follows that
(94)
u
m
(
t
,
·
)
-
v
m
(
t
,
·
)
H
s
2
(95)
=
∑
ξ
=
-
∞
+
∞
‍
1
+
ξ
2
s
u
m
^
(
t
,
ξ
)
-
v
m
^
(
t
,
ξ
)
2
=
4
tanh
2
(
π
)
∑
ξ
=
-
∞
+
∞
‍
1
+
ξ
2
s
-
2
hhhhhhhhhhihh
×
a
m
1
/
n
e
-
i
a
m
t
ξ
-
c
m
1
/
n
e
-
i
c
m
t
ξ
2
=
u
m
(
0
,
x
)
-
v
m
(
0
,
x
)
H
s
2
+
8
tanh
2
(
π
)
a
m
c
m
1
/
n
∑
ξ
=
-
∞
+
∞
‍
1
-
cos
⁡
⁡
a
m
-
c
m
t
ξ
1
+
ξ
2
2
-
s
≥
8
tanh
2
(
π
)
a
m
c
m
1
/
n
∑
ξ
=
-
∞
+
∞
‍
1
-
cos
⁡
(
(
a
m
-
c
m
)
t
ξ
)
1
+
ξ
2
2
-
s
.
Note that (
1
-
cos
⁡
(
(
a
m
-
c
m
)
t
ξ
)
)
/
1
+
ξ
2
2
-
s
≥
0. For any fixed t
>
0, we choose m
0
∈
N large enough such that m
0
π
/
3
t
>
1 and take c
m
=
m
2
n
(
1
+
|
s
|
), a
m
=
c
m
+
1
/
m, m
≥
m
0. Let ξ
1
=
m
π
/
2
t
+
1, ξ
2
=
3
m
π
/
2
t; then we have ξ
2
-
ξ
1
≥
m
π
/
3
t
>
0 and
(96)
8
tanh
2
(
π
)
a
m
c
m
1
/
n
×
∑
ξ
=
-
∞
+
∞
‍
1
-
cos
⁡
(
(
a
m
-
c
m
)
t
ξ
)
1
+
ξ
2
2
-
s
≥
8
tanh
2
(
π
)
a
m
c
m
1
/
n
×
∑
ξ
=
ξ
1
ξ
2
‍
1
-
cos
⁡
(
(
a
m
-
c
m
)
t
ξ
)
1
+
ξ
2
2
-
s
≥
8
tanh
2
(
π
)
m
4
(
1
+
|
s
|
)
×
∑
ξ
=
m
π
/
2
t
+
1
3
m
π
/
2
t
‍
1
-
cos
⁡
(
t
ξ
/
m
)
1
+
ξ
2
2
-
s
≥
8
tanh
2
(
π
)
m
4
(
1
+
|
s
|
)
×
∑
ξ
=
m
π
/
2
t
+
1
3
m
π
/
2
t
‍
1
2
ξ
2
2
-
s
≥
2
5
-
s
tanh
2
(
π
)
3
5
-
2
s
π
3
-
2
s
t
3
-
2
s
m
2
s
+
4
|
s
|
+
1
=
C
2
s
2
t
3
-
2
s
m
2
s
+
4
|
s
|
+
1
,
where C
2
(
s
)
=
2
5
-
s
tanh
2
(
π
)
/
3
5
-
2
s
π
3
-
2
s. Therefore, from (95), we have
(97)
u
m
(
t
,
·
)
-
v
m
(
t
,
·
)
H
s
≥
C
2
(
s
)
t
3
/
2
-
s
m
s
+
2
|
s
|
+
1
/
2
.
Since
(98)
a
m
-
c
m
=
a
m
1
/
n
-
c
m
1
/
n
∑
j
=
0
n
-
1
‍
a
m
(
n
-
1
-
j
)
/
n
c
m
j
/
n
,
we have
(99)
a
m
1
/
n
-
c
m
1
/
n
=
a
m
-
c
m
∑
j
=
0
n
-
1
a
m
(
n
-
1
-
j
)
/
n
c
m
j
/
n
≤
1
/
m
n
c
m
(
n
-
1
)
/
n
<
1
m
.
Hence we can infer from (93) that
(100)
u
m
(
0
,
·
)
-
v
m
(
0
,
·
)
H
s
2
≤
4
tanh
2
(
π
)
m
2
∑
ξ
=
-
∞
+
∞
‍
1
+
ξ
2
s
-
2
=
C
1
s
2
m
2
.
Combining (97) and (100), we complete the proof of Theorem 5.Remark 24.
From Theorems2 and 5, we see s
=
3
/
2 is the critical index of regularity for well-posedness in Sobolev space H
s for (1) and (2).Remark 25.
From the dynamical system point of view, whens
<
3
/
2, we can deduce that none of the periodic peakons is Lyapunov stable. In fact, for all ε
>
0, for any periodic peakon u
c
(
t
,
x
) with c
>
0, following (93) and (95), we see that, for η
>
0,
(101)
u
c
+
η
(
0
,
x
)
-
u
c
(
0
,
x
)
H
s
2
≤
4
tanh
2
(
π
)
∑
ξ
=
-
∞
+
∞
‍
1
+
ξ
2
s
-
2
η
2
,
u
c
+
η
(
t
,
x
)
-
u
c
(
t
,
x
)
H
s
2
≥
8
tanh
2
(
π
)
c
2
/
n
∑
ξ
=
-
∞
+
∞
‍
1
-
cos
⁡
(
η
t
ξ
)
1
+
ξ
2
2
-
s
≥
2
5
-
s
tanh
2
(
π
)
3
5
-
2
s
π
3
-
2
s
t
3
-
2
s
η
3
-
2
s
c
2
/
n
.
If we let η
=
η
(
s
,
ε
)
>
0 and t
=
t
η such that
(102)
η
2
<
ε
2
4
tanh
2
(
π
)
∑
ξ
=
-
∞
+
∞
1
+
ξ
2
s
-
2
,
t
=
t
η
=
π
η
,
then we have
(103)
u
c
+
η
(
t
,
x
)
-
u
c
(
t
,
x
)
H
s
2
<
ε
2
,
u
c
+
η
(
t
η
,
x
)
-
u
c
(
t
η
,
x
)
H
s
2
≥
2
5
-
s
tanh
2
(
π
)
3
5
-
2
s
c
2
/
n
.
Hence u
c is indeed unstable in the sense of Lyapunov.


## 5.1. Existence of Periodic Peaked Solutions

Proposition 23.
Letc
>
0 and θ
=
x
-
c
t
-
2
π
(
x
-
c
t
)
/
2
π
-
π. Then u
(
t
,
x
)
=
γ
cosh
⁡
θ is a periodic peaked solution to (1) if and only if γ
=
c
1
/
n
sech
⁡
π. Moreover, u
(
t
,
·
)
∈
H
s with s
<
3
/
2.Proof.
LetH
1
=
∂
t
(
1
-
∂
x
2
)
u, H
2
=
(
1
/
n
+
1
)
∂
x
(
1
-
∂
x
2
)
u
n
+
1, H
3
=
∂
x
u
n
+
1
+
(
(
2
n
-
1
)
/
2
)
u
n
-
1
u
x
2, and H
4
=
(
(
n
-
1
)
/
2
)
u
n
-
2
u
x
3. Then (1) can be expressed as ∑
i
=
1
4
‍
H
i
=
0. Assuming that u
(
x
,
t
)
=
γ
cosh
⁡
θ is the periodic solution to (1), we have
(80)
u
x
=
γ
sinh
⁡
θ
,
(81)
u
x
x
=
γ
cosh
⁡
θ
-
2
γ
sinh
⁡
π
δ
c
t
,
where δ
c
t is the periodic Dirac delta function at x
=
c
t (mod 2
π). Thus
(82)
H
1
=
-
2
c
γ
sinh
⁡
π
δ
c
t
′
.
Direct computation shows that
(83)
u
n
+
1
x
x
=
(
n
+
1
)
γ
n
+
1
×
cosh
⁡
n
+
1
(
θ
)
+
n
cosh
⁡
n
-
1
(
θ
)
sinh
⁡
2
(
θ
)
h
h
h
h
h
-
2
cosh
⁡
n
(
π
)
sinh
⁡
π
δ
c
t
′
cosh
⁡
n
+
1
(
θ
)
+
n
cosh
⁡
n
-
1
(
θ
)
sinh
⁡
2
(
θ
)
=
(
n
+
1
)
γ
n
+
1
×
(
n
+
1
)
cosh
⁡
n
+
1
(
θ
)
-
n
cosh
⁡
n
-
1
(
θ
)
h
h
h
h
h
-
2
cosh
⁡
n
(
π
)
sinh
⁡
π
δ
c
t
′
(
n
+
1
)
cosh
⁡
n
+
1
(
θ
)
-
n
cosh
⁡
n
-
1
(
θ
)
.
Therefore, we have
(84)
H
2
=
1
n
+
1
∂
x
u
n
+
1
-
u
n
+
1
x
x
=
γ
n
+
1
(
-
n
2
-
2
n
)
cosh
⁡
n
(
θ
)
sinh
⁡
θ
h
h
h
h
h
i
+
(
n
2
-
n
)
cosh
⁡
n
-
2
(
θ
)
sinh
⁡
θ
+
2
γ
n
+
1
cosh
⁡
n
-
1
(
π
)
sinh
⁡
π
δ
c
t
′
.
Using (80), we can compute that
(85)
H
3
=
γ
n
+
1
(
n
-
1
)
(
2
n
-
1
)
2
3
n
cosh
⁡
n
(
θ
)
sinh
⁡
θ
h
h
h
h
i
i
+
(
n
-
1
)
(
2
n
-
1
)
2
cosh
⁡
n
-
2
(
θ
)
sinh
⁡
3
(
θ
)
=
γ
n
+
1
2
n
2
+
3
n
-
1
2
cosh
⁡
n
(
θ
)
sinh
⁡
θ
h
h
h
h
i
i
-
(
n
-
1
)
(
2
n
-
1
)
2
cosh
⁡
n
-
2
(
θ
)
sinh
⁡
θ
2
n
2
+
3
n
-
1
2
cosh
⁡
n
(
θ
)
sinh
⁡
θ
.
Similarly,
(86)
H
4
=
n
-
1
2
γ
n
+
1
cosh
⁡
n
-
2
(
θ
)
sinh
⁡
3
(
θ
)
=
n
-
1
2
γ
n
+
1
cosh
⁡
n
(
θ
)
sinh
⁡
θ
-
cosh
⁡
n
-
2
(
θ
)
sinh
⁡
θ
.
Putting these results together, we see that
(87)
∑
i
=
1
4
‍
H
i
=
2
γ
n
+
1
cosh
⁡
n
(
π
)
sinh
⁡
π
δ
c
t
′
-
2
c
γ
sinh
⁡
π
δ
c
t
′
=
0
,
which implies γ
=
c
1
/
n
sech
⁡
π.
Obviously, ifu
(
t
,
x
)
=
c
1
/
n
sech
⁡
π
cosh
⁡
(
x
-
c
t
-
2
π
(
x
-
c
t
)
/
2
π
-
π
), we have
(88)
u
^
t
,
ξ
=
c
1
/
n
sech
⁡
π
×
∫
T
‍
e
-
i
x
ξ
cosh
⁡
(
x
-
c
t
-
2
π
x
-
c
t
2
π
-
π
)
d
x
=
2
c
1
/
n
sech
⁡
π
·
e
-
i
c
t
ξ
×
∫
0
2
π
‍
e
(
1
-
i
ξ
)
x
-
π
+
e
-
(
1
+
i
ξ
)
x
+
π
d
x
=
2
c
1
/
n
tanh
π
·
e
-
i
c
t
ξ
1
+
ξ
2
.
Therefore, when s
<
3
/
2, we have
(89)
u
H
s
2
=
∑
ξ
=
-
∞
+
∞
‍
1
+
ξ
2
s
u
^
2
=
4
tanh
2
(
π
)
c
2
/
n
∑
ξ
=
∞
+
∞
‍
1
+
ξ
2
s
-
2
<
∞
.
Hence u
(
t
,
x
)
=
c
1
/
n
sech
⁡
π
cosh
⁡
θ
∈
H
s with s
<
3
/
2.


## 5.2. Proof of Theorem5

By Proposition23, we know that (1) has two sequences of periodic peakon (weak) solutions:
(90)
u
m
(
t
,
x
)
=
a
m
1
/
n
sech
⁡
π
cosh
⁡
x
-
a
m
t
-
2
π
x
-
a
m
t
2
π
-
π
,
v
m
(
t
,
x
)
=
c
m
1
/
n
sech
⁡
π
cosh
⁡
x
-
c
m
t
-
2
π
x
-
c
m
t
2
π
-
π
,
where a
m
,
c
m
∈
R are constants velocity which will be specified later. By (88), we know
(91)
u
m
^
(
t
,
ξ
)
=
2
a
m
1
/
n
tanh
(
π
)
·
e
-
i
a
m
t
ξ
1
+
ξ
2
,
v
m
^
(
t
,
ξ
)
=
2
c
m
1
/
n
tanh
(
π
)
·
e
-
i
c
m
t
ξ
1
+
ξ
2
.Notes
<
3
/
2; when t
=
0, we have
(92)
u
m
(
0
,
·
)
-
v
m
(
0
,
·
)
H
s
2
(93)
=
∑
ξ
=
-
∞
+
∞
‍
1
+
ξ
2
s
u
m
^
(
0
,
ξ
)
-
v
m
^
(
0
,
ξ
)
2
=
4
tanh
2
(
π
)
a
m
1
/
n
-
c
m
1
/
n
2
∑
ξ
=
-
∞
+
∞
‍
1
+
ξ
2
s
-
2
=
C
1
s
2
a
m
1
/
n
-
c
m
1
/
n
2
,
where C
1
(
s
)
=
2
tanh
(
π
)
∑
ξ
=
-
∞
+
∞
‍
1
+
ξ
2
s
-
2
1
/
2
<
+
∞.Whent
>
0, it follows that
(94)
u
m
(
t
,
·
)
-
v
m
(
t
,
·
)
H
s
2
(95)
=
∑
ξ
=
-
∞
+
∞
‍
1
+
ξ
2
s
u
m
^
(
t
,
ξ
)
-
v
m
^
(
t
,
ξ
)
2
=
4
tanh
2
(
π
)
∑
ξ
=
-
∞
+
∞
‍
1
+
ξ
2
s
-
2
hhhhhhhhhhihh
×
a
m
1
/
n
e
-
i
a
m
t
ξ
-
c
m
1
/
n
e
-
i
c
m
t
ξ
2
=
u
m
(
0
,
x
)
-
v
m
(
0
,
x
)
H
s
2
+
8
tanh
2
(
π
)
a
m
c
m
1
/
n
∑
ξ
=
-
∞
+
∞
‍
1
-
cos
⁡
⁡
a
m
-
c
m
t
ξ
1
+
ξ
2
2
-
s
≥
8
tanh
2
(
π
)
a
m
c
m
1
/
n
∑
ξ
=
-
∞
+
∞
‍
1
-
cos
⁡
(
(
a
m
-
c
m
)
t
ξ
)
1
+
ξ
2
2
-
s
.
Note that (
1
-
cos
⁡
(
(
a
m
-
c
m
)
t
ξ
)
)
/
1
+
ξ
2
2
-
s
≥
0. For any fixed t
>
0, we choose m
0
∈
N large enough such that m
0
π
/
3
t
>
1 and take c
m
=
m
2
n
(
1
+
|
s
|
), a
m
=
c
m
+
1
/
m, m
≥
m
0. Let ξ
1
=
m
π
/
2
t
+
1, ξ
2
=
3
m
π
/
2
t; then we have ξ
2
-
ξ
1
≥
m
π
/
3
t
>
0 and
(96)
8
tanh
2
(
π
)
a
m
c
m
1
/
n
×
∑
ξ
=
-
∞
+
∞
‍
1
-
cos
⁡
(
(
a
m
-
c
m
)
t
ξ
)
1
+
ξ
2
2
-
s
≥
8
tanh
2
(
π
)
a
m
c
m
1
/
n
×
∑
ξ
=
ξ
1
ξ
2
‍
1
-
cos
⁡
(
(
a
m
-
c
m
)
t
ξ
)
1
+
ξ
2
2
-
s
≥
8
tanh
2
(
π
)
m
4
(
1
+
|
s
|
)
×
∑
ξ
=
m
π
/
2
t
+
1
3
m
π
/
2
t
‍
1
-
cos
⁡
(
t
ξ
/
m
)
1
+
ξ
2
2
-
s
≥
8
tanh
2
(
π
)
m
4
(
1
+
|
s
|
)
×
∑
ξ
=
m
π
/
2
t
+
1
3
m
π
/
2
t
‍
1
2
ξ
2
2
-
s
≥
2
5
-
s
tanh
2
(
π
)
3
5
-
2
s
π
3
-
2
s
t
3
-
2
s
m
2
s
+
4
|
s
|
+
1
=
C
2
s
2
t
3
-
2
s
m
2
s
+
4
|
s
|
+
1
,
where C
2
(
s
)
=
2
5
-
s
tanh
2
(
π
)
/
3
5
-
2
s
π
3
-
2
s. Therefore, from (95), we have
(97)
u
m
(
t
,
·
)
-
v
m
(
t
,
·
)
H
s
≥
C
2
(
s
)
t
3
/
2
-
s
m
s
+
2
|
s
|
+
1
/
2
.
Since
(98)
a
m
-
c
m
=
a
m
1
/
n
-
c
m
1
/
n
∑
j
=
0
n
-
1
‍
a
m
(
n
-
1
-
j
)
/
n
c
m
j
/
n
,
we have
(99)
a
m
1
/
n
-
c
m
1
/
n
=
a
m
-
c
m
∑
j
=
0
n
-
1
a
m
(
n
-
1
-
j
)
/
n
c
m
j
/
n
≤
1
/
m
n
c
m
(
n
-
1
)
/
n
<
1
m
.
Hence we can infer from (93) that
(100)
u
m
(
0
,
·
)
-
v
m
(
0
,
·
)
H
s
2
≤
4
tanh
2
(
π
)
m
2
∑
ξ
=
-
∞
+
∞
‍
1
+
ξ
2
s
-
2
=
C
1
s
2
m
2
.
Combining (97) and (100), we complete the proof of Theorem 5.Remark 24.
From Theorems2 and 5, we see s
=
3
/
2 is the critical index of regularity for well-posedness in Sobolev space H
s for (1) and (2).Remark 25.
From the dynamical system point of view, whens
<
3
/
2, we can deduce that none of the periodic peakons is Lyapunov stable. In fact, for all ε
>
0, for any periodic peakon u
c
(
t
,
x
) with c
>
0, following (93) and (95), we see that, for η
>
0,
(101)
u
c
+
η
(
0
,
x
)
-
u
c
(
0
,
x
)
H
s
2
≤
4
tanh
2
(
π
)
∑
ξ
=
-
∞
+
∞
‍
1
+
ξ
2
s
-
2
η
2
,
u
c
+
η
(
t
,
x
)
-
u
c
(
t
,
x
)
H
s
2
≥
8
tanh
2
(
π
)
c
2
/
n
∑
ξ
=
-
∞
+
∞
‍
1
-
cos
⁡
(
η
t
ξ
)
1
+
ξ
2
2
-
s
≥
2
5
-
s
tanh
2
(
π
)
3
5
-
2
s
π
3
-
2
s
t
3
-
2
s
η
3
-
2
s
c
2
/
n
.
If we let η
=
η
(
s
,
ε
)
>
0 and t
=
t
η such that
(102)
η
2
<
ε
2
4
tanh
2
(
π
)
∑
ξ
=
-
∞
+
∞
1
+
ξ
2
s
-
2
,
t
=
t
η
=
π
η
,
then we have
(103)
u
c
+
η
(
t
,
x
)
-
u
c
(
t
,
x
)
H
s
2
<
ε
2
,
u
c
+
η
(
t
η
,
x
)
-
u
c
(
t
η
,
x
)
H
s
2
≥
2
5
-
s
tanh
2
(
π
)
3
5
-
2
s
c
2
/
n
.
Hence u
c is indeed unstable in the sense of Lyapunov.



---

*Source: 975925-2014-12-25.xml*