# Wave-Breaking Phenomena and Existence of Peakons for a Generalized Compressible Elastic-Rod Equation

Authors: Xiaolian Ai; Lingyu Jiang; Ting Yi
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

Consideration in this paper is the Cauchy problem of a generalized hyperelastic-rod wave equation. We first derive a wave-breaking mechanism for strong solutions, which occurs in finite time for certain initial profiles. In addition, we determine the existence of some new peaked solitary wave solutions.


---

## Body

## 1. Introduction

Consideration herein is the following generalized hyperelastic-rod wave equation:(1)
u
t
-
u
t
x
x
+
(
g
(
u
)
2
)
x
=
γ
(
2
u
x
u
x
x
+
u
u
x
x
x
)
,
t
>
0
,
x
∈
ℝ
,
u
(
0
,
x
)
=
u
0
(
x
)
,
x
∈
ℝ
,
where g
:
ℝ
→
ℝ is a given C
∞-function and the real number γ with γ
≠
0 is a given parameter. In general, the constant γ is given in terms of the material constants and the prestress of the rod. In fact, the authors [1, 2] studied the special compressible materials which lead to values of γ ranging from -
29.4760 to 3.4174.Whenγ
=
1, g
(
u
)
=
3
u
2, (1) becomes the classical Camassa-Holm (CH) equation [3]
(2)
u
t
-
u
x
x
t
+
3
u
u
x
+
k
u
x
=
2
u
x
u
x
x
+
u
u
x
x
x
with initial data u
(
0
,
x
)
=
u
0
(
x
). While for γ
=
0 and g
(
u
)
=
3
u
2, (1) becomes the Benjamin-Bona-Mahony equation [4]. And when g
(
u
)
=
3
u
2, γ
∈
(
-
29.4760,3.4174
), (1) is so-called the hyperelastic-rod wave equation [2].The classical CH equation (2) was originally proposed as a model for surface waves [3] and has been studied extensively in the last twenty years because of its many remarkable properties: infinity of conservation laws and complete integrability [3], existence of peaked solitons and multipeakons [3, 5, 6] (with k
=
0), well-posedness and breaking waves, and meaning solutions that remain bounded, while its slope becomes unbounded in finite time [7, 8]. On the other hand, if k
=
0, the CH equation (2) admits peaked solitary wave solutions (called peakons) which possess the form u
(
t
,
x
)
=
c
e
-
|
x
-
c
t
| with speed c
∈
ℝ,  c
≠
0, and their stability was studied in [9]. Recently, Gui et al. [10] proved that there exist some peaked functions which are global weak solutions to a modified Camassa-Holm equation. We should mention that the solutions to the CH equation (2) can be uniquely continued after wave-breaking as either a global conservative or global dissipative weak solution [11–14]. It is worth pointing out that there exists a global-in-time weak solution to the CH equation in the energy space [15].Forγ
∈
ℝ
∖
{
0
}, g
(
u
)
=
2
k
u
+
(
3
/
2
)
u
2, (1) serves as a model equation for mechanical vibrations in a hyperelastic rod [1, 2]. Similar to CH equation, stability of solitary wave solutions has been studied in [16]. In addition, the solutions to the hyperelastic-rod wave equation can be uniquely continued after wave breaking as a global conservative weak solution [17]. Moreover, there exists a global-in-time weak solution to the hyperelastic-rod wave equation in the energy space [18].Motivated by the approaches in [9, 10], our goals in this paper are concerned with the wave-breaking phenomena and the existence of some new peakons of (1) with g
(
u
)
=
a
u
2
+
2
k
u  (
a
,
k
∈
ℝ
,
a
>
0
),  γ
≠
0. In this case, (1) may be read as
(3)
u
t
-
u
t
x
x
+
(
a
2
u
2
)
x
+
k
u
x
=
γ
(
2
u
x
u
x
x
+
u
u
x
x
x
)
,
t
>
0
,
x
∈
ℝ
,
u
(
0
,
x
)
=
u
0
(
x
)
,
x
∈
ℝ
.
Introducing the momentum m
:
=
u
-
u
x
x, we get from (3) that
(4)
m
t
+
γ
u
m
x
+
2
γ
u
x
m
+
(
a
-
3
γ
)
u
u
x
+
k
u
x
=
0
,
t
>
0
,
x
∈
ℝ
,
m
(
0
,
x
)
=
m
0
(
x
)
:
=
u
0
-
u
0
x
x
,
x
∈
ℝ
.
Note that if p
(
x
)
:
=
(
1
/
2
)
e
-
|
x
|
,
x
∈
ℝ, then (
1
-
∂
x
2
)
-
1
f
=
p
*
f for all f
∈
L
2
(
ℝ
) and p
*
(
u
-
u
x
x
)
=
u, where * denotes convolution with respect to the spatial variable x. Therefore, (3) can also be rewritten as the following equivalent form
(5)
u
t
+
γ
u
u
x
=
-
∂
x
p
*
(
a
-
γ
2
u
2
+
γ
2
u
x
2
+
k
u
)
,
t
>
0
,
x
∈
ℝ
,
u
(
0
,
x
)
=
u
0
(
x
)
,
x
∈
ℝ
.We are now in a position to give the notions of strong and weak solutions.Definition 1.
Ifu
∈
𝒞
(
[
0
,
T
]
;
H
s
)
∩
𝒞
1
(
[
0
,
T
]
;
H
s
-
1
) with s
>
3
/
2 and some T
>
0 satisfies (5), then u is called a strong solution on [
0
,
T
]. If u is a strong solution on [
0
,
T
] for every T
>
0, then it is called a global strong solution.Definition 2.
Given initial datau
0
∈
H
1
(
ℝ
), the function u
∈
L
loc
⁡
∞
(
[
0
,
T
);  H
loc
⁡
1
(
ℝ
)
) is said to be a weak solution to the initial-value problem (5) if it satisfies the following identity:
(6)
∫
0
T
∫
ℝ
[
u
φ
t
+
γ
2
u
2
φ
x
+
p
*
(
a
-
γ
2
u
2
+
γ
2
u
x
2
+
k
u
)
∂
x
φ
]
d
x
d
t
+
∫
ℝ
‍
u
0
(
x
)
φ
(
0
,
x
)
d
x
=
0
,
for any smooth test function φ
(
t
,
x
)
∈
C
c
∞
(
[
0
,
T
)
×
ℝ
). If u is a weak solution on [
0
,
T
) for every T
>
0, then it is called a global weak solution.Our main results of the present paper are Theorem9 (wave breaking) and Theorem 10 (existence of peakons).The remainder of the paper is organized as follows. In Section2, the results of blow-up to strong solutions are presented in detail. It is shown that the solutions of (3) can only have singularities which correspond to wave breaking (Theorems 9). In Section 3, the existence of some new peaked solutions of (3) is verified. From this, we know that there exist some peaked solitary wave solutions for the case k
≠
0 (compared to the case in the Camassa-Holm equation; see Remark 11).Notation
As above and henceforth, we denote the norm of the Lebesgue spaceL
p
(
ℝ
)  (
1
≤
p
≤
∞
) by ∥
·
∥
L
p and the norm of the Sobolev space H
s
(
ℝ
)  (
s
∈
ℝ
) by ∥
·
∥
H
s. We denote by * the spatial convolution on ℝ.


## 2.  Wave-Breaking Phenomena

By using the Kato's method [19], we may easily get the following results about the local well-posedness and blow-up criterion of strong solutions to (5), of which proofs are similar to the one as in the CH equation in [7] (up to a slight modification) and we omit it here.Theorem 3.
Letu
0
∈
H
s
(
ℝ
) with s
>
3
/
2. Then there exists a time T
>
0 such that the initial-value problem (5) has a unique strong solution u
∈
𝒞
(
[
0
,
T
]
;
H
s
)
∩
𝒞
1
(
[
0
,
T
]
;
H
s
-
1
) and the map u
0
↦
u is continuous from a neighborhood of u
0 in H
s into 𝒞
(
[
0
,
T
]
;
H
s
)
∩
𝒞
1
(
[
0
,
T
]
;
H
s
-
1
).We are now in a position to state a blow-up criterion for (5).Theorem 4.
Letu
0
∈
H
s
(
ℝ
) be as in Theorem 3 with s
>
3
/
2. Let u be the corresponding solution to (5). Assume that T
u
0
*
>
0 is the maximum time of existence. Then
(7)
T
u
0
*
<
∞
⟹
∫
0
T
u
0
*
‍
∥
u
x
(
τ
)
∥
L
∞
d
τ
=
∞
.Remark 5.
The blow-up criterion (7) implies that the lifespan T
u
0
* does not depend on the regularity index s of the initial data u
0. Indeed, let u
0 be in H
s for some s
>
3
/
2 and consider some s
′
∈
(
3
/
2
,
s
). Denote by u
s (resp., u
s
′) the corresponding maximal H
s (resp., H
s
′) solution given by the above theorem. Denote by T
s
* (resp., T
s
′
*) the lifespan of u
s (resp., u
s
′). Since H
s
↪
H
s
′, uniqueness ensures that T
s
*
≤
T
s
′
* and that u
s
≡
u
s
′ on [
0
,
T
s
*
). Now, if T
s
*
<
T
s
′
*, then we must have u
s
′ in 𝒞
(
[
0
,
T
s
*
]
;
H
s
′
). Hence, ∂
x
u
s
′
∈
L
1
(
[
0
,
T
s
*
]
;
L
∞
), which contradicts the above blow-up criterion (7). Therefore, T
s
*
=
T
s
′
*.Remark 6.
For a strong solutionu in Theorem 3, the Hamiltonian functionals are conserved; that is
(8)
d
d
t
∫
ℝ
‍
(
u
2
+
u
x
2
)
d
x
=
0
,
d
d
t
∫
ℝ
‍
(
a
3
u
3
+
γ
u
u
x
2
+
k
u
2
)
d
x
=
0
,
for all t
∈
[
0
,
T
).The following blow-up criterion shows that the wave-breaking depends only on the infimum ofγ
u
x.Theorem 7.
Letu
0
∈
H
s
(
ℝ
) be as in Theorem 3 with s
>
3
/
2. Then the corresponding solution u to (5) blows up in finite time T
u
0
*
>
0 if and only if
(9)
liminf
⁡
t
↑
T
u
0
*
inf
⁡
x
∈
ℝ
{
γ
u
x
(
t
,
x
)
}
=
-
∞
.Proof.
Since, in view of Remark5, the existence time T
u
0
* is independent of the choice of s, we need only to consider the case s
=
3, which relies on a simple density argument.
Multiplying (3) by m and integrating over ℝ with respect to x and then integration by parts produce
(10)
1
2
d
d
t
∫
ℝ
‍
m
2
d
x
=
-
a
2
∫
ℝ
‍
(
u
2
)
x
m
d
x
+
γ
∫
ℝ
‍
m
(
2
u
x
u
x
x
+
u
u
x
x
x
)
d
x
=
-
a
∫
ℝ
‍
u
u
x
m
d
x
+
2
γ
∫
ℝ
‍
u
x
(
u
-
m
)
m
d
x
+
γ
∫
ℝ
‍
u
(
u
x
-
m
x
)
m
d
x
=
(
3
γ
-
a
)
∫
ℝ
‍
u
u
x
m
d
x
-
3
2
γ
∫
ℝ
‍
u
x
m
2
d
x
.
We next differentiate (3) with respect to x to get
(11)
m
x
t
=
(
3
γ
-
a
)
(
u
x
2
+
u
2
-
u
m
)
-
3
γ
u
x
m
x
-
2
γ
u
m
-
γ
u
m
x
x
+
k
(
m
-
u
)
.
Multiplying by m
x then integrating over ℝ with respect to x lead to
(12)
1
2
d
d
t
∫
ℝ
‍
m
x
2
d
x
=
-
2
(
3
γ
-
a
)
∫
ℝ
‍
u
x
(
u
-
m
)
m
d
x
-
2
(
3
γ
-
a
)
∫
ℝ
‍
u
x
u
m
d
x
-
(
3
γ
-
a
)
∫
ℝ
‍
u
m
m
x
d
x
-
3
γ
∫
ℝ
‍
u
x
m
x
2
d
x
-
2
γ
∫
ℝ
‍
u
m
m
x
d
x
+
γ
2
∫
ℝ
‍
u
x
m
x
2
d
x
=
-
4
(
3
γ
-
a
)
∫
ℝ
‍
u
x
u
m
d
x
+
(
5
a
-
17
γ
)
∫
ℝ
‍
u
m
m
x
d
x
-
5
2
γ
∫
ℝ
‍
u
x
m
x
2
d
x
.
Therefore,
(13)
d
d
t
∫
ℝ
‍
(
m
2
+
m
x
2
)
d
x
=
-
6
(
3
γ
-
a
)
∫
ℝ
‍
u
x
u
m
d
x
+
2
(
5
a
-
17
γ
)
∫
ℝ
‍
u
m
m
x
d
x
-
∫
ℝ
‍
(
γ
u
x
)
(
3
m
2
+
5
m
x
2
)
d
x
.
If γ
u
x is bounded from below on [
0
,
T
m
0
*
)
×
ℝ, that is, there exists a positive constant C
1
>
0 such that γ
u
x
≥
-
C
1 on [
0
,
T
m
0
*
)
×
ℝ, then the above estimate implies
(14)
d
d
t
∫
ℝ
‍
(
m
2
+
m
x
2
)
d
x
≤
5
C
1
∫
ℝ
‍
(
m
2
+
m
x
2
)
d
x
+
|
2
(
3
γ
-
a
)
|
∥
u
(
t
,
·
)
∥
L
∞
∫
ℝ
‍
(
u
x
2
+
m
2
)
d
x
+
|
5
a
-
17
γ
|
∥
u
(
t
,
·
)
∥
L
∞
∫
ℝ
‍
(
m
2
+
m
x
2
)
d
x
≤
C
2
(
a
,
γ
,
∥
u
0
∥
H
1
)
∫
ℝ
‍
(
m
2
+
m
x
2
)
d
x
,
where we used (8) so that ∥
u
(
t
,
·
)
∥
L
∞
2
≤
(
1
/
2
)
E
(
u
(
t
,
·
)
)
=
(
1
/
2
)
E
(
u
0
)
=
(
1
/
2
)
∥
u
0
∥
H
1
2. Applying Gronwall's inequality then yields
(15)
∥
m
(
t
)
∥
H
1
2
=
∫
ℝ
‍
(
m
2
+
m
x
2
)
d
x
≤
e
C
2
t
∥
m
0
∥
H
1
2
for t
∈
[
0
,
T
m
0
*
), which ensures that the solution m
(
t
,
x
) does not blow up in finite time.
On the other hand, if(16)
liminf
⁡
t
↑
T
u
0
*
[
inf
⁡
x
∈
ℝ
(
γ
u
x
(
t
,
x
)
)
]
=
-
∞
,
by Theorem 3 for the existence of local strong solutions and the Sobolev embedding theorem, we infer that the solution will blow up in finite time. The proof of Theorem 7 is then complete.Forγ
≠
0, we define
(17)
𝔰
(
t
)
=
def
sign
⁡
{
γ
}
inf
⁡
x
∈
ℝ
[
u
x
(
t
,
x
)
sign
⁡
{
γ
}
]
,
t
≥
0
,
where sign
⁡
(
γ
) is the sign function of γ
∈
ℝ, and set 𝔰
0
:
=
𝔰
(
0
). Then, thanks to Theorem 3, for every t
∈
[
0
,
T
) there exists at least one point ξ
(
t
)
∈
ℝ with 𝔰
(
t
)
=
u
x
(
t
,
ξ
(
t
)
). Just as the proof given in [8], one can show the following property of 𝔰
(
t
).Lemma 8.
Letu
(
t
) be the solution to (5) on [
0
,
T
) with initial data u
0
∈
H
2
(
ℝ
), as given by Theorem 3. Then the function 𝔰
(
t
) is almost everywhere differentiable on [
0
,
T
), with
(18)
d
𝔰
d
t
=
u
t
x
(
t
,
ξ
(
t
)
)
a
.
e
.
o
n
(
0
,
T
)
.With Lemma8 in hand, we may get the following result about the wave-breaking phenomena of solutions to the generalized hyperelastic-rod wave equation (3).Theorem 9.
Leta
>
0,  γ
≠
0, and  u
0
∈
H
s
(
ℝ
) with s
>
3
/
2. Assume that
(19)
inf
⁡
x
∈
ℝ
{
u
0
′
(
x
)
}
<
-
K
i
f
γ
>
0
,
o
r
sup
⁡
x
∈
ℝ
{
u
0
′
(
x
)
}
>
K
i
f
γ
<
0
,
where
(20)
K
:
=
{
|
a
-
γ
|
2
|
γ
|
∥
u
0
∥
H
1
,
i
f
γ
∈
(
-
∞
,
0
)
∪
(
a
,
+
∞
)
,
a
-
(
1
+
α
0
)
γ
2
γ
∥
u
0
∥
H
1
,
i
f
γ
∈
(
0
,
a
1
+
α
0
]
,
0
,
i
f
γ
∈
(
a
1
+
α
0
,
a
)
with α
0
=
(
4
a
-
3
γ
)
/
4
γ
-
(
1
/
2
) for γ
∈
(
0
,
a
). Then the corresponding solution to (5) blows up in finite time. Moreover, the maximal time of existence is estimated above by
(21)
0
<
T
≤
-
2
inf
⁡
x
∈
ℝ
{
sign
⁡
(
γ
)
u
0
′
(
x
)
}
|
γ
|
[
(
inf
⁡
x
∈
ℝ
{
sign
⁡
(
γ
)
u
0
′
(
x
)
}
)
2
-
K
2
]
<
+
∞
.Proof.
LetT
>
0 be the maximal existence time of the solution u
(
t
,
x
) to (5) with initial data u
0 as stated in Theorem 3. Applying Theorem 3, Remark 5, and a simple density argument, we need only to show that Theorem 9 holds with some s
≥
5. Here we assume s
=
5 to prove the theorem.
Firstly, thanks to Lemma8 and the definition of 𝔰
(
t
), there are x
0
∈
ℝ and ξ
(
t
)
∈
ℝ such that u
0
′
(
x
0
)
=
inf
⁡
x
∈
ℝ
u
0
′
(
x
), (18) holds, and
(22)
u
x
x
(
t
,
ξ
(
t
)
)
=
0
,
∀
t
∈
[
0
,
T
)
.
Differentiating (5) with respect to x and using the identity ∂
x
2
p
*
f
=
p
*
f
-
f, we obtain
(23)
u
x
t
=
-
γ
2
u
x
2
-
γ
u
u
x
x
+
a
-
γ
2
u
2
-
p
*
(
a
-
γ
2
u
2
+
γ
2
u
x
2
)
-
k
u
x
x
,
which along with (22) implies
(24)
d
d
t
𝔰
(
t
)
=
-
γ
2
𝔰
(
t
)
2
+
a
-
γ
2
u
2
(
t
,
ξ
(
t
)
)
-
p
*
(
a
-
γ
2
u
2
+
γ
2
u
x
2
)
(
t
,
ξ
(
t
)
)
a
.
e
.
on
⁡
(
0
,
T
)
.
For the caseγ
≥
a, thanks to the Young's inequality and the Sobolev inequality, we get from Remark 6 that, for all t
∈
[
0
,
T
),
(25)
∥
p
*
a
-
γ
2
u
2
∥
L
∞
≤
γ
-
a
2
∥
u
(
t
)
∥
L
∞
2
≤
γ
-
a
4
∥
u
(
t
)
∥
H
1
2
=
γ
-
a
4
∥
u
0
∥
H
1
2
.
Note that p
*
(
(
γ
/
2
)
u
x
2
)
≥
0 and (
(
a
-
γ
)
/
2
)
u
2
(
t
,
ξ
(
t
)
)
≤
0, we infer that
(26)
d
d
t
𝔰
(
t
)
≤
-
γ
2
𝔰
2
(
t
)
+
γ
-
a
4
∥
u
0
∥
H
1
2
a
.
e
.
on
⁡
(
0
,
T
)
.
Set K
:
=
(
γ
-
a
)
/
2
γ
∥
u
0
∥
H
1; we have
(27)
d
d
t
𝔰
(
t
)
≤
-
γ
2
𝔰
2
(
t
)
+
γ
2
K
2
a
.
e
.
on
⁡
(
0
,
T
)
.
We now claim that if𝔰
(
0
)
<
-
K, then 𝔰
(
t
)
<
-
K for all t
∈
(
0
,
T
).
In fact, assuming that the contrary would, in view of𝔰
(
t
)
(
t
) being continuous, ensure the existence of some t
0
∈
(
0
,
T
), such that 𝔰
(
t
)
<
-
K in (
0
,
t
0
) and 𝔰
(
t
0
)
=
-
K. Combining this with (27) would give (
d
/
d
t
)
𝔰
(
t
)
<
0
a
.
e
.
on
⁡
[
0
,
t
0
). Since 𝔰
(
t
) is absolutely continuous on [
0
,
t
0
], an integration of this inequality would give the following inequality and we get the contradiction 𝔰
(
t
0
)
<
𝔰
(
0
)
<
-
K. This proves the previous claim.
Therefore it follows from (27) that (
d
/
d
t
)
𝔰
(
t
)
<
0  a
.
e
.
on
⁡  (
0
,
T
) which implies that 𝔰
(
t
) is strictly decreasing on [
0
,
T
).
Setδ
:
=
(
γ
/
2
)
(
1
-
(
K
/
u
0
′
(
x
0
)
)
2
)
∈
(
0
,
γ
/
2
). Then we have K
2
/
(
1
-
(
2
/
γ
)
δ
)
=
(
u
0
′
(
x
0
)
)
2
<
𝔰
2
(
t
), that is, K
2
<
(
1
-
(
2
/
γ
)
δ
)
𝔰
2
(
t
). Therefore it follows that
(28)
d
d
t
𝔰
(
t
)
≤
-
γ
2
𝔰
2
(
t
)
[
1
-
(
1
-
2
γ
δ
)
]
=
-
δ
𝔰
2
(
t
)
,
a
.
e
.
on
⁡
(
0
,
T
)
,
which leads to
(29)
𝔰
(
t
)
≤
u
0
′
(
x
0
)
1
+
δ
t
u
0
′
(
x
0
)
⟶
-
∞
,
as
t
⟶
-
1
δ
u
0
′
(
x
0
)
.
This implies that the maximal time of existence T satisfies
(30)
T
≤
-
1
δ
u
0
′
(
x
0
)
=
2
inf
⁡
x
∈
ℝ
u
0
′
(
x
)
γ
[
K
2
-
(
inf
⁡
x
∈
ℝ
u
0
′
(
x
)
)
2
]
<
+
∞
.
In consequence, we have
(31)
liminf
⁡
t
↑
T
(
inf
⁡
x
∈
ℝ
γ
u
x
(
t
,
x
)
)
=
-
∞
,
which along with Theorem 7 applied completes the proof of Theorem 9 for the case γ
≥
a.
Next, to deal with the case0
<
γ
<
a, we need first to verify the following inequality for any positive real number α:
(32)
p
*
(
(
α
2
+
α
)
u
2
+
u
x
2
)
(
x
)
≥
α
u
2
(
x
)
for
∀
u
∈
H
3
(
ℝ
)
.
Indeed, observe that the inequality
(33)
e
-
x
∫
-
∞
x
‍
e
y
(
α
2
u
2
+
u
x
2
)
(
y
)
d
y
≥
2
α
e
-
x
∫
-
∞
x
‍
e
y
u
(
y
)
u
x
(
y
)
d
y
=
α
e
-
x
∫
-
∞
x
‍
e
y
d
d
y
(
u
2
(
y
)
)
d
y
=
α
u
2
(
x
)
-
α
e
-
x
∫
-
∞
x
‍
e
y
u
2
(
y
)
d
y
yields
(34)
e
-
x
∫
-
∞
x
‍
e
y
(
(
α
2
+
α
)
u
2
+
u
x
2
)
(
y
)
d
y
≥
α
u
2
(
x
)
.
Whereas the inequality
(35)
e
x
∫
x
∞
‍
e
-
y
(
α
2
u
2
+
u
x
2
)
(
y
)
d
y
≥
-
2
α
e
x
∫
x
∞
‍
e
-
y
u
(
y
)
u
x
(
y
)
d
y
=
-
α
e
x
∫
x
∞
‍
e
-
y
d
d
y
(
u
2
(
y
)
)
d
y
=
α
u
2
(
x
)
-
α
e
x
∫
x
∞
‍
e
-
y
u
2
(
y
)
d
y
leads to
(36)
e
x
∫
x
∞
‍
e
-
y
(
(
α
2
+
α
)
u
2
+
u
x
2
)
(
y
)
d
y
≥
α
u
2
(
x
)
,
which along with (34) gives rise to (32).
Notice thatα
0
=
(
4
a
-
3
γ
)
/
4
γ
-
(
1
/
2
) is the only positive root of α
2
+
α
=
(
a
-
γ
)
/
γ; we get from (32) that
(37)
p
*
(
a
-
γ
2
u
2
+
γ
2
u
x
2
)
(
t
,
ξ
(
t
)
)
≥
γ
α
0
2
u
x
2
(
t
,
ξ
(
t
)
)
,
which together with (24) implies that
(38)
d
d
t
𝔰
(
t
)
≤
-
γ
2
𝔰
(
t
)
2
+
a
-
(
1
+
α
0
)
γ
2
u
2
(
t
,
ξ
(
t
)
)
a
.
e
.
on
⁡
(
0
,
T
)
.
Thanks to the Sobolev inequality and Remark 6, we get that, for any t
∈
(
0
,
T
),  ∥
u
(
t
)
∥
L
∞
2
≤
(
1
/
2
)
∥
u
(
t
)
∥
H
1
2
=
(
1
/
2
)
∥
u
0
∥
H
1
2, which along with (38) implies that (27) holds for
(39)
K
:
=
{
a
-
(
1
+
α
0
)
γ
2
γ
∥
u
0
∥
H
1
,
if
γ
∈
(
0
,
a
1
+
α
0
]
,
0
,
if
γ
∈
(
a
1
+
α
0
,
a
)
.
Repeating the above argument in the proof of (31) for the case γ
≥
a leads to (31), and then the proof of Theorem 9 is complete for the case 0
<
γ
<
a.
Let us now consider the caseγ
<
0. It follows from (24) that
(40)
d
d
t
𝔰
(
t
)
≥
|
γ
|
2
𝔰
2
(
t
)
-
a
-
γ
4
∥
u
0
∥
H
1
2
a
.
e
.
on
⁡
(
0
,
T
)
,
where we used Young's inequality so that, for allt
∈
[
0
,
T
),
(41)
∥
p
*
u
2
∥
L
∞
≤
∥
u
(
t
)
∥
L
∞
2
≤
1
2
∥
u
(
t
)
∥
H
1
2
.
Therefore, set K
:
=
(
a
-
γ
)
/
-
2
γ
∥
u
0
∥
H
1; we may get
(42)
d
d
t
𝔰
(
t
)
≥
|
γ
|
2
𝔰
2
(
t
)
-
|
γ
|
2
K
2
a
.
e
.
on
⁡
(
0
,
T
)
.
Repeating the above argument in the proof of (31) for the case γ
≥
a again ends the proof of Theorem 9.


## 3. Peaked Solitary Wave Solution

In this section we consider the existence of peaked solitary wave solutions of (3) in the case of g
(
u
)
=
a
u
2
+
2
k
u (a
,
k
∈
ℝ and k
≠
0), which can be understood as global weak solutions.Theorem 10.
For anyγ
∈
ℝ
∖
{
a
/
3
,
0
}, a
γ
>
0, and k
≠
0, the peaked function of the form
(43)
u
¯
c
(
t
,
x
)
=
c
γ
e
-
a
/
3
γ
|
x
-
c
t
|
,
w
i
t
h
c
=
3
k
γ
3
γ
-
a
is a global weak solution to (3) in the sense of Definition (2).Remark 11.
It is known that for the Camassa-Holm equation (that is, (5) with γ
=
1 and a
=
3), for any c
≠
0, the peaked function of the form u
c
(
t
,
x
)
=
c
e
-
|
x
-
c
t
| is a global weak solution to the Camassa-Holm equation with the case k
=
0 in the sense of Definition (2), and if k
≠
0, the Camassa-Holm equation has no peaked solution. Here Theorem 10 implies that, if k
≠
0, (5) still has some peaked solutions, as long as γ
≠
a
/
3 and 0.Proof of Theorem10.
The proof of the theorem is motivated by the method in [10]. First, we can reduce the result to the one with the case γ
=
1. In fact, multiplying γ with (1) and setting u
¯
=
γ
u, (1) become the following form:
(44)
u
¯
t
-
u
¯
t
x
x
+
(
a
2
γ
u
¯
2
+
k
u
¯
)
x
=
2
u
¯
x
u
¯
x
x
+
u
¯
u
¯
x
x
x
,
t
>
0
,
x
∈
ℝ
,
(45)
u
¯
(
0
,
x
)
=
u
¯
0
(
x
)
,
x
∈
ℝ
.
We claim that, for allt
∈
ℝ
+,
(46)
∂
x
u
¯
c
(
t
,
x
)
=
-
a
3
γ
sign
⁡
(
x
-
c
t
)
u
¯
c
(
t
,
x
)
∈
L
∞
(
ℝ
)
,
in the sense of distribution 𝒮
'
(
ℝ
).
Indeed, for any test functionφ
(
·
)
∈
C
c
∞
(
ℝ
), we may get by using integration by parts that
(47)
∫
ℝ
‍
sign
⁡
(
y
)
e
-
a
/
3
γ
|
y
|
φ
(
y
)
d
y
=
-
∫
-
∞
0
‍
e
a
/
3
γ
y
φ
(
y
)
d
y
+
∫
0
+
∞
‍
e
-
a
/
3
γ
y
φ
(
y
)
d
y
=
-
φ
(
0
)
+
3
γ
a
∫
-
∞
0
‍
e
a
/
3
γ
y
φ
′
(
y
)
d
y
+
φ
(
0
)
+
3
γ
a
∫
0
+
∞
‍
e
-
a
/
3
γ
y
φ
′
(
y
)
d
y
=
3
γ
a
∫
ℝ
‍
e
-
a
/
3
γ
|
y
|
φ
′
(
y
)
d
y
,
which along with the fact that sign
⁡
(
·
-
c
t
)
u
¯
c
(
t
,
·
)
∈
L
∞
(
ℝ
) ends the proof of the claim.
Let us now setu
¯
0
,
c
(
x
)
:
=
u
¯
c
(
0
,
x
) for x
∈
ℝ. Then we have
(48)
lim
⁡
t
→
0
+
∥
u
¯
c
(
t
,
·
)
-
u
¯
0
,
c
(
·
)
∥
W
1
,
∞
=
0
.
Similar to the proof of (46), we get that, for all t
≥
0,
(49)
∂
t
u
¯
c
(
t
,
x
)
=
c
sign
⁡
(
x
-
c
t
)
u
¯
c
(
t
,
x
)
∈
L
∞
(
ℝ
)
.
Therefore, from (46), (48), (49), and c
=
3
k
γ
/
(
3
γ
-
a
) and γ
=
1, together with integration by parts applied, we obtain that, for every test function ϕ
(
t
,
x
)
∈
C
c
∞
(
[
0
,
+
∞
)
×
ℝ
),
(50)
∫
0
+
∞
‍
∫
ℝ
‍
(
u
¯
d
∂
t
ϕ
+
1
2
u
¯
c
2
∂
x
ϕ
)
d
x
d
t
+
∫
ℝ
‍
u
¯
c
(
0
,
x
)
ϕ
(
0
)
d
x
d
t
=
-
∫
0
+
∞
‍
∫
ℝ
‍
ϕ
(
∂
t
u
¯
c
+
u
¯
c
∂
x
u
¯
c
)
d
x
d
t
=
-
∫
0
+
∞
‍
∫
ℝ
‍
ϕ
(
c
a
3
sign
⁡
(
x
-
c
t
)
c
e
-
a
/
3
|
x
-
c
t
|
-
a
3
sign
⁡
(
x
-
c
t
)
×
c
2
e
-
2
a
/
3
|
x
-
c
t
|
c
a
3
)
d
x
d
t
=
-
∫
0
+
∞
‍
∫
ℝ
‍
ϕ
c
2
a
3
sign
⁡
(
x
-
c
t
)
×
(
e
-
a
/
3
|
x
-
c
t
|
-
e
-
2
a
/
3
|
x
-
c
t
|
)
d
x
d
t
,
(51)
∫
0
+
∞
‍
∫
ℝ
‍
p
*
(
a
-
1
2
u
¯
c
2
+
1
2
(
∂
x
u
¯
c
)
2
+
k
u
¯
c
)
ϕ
x
d
x
d
t
=
-
∫
0
+
∞
‍
∫
ℝ
‍
ϕ
∂
x
p
*
(
a
-
1
2
u
¯
c
2
+
1
2
(
∂
x
u
¯
c
)
2
+
k
u
¯
c
)
d
x
d
t
.
Notice from (46) that ∂
x
p
=
-
(
1
/
2
)
sign
⁡
(
x
)
e
-
|
x
| for x
∈
ℝ; we have
(52)
∂
x
p
*
(
a
-
1
2
u
¯
c
2
+
1
2
(
∂
x
u
¯
c
)
2
+
k
u
¯
c
)
=
-
1
2
∫
-
∞
+
∞
‍
sign
⁡
(
x
-
y
)
e
-
|
x
-
y
|
×
(
4
a
-
3
6
c
2
e
-
2
a
/
3
|
y
-
c
t
|
+
k
c
e
-
a
/
3
|
y
-
c
t
|
)
d
y
.
When x
>
c
t, we split the right hand side of (52) into the following three parts:
(53)
∂
x
p
*
(
a
-
1
2
u
¯
c
2
+
1
2
(
∂
x
u
¯
c
)
2
+
k
u
¯
c
)
=
-
1
2
(
∫
-
∞
c
t
‍
+
∫
c
t
x
‍
+
∫
x
+
∞
‍
)
sign
⁡
(
x
-
y
)
e
-
|
x
-
y
|
×
(
4
a
-
3
6
c
2
e
-
2
a
/
3
|
y
-
c
t
|
+
k
c
e
-
a
/
3
|
y
-
c
t
|
)
d
y
=
:
I
1
+
I
2
+
I
3
.
We directly compute I
1 as follows:
(54)
I
1
=
-
1
2
∫
-
∞
c
t
‍
e
-
(
x
-
y
)
(
4
a
-
3
6
c
2
e
2
a
/
3
(
y
-
c
t
)
+
k
c
e
a
/
3
(
y
-
c
t
)
)
d
y
=
(
-
4
a
-
3
12
(
1
+
2
a
/
3
)
d
2
-
k
c
2
(
1
+
a
/
3
)
)
e
-
(
x
-
c
t
)
.
Similarly, we have
(55)
I
2
=
(
-
4
a
-
3
12
(
1
-
2
a
/
3
)
c
2
)
e
-
2
a
/
3
(
x
-
c
t
)
+
(
-
k
c
2
(
1
-
a
/
3
)
)
e
-
a
/
3
(
x
-
c
t
)
+
(
4
a
-
3
12
(
1
-
2
a
/
3
)
c
2
+
k
c
2
(
1
-
a
/
3
)
)
e
-
(
x
-
c
t
)
and I
3
=
(
(
4
a
-
3
)
/
(
12
(
1
+
2
a
/
3
)
)
)
c
2
e
-
2
a
/
3
(
x
-
c
t
)
+
(
k
c
/
(
2
(
1
+
a
/
3
)
)
)
e
-
a
/
3
(
x
-
c
t
). From this, we deduce that, for x
>
c
t,
(56)
∂
x
p
*
(
a
-
1
2
u
¯
c
2
+
1
2
(
∂
x
u
¯
c
)
2
+
k
u
¯
c
)
=
a
3
c
2
(
e
-
2
a
/
3
(
x
-
c
t
)
-
e
-
a
/
3
(
x
-
c
t
)
)
.
And for the case x
≤
c
t, we spilt the right hand side of (51) into the following three parts:
(57)
∂
x
p
*
(
a
-
1
2
u
¯
c
2
+
1
2
(
∂
x
u
¯
c
)
2
+
k
u
¯
c
)
=
-
1
2
(
∫
-
∞
x
‍
+
∫
x
c
t
‍
+
∫
c
t
+
∞
‍
)
sign
⁡
(
x
-
y
)
e
-
|
x
-
y
|
×
(
4
a
-
3
6
c
2
e
-
2
a
/
3
|
y
-
c
t
|
+
k
c
e
-
a
/
3
|
y
-
c
t
|
)
d
y
=
:
I
I
1
+
I
I
2
+
I
I
3
.
We compute
(58)
I
I
1
=
-
1
2
∫
-
∞
x
‍
e
-
(
x
-
y
)
(
4
a
-
3
6
c
2
e
2
a
/
3
(
y
-
c
t
)
+
k
c
e
a
/
3
(
y
-
c
t
)
4
a
-
3
6
)
d
y
=
-
4
a
-
3
12
(
1
+
2
a
/
3
)
c
2
e
2
a
/
3
(
x
-
c
t
)
-
k
c
2
(
1
+
a
/
3
)
e
a
/
3
(
x
-
c
t
)
.
Similarly, we have
(59)
I
I
2
=
4
a
-
3
12
(
2
a
/
3
-
1
)
c
2
(
e
x
-
c
t
-
e
2
a
/
3
(
x
-
c
t
)
)
+
k
c
2
(
a
/
3
-
1
)
(
e
x
-
c
t
-
e
a
/
3
(
x
-
c
t
)
)
and I
I
3
=
(
(
(
4
a
-
3
)
/
(
12
(
1
+
2
a
/
3
)
)
)
c
2
+
k
c
/
(
2
(
1
+
a
/
3
)
)
)
e
x
-
c
t, which along with (57) and (58) implies that, for x
≤
c
t,
(60)
∂
x
p
*
(
a
-
1
2
u
¯
c
2
+
1
2
(
∂
x
u
¯
c
)
2
+
k
u
¯
c
)
=
a
3
c
2
(
e
a
/
3
(
x
-
c
t
)
-
e
2
a
/
3
(
x
-
c
t
)
)
.
From (56) and (60), we obtain
(61)
∫
0
+
∞
‍
∫
ℝ
‍
p
*
(
a
-
1
2
u
¯
c
2
+
1
2
(
∂
x
u
¯
c
)
2
+
k
u
¯
c
)
ϕ
x
d
x
d
t
=
{
-
a
3
c
2
×
∫
0
+
∞
‍
∫
ℝ
‍
ϕ
(
e
-
2
a
/
3
(
x
-
c
t
)
-
e
-
a
/
3
(
x
-
c
t
)
)
d
x
d
t
,
x
>
c
t
,
-
a
3
c
2
×
∫
0
+
∞
‍
∫
ℝ
‍
ϕ
(
e
a
/
3
(
x
-
c
t
)
-
e
2
a
/
3
(
x
-
c
t
)
)
d
x
d
t
,
x
≤
c
t
.
On the other hand, by using the form of u
¯
c, we get from (50) that
(62)
∫
0
+
∞
‍
∫
ℝ
‍
(
u
¯
c
∂
t
ϕ
+
1
2
u
¯
c
2
∂
x
ϕ
)
d
x
d
t
+
∫
ℝ
‍
u
¯
d
(
0
,
x
)
ϕ
(
0
)
d
x
d
t
=
{
-
a
3
c
2
×
∫
0
+
∞
‍
∫
ℝ
‍
ϕ
(
e
-
a
/
3
(
x
-
c
t
)
-
e
-
2
a
/
3
(
x
-
c
t
)
)
d
x
d
t
,
x
>
c
t
,
a
3
c
2
×
∫
0
+
∞
‍
∫
ℝ
‍
ϕ
(
e
a
/
3
(
x
-
c
t
)
-
e
2
a
/
3
(
x
-
c
t
)
)
d
x
d
t
,
x
≤
c
t
.
Therefore, for every test function ϕ
(
t
,
x
)
∈
C
c
∞
(
[
0
,
+
∞
)
×
ℝ
), we conclude that
(63)
∫
0
+
∞
‍
∫
ℝ
‍
[
u
¯
c
∂
t
ϕ
+
1
2
u
¯
c
2
∂
x
ϕ
+
p
*
(
a
-
1
2
u
¯
c
2
+
1
2
(
∂
x
u
¯
c
)
2
+
k
u
¯
c
)
∂
x
ϕ
]
d
x
d
t
+
∫
ℝ
‍
u
¯
c
(
0
,
x
)
ϕ
(
0
,
x
)
d
x
=
0
,
which completes the proof of Theorem 10.



---

*Source: 876849-2014-02-26.xml*