# On the Existence of Global Weak Solutions for a Weakly Dissipative Hyperelastic Rod Wave Equation

Authors: Haibo Yan; Ls Yong; Yu Yang; Yang Wang
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

Assuming that the initial valuev
0
(
x
) belongs to the space H
1
(
R
), we prove the existence of global weak solutions for a weakly dissipative hyperelastic rod wave equation in the space C
(
[
0
,
∞
)
×
R
)
⋂
‍
L
∞
(
[
0
,
∞
)
;
H
1
(
R
)
). The limit of the viscous approximation for the equation is used to establish the existence.


---

## Body

## 1. Introduction

This paper focuses on the study of the weakly dissipative model(1)
v
t
-
v
t
x
x
+
∂
x
f
(
v
)
+
λ
(
v
-
v
x
x
)
=
2
α
v
x
v
x
x
+
α
v
v
x
x
x
+
β
v
2
m
v
x
x
,
where α
>
0
,
β
≥
0
,
λ
≥
0, f
(
v
) is a polynomial with order n, and m is a nonnegative integer. If f
(
v
)
=
2
k
v
+
(
3
/
2
)
v
2
,
α
=
1
,
β
=
0,  and  λ
=
0, (1) becomes the Camassa-Holm equation [1]. When f
(
v
)
=
(
3
/
2
)
v
2   and   λ
=
β
=
0, (1) is turned into the hyperelastic rod wave equation [2]. Since the terms λ
(
v
-
v
x
x
) and β
v
2
m
v
x
x appear in the equation, here we call (1) a weakly dissipative hyperelastic rod wave equation.In order to link with previous researches in the field, we state here several works on the existence of global weak solution for the Camassa-Holm equation (CH) and a generalized hyperelastic rod wave equation (or a generalized Camassa-Holm equation). Under the sign condition of the initial value, Constantin and Escher [3] obtained the existence and uniqueness results for the global weak solutions of Camassa-Holm equation (also see Constantin and Molinet [4]). Xin and Zhang [5] proved the existence of the global weak solution for the Camassa-Holm equation in the energy space H
1
(
R
) without adding the sign conditions on the initial value. Coclite et al. [6] used the analysis in [5] and established the existence of global weak solutions for a generalized hyperelastic rod wave equation (or a generalized Camassa-Holm equation); namely, λ
=
β
=
0 in (1). For the global or local solutions of the Camassa-Holm equation and other partial differential equations, the reader is referred to [7–12] and the references therein.The objective of this paper is to extend parts of works made by Coclite et al. [6]. We investigate the existence of global weak solutions for the weakly dissipative hyperelastic rod wave equation (1) in the space C
(
[
0
,
∞
)
×
R
)
⋂
‍
L
∞
(
[
0
,
∞
)
;
H
1
(
R
)
). Using the limit of viscous approximations for the equation and several estimates derived from the equation itself, we establish the existence of the global weak solution by merely assuming the initial value in the space H
1
(
R
).This paper is organized as follows. The main result is stated in Section2. In Section 3, we present the viscous problem and establish several estimates of the problem. Namely, we give proofs of an upper bound, a higher integrability estimate, and some basic compactness properties for the viscous approximations. Strong compactness about the derivative of solutions for the viscous approximations is proved in Section 4, where the proof of the main result is finished.


## 2. Main Result

We write the Cauchy problem for (1):
(2)
v
t
-
v
t
x
x
+
∂
x
f
(
v
)
+
λ
(
v
-
v
x
x
)
=
2
α
v
x
v
x
x
+
α
v
v
x
x
x
+
β
v
2
m
v
x
x
,
v
(
0
,
x
)
=
v
0
(
x
)
,
which is equal to the form
(3)
v
t
+
α
v
v
x
+
λ
v
+
∂
Q
∂
x
=
0
,
∂
Q
∂
x
=
Λ
-
2
∂
x
[
f
(
v
)
-
α
2
v
2
+
α
2
v
x
2
-
β
v
2
m
v
x
]
+
2
m
β
Λ
-
2
[
v
2
m
-
1
v
x
2
]
,
v
(
0
,
x
)
=
v
0
(
x
)
,
where the operator Λ
2
=
1
-
(
∂
2
/
∂
x
2
) and
(4)
Λ
-
2
h
(
x
)
=
1
2
∫
R
‍
e
-
|
x
-
y
|
h
(
y
)
d
y
for
h
(
x
)
∈
L
2
(
R
)
.
Using problem (2), we derive the conservation law
(5)
∫
R
‍
(
v
2
+
v
x
2
)
d
x
+
2
∫
0
t
‍
∫
R
‍
[
β
(
2
m
+
1
)
v
2
m
v
x
2
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
l
+
λ
(
v
2
+
v
x
2
)
]
d
x
d
t
=
∫
R
‍
(
v
0
2
+
v
0
x
2
)
d
x
.Here we adopt the definition of global weak solution presented in [5].Definition 1.
A continuous functionv
:
[
0
,
∞
)
×
R
→
R is said to be a global weak solution for the Cauchy problem (2) or (3) if(a)
v
∈
C
(
[
0
,
∞
)
×
R
)
⋂
‍
L
∞
(
[
0
,
∞
)
;
H
1
(
R
)
);
(b)
∥
v
(
t
,
·
)
∥
H
1
(
R
)
≤
∥
v
0
∥
H
1
(
R
);
(c)
v
=
v
(
t
,
x
) satisfies (3) in the sense of distributions and takes on the initial value pointwise.Now we state the main result of this paper.Theorem 2.
Ifv
0
(
x
)
∈
H
1
(
R
), then the Cauchy problem (2) or (3) has a global weak solution v
(
t
,
x
) in the sense of Definition 1. In addition, the weak solution has the following properties.
(
1
) There exists a positive constant c
0 depending on ∥
v
0
∥
H
1
(
R
) and the coefficients of (1) such that
(6)
∂
v
(
t
,
x
)
∂
x
≤
4
α
t
+
c
0
,
f
o
r
(
t
,
x
)
∈
[
0
,
∞
)
×
R
.
(
2
) Assume that 0
<
ρ
<
1, T
>
0, and a
,
b
∈
R
,
a
<
b. Then there exists a positive constant c
1 depending only on ∥
v
0
∥
H
1
(
R
), ρ, T, a, b, and the coefficients of (1) such that
(7)
∫
0
t
‍
∫
a
b
‍
|
∂
v
(
t
,
x
)
∂
x
|
2
+
ρ
d
x
≤
c
1
.


## 3. Viscous Approximations

Consider(8)
ϕ
(
x
)
=
{
e
1
/
(
x
2
-
1
)
,
|
x
|
<
1
,
0
,
|
x
|
≥
1
.
We choose the mollifier ϕ
ɛ
(
x
)
=
ɛ
-
1
/
4
ϕ
(
ɛ
-
1
/
4
x
) with 0
<
ɛ
<
1
/
4 and v
ɛ
,
0
=
ϕ
ɛ
⋆
v
0. We know that v
ɛ
,
0
∈
C
∞ for any v
0
∈
H
s, s
>
0 (see [7]). In fact, we have
(9)
∥
v
ɛ
,
0
∥
H
1
(
R
)
≤
c
∥
v
0
∥
H
1
(
R
)
,
v
ɛ
,
0
⟶
v
0
in
H
1
(
R
)
,
where c is a positive constant.To prove the existence of global weak solutions to the Cauchy problem (3), we will establish compactness of a sequence of smooth functions {
v
ɛ
}
ɛ
>
0 satisfying the viscous problem
(10)
∂
v
ɛ
∂
t
+
α
v
ɛ
∂
v
ɛ
∂
x
+
λ
v
ɛ
+
∂
Q
ɛ
∂
x
=
ɛ
∂
2
v
ɛ
∂
x
2
,
∂
Q
ɛ
∂
x
=
Λ
-
2
∂
x
[
f
(
v
ɛ
)
-
α
2
v
ɛ
2
+
α
2
(
∂
v
ɛ
∂
x
)
2
-
β
v
ɛ
2
m
∂
v
ɛ
∂
x
]
+
2
m
β
Λ
-
2
[
v
ɛ
2
m
-
1
(
∂
v
ɛ
∂
x
)
2
]
,
v
ɛ
(
0
,
x
)
=
v
ɛ
,
0
(
x
)
.For simplicity, throughout this paper, we letc denote any positive constant which is independent of parameter ɛ.Now we give the well-posedness result to problem (10).Lemma 3.
Ifv
0
∈
H
1
(
R
), for any σ
≥
1, there is a unique solution v
ɛ
∈
C
(
[
0
,
∞
)
×
R
) to the Cauchy problem (10). In addition, it holds that
(11)
∫
R
‍
(
v
ɛ
2
+
(
∂
v
ɛ
∂
x
)
2
)
d
x
+
2
∫
0
t
‍
∫
R
‍
[
β
(
2
m
+
1
)
v
ɛ
2
m
(
∂
v
ɛ
∂
x
)
2
h
h
h
h
h
h
h
h
h
h
h
+
λ
(
v
ɛ
2
+
(
∂
v
ɛ
∂
x
)
2
)
]
d
x
d
t
+
2
ɛ
∫
0
t
‍
∫
R
‍
(
(
∂
v
ɛ
∂
x
)
2
+
(
∂
2
v
ɛ
∂
x
2
)
2
)
(
s
,
x
)
d
x
d
s
=
∥
v
ɛ
,
0
∥
H
1
(
R
)
2
,
or
(12)
∥
v
ɛ
(
t
,
·
)
∥
H
1
(
R
)
2
+
2
ɛ
∫
0
t
‍
∥
∂
v
ɛ
∂
x
(
s
,
·
)
∥
H
1
(
R
)
2
d
s
+
2
λ
∫
0
t
‍
∥
v
ɛ
(
s
,
·
)
∥
H
1
(
R
)
2
d
s
+
2
β
(
2
m
+
1
)
∫
0
t
‍
∫
R
‍
v
ɛ
2
m
(
∂
v
ɛ
∂
x
)
2
d
x
=
∥
v
ɛ
,
0
∥
H
1
(
R
)
2
.Proof .
Ifσ
≥
1 and v
0
∈
H
1
(
R
), we know v
ɛ
,
0
∈
C
(
[
0
,
∞
)
;
H
∞
(
R
)
). Using Theorem 2.1 in [6] or Theorem 2.3 in [10], we derive that problem (10) has a unique solution v
ɛ
∈
C
(
[
0
,
∞
)
;
H
∞
(
R
)
).
Applying the first equation in problem (10), we derive
(13)
∂
v
ɛ
∂
t
-
∂
3
v
ɛ
∂
t
x
2
+
∂
f
(
v
ɛ
)
∂
x
+
λ
(
v
ɛ
-
∂
2
v
ɛ
∂
x
2
)
=
2
α
∂
v
ɛ
∂
x
∂
2
v
ɛ
∂
x
2
+
α
v
ɛ
∂
3
v
ɛ
∂
x
3
+
β
v
ɛ
2
m
∂
2
v
ɛ
∂
x
2
+
ɛ
(
∂
2
v
ɛ
∂
x
2
-
∂
4
v
ɛ
∂
x
4
)
,
from which we obtain
(14)
1
2
d
d
t
∫
R
‍
(
v
ɛ
2
+
(
∂
v
ɛ
∂
x
)
2
)
d
x
+
λ
∫
R
‍
[
v
ɛ
2
+
∂
2
v
ɛ
∂
x
2
]
d
x
+
β
(
2
m
+
1
)
∫
R
‍
v
ɛ
2
m
(
∂
v
ɛ
∂
x
)
2
d
x
+
ɛ
∫
R
‍
(
(
∂
v
ɛ
∂
x
)
2
+
(
∂
2
v
ɛ
∂
x
2
)
2
)
d
x
=
0
.
The proof is completed.Using Lemma3, we get
(15)
∥
v
ɛ
∥
L
∞
(
R
)
≤
∥
v
ɛ
∥
H
1
(
R
)
≤
∥
v
ɛ
,
0
∥
H
1
(
R
)
≤
c
∥
v
0
∥
H
1
(
R
)
.
Differentiating the first equation of problem (10) with respect to x and writing ∂
v
ɛ
/
∂
x
=
q
ɛ, we obtain
(16)
∂
q
ɛ
∂
t
+
α
v
ɛ
∂
q
ɛ
∂
x
-
ɛ
∂
2
q
ɛ
∂
x
2
+
α
2
q
ɛ
2
+
β
(
v
ɛ
)
2
m
q
ɛ
+
λ
q
ɛ
=
f
(
v
ɛ
)
-
α
2
v
ɛ
2
-
Λ
-
2
[
f
(
v
ɛ
)
-
α
2
(
v
ɛ
2
-
q
2
)
-
β
(
v
ɛ
)
2
m
q
ɛ
hhhhh
α
2
+
2
m
β
∂
x
[
(
v
ɛ
)
2
m
-
1
q
ɛ
2
]
]
=
P
ɛ
(
t
,
x
)
.Lemma 4.
Let0
<
ρ
<
1, T
>
0, and a
,
b
∈
R
,
a
<
b. Then there exists a positive constant c
1 depending only on ∥
v
0
∥
H
1
(
R
), ρ
,
T
,
a, b, and the coefficients of (1), but independent of ɛ, such that the space higher integrability estimate holds:
(17)
∫
0
T
‍
∫
a
b
‍
|
∂
v
ɛ
(
t
,
x
)
∂
x
|
2
+
ρ
d
x
≤
c
1
,
where v
ɛ
=
v
ɛ
(
t
,
x
) is the unique solution of problem (10).Proof.
The proof is a variant of the proof presented in Xin and Zhang [5] (or see Coclite et al. [6]). Let χ
∈
C
∞
(
R
) be a cut-off function such that 0
≤
χ
≤
1 and
(18)
χ
(
x
)
=
{
1
,
if
x
∈
[
a
,
b
]
,
0
,
if
x
∈
(
-
∞
,
a
-
1
]
∪
[
b
+
1
,
∞
)
.
Considering the map ω
(
ξ
)
:
=
ξ
(
1
+
|
ξ
|
)
ρ
,
ξ
∈
R
,
0
<
ρ
<
1, and observing that
(19)
ω
′
(
ξ
)
=
(
1
+
(
1
+
ρ
)
|
ξ
|
)
(
1
+
|
ξ
|
)
ρ
-
1
,
ω
′′
(
ξ
)
=
ρ
sign
⁡
(
ξ
)
(
1
+
|
ξ
|
)
ρ
-
2
(
2
+
(
1
+
ρ
)
|
ξ
|
)
,
=
ρ
(
1
+
ρ
)
sign
⁡
(
ξ
)
(
1
+
|
ξ
|
)
ρ
-
1
+
(
1
-
ρ
)
ρ
sign
⁡
(
ξ
)
(
1
+
|
ξ
|
)
ρ
-
2
,
we have
(20)
|
ω
(
ξ
)
|
≤
|
ξ
|
+
|
ξ
|
1
+
ρ
,
|
ω
′
(
ξ
)
|
≤
1
+
(
1
+
ρ
)
|
ξ
|
,
|
ω
′′
(
ξ
)
|
≤
2
ρ
,
(21)
ξ
ω
(
ξ
)
-
1
2
ξ
2
ω
′
(
ξ
)
=
1
-
ρ
2
ξ
2
(
1
+
|
ξ
|
)
ρ
+
ρ
2
ξ
2
(
1
+
|
ξ
|
)
ρ
-
1
≥
1
-
ρ
2
ξ
2
(
1
+
|
ξ
|
)
ρ
.
Differentiating the first equation of problem (10) with respect to x and setting v
=
v
ɛ and ∂
v
ɛ
/
∂
x
=
q
ɛ
=
q for simplicity, we obtain
(22)
∂
q
∂
t
+
α
v
∂
q
∂
x
-
ɛ
∂
2
q
∂
x
2
+
α
2
q
2
+
β
(
v
ɛ
)
2
m
q
+
λ
q
=
f
(
v
)
-
α
2
v
2
-
Λ
-
2
[
f
(
v
)
-
α
2
(
v
2
-
q
2
)
-
β
v
2
m
q
hhhhhhhhhhhhhhlhhhh
+
2
m
β
∂
x
[
v
2
m
-
1
v
x
2
]
α
2
]
=
P
ɛ
(
t
,
x
)
.
Multiplying (22) by χ
ω
′
(
q
) and integrating over Π
T
:
=
[
0
,
T
]
×
R, we get
(23)
∫
Π
T
‍
α
χ
(
x
)
q
ω
(
q
)
d
t
d
x
-
α
2
∫
Π
T
‍
q
2
χ
(
x
)
ω
′
(
q
)
d
t
d
x
=
∫
R
‍
χ
(
x
)
(
ω
(
q
(
t
,
x
)
)
-
ω
(
q
(
0
,
x
)
)
)
d
x
-
∫
Π
T
‍
α
v
χ
′
(
x
)
ω
(
q
)
d
t
d
x
+
ɛ
∫
Π
T
‍
∂
q
∂
x
χ
′
(
x
)
ω
′
(
q
)
d
t
d
x
+
ɛ
∫
Π
T
‍
(
∂
q
∂
x
)
2
χ
(
x
)
ω
′′
(
q
)
d
t
d
x
+
∫
Π
T
‍
[
β
v
2
m
+
λ
]
q
χ
(
x
)
ω
′
(
q
)
d
t
d
x
-
∫
Π
T
‍
Q
ɛ
(
t
,
x
)
χ
(
x
)
ω
′
(
q
)
d
t
d
x
.
Using (21) yields
(24)
∫
Π
T
‍
α
χ
(
x
)
q
ω
(
q
)
d
t
d
x
-
α
2
∫
Π
T
‍
q
2
χ
(
x
)
ω
′
(
q
)
d
t
d
x
=
∫
Π
T
‍
α
χ
(
x
)
(
q
ω
(
q
)
-
1
2
q
2
ω
′
(
q
)
)
d
t
d
x
≥
α
(
1
-
ρ
)
2
∫
Π
T
‍
χ
(
x
)
q
2
(
1
+
|
q
|
)
ρ
d
t
d
x
.
Applying the Hölder inequality, (15) and (20) give rise to
(25)
|
∫
R
‍
χ
(
x
)
ω
(
q
)
d
x
|
≤
∫
R
‍
χ
(
x
)
(
|
q
|
1
+
ρ
+
|
q
|
)
d
x
≤
∥
χ
∥
L
2
/
(
1
-
ρ
)
(
R
)
∥
q
∥
L
2
(
R
)
1
+
ρ
+
∥
χ
∥
L
2
(
R
)
∥
q
∥
L
2
(
R
)
≤
(
b
-
a
+
2
)
(
1
-
ρ
)
/
2
∥
v
0
∥
H
1
(
R
)
1
+
ρ
+
(
b
-
a
+
2
)
1
/
2
∥
v
0
∥
H
1
(
R
)
,
|
∫
Π
T
‍
α
v
χ
′
(
x
)
ω
(
q
)
d
t
d
x
|
≤
∫
Π
T
‍
α
|
v
|
|
χ
′
(
x
)
|
(
|
q
|
1
+
ρ
+
|
q
|
)
d
t
d
x
≤
α
∥
v
0
∥
H
1
(
R
)
×
∫
0
T
‍
(
∥
χ
′
∥
L
2
/
(
1
-
ρ
)
(
R
)
∥
q
∥
L
2
(
R
)
1
+
ρ
+
∥
χ
′
∥
L
2
(
R
)
∥
q
∥
L
2
(
R
)
)
d
t
≤
c
T
(
∥
χ
′
∥
L
2
/
(
1
-
ρ
)
(
R
)
∥
v
0
∥
H
1
(
R
)
1
+
ρ
+
∥
χ
′
∥
L
2
(
R
)
∥
v
0
∥
H
1
(
R
)
)
.
Integrating by parts, we have
(26)
∫
Π
T
‍
∂
q
∂
x
χ
′
(
x
)
ω
′
(
q
)
d
t
d
x
=
-
∫
Π
T
‍
ω
(
q
)
χ
′′
(
x
)
d
t
d
x
.
From (20), (26), and the Hölder inequality, we have
(27)
|
∫
Π
T
‍
∂
q
∂
x
χ
′
(
x
)
ω
′
(
q
)
d
t
d
x
|
≤
ɛ
∫
Π
T
‍
|
ω
(
q
)
|
|
χ
′′
(
x
)
|
d
t
d
x
≤
ɛ
∫
Π
T
‍
|
χ
′′
(
x
)
|
(
|
q
|
1
+
ρ
+
|
q
|
)
d
t
d
x
≤
ɛ
∫
0
T
‍
(
∥
χ
′′
∥
L
2
/
(
1
-
ρ
)
(
R
)
∥
q
∥
L
2
(
R
)
1
+
ρ
+
∥
χ
′′
∥
L
2
(
R
)
∥
q
∥
L
2
(
R
)
)
d
t
≤
ɛ
T
(
∥
χ
′′
∥
L
2
/
(
1
-
ρ
)
(
R
)
∥
v
0
∥
H
1
(
R
)
1
+
ρ
+
∥
χ
′′
∥
L
2
(
R
)
∥
v
0
∥
H
1
(
R
)
)
.
Using (20) and Lemma 3 derives
(28)
ɛ
|
∫
Π
T
‍
(
∂
q
∂
x
)
2
χ
(
x
)
ω
′′
(
q
)
d
t
d
x
|
≤
2
γ
ɛ
∫
Π
T
‍
(
∂
q
∂
x
)
2
d
t
d
x
≤
γ
T
∥
v
0
∥
H
1
(
R
)
2
.
It follows from (20) that
(29)
|
∫
Π
T
‍
[
β
v
2
m
+
λ
]
q
χ
(
x
)
θ
′
(
q
)
d
t
d
x
|
≤
c
∫
Π
T
‍
|
χ
(
x
)
|
(
(
1
+
ρ
)
q
2
+
|
q
|
)
d
t
d
x
≤
β
c
(
(
1
+
ρ
)
∥
χ
∥
L
∞
∥
v
0
∥
H
1
(
R
)
2
hhhhhhh
+
∥
χ
(
x
)
∥
L
2
(
R
)
∥
v
0
∥
H
1
(
R
)
)
.
In fact, we have
(30)
|
Λ
-
2
[
v
2
m
-
1
v
x
2
]
x
|
=
|
1
2
e
-
x
∫
-
∞
x
‍
e
y
[
v
2
m
-
1
v
y
2
]
y
d
y
+
1
2
e
x
∫
x
∞
‍
e
-
y
[
v
2
m
-
1
v
y
2
]
y
d
y
|
=
|
-
1
2
e
-
x
∫
-
∞
x
‍
e
y
v
2
m
-
1
v
y
2
d
y
+
1
2
e
x
∫
x
∞
‍
v
2
m
-
1
v
y
2
d
y
|
≤
2
∫
-
∞
∞
‍
e
-
|
x
-
y
|
|
v
2
m
-
1
|
v
y
2
d
y
≤
2
∥
v
∥
L
∞
(
R
)
2
m
-
1
∥
v
∥
H
1
(
R
)
≤
2
∥
v
0
∥
L
∞
(
R
)
2
m
-
1
∥
v
0
∥
H
1
(
R
)
.
Applying (15), the Hölder inequality, Lemma 3, and ∫
R
‍
e
-
|
x
-
y
|
d
y
=
2, we have
(31)
|
f
(
v
)
-
α
2
v
2
-
Λ
-
2
[
f
(
v
)
-
α
2
(
v
2
-
v
x
2
)
-
β
v
2
m
v
x
]
|
≤
3
(
∥
f
(
v
)
∥
L
∞
(
R
)
+
α
2
∥
v
∥
L
∞
(
R
)
2
)
+
α
2
∥
v
∥
H
1
(
R
)
2
+
β
∥
v
∥
L
∞
(
R
)
2
m
-
1
∥
v
∥
L
2
(
R
)
∥
v
x
∥
L
2
(
R
)
≤
c
,
where c is a constant independent of ɛ.
From (30) and (31), we know that there exists a positive constant c depending on ∥
v
0
∥
H
1
(
R
), but independent of ɛ, such that
(32)
∥
P
ɛ
(
t
,
x
)
∥
L
∞
(
R
)
≤
c
,
which results in
(33)
|
∫
Π
T
‍
P
ɛ
(
t
,
x
)
χ
(
x
)
ω
′
(
q
)
d
t
d
x
|
≤
c
∫
Π
T
‍
|
χ
(
x
)
|
(
(
1
+
ρ
)
|
q
|
+
1
)
d
t
d
x
≤
c
T
(
(
1
+
ρ
)
∥
χ
(
x
)
∥
L
2
(
R
)
∥
v
0
∥
H
1
(
R
)
+
∫
R
‍
|
χ
(
x
)
|
d
x
)
.
From inequalities (22)–(29) and (33), we obtain (17).Lemma 5.
There exists a positive constantC depending only on ∥
v
0
∥
H
1
(
R
) and the coefficients of (1) such that
(34)
∥
P
ɛ
(
t
,
·
)
∥
L
∞
(
R
)
≤
c
,
(35)
∥
P
ɛ
(
t
,
·
)
∥
L
1
(
R
)
≤
c
,
(36)
∥
P
ɛ
(
t
,
·
)
∥
L
2
(
R
)
≤
c
.Proof.
Writingv
(
t
,
x
)
=
v
ɛ
(
t
,
x
), we get
(37)
P
ɛ
(
t
,
x
)
=
f
(
v
)
-
α
2
v
2
-
Λ
-
2
[
f
(
v
)
-
α
2
(
v
2
-
v
x
2
)
-
β
v
2
m
v
x
hhhhhhhhhhhhhhhhhhhhhl
+
2
m
β
∂
x
[
v
2
m
-
1
v
x
2
]
α
2
]
.
Inequality (34) is proved in Lemma 4 (see (32)). Now we prove (35). Since
(38)
∫
R
‍
|
Λ
-
2
[
v
2
m
-
1
v
x
2
]
x
|
d
x
=
∫
R
‍
|
1
2
e
-
x
∫
-
∞
x
‍
e
y
[
v
2
m
-
1
v
y
2
]
y
d
y
hhhh
+
1
2
e
x
∫
x
∞
‍
e
-
y
[
v
2
m
-
1
v
y
2
]
y
d
y
|
d
x
,
=
∫
R
‍
|
-
1
2
∫
-
∞
x
‍
e
y
v
2
m
-
1
v
y
2
d
y
hhhhl
+
1
2
∫
x
∞
‍
v
2
m
-
1
v
y
2
d
y
|
d
x
,
≤
∫
∫
R
‍
e
-
|
x
-
y
|
|
v
2
m
-
1
|
v
y
2
d
y
d
x
≤
2
∥
v
∥
L
∞
(
R
)
2
m
-
1
∥
v
∥
H
1
(
R
)
≤
2
∥
v
0
∥
L
∞
(
R
)
2
m
-
1
∥
v
0
∥
H
1
(
R
)
,
∥
f
(
v
)
-
α
2
v
2
∥
L
1
(
R
)
≤
c
0
∥
v
∥
L
∞
(
R
)
n
-
2
∥
v
∥
L
2
(
R
)
2
+
α
2
∥
v
∥
L
2
(
R
)
2
≤
c
0
(
∥
v
0
∥
H
1
(
R
)
n
+
∥
v
0
∥
H
1
(
R
)
2
)
,
∫
R
‍
|
Λ
-
2
[
f
(
v
)
-
α
2
v
2
+
α
2
v
x
2
-
β
v
2
m
v
x
]
|
d
x
≤
∫
R
‍
|
∫
R
‍
e
-
|
x
-
y
|
[
f
(
v
)
-
α
2
v
2
+
α
2
v
y
2
-
β
v
2
m
v
y
]
d
y
|
d
x
≤
∫
R
‍
e
-
|
x
-
y
|
d
x
∫
R
‍
|
f
(
v
)
-
α
2
v
2
+
α
2
v
y
2
-
β
v
2
m
v
y
|
d
y
≤
2
c
0
(
∥
v
0
∥
H
1
(
R
)
2
+
∥
v
0
∥
H
1
(
R
)
n
+
∥
v
∥
H
1
(
R
)
2
+
∥
v
∥
L
∞
(
R
)
2
m
-
1
∥
v
∥
H
1
(
R
)
2
)
≤
c
0
(
∥
v
0
∥
H
1
(
R
)
2
+
∥
v
0
∥
H
1
(
R
)
n
+
∥
v
0
∥
H
1
(
R
)
2
m
+
1
)
,
we know that (35) holds.
Applying the Tonelli theorem, (34), and (35), we get
(39)
∥
P
ɛ
(
t
,
·
)
∥
L
2
(
R
)
2
≤
∥
P
ɛ
(
t
,
·
)
∥
L
∞
∥
P
ɛ
(
t
,
·
)
∥
L
1
(
R
)
≤
c
.
The proof of Lemma 5 is completed.Lemma 6.
Assume thatv
ɛ
=
v
ɛ
(
t
,
x
) is the unique solution of (10). For an arbitrary T
>
0, there exists a positive constant C depending only on ∥
v
0
∥
H
1
(
R
) and the coefficients of (1) such that the following one-sided L
∞ norm estimate on the first order spatial derivative holds:
(40)
∂
v
ɛ
(
t
,
x
)
∂
x
≤
4
α
t
+
C
,
f
o
r
(
t
,
x
)
∈
[
0
,
∞
)
×
R
.Proof.
From (16) and Lemma 5, we know that there exists a positive constant C depending only on ∥
v
0
∥
H
1
(
R
) and the coefficients of (1) such that ∥
P
ɛ
(
t
,
x
)
∥
L
∞
(
R
)
≤
C. Therefore,
(41)
∂
q
ɛ
∂
t
+
α
v
ɛ
∂
q
ɛ
∂
x
+
α
2
q
ɛ
2
+
β
(
v
ɛ
)
2
m
q
ɛ
+
λ
q
ɛ
-
ɛ
∂
2
q
ɛ
∂
x
2
=
P
ɛ
(
t
,
x
)
≤
C
.
Letg
=
g
(
t
) be the solution of
(42)
d
g
d
t
+
α
2
g
2
+
β
(
v
*
)
2
m
g
+
λ
g
=
C
,
t
>
0
,
g
(
0
)
=
∥
∂
v
ɛ
,
0
∂
x
∥
L
∞
(
R
)
,
where v
ɛ
* is the value of v
ɛ
(
t
,
x
) when sup
⁡
x
∈
R
q
ɛ
(
t
,
x
)
=
g
(
t
). From the comparison principle for parabolic equations, we get
(43)
q
ɛ
(
t
,
x
)
≤
g
(
t
)
.
Using (15) and suitably choosing C
1, we have -
β
(
u
*
)
2
m
g
≤
(
α
/
8
)
g
2
+
C
1 and -
λ
g
≤
(
α
/
8
)
g
2
+
C
1. Furthermore, we have
(44)
d
g
d
t
=
C
-
α
2
g
2
-
β
(
v
*
)
2
m
g
-
λ
g
≤
C
-
α
2
g
2
+
α
4
g
2
+
2
C
1
≤
C
-
α
4
g
2
+
2
C
1
.
Setting M
0
=
C
+
2
C
1, we obtain
(45)
d
g
d
t
+
α
4
g
2
≤
M
0
.
Letting K
(
t
)
=
4
/
α
t
+
(
4
/
α
)
M
0, we have d
K
(
t
)
/
d
t
+
(
α
/
4
)
K
2
(
t
)
-
M
0
=
4
M
0
/
α
/
t
>
0. From the comparison principle for ordinary differential equations, we get g
(
t
)
≤
K
(
t
) for all t
>
0. Therefore, by this and (43), the estimate (40) is proved.Lemma 7.
There exists a sequence{
ɛ
j
}
j
∈
N tending to zero and a function v
∈
L
∞
(
[
0
,
∞
)
;
H
1
(
R
)
)
⋂
‍
H
1
(
[
0
,
T
]
×
R
) such that, for each T
>
0, it holds that
(46)
v
ɛ
j
⇀
v
i
n
H
1
(
[
0
,
T
]
×
R
)
,
f
o
r
e
a
c
h
T
>
0
,
v
ɛ
j
⟶
v
i
n
L
loc
⁡
∞
(
[
0
,
∞
)
×
R
)
,
where v
ɛ
=
v
ɛ
(
t
,
x
) is the unique solution of (10).The proof of this lemma is fully similar to that of Lemma 5.2 in [6]. Here we omit its proof.Lemma 8.
There exists a sequence{
ɛ
j
}
j
∈
N tending to zero and a function P
∈
L
∞
(
[
0
,
∞
)
×
R
) such that for each 1
<
p
<
∞
(47)
P
ɛ
j
⟶
P
s
t
r
o
n
g
l
y
i
n
L
loc
⁡
p
(
[
0
,
∞
)
×
R
)
.Proof.
Using the same arguments presented in Coclite et al. [6], we obtain ∫
R
‍
|
d
P
ɛ
/
d
t
|
d
x
≤
c which together with Lemma 5 derives that (47) holds.In this paper we use overbars to denote weak limits which are taken inL
loc
⁡
p
(
[
0
,
∞
)
×
R
) with 1
<
p
<
3.Lemma 9.
There exists a sequence{
ɛ
j
}
j
∈
N tending to zero and two functions q
∈
L
loc
⁡
p
(
[
0
,
∞
)
×
R
), q
2
¯
∈
L
loc
⁡
r
(
[
0
,
∞
)
×
R
) such that
(48)
q
ɛ
j
⇀
q
i
n
L
loc
⁡
p
(
[
0
,
∞
)
×
R
)
,
q
ɛ
j
⇀
⋆
q
i
n
L
loc
⁡
∞
(
[
0
,
∞
)
;
L
2
(
R
)
)
,
(49)
q
ɛ
j
2
⇀
q
2
¯
i
n
L
loc
⁡
r
(
[
0
,
∞
)
×
R
)
,
for each 1
<
p
<
3 and 1
<
r
<
3
/
2. Moreover,
(50)
q
2
(
t
,
x
)
≤
q
2
¯
(
t
,
x
)
f
o
r
a
l
m
o
s
t
e
v
e
r
y
(
t
,
x
)
∈
[
0
,
∞
)
×
R
,
(51)
∂
v
∂
x
=
q
i
n
t
h
e
s
e
n
s
e
o
f
d
i
s
t
r
i
b
u
t
i
o
n
s
o
n
[
0
,
∞
)
×
R
.Proof.
Using Lemmas3 and 4 derives that (48) and (49) hold. It follows from (49) that (50) holds. From Lemma 7 and (48), we know that (51) is valid.Using (48), for any convex function ψ
∈
C
1
(
R
) with ψ
′ being bounded and Lipschitz continuous on R and for any 1
<
p
<
3, we conclude that
(52)
ψ
(
q
ɛ
)
⇀
ψ
(
q
)
¯
in
L
loc
⁡
p
(
[
0
,
∞
)
×
R
)
,
ψ
(
q
ɛ
)
⇀
⋆
ψ
(
q
)
¯
in
L
loc
⁡
∞
(
[
0
,
∞
)
;
L
2
(
R
)
)
.
Multiplying (16) by ψ
′
(
q
ɛ
) yields
(53)
∂
∂
t
ψ
(
q
ɛ
)
+
α
∂
∂
x
(
v
ɛ
ψ
(
q
ɛ
)
)
-
ɛ
∂
2
∂
x
2
ψ
(
q
ɛ
)
+
ɛ
ψ
′′
(
q
ɛ
)
(
∂
q
ɛ
∂
x
)
2
=
α
q
ɛ
ψ
(
q
ɛ
)
-
α
2
ψ
′
(
q
ɛ
)
q
ɛ
2
-
[
β
(
v
ɛ
)
2
m
+
λ
]
q
ɛ
ψ
′
(
q
ɛ
)
+
P
ɛ
(
t
,
x
)
ψ
′
(
q
ɛ
)
.Lemma 10.
For the convexψ
∈
C
1
(
R
) with ψ
′ being bounded and Lipschitz continuous on R, it has
(54)
∂
ψ
(
q
)
∂
t
¯
+
∂
∂
x
(
α
v
ψ
(
q
)
¯
)
≤
α
q
ψ
(
q
)
¯
-
α
2
ψ
′
(
q
)
q
2
¯
-
[
β
v
2
m
+
λ
]
q
ψ
′
(
q
)
¯
+
P
(
t
,
x
)
ψ
′
(
q
)
¯
in the sense of distributions on [
0
,
∞
)
×
R. Here q
ψ
(
q
)
¯ and ψ
′
(
q
)
q
2
¯ denote the weak limits of q
ɛ
ψ
(
q
ɛ
) and q
ɛ
2
ψ
′
(
q
ɛ
) in L
loc
⁡
r
(
[
0
,
∞
)
×
R
), 1
<
r
<
3
/
2, respectively.Proof .
In (53), by the convexity of ψ, (15), Lemmas 7, 8, and 9, taking limit for ɛ
→
0 gives rise to the desired result.Remark 11.
Using Lemma9, we know that
(55)
q
=
q
+
+
q
-
=
q
+
¯
+
q
-
¯
,
q
2
=
(
q
+
)
2
+
(
q
-
)
2
,
q
2
¯
=
(
q
+
)
2
¯
+
(
q
-
)
2
¯
almost everywhere in [
0
,
∞
)
×
R, where ξ
+
:
=
ξ
χ
[
0
,
+
∞
)
(
ξ
)
,
ξ
-
:
=
ξ
χ
(
-
∞
,
0
]
(
ξ
) for ξ
∈
R. From Lemma 6 and (48), we have
(56)
q
ɛ
(
t
,
x
)
,
q
(
t
,
x
)
≤
4
α
t
+
C
,
for
t
>
0
,
x
∈
R
,
where C is a constant depending only on ∥
v
0
∥
H
1
(
R
) and the coefficients of (1).Lemma 12.
The identity(57)
∂
q
∂
t
+
∂
∂
x
(
α
v
q
)
=
α
2
q
2
¯
-
[
β
v
2
m
+
λ
]
q
+
P
(
t
,
x
)
holds in the sense of distributions on [
0
,
∞
)
×
R.Proof .
Applying (16), Lemmas 7 and 8, (48), (49), and (51), the conclusion (57) holds by taking limit for ɛ
→
0 in (16).Lemma 13.
Ifψ
∈
C
1
(
R
) with ψ
∈
L
∞
(
R
), then the identity
(58)
∂
ψ
(
q
)
∂
t
+
∂
∂
x
(
α
v
ψ
(
q
)
)
=
α
q
ψ
(
q
)
+
(
α
2
q
2
¯
-
α
q
2
)
ψ
′
(
q
)
-
[
β
v
2
m
+
λ
]
q
ψ
′
(
q
)
+
P
(
t
,
x
)
ψ
′
(
q
)
holds in the sense of distributions on [
0
,
∞
)
×
R.Proof .
Let{
ω
δ
}
δ be a family of mollifiers defined on R. Consider q
δ
(
t
,
x
)
:
=
(
q
(
t
,
·
)
⋆
ω
δ
)
(
x
) where the ⋆ is the convolution with respect to x variable. Multiplying (57) by ψ
′
(
q
δ
) gives rise to
(59)
∂
ψ
(
q
δ
)
∂
t
=
ψ
′
(
q
δ
)
∂
q
δ
∂
t
=
ψ
′
(
q
δ
)
[
α
2
q
2
¯
⋆
ω
δ
-
[
β
v
2
m
+
λ
]
q
δ
hhhhhhhhhh
+
P
(
t
,
x
)
⋆
ω
δ
-
α
q
2
hhhhhhhhhh
⋆
ω
δ
-
α
v
∂
q
∂
x
⋆
ω
δ
]
,
∂
∂
x
(
α
v
ψ
(
q
δ
)
)
=
α
q
ψ
(
q
δ
)
+
α
v
ψ
′
(
q
δ
)
(
∂
q
δ
∂
x
)
.
Applying the boundedness ofψ, ψ
′ and letting δ
→
0 in (59), we derive that (58) holds.


## 4. Strong Convergence ofq
ɛ and Existence of Global Weak Solutions

Using the methods in [5] or [6], in this section, we will prove that the weak convergence of q
ɛ in (48) is strong convergence. This results in the existence of global weak solutions for problem (10).Lemma 14.
Assume thatv
0
∈
H
1
(
R
). Then
(60)
lim
⁡
t
→
0
∫
R
‍
q
2
(
t
,
x
)
d
x
=
lim
⁡
t
→
0
∫
R
‍
q
2
¯
(
t
,
x
)
d
x
=
∫
R
‍
(
∂
v
0
∂
x
)
2
d
x
.Lemma 15.
Ifv
0
∈
H
1
(
R
), for constant M
>
0, then
(61)
lim
⁡
t
→
0
∫
R
‍
(
ψ
M
±
(
q
)
¯
(
t
,
x
)
-
ψ
M
±
(
q
(
t
,
x
)
)
)
d
x
=
0
,
where
(62)
ψ
M
(
ξ
)
:
=
{
1
2
ξ
2
,
i
f
|
ξ
|
≤
M
,
M
|
ξ
|
-
1
2
M
2
,
i
f
|
ξ
|
>
M
,
and ψ
M
+
(
ξ
)
:
=
ψ
M
(
ξ
)
χ
[
0
,
+
∞
)
(
ξ
)
,
ψ
M
-
(
ξ
)
:
=
ψ
M
(
ξ
)
χ
(
-
∞
,
0
]
(
ξ
), ξ
∈
R.Lemma 16.
Let constantM
>
0. Then for each ξ
∈
R
(63)
ψ
M
(
ξ
)
=
1
2
ξ
2
-
1
2
(
M
-
|
ξ
|
)
2
χ
(
-
∞
,
-
M
)
⋂
‍
(
M
,
∞
)
(
ξ
)
,
ψ
M
′
(
ξ
)
ξ
=
ξ
+
(
M
-
|
ξ
|
)
sign
⁡
(
ξ
)
χ
(
-
∞
,
-
M
)
⋂
‍
(
M
,
∞
)
(
ξ
)
,
ψ
M
+
(
ξ
)
=
1
2
(
ξ
+
)
2
-
1
2
(
M
-
ξ
)
2
χ
(
M
,
∞
)
(
ξ
)
,
(
ψ
M
+
)
′
(
ξ
)
=
ξ
+
+
(
M
-
ξ
)
χ
(
M
,
∞
)
(
ξ
)
,
ψ
M
-
(
ξ
)
=
1
2
(
ξ
-
)
2
-
1
2
(
M
+
ξ
)
2
χ
(
-
∞
,
-
M
)
(
ξ
)
,
(
ψ
M
-
)
′
(
ξ
)
=
ξ
-
-
(
M
+
ξ
)
χ
(
-
∞
,
-
M
)
(
ξ
)
.The proofs of Lemmas14, 15, and 16 can be found in [5] or [6].Lemma 17.
Assume thatv
0
∈
H
1
(
R
). Then for almost all t
>
0
(64)
1
2
∫
R
‍
(
(
q
+
)
2
¯
-
q
+
2
)
(
t
,
x
)
d
x
≤
∫
0
t
‍
∫
R
‍
P
(
s
,
x
)
[
q
+
¯
(
s
,
x
)
-
q
+
(
s
,
x
)
]
d
s
d
x
.Lemma 18.
For anyt
>
0 and constant M
>
0, it holds that
(65)
∫
R
‍
(
ψ
M
-
(
q
)
¯
-
ψ
M
-
(
q
)
)
(
t
,
x
)
d
x
≤
α
M
2
2
∫
0
t
‍
∫
R
‍
v
(
M
+
q
)
χ
(
-
∞
,
-
M
)
(
q
)
¯
d
s
d
x
-
α
M
2
2
∫
0
t
‍
∫
R
‍
v
(
M
+
q
)
χ
(
-
∞
,
-
M
)
(
q
)
d
s
d
x
+
α
M
∫
0
t
‍
∫
R
‍
v
[
ψ
M
-
(
q
)
¯
-
ψ
M
-
(
q
)
]
d
s
d
x
+
α
M
2
∫
0
t
‍
∫
R
‍
v
(
q
+
2
¯
-
q
+
2
)
d
s
d
x
+
∫
0
t
‍
∫
R
‍
P
(
t
,
x
)
(
(
ψ
M
-
)
′
(
q
)
¯
-
(
ψ
M
-
)
′
(
q
)
)
d
s
d
x
.Using the same techniques in [5] or [6], because of (54), (58), and Lemma 6, we can prove Lemmas 17 and 18. Here we omit their proofs.Lemma 19.
It holds that(66)
q
2
¯
=
q
2
a
l
m
o
s
t
e
v
e
r
y
w
h
e
r
e
i
n
[
0
,
∞
)
×
(
-
∞
,
∞
)
.Directly applying the approaches in [5] or [6] and Lemmas 17 and 18, we obtain that Lemma 19 holds. Here we omit its proof.Proof of the Main Result.
From (9), (11), and Lemma 7, we conclude that conditions (a) and (b) in Definition 1 are satisfied. We need to prove (c). Using Lemma 19, we get
(67)
q
ɛ
⟶
q
in
L
loc
⁡
2
(
[
0
,
∞
)
×
R
)
.
From Lemma 7, (47), and (67), we know that v is a distributional solution to problem (3). The inequalities (6) and (7) are consequences of Lemmas 4 and 6. The proof is completed.



---

*Source: 264162-2014-09-02.xml*