Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 247841, 5 pages
http://dx.doi.org/10.1155/2013/247841
Research Article
Several Dynamic Properties of Solutions to a Generalized
Camassa-Holm Equation
Zheng Yin
Department of Mathematics, Southwestern University of Finance and Economics, Chengdu 610074, China
Correspondence should be addressed to Zheng Yin; yinzheng@swufe.edu.cn
Received 31 March 2013; Accepted 12 May 2013
Academic Editor: Shaoyong Lai
Copyright © 2013 Zheng Yin. This is an open access article distributed under the Creative Commons Attribution License, which
permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
For a nonlinear generalization of the Camassa-Holm equation, we investigate the dynamic properties of solutions for the equation
under the assumption that the initial value 𝑢0 (𝑥) lies in the space 𝐻1 (𝑅). A one-sided upper bound estimate on the first-order
spatial derivative, 𝐿𝑝 bound estimate, and a space-time higher-norm estimate for the solutions are obtained.
1. Introduction
Hakkaev and Kirchev [1] investigated the following generalized Camassa-Holm equation:
𝑢𝑡 − 𝑢𝑡𝑥𝑥 + (2𝑘𝑢 +
2
𝑢
𝑛 + 2 𝑛+1
𝑢 ) = (𝑛𝑢𝑛−1 𝑥 + 𝑢𝑛 𝑢𝑥𝑥 ) ,
2
2
𝑥
𝑥
(1)
where 𝑛 is an integer. When 𝑛 = 1, (1) becomes the
Camassa-Holm model (see [2]). The local well-posedness
in the Sobolev space 𝐻𝑠 with 𝑠 > 3/2 is established, and
sufficient conditions for the stability and instability of the
solitary wave solutions are given in [1]. However, the 𝐿𝑝
estimate of strong solutions and one-sided upper bound
estimate on the first-order spatial derivative for the solutions
are not discussed in [1]. This constitutes the objective of this
work.
Like the Camassa-Holm equation (see [1, 2]), (1) has the
conservation law
∫ (𝑢2 (𝑡, 𝑥) + 𝑢𝑥2 (𝑡, 𝑥)) 𝑑𝑥 = ∫ (𝑢2 (0, 𝑥) + 𝑢𝑥2 (0, 𝑥)) 𝑑𝑥,
𝑅
𝑅
(2)
which plays an important role in our further investigations.
In fact, many scholars have paid their attentions to the
study of the Camassa-Holm equation. The existence of global
weak solutions is established in Constantin and Escher [3],
Constantin and Molinet [4], Xin and Zhang [5], and Coclite
et al. [6]. It was shown in Constantin and Escher [7] 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 is continued
uniquely either as a global conservative weak solution [8, 9]
or as a global dissipative solution [10, 11]. Exact traveling wave
solutions for the Camassa-Holm equation are presented in
[12]. For other methods to investigate the problems involving
various dynamic properties of the Camassa-Holm equation,
the reader is referred to [13–16] and the references therein.
In this paper, we investigate several dynamic properties of
strong solutions for the generalized Camassa-Holm equation
(1) in the case where 𝑛 is an odd natural number and the
assumption 𝑢0 (𝑥) ∈ 𝐻1 (𝑅). The results obtained in this
work include a one-sided upper bound estimate on the firstorder derivatives of the solution, a space-time higher-norm
estimate, and the 𝐿𝑝 (2 ≤ 𝑝 < ∞) bound estimate.
The rest of this paper is organized as follows. Section 2
states the main result. Several lemmas are given in Section 3
where the proof of main result is completed.
2. Main Result
Let 𝑚 be a nonnegative integer and 𝑛 = 2𝑚 + 1. In this case,
the Cauchy problem for (1) is written in the form
𝑢𝑡 − 𝑢𝑡𝑥𝑥 + 𝜕𝑥 (2𝑘𝑢 +
2𝑚 + 3 2𝑚+2
)
𝑢
2
2
Abstract and Applied Analysis
=
1
𝑢2𝑚+2 𝜕𝑥3 𝑢2𝑚+2
2𝑚 + 2
−
Differentiating the first equation of problem (4) with respect
to 𝑥 and writing 𝜕𝑢/𝜕𝑥 = 𝑞, we obtain
2𝑚 + 1
𝜕𝑥 (𝑢2𝑚 𝑢𝑥2 ) ,
2
𝜕𝑞 2𝑚 + 1 2𝑚 2
𝜕𝑞
+ 𝑢2𝑚+1
+
𝑢 𝑞
𝜕𝑡
𝜕𝑥
2
𝑢 (0, 𝑥) = 𝑢0 (𝑥) ,
(3)
which is equivalent to
𝑢𝑡 + 𝑢
2𝑚+1
𝜕𝐻
𝑢𝑥 +
= 0,
𝜕𝑥
𝑢 (0, 𝑥) = 𝑢0 (𝑥) ,
(4)
where the operator Λ2 = 1 − 𝜕2 /𝜕𝑥2 and
1
∫ 𝑒−|𝑥−𝑦| 𝑓 (𝑦) 𝑑𝑦
2 𝑅
for 𝑓 (𝑥) ∈ 𝐿2 (𝑅) . (5)
We introduce a result presented in [1] for problem (3).
Lemma 1 (see [1]). Suppose that 𝑢0 ∈ 𝐻𝑠 with constant 𝑠 >
3/2. Then there is a real number 𝑇 > 0 such that the problem
(3) has a unique solution 𝑢(𝑡, 𝑥) satisfying
𝑠
1
𝑢 (𝑡, 𝑥) ∈ 𝐶 ([0, 𝑇) ; 𝐻 (𝑅)) ⋂ 𝐶 ([0, 𝑇) ; 𝐻
2𝑚2 + 5𝑚 + 2 2𝑚+2
𝑢
2 (𝑚 + 1)
− Λ−2 [2𝑘𝑢 +
2𝑚2 + 5𝑚 + 2 2𝑚+2 2𝑚 + 1 2𝑚 2
𝜕𝐻
𝑢
+
= Λ−2 𝜕𝑥 [2𝑘𝑢 +
𝑢 𝑢𝑥 ] ,
𝜕𝑥
2 (𝑚 + 1)
2
Λ−2 𝑓 (𝑥) =
= 2𝑘𝑢 +
𝑠−1
(𝑅)) . (6)
2𝑚2 + 5𝑚 + 2 2𝑚+2 2𝑚 + 1 2𝑚 2
+
𝑢
𝑢 𝑞]
2 (𝑚 + 1)
2
= 𝑄 (𝑡, 𝑥) .
(11)
Lemma 3. Let 0 < 𝛾 < 1, 𝑇 > 0, and 𝑎, 𝑏 ∈ 𝑅, 𝑎 <
𝑏. Then there exists a positive constant 𝑐1 depending only
on ‖𝑢0 ‖𝐻1 (𝑅) , 𝛾, 𝑇, 𝑎, 𝑏, and 𝑚, such that the space higher
integrability estimate holds
󵄨󵄨 𝜕𝑢 (𝑡, 𝑥) 󵄨󵄨2+𝛾
𝑇 𝑏
󵄨
󵄨󵄨
∫ ∫ 𝑢2𝑚 󵄨󵄨󵄨
󵄨 𝑑𝑥 ≤ 𝑐1 ,
󵄨󵄨 𝜕𝑥 󵄨󵄨󵄨
0 𝑎
(12)
where 𝑢 = 𝑢(𝑡, 𝑥) is the unique solution of problem (3).
Proof. The proof is a variant of the proof presented in Xin and
Zhang [5] (or see Coclite et al. [6]). Let 𝜙 ∈ 𝐶∞ (𝑅) be a cut-off
function such that 0 < 𝜙 < 1 and
Now we state the main result of this paper.
Theorem 2. Let 𝑢0 ∈ 𝐻𝑠 (𝑅) with 𝑠 > 3. Then the solution of
problem (3) has the following properties.
(a) There exists a positive constant 𝑐0 depending on 𝑚 and
‖𝑢0 ‖𝐻1 (𝑅) such that the one-sided 𝐿∞ norm estimate on
the first-order spatial derivative holds
𝜕𝑢 (𝑡, 𝑥)
≤ 𝑐0 (1 + 𝑡) ,
𝜕𝑥
for (𝑡, 𝑥) ∈ [0, ∞) × 𝑅.
(7)
(b) Let 0 < 𝛾 < 1, 𝑇 > 0, and 𝑎, 𝑏 ∈ 𝑅, 𝑎 < 𝑏.
Then there exists a positive constant 𝑐1 depending only
on ‖𝑢0 ‖𝐻1 (𝑅) , 𝛾, 𝑇, 𝑎, 𝑏, and 𝑚 such that the following
estimate holds:
󵄨󵄨 𝜕𝑢 (𝑡, 𝑥) 󵄨󵄨2+𝛾
𝑡 𝑏
󵄨
󵄨󵄨
(8)
∫ ∫ 𝑢2𝑚 󵄨󵄨󵄨
󵄨󵄨 𝑑𝑥 ≤ 𝑐1 .
󵄨
󵄨󵄨
𝜕𝑥
0 𝑎
󵄨
(c) There exists a constant 𝑐2 depending only on
‖𝑢0 ‖𝐻1 (𝑅) , and 𝑚, 𝑝 such that
‖𝑢‖𝐿𝑝 (𝑅) ≤ 𝑐2 (1 + 𝑡) ,
𝑡 ∈ [0, ∞) , 2 ≤ 𝑝 < ∞.
(9)
if 𝑥 ∈ [𝑎, 𝑏] ,
if 𝑥 ∈ (−∞, 𝑎 − 1] ⋃ [𝑏 + 1, ∞) .
(13)
Letting 𝜃(𝜉) = 𝜉(1 + |𝜉|)𝛾 , 𝜉 ∈ 𝑅, and 0 < 𝛾 < 1 yields
󵄨 󵄨 𝛾−1
󵄨 󵄨
𝜃󸀠 (𝜉) = (1 + (1 + 𝛾) 󵄨󵄨󵄨𝜉󵄨󵄨󵄨) (1 + 󵄨󵄨󵄨𝜉󵄨󵄨󵄨) , 𝜃󸀠󸀠 (𝜉)
󵄨 󵄨 𝛾−2
󵄨 󵄨
= 𝛾 sign (𝜉) (1 + 󵄨󵄨󵄨𝜉󵄨󵄨󵄨) (2 + (1 + 𝛾) 󵄨󵄨󵄨𝜉󵄨󵄨󵄨)
󵄨 󵄨 𝛾−1
= 𝛾 (1 + 𝛾) sign (𝜉) (1 + 󵄨󵄨󵄨𝜉󵄨󵄨󵄨)
(14)
󵄨 󵄨 𝛾−2
+ (1 − 𝛾) 𝛾 sign (𝜉) (1 + 󵄨󵄨󵄨𝜉󵄨󵄨󵄨) ,
from which we get
󵄨 󵄨 󵄨 󵄨 󵄨1+𝛾
󵄨󵄨
󵄨󵄨󵄨𝜃󸀠 (𝜉)󵄨󵄨󵄨 ≤ 1 + (1 + 𝛾) 󵄨󵄨󵄨𝜉󵄨󵄨󵄨 ,
󵄨󵄨𝜃 (𝜉)󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨𝜉󵄨󵄨󵄨 + 󵄨󵄨󵄨𝜉󵄨󵄨󵄨 ,
󵄨󵄨
󵄨󵄨
󵄨 󵄨
󵄨󵄨 󸀠󸀠 󵄨󵄨
󵄨󵄨𝜃 (𝜉)󵄨󵄨 ≤ 2𝛾,
󵄨
󵄨
(15)
1−𝛾 2
1
󵄨 󵄨𝛾 𝛾
󵄨 󵄨 𝛾−1
𝜉𝜃 (𝜉) − 𝜉2 𝜃󸀠 (𝜉) =
𝜉 (1 + 󵄨󵄨󵄨𝜉󵄨󵄨󵄨) + 𝜉2 (1 + 󵄨󵄨󵄨𝜉󵄨󵄨󵄨)
2
2
2
3. Proof of Main Result
From the conservation law (2), we have
󵄩 󵄩
‖𝑢‖𝐿∞ (𝑅) ≤ ‖𝑢‖𝐻1 (𝑅) = 󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐻1 (𝑅) .
1,
𝜙 (𝑥) = {
0,
≥
(10)
1−𝛾 2
󵄨 󵄨𝛾
𝜉 (1 + 󵄨󵄨󵄨𝜉󵄨󵄨󵄨) .
2
(16)
Abstract and Applied Analysis
3
Multiplying (11) by 𝜙𝜃󸀠 (𝑞), using the chain rule, and
integrating over 𝜋𝑇 := [0, 𝑇] × 𝑅, we have
(2𝑚 + 1) ∫ 𝜙 (𝑥) 𝑢2𝑚 𝑞𝜃 (𝑞) 𝑑𝑡 𝑑𝑥
𝜋𝑇
−
2𝑚 + 1
∫ 𝜙 (𝑥) 𝑢2𝑚 𝑞2 𝜃󸀠 (𝑞) 𝑑𝑡 𝑑𝑥
2
𝜋𝑇
= ∫ 𝜙 (𝑥) (𝜃 (𝑞 (𝑡, 𝑥)) − 𝜃 (𝑞 (0, 𝑥))) 𝑑𝑥
𝑅
(17)
− ∫ 𝑢2𝑚+1 𝜙󸀠 (𝑥) 𝜃 (𝑞) 𝑑𝑡 𝑑𝑥
𝜋𝑇
Using (10), we obtain
󵄩
󵄩󵄩
2𝑚2 + 5𝑚 + 2 2𝑚+2 󵄩󵄩󵄩
󵄩󵄩
󵄩󵄩 ≤ 𝑐,
󵄩󵄩2𝑘𝑢 +
𝑢
󵄩󵄩 ∞
󵄩󵄩
2(𝑚 + 1)
󵄩𝐿
󵄩
󵄩󵄩
2
󵄩󵄩󵄩 −2
󵄩󵄩Λ [2𝑘𝑢 + 2𝑚 + 5𝑚 + 2 𝑢2𝑚+2 ]󵄩󵄩󵄩 ≤ 𝑐.
󵄩󵄩
󵄩󵄩
2 (𝑚 + 1)
󵄩󵄩
󵄩󵄩𝐿∞
Applying (10), the Hölder inequality, and ∫𝑅 𝑒−|𝑥−𝑦| 𝑑𝑦 = 2, we
have
󵄨
󵄨
󵄨󵄨󵄨Λ−2 [𝑢2𝑚 𝑞2 ]󵄨󵄨󵄨 = 󵄨󵄨󵄨󵄨∫ 1 𝑒−|𝑥−𝑦| 𝑢2𝑚 𝑢2 𝑑𝑦󵄨󵄨󵄨󵄨 ≤ 𝑐‖𝑢‖ 1 ≤ 𝑐,
󵄨󵄨
󵄨󵄨 󵄨󵄨
𝐻 (𝑅)
𝑦
󵄨󵄨
󵄨 𝑅2
󵄨
(21)
where 𝑐 is a constant depending on ‖𝑢0 ‖𝐻1 (𝑅) and 𝑚.
From (20) and (21), we deduce that there exists a positive
constant 𝑐 depending on ‖𝑢0 ‖𝐻1 (𝑅) and 𝑚, such that
− ∫ 𝑄 (𝑡, 𝑥) 𝜙 (𝑥) 𝜃󸀠 (𝑞) 𝑑𝑡 𝑑𝑥.
𝜋𝑇
‖𝑄(𝑡, 𝑥)‖𝐿∞ (𝑅) ≤ 𝑐,
From (16), we get
𝜋𝑇
2𝑚 + 1
∫ 𝜙 (𝑥) 𝑢2𝑚 𝑞2 𝜃󸀠 (𝑞) 𝑑𝑡 𝑑𝑥
2
𝜋𝑇
= ∫ (2𝑚 + 1) 𝜙 (𝑥) 𝑢
2𝑚
𝜋𝑇
≥
󵄨 󵄨
󵄨
󵄨
≤ 𝑐 ∫ 󵄨󵄨󵄨𝜙 (𝑥)󵄨󵄨󵄨 ((1 + 𝛾) 󵄨󵄨󵄨𝑞󵄨󵄨󵄨 + 1) 𝑑𝑡 𝑑𝑥
𝜋
𝑇
1
(𝑞𝜃 (𝑞) − 𝑞2 𝜃󸀠 (𝑞)) 𝑑𝑡 𝑑𝑥
2
(1 − 𝛾)
󵄨 󵄨𝛾
∫ 𝜙 (𝑥) 𝑢2𝑚 𝑞2 (1 + 󵄨󵄨󵄨𝑞󵄨󵄨󵄨) 𝑑𝑡 𝑑𝑥.
2
𝜋𝑇
(18)
From inequalities (18)–(19) and (23), we obtain (12).
Lemma 4. There exists a positive constant 𝑐 depending only on
‖𝑢0 ‖𝐻1 (𝑅) and 𝑚 such that
󵄩󵄩 𝜕𝐻(𝑡, ⋅) 󵄩󵄩
󵄩󵄩
󵄩󵄩
≤ 𝑐.
‖𝑄(𝑡, ⋅)‖𝐿∞ (𝑅) ≤ 𝑐,
󵄩󵄩
󵄩󵄩
(24)
󵄩󵄩 𝜕𝑥 󵄩󵄩𝐿∞ (𝑅)
󵄨󵄨
󵄨󵄨
󵄨󵄨
󵄨
󵄨󵄨∫ 𝜙 (𝑥) 𝜃 (𝑞) 𝑑𝑥󵄨󵄨󵄨
󵄨󵄨 𝑅
󵄨󵄨
If 𝑘 = 0, it holds that
󵄨 󵄨1+𝛾 󵄨 󵄨
≤ ∫ 𝜙 (𝑥) (󵄨󵄨󵄨𝑞󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑞󵄨󵄨󵄨) 𝑑𝑥
𝑅
‖𝑄(𝑡, ⋅)‖𝐿1 (𝑅) ≤ 𝑐,
󵄩 󵄩
󵄩 󵄩1+𝛾
󵄩 󵄩
󵄩 󵄩
≤ 󵄩󵄩󵄩𝜙󵄩󵄩󵄩𝐿2/(1−𝛾) (𝑅) 󵄩󵄩󵄩𝑞󵄩󵄩󵄩𝐿2 (𝑅) + 󵄩󵄩󵄩𝜙󵄩󵄩󵄩𝐿2 (𝑅) 󵄩󵄩󵄩𝑞󵄩󵄩󵄩𝐿2 (𝑅)
‖𝑄(𝑡, ⋅)‖𝐿2 (𝑅) ≤ 𝑐,
󵄩 󵄩1+𝛾
󵄩 󵄩
≤ (𝑏 − 𝑎 + 2)(1−𝛾)/2 󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐻1 (𝑅) + (𝑏 − 𝑎 + 2)1/2 󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐻1 (𝑅) ,
󵄨 󵄨 󵄨1+𝛾 󵄨 󵄨
󵄨
≤ ∫ |𝑢|2𝑚+1 󵄨󵄨󵄨󵄨𝜙󸀠 (𝑥)󵄨󵄨󵄨󵄨 (󵄨󵄨󵄨𝑞󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑞󵄨󵄨󵄨) 𝑑𝑡 𝑑𝑥
𝜋𝑇
𝑇
󵄩 󵄩
󵄩 󵄩2𝑚+1
󵄩 󵄩1+𝛾
≤ 󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐻1 (𝑅) ∫ (󵄩󵄩󵄩󵄩𝜙󸀠 󵄩󵄩󵄩󵄩𝐿2/(1−𝛾) (𝑅) 󵄩󵄩󵄩𝑞󵄩󵄩󵄩𝐿2 (𝑅)
0
󵄩 󵄩
󵄩 󵄩
+󵄩󵄩󵄩󵄩𝜙󸀠 󵄩󵄩󵄩󵄩𝐿2 (𝑅) 󵄩󵄩󵄩𝑞󵄩󵄩󵄩𝐿2 (𝑅) ) 𝑑𝑡
󵄩 󵄩
󵄩 󵄩
󵄩 󵄩1+𝛾
󵄩 󵄩
≤ 𝑐𝑇 (󵄩󵄩󵄩󵄩𝜙󸀠 󵄩󵄩󵄩󵄩𝐿2/(1−𝛾) (𝑅) 󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐻1 (𝑅) + 󵄩󵄩󵄩󵄩𝜙󸀠 󵄩󵄩󵄩󵄩𝐿2 (𝑅) 󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐻1 (𝑅) ) .
(23)
󵄩
󵄩
󵄩 󵄩
≤ 𝑐𝑇 ( (1 + 𝛾) 󵄩󵄩󵄩𝜙(𝑥)󵄩󵄩󵄩𝐿2 (𝑅) 󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐻1 (𝑅)
󵄨
󵄨
+ ∫ 󵄨󵄨󵄨𝜙 (𝑥)󵄨󵄨󵄨 𝑑𝑥) .
𝑅
Using the Hölder inequality, (2), and (15) yields
󵄨󵄨
󵄨󵄨
󵄨󵄨
󵄨
󵄨󵄨∫ 𝑢2𝑚+1 𝜙󸀠 (𝑥) 𝜃 (𝑞) 𝑑𝑡 𝑑𝑥󵄨󵄨󵄨
󵄨󵄨 𝜋
󵄨󵄨
󵄨 𝑇
󵄨
(22)
from which we get
󵄨󵄨
󵄨󵄨
󵄨󵄨
󵄨
󵄨󵄨∫ 𝑄 (𝑡, 𝑥) 𝜙 (𝑥) 𝜃󸀠 (𝑞) 𝑑𝑡 𝑑𝑥󵄨󵄨󵄨
󵄨󵄨 𝜋
󵄨󵄨
󵄨 𝑇
󵄨
(2𝑚 + 1) ∫ 𝜙 (𝑥) 𝑢2𝑚 𝑞𝜃 (𝑞) 𝑑𝑡 𝑑𝑥
−
(20)
󵄩󵄩 𝜕𝐻(𝑡, ⋅) 󵄩󵄩
󵄩󵄩
󵄩󵄩
≤ 𝑐,
󵄩󵄩
󵄩󵄩
󵄩󵄩 𝜕𝑥 󵄩󵄩𝐿1 (𝑅)
󵄩󵄩󵄩 𝜕𝐻(𝑡, ⋅) 󵄩󵄩󵄩
󵄩󵄩
󵄩
󵄩󵄩 𝜕𝑥 󵄩󵄩󵄩 2 ≤ 𝑐.
󵄩
󵄩𝐿 (𝑅)
(25)
(26)
Proof. We have
(19)
𝑄 (𝑡, 𝑥)
= 2𝑘𝑢 +
2𝑚2 + 5𝑚 + 2 2𝑚+2
𝑢
2 (𝑚 + 1)
− Λ−2 [2𝑘𝑢 +
2𝑚2 + 5𝑚 + 2 2𝑚+2 2𝑚 + 1 2𝑚 2
+
𝑢
𝑢 𝑞 ],
2 (𝑚 + 1)
2
𝜕𝐻
2𝑚2 + 5𝑚 + 2 2𝑚+2 2𝑚 + 1 2𝑚 2
+
= Λ−2 𝜕𝑥 [2𝑘𝑢 +
𝑢
𝑢 𝑢𝑥 ] .
𝜕𝑥
2 (𝑚 + 1)
2
(27)
4
Abstract and Applied Analysis
The first inequality of (24) is proved in Lemma 3 (see (22)).
For the second inequality in (24), we have
󵄩
󵄩󵄩 −2
󵄩󵄩Λ [𝜕𝑥 𝑢2𝑚+2 ]󵄩󵄩󵄩 1 ≤ ∬ 𝑒−|𝑥−𝑦| 𝑢2𝑚+2 𝑑𝑥 𝑑𝑦
󵄩𝐿 (𝑅)
󵄩
𝑅
󵄩 󵄩2
≤ 𝑐‖𝑢‖2𝐻1 (𝑅) ≤ 𝑐󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐻1 (𝑅) ,
󵄨󵄨 𝜕𝐻 (𝑡, ⋅) 󵄨󵄨
󵄨󵄨
󵄨󵄨
󵄨
󵄨󵄨
󵄨󵄨 𝜕𝑥 󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨 1
−|𝑥−𝑦|
= 󵄨󵄨󵄨 ∫ 𝑒
󵄨󵄨 2 𝑅
󵄨
󵄩
󵄩󵄩 −2
󵄩󵄩Λ 𝜕𝑥 [𝑢2𝑚 𝑢𝑥2 ]󵄩󵄩󵄩 1 ≤ ∬ 𝑒−|𝑥−𝑦| 𝑢2𝑚 𝑢𝑦2 𝑑𝑥 𝑑𝑦
󵄩𝐿 (𝑅)
󵄩
𝑅
󵄩 󵄩2
≤ 𝑐‖𝑢‖2𝐻1 (𝑅) ≤ 𝑐󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐻1 (𝑅) .
× [2𝑘𝑢 +
2𝑚2 + 5𝑚 + 2 2𝑚+2
𝑢
2 (𝑚 + 1)
From (3)–(31), we obtain (25).
Using
󵄨󵄨
2𝑚 + 1 2𝑚 2
󵄨󵄨
+
𝑢 𝑢𝑦 ] 𝑑𝑦󵄨󵄨󵄨
󵄨󵄨
2
𝑦
󵄨
‖𝑄(𝑡, ⋅)‖2𝐿2 (𝑅) ≤ ‖𝑄(𝑡, ⋅)‖𝐿∞ ‖𝑄(𝑡, ⋅)‖𝐿1 (𝑅) ≤ 𝑐,
󵄩󵄩 𝜕𝐻(𝑡, ⋅) 󵄩󵄩2
󵄩󵄩 𝜕𝐻(𝑡, ⋅) 󵄩󵄩 󵄩󵄩 𝜕𝐻(𝑡, ⋅) 󵄩󵄩
󵄩󵄩
󵄩
󵄩󵄩
󵄩󵄩 󵄩󵄩
󵄩󵄩
≤ 󵄩󵄩󵄩
≤𝑐
󵄩󵄩
󵄩󵄩
󵄩 󵄩
󵄩
󵄩󵄩 𝜕𝑥 󵄩󵄩𝐿2 (𝑅) 󵄩󵄩 𝜕𝑥 󵄩󵄩󵄩𝐿∞ 󵄩󵄩󵄩 𝜕𝑥 󵄩󵄩󵄩𝐿1 (𝑅)
󵄨󵄨
𝑥
2𝑚2 + 5𝑚 + 2 2𝑚+2
󵄨1
= 󵄨󵄨󵄨󵄨 𝑒−𝑥 ∫ 𝑒𝑦 [2𝑘𝑢 +
𝑢
2 (𝑚 + 1)
󵄨󵄨 2
−∞
+
completes the proof of (26).
From Lemma 1, we know that for any 𝑢0 ∈ 𝐻𝑠 (𝑅) with
𝑠 > 3/2, there exist a maximal 𝑇 = 𝑇(𝑢0 ) > 0 and a unique
strong solution 𝑢 to problem (3) such that
2𝑚 + 1 2𝑚 2
𝑢 𝑢𝑦 ] 𝑑𝑦
2
𝑦
∞
1
2𝑚2 + 5𝑚 + 2 2𝑚+2
𝑢
+ 𝑒𝑥 ∫ 𝑒−𝑦 [2𝑘𝑢 +
2
2 (𝑚 + 1)
𝑥
󵄨󵄨
󵄨󵄨
2𝑚 + 1 2𝑚 2
+
𝑢 𝑢𝑦 ] 𝑑𝑦󵄨󵄨󵄨
󵄨󵄨
2
𝑦
󵄨
󵄨󵄨
𝑥
2𝑚2 + 5𝑚 + 2 2𝑚+2
󵄨 1
= 󵄨󵄨󵄨󵄨− 𝑒−𝑥 ∫ 𝑒𝑦 [2𝑘𝑢 +
𝑢
2 (𝑚 + 1)
󵄨󵄨 2
−∞
+
2𝑚 + 1 2𝑚 2
𝑢 𝑢𝑦 ] 𝑑𝑦
2
∞
1
2𝑚2 + 5𝑚 + 2 2𝑚+2
𝑢
+ 𝑒𝑥 ∫ 𝑒−𝑦 [2𝑘𝑢 +
2
2 (𝑚 + 1)
𝑥
+
󵄨󵄨
2𝑚 + 1 2𝑚 2
󵄨
𝑢 𝑢𝑦 ] 𝑑𝑦󵄨󵄨󵄨󵄨
2
󵄨󵄨
󵄨󵄨
2𝑚2 + 5𝑚 + 2 2𝑚+2
󵄨
≤ ∫ 𝑒−|𝑥−𝑦| 󵄨󵄨󵄨󵄨2𝑘𝑢 +
𝑢
2 (𝑚 + 1)
󵄨󵄨
𝑅
󵄨
2𝑚 + 1 2𝑚 2 󵄨󵄨󵄨
+
𝑢 𝑢𝑦 󵄨󵄨󵄨 𝑑𝑦
2
󵄨󵄨
𝑢 ∈ 𝐶 ([0, 𝑇) ; 𝐻𝑠 (𝑅)) 𝐶1 ([0, 𝑇) ; 𝐻𝑠−1 (𝑅)) .
which together with (22) results in (24).
In fact, we have
󵄩󵄩 2𝑚+2 󵄩󵄩
󵄩󵄩 1 ≤ 𝑐 ‖𝑢‖2𝐿2 (𝑅) ≤ 𝑐‖𝑢‖2𝐻1 (𝑅) ≤ 𝑐󵄩󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩󵄩2𝐻1 (𝑅) ,
󵄩󵄩𝑢
󵄩𝐿 (𝑅)
󵄩
󵄩󵄩
󵄩
󵄩󵄩 −2 2𝑚2 + 5𝑚 + 2 2𝑚+2 2𝑚 + 1 2𝑚 2 󵄩󵄩󵄩
󵄩󵄩Λ [
󵄩󵄩
+
𝑞
]
𝑢
𝑢
󵄩󵄩
󵄩󵄩 1
2 (𝑚 + 1)
2
󵄩
󵄩𝐿 (𝑅)
󵄩 󵄩2
≤ 𝑐‖𝑢‖2𝐻1 (𝑅) ≤ 𝑐󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐻1 (𝑅) ,
(30)
Consider the differential equation
(28)
𝜉𝑡 = 𝑢2𝑚+1 (𝑡, 𝜉) ,
𝑡 ∈ [0, 𝑇) ,
𝜉 (0, 𝑥) = 𝑥.
(31)
Lemma 5. Assume 𝑢0 ∈ 𝐻𝑠 , 𝑠 > 3, and let 𝑇 > 0 be
the maximal existence time of the solution to problem (3).
Then there exists a unique solution 𝜉 ∈ 𝐶1 ([0, 𝑇) × 𝑅, 𝑅)
to problem (31). In addition, the map 𝑝(𝑡, ⋅) is an increasing
diffeomorphism of 𝑅 with 𝜉𝑥 (𝑡, 𝑥) > 0 for (𝑡, 𝑥) ∈ [0, 𝑇) × 𝑅.
Proof. Using Lemma 1, we obtain 𝑢 ∈ 𝐶1 ([0, 𝑇); 𝐻𝑠−1 (𝑅)) and
𝐻𝑠−1 ∈ 𝐶1 (𝑅). Therefore, we know that functions 𝑢(𝑡, 𝑥) and
𝑢𝑥 (𝑡, 𝑥) are bounded, Lipschitz in space, and 𝐶1 in time. The
existence and uniqueness theorem for differential equations
guarantees that problem (31) has a unique solution 𝜉 ∈
𝐶1 ([0, 𝑇) × 𝑅, 𝑅).
From (31), we get
𝑑
𝜉 = (2𝑚 + 1) 𝑢2𝑚 𝑢𝜉 (𝑡, 𝜉) 𝜉𝑥 ,
𝑑𝑡 𝑥
𝑡 ∈ [0, 𝑇) ,
(32)
𝑝𝑥 (0, 𝑥) = 1.
≤ 𝑐‖𝑢‖𝐻1 (𝑅)
󵄩 󵄩
≤ 𝑐󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐻1 (𝑅) ,
(29)
Furthermore,
𝑡
𝜉𝑥 (𝑡, 𝑥) = exp (∫ (2𝑚 + 1) 𝑢2𝑚 𝑢𝜉 (𝜏, 𝜉 (𝜏, 𝑥)) 𝑑𝜏) .
0
(33)
For every 𝑇󸀠 < 𝑇, the Sobolev imbedding theorem gives rise
to
󵄨󵄨
󵄨󵄨
sup
󵄨󵄨𝑢𝑥 (𝜏, 𝑥)󵄨󵄨 < ∞.
(34)
󸀠
(𝜏,𝑥)∈[0,𝑇 )×𝑅
Therefore, there exists a constant 𝐾0 > 0 such that 𝜉𝑥 (𝑡, 𝑥) ≥
𝑒−𝐾0 𝑡 for (𝑡, 𝑥) ∈ [0, 𝑇) × 𝑅. The proof is completed.
Abstract and Applied Analysis
5
Acknowledgment
Using (11) and (32), we get
𝑑𝑢𝜉 (𝑡, 𝜉 (𝑡, 𝑥))
𝑑𝑡
=
=
𝑑𝑢𝜉 (𝑡, 𝜉 (𝑡, 𝑥))
𝑑𝑡
𝑑𝑢𝜉 (𝑡, 𝜉 (𝑡, 𝑥))
𝑑𝑡
+
+
𝑑𝑢𝜉 (𝑡, 𝜉 (𝑡, 𝑥))
𝑑𝜉
𝑑𝑢𝜉 (𝑡, 𝜉 (𝑡, 𝑥))
𝑑𝜉
𝜉𝑡
This work is supported by the Fundamental Research Funds
for the Central Universities (JBK120504).
𝑢2𝑚+1
References
2𝑚 + 1 2𝑚 2
= 𝑄 (𝑡, 𝜉 (𝑡, 𝑥)) −
𝑢 𝑞
2
≤ ‖ 𝑄 (𝑡, 𝑥) ‖𝐿∞
≤ 𝑐.
(35)
It follows from (35) that
𝑢𝜉 (𝑡, 𝜉 (𝑡, 𝑥)) ≤ 𝑐 + 𝑐𝑡.
(36)
Using Lemma 5 and (36), we have
𝑢𝑥 (𝑡, 𝑥) ≤ 𝑐 + 𝑐𝑡,
𝑡 ∈ [0, ∞) .
(37)
Using the first equation of problem (4) and Lemma 4, for
an arbitrary integer 𝐾 ≥ 2, we have
𝑑
∫ 𝑢2𝐾 𝑑𝑥
𝑑𝑡 𝑅
= 2𝐾 ∫ 𝑢2𝐾−1
𝑅
𝑑𝑢
𝑑𝑥
𝑑𝑡
= 2𝐾 ∫ 𝑢2𝐾−1 [−𝑢2𝑚+1 𝑢𝑥 −
𝑅
𝜕𝐻
] 𝑑𝑥
𝜕𝑥
= −2𝐾 ∫ 𝑢2𝐾+2𝑚 𝑢𝑥 𝑑𝑥 − 2𝐾 ∫ 𝑢2𝐾−1
𝑅
𝑅
𝜕𝐻
𝑑𝑥
𝜕𝑥
(38)
𝜕𝐻
𝑑𝑥
𝜕𝑥
𝑅
󵄩󵄩 𝜕𝐻 󵄩󵄩
󵄩󵄩
󵄩󵄩
2
≤ 2𝐾‖𝑢‖2𝐾−3
𝐿∞ (𝑅) 󵄩
󵄩󵄩 𝜕𝑥 󵄩󵄩󵄩 ∞ ‖𝑢‖𝐿2 (𝑅)
󵄩
󵄩𝐿 (𝑅)
󵄩󵄩 󵄩󵄩
≤ 𝑐󵄩󵄩𝑢0 󵄩󵄩𝐻1 (𝑅)
= −2𝐾 ∫ 𝑢2𝐾−1
≤ 𝑐,
from which we get
󵄩 󵄩2
󵄩 󵄩
∫ 𝑢2𝐾 𝑑𝑥 ≤ 𝑐𝑡 + ∫ 𝑢02𝐾 𝑑𝑥 ≤ 𝑐𝑡 + 𝑐󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐿∞ (𝑅) 󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐻1 (𝑅) . (39)
𝑅
𝑅
By the 𝐿𝑝 interpolation theorem, for all 2 ≤ 𝑝 < ∞, we
obtain
‖𝑢‖𝐿𝑝 (𝑅) ≤ 𝑐 (1 + 𝑡) ,
𝑡 ∈ [0, ∞) ,
(40)
where 𝑐 depends on ‖𝑢0 ‖𝐻1 (𝑅) and 𝑝.
From Lemma 3 and (37) and (40), we complete the proof
of Theorem 2.
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