Hindawi Publishing Corporation
Advances in Mathematical Physics
Volume 2013, Article ID 810725, 11 pages
http://dx.doi.org/10.1155/2013/810725
Research Article
On the Cauchy Problem for the Two-Component
Novikov Equation
Yongsheng Mi,1,2 Chunlai Mu,1 and Weian Tao2
1
2
College of Mathematics and Statistics, Chongqing University, Chongqing 400044, China
College of Mathematics and Computer Sciences, Yangtze Normal University, Chongqing, Fuling 408100, China
Correspondence should be addressed to Yongsheng Mi; miyongshen@163.com
Received 9 January 2013; Accepted 27 March 2013
Academic Editor: M. Lakshmanan
Copyright © 2013 Yongsheng Mi et al. 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.
We are concerned with the Cauchy problem of two-component Novikov equation, which was proposed by Geng and Xue (2009). We
establish the local well-posedness in a range of the Besov spaces by using Littlewood-Paley decomposition and transport equation
theory which is motivated by that in Danchin’s cerebrated paper (2001). Moreover, with analytic initial data, we show that its
solutions are analytic in both variables, globally in space and locally in time, which extend some results of Himonas (2003) to
more general equations.
1. Introduction
In this paper, we consider the following Cauchy problem of
the two-component Novikov equation
𝑚𝑡 + 𝑢V𝑚𝑥 + 3V𝑢𝑥 𝑚 = 0,
𝑛𝑡 + 𝑢V𝑛𝑥 + 3𝑢V𝑥 𝑛 = 0,
𝑚 = 𝑢 − 𝑢𝑥𝑥 ,
𝑛 = V − V𝑥𝑥 ,
𝑢 (0, 𝑥) = 𝑢0 (𝑥) ,
𝑡 > 0, 𝑥 ∈ R,
𝑡 > 0, 𝑥 ∈ R,
𝑡 > 0, 𝑥 ∈ R,
V (0, 𝑡) = V0 (𝑥) ,
(1)
𝑥 ∈ R.
The two-component system in (1) was found by Geng and
Xue [1]. It was shown in [1] that the system (1) is exactly a
negative flow in the hierarchy and admits exact solutions with
𝑁-peakons and an infinite sequence of conserved quantities.
Moreover, a reduction of this hierarchy and its Hamiltonian
structures are discussed.
For V = 1, (1) becomes the Degasperis-Procesi equation
𝑚𝑡 + 𝑢𝑚𝑥 + 3𝑢𝑥 𝑚 = 0,
𝑚 = 𝑢 − 𝑢𝑥𝑥 ,
𝑡 > 0, 𝑥 ∈ R,
𝑡 > 0, 𝑥 ∈ R,
𝑢 (0, 𝑥) = 𝑢0 (𝑥) ,
𝑥 ∈ R.
(2)
Degasperis et al. [2] proved the formal integrability of
(2) by constructing a Lax pair. They also showed that it
has bi-Hamiltonian structure and an infinite sequence of
conserved quantities and admits exact peakon solutions. The
direct and inverse scattering approach to pursue it can be
seen in [3]. Moreover, in [4], they also presented that the
Degasperis-Procesi equation has a bi-Hamiltonian structure
and an infinite number of conservation laws and admits
exact peakon solutions which are analogous to the CamassaHolm peakons. It is worth pointing out that solutions of this
type are not mere abstractizations: the peakons replicate a
feature that is characteristic for the waves of great heightwaves of the largest amplitude that are exact solutions of
the governing equations for irrotational water waves (cf. the
papers [5–7]). The Degasperis-Procesi equation is a model
for nonlinear shallow water dynamics (cf. the discussion
in [8]). The numerical stability of solitons and peakons,
the multisoliton solutions, and their peakon limits, together
with an inverse scattering method to compute 𝑁-peakon
solutions to Degasperis-Procesi equation, have been investigated, respectively, in [9–11]. Furthermore, the traveling wave
solutions and the classification of all weak traveling wave
solutions to Degasperis-Procesi equation were presented in
[12, 13].
2
Advances in Mathematical Physics
𝑠
𝑠
(𝑇) × 𝐸𝑝,𝑟
(𝑇) and the map (𝑢0 , V0 ) 󳨃→ (𝑢, V) is
(𝑢, V) ∈ 𝐸𝑝,𝑟
𝑠
𝑠
continuous from a neighborhood of (𝑢0 , V0 ) in 𝐵𝑝,𝑟
× 𝐵𝑝,𝑟
into
For 𝑢 = V, (1) becomes the Novikov equation
𝑚𝑡 + 𝑢2 𝑚𝑥 + 3𝑢𝑢𝑥 𝑚 = 0,
𝑚 = 𝑢 − 𝑢𝑥𝑥 ,
𝑡 > 0, 𝑥 ∈ R,
𝑡 > 0, 𝑥 ∈ R,
𝑢 (0, 𝑥) = 𝑢0 (𝑥) ,
(3)
𝑥 ∈ R,
which has been recently discovered by Vladimir Novikov in
a symmetry classification of nonlocal PDEs with quadratic or
cubic nonlinearity [14]. The perturbative symmetry approach
yields necessary conditions for a PDE to admit infinitely
many symmetries. Using this approach, Novikov was able
to isolate (3) and find its first few symmetries, and he
subsequently found a scalar Lax pair for it, then proved
that the equation is integrable, which can be thought as a
generalization of the Camassa-Holm equation. In [15], it is
shown that the Novikov equation admits peakon solutions
like the Camassa-Holm. Also, it has a Lax pair in matrix form
and a bi-Hamiltonian structure. Furthermore, it has infinitely
many conserved quantities, like Camassa-Holm. The most
important quantity conserved by a solution 𝑢 to Novikov
equation is its 𝐻1 -norm ‖𝑢‖2𝐻1 = ∫𝑅 (𝑢2 + 𝑢𝑥2 ), which plays
an important role in the study of (1). In [16–19], the authors
study well-posedness and dependence on initial data for the
Cauchy problem for Novikov equation. Recently, in [20], a
global existence result and conditions on the initial data were
considered. Existence and uniqueness of global weak solution
to Novikov equation with initial data under some conditions
was proved in [21]. The Novikov equation with dissipative
term was considered in [22]. Multipeakon solutions were
studied in [15, 23]. The Cauchy problem of the Novikov
equation on the circle was investigated in [24]. An alternative
modified Camassa-Holm equation was introduced in [25].
Motivated by the references cited above, the goal of the
present paper is to establish the local well-posedness for
the strong solutions to the Cauchy problem (1). The proof
of the local well-posedness is inspired by the argument
of approximate solutions by Danchin [26] in the study of
the local wellposedness to the Camassa-Holm equation.
However, one problematic issue is that we here deal with
two-component system with a higher order nonlinearity
in the Besov spaces, making the proof of several required
nonlinear estimates somewhat delicate. These difficulties are
nevertheless overcome by careful estimates for each iterative
approximation of solutions to (1). Moreover, we also prove the
analyticity of its solutions 𝑢 = 𝑢(𝑡, 𝑥) in both variables, with
𝑥 in R and 𝑡 in an interval around zero, provided that the
initial profile 𝑢0 is an analytic function on the real line. Hence,
this analytic result can be viewed as a Cauchy-Kowalevski
theorem for (1).
Now we are in the position to state the local existence
result and analyticity result, where the definition of Besov
𝑠
𝑠
, 𝐸𝑝,𝑟
(𝑇), and and 𝐸𝑠0 will be given in Sections 2
spaces 𝐵𝑝,𝑟
and 3.
Theorem 1. Let 𝑝, 𝑟 ∈ [1, ∞] and 𝑠 > max{5/2, 2 + (1/𝑝)}.
𝑠
𝑠
× 𝐵𝑝,𝑟
. There exists a time 𝑇 > 0
Assume that (𝑢0 , V0 ) ∈ 𝐵𝑝,𝑟
such that the initial-value problem (1) has a unique solution
󸀠
󸀠
𝑠
𝑠 −1
𝐶 ([0, 𝑇] ; 𝐵𝑝,𝑟
) ∩ 𝐶1 ([0, 𝑇] ; 𝐵𝑝,𝑟
)
󸀠
󸀠
𝑠
𝑠 −1
× 𝐶 ([0, 𝑇] ; 𝐵𝑝,𝑟
) ∩ 𝐶1 ([0, 𝑇] ; 𝐵𝑝,𝑟
)
(4)
for every 𝑠󸀠 < 𝑠 when 𝑟 = ∞ and 𝑠󸀠 = 𝑠 whereas 𝑟 < ∞.
Theorem 2. If the initial data ( 𝑢V00 ) is real analytic on the line
R and belongs to a space 𝐸𝑠0 , for some 0 < 𝑠0 ≤ 1, then there
exist an 𝜀 > 0 and a unique solution ( 𝑢V ) to the Cauchy problem
(1) that is analytic on (−𝜀, 𝜀) × R.
The rest of this paper is organized as follows. In Section 2,
we prove the local well-posedness of the initial value problem
(1) in the Besov space. Section 3 is devoted to the study of the
analyticity of the Cauchy problem (1) based on a contraction
type argument in a suitably chosen scale of the Banach spaces.
2. Local Well-Posedness in the Besov Spaces
In this section, we will establish local well-posedness of the
initial value problem (1) in the Besov spaces.
First, for the convenience of the readers, we recall some
facts on the Littlewood-Paley decomposition and some useful
lemmas.
Notation. S stands for the Schwartz space of smooth functions over R𝑑 whose derivatives of all order decay at infinity.
The set S󸀠 of temperate distributions is the dual set of S for
the usual pairing. We denote the norm of the Lebesgue space
𝐿𝑝 (R) by ‖ ⋅ ‖𝐿𝑝 with 1 ≤ 𝑝 ≤ ∞, and the norm in the Sobolev
space 𝐻𝑠 (R) with 𝑠 ∈ R by ‖ ⋅ ‖𝐻𝑠 .
Proposition 3 (Littlewood-Paley decomposition [27]). Let
B ≐ {𝜉 ∈ R𝑑 , |𝜉| ≤ 4/3} and C ≐ {𝜉 ∈ R𝑑 , 4/3 ≤ |𝜉| ≤ 8/3}.
There exist two radial functions 𝜒 ∈ 𝐶𝑐∞ (B) and 𝜑 ∈ 𝐶𝑐∞ (C)
such that
𝜒 (𝜉) + ∑ 𝜑 (2−𝑞 𝜉) = 1,
∀𝜉 ∈ R𝑑 ,
𝑞≥0
󸀠
󵄨󵄨
󵄨
󵄨󵄨𝑞 − 𝑞󸀠 󵄨󵄨󵄨 ≥ 2 󳨐⇒ Supp 𝜑 (2−𝑞 ⋅) ∩ Supp 𝜑 (2−𝑞 ⋅) = 0,
󵄨
󵄨
󸀠
𝑞 ≥ 1 󳨐⇒ Supp 𝜒 (⋅) ∩ Supp 𝜑 (2−q ⋅) = 0,
1
2
≤ 𝜒(𝜉)2 + ∑ 𝜑(2−𝑞 𝜉) ≤ 1,
3
𝑞≥0
(5)
∀𝜉 ∈ R𝑑 .
Furthermore, let ℎ ≐ F−1 𝜑 and ̃ℎ ≐ F−1 𝜒. Then for all
𝑓 ∈ S󸀠 (R𝑑 ), the dyadic operators Δ 𝑞 and 𝑆𝑞 can be defined as
follows:
Advances in Mathematical Physics
3
Δ 𝑞 𝑓 ≐ 𝜑 (2−𝑞 𝐷) 𝑓
𝑠
⇔ 1 ≤ 𝑝, 𝑟 ≤ ∞.
(2) Density: 𝐶𝑐∞ is dense in 𝐵𝑝,𝑟
= 2𝑞𝑑 ∫ ℎ (2𝑞 𝑦) 𝑓 (𝑥 − 𝑦) 𝑑𝑦
𝑆𝑞 𝑓 ≐ 𝜒 (2−𝑞 𝐷) 𝑓
=
𝑠−𝑛(1/𝑝1 )−(1/𝑝2 )
(3) Embedding: 𝐵𝑝𝑠 1 ,𝑟1 󳨅→ 𝐵𝑝2 ,𝑟2
𝑟1 ≤ 𝑟2 .
for 𝑞 ≥ 0,
R𝑑
(6)
Δ 𝑘 = 2 ∫ ̃ℎ (2𝑞 𝑦) 𝑓 (𝑥 − 𝑦) 𝑑𝑦,
𝑞𝑑
∑
−1≤𝑘≤𝑞−1
R𝑑
Δ −1 𝑓 ≐ 𝑆0 𝑓,
Δ 𝑞𝑓 ≐ 0
for 𝑞 ≤ −2.
𝑓 = ∑Δ 𝑞𝑓
𝑑
in S (R ) ,
𝑞≥0
𝑝,𝑟
(7)
Supp 𝑢̂ ⊂ 𝜆B
󵄩 󵄩
󳨐⇒ sup 󵄩󵄩󵄩𝜕𝛼 𝑢󵄩󵄩󵄩𝐿𝑎 ≤ 𝐶𝑘+1 𝜆𝑘+𝑑((1/𝑎)−(1/𝑏)) ‖𝑢‖𝐿𝑎 ,
|𝛼|=𝑘
Supp 𝑢̂ ⊂ 𝜆C
󵄩 󵄩
󳨐⇒ 𝐶−𝑘−1 𝜆𝑘 ‖𝑢‖𝐿𝑎 ≤ sup 󵄩󵄩󵄩𝜕𝛼 𝑢󵄩󵄩󵄩𝐿𝑎 ≤ 𝐶𝑘+1 𝜆𝑘 ‖𝑢‖𝐿𝑎 ,
(8)
|𝛼|=𝑘
Supp 𝑢̂ ⊂ 𝜆C
󳨐⇒ ‖𝜎 (𝐷) 𝑢‖𝐿𝑏 ≤ 𝐶𝜎,𝑚 𝜆𝑚+𝑑((1/𝑎)−(1/𝑏)) ‖𝑢‖𝐿𝑎 ,
for any function 𝑢 ∈ 𝐿𝑎 .
Definition 5 (Besov space). Let 𝑠 ∈ R, 1 ≤ 𝑝, 𝑟 ≤ ∞.
𝑠
𝑠
The inhomogeneous Besov space 𝐵𝑝,𝑟
(R𝑑 ) (𝐵𝑝,𝑟
for short) is
defined by
󵄩 󵄩
≐ {𝑓 ∈ S (R ) ; 󵄩󵄩󵄩𝑓󵄩󵄩󵄩𝐵𝑠 < ∞} ,
𝑝,𝑟
󸀠
𝑑
for 𝑟 < ∞,
(11)
󵄩 󵄩
𝑠
≤ lim inf 󵄩󵄩󵄩𝑢𝑛 󵄩󵄩󵄩𝐵𝑠 .
‖𝑢‖𝐵𝑝,𝑟
𝑛→∞
𝑝,𝑟
(12)
(7) Let 𝑚 ∈ R and f be an 𝑆𝑚 -multiplier (i.e., 𝑓 : R𝑑 → R
is smooth and satisfies that for all 𝛼 ∈ N𝑑 , there exists
a constant 𝐶𝛼 , s.t. |𝜕𝛼 𝑓(𝜉)| ≤ 𝐶𝛼 (1 + |𝜉|𝑚−|𝛼| ) for all
𝜉 ∈ R𝑑 ). Then the operator 𝑓(𝐷) is continuous from
𝑠
𝑠−𝑚
to 𝐵𝑝,𝑟
.
𝐵𝑝,𝑟
Now we state some useful results in the transport equation theory, which are crucial to the proofs of our main
theorems later.
Lemma 7 (see [26, 28]). Suppose that (𝑝, 𝑟) ∈ [1, +∞]2 and
𝑠 > −(𝑑/𝑝). Let V be a vector field such that ∇V belongs to
𝑑/𝑝
𝑠−1
𝐿1 ([0, 𝑇]; 𝐵𝑝,𝑟
) if 𝑠 > 1 + (𝑑/𝑝) or to 𝐿1 ([0, 𝑇]; 𝐵𝑝,𝑟 ∩ 𝐿∞ )
𝑠
𝑠
otherwise. Suppose also that 𝑓0 ∈ 𝐵𝑝,𝑟
, 𝐹 ∈ 𝐿1 ([0, 𝑇]; 𝐵𝑝,𝑟
)
∞ 1
𝑠
󸀠
and that 𝑓 ∈ 𝐿 (𝐿 ([0, 𝑇]; 𝐵𝑝,𝑟 ) ∩ 𝐶([0, 𝑇]; S ) solves the 𝑑dimensional linear transport equations
𝜕𝑡 𝑓 + V ⋅ ∇𝑓 = 𝐹,
󵄨
𝑓󵄨󵄨󵄨𝑡=0 = 𝑓0 .
(𝑇)
Then there exists a constant 𝐶 depending only on 𝑠, 𝑝, and 𝑑
such that the following statements hold.
1/𝑟
{
󵄩󵄩𝑟
𝑞𝑠𝑟 󵄩
{
󵄩
{
{( ∑ 2 󵄩󵄩󵄩Δ 𝑞 𝑓󵄩󵄩󵄩𝐿 ) ,
𝑝
≐ { 𝑞∈Z
{
󵄩󵄩
󵄩󵄩
𝑞𝑠
{
{sup2 󵄩󵄩Δ 𝑞 𝑓󵄩󵄩 ,
󵄩
󵄩𝐿 𝑝
{ 𝑞∈Z
𝑝,𝑟
𝑠1
𝑠2
∀𝑢 ∈ 𝐵𝑝,𝑟
∩ 𝐵𝑝,𝑟
, ∀𝜃 ∈ [0, 1] .
𝑠
(6) Fatou lemma: If (𝑢𝑛 )𝑛∈N is bounded in 𝐵𝑝,𝑟
and 𝑢𝑛 →
󸀠
𝑠
𝑢 in S , then 𝑢 ∈ 𝐵𝑝,𝑟 and
(9)
where
𝑝,𝑟
𝑠 ,
≤ 𝐶‖𝑢‖𝜃𝐵𝑠1 ‖𝑢‖1−𝜃
𝐵2
𝑝,𝑟
Lemma 4 (Bernstein’s inequality [28]). Let B be a ball with
center 0 in R𝑑 and C a ring with center 0 in R𝑑 . A constant 𝐶
exists so that, for any positive real number 𝜆, any nonnegative
integer 𝑘, any smooth homogeneous function 𝜎 of degree 𝑚, and
any couple of real numbers (𝑎, 𝑏) with 𝑏 ≥ 𝑎 ≥ 1, there hold
󵄩󵄩 󵄩󵄩
󵄩󵄩𝑓󵄩󵄩𝐵𝑠
𝑠
∩ 𝐿∞ is an
(4) Algebraic properties: for all 𝑠 > 0, 𝐵𝑝,𝑟
𝑠
algebra. Moreover, 𝐵𝑝,𝑟 is an algebra, provided that
𝑠 > 𝑛/𝑝 or 𝑠 ≥ 𝑛/𝑝 and 𝑟 = 1.
‖𝑢‖𝐵𝜃𝑠1 +(1−𝜃)𝑠2
where the right-hand side is called the nonhomogeneous
Littlewood-Paley decomposition of 𝑓.
𝑠
𝐵𝑝,𝑟
𝑠2
𝑠1
𝐵𝑝,𝑟
󳨅→ 𝐵𝑝,𝑟
locally compact, if 𝑠1 < 𝑠2 .
2
1
(5) Complex interpolation:
Hence,
󸀠
, if 𝑝1 ≤ 𝑝2 and
(1) If 𝑟 = 1 or 𝑠 ≠ 1 + (𝑑/𝑝), then
(10)
𝑡
󵄩󵄩 󵄩󵄩
󵄩 󵄩
𝑠 𝑑𝜏
≤ 󵄩󵄩󵄩𝑓0 󵄩󵄩󵄩𝐵𝑠 + ∫ ‖𝐹 (𝜏)‖𝐵𝑝,𝑟
󵄩󵄩𝑓󵄩󵄩𝐵𝑝,𝑟
𝑠
𝑝,𝑟
0
for 𝑟 = ∞.
𝑡
∞
𝑠
If 𝑠 = ∞, 𝐵𝑝,𝑟
≐ ∩𝑠∈R 𝐵𝑝,𝑟
.
Proposition 6 (see [28]). Suppose that 𝑠 ∈ R, 1 ≤ 𝑝, 𝑟, 𝑝𝑖 , 𝑟𝑖 ≤
∞ (𝑖 = 1, 2). One has the following.
𝑠
is a Banach space which is
(1) Topological properties: 𝐵𝑝,𝑟
continuously embedded in S󸀠 .
(13)
󸀠
𝑠 𝑑𝜏,
+ 𝐶 ∫ 𝑉 (𝜏) ‖𝐹 (𝜏)‖𝐵𝑝,𝑟
0
or
𝑡
󵄩󵄩 󵄩󵄩
󵄩 󵄩
(14)
𝑠 𝑑𝜏)
≤ 𝑒𝐶𝑉(𝑡) 𝐶 (󵄩󵄩󵄩𝑓0 󵄩󵄩󵄩𝐵𝑠 + ∫ 𝑒−𝐶𝑉(𝜏) ‖𝐹 (𝜏)‖𝐵𝑝,𝑟
󵄩󵄩𝑓󵄩󵄩𝐵𝑝,𝑟
𝑠
𝑝,𝑟
0
4
Advances in Mathematical Physics
𝑡
hold, where 𝑉(𝑡) = ∫0 ‖∇V(𝜏)‖𝐵𝑑/𝑝 ∩𝐿∞ 𝑑𝜏 if 𝑠 < 1+(𝑑/𝑝)
𝑝,𝑟
𝑡
𝑠−1 𝑑𝜏 else.
and 𝑉(𝑡) = ∫0 ‖∇V(𝜏)‖𝐵𝑝,𝑟
(2) If 𝑠 ≤ 1 + (𝑑/𝑝) and ∇𝑓0 ∈ 𝐿∞ , ∇𝑓 ∈ 𝐿∞ ([0, 𝑇] × R𝑑 )
and ∇𝐹 ∈ 𝐿1 ([0, 𝑇]; 𝐿∞ ), then
󵄩󵄩 󵄩󵄩
󵄩 󵄩
+ 󵄩󵄩󵄩∇𝑓󵄩󵄩󵄩𝐿∞
󵄩󵄩𝑓󵄩󵄩𝐵𝑝,𝑟
𝑠
󵄩 󵄩
󵄩 󵄩
≤ 𝑒𝐶𝑉(𝑡) (󵄩󵄩󵄩𝑓0 󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩∇𝑓0 󵄩󵄩󵄩𝐿∞
𝑝,𝑟
(15)
𝑡
𝑠
+ ∫ 𝑒−𝐶𝑉(𝜏) ‖𝐹 (𝜏)‖𝐵𝑝,𝑟
+‖∇𝐹 (𝜏)‖𝐿∞ 𝑑𝜏)
0
In the following, we denote 𝐶 > 0 a generic constant
only depending on 𝑝, 𝑟, 𝑠. Uniqueness and continuity with
respect to the initial data are an immediate consequence of
the following result.
Proposition 11. Let 1 ≤ 𝑝, 𝑟 ≤ +∞ and 𝑠 > max{5/2, 2 +
𝑠
) ∩ 𝐶([0, 𝑇];
(1/𝑝)}. Suppose that (𝑢(𝑖) ; V(𝑖) ) ∈ {𝐿∞ ([0, 𝑇]; 𝐵𝑝,𝑟
󸀠 2
S )} (𝑖 = 1, 2) be two given solutions of the initial-value
𝑠
𝑠
problem (1) with the initial data (𝑢0(𝑖) ; V0(𝑖) ) ∈ 𝐵𝑝,𝑟
× 𝐵𝑝,𝑟
(𝑖 =
1; 2). Then for every 𝑡 ∈ [0; 𝑇], one has
󵄩
󵄩
󵄩
󵄩󵄩 (1)
󵄩󵄩𝑢 (𝑡) − 𝑢(2) (𝑡)󵄩󵄩󵄩 𝑠−1 + 󵄩󵄩󵄩V(1) (𝑡) − V(2) (𝑡)󵄩󵄩󵄩 𝑠−1
󵄩𝐵𝑝,𝑟 󵄩
󵄩𝐵𝑝,𝑟
󵄩
󵄩
󵄩
󵄩
󵄩
≤ (󵄩󵄩󵄩󵄩𝑢0(1) − 𝑢0(2) 󵄩󵄩󵄩󵄩𝐵𝑠−1 + 󵄩󵄩󵄩󵄩V0(1) − V0(2) 󵄩󵄩󵄩󵄩𝐵𝑠−1 )
𝑝,𝑟
𝑝,𝑟
𝑡
with 𝑉(𝑡) = ∫0 ‖∇V(𝜏)‖𝐵𝑑/𝑝 ∩𝐿∞ 𝑑𝜏.
𝑝,𝑟
(3) If 𝑓 = V, then for all 𝑠 > 0, the estimate (14) holds with
𝑡
𝑠−1 𝑑𝜏.
𝑉(𝑡) = ∫0 ‖∇V(𝜏)‖𝐵𝑝,𝑟
𝑡
󵄩2
󵄩2
󵄩
󵄩
× exp {𝐶 ∫ (󵄩󵄩󵄩󵄩𝑢(1) (𝜏)󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩𝑢(2) (𝜏)󵄩󵄩󵄩󵄩𝐵𝑠
𝑝,𝑟
𝑝,𝑟
0
𝑠
). If 𝑟 = +∞, then
(4) If 𝑟 < +∞, then 𝑓 ∈ 𝐶([0, 𝑇]; 𝐵𝑝,𝑟
󵄩2
󵄩2
󵄩
󵄩
+󵄩󵄩󵄩󵄩V(1) (𝜏)󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩V(2) (𝜏)󵄩󵄩󵄩󵄩𝐵𝑠 ) 𝑑𝜏} .
𝑝,𝑟
𝑝,𝑟
(20)
󸀠
𝑠
𝑓 ∈ 𝐶([0, 𝑇]; 𝐵𝑝,𝑟
) for all 𝑠󸀠 < 𝑠.
Lemma 8 (existence and uniqueness see [26, 28]). Let
(𝑝, 𝑝1 , 𝑟) ∈ [1, +∞]3 and 𝑠 > −𝑑 min{1/𝑝1 , 1/𝑝󸀠 } with 𝑝󸀠 ≐
𝑠
𝑠
(1 − (1/𝑝))−1 . Assume that 𝑓0 ∈ 𝐵𝑝,𝑟
, 𝐹 ∈ 𝐿1 ([0, 𝑇]; 𝐵𝑝,𝑟
). Let V
𝜌
−𝑀
)
be a time-dependent vector field such that V ∈ 𝐿 ([0, 𝑇]; 𝐵∞,∞
𝑑/𝑝
1
∞
for some 𝜌 > 1, 𝑀 > 0 and ∇V ∈ 𝐿 ([0, 𝑇]; 𝐵𝑝,𝑟 ∩ 𝐿 ) if
𝑠 < 1 + (𝑑/𝑝1 ) and ∇V ∈ 𝐿1 ([0, 𝑇]; 𝐵𝑝𝑠−1
) if 𝑠 > 1 + (𝑑/𝑝) or
1 ,𝑟
𝑠 = 1+(𝑑/𝑝1 ) and 𝑟 = 1. Then the transport equations (𝑇) have
𝑠
𝑠󸀠
a unique solution 𝑓 ∈ 𝐿∞ ([0, 𝑇]; 𝐵𝑝,𝑟
) ∩ (∩𝑠󸀠 <𝑠 𝐶[0, 𝑇]; 𝐵𝑝,1
)
and the inequalities in Lemma 7 hold true. Moreover, 𝑟 < ∞,
𝑠
).
then one has 𝑓 ∈ 𝐶[0, 𝑇]; 𝐵𝑝,1
Lemma 9 (1-𝐷 Morse-type estimates [26, 28]). Assume that
1 ≤ 𝑝, 𝑟 ≤ +∞, the following estimates hold.
Proof. Denote 𝑢(12) = 𝑢(2) − 𝑢(1) , V(12) = V(2) − V(1) , 𝑚(12) =
𝑚(2) − 𝑚(1) and 𝑛(12) = 𝑛(2) − 𝑛(1) . It is obvious that
𝑠
) ∩ 𝐶 ([0, 𝑇] ; S󸀠 ) ,
𝑢(12) , V(12) ∈ 𝐿∞ ([0, 𝑇] ; 𝐵𝑝,𝑟
𝑠−1
which implies that 𝑢(12) , V(12) ∈ 𝐶([0, 𝑇]; 𝐵𝑝,𝑟
) and (𝑢(12) ,
V(12) , 𝑚(12) , 𝑚(12) ) solves the transport equations
𝑚𝑡(12) + 𝑢(1) V(1) 𝑚𝑥(12) = 𝐹,
𝑛𝑡(12) + 𝑢(1) V(1) 𝑛𝑥(12) = 𝐺,
𝐹 = − (𝑢(2) V(12) + 𝑢(12) V(1) ) 𝑚𝑥(2)
− 3 (𝑢𝑥(2) V(12) 𝑚(2) + 𝑢𝑥(2) V(1) 𝑚(12) + 𝑢𝑥(12) V(1) 𝑚(1) ) ,
(16)
(ii) For all 𝑠1 ≤ 1/𝑝 < 𝑠2 (𝑠2 ≥ 1/𝑝 if 𝑟 = 1) and 𝑠1 +𝑠2 > 0,
one has
󵄩󵄩 󵄩󵄩 𝑠
󵄩󵄩 󵄩󵄩 𝑠 󵄩󵄩 󵄩󵄩 𝑠
1 󵄩
(17)
󵄩𝑔󵄩󵄩𝐵𝑝,𝑟2 .
󵄩󵄩𝑓𝑔󵄩󵄩𝐵𝑝,𝑟1 ≤ 𝐶󵄩󵄩𝑓󵄩󵄩𝐵𝑝,𝑟
𝑠
, one has for 𝑠 > 0,
(iii) In Sobolev spaces 𝐻𝑠 = 𝐵2,2
󵄩󵄩
󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩
󵄩 󵄩 󵄩 󵄩
󵄩󵄩
󵄩󵄩𝑓𝜕𝑥 𝑔󵄩󵄩𝐻𝑠 ≤ 𝐶 (󵄩󵄩𝑓󵄩󵄩𝐻𝑠+1 󵄩󵄩𝑔󵄩󵄩𝐿∞ + 󵄩󵄩󵄩𝜕𝑥 𝑔󵄩󵄩󵄩𝐻𝑠 󵄩󵄩󵄩𝑓󵄩󵄩󵄩𝐿∞ ) ,
𝐺 = − (V(2) 𝑢(12) + V(12) 𝑢(1) ) 𝑛𝑥(2)
According to Lemma 7, we have
𝑡
𝑝,𝑟
󵄩
󵄩
≤ 󵄩󵄩󵄩󵄩𝑚0(12) 󵄩󵄩󵄩󵄩𝐵𝑠−3
𝑝,𝑟
𝑡
≜
𝑠
∩𝑇>0 𝐸𝑝,𝑟
𝜏
−𝐶 ∫0 ‖𝜕𝑥 (𝑢(1) V(1) )(𝜏󸀠 )‖𝐵𝑠−2 𝑑𝜏󸀠
+ 𝐶∫ 𝑒
𝑝,𝑟
0
𝑡
if 𝑟 < +∞,
−𝐶 ∫0 ‖𝜕𝑥 (𝑢(1) V(1) )(𝜏󸀠 )‖𝐵𝑠−2 𝑑𝜏󸀠 󵄩
󵄩 (12)
𝑝,𝑟
󵄩󵄩𝑛
󵄩
𝑒
𝑠−3 ) 𝑑𝜏,
(‖𝐹‖𝐵𝑝,𝑟
(24)
󵄩
(𝑡)󵄩󵄩󵄩󵄩𝐵𝑠−3
𝑝,𝑟
󵄩
󵄩
≤ 󵄩󵄩󵄩󵄩𝑛0(12) 󵄩󵄩󵄩󵄩𝐵𝑠−3
𝑠
𝑠
𝑠−1
) ∩ lip1 ([0, 𝑇] ; 𝐵𝑝,∞
),
𝐸𝑝,∞
(𝑇) ≜ 𝐿∞ ([0, 𝑇] ; 𝐵𝑝,∞
𝑠
𝐸𝑝,𝑟
󵄩
󵄩󵄩𝑚(12) (𝑡)󵄩󵄩󵄩 𝑠−3
󵄩𝐵𝑝,𝑟
󵄩
−𝐶 ∫0 ‖𝜕𝑥 (𝑢(1) V(1) )(𝜏󸀠 )‖𝐵𝑠−2 𝑑𝜏󸀠 󵄩
󵄩
𝑒
where 𝐶 is a positive constant independent of 𝑓 and 𝑔.
𝑠
𝑠
𝑠−1
) ∩ 𝐶1 ([0, 𝑇] ; 𝐵𝑝,𝑟
)
𝐸𝑝,𝑟
(𝑇) ≜ 𝐶 ([0, 𝑇] ; 𝐵𝑝,𝑟
𝑝,𝑟
(𝑇) .
𝑡
(19)
(23)
− 3 (V𝑥(2) 𝑢(12) 𝑛(2) + V𝑥(2) 𝑢(1) 𝑛(12) + V𝑥(12) 𝑢(1) 𝑛(1) ) .
(18)
Definition 10. For 𝑇 > 0, 𝑠 ∈ R and 1 ≤ 𝑝 ≤ +∞, we set
(22)
with
(i) For 𝑠 > 0,
󵄩󵄩 󵄩󵄩
󵄩 󵄩 󵄩 󵄩
󵄩 󵄩 󵄩 󵄩
≤ 𝐶 (󵄩󵄩󵄩𝑓󵄩󵄩󵄩𝐵𝑠 󵄩󵄩󵄩𝑔󵄩󵄩󵄩𝐿∞ + 󵄩󵄩󵄩𝑔󵄩󵄩󵄩𝐵𝑠 󵄩󵄩󵄩𝑓󵄩󵄩󵄩𝐿∞ ) .
󵄩󵄩𝑓𝑔󵄩󵄩𝐵𝑝,𝑟
𝑠
𝑝,𝑟
𝑝,𝑟
(21)
0
𝜏
−𝐶 ∫0 ‖𝜕𝑥 (𝑢(1) V(1) )(𝜏󸀠 )‖𝐵𝑠−2 𝑑𝜏󸀠
+ 𝐶∫ 𝑒
𝑝,𝑟
𝑠−3 ) 𝑑𝜏,
(‖𝐺‖𝐵𝑝,𝑟
Advances in Mathematical Physics
5
󵄩
󵄩
󵄩
󵄩
× (󵄩󵄩󵄩󵄩V(12) 󵄩󵄩󵄩󵄩𝐵𝑠−1 + 󵄩󵄩󵄩󵄩𝑢(12) 󵄩󵄩󵄩󵄩𝐵𝑠−1 )
𝑝,𝑟
𝑝,𝑟
For 𝑠 > max{5/2, 2 + (1/𝑝)}, by Lemma 9, we have
󵄩 󵄩2
󵄩 󵄩2
󵄩 󵄩2
󵄩 󵄩2
× (󵄩󵄩󵄩󵄩V(1) 󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩𝑢(1) 󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩V(2) 󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩𝑢(2) 󵄩󵄩󵄩󵄩𝐵𝑠 ) 𝑑𝜏.
𝑝,𝑟
𝑝,𝑟
𝑝,𝑟
𝑝,𝑟
𝑠−3
‖𝐹‖𝐵𝑝,𝑟
󵄩
= 󵄩󵄩󵄩󵄩− (𝑢(2) V(12) + 𝑢(12) V(1) ) 𝑚𝑥(2)
(26)
󵄩
+3 (𝑢𝑥(2) V(12) 𝑚(2) + 𝑢𝑥(2) V(1) 𝑚(12) + 𝑢𝑥(12) V(1) 𝑚(1) )󵄩󵄩󵄩󵄩𝐵𝑠−3
𝑝,𝑟
󵄩 󵄩
󵄩
󵄩
≤ 𝐶󵄩󵄩󵄩󵄩𝑢(2) V(12) + 𝑢(12) V(1) 󵄩󵄩󵄩󵄩𝐵𝑠−3 󵄩󵄩󵄩󵄩𝑚(2) 󵄩󵄩󵄩󵄩𝐵𝑠−2
𝑝,𝑟
𝑝,𝑟
󵄩
󵄩 󵄩 󵄩
+ 𝐶󵄩󵄩󵄩󵄩𝑢(2) 󵄩󵄩󵄩󵄩𝐵𝑠−2 󵄩󵄩󵄩󵄩V(12) 𝑚(2) 󵄩󵄩󵄩󵄩𝐵𝑠−3
𝑝,𝑟
𝑝,𝑟
󵄩 󵄩
󵄩
󵄩
+ 𝐶󵄩󵄩󵄩󵄩𝑢(12) 󵄩󵄩󵄩󵄩𝐵𝑠−2 󵄩󵄩󵄩󵄩V(1) 𝑚(1) 󵄩󵄩󵄩󵄩𝐵𝑠−3
𝑝,𝑟
󵄩
󵄩
󵄩
󵄩
≤ 𝐶 (󵄩󵄩󵄩󵄩𝑢(12) 󵄩󵄩󵄩󵄩𝐵𝑠−1 + 󵄩󵄩󵄩󵄩V(12) 󵄩󵄩󵄩󵄩𝐵𝑠−1 )
𝑝,𝑟
𝑝,𝑟
Lemma 12. Assume that 𝑢(0) = V(0) = 0. Let 1 ≤ 𝑝, 𝑟 ≤ +∞,
𝑠
. Then there exists
𝑠 > max{5/2, 2 + (1/𝑝)} and 𝑢0 , V0 ∈ 𝐵𝑝,𝑟
󵄩 󵄩2
󵄩 󵄩2
󵄩 󵄩2
󵄩 󵄩2
× (󵄩󵄩󵄩󵄩𝑢(1) 󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩V(1) 󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩𝑢(2) 󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩V(2) 󵄩󵄩󵄩󵄩𝐵𝑠 ) ,
𝑝,𝑟
𝑝,𝑟
𝑝,𝑟
𝑝,𝑟
∞ 2
a sequence of smooth functions (𝑢(𝑙) , V(𝑙) )𝑙∈N ∈ 𝐶(𝑅+ ; 𝐵𝑝,𝑟
)
solving the following linear transport equation by induction:
𝑠−3
‖𝐺‖𝐵𝑝,𝑟
󵄩
= 󵄩󵄩󵄩󵄩− (V(2) 𝑢(12) + V(12) 𝑢(1) ) 𝑛𝑥(2)
(𝜕𝑡 + (𝑢(𝑙) V(𝑙) ) 𝜕𝑥 ) 𝑚(𝑙+1)
󵄩
+3 (V𝑥(2) 𝑢(12) 𝑛(2) + V𝑥(2) 𝑢(1) 𝑛(12) + V𝑥(12) 𝑢(1) 𝑛(1) )󵄩󵄩󵄩󵄩𝐵𝑠−3
= −3V(𝑙) 𝑢𝑥(𝑙) 𝑚(𝑙) ,
𝑝,𝑟
󵄩 󵄩 󵄩
󵄩
≤ 𝐶󵄩󵄩󵄩󵄩V(2) 𝑢(12) + V(12) 𝑢(1) 󵄩󵄩󵄩󵄩𝐵𝑠−3 󵄩󵄩󵄩󵄩𝑛(2) 󵄩󵄩󵄩󵄩𝐵𝑠−2
𝑝,𝑟
=
𝑝,𝑟
󵄩
󵄩 󵄩 󵄩
+ 𝐶󵄩󵄩󵄩󵄩V(2) 󵄩󵄩󵄩󵄩𝐵𝑠−2 󵄩󵄩󵄩󵄩𝑢(1) 𝑛(12) 󵄩󵄩󵄩󵄩𝐵𝑠−3
𝑝,𝑟
𝑝,𝑟
󵄩 󵄩
󵄩
󵄩
+ 𝐶󵄩󵄩󵄩󵄩V(12) 󵄩󵄩󵄩󵄩𝐵𝑠−2 󵄩󵄩󵄩󵄩𝑢(1) 𝑛(1) 󵄩󵄩󵄩󵄩𝐵𝑠−3
𝑝,𝑟
𝑝,𝑟
(25)
Therefore, inserting the above estimates to (24), we obtain
𝑡
𝑝,𝑟
󵄩
󵄩
󵄩
󵄩
× (󵄩󵄩󵄩󵄩𝑢(12) (𝑡)󵄩󵄩󵄩󵄩𝐵𝑠−1 + 󵄩󵄩󵄩󵄩V(12) (𝑡)󵄩󵄩󵄩󵄩𝐵𝑠−1 )
𝑝,𝑟
𝑝,𝑟
󵄩
󵄩
󵄩
󵄩
≤ 󵄩󵄩󵄩󵄩𝑢0(12) 󵄩󵄩󵄩󵄩𝐵𝑠−1 + 󵄩󵄩󵄩󵄩V0(12) 󵄩󵄩󵄩󵄩𝐵𝑠−1
𝑝,𝑟
𝑝,𝑟
𝑡
0
𝑡 > 0, 𝑥 ∈ R
𝑢(𝑙+1) (𝑥, 0) = 𝑢0(𝑙+1) (𝑥) = 𝑆𝑙+1 𝑢0 ,
𝑥 ∈ R,
V(𝑙+1) (𝑥, 0) = V0(𝑙+1) (𝑥) = 𝑆𝑙+1 V0 ,
𝑥 ∈ R.
𝑠−1
)×
(ii) (𝑢(𝑙) , V(𝑙) )𝑙∈N is a Cauchy sequence in 𝐶([0, 𝑇]; 𝐵𝑝,𝑟
𝑠−1
𝐶([0, 𝑇]; 𝐵𝑝,𝑟 ).
∞
Proof. Since all the data 𝑆𝑛+1 𝑢0 and 𝑆𝑛+1 V0 belong to 𝐵𝑝,𝑟
,
Lemma 8 enables us to show by induction that for all 𝑙 ∈
∞ 2
).
N, (28) has a global solution which belongs to 𝐶(𝑅+ ; 𝐵𝑝,𝑟
Thanks to Lemma 7 and the proof of Proposition 11, we have
the following inequality for all 𝑙 ∈ N:
𝑡
󵄩
󵄩󵄩𝑚(𝑙+1) (𝑡)󵄩󵄩󵄩 𝑠−2
󵄩𝐵𝑝,𝑟
󵄩
−𝐶 ∫0 ‖𝜕𝑥 (𝑢(𝑙) V(𝑙) )(𝜏󸀠 )‖𝐵𝑠−1 𝑑𝜏󸀠 󵄩
󵄩
𝑒
𝑝,𝑟
𝑡
𝜏
−𝐶 ∫ ‖𝜕𝑥 (𝑢
󵄩 󵄩
≤ 󵄩󵄩󵄩𝑚0 󵄩󵄩󵄩𝐵𝑠−2 + 𝐶 ∫ 𝑒 0
𝑝,𝑟
(1) (1)
V )(𝜏󸀠 )‖𝐵𝑠−1 𝑑𝜏󸀠
𝑝,𝑟
0
𝜏
−𝐶 ∫0 ‖𝜕𝑥 (𝑢(1) V(1) )(𝜏󸀠 )‖𝐵𝑠−2 𝑑𝜏󸀠
+ 𝐶∫ 𝑒
(28)
𝑠
𝑠
(𝑇) × 𝐸𝑝,𝑟
(𝑇),
(i) (𝑢(𝑙) , V(𝑙) )𝑙∈N is uniformly bounded in 𝐸𝑝,𝑟
󵄩 󵄩2
󵄩 󵄩2
󵄩 󵄩2
󵄩 󵄩2
× (󵄩󵄩󵄩󵄩V(1) 󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩𝑢(1) 󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩V(2) 󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩𝑢(2) 󵄩󵄩󵄩󵄩𝐵𝑠 ) .
𝑝,𝑟
𝑝,𝑟
𝑝,𝑟
𝑝,𝑟
−𝐶 ∫0 ‖𝜕𝑥 (𝑢(1) V(1) )(𝜏󸀠 )‖𝐵𝑠−2 𝑑𝜏󸀠
−3𝑢(𝑙) V𝑥(𝑙) 𝑛(𝑙) ,
Moreover, there is a positive 𝑇 such that the solutions satisfying
the following properties:
󵄩
󵄩
󵄩
󵄩
≤ 𝐶 (󵄩󵄩󵄩󵄩V(12) 󵄩󵄩󵄩󵄩𝐵𝑠−1 + 󵄩󵄩󵄩󵄩𝑢(12) 󵄩󵄩󵄩󵄩𝐵𝑠−1 )
𝑝,𝑟
𝑝,𝑟
𝑒
𝑡 > 0, 𝑥 ∈ R
(𝜕𝑡 + (𝑢(𝑙) V(𝑙) ) 𝜕𝑥 ) 𝑛(𝑙+1)
𝑝,𝑟
󵄩
󵄩 󵄩 󵄩
+ 𝐶󵄩󵄩󵄩󵄩V(2) 󵄩󵄩󵄩󵄩𝐵𝑠−2 󵄩󵄩󵄩󵄩𝑢(12) 𝑚(2) 󵄩󵄩󵄩󵄩𝐵𝑠−3
𝑝,𝑟
𝑝,𝑟
(27)
Now let us start the proof of Theorem 1, which is motivated by the proof of local existence theorem about the
Camassa-Holm equation in [26]. Firstly, we will use the
classical Friedrichs’ regularization method to construct the
approximate solutions to the Cauchy problem (14).
𝑝,𝑟
𝑝,𝑟
󵄩󵄩
󵄩
󵄩 󵄩2
󵄩 󵄩2
󵄩󵄩𝜕𝑥 (𝑢(1) V(1) )󵄩󵄩󵄩 𝑠−2 ≤ 𝐶 (󵄩󵄩󵄩𝑢(1) 󵄩󵄩󵄩 𝑠 + 󵄩󵄩󵄩V(1) 󵄩󵄩󵄩 𝑠 ) ,
󵄩
󵄩𝐵𝑝,𝑟
󵄩 󵄩𝐵𝑝,𝑟 󵄩 󵄩𝐵𝑝,𝑟
and then applying the Gronwall’s inequality, we reach (20).
󵄩
󵄩 󵄩 󵄩
+ 𝐶󵄩󵄩󵄩󵄩𝑢(2) 󵄩󵄩󵄩󵄩𝐵𝑠−2 󵄩󵄩󵄩󵄩V(1) 𝑚(12) 󵄩󵄩󵄩󵄩𝐵𝑠−3
𝑝,𝑟
Hence, thanks to
󵄩
󵄩
× 󵄩󵄩󵄩󵄩3V(𝑙) 𝑢𝑥(𝑙) 𝑚(𝑙) 󵄩󵄩󵄩󵄩𝐵𝑠−2 𝑑𝜏,
𝑝,𝑟
6
Advances in Mathematical Physics
𝑡
−𝐶 ∫0 ‖𝜕𝑥 (𝑢(𝑙) V(𝑙) )(𝜏󸀠 )‖𝐵𝑠−1 𝑑𝜏󸀠 󵄩
󵄩 (𝑙+1)
𝑝,𝑟
󵄩󵄩𝑛
󵄩
𝑒
𝑡
󵄩
(𝑡)󵄩󵄩󵄩󵄩𝐵𝑠−2
𝑠−1
is an algebra, one obtains from (32) that for
Indeed, since 𝐵𝑝,𝑟
any 0 < 𝜏 < 𝑡
𝑝,𝑟
𝜏
−𝐶 ∫0 ‖𝜕𝑥 (𝑢(1) V(1) )(𝜏󸀠 )‖𝐵𝑠−1 𝑑𝜏󸀠
𝑝,𝑟
󵄩 󵄩
≤ 󵄩󵄩󵄩𝑚0 󵄩󵄩󵄩𝐵𝑠−2 + 𝐶 ∫ 𝑒
𝑝,𝑟
0
𝑡
󵄩
󵄩
𝐶 ∫ 󵄩󵄩󵄩󵄩𝜕𝑥 (𝑢(𝑙) V(𝑙) ) (𝜏󸀠 )󵄩󵄩󵄩󵄩𝐵𝑠−1 𝑑𝜏󸀠
𝑝,𝑟
𝜏
󵄩
󵄩
× 󵄩󵄩󵄩󵄩3𝑢(𝑙) V𝑥(𝑙) 𝑛(𝑙) 󵄩󵄩󵄩󵄩𝐵𝑠−2 𝑑𝜏.
𝑝,𝑟
𝑡
2
󵄩
󵄩
󵄩
󵄩
≤ 𝐶 ∫ (󵄩󵄩󵄩󵄩𝑢(𝑙) (𝑡)󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩V(𝑙) (𝑡)󵄩󵄩󵄩󵄩𝐵𝑠 ) 𝑑𝜏
𝑝,𝑟
𝑝,𝑟
𝜏
(29)
2
󵄩 󵄩
󵄩 󵄩
(󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠 )
𝑝,𝑟
𝑝,𝑟
≤ 𝐶∫
𝑑𝜏
2
󵄩󵄩 󵄩󵄩
󵄩󵄩 󵄩󵄩
𝜏
1 − 4𝐶(󵄩󵄩𝑢0 󵄩󵄩𝐵𝑠 + 󵄩󵄩V0 󵄩󵄩𝐵𝑠 ) 𝑡
𝑝,𝑟
𝑝,𝑟
𝑡
𝑠−2
Thanks to 𝑠 > max{5/2, 2 + (1/𝑝)}, we find 𝐵𝑝,𝑟
is an algebra.
From this, one obtains
−
3
󵄩 󵄩
󵄩 󵄩
≤ 𝐶(󵄩󵄩󵄩󵄩𝑢(𝑙) 󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩V(𝑙) 󵄩󵄩󵄩󵄩𝐵𝑠 ) ,
𝑝,𝑟
𝑝,𝑟
And then inserting (33) and (32) into (31) leads to
3
󵄩 󵄩
󵄩 󵄩
≤ (󵄩󵄩󵄩󵄩V(𝑙) 󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩𝑢(𝑙) 󵄩󵄩󵄩󵄩𝐵𝑠 ) ,
𝑝,𝑟
𝑝,𝑟
󵄩
󵄩󵄩 (𝑙+1) 󵄩󵄩
󵄩
󵄩󵄩𝑢
(𝑡)󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩V(𝑙+1) (𝑡)󵄩󵄩󵄩󵄩𝐵𝑠
󵄩
𝑝,𝑟
𝑝,𝑟
≤
which along with the above inequality leads to
𝑡
󵄩󵄩 󵄩󵄩
󵄩 󵄩
+ 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠
󵄩󵄩𝑢0 󵄩󵄩𝐵𝑝,𝑟
𝑠
𝑝,𝑟
2
󵄩 󵄩
󵄩 󵄩
(1 − 4𝐶(󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠 ) 𝑡)
𝑝,𝑟
𝑝,𝑟
+
−𝐶 ∫0 ‖𝜕𝑥 (𝑢(𝑙) V(𝑙) )(𝜏󸀠 )‖𝐵𝑠−1 𝑑𝜏󸀠
𝑝,𝑟
󵄩
󵄩
󵄩
󵄩
× (󵄩󵄩󵄩󵄩𝑢(𝑙+1) (𝑡)󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩V(𝑙+1) (𝑡)󵄩󵄩󵄩󵄩𝐵𝑠 )
𝑝,𝑟
𝑝,𝑟
𝐶
2
󵄩 󵄩
󵄩 󵄩
(1 − 4𝐶(󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠 ) 𝑡)
𝑝,𝑟
𝑝,𝑟
(31)
1/4
2
󵄩 󵄩
󵄩 󵄩
(󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠 )
𝑝,𝑟
𝑝,𝑟
𝜏
×
𝑝,𝑟
0
3
󵄩 󵄩
󵄩 󵄩
× (󵄩󵄩󵄩󵄩𝑢(𝑙) 󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩V(𝑙) 󵄩󵄩󵄩󵄩𝐵𝑠 ) 𝑑𝜏.
𝑝,𝑟
𝑝,𝑟
≤
2
𝑠 + ‖ V0 ‖𝐵𝑠 ) 𝑇
Let us choose a 𝑇 > 0 such that 4𝐶(‖ 𝑢0 ‖𝐵𝑝,𝑟
𝑝,𝑟
< 1, and suppose by induction that for all 𝑡 ∈ [0, 𝑇]
2
󵄩 󵄩
󵄩 󵄩
(1 − 4𝐶(󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠 ) 𝜏)
𝑝,𝑟
𝑝,𝑟
󵄩󵄩 (𝑙) 󵄩󵄩
󵄩
󵄩
󵄩󵄩𝑢 (𝑡)󵄩󵄩 𝑠 + 󵄩󵄩󵄩V(𝑙) (𝑡)󵄩󵄩󵄩 𝑠
󵄩
󵄩𝐵𝑝,𝑟 󵄩
󵄩𝐵𝑝,𝑟
2
󵄩 󵄩
󵄩 󵄩
(1 − 4𝐶(󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠 ) 𝑡)
𝑝,𝑟
𝑝,𝑟
× (1 + 𝐶 ∫
≤
=
2
󵄩 󵄩
󵄩 󵄩
(1 − 4𝐶(󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠 ) 𝑡)
𝑝,𝑟
𝑝,𝑟
1/2
.
(32)
3/2
𝑑𝜏
󵄩󵄩 󵄩󵄩
󵄩 󵄩
+ 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠
󵄩󵄩𝑢0 󵄩󵄩𝐵𝑝,𝑟
𝑠
𝑝,𝑟
𝑡
0
󵄩󵄩 󵄩󵄩
󵄩 󵄩
+ 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠
󵄩󵄩𝑢0 󵄩󵄩𝐵𝑝,𝑟
𝑠
𝑝,𝑟
1/4
2
󵄩 󵄩
󵄩 󵄩
× ∫ (1 − 4𝐶(󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠 ) 𝜏)
𝑝,𝑟
𝑝,𝑟
0
−𝐶 ∫0 ‖𝜕𝑥 (𝑢(1) V(1) )(𝜏󸀠 )‖𝐵𝑠−1 𝑑𝜏󸀠
+ 𝐶∫ 𝑒
1/4
𝑡
󵄩 󵄩
󵄩 󵄩
≤ 󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠
𝑝,𝑟
𝑝,𝑟
𝑡
2
1
󵄩 󵄩
󵄩 󵄩
ln (1 − 4𝐶(󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠 ) 𝑡) .
𝑝,𝑟
𝑝,𝑟
4
(30)
󵄩󵄩 (𝑙) (𝑙) (𝑙) 󵄩󵄩
󵄩 󵄩 󵄩 󵄩 󵄩 󵄩
󵄩󵄩𝑢 V𝑥 𝑛 󵄩󵄩 𝑠−2 ≤ 𝐶󵄩󵄩󵄩𝑢(𝑙) 󵄩󵄩󵄩 𝑠−2 󵄩󵄩󵄩𝑛(𝑙) 󵄩󵄩󵄩 𝑠−2 󵄩󵄩󵄩V𝑥(𝑙) 󵄩󵄩󵄩 𝑠−2
󵄩
󵄩𝐵𝑝,𝑟
󵄩 󵄩𝐵𝑝,𝑟 󵄩 󵄩𝐵𝑝,𝑟 󵄩 󵄩𝐵𝑝,𝑟
𝑒
2
1
󵄩 󵄩
󵄩 󵄩
ln (1 − 4𝐶(󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠 ) 𝜏)
𝑝,𝑟
𝑝,𝑟
4
=
󵄩 󵄩 󵄩
󵄩 󵄩 󵄩
󵄩󵄩 (𝑙) (𝑙) (𝑙) 󵄩󵄩
󵄩󵄩V 𝑢𝑥 𝑚 󵄩󵄩 𝑠−2 ≤ 𝐶󵄩󵄩󵄩V(𝑙) 󵄩󵄩󵄩 𝑠−2 󵄩󵄩󵄩𝑚(𝑙) 󵄩󵄩󵄩 𝑠−2 󵄩󵄩󵄩𝑢𝑥(𝑙) 󵄩󵄩󵄩 𝑠−2
󵄩𝐵𝑝,𝑟
󵄩𝐵𝑝,𝑟 󵄩 󵄩𝐵𝑝,𝑟
󵄩 󵄩𝐵𝑝,𝑟 󵄩
󵄩
(33)
1/4
2
󵄩 󵄩
󵄩 󵄩
(󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠 )
𝑝,𝑟
𝑝,𝑟
2
󵄩 󵄩
󵄩 󵄩
(1 − 4𝐶(󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠 ) 𝑡)
𝑝,𝑟
𝑝,𝑟
󵄩󵄩 󵄩󵄩
󵄩 󵄩
+ 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠
󵄩󵄩𝑢0 󵄩󵄩𝐵𝑝,𝑟
𝑠
𝑝,𝑟
2
󵄩 󵄩
󵄩 󵄩
(1 − 4𝐶(󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠 ) 𝑡)
𝑝,𝑟
𝑝,𝑟
1/2
5/4
𝑑𝜏)
.
(34)
Advances in Mathematical Physics
7
Similar to the proof of Proposition 11, then for every 𝑡 ∈
[0, 𝑇], we obtain
Hence, one can see that
󵄩󵄩 (𝑙+1) 󵄩󵄩
󵄩
󵄩
󵄩󵄩𝑢
(𝑡)󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩V(𝑙+1) (𝑡)󵄩󵄩󵄩󵄩𝐵𝑠
󵄩
𝑝,𝑟
𝑝,𝑟
≤
𝑡
−𝐶 ∫0 ‖𝜕𝑥 (𝑢(𝑘+𝑙) V(𝑘+𝑙) )(𝜏󸀠 )‖𝐵𝑠−2 𝑑𝜏󸀠
󵄩󵄩 󵄩󵄩
󵄩 󵄩
+ 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠
󵄩󵄩𝑢0 󵄩󵄩𝐵𝑝,𝑟
𝑠
𝑝,𝑟
2
󵄩 󵄩
󵄩 󵄩
(1 − 4𝐶(󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩V0 󵄩󵄩󵄩𝐵𝑠 ) 𝑡)
𝑝,𝑟
𝑝,𝑟
1/2
,
(35)
𝑒
𝑝,𝑟
󵄩
󵄩
× (󵄩󵄩󵄩󵄩(𝑢(𝑘+𝑙+1) − 𝑢(𝑙+1) ) (𝑡)󵄩󵄩󵄩󵄩𝐵𝑠−1
𝑝,𝑟
which implies that (𝑢(𝑙) , V(𝑙) )𝑙∈N is uniformly bounded in
𝑠
𝑠
) × 𝐶([0; 𝑇]; 𝐵𝑝,𝑟
). Using the Moser-type esti𝐶([0; 𝑇]; 𝐵𝑝,𝑟
mates (see Lemma 9), one finds that
󵄩
󵄩
󵄩󵄩 (𝑙) (𝑙)
󵄩
󵄩 󵄩2
󵄩 󵄩2
󵄩󵄩𝑢 V 𝜕𝑥 𝑚(𝑙+1) 󵄩󵄩󵄩 𝑠−3 ≤ 𝐶󵄩󵄩󵄩𝑢(𝑙+1) 󵄩󵄩󵄩 𝑠 (󵄩󵄩󵄩𝑢(𝑙) 󵄩󵄩󵄩 𝑠 + 󵄩󵄩󵄩V(𝑙) 󵄩󵄩󵄩 𝑠 ) ,
󵄩𝐵𝑝,𝑟
󵄩𝐵𝑝,𝑟 󵄩 󵄩𝐵𝑝,𝑟 󵄩 󵄩𝐵𝑝,𝑟
󵄩
󵄩
󵄩
󵄩
󵄩 󵄩2
󵄩 󵄩2
󵄩󵄩 (𝑙) (𝑙) (𝑙+1) 󵄩󵄩
󵄩󵄩 𝑠−3 ≤ 𝐶󵄩󵄩󵄩V(𝑙+1) 󵄩󵄩󵄩 𝑠 (󵄩󵄩󵄩𝑢(𝑙) 󵄩󵄩󵄩 𝑠 + 󵄩󵄩󵄩V(𝑙) 󵄩󵄩󵄩 𝑠 ) ,
󵄩󵄩V 𝑢 𝜕𝑥 𝑛
󵄩𝐵𝑝,𝑟
󵄩𝐵𝑝,𝑟 󵄩 󵄩𝐵𝑝,𝑟 󵄩 󵄩𝐵𝑝,𝑟
󵄩
󵄩
󵄩
󵄩
+󵄩󵄩󵄩󵄩(V(𝑘+𝑙+1) − V𝑙+1 ) (𝑡)󵄩󵄩󵄩󵄩𝐵𝑠−1 )
𝑝,𝑟
󵄩
󵄩
󵄩
󵄩
≤ 󵄩󵄩󵄩󵄩𝑢0(𝑘+𝑙+1) − 𝑢0(𝑙+1) 󵄩󵄩󵄩󵄩𝐵𝑠−1 + 󵄩󵄩󵄩󵄩𝑢0(𝑘+𝑙+1) − 𝑢0(𝑙+1) 󵄩󵄩󵄩󵄩𝐵𝑠−1
𝑝,𝑟
𝑝,𝑟
𝑡
󵄩
󵄩
󵄩
󵄩
× (󵄩󵄩󵄩󵄩V(𝑘+𝑙) − V(𝑙) 󵄩󵄩󵄩󵄩𝐵𝑠−1 + 󵄩󵄩󵄩󵄩V(𝑘+𝑙) − V(𝑙) 󵄩󵄩󵄩󵄩𝐵𝑠−1 )
𝑝,𝑟
𝑝,𝑟
󵄩2
󵄩 󵄩2
󵄩
󵄩 󵄩2
× (󵄩󵄩󵄩󵄩V(𝑙) 󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩𝑢(𝑙) 󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩V(𝑙+𝑘) 󵄩󵄩󵄩󵄩𝐵𝑠
𝑝,𝑟
𝑝,𝑟
𝑝,𝑟
3
󵄩 󵄩
󵄩 󵄩
≤ 𝐶(󵄩󵄩󵄩󵄩𝑢(𝑙) 󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩V(𝑙) 󵄩󵄩󵄩󵄩𝐵𝑠 ) ,
𝑝,𝑟
𝑝,𝑟
󵄩2
󵄩2
󵄩2
󵄩
󵄩
󵄩
+󵄩󵄩󵄩󵄩𝑢(𝑘+𝑙) 󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩𝑢(𝑙+1) 󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩V(𝑙+1) 󵄩󵄩󵄩󵄩𝐵𝑠 ) 𝑑𝜏.
𝑝,𝑟
𝑝,𝑟
𝑝,𝑟
(40)
󵄩 󵄩 󵄩 󵄩 󵄩 󵄩
󵄩󵄩 (𝑙) (𝑙) (𝑙) 󵄩󵄩
󵄩󵄩𝑢 V𝑥 𝑛 󵄩󵄩 𝑠−2 ≤ 𝐶󵄩󵄩󵄩𝑢(𝑙) 󵄩󵄩󵄩 𝑠−2 󵄩󵄩󵄩𝑛(𝑙) 󵄩󵄩󵄩 𝑠−2 󵄩󵄩󵄩V𝑥(𝑙) 󵄩󵄩󵄩 𝑠−2
󵄩𝐵𝑝,𝑟
󵄩 󵄩𝐵𝑝,𝑟 󵄩 󵄩𝐵𝑝,𝑟 󵄩 󵄩𝐵𝑝,𝑟
󵄩
𝑠
𝑠
Since (𝑢(𝑙) , V(𝑙) )𝑙∈N is uniformly bounded in 𝐸𝑝,𝑟
(𝑇) × 𝐸𝑝,𝑟
(𝑇)
and
3
󵄩 󵄩
󵄩 󵄩
≤ 𝐶(󵄩󵄩󵄩󵄩V(𝑙) 󵄩󵄩󵄩󵄩𝐵𝑠 + 󵄩󵄩󵄩󵄩𝑢(𝑙) 󵄩󵄩󵄩󵄩𝐵𝑠 ) .
𝑝,𝑟
𝑝,𝑟
(36)
Hence, using (28), we have
uniformly bounded, which yields that the sequence
𝑠
𝑠
(𝑢(𝑙) , V(𝑙) )∈N is uniformly bounded in 𝐸𝑝,𝑟
(𝑇) × 𝐸𝑝,𝑟
(𝑇).
(𝑙)
(𝑙)
Now, it suffices to show that (𝑢 , V )𝑙∈N is a Cauchy
𝑠−1
𝑠−1
sequence in 𝐶([0; 𝑇]; 𝐵𝑝,𝑟
) × 𝐶([0; 𝑇]; 𝐵𝑝,𝑟
). In fact, for all
𝑙, 𝑘 ∈ N, from (28), we have
(𝜕𝑡 + (𝑢(𝑙+𝑘) V(𝑙+𝑘) ) 𝜕𝑥 ) (𝑚(𝑙+𝑘+1) − 𝑚(𝑙+1) ) = 𝐹󸀠 ,
V
𝑙+𝑘
𝑢0(𝑙+𝑘+1) − 𝑢0(𝑙+1) = 𝑆𝑘+𝑙+1 𝑢0 − 𝑆𝑙+1 𝑢0 = ∑ Δ 𝑞 𝑢0 ,
𝑞=𝑙+1
𝑠−1
𝑠−1
(𝜕𝑥 𝑢(𝑙+1) , 𝜕𝑥 V(𝑙+1) )𝑙∈N ∈ 𝐶 ([0; 𝑇] ; 𝐵𝑝,𝑟
) × 𝐶 ([0; 𝑇] ; 𝐵𝑝,𝑟
)
(37)
(𝜕𝑡 + (𝑢
𝑝,𝑟
0
󵄩 󵄩 󵄩
󵄩 󵄩 󵄩
󵄩󵄩 (𝑙) (𝑙) (𝑙) 󵄩󵄩
󵄩󵄩V 𝑢𝑥 𝑚 󵄩󵄩 𝑠−2 ≤ 𝐶󵄩󵄩󵄩V(𝑙) 󵄩󵄩󵄩 𝑠−2 󵄩󵄩󵄩𝑚(𝑙) 󵄩󵄩󵄩 𝑠−2 󵄩󵄩󵄩𝑢𝑥(𝑙) 󵄩󵄩󵄩 𝑠−2
󵄩𝐵𝑝,𝑟
󵄩𝐵𝑝,𝑟 󵄩 󵄩𝐵𝑝,𝑟
󵄩 󵄩𝐵𝑝,𝑟 󵄩
󵄩
(𝑙+𝑘) (𝑙+𝑘)
𝜏
−𝐶 ∫0 ‖𝜕𝑥 (𝑢(𝑙) V(𝑙) )(𝜏󸀠 )‖𝐵𝑠−2 𝑑𝜏󸀠
+ 𝐶∫ 𝑒
) 𝜕𝑥 ) (𝑛
(𝑙+𝑘+1)
−𝑛
(𝑙+1)
󸀠
)=𝐺,
(38)
with
𝑙+𝑘
V0(𝑙+𝑘+1) − V0(𝑙+1) = 𝑆𝑘+𝑙+1 V0 − 𝑆𝑙+1 V0 = ∑ Δ 𝑞 V0 ,
𝑞=𝑙+1
we get a constant 𝐶𝑇 independent of 𝑙, 𝑘 such that for all 𝑡 ∈
[0, 𝑇],
󵄩
󵄩󵄩 (𝑘+𝑙+1)
󵄩󵄩(𝑢
− 𝑢(𝑙+1) ) (𝑡)󵄩󵄩󵄩󵄩𝐵𝑠−1
󵄩
𝑝,𝑟
󵄩
󵄩
+ 󵄩󵄩󵄩󵄩(V(𝑘+𝑙+1) − V(𝑙+1) ) (𝑡)󵄩󵄩󵄩󵄩𝐵𝑠−1
𝑝,𝑟
−
+
(V
(𝑘+𝑙)
𝑢𝑥(𝑘+𝑙) V(𝑙)
(𝑙)
󵄩
󵄩
≤ 𝐶𝑇 (2−𝑛 + ∫ (󵄩󵄩󵄩󵄩(𝑢(𝑘+𝑙) − 𝑢(𝑙) ) (𝜏)󵄩󵄩󵄩󵄩𝐵𝑠−1
𝑝,𝑟
0
󵄩
󵄩
+󵄩󵄩󵄩󵄩(V(𝑘+𝑙) − 𝑢(𝑙) ) (𝜏)󵄩󵄩󵄩󵄩𝐵𝑠−1 ) 𝑑𝜏) .
(𝑚
Arguing by induction with respect to the index 𝑙, one can
easily prove that
󵄩󵄩󵄩(𝑢(𝑘+𝑙+1) − 𝑢(𝑙+1) ) (𝑡)󵄩󵄩󵄩
󵄩󵄩
󵄩󵄩𝐿∞ (𝐵𝑝,𝑟
𝑠−1 )
𝑇
(𝑙)
−𝑚 )
+ (𝑢𝑥(𝑘+𝑙) − 𝑢𝑥(𝑙) ) V(𝑙) 𝑚(𝑙) ) ,
󸀠
𝐺 = − (V
(𝑙+𝑘)
(𝑢
(𝑙+𝑘)
𝑝,𝑟
(𝑘+𝑙)
− V )𝑚
(𝑘+𝑙)
(𝑙)
− 𝑢 ) + (V
(𝑙+𝑘)
− 3 (V𝑥(𝑘+𝑙) (𝑢(𝑘+𝑙) − 𝑢(𝑙) ) 𝑛(𝑘+𝑙)
+ V𝑥(𝑘+𝑙) 𝑢(𝑙) (𝑛(𝑘+𝑙) − 𝑛(𝑙) )
+ (V𝑥(𝑘+𝑙) − V𝑥(𝑙) ) 𝑢(𝑙) 𝑛(𝑙) ) .
(42)
𝑡
𝐹󸀠 = − (𝑢(𝑙+𝑘) (V(𝑙+𝑘) − V(𝑙) ) + (𝑢(𝑙+𝑘) − 𝑢(𝑙) ) V(𝑙) ) 𝑚𝑥(𝑙+1)
3 (𝑢𝑥(𝑘+𝑙)
(41)
(𝑙)
− V )𝑢
(𝑙)
) 𝑛𝑥(𝑙+1)
󵄩
󵄩
+ 󵄩󵄩󵄩󵄩(V(𝑘+𝑙+1) − V(𝑙+1) ) (𝑡)󵄩󵄩󵄩󵄩𝐿∞ (𝐵𝑠−1 )
(39)
𝑇
𝑙+1
≤
𝑝,𝑟
(𝑇𝐶𝑇 )
󵄩 󵄩
󵄩󵄩 𝑘 󵄩󵄩
󵄩V 󵄩󵄩 ∞ 𝑠−1 )
(󵄩󵄩󵄩𝑢𝑘 󵄩󵄩󵄩
𝑠−1 ) + 󵄩
󵄩 󵄩𝐿 𝑇 (𝐵𝑝,𝑟 )
(𝑙 + 1)! 󵄩 󵄩𝐿∞𝑇 (𝐵𝑝,𝑟
𝑙
+ 𝐶𝑇 ∑ 2𝑞−𝑙
𝑞=0
𝑞
(𝑇𝐶𝑇 )
.
𝑞!
(43)
8
Advances in Mathematical Physics
(𝑘)
𝑠−1 ) , ‖ V
𝑠−1 ) , and 𝐶 are bounded indepenAs ‖ 𝑢(𝑘) ‖𝐿∞𝑇 (𝐵𝑝,𝑟
‖𝐿∞𝑇 (𝐵𝑝,𝑟
𝐶𝑇󸀠
dently of 𝑘, there exists constant
independent of 𝑙, 𝑘 such
that
󵄩
󵄩󵄩 (𝑘+𝑙+1)
− 𝑢(𝑙+1) ) (𝑡)󵄩󵄩󵄩󵄩𝐿∞ (𝐵𝑠−1 )
󵄩󵄩󵄩(𝑢
𝑝,𝑟
𝑇
(44)
󵄩
󵄩󵄩 (𝑘+𝑙+1)
(𝑙+1)
+ 󵄩󵄩󵄩(V
−V
) (𝑡)󵄩󵄩󵄩󵄩𝐿∞ (𝐵𝑠−1 ) ≤ 𝐶𝑇󸀠 2−𝑛 .
𝑇
𝑝,𝑟
𝑠−1
Thus, (𝑢(𝑙) , V(𝑙) )𝑛∈N is a Cauchy sequence in 𝐶([0, 𝑇]; 𝐵𝑝,𝑟
)×
𝑠−1
([0, 𝑇]; 𝐵𝑝,𝑟 ).
Proof of Theorem 1. Thanks to Lemma 12, we obtain that
𝑠−1
(𝑢(𝑙) , V(𝑙) )𝑙∈N is a Cauchy sequence in 𝐶([0, 𝑇]; 𝐵𝑝,𝑟
) × 𝐶([0,
𝑠−1
𝑇]; 𝐵𝑝,𝑟 ); so, it converges to some function (𝑢, V) ∈ 𝐶([0,
𝑠−1
𝑠−1
𝑇]; 𝐵𝑝,𝑟
) × 𝐶([0, 𝑇]; 𝐵𝑝,𝑟
). We now have to check that (𝑢, V)
𝑠
𝑠
belongs to 𝐸𝑝,𝑟 (𝑇) × 𝐸𝑝,𝑟 (𝑇) and solves the Cauchy prob-
lem (1). Since (𝑢(𝑙) , V(𝑙) )𝑙∈N is uniformly bounded in 𝐿∞ ([0,
𝑠
𝑠
) × 𝐿∞ ([0, 𝑇]; 𝐵𝑝,𝑟
), according to Lemma 12, the Fatou
𝑇]; 𝐵𝑝,𝑟
property for the Besov spaces (Proposition 6) guarantees that
(𝑢, V) also belongs to
𝑠
𝑠
𝐿∞ ([0, 𝑇] ; 𝐵𝑝,𝑟
) × 𝐿∞ ([0, 𝑇] ; 𝐵𝑝,𝑟
).
󸀠
𝑠
ment ensures that the convergence holds in 𝐶([0, 𝑇]; 𝐵𝑝,𝑟
)×
󸀠
𝑠
𝐶([0, 𝑇]; 𝐵𝑝,𝑟
), for any 𝑠󸀠 < 𝑠. It is then easy to pass to the
limit in (28) and to conclude that (𝑢, V) is indeed a solution to
the Cauchy problem (1). Thanks to the fact that 𝑢 belongs to
𝑠
𝑠
) × 𝐿∞ ([0, 𝑇]; 𝐵𝑝,𝑟
), the right-hand side of the
𝐿∞ ([0, 𝑇]; 𝐵𝑝,𝑟
equation
where 𝐻2 (R) is the Sobolev space of order two on the real
line R and N0 is the set of nonnegative integers. One can easily
verify that 𝐸𝑠 equipped with the norm |||⋅|||𝑠 is a Banach space
and that, for any 0 < 𝑠󸀠 < 𝑠, 𝐸𝑠 is continuously embedded in
𝐸𝑠󸀠 with
|||𝑢|||𝑠󸀠 ≤ |||𝑢|||𝑠 .
(47)
𝑠
belongs to 𝐿∞ ([0, 𝑇]; 𝐵𝑝,𝑟
). In particular, for the case 𝑟 < ∞,
󸀠
𝑠
Lemma 8 enables us to conclude that (𝑢, V) ∈ 𝐶([0, 𝑇]; 𝐵𝑝,𝑟
)×
Lemma 13 (see [29]). Let 0 < 𝑠 < 1. There is a constant 𝐶 > 0,
independent of 𝑠, such that for any 𝑢 and V in 𝐸𝑠 , one has
|||𝑢V|||𝑠 ≤ 𝐶|||𝑢|||𝑠 |||V|||𝑠 .
Theorem 15 (see [30]). Let {𝑋𝑠 }0<𝑠<1 be a scale of decreasing
Banach spaces; namely, for any 𝑠󸀠 < 𝑠, one has 𝑋𝑠 ⊂ 𝑋𝑠󸀠 and
||| ⋅ |||𝑠󸀠 ≤ ||| ⋅ |||𝑠 . Consider the Cauchy problem
𝑑𝑢
= 𝐹 (𝑡, 𝑢 (𝑡)) ,
𝑑𝑡
Let 𝑇, 𝐻, and 𝐶 be positive constants, and assume that 𝐹
satisfies the following conditions.
(1) If for 0 < 𝑠󸀠 < 𝑠 < 1 the function 𝑡 󳨃→ 𝑢(𝑡) is
holomorphic in |𝑡| < 𝑇 and continuous on |𝑡| ≤ 𝑇 with
values in 𝑋𝑠 and
sup |||𝑢 (𝑡)|||𝑠 < 𝐻,
󸀠
3. Analyticity of Solutions
In this section, we will show the existence and uniqueness of
analytic solutions to the system (1) on the line R.
(51)
𝑢 (0) = 0.
󸀠
󸀠
(50)
Lemma 14 (see [29]). There is a constant 𝑐 > 0 such that for
any 0 < 𝑠󸀠 < 𝑠 < 1, one has |||𝜕𝑥 𝑢|||𝑠󸀠 ≤ (𝐶/(𝑠 − 𝑠󸀠 ))|||𝑢|||𝑠 ,
|||(1 − 𝜕𝑥2 )−1 𝑢|||𝑠󸀠 ≤ |||𝑢|||𝑠 , and |||(1 − 𝜕𝑥2 )−1 𝜕𝑥 𝑢|||𝑠󸀠 ≤ |||𝑢|||𝑠 .
𝑠
𝐶([0, 𝑇]; 𝐵𝑝,𝑟
) for any 𝑠󸀠 < 𝑠. Finally, using the equation
𝑠
𝑠
again, we see that (𝜕𝑡 𝑢, 𝜕𝑡 V) ∈ 𝐶([0, 𝑇]; 𝐵𝑝,𝑟
) × 𝐶([0, 𝑇]; 𝐵𝑝,𝑟
)
∞
𝑠−1
∞
𝑠−1
if 𝑟 < ∞, and 𝐿 ([0, 𝑇]; 𝐵𝑝,𝑟 ) × 𝐿 ([0, 𝑇]; 𝐵𝑝,𝑟 ) otherwise.
𝑠
𝑠
(𝑇) × 𝐸𝑝,𝑟
(𝑇). Moreover, a
Therefore, (𝑢, V) belongs to 𝐸𝑝,𝑟
standard use of a sequence of viscosity approximate solutions
(𝑢𝜀 , V𝜀 )𝜀>0 for the Cauchy problem (1) which converges uni𝑠
𝑠−1
𝑠
) ∩ 𝐶1 ([0, 𝑇]; 𝐵𝑝,𝑟
) × 𝐶([0, 𝑇]; 𝐵𝑝,𝑟
)∩
formly in 𝐶([0, 𝑇]; 𝐵𝑝,𝑟
1
𝑠−1
𝐶 ([0, 𝑇]; 𝐵𝑝,𝑟 ) gives the continuity of the solution (𝑢, V) in
𝑠
𝑠
× 𝐸𝑝,𝑟
. The proof of Theorem 1 is complete.
𝐸𝑝,𝑟
(49)
Another simple consequence of the definition is that any 𝑢 in
𝐸𝑠 is a real analytic function on R. Crucial for our purposes
is the fact that each 𝐸𝑠 forms an algebra under pointwise
multiplication of functions.
(46)
𝑠
belongs to 𝐿∞ ([0, 𝑇]; 𝐵𝑝,𝑟
), and the right-hand side of the
equation
𝜕𝑡 𝑚 + 𝑢V𝜕𝑥 𝑛 = −3𝑢V𝑥 𝑛
󵄩 󵄩
𝑠𝑘 󵄩󵄩󵄩󵄩𝜕𝑘 𝑢󵄩󵄩󵄩󵄩𝐻2
{
}
∞
< ∞} ,
𝐸𝑠 = {𝑢 ∈ 𝐶 (R) : |||𝑢|||𝑠 = sup
2
𝑘∈N0 𝑘!/(𝑘 + 1)
{
}
(48)
(45)
On the other hand, as (𝑢(𝑙) , V(𝑙) )𝑙∈N converges to (𝑢, V)
𝑠−1
𝑠−1
in 𝐶([0, 𝑇]; 𝐵𝑝,𝑟
) × 𝐶([0, 𝑇]; 𝐵𝑝,𝑟
), an interpolation argu-
𝜕𝑡 𝑚 + 𝑢V𝜕𝑥 𝑚 = −3V𝑢𝑥 𝑚
First, we will need a suitable scale of Banach spaces as
follows. For any 𝑠 > 0, we set
|𝑡|≤𝑇
(52)
then 𝑡 󳨃→ 𝐹(𝑡, 𝑢(𝑡)) is a holomorphic function on |𝑡| <
𝑇 with values in 𝑋𝑠󸀠 .
(2) For any 0 < 𝑠󸀠 < 𝑠 < 1 and any 𝑢, V ∈ 𝐵(0, 𝐻) ⊂ 𝑋𝑠 ,
that is, |||𝑢|||𝑠 < 𝐻, |||V|||𝑠 < 𝐻, one has
sup |||𝐹 (𝑡, 𝑢) − 𝐹 (𝑡, V)|||𝑠󸀠 ≤
|𝑡|≤𝑇
𝐶
|||𝑢 − V|||𝑠 .
𝑠 − 𝑠󸀠
(53)
(3) There exists 𝑀 > 0 such that for any 0 < 𝑠 < 1,
sup |||𝐹 (𝑡, 0)|||𝑠 ≤
|𝑡|≤𝑇
𝑀
.
1−𝑠
(54)
Advances in Mathematical Physics
9
Then, there exists a 𝑇0 ∈ (0, 𝑇) and a unique function 𝑢(𝑡),
which for every 𝑠 ∈ (0, 1) is holomorphic in |𝑡| < (1 − 𝑠)𝑇0 with
values in 𝑋𝑠 and is a solution to the Cauchy problem (51).
−1
𝜕𝑡 𝑢4 = − (𝑢4 𝑢1 𝑢4 + 𝑢3 𝜕𝑥 (𝑢1 𝑢4 )) − (1 − 𝜕𝑥2 )
× (𝑢4 𝜕𝑥 (𝑢4 𝑢1 + 𝑢3 𝑢2 ) − 𝑢42 𝑢1 − 𝑢3 𝑢2 𝑢4
−2𝑢1 𝜕𝑥 𝑢42 − 𝑢3 𝑢2 𝜕𝑥 𝑢4 + 3𝑢4 𝑢1 𝑢3 )
Next, we restate the Cauchy problem (1) in a more
convenient form, and we can rewrite the Cauchy problem (1)
as follows:
= 𝐹4 (𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 ) ,
𝑢𝑡 − 𝑢𝑥𝑥𝑡 + 4𝑢V𝑢𝑥 − 3𝑢𝑥 𝑢𝑥𝑥 V − 𝑢𝑢𝑥𝑥𝑥 V = 0,
𝑢1 (0, 𝑥) = 𝑢0 (𝑥) ,
V𝑡 − V𝑥𝑥𝑡 + 4V𝑢V𝑥 − 3V𝑥 V𝑥𝑥 𝑢 − VV𝑥𝑥𝑥 𝑢 = 0,
𝑢 (0, 𝑥) = 𝑢0 (𝑥) ,
𝑢2 (0, 𝑥) = 𝑢0󸀠 (𝑥) ,
(55)
V (0, 𝑡) = V0 (𝑥) .
𝑢3 (0, 𝑥) = V0 (𝑥) ,
Applying the operator (1 − 𝜕𝑥2 )−1 to both sides of the first
equation and second equation in (55), we obtain
𝑢𝑡 + 𝑢V𝑢𝑥 + (1 −
𝑢4 (0, 𝑥) = V0󸀠 (𝑥) .
(57)
−1
𝜕𝑥2 )
× ((𝑢𝑥 V + 𝑢V𝑥 )𝑥 𝑢𝑥 − 𝑢𝑥2 V − 𝑢V𝑥 𝑢𝑥
Define
2
−2V(𝑢𝑥 )𝑥 − 𝑢V𝑥 𝑢𝑥𝑥 + 3𝑢𝑥 V𝑢) = 0,
−1
V𝑡 + V𝑢V𝑥 + (1 − 𝜕𝑥2 )
𝑈 ≡ (𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 ) ,
(56)
× ((V𝑥 𝑢 + V𝑢𝑥 )𝑥 V𝑥 −
V𝑥2 𝑢
𝐹 (𝑈) = 𝐹 (𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 )
− V𝑢𝑥 V𝑥
≡ (𝐹1 (𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 ) , 𝐹2 (𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 ) ,
2
−2𝑢(V𝑥 )𝑥 − V𝑢𝑥 V𝑥𝑥 + 3V𝑥 𝑢V) = 0,
𝑢 (0, 𝑥) = 𝑢0 (𝑥) ,
𝐹3 (𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 ) , 𝐹4 (𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 )) .
V (0, 𝑡) = V0 (𝑥) .
Differentiating with respect to 𝑥 on both sides of the previous
equation and letting 𝑢1 = 𝑢, 𝑢2 = 𝑢𝑥 , 𝑢3 = V, and 𝑢4 = V𝑥 ,
then the problem (56) can be written as a system for 𝑢1 , 𝑢2 ,
𝑢2 , 𝑢4 as follows:
Then, we have
𝑑𝑈
= 𝐹 (𝑡, 𝑈 (𝑡)) ,
𝑑𝑡
−1
𝜕𝑡 𝑢1 = −𝑢1 𝑢3 𝑢2 − (1 − 𝜕𝑥2 )
𝑈 (0) =
× (𝑢2 𝜕𝑥 (𝑢2 𝑢3 + 𝑢1 𝑢4 ) − 𝑢22 𝑢3 − 𝑢1 𝑢4 𝑢2
−2𝑢3 𝜕𝑥 𝑢22 − 𝑢1 𝑢4 𝜕𝑥 𝑢2 + 3𝑢2 𝑢3 𝑢1 )
= 𝐹1 (𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 ) ,
−1
𝜕𝑡 𝑢2 = − (𝑢2 𝑢3 𝑢2 + 𝑢1 𝜕𝑥 (𝑢3 𝑢2 )) − 𝜕𝑥 (1 − 𝜕𝑥2 )
× ((𝑢2 𝑢3 + 𝑢1 𝑢4 )𝑥 𝑢2 − 𝑢22 𝑢3 − 𝑢1 𝑢4 𝑢2
−2𝑢3 𝜕𝑥 𝑢22 − 𝑢1 𝑢4 𝜕𝑥 𝑢2 + 3𝑢2 𝑢3 𝑢1 )
= 𝐹2 (𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 ) ,
−1
𝜕𝑡 𝑢3 = −𝑢3 𝑢1 𝑢4 − (1 − 𝜕𝑥2 )
× (𝑢4 𝜕𝑥 (𝑢4 𝑢1 + 𝑢3 𝑢2 ) − 𝑢42 𝑢1 − 𝑢3 𝑢2 𝑢4
−2𝑢1 𝜕𝑥 𝑢42
(58)
− 𝑢3 𝑢2 𝜕𝑥 𝑢4 + 3𝑢4 𝑢1 𝑢3 )
= 𝐹3 (𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 ) ,
(59)
(𝑢0 , 𝑢0󸀠 , V0 , V0󸀠 ) .
Proof of Theorem 2. Theorem 2 is a straightforward consequence of the abstract Cauchy-Kowalevski Theorem 15.
We only need to verify conditions (1)-(2) in the statement of the abstract Cauchy-Kowalevski Theorem 15 for
𝐹𝑖 (𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 ) (𝑖 = 1, 2, 3, 4) in the system (57) since
𝐹𝑖 (𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 ), (𝑖 = 1, 2, 3, 4) do not depend on 𝑡 explicitly.
We observe that, for 0 < 𝑠󸀠 < 𝑠 < 1, by the estimates in
Lemmas 13 and 14, condition (1) holds
Next, we verify the second condition. For any 𝑢𝑗 and V𝑗 ∈
𝐵(0, 𝐻) ⊂ 𝐸𝑠 (𝑗 = 1, 2, 3, 4), we have
󵄨󵄨󵄨󵄨󵄨󵄨
󵄨󵄨󵄨
󵄨󵄨󵄨󵄨󵄨󵄨𝐹 (𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 ) − 𝐹 (V1 , V2 , V3 , V4 )󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠󸀠
4
󵄨󵄨󵄨
󵄨󵄨󵄨
= ∑󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝐹𝑖 (𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 ) − 𝐹𝑖 (V1 , V2 , V3 , V4 )󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠󸀠
𝑖=1
≡ 𝐼1 + 𝐼2 + 𝐼3 + 𝐼4 .
(60)
10
Advances in Mathematical Physics
We will estimate 𝐼1 , 𝐼2 , 𝐼3 , and 𝐼4 , respectively, where
󵄨󵄨󵄨
󵄨󵄨󵄨
𝐼1 ≤ 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢1 𝑢3 𝑢2 − V1 V3 V2 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠󸀠
󵄨󵄨󵄨󵄨󵄨󵄨
−1
+ 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨(1 − 𝜕𝑥2 ) (𝑢2 𝜕𝑥 (𝑢2 𝑢3 + 𝑢1 𝑢4 )
󵄨󵄨󵄨
󵄨󵄨󵄨
−V2 𝜕𝑥 (V2 V3 + V1 V4 ))󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠󸀠
−1
󵄨󵄨󵄨󵄨󵄨󵄨
󵄨󵄨󵄨󵄨󵄨󵄨
+ 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨(1 − 𝜕𝑥2 ) (𝑢1 𝑢4 𝑢2 − V1 V4 V2 )󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 󸀠
󵄨󵄨󵄨
󵄨󵄨󵄨𝑠
󵄨󵄨󵄨󵄨󵄨󵄨
󵄨󵄨󵄨󵄨󵄨󵄨
−1
+ 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨(1 − 𝜕𝑥2 ) (2𝑢3 𝜕𝑥 𝑢22 − 2V3 𝜕𝑥 V22 )󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 󸀠
󵄨󵄨󵄨
󵄨󵄨󵄨𝑠
󵄨󵄨󵄨󵄨󵄨󵄨
󵄨󵄨󵄨󵄨󵄨󵄨
−1
+ 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨(1 − 𝜕𝑥2 ) (𝑢1 𝑢4 𝜕𝑥 𝑢2 − V1 V4 𝜕𝑥 V2 )󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 󸀠
󵄨󵄨󵄨
󵄨󵄨󵄨𝑠
−1
󵄨󵄨󵄨󵄨󵄨󵄨
󵄨󵄨󵄨󵄨󵄨󵄨
+ 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨(1 − 𝜕𝑥2 ) (3𝑢2 𝑢3 𝑢1 − 3V2 V3 V1 )󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 󸀠
󵄨󵄨󵄨
󵄨󵄨󵄨𝑠
󵄨󵄨󵄨󵄨󵄨󵄨
󵄨󵄨󵄨󵄨󵄨󵄨
≤ 󵄨󵄨󵄨󵄨󵄨󵄨𝑢1 𝑢3 𝑢2 − V1 V3 V2 󵄨󵄨󵄨󵄨󵄨󵄨𝑠󸀠
󵄨󵄨󵄨
󵄨󵄨󵄨
+ 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢2 𝜕𝑥 (𝑢2 𝑢3 + 𝑢1 𝑢4 ) − V2 𝜕𝑥 (V2 V3 + V1 V4 )󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠󸀠
󵄨󵄨󵄨
󵄨󵄨󵄨
+ 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢1 𝑢4 𝑢2 − V1 V4 V2 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠󸀠
󵄨󵄨󵄨
󵄨󵄨󵄨
+ 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨2𝑢3 𝜕𝑥 𝑢22 − 2V3 𝜕𝑥 V22 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠󸀠
󵄨󵄨󵄨
󵄨󵄨󵄨
+ 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢1 𝑢4 𝜕𝑥 𝑢2 − V1 V4 𝜕𝑥 V2 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠󸀠
󵄨󵄨󵄨
󵄨󵄨󵄨
+ 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨3𝑢2 𝑢3 𝑢1 − 3V2 V3 V1 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠󸀠
󵄨󵄨󵄨
󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨
≤ 𝐶 (󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢1 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢2 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢3 − V3 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠
󵄨󵄨󵄨
󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨
+ 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢1 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨V3 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢2 − V2 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠
󵄨󵄨󵄨
󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨
+ 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨V2 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨V3 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢1 − V1 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠
󵄨󵄨󵄨
󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨
+ 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢1 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢2 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢4 − V4 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠
󵄨󵄨󵄨
󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨
+ 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢1 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨V4 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢2 − V2 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠
󵄨󵄨󵄨
󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨
+󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨V2 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨V4 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢1 − V1 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 )
+
≤
𝐼4 ≤
(62)
where the constant 𝐶 depends only on 𝐻. This implies that
condition (2) also holds. Conditions (1) through (3) are now
easily verified once our system (57) is transformed into a
new system with zero initial data as in (59). The proof of
Theorem 2 is complete.
Acknowledgments
This work was partially supported by NSF of China
(11071266), partially supported by Scholarship Award for
Excellent Doctoral Student granted by the Ministry of Education, and partially supported by the Educational Science
Foundation of Chongqing China (KJ121302).
References
𝐶 󵄨󵄨󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨󵄨󵄨
󵄨󵄨󵄨
(󵄨󵄨󵄨𝑢 󵄨󵄨󵄨 󵄨󵄨󵄨𝑢 󵄨󵄨󵄨 󵄨󵄨󵄨𝑢 − V4 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠
𝑠 − 𝑠󸀠 󵄨󵄨󵄨 1 󵄨󵄨󵄨𝑠 󵄨󵄨󵄨 2 󵄨󵄨󵄨𝑠 󵄨󵄨󵄨 4
󵄨󵄨󵄨
󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨
+ 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢1 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨V4 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢2 − V2 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠
󵄨󵄨󵄨
󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨
+ 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨V2 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨V4 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢1 − V1 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠
󵄨󵄨󵄨 󵄨󵄨󵄨
󵄨󵄨󵄨
󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨
+ 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢3 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢2 + V2 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢2 − V2 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠
󵄨󵄨󵄨
󵄨󵄨󵄨 󵄨󵄨󵄨2 󵄨󵄨󵄨
+󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢2 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑢3 − V3 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 )
𝐶 󵄨󵄨󵄨󵄨󵄨󵄨
󵄨󵄨󵄨
󵄨󵄨󵄨(𝑢 , 𝑢 , 𝑢 , 𝑢 ) − (𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 )󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 .
𝑠 − 𝑠󸀠 󵄨󵄨󵄨 1 2 3 4
(61)
In a similar way to what we just did, we can show that the
following estimates hold:
𝐼2 ≤
𝐶 󵄨󵄨󵄨󵄨󵄨󵄨
󵄨󵄨󵄨
󵄨󵄨󵄨(𝑢 , 𝑢 , 𝑢 , 𝑢 ) − (𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 )󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 ,
𝑠 − 𝑠󸀠 󵄨󵄨󵄨 1 2 3 4
𝐼1 ≤
𝐶 󵄨󵄨󵄨󵄨󵄨󵄨
󵄨󵄨󵄨
󵄨󵄨󵄨(𝑢 , 𝑢 , 𝑢 , 𝑢 ) − (𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 )󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 ,
𝑠 − 𝑠󸀠 󵄨󵄨󵄨 1 2 3 4
𝐶 󵄨󵄨󵄨󵄨󵄨󵄨
󵄨󵄨󵄨
󵄨󵄨󵄨(𝑢 , 𝑢 , 𝑢 , 𝑢 ) − (𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 )󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨𝑠 ,
𝑠 − 𝑠󸀠 󵄨󵄨󵄨 1 2 3 4
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