Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 619068, 5 pages
http://dx.doi.org/10.1155/2013/619068
Research Article
The Local Strong Solutions and Global Weak Solutions for
a Nonlinear Equation
Meng Wu
Department of Mathematics, Southwestern University of Finance and Economics, Chengdu 610074, China
Correspondence should be addressed to Meng Wu; wumeng@swufe.edu.cn
Received 31 March 2013; Accepted 26 April 2013
Academic Editor: Shaoyong Lai
Copyright © 2013 Meng Wu. 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.
The existence and uniqueness of local strong solutions for a nonlinear equation are investigated in the Sobolev space 𝐶([0, 𝑇);
𝐻𝑠 (𝑅)) ∩ 𝐶1 ([0, 𝑇); 𝐻𝑠−1 (𝑅)) provided that the initial value lies in 𝐻𝑠 (𝑅) with 𝑠 > 3/2. Meanwhile, we prove the existence of global
weak solutions in 𝐿∞ ([0, ∞); 𝐿2 (𝑅)) for the equation.
1. Introduction
Coclite and Karlsen [1] investigated the well posedness
in classes of discontinuous functions for the generalized
Degasperis-Procesi equation:
𝑢𝑡 − 𝑢𝑡𝑥𝑥 + 4ℎ󸀠 (𝑢) 𝑢𝑥
󸀠󸀠󸀠
=ℎ
(𝑢) 𝑢𝑥3
󸀠󸀠
󸀠
+ 3ℎ (𝑢) 𝑢𝑥 𝑢𝑥𝑥 + ℎ (𝑢) 𝑢𝑥𝑥𝑥 ,
(1)
|ℎ (𝑢)| ≤ 𝑐|𝑢|2 ,
𝑢𝑡 − 𝑢𝑡𝑥𝑥 + 3𝑚𝑢2 𝑢𝑥 = 6𝑢𝑥3 + 18𝑢𝑢𝑥 𝑢𝑥𝑥 + 3𝑢2 𝑢𝑥𝑥𝑥 ,
𝑢 (0, 𝑥) = 𝑢0 (𝑥) .
which is subject to the condition
󵄨󵄨 󸀠 󵄨󵄨
󵄨󵄨ℎ (𝑢)󵄨󵄨 ≤ 𝑐 |𝑢| ,
󵄨
󵄨
(2) and (3) do not include the case ℎ(𝑢) = 𝑢3 . In this paper, we
will study the case ℎ(𝑢) = 𝑢3 , and 𝑚 is an arbitrary positive
constant.
In fact, the Cauchy problem of (4) in the case ℎ(𝑢) = 𝑢3
is equivalent to the following system:
(5)
(2)
Using the operator (1 − 𝜕𝑥2 )−1 to multiply the first equation of
the problem (5), we obtain
(3)
𝑢𝑡 + 3𝑢2 𝑢𝑥 + (𝑚 − 1) (1 − 𝜕𝑥2 ) 𝜕𝑥 (𝑢3 ) = 0,
or
󵄨󵄨 󸀠 󵄨󵄨
󵄨󵄨ℎ (𝑢)󵄨󵄨 ≤ 𝑐,
󵄨
󵄨
|ℎ (𝑢)| ≤ 𝑐 |𝑢| ,
where 𝑐 is a positive constant. The existence and 𝐿1 stability of
entropy weak solutions belonging to the class 𝐿1 (𝑅) ∩ 𝐵𝑉(𝑅)
are established for (1) in paper [1].
In this work, we study the following model:
𝑢𝑡 − 𝑢𝑡𝑥𝑥 + 𝑚ℎ󸀠 (𝑢) 𝑢𝑥
= ℎ󸀠󸀠󸀠 (𝑢) 𝑢𝑥3 + 3ℎ󸀠󸀠 (𝑢) 𝑢𝑥 𝑢𝑥𝑥 + ℎ󸀠 (𝑢) 𝑢𝑥𝑥𝑥 ,
(4)
where 𝑚 is a positive constant and ℎ(𝑢) ∈ 𝐶3 . If 𝑚 = 4
and ℎ(𝑢) = 𝑢2 /2, (4) reduces to the classical DegasperisProcesi model (see [2–13]). Here, we notice that assumptions
−1
(6)
𝑢 (0, 𝑥) = 𝑢0 (𝑥) .
It is shown in this work that there exists a unique
local strong solution in the Sobolev space 𝐶([0, 𝑇); 𝐻𝑠 (𝑅)) ∩
𝐶1 ([0, 𝑇); 𝐻𝑠−1 (𝑅)) by assuming that the initial value 𝑢0 (𝑥)
belongs to 𝐻𝑠 (𝑅) with 𝑠 > 3/2. In addition, we prove the
existence of global weak solutions in 𝐿∞ ([0, ∞); 𝐿2 (𝑅)) for
the system (6).
This paper is organized as follows. Section 2 investigates
the existence and uniqueness of local strong solutions. The
result about global weak solution is given in Section 3.
2
Abstract and Applied Analysis
2. Local Existence
In this section, we will use the Kato theorem in [14] for
abstract differential equation to establish the existence of
local strong solution for the problem (6). Let us consider the
following problem:
𝑑V
+ 𝐻 (V) V = 𝑔 (V) ,
𝑡 ≥ 0,
V (0) = V0 . (7)
𝑑𝑡
Let 𝑋 and 𝑌 be Hilbert spaces such that 𝑌 is continuously and
densely embedded in 𝑋, and let 𝑄 : 𝑌 → 𝑋 be a topological
isomorphism. Let 𝐿(𝑌, 𝑋) be the space of all bounded linear
operators from 𝑌 to 𝑋. In the case of 𝑋 = 𝑌, we denote
this space by 𝐿(𝑋). We illustrate the following conditions in
which 𝜎1 , 𝜎2 , 𝜎3 , and 𝜎4 are constants depending only on
max{‖𝑦‖𝑌 , ‖𝑧‖𝑌 }.
(i) 𝐻(𝑦) ∈ 𝐿(𝑌, 𝑋) for 𝑦 ∈ 𝑋 with
󵄩
󵄩
󵄩󵄩
󵄩
󵄩󵄩(𝐻 (𝑦) − 𝐻 (𝑧)) 𝑤󵄩󵄩󵄩𝑋 ≤ 𝜎1 󵄩󵄩󵄩𝑦 − 𝑧󵄩󵄩󵄩𝑋 ‖𝑤‖𝑌 ,
𝑦, 𝑧, 𝑤 ∈ 𝑌,
(8)
Lemma 3. For 𝑠 > 3/2, 𝑢, 𝑧 ∈ 𝐻𝑠 (𝑅) and 𝑤 ∈ 𝐻𝑠−1 , it holds
that 𝐴(𝑢) = [Λ𝑠 , 3𝑢2 𝜕𝑥 ]Λ−𝑠 ∈ 𝐿(𝐻𝑠−1 ) for 𝑢 ∈ 𝐻𝑠 and
‖(𝐴(𝑢) − 𝐴(𝑧))𝑤‖𝐻𝑠−1 ≤ 𝜎2 ‖𝑢 − 𝑧‖𝐻𝑠 ‖𝑤‖𝐻𝑠−1 .
The above three Lemmas can be found in Ni and Zhou
[15].
Lemma 4. Let 𝑢, 𝑧 ∈ 𝐻𝑠 with 𝑠 > 3/2 and 𝑔(𝑢) = (𝑚 −
1)Λ−2 𝜕𝑥 (𝑢3 ). Then, 𝑔 is bounded on bounded sets in 𝐻𝑠 and
satisfies
󵄩󵄩󵄩𝑔 (𝑢) − 𝑔 (𝑧)󵄩󵄩󵄩 𝑠 ≤ 𝜎3 ‖𝑢 − 𝑧‖𝐻𝑠 ,
󵄩𝐻
󵄩
(16)
󵄩󵄩
󵄩󵄩
󵄩󵄩𝑔 (𝑢) − 𝑔 (𝑧)󵄩󵄩𝐻𝑠−1 ≤ 𝜎4 ‖𝑢 − 𝑧‖𝐻𝑠−1 .
Proof. For 𝑠0 > 1/2, we know that ‖𝑢V‖𝐻𝑠0 (𝑅)
𝑐‖𝑢‖𝐻𝑠0 (𝑅) ‖V‖𝐻𝑠0 (𝑅) . Consequently, we have
󵄩 3
󵄩
󵄩󵄩
3󵄩
󵄩󵄩𝑔(𝑢) − 𝑔(𝑧)󵄩󵄩󵄩𝐻𝑠 ≤ 𝑐󵄩󵄩󵄩󵄩𝑢 − 𝑧 󵄩󵄩󵄩󵄩𝐻𝑠−1
(ii) 𝑄𝐻(𝑦)𝑄 = 𝐻(𝑦) + 𝐴(𝑦), where 𝐴(𝑦) ∈ 𝐿(𝑋) is
bounded, uniformly on bounded sets in 𝑌. Moreover,
󵄩
󵄩
󵄩󵄩
󵄩
󵄩󵄩(𝐴 (𝑦) − 𝐴 (𝑧)) 𝑤󵄩󵄩󵄩𝑋 ≤ 𝜎2 󵄩󵄩󵄩𝑦 − 𝑧󵄩󵄩󵄩𝑌 ‖𝑤‖𝑋 , 𝑦, 𝑧 ∈ 𝑌, 𝑤 ∈ 𝑋.
(9)
(iii) 𝑔 : 𝑌 → 𝑌 extends to a map from 𝑋 into 𝑋, is
bounded on bounded sets in 𝑌, and satisfies
󵄩
󵄩
󵄩󵄩
󵄩
(10)
󵄩󵄩𝑔(𝑦) − 𝑔(𝑧)󵄩󵄩󵄩𝑌 ≤ 𝜎3 󵄩󵄩󵄩𝑦 − 𝑧󵄩󵄩󵄩𝑌 , 𝑦, 𝑧 ∈ 𝑌,
󵄩
󵄩
󵄩󵄩
󵄩
(11)
󵄩󵄩𝑔(𝑦) − 𝑔(𝑧)󵄩󵄩󵄩𝑋 ≤ 𝜎4 󵄩󵄩󵄩𝑦 − 𝑧󵄩󵄩󵄩𝑋 , 𝑦, 𝑧 ∈ 𝑋.
Kato Theorem (see [14]). Assume that conditions (i), (ii), and
(iii) hold. If V0 ∈ 𝑌, there is a maximal 𝑇 > 0 depending only
on ‖V0 ‖𝑌 and a unique solution V to the problem (7) such that
V = V (⋅, V0 ) ∈ 𝐶 ([0, 𝑇) ; 𝑌) ∩ 𝐶1 ([0, 𝑇) ; 𝑋) .
(12)
Moreover, the map V0 → V(⋅, V0 ) is a continuous map from 𝑌
to the following space:
𝐶 ([0, 𝑇) ; 𝑌) ∩ 𝐶1 ([0, 𝑇) ; 𝑋) .
(13)
In order to apply the Kato theorem to establish the local
well posedness for the problem (6), we let 𝐻(𝑢) = 3𝑢2 𝜕𝑥 , 𝑌 =
𝐻𝑠 (𝑅), 𝑋 = 𝐻𝑠−1 (𝑅), Λ = (1−𝜕𝑥2 )1/2 , 𝑔(𝑢) = (𝑚−1)Λ−2 𝜕𝑥 (𝑢3 ),
and 𝑄 = Λ𝑠 . We know that 𝑄 is an isomorphism of 𝐻𝑠 onto
𝐻𝑠−1 . Now, we cite the following Lemmas.
2
𝑠
Lemma 1. The operator 𝐴(𝑢) = 𝑢 𝜕𝑥 with 𝑢 ∈ 𝐻 (𝑅), 𝑠 > 3/2
belongs to 𝐺(𝐻𝑠−1 (𝑅), 1, 𝛽).
Lemma 2. Assume that 𝐻(𝑢) = 3𝑢2 𝜕𝑥 with 𝑢 ∈ 𝐻𝑠 (𝑅) and
𝑠 > 3/2. Then, 𝐻(𝑢) ∈ 𝐿(𝐻𝑠 (𝑅), 𝐻𝑠−1 (𝑅)) for all 𝑢 ∈ 𝐻𝑠 (𝑅).
Moreover,
‖(𝐻 (𝑢) − 𝐻 (𝑧)) 𝑤‖𝐻𝑠−1 ≤ 𝜎1 ‖𝑢 − 𝑧‖𝐻𝑠−1 ‖𝑤‖𝐻𝑠 ,
𝑢, 𝑧, 𝑤 ∈ 𝐻𝑠 (𝑅) .
(14)
≤
≤ 𝑐‖𝑢 − 𝑧‖𝐻𝑠−1 (‖𝑢‖2𝐻𝑠−1 + ‖V‖2𝐻𝑠−1 )
and 𝐻(𝑦) ∈ 𝐺(𝑋, 1, 𝛽) (i.e., 𝐻(𝑦) is quasi-maccretive), uniformly on bounded sets in 𝑌.
−1
(15)
≤ 𝜎3 ‖𝑢 − 𝑧‖𝐻𝑠 ,
󵄩 3
󵄩
󵄩󵄩
3󵄩
󵄩󵄩𝑔 (𝑢) − 𝑔 (𝑧)󵄩󵄩󵄩𝐻𝑠−1 ≤ 𝑐󵄩󵄩󵄩󵄩𝑢 − 𝑧 󵄩󵄩󵄩󵄩𝐻𝑠−2
󵄩
󵄩
≤ 𝑐󵄩󵄩󵄩󵄩𝑢3 − 𝑧3 󵄩󵄩󵄩󵄩𝐻𝑠−1
≤ 𝑐‖𝑢 − 𝑧‖𝐻𝑠−1 (‖𝑢‖2𝐻𝑠−1 + ‖V‖2𝐻𝑠−1 )
≤ 𝜎4 ‖𝑢 − 𝑧‖𝐻𝑠−1 .
(17)
Using the Kato Theorem, Lemmas 1–4, we immediately
obtain the local well-posedness theorem.
Theorem 5. Assume that 𝑢0 ∈ 𝐻𝑠 (𝑅) with 𝑠 > 3/2. Then,
there exists a 𝑇 > 0 such that the system (5) or the problem (6)
has a unique solution 𝑢(𝑡, 𝑥) satisfying
𝑢 (𝑡, 𝑥) ∈ 𝐶 ([0, 𝑇) ; 𝐻𝑠 (𝑅)) ∩ 𝐶1 ([0, 𝑇) ; 𝐻𝑠−1 (𝑅)) . (18)
3. Weak Solutions
In this section, our aim is to establish the existence of global
weak solutions for the system (6). Firstly, we prove that the
solution of the problem (5) is bounded in the space 𝐿2 (𝑅) and
𝐿∞ (𝑅).
Lemma 6. The solution of the problem (5) with 𝑚 > 0 satisfies
∫ 𝐾1 𝐾 𝑑𝑥 = ∫
𝑅
𝑅
1 + 𝜉2 󵄨󵄨
1 + 𝜉2 󵄨󵄨
󵄨2
󵄨󵄨2
̂
𝑢
(𝜉)
𝑢̂ (𝜉)󵄨󵄨󵄨 𝑑𝜉,
𝑑𝜉
=
∫
󵄨
󵄨
󵄨
󵄨
2 󵄨󵄨 0
𝑚 + 𝜉2
𝑚
+
𝜉
𝑅
(19)
2
2 −1
𝑢, and 𝐾 = (𝑚 − 𝜕𝑥𝑥
) 𝑢. Moreover, there
where 𝐾1 = 𝑢 − 𝜕𝑥𝑥
exist two constants 𝑐1 > 0 and 𝑐2 > 0 depending only on 𝑚 such
that
󵄩 󵄩
󵄩 󵄩
𝑐1 󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐿2 (𝑅) ≤ 𝑐1 ‖𝑢‖𝐿2 (𝑅) ≤ 𝑐2 󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐿2 (𝑅) .
(20)
Abstract and Applied Analysis
3
2
2 −1
Proof. Setting 𝐾1 = 𝑢 − 𝜕𝑥𝑥
𝑢 and 𝐾 = (𝑚 − 𝜕𝑥𝑥
) 𝑢 and using
the first equation of the problem (5), we obtain 𝑢 = 𝑚𝑦 − 𝑦𝑥𝑥
and
𝑅
𝑡
𝑝𝑥 (𝑡, 𝑥) = exp (∫ 6𝑢𝑢𝑥 (𝜏, 𝑝 (𝜏, 𝑥)) 𝑑𝜏) .
0
𝑑
∫ 𝐾 𝐾 𝑑𝑥
𝑑𝑡 𝑅 1
=∫
from which we obtain
For every 𝑇󸀠 < 𝑇, using the Sobolev imbedding theorem
yields that
𝜕𝐾1
𝜕𝐾1
𝜕𝐾
𝐾 𝑑𝑥 + ∫ 𝐾1
𝑑𝑥 = 2 ∫
𝐾 𝑑𝑥
𝜕𝑡
𝜕𝑡
𝑅
𝑅 𝜕𝑡
2
= 2 ∫ [−3𝑚𝑢 𝑢𝑥 +
𝑅
6𝑢𝑥3
(25)
󵄨
󵄨󵄨
sup
󵄨󵄨𝑢𝑥 (𝜏, 𝑥)󵄨󵄨󵄨 < ∞.
(𝜏,𝑥)∈[0,𝑇󸀠 )×𝑅
2
+ 18𝑢𝑥 𝑢𝑥𝑥 + 3𝑢 𝑢𝑥𝑥𝑥 ] 𝐾 𝑑𝑥
(26)
It is inferred that there exists a constant 𝐾0 > 0 such that
𝑝𝑥 (𝑡, 𝑥) ≥ 𝑒−𝐾0 𝑡 for (𝑡, 𝑥) ∈ [0, 𝑇) × 𝑅. It completes the proof.
= 2 ∫ [−𝑚𝜕𝑥 (𝑢3 ) + (𝑢3 )𝑥𝑥𝑥 ] 𝐾 𝑑𝑥
𝑅
= ∫ (𝑚𝑢3 ) 𝐾𝑥 − 𝑢3 𝐾𝑥𝑥𝑥 𝑑𝑥
Lemma 8. Assume that 𝑢0 ∈ 𝐻𝑠 (𝑅), 𝑠 > 3/2. Let 𝑇 be the
maximal existence time of the solution 𝑢 to the problem (6).
Then, it has
𝑅
= ∫ [𝑚𝑢3 ] 𝐾𝑥 − 𝑢3 (𝑚𝐾𝑥 − 𝑢𝑥 ) 𝑑𝑥
𝑅
󵄩 󵄩
‖𝑢 (𝑡, 𝑥)‖𝐿∞ ≤ 󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐿∞ 𝑒𝑐𝑡
= ∫ 𝑢3 𝑢𝑥 𝑑𝑥,
𝑅
= 0.
∀𝑡 ∈ [0, 𝑇] ,
(27)
where 𝑐 > 0 is a constant independent of 𝑡.
(21)
Using the Parseval identity and (21), we obtain (19) and (20).
From Theorem 5, we know that for any 𝑢0 ∈ 𝐻𝑠 (𝑅) with
𝑠 > 3/2, there exists a maximal 𝑇 = 𝑇(𝑢0 ) > 0 and a unique
strong solution 𝑢 to the problem (6) such that
𝑢 ∈ 𝐶 ([0, 𝑇) ; 𝐻𝑠 (𝑅)) ∩ 𝐶1 ([0, 𝑇) ; 𝐻𝑠−1 (𝑅)) .
(22)
Firstly, we study the following differential equation:
2
𝑝𝑡 = 3𝑢 (𝑡, 𝑝) ,
𝑡 ∈ [0, 𝑇) ,
𝑝 (0, 𝑥) = 𝑥.
(23)
Proof. Using Theorem 5, 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.
Using the existence and uniqueness theorem for ordinary
differential equations derives that the problem (23) has a
unique solution 𝑝 ∈ 𝐶1 ([0, 𝑇) × 𝑅, 𝑅).
Differentiating (23) with respect to 𝑥 gives rise to the
following:
𝑝𝑥 (0, 𝑥) = 1,
𝑡 ∈ [0, 𝑇) ,
𝑢𝑡 + 3𝑢2 𝑢𝑥 = − (𝑚 − 1) 𝜉 ⋆ (3𝑢2 𝑢𝑥 ) .
(28)
Since
Lemma 7. Let 𝑢0 ∈ 𝐻𝑠 , 𝑠 > 3, and let 𝑇 > 0 be the maximal
existence time of the solution to the problem (6). Then, the
problem (23) has a unique solution 𝑝 ∈ 𝐶1 ([0, 𝑇) × 𝑅, 𝑅).
Moreover, the map 𝑝(𝑡, ⋅) is an increasing diffeomorphism of
𝑅 with 𝑝𝑥 (𝑡, 𝑥) > 0 for (𝑡, 𝑥) ∈ [0, 𝑇) × 𝑅.
𝑑
𝑝 = 6𝑢𝑢𝑥 (𝑡, 𝑝) 𝑝𝑥 ,
𝑑𝑡 𝑥
Proof. Let 𝜉(𝑥) = (1/2)𝑒−|𝑥| , we have (1 − 𝜕𝑥2 )−1 𝑔 = 𝜉 ⋆ 𝑓 for
all 𝑔 ∈ 𝐿2 (𝑅) and 𝑢 = 𝜉 ⋆ 𝐾1 (𝑡, 𝑥). Using a simple density
argument presented in [7], it suffices to consider 𝑠 = 3 to
prove this lemma. Let 𝑇 be the maximal existence time of the
solution 𝑢 to the problem (6) with the initial value 𝑢0 ∈ 𝐻3 (𝑅)
such that 𝑢 ∈ 𝐶([0, 𝑇), 𝐻3 (𝑅)) ∩ 𝐶1 ([0, 𝑇), 𝐻2 (𝑅)). From (6),
we have
(24)
1 ∞
−𝜉 ⋆ (3𝑢2 𝑢𝑥 ) = − ∫ 𝑒−|𝑥−𝜂| 3𝑢2 𝑢𝜂 𝑑𝜂
2 −∞
3 +∞
3 𝑥
= − ∫ 𝑒−𝑥+𝜂 𝑢2 𝑢𝜂 𝑑𝜂 − ∫ 𝑒𝑥−𝜂 𝑢2 𝑢𝜂 𝑑𝜂
2 −∞
2 𝑥
=
1 𝑥 −|𝑥−𝜂| 3
1 ∞
𝑢 𝑑𝜂 − ∫ 𝑒−|𝑥−𝜂| 𝑢3 𝑑𝜂,
∫ 𝑒
2 ∞
2 𝑥
𝑑𝑢 (𝑡, 𝑝 (𝑡, 𝑥))
𝑑𝑝 (𝑡, 𝑥)
= 𝑢𝑡 (𝑡, 𝑝 (𝑡, 𝑥)) + 𝑢𝑥 (𝑡, 𝑝 (𝑡, 𝑥))
𝑑𝑡
𝑑𝑡
= (𝑢𝑡 + 3𝑢2 𝑢𝑥 ) (𝑡, 𝑝 (𝑡, 𝑥)) ,
(29)
from (29), we have
𝑑𝑢 (𝑡, 𝑝 (𝑡, 𝑥)) 𝑚 − 1 𝑝(𝑡,𝑥) −|𝑝(𝑡,𝑥)−𝜂| 3
𝑒
𝑢 𝑑𝜂
=
∫
𝑑𝑡
2
−∞
𝑚 − 1 ∞ −|𝑝(𝑡,𝑥)−𝜂| 3
−
𝑒
𝑢 𝑑𝜂.
∫
2
𝑝(𝑡,𝑥)
(30)
4
Abstract and Applied Analysis
Sending 𝑡 → 𝑇, we know that inequalities (37) are still valid.
This means that for 𝑡 ∈ [0, ∞), (37) hold.
Now, we state the concepts of weak solutions.
Using Lemma 6 and (30) derives that
󵄨󵄨 𝑑𝑢 (𝑡, 𝑝 (𝑡, 𝑥)) 󵄨󵄨
󵄨󵄨
󵄨󵄨 |𝑚 − 1| ∞ −|𝑝(𝑡,𝑥)−𝜂| 2
󵄨󵄨
󵄨󵄨 ≤
𝑢 𝑑𝜂
∫ 𝑒
󵄨󵄨
󵄨󵄨
𝑑𝑡
2
−∞
󵄨
󵄨
≤
Definition 9 (weak solution). We call a function 𝑢 : 𝑅+ ×𝑅 →
𝑅 a weak solution of the Cauchy problem (5) provided that
|𝑚 − 1| ∞ 3
∫ 𝑢 𝑑𝜂
2
−∞
(31)
|𝑚 − 1| 2
≤
‖𝑢‖𝐿2 ‖𝑢‖𝐿∞
2
󵄩 󵄩
≤ 𝑐󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐿2 (𝑅) ‖𝑢‖𝐿∞
where 𝑐 is a positive constant independent of 𝑡. Using (31)
results in the following:
𝑡
𝑡
0
0
𝑅+
𝑅
(38)
+ ∫ 𝑢0 (𝑥) 𝜑 (0, 𝑥) 𝑑𝑥 = 0.
𝑅
−𝑐 ∫ ‖𝑢‖𝐿∞ (𝑅) 𝑑𝑡 + 𝑢0 ≤ 𝑢 (𝑡, 𝑝 (𝑡, 𝑥)) ≤ 𝑐 ∫ ‖𝑢‖𝐿∞ (𝑅) 𝑑𝑡 + 𝑢0 .
(32)
Theorem 10. Let 𝑢0 (𝑥) ∈ 𝐿2 (𝑅). Then, there exists a weak
solution 𝑢(𝑡, 𝑥) ∈ 𝐿∞ ([0, ∞); 𝐿2 (𝑅)) to the problem (5).
(33)
Proof. Consider the problem (36). For an arbitrary 𝑇 > 0,
choosing a subsequence 𝜀𝑛 → 0, from (37), we know that
𝑢𝜀𝑛 is bounded in 𝐿∞ and ‖𝑢𝜀𝑛 ‖𝐿2 (𝑅) is uniformly bounded
Therefore,
󵄩
󵄨 󵄩
󵄨󵄨
󵄨󵄨𝑢 (𝑡, 𝑝 (𝑡, 𝑥))󵄨󵄨󵄨 ≤ 󵄩󵄩󵄩𝑢 (𝑡, 𝑝 (𝑡, 𝑥))󵄩󵄩󵄩𝐿∞
𝑡
󵄩 󵄩
≤ 󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐿∞ + 𝑐 ∫ ‖𝑢‖𝐿∞ (𝑅) 𝑑𝑡.
0
Using the Sobolev embedding theorem to ensure the uniform
boundedness of 𝑢𝑥 (𝑠, 𝜂) for (𝑠, 𝜂) ∈ [0, 𝑡] × 𝑅 with 𝑡 ∈ [0, 𝑇󸀠 ),
from Lemma 7, for every 𝑡 ∈ [0, 𝑇󸀠 ), we get a constant 𝐶(𝑡)
such that
𝑥 ∈ 𝑅.
(34)
We deduce from (34) that the function 𝑝(𝑡, ⋅) is strictly
increasing on 𝑅 with lim𝑥 → ±∞ 𝑝(𝑡, 𝑥) = ±∞ as long as 𝑡 ∈
[0, 𝑇󸀠 ). It follows from (33) that
𝑡
󵄩
󵄩
󵄩 󵄩
‖𝑢 (𝑡, 𝑥)‖𝐿∞ = 󵄩󵄩󵄩𝑢 (𝑡, 𝑝 (𝑡, 𝑥))󵄩󵄩󵄩𝐿∞ ≤ 󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐿∞ + 𝑐 ∫ ‖𝑢‖𝐿∞ (𝑅) 𝑑𝑡.
0
(35)
Using the Gronwall inequality and (35) derives that (27)
holds.
For a real number 𝑠 with 𝑠 > 0, suppose that the function
𝑢0 (𝑥) is in 𝐻𝑠 (𝑅), and let 𝑢𝜀0 be the convolution 𝑢𝜀0 = 𝜙𝜀 ⋆ 𝑢0
of the function 𝜙𝜀 (𝑥) = 𝜀−1/4 𝜙(𝜀−1/4 𝑥) and 𝑢0 such that the
̂ ≥ 0, and
Fourier transform 𝜙̂ of 𝜙 satisfies 𝜙̂ ∈ 𝐶0∞ , 𝜙(𝜉)
̂ = 1 for any 𝜉 ∈ (−1, 1). Then, we have 𝑢 (𝑥) ∈ 𝐶∞ . It
𝜙(𝜉)
𝜀0
follows from Theorem 5 that for each 𝜀 satisfying 0 < 𝜀 < 1/2,
the Cauchy problem,
𝑢𝑡 − 𝑢𝑡𝑥𝑥 + 3𝑚𝑢2 𝑢𝑥 = 6𝑢𝑥3 + 18𝑢𝑢𝑥 𝑢𝑥𝑥 + 3𝑢2 𝑢𝑥𝑥𝑥 ,
𝑢 (0, 𝑥) = 𝑢𝜀0 (𝑥) ,
(ii) 𝑢𝑡 − 𝑢𝑡𝑥𝑥 + 3𝑚𝑢2 𝑢𝑥 = 6𝑢𝑥3 + 18𝑢𝑢𝑥 𝑢𝑥𝑥 + 3𝑢2 𝑢𝑥𝑥𝑥 in
𝐷󸀠 ([0, ∞) × 𝑅), that is, for all 𝜑 ∈ 𝐶0∞ ([0, ∞) × 𝑅)
there holds the following identity:
∫ ∫ (𝑢 (𝜑𝑡 − 𝜑𝑡𝑥𝑥 ) + 𝑚𝑢3 𝜑𝑥 − 𝑢3 𝜑𝑥𝑥𝑥 ) 𝑑𝑥 𝑑𝑡
≤ 𝑐‖𝑢‖𝐿∞ ,
𝑒−𝐶(𝑡) ≤ 𝑝𝑥 (𝑡, 𝑥) ≤ 𝑒𝐶(𝑡) ,
(i) 𝑢 ∈ 𝐿∞ (𝑅+ ; 𝐿2 (𝑅));
(36)
has a unique solution 𝑢𝜀 (𝑡, 𝑥) ∈ 𝐶∞ ([0, 𝑇); 𝐻∞ ). Using
Lemmas 6 and 8, for every 𝑡 ∈ [0, 𝑇), we obtain
󵄩
󵄩
󵄩󵄩
󵄩
󵄩 󵄩
󵄩󵄩𝑢𝜀 (𝑡, 𝑥)󵄩󵄩󵄩𝐿2 (𝑅) ≤ 𝑐󵄩󵄩󵄩𝑢𝜀 (0, 𝑥)󵄩󵄩󵄩𝐿2 (𝑅) ≤ 𝑐󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐿2 (𝑅) ,
(37)
󵄩󵄩󵄩𝑢𝜀 (𝑡, 𝑥)󵄩󵄩󵄩 ∞ ≤ 󵄩󵄩󵄩𝑢𝜀 (0, 𝑥)󵄩󵄩󵄩 ∞ 𝑒𝑐𝑡 ≤ 𝑐󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩 ∞ 𝑒𝑐𝑡 .
󵄩𝐿
󵄩𝐿
󵄩
󵄩 󵄩𝐿
󵄩
in 𝐿2 (𝑅). Therefore, we obtain that 𝑢𝜀3𝑛 is bounded in 𝐿2 (𝑅).
Therefore, there exist subsequences {𝑢𝜀𝑛 } and {𝑢𝜀3𝑛 }, still
denoted by {𝑢𝜀𝑛 } and {𝑢𝜀3𝑛 }, are weakly convergent to V in
𝐿2 (𝑅). Noticing (38) completes the proof.
Acknowledgment
This work is supported by the Fundamental Research Funds
for the Central Universities (JBK120504).
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