Volume 2013, Article ID 619068,5pages 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; [email protected] 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 space𝐶([0, 𝑇);
𝐻𝑠(𝑅)) ∩ 𝐶1([0, 𝑇); 𝐻𝑠−1(𝑅))provided that the initial value lies in𝐻𝑠(𝑅)with𝑠 > 3/2. Meanwhile, we prove the existence of global weak solutions in𝐿∞([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) 𝑢3𝑥+ 3ℎ(𝑢) 𝑢𝑥𝑢𝑥𝑥+ ℎ(𝑢) 𝑢𝑥𝑥𝑥, (1) which is subject to the condition
ℎ(𝑢) ≤ 𝑐 |𝑢| , |ℎ (𝑢)| ≤ 𝑐|𝑢|2, (2) or
ℎ(𝑢) ≤ 𝑐, |ℎ (𝑢)| ≤ 𝑐 |𝑢| , (3) where𝑐is a positive constant. The existence and𝐿1stability of entropy weak solutions belonging to the class𝐿1(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) 𝑢3𝑥+ 3ℎ(𝑢) 𝑢𝑥𝑢𝑥𝑥+ ℎ(𝑢) 𝑢𝑥𝑥𝑥, (4) where 𝑚is a positive constant andℎ(𝑢) ∈ 𝐶3. If 𝑚 = 4 and ℎ(𝑢) = 𝑢2/2, (4) reduces to the classical Degasperis- Procesi model (see [2–13]). Here, we notice that assumptions
(2) and (3) do not include the caseℎ(𝑢) = 𝑢3. In this paper, we will study the caseℎ(𝑢) = 𝑢3, and𝑚is an arbitrary positive constant.
In fact, the Cauchy problem of (4) in the caseℎ(𝑢) = 𝑢3 is equivalent to the following system:
𝑢𝑡− 𝑢𝑡𝑥𝑥+ 3𝑚𝑢2𝑢𝑥= 6𝑢3𝑥+ 18𝑢𝑢𝑥𝑢𝑥𝑥+ 3𝑢2𝑢𝑥𝑥𝑥,
𝑢 (0, 𝑥) = 𝑢0(𝑥) . (5)
Using the operator(1 − 𝜕2𝑥)−1to multiply the first equation of the problem (5), we obtain
𝑢𝑡+ 3𝑢2𝑢𝑥+ (𝑚 − 1) (1 − 𝜕𝑥2)−1𝜕𝑥(𝑢3) = 0,
𝑢 (0, 𝑥) = 𝑢0(𝑥) . (6)
It is shown in this work that there exists a unique local strong solution in the Sobolev space𝐶([0, 𝑇); 𝐻𝑠(𝑅)) ∩ 𝐶1([0, 𝑇); 𝐻𝑠−1(𝑅))by assuming that the initial value 𝑢0(𝑥) belongs to𝐻𝑠(𝑅)with𝑠 > 3/2. In addition, we prove the existence of global weak solutions in𝐿∞([0, ∞); 𝐿2(𝑅))for the system (6).
This paper is organized as follows.Section 2investigates the existence and uniqueness of local strong solutions. The result about global weak solution is given inSection 3.
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
𝑑𝑡 + 𝐻 (V)V= 𝑔 (V) , 𝑡 ≥ 0, V(0) =V0. (7) Let𝑋and𝑌be Hilbert spaces such that𝑌is continuously and densely embedded in𝑋, and let𝑄 : 𝑌 → 𝑋be a topological isomorphism. Let𝐿(𝑌, 𝑋)be the space of all bounded linear operators from𝑌to 𝑋. In the case of𝑋 = 𝑌, we denote this space by𝐿(𝑋). We illustrate the following conditions in which𝜎1, 𝜎2, 𝜎3, and𝜎4 are constants depending only on max{‖𝑦‖𝑌, ‖ 𝑧‖𝑌}.
(i)𝐻(𝑦) ∈ 𝐿(𝑌, 𝑋)for𝑦 ∈ 𝑋with
(𝐻(𝑦) − 𝐻(𝑧))𝑤𝑋≤ 𝜎1𝑦 − 𝑧𝑋‖𝑤‖𝑌, 𝑦, 𝑧, 𝑤 ∈ 𝑌, (8) and 𝐻(𝑦) ∈ 𝐺(𝑋, 1, 𝛽) (i.e., 𝐻(𝑦) is quasi-m- accretive), uniformly on bounded sets in𝑌.
(ii)𝑄𝐻(𝑦)𝑄−1 = 𝐻(𝑦) + 𝐴(𝑦), where𝐴(𝑦) ∈ 𝐿(𝑋)is bounded, uniformly on bounded sets in𝑌. Moreover,
(𝐴(𝑦) − 𝐴(𝑧))𝑤𝑋≤ 𝜎2𝑦 − 𝑧𝑌‖𝑤‖𝑋, 𝑦, 𝑧 ∈ 𝑌, 𝑤 ∈ 𝑋.
(9) (iii)𝑔 : 𝑌 → 𝑌extends to a map from𝑋 into𝑋, is
bounded on bounded sets in𝑌, and satisfies
𝑔(𝑦) − 𝑔(𝑧)𝑌≤ 𝜎3𝑦 − 𝑧𝑌, 𝑦, 𝑧 ∈ 𝑌, (10)
𝑔(𝑦) − 𝑔(𝑧)𝑋≤ 𝜎4𝑦 − 𝑧𝑋, 𝑦, 𝑧 ∈ 𝑋. (11) Kato Theorem (see [14]). Assume that conditions (i), (ii), and (iii) hold. IfV0 ∈ 𝑌, there is a maximal𝑇 > 0depending only on‖V0‖𝑌and a unique solutionVto the problem(7)such that
V=V(⋅,V0) ∈ 𝐶 ([0, 𝑇) ; 𝑌) ∩ 𝐶1([0, 𝑇) ; 𝑋) . (12) Moreover, the mapV0 → V(⋅,V0)is a continuous map from𝑌 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 let𝐻(𝑢) = 3𝑢2𝜕𝑥,𝑌 = 𝐻𝑠(𝑅),𝑋 = 𝐻𝑠−1(𝑅),Λ = (1−𝜕𝑥2)1/2,𝑔(𝑢) = (𝑚−1)Λ−2𝜕𝑥(𝑢3), and𝑄 = Λ𝑠. We know that𝑄is an isomorphism of𝐻𝑠onto 𝐻𝑠−1. Now, we cite the following Lemmas.
Lemma 1. The operator𝐴(𝑢) = 𝑢2𝜕𝑥with𝑢 ∈ 𝐻𝑠(𝑅),𝑠 > 3/2 belongs to𝐺(𝐻𝑠−1(𝑅), 1, 𝛽).
Lemma 2. Assume that𝐻(𝑢) = 3𝑢2𝜕𝑥with𝑢 ∈ 𝐻𝑠(𝑅)and 𝑠 > 3/2. Then,𝐻(𝑢) ∈ 𝐿(𝐻𝑠(𝑅), 𝐻𝑠−1(𝑅))for all𝑢 ∈ 𝐻𝑠(𝑅).
Moreover,
‖(𝐻 (𝑢) − 𝐻 (𝑧)) 𝑤‖𝐻𝑠−1 ≤ 𝜎1‖𝑢 − 𝑧‖𝐻𝑠−1‖𝑤‖𝐻𝑠,
𝑢, 𝑧, 𝑤 ∈ 𝐻𝑠(𝑅) . (14)
Lemma 3. For𝑠 > 3/2,𝑢, 𝑧 ∈ 𝐻𝑠(𝑅)and𝑤 ∈ 𝐻𝑠−1, it holds that𝐴(𝑢) = [Λ𝑠, 3𝑢2𝜕𝑥]Λ−𝑠∈ 𝐿(𝐻𝑠−1)for𝑢 ∈ 𝐻𝑠and
‖(𝐴(𝑢) − 𝐴(𝑧))𝑤‖𝐻𝑠−1 ≤ 𝜎2‖𝑢 − 𝑧‖𝐻𝑠‖𝑤‖𝐻𝑠−1. (15) The above three Lemmas can be found in Ni and Zhou [15].
Lemma 4. Let 𝑢, 𝑧 ∈ 𝐻𝑠 with𝑠 > 3/2 and𝑔(𝑢) = (𝑚 − 1)Λ−2𝜕𝑥(𝑢3). Then,𝑔is bounded on bounded sets in𝐻𝑠 and satisfies
𝑔(𝑢) − 𝑔(𝑧)𝐻𝑠≤ 𝜎3‖𝑢 − 𝑧‖𝐻𝑠,
𝑔(𝑢) − 𝑔(𝑧)𝐻𝑠−1 ≤ 𝜎4‖𝑢 − 𝑧‖𝐻𝑠−1. (16) Proof. For 𝑠0 > 1/2, we know that ‖𝑢V‖𝐻𝑠0(𝑅) ≤ 𝑐‖𝑢‖𝐻𝑠0(𝑅)‖V‖𝐻𝑠0(𝑅). Consequently, we have
𝑔(𝑢) − 𝑔(𝑧)𝐻𝑠≤ 𝑐𝑢3− 𝑧3𝐻𝑠−1
≤ 𝑐‖𝑢 − 𝑧‖𝐻𝑠−1(‖𝑢‖2𝐻𝑠−1+ ‖V‖2𝐻𝑠−1)
≤ 𝜎3‖𝑢 − 𝑧‖𝐻𝑠,
𝑔(𝑢) − 𝑔(𝑧)𝐻𝑠−1 ≤ 𝑐𝑢3− 𝑧3𝐻𝑠−2
≤ 𝑐𝑢3− 𝑧3𝐻𝑠−1
≤ 𝑐‖𝑢 − 𝑧‖𝐻𝑠−1(‖𝑢‖2𝐻𝑠−1+ ‖V‖2𝐻𝑠−1)
≤ 𝜎4‖𝑢 − 𝑧‖𝐻𝑠−1.
(17) Using the Kato Theorem, Lemmas1–4, we immediately obtain the local well-posedness theorem.
Theorem 5. Assume that𝑢0 ∈ 𝐻𝑠(𝑅)with𝑠 > 3/2. Then, there exists a𝑇 > 0such that the system(5)or the problem(6) has a unique solution𝑢(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 space𝐿2(𝑅)and 𝐿∞(𝑅).
Lemma 6. The solution of the problem(5)with𝑚 > 0satisfies
∫𝑅𝐾1𝐾 𝑑𝑥 = ∫
𝑅
1 + 𝜉2
𝑚 + 𝜉2̂𝑢(𝜉)2𝑑𝜉 = ∫
𝑅
1 + 𝜉2
𝑚 + 𝜉2̂𝑢0(𝜉)2𝑑𝜉, (19) where𝐾1= 𝑢 − 𝜕𝑥𝑥2 𝑢, and𝐾 = (𝑚 − 𝜕𝑥𝑥2 )−1𝑢. Moreover, there exist two constants𝑐1> 0and𝑐2> 0depending only on𝑚such that
𝑐1𝑢0𝐿2(𝑅) ≤ 𝑐1‖𝑢‖𝐿2(𝑅)≤ 𝑐2𝑢0𝐿2(𝑅). (20)
Proof. Setting𝐾1= 𝑢 − 𝜕2𝑥𝑥𝑢and𝐾 = (𝑚 − 𝜕𝑥𝑥2 )−1𝑢and using the first equation of the problem (5), we obtain𝑢 = 𝑚𝑦 − 𝑦𝑥𝑥 and
𝑑 𝑑𝑡∫
𝑅𝐾1𝐾 𝑑𝑥
= ∫𝑅
𝜕𝐾1
𝜕𝑡 𝐾 𝑑𝑥 + ∫
𝑅𝐾1𝜕𝐾
𝜕𝑡𝑑𝑥 = 2 ∫
𝑅
𝜕𝐾1
𝜕𝑡 𝐾 𝑑𝑥
= 2 ∫
𝑅[−3𝑚𝑢2𝑢𝑥+ 6𝑢3𝑥+ 18𝑢𝑥𝑢𝑥𝑥+ 3𝑢2𝑢𝑥𝑥𝑥] 𝐾 𝑑𝑥
= 2 ∫
𝑅[−𝑚𝜕𝑥(𝑢3) + (𝑢3)𝑥𝑥𝑥] 𝐾 𝑑𝑥
= ∫𝑅(𝑚𝑢3) 𝐾𝑥− 𝑢3𝐾𝑥𝑥𝑥𝑑𝑥
= ∫𝑅[𝑚𝑢3] 𝐾𝑥− 𝑢3(𝑚𝐾𝑥− 𝑢𝑥) 𝑑𝑥
= ∫𝑅𝑢3𝑢𝑥𝑑𝑥,
= 0.
(21) Using the Parseval identity and (21), we obtain (19) and (20).
FromTheorem 5, we know that for any𝑢0 ∈ 𝐻𝑠(𝑅)with 𝑠 > 3/2, there exists a maximal𝑇 = 𝑇(𝑢0) > 0and a unique strong solution𝑢to 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:
𝑝𝑡= 3𝑢2(𝑡, 𝑝) , 𝑡 ∈ [0, 𝑇) ,
𝑝 (0, 𝑥) = 𝑥. (23)
Lemma 7. Let𝑢0 ∈ 𝐻𝑠,𝑠 > 3, and let𝑇 > 0be the maximal existence time of the solution to the problem(6). Then, the problem(23) has a unique solution𝑝 ∈ 𝐶1([0, 𝑇) × 𝑅, 𝑅).
Moreover, the map𝑝(𝑡, ⋅)is an increasing diffeomorphism of 𝑅with𝑝𝑥(𝑡, 𝑥) > 0for(𝑡, 𝑥) ∈ [0, 𝑇) × 𝑅.
Proof. UsingTheorem 5, we obtain𝑢 ∈ 𝐶1([0, 𝑇); 𝐻𝑠−1(𝑅)) and𝐻𝑠−1 ∈ 𝐶1(𝑅). Therefore, we know that functions𝑢(𝑡, 𝑥) and𝑢𝑥(𝑡, 𝑥)are bounded, Lipschitz in space, and𝐶1in time.
Using the existence and uniqueness theorem for ordinary differential equations derives that the problem (23) has a unique solution𝑝 ∈ 𝐶1([0, 𝑇) × 𝑅, 𝑅).
Differentiating (23) with respect to 𝑥 gives rise to the following:
𝑑
𝑑𝑡𝑝𝑥= 6𝑢𝑢𝑥(𝑡, 𝑝) 𝑝𝑥, 𝑡 ∈ [0, 𝑇) , 𝑝𝑥(0, 𝑥) = 1,
(24)
from which we obtain 𝑝𝑥(𝑡, 𝑥) =exp(∫𝑡
06𝑢𝑢𝑥(𝜏, 𝑝 (𝜏, 𝑥)) 𝑑𝜏) . (25) For every𝑇 < 𝑇, using the Sobolev imbedding theorem yields that
sup
(𝜏,𝑥)∈[0,𝑇)×𝑅𝑢𝑥(𝜏, 𝑥) < ∞. (26) It is inferred that there exists a constant𝐾0> 0such that 𝑝𝑥(𝑡, 𝑥) ≥ 𝑒−𝐾0𝑡for(𝑡, 𝑥) ∈ [0, 𝑇) × 𝑅. It completes the proof.
Lemma 8. Assume that𝑢0 ∈ 𝐻𝑠(𝑅), 𝑠 > 3/2. Let𝑇be the maximal existence time of the solution𝑢to the problem(6).
Then, it has
‖𝑢 (𝑡, 𝑥)‖𝐿∞ ≤ 𝑢0𝐿∞𝑒𝑐𝑡 ∀𝑡 ∈ [0, 𝑇] , (27) where𝑐 > 0is a constant independent of𝑡.
Proof. Let𝜉(𝑥) = (1/2)𝑒−|𝑥|, we have(1 − 𝜕𝑥2)−1𝑔 = 𝜉 ⋆ 𝑓for all𝑔 ∈ 𝐿2(𝑅)and𝑢 = 𝜉 ⋆ 𝐾1(𝑡, 𝑥). Using a simple density argument presented in [7], it suffices to consider𝑠 = 3 to prove this lemma. Let𝑇be the maximal existence time of the solution𝑢to the problem (6) with the initial value𝑢0∈ 𝐻3(𝑅) such that𝑢 ∈ 𝐶([0, 𝑇), 𝐻3(𝑅)) ∩ 𝐶1([0, 𝑇), 𝐻2(𝑅)). From (6), we have
𝑢𝑡+ 3𝑢2𝑢𝑥= − (𝑚 − 1) 𝜉 ⋆ (3𝑢2𝑢𝑥) . (28) Since
−𝜉 ⋆ (3𝑢2𝑢𝑥) = −1 2∫∞
−∞𝑒−|𝑥−𝜂|3𝑢2𝑢𝜂𝑑𝜂
= −3 2∫𝑥
−∞𝑒−𝑥+𝜂𝑢2𝑢𝜂𝑑𝜂 −3 2∫+∞
𝑥 𝑒𝑥−𝜂𝑢2𝑢𝜂𝑑𝜂
=1 2∫𝑥
∞𝑒−|𝑥−𝜂|𝑢3𝑑𝜂 − 1 2∫∞
𝑥 𝑒−|𝑥−𝜂|𝑢3𝑑𝜂, 𝑑𝑢 (𝑡, 𝑝 (𝑡, 𝑥))
𝑑𝑡 = 𝑢𝑡(𝑡, 𝑝 (𝑡, 𝑥)) + 𝑢𝑥(𝑡, 𝑝 (𝑡, 𝑥))𝑑𝑝 (𝑡, 𝑥) 𝑑𝑡
= (𝑢𝑡+ 3𝑢2𝑢𝑥) (𝑡, 𝑝 (𝑡, 𝑥)) ,
(29) from (29), we have
𝑑𝑢 (𝑡, 𝑝 (𝑡, 𝑥))
𝑑𝑡 = 𝑚 − 1
2 ∫𝑝(𝑡,𝑥)
−∞ 𝑒−|𝑝(𝑡,𝑥)−𝜂|𝑢3𝑑𝜂
−𝑚 − 1 2 ∫∞
𝑝(𝑡,𝑥)𝑒−|𝑝(𝑡,𝑥)−𝜂|𝑢3𝑑𝜂.
(30)
UsingLemma 6and (30) derives that
𝑑𝑢 (𝑡, 𝑝 (𝑡, 𝑥))
𝑑𝑡 ≤ |𝑚 − 1|
2 ∫∞
−∞𝑒−|𝑝(𝑡,𝑥)−𝜂|𝑢2𝑑𝜂
≤ |𝑚 − 1|
2 ∫∞
−∞𝑢3𝑑𝜂
≤ |𝑚 − 1|
2 ‖𝑢‖2𝐿2‖𝑢‖𝐿∞
≤ 𝑐𝑢0𝐿2(𝑅)‖𝑢‖𝐿∞
≤ 𝑐‖𝑢‖𝐿∞,
(31)
where𝑐 is a positive constant independent of𝑡. Using (31) results in the following:
−𝑐 ∫𝑡
0‖𝑢‖𝐿∞(𝑅)𝑑𝑡 + 𝑢0≤ 𝑢 (𝑡, 𝑝 (𝑡, 𝑥)) ≤ 𝑐 ∫𝑡
0‖𝑢‖𝐿∞(𝑅)𝑑𝑡 + 𝑢0. (32) Therefore,
𝑢(𝑡,𝑝(𝑡,𝑥)) ≤ 𝑢(𝑡,𝑝(𝑡,𝑥))𝐿∞
≤ 𝑢0𝐿∞+ 𝑐 ∫𝑡
0‖𝑢‖𝐿∞(𝑅)𝑑𝑡. (33) Using the Sobolev embedding theorem to ensure the uniform boundedness of𝑢𝑥(𝑠, 𝜂)for(𝑠, 𝜂) ∈ [0, 𝑡] × 𝑅with𝑡 ∈ [0, 𝑇), fromLemma 7, for every𝑡 ∈ [0, 𝑇), we get a constant𝐶(𝑡) such that
𝑒−𝐶(𝑡)≤ 𝑝𝑥(𝑡, 𝑥) ≤ 𝑒𝐶(𝑡), 𝑥 ∈ 𝑅. (34) We deduce from (34) that the function 𝑝(𝑡, ⋅) is strictly increasing on𝑅with lim𝑥 → ±∞𝑝(𝑡, 𝑥) = ±∞as long as𝑡 ∈ [0, 𝑇). It follows from (33) that
‖𝑢 (𝑡, 𝑥)‖𝐿∞ = 𝑢 (𝑡, 𝑝 (𝑡, 𝑥))𝐿∞ ≤ 𝑢0𝐿∞+ 𝑐 ∫𝑡
0‖𝑢‖𝐿∞(𝑅)𝑑𝑡.
(35) Using the Gronwall inequality and (35) derives that (27) holds.
For a real number𝑠with𝑠 > 0, suppose that the function 𝑢0(𝑥)is in𝐻𝑠(𝑅), and let𝑢𝜀0be the convolution𝑢𝜀0 = 𝜙𝜀⋆ 𝑢0 of the function𝜙𝜀(𝑥) = 𝜀−1/4𝜙(𝜀−1/4𝑥)and𝑢0such that the Fourier transform ̂𝜙of𝜙satisfies ̂𝜙 ∈ 𝐶∞0 ,𝜙(𝜉) ≥ 0, and̂ 𝜙(𝜉) = 1̂ for any𝜉 ∈ (−1, 1). Then, we have𝑢𝜀0(𝑥) ∈ 𝐶∞. It follows fromTheorem 5that for each𝜀satisfying0 < 𝜀 < 1/2, the Cauchy problem,
𝑢𝑡− 𝑢𝑡𝑥𝑥+ 3𝑚𝑢2𝑢𝑥= 6𝑢3𝑥+ 18𝑢𝑢𝑥𝑢𝑥𝑥+ 3𝑢2𝑢𝑥𝑥𝑥,
𝑢 (0, 𝑥) = 𝑢𝜀0(𝑥) , (36)
has a unique solution 𝑢𝜀(𝑡, 𝑥) ∈ 𝐶∞([0, 𝑇); 𝐻∞). Using Lemmas6and8, for every𝑡 ∈ [0, 𝑇), we obtain
𝑢𝜀(𝑡, 𝑥)𝐿2(𝑅)≤ 𝑐𝑢𝜀(0, 𝑥)𝐿2(𝑅)≤ 𝑐𝑢0𝐿2(𝑅),
𝑢𝜀(𝑡, 𝑥)𝐿∞ ≤ 𝑢𝜀(0, 𝑥)𝐿∞𝑒𝑐𝑡≤ 𝑐𝑢0𝐿∞𝑒𝑐𝑡. (37)
Sending𝑡 → 𝑇, we know that inequalities (37) are still valid.
This means that for𝑡 ∈ [0, ∞), (37) hold.
Now, we state the concepts of weak solutions.
Definition 9(weak solution). We call a function𝑢 : 𝑅+×𝑅 → 𝑅a weak solution of the Cauchy problem (5) provided that
(i)𝑢 ∈ 𝐿∞(𝑅+; 𝐿2(𝑅));
(ii)𝑢𝑡− 𝑢𝑡𝑥𝑥+ 3𝑚𝑢2𝑢𝑥 = 6𝑢3𝑥+ 18𝑢𝑢𝑥𝑢𝑥𝑥+ 3𝑢2𝑢𝑥𝑥𝑥in 𝐷([0, ∞) × 𝑅), that is, for all𝜑 ∈ 𝐶∞0 ([0, ∞) × 𝑅) there holds the following identity:
∫𝑅+∫
𝑅(𝑢 (𝜑𝑡− 𝜑𝑡𝑥𝑥) + 𝑚𝑢3𝜑𝑥− 𝑢3𝜑𝑥𝑥𝑥) 𝑑𝑥 𝑑𝑡 + ∫𝑅𝑢0(𝑥) 𝜑 (0, 𝑥) 𝑑𝑥 = 0.
(38)
Theorem 10. Let𝑢0(𝑥) ∈ 𝐿2(𝑅). Then, there exists a weak solution𝑢(𝑡, 𝑥) ∈ 𝐿∞([0, ∞); 𝐿2(𝑅))to the problem(5).
Proof. Consider the problem (36). For an arbitrary𝑇 > 0, choosing a subsequence𝜀𝑛 → 0, from (37), we know that 𝑢𝜀𝑛 is bounded in𝐿∞ and‖𝑢𝜀𝑛‖𝐿2(𝑅) is uniformly bounded in𝐿2(𝑅). Therefore, we obtain that𝑢3𝜀𝑛is bounded in𝐿2(𝑅).
Therefore, there exist subsequences {𝑢𝜀𝑛} and {𝑢3𝜀𝑛}, still denoted by {𝑢𝜀𝑛} and {𝑢3𝜀𝑛}, are weakly convergent toV 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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