The Uniqueness of Strong Solutions for the Camassa-Holm Equation

Hindawi Publishing Corporation Journal of Function Spaces and Applications Volume 2013, Article ID 409760, 7 pages http://dx.doi.org/10.1155/2013/409760

Research Article The Uniqueness of Strong Solutions for the Camassa-Holm Equation Meng Wu1 and Chong Lai2 1 2

The School of Mathematics, Southwestern University of Finance and Economics, Chengdu 610074, China The School of Finance, Southwestern University of Finance and Economics, Chengdu 610074, China

Correspondence should be addressed to Meng Wu; [email protected] Received 28 February 2013; Accepted 10 May 2013 Academic Editor: Janusz Matkowski Copyright © 2013 M. Wu and C. Lai. 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. Assume that there exists a strong solution of the Camassa-Holm equation and the initial value of the solution belongs to the Sobolev space 𝐻1 (𝑅). We provide a new proof of the uniqueness of the strong solution for the equation.

1. Introduction

We consider the equivalent form of the Cauchy problem for (1)

The integrable Camassa-Holm model [1]

𝑢𝑡 − 𝑢𝑡𝑥𝑥 + 3𝑢𝑢𝑥 = 2𝑢𝑥 𝑢𝑥𝑥 + 𝑢𝑢𝑥𝑥𝑥

1 𝑢𝑡 + (𝑢2 )𝑥 + 𝜕𝑥 𝑃𝑢 (𝑡, 𝑥) = 0, 2

(2)

𝑢 (0, 𝑥) = 𝑢0 ,

(1)

∞

has been investigated by many scholars. Equation (1) has peaked solitary wave solutions, which takes the form 𝑐𝑒−|𝑥−𝑐𝑡| , 𝑐 ∈ 𝑅. The existence and uniqueness of the global weak solutions for (1) have been given by Constantin and Escher [2] and Constantin and Molinet [3] in which the 𝑚 = 𝑢0 − 𝑢0𝑥𝑥 is a positive (or negative) Radon measure. The local wellposedness of strong solutions for the Camassa-Holm model and its various generalized forms are provided in [4–8]. For the initial value 𝑢0 satisfying 𝑢0 − 𝑢0𝑥𝑥 ≥ 0 or 𝑢0 − 𝑢0𝑥𝑥 ≤ 0, it is shown in [9] that the Camassa-Holm equation has unique global strong solutions in the Sobolev space 𝐻𝑠 (𝑅) with 𝑠 > 3/2. If the initial data satisfy certain conditions, we know that the local strong solutions blow up in finite time [10, 11]. It means that the slope of the solution becomes unbounded while the solution itself remains bounded. For other techniques to obtain the dynamic properties for the Camassa-Holm equation and other related shallow water equations, the reader is referred to [12–16] and the references therein.

where 𝑃𝑢 (𝑡, 𝑥) = Λ−2 (𝑢2 + (1/2)𝑢𝑥2 ) = (1/2) ∫−∞ 𝑒|𝑥−𝑦| (𝑢2 (𝑡, 𝑦)+ (1/2)𝑢𝑦2 (𝑡, 𝑦))𝑑𝑦. If (1) has a suitable smooth strong solution, we have the conservation law ∫

∞

−∞

(𝑢2 (𝑡, 𝑥) + 𝑢𝑥2 (𝑡, 𝑥)) 𝑑𝑥 ∞

=∫

−∞

(3) 2

(𝑢 (0, 𝑥) +

𝑢𝑥2

(0, 𝑥)) 𝑑𝑥,

which derives 󵄩 󵄩 ‖𝑢‖𝐿∞ (𝑅) ≤ ‖𝑢‖𝐻1 (𝑅) = 󵄩󵄩󵄩𝑢0 󵄩󵄩󵄩𝐻1 (𝑅) .

(4)

The objective of this work is to give a new proof of the uniqueness for the solutions of the Camassa-Holm equation (1). Firstly, we establish the following inequality: ‖𝑢 (𝑡, ⋅) − V (𝑡, ⋅)‖𝐿1 (𝑅) ≤ 𝑐𝑒𝑐𝑡 (∫

∞

−∞

∞

󵄨󵄨 2 2 󵄨󵄨 󵄨󵄨𝑢0𝑥 − V0𝑥 󵄨󵄨 𝑑𝑥) , 󵄨 󵄨 −∞ (5)

󵄨 󵄨󵄨 󵄨󵄨𝑢0 (𝑥) − V0 (𝑥)󵄨󵄨󵄨 𝑑𝑥 + ∫

2

Journal of Function Spaces and Applications

where 𝑡 ∈ [0, 𝑇0 ), functions 𝑢 and V are two local or global strong solutions of problem (2) with initial data 𝑢(0, ⋅) = 𝑢0 ∈ 𝐻1 (𝑅) and V(0, ⋅) = V0 ∈ 𝐻1 (𝑅), respectively. Constant 𝑐 depends on ‖𝑢0 ‖𝐻1 (𝑅) , ‖V0 ‖𝐻1 (𝑅) , and the maximum existence time 𝑇0 . Secondly, from (5), we immediately arrive at the goal of the uniqueness. Here we state that the approach to establish (5) is the device of doubling variables which was presented in Kruzkov’s paper [17]. This paper is organized as follows. Several lemmas are given in Section 2, while the proofs of the main results are established in Section 3.

Lemma 2 (see [17]). If the function |𝜕𝐹(𝑢)/𝜕𝑢| is bounded, then the function 𝐻(𝑢, V) = sign(𝑢 − V)(𝐹(𝑢) − 𝐹(V)) satisfies the Lipschitz condition in 𝑢 and V, respectively. Lemma 3. Let 𝑢0 ∈ 𝐻1 (𝑅). It holds that the function 𝑄𝑢 (𝑡, 𝑥) = 𝜕𝑥 𝑃𝑢 (𝑡, 𝑥) = 𝜕𝑥 Λ−2 (𝑢2 + (1/2)𝑢𝑥2 ) satisfies 󵄩 󵄩󵄩 󵄩󵄩𝑃𝑢 (𝑡, 𝑥)󵄩󵄩󵄩𝐿∞ (𝑅) < ∞, 󵄩󵄩 󵄩 󵄩󵄩𝑃𝑢 (𝑡, 𝑥)󵄩󵄩󵄩𝐿1 (𝑅) < ∞, 󵄩󵄩󵄩𝑃𝑢 (𝑡, 𝑥)󵄩󵄩󵄩 2 < ∞, 󵄩 󵄩𝐿 (𝑅) 󵄩󵄩 󵄩󵄩 󵄩󵄩𝑄𝑢 (𝑡, 𝑥)󵄩󵄩𝐿∞ (𝑅) < ∞, 󵄩󵄩 󵄩 󵄩󵄩𝑄𝑢 (𝑡, 𝑥)󵄩󵄩󵄩𝐿1 (𝑅) < ∞,

2. Notations and Several Lemmas Set 𝜉𝑇 = [0, 𝑇] × 𝑅 for an arbitrary 𝑇 > 0. The space of all infinitely differentiable functions 𝑓(𝑡, 𝑥) with compact support in [0, 𝑇] × 𝑅 is denoted by 𝐶0∞ (𝜉𝑇 ). We define 𝜌(𝜎) as a function which is infinitely differentiable on (−∞, +∞) such ∞ that 𝜌(𝜎) ≥ 0, 𝜌(𝜎) = 0 for |𝜎| ≥ 1 and ∫−∞ 𝜌(𝜎)𝑑𝜎 = 1. For any number ℎ > 0, we let 𝜌ℎ (𝜎) = 𝜌(ℎ−1 𝜎)/ℎ. Then we have that 𝜌ℎ (𝜎) is a function in 𝐶∞ (−∞, ∞) and 𝜌ℎ (𝜎) ≥ 0,

𝜌ℎ (𝜎) = 0 if |𝜎| ≥ ℎ,

󵄨 𝑐 󵄨󵄨 󵄨󵄨𝜌ℎ (𝜎)󵄨󵄨󵄨 ≤ , ℎ

∫

∞

−∞

𝜌ℎ (𝜎) = 1.

(6)

󵄩 󵄩󵄩 󵄩󵄩𝑄𝑢 (𝑡, 𝑥)󵄩󵄩󵄩𝐿2 (𝑅) < ∞. The proof of Lemma 3 can be found in [13, 15] (see [13, Lemma 5.1]). Lemma 4. Let 𝑢 be the strong solution of problem (2), 𝑓(𝑡, 𝑥) ∈ 𝐶0∞ (𝜉𝑇 ), and 𝑓(0, 𝑥) = 0. Then ∬ {|𝑢 − 𝑘| 𝑓𝑡 + sign (𝑢 − 𝑘)

Assume that the function V(𝑥) is locally integrable in (−∞, ∞). We define the approximation of function V(𝑥) as Vℎ (𝑥) =

𝑥−𝑦 1 ∞ ) V (𝑦) 𝑑𝑦, ∫ 𝜌( ℎ −∞ ℎ

ℎ > 0.

lim

1

󵄨󵄨 󵄨 󵄨󵄨V (𝑥) − V (𝑥0 )󵄨󵄨󵄨 𝑑𝑥 = 0. |𝑥−𝑥 |≤ℎ

∫

(8)

0

Lemma 1 (see [17]). Let the function V(𝑡, 𝑥) be bounded and measurable in cylinder Ω = [0, 𝑇] × 𝐻𝑟 . If 𝛿 ∈ (0, min[𝑟, 𝑇]) and any number ℎ ∈ (0, 𝛿), then the function 1 󵄨 󵄨 ∭∫ |(𝑡−𝜏)/2|≤ℎ, 󵄨󵄨󵄨V (𝑡, 𝑥) − V (𝜏, 𝑦)󵄨󵄨󵄨 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝜏 ℎ2 𝛿≤(𝑡+𝜏)/2≤𝑇−𝛿, |(𝑥−𝑦)/2|≤ℎ, |(𝑥+𝑦)/2|≤𝑟−𝛿 (9)

satisfies limℎ → 0 𝑉ℎ = 0.

1 2 [𝑢 − 𝑘2 ] 𝑓𝑥 2

(11)

− sign (𝑢 − 𝑘) 𝑄𝑢 (𝑡, 𝑥) 𝑓} 𝑑𝑥 𝑑𝑡 = 0, where 𝑘 is an arbitrary constant.

At any Lebesgue point 𝑥0 of the function V(𝑥), we have limℎ → 0 Vℎ (𝑥0 ) = V(𝑥0 ). Since the set of points which are not Lebesgue points of V(𝑥) has measured zero, we get Vℎ (𝑥) → V(𝑥) as ℎ → 0 almost everywhere. We introduce notation connected with the concept of a characteristic cone. For any 𝑀0 > 0, we define 𝑁 > sup𝑡∈[0,∞) ‖𝑢‖𝐿∞ (𝑅) < ∞. Let ℧ designate the cone {(𝑡, 𝑥) : |𝑥| ≤ 𝑀0 − 𝑁𝑡, 0 < 𝑡 < 𝑇0 = min(T, 𝑀0 𝑁−1 )}. We let 𝑆𝜏 designate the cross section of the cone ℧ by the plane 𝑡 = 𝜏, 𝜏 ∈ [0, 𝑇0 ]. Let 𝐻𝑟 = {𝑥 : |𝑥| ≤ 𝑟}, where 𝑟 > 0.

𝑉ℎ =

𝜉𝑇

(7)

We call 𝑥0 a Lebesgue point of function V(𝑥) if ℎ→0ℎ

(10)

Proof. Let Φ(𝑢) be an arbitrary twice smooth function on the line −∞ < 𝑢 < ∞. We multiply (2) by the function Φ󸀠 (𝑢)𝑓(𝑡, 𝑥), where 𝑓(𝑡, 𝑥) ∈ 𝐶0∞ (𝜉𝑇 ). Integrating over 𝜉𝑇 and transferring the derivatives with respect to 𝑡 and 𝑥 to the test function 𝑓, for any constant 𝑘, we obtain 𝑢

∬ {Φ (𝑢) 𝑓𝑡 + [∫ Φ󸀠 (𝑧) 𝑧 𝑑𝑧] 𝑓𝑥 𝑘

𝜉𝑇

(12)

󸀠

− Φ (𝑢) 𝑄𝑢 (𝑡, 𝑥) 𝑓} 𝑑𝑥 𝑑𝑡 = 0, ∞

𝑢

in which we have used ∫−∞ [∫𝑘 Φ󸀠 (𝑧)𝑧 𝑑𝑧]𝑓𝑥 𝑑𝑥 ∞

=

󸀠

− ∫−∞ [𝑓Φ (𝑢)𝑢𝑢𝑥 ]𝑑𝑥. We have ∫

∞

−∞

𝑢

[∫ Φ󸀠 (𝑧) 𝑧 𝑑𝑧] 𝑓𝑥 𝑑𝑥 𝑘

∞

1 1 [Φ󸀠 (𝑢) ( 𝑢2 − 𝑘2 ) 2 2 −∞

=∫

−

1 𝑢 2 ∫ (𝑧 − 𝑘2 ) Φ󸀠󸀠 (𝑧) 𝑑𝑧] 𝑓𝑥 𝑑𝑥. 2 𝑘

(13)

Journal of Function Spaces and Applications

3

Let Φℎ (𝑢) be an approximation of the function |𝑢 − 𝑘| and set Φ(𝑢) = Φℎ (𝑢). Using the properties of the sign(𝑢 − 𝑘), (12), and (13) and sending ℎ → 0, we have ∬ {|𝑢 − 𝑘| 𝑓𝑡 + sign (𝑢 − 𝑘) 𝜉𝑇

1 2 [𝑢 − 𝑘2 ] 𝑓𝑥 2

∞ 󵄨󵄨 2 󵄨 󵄨 2 2 󵄨󵄨 󵄨󵄨𝑢0 − V02 󵄨󵄨󵄨 𝑑𝑥 + ∫ 󵄨󵄨󵄨𝑢0𝑥 󵄨󵄨 𝑑𝑥 − V0𝑥 󵄨 󵄨 󵄨 󵄨 −∞ −∞

≤ 𝑐 (∫

∞

+∫

∞

−∞

|𝑢 − V| 𝑑𝑥) , (16)

(14) in which we have used the Fubini theorem, ‖𝑢‖𝐿∞ ≤ ‖𝑢0 ‖𝐻1 (𝑅) and ‖V‖𝐿∞ ≤ ‖V0 ‖𝐻1 (𝑅) . The proof is completed.

− sign (𝑢 − 𝑘) 𝑄𝑢 (𝑡, 𝑥) 𝑓} 𝑑𝑥 𝑑𝑡 = 0, which completes the proof.

3. Main Results In fact, the proof of (11) can also be found in [17]. Lemma 5. Assume 𝑢 and V are two strong solutions of problem (2). It has 󵄨󵄨 ∞ 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨∫ sign (𝑢 − V) [𝑄𝑢 (𝑡, 𝑥) − 𝑄V (𝑡, 𝑥)] 𝑓 𝑑𝑥󵄨󵄨󵄨 󵄨󵄨 −∞ 󵄨󵄨 ≤ 𝑐∫

∞

−∞

󵄨 󵄨 2 󵄨 2 󵄨󵄨 󵄨󵄨 + |𝑢 − V|) 𝑑𝑥. (󵄨󵄨󵄨󵄨𝑢02 − V02 󵄨󵄨󵄨󵄨 + 󵄨󵄨󵄨󵄨𝑢0𝑥 − V0𝑥 󵄨

(15)

∞

−∞

(𝑢𝑥2 − V𝑥2 ) 𝑑𝑥 = ∫

∞

−∞

2 2 [(𝑢02 + 𝑢0𝑥 ) − (V02 + V0𝑥 )] 𝑑𝑥 ∞

−∫

−∞

(𝑢2 − V2 ) 𝑑𝑥,

󵄨󵄨 ∞ 1 2 1 2 󵄨󵄨 −2 2 2 󵄨󵄨∫ Λ (𝑢 + 𝑢𝑥 − V − V𝑥 ) 󵄨󵄨 −∞ 2 2

󵄨󵄨 󵄨 × 𝜕𝑥 [sign [𝑢 (𝑡, 𝑥) − V (𝑡, 𝑥)] 𝑓 (𝑡, 𝑥)] 𝑑𝑥󵄨󵄨󵄨 󵄨󵄨

󵄨󵄨 1 ∞ 1 1 󵄨 = 󵄨󵄨󵄨 ∬ 𝑒−|𝑥−𝑦| (𝑢2 + 𝑢𝑦2 − V2 − V𝑦2 ) 𝑑𝑦 󵄨󵄨 2 −∞ 2 2 󵄨󵄨 󵄨 × 𝜕𝑥 [sign [𝑢 (𝑡, 𝑥) − V (𝑡, 𝑥)] 𝑓 (𝑡, 𝑥)] 𝑑𝑥󵄨󵄨󵄨 󵄨󵄨 ≤

󵄨󵄨 1 ∞ 󵄨󵄨󵄨󵄨 ∞ 1 1 󵄨 ∫ 󵄨󵄨∫ (𝑢2 + 𝑢𝑦2 − V2 − V𝑦2 ) 𝑑𝑦󵄨󵄨󵄨 󵄨󵄨 2 −∞ 󵄨󵄨 −∞ 2 2 󵄨 󵄨 × 𝑒−|𝑥−𝑦| 󵄨󵄨󵄨𝜕𝑥 [sign [𝑢 (𝑡, 𝑥) − V (𝑡, 𝑥)] 𝑓 (𝑡, 𝑥)]󵄨󵄨󵄨 𝑑𝑥

≤

󵄨󵄨 1 󵄨󵄨󵄨󵄨 ∞ 1 2 1 2 󵄨 2 2 󵄨󵄨∫ (𝑢 + 𝑢𝑦 − V − V𝑦 ) 𝑑𝑦󵄨󵄨󵄨 󵄨󵄨 2 󵄨󵄨 −∞ 2 2 ×∫

∞

−∞

‖𝑢 (𝑡, ⋅) − V (𝑡, ⋅)‖𝐿1 (𝑅) ≤ 𝑒𝑐𝑡 𝑐 (1 + 𝑇0 ) ∞

∞ 󵄨 2 󵄨 󵄨󵄨 2 󵄨 󵄨󵄨𝑢0 (𝑥) − V0 (𝑥)󵄨󵄨󵄨 𝑑𝑥 + ∫ 󵄨󵄨󵄨󵄨𝑢0𝑥 − V0𝑥 󵄨󵄨󵄨󵄨 𝑑𝑥) , −∞ −∞ (17)

× (∫

Proof. We have ∫

Theorem 6. Let 𝑢 and V be two local or global strong solutions of problem (2) with initial data 𝑢(0, ⋅) = 𝑢0 ∈ 𝐻1 (𝑅) and V(0, ⋅) = V0 ∈ 𝐻1 (𝑅), respectively. Let 𝑇0 be the maximum existence time of solutions 𝑢 and V. For any 𝑡 ∈ [0, 𝑇0 ), it holds that

󵄨󵄨 󵄨 󵄨󵄨𝜕𝑥 [sign [𝑢 (𝑡, 𝑥) − V (𝑡, 𝑥)] 𝑓 (𝑡, 𝑥)]󵄨󵄨󵄨 𝑑𝑥

󵄨󵄨 ∞ 󵄨󵄨 1 1 󵄨 󵄨 ≤ 𝑐 󵄨󵄨󵄨∫ (𝑢2 + 𝑢𝑦2 − V2 − V𝑦2 ) 𝑑𝑦󵄨󵄨󵄨 󵄨󵄨 −∞ 󵄨󵄨 2 2

where 𝑐 depends on ‖𝑢0 ‖𝐻1 (𝑅) and ‖V0 ‖𝐻1 (𝑅) . From Theorem 6, we immediately obtain the uniqueness result. Theorem 7. Let 𝑢(𝑡, 𝑥) be a strong solution of (1) with 𝑢0 ∈ 𝐻1 (𝑅), and let 𝑇0 be the maximum existence time of solution 𝑢. Then any strong solution of (1) is unique. Proof of Theorem 6. For an arbitrary 𝑇 > 0, set 𝜉𝑇 = [0, 𝑇]×𝑅. Let 𝑓(𝑡, 𝑥) ∈ 𝐶0∞ (𝜉𝑇 ). We assume that 𝑓(𝑡, 𝑥) = 0 outside some cylinder z = {(𝑡, 𝑥)} = [𝛿, 𝑇 − 2𝛿] × 𝐻𝑟−2𝛿 ,

0 < 2𝛿 ≤ min (𝑇, 𝑟) . (18)

We define 𝑔 = 𝑓(

𝑥−𝑦 𝑡−𝜏 𝑡+𝜏 𝑥+𝑦 , ) 𝜌ℎ ( ) 𝜌ℎ ( ) 2 2 2 2

(19)

= 𝑓 (⋅ ⋅ ⋅) 𝜆 ℎ (∗) , where (⋅ ⋅ ⋅) = ((𝑡 + 𝜏)/2, (𝑥 + 𝑦)/2) and (∗) = ((𝑡 − 𝜏)/2, (𝑥 − 𝑦)/2). The function 𝜌(𝜎) is defined in (6). Note that 𝑔𝑡 + 𝑔𝜏 = 𝑓𝑡 (⋅ ⋅ ⋅) 𝜆 ℎ (∗) , 𝑔𝑥 + 𝑔𝑦 = 𝑓𝑥 (⋅ ⋅ ⋅) 𝜆 ℎ (∗) .

(20)

4

Journal of Function Spaces and Applications

Taking 𝑘 = V(𝜏, 𝑦) and assuming 𝑓(𝑡, 𝑥) = 0 outside the cylinder z, from Lemma 4, we have ∭∫

𝜉𝑇 ×𝜉𝑇

󵄨󵄨 󵄨 + 󵄨󵄨󵄨󵄨∬ sign (𝑢 (𝑡, 𝑥) − V (𝑡, 𝑥)) 󵄨󵄨 𝜉𝑇

󵄨󵄨 󵄨 × [𝑄𝑢 (𝑡, 𝑥) − 𝑄V (𝑡, 𝑥)] 𝑓 𝑑𝑥 𝑑𝑡󵄨󵄨󵄨󵄨 . 󵄨󵄨

󵄨 󵄨 { 󵄨󵄨󵄨𝑢 (𝑡, 𝑥) − V (𝜏, 𝑦)󵄨󵄨󵄨 𝑔𝑡 + sign (𝑢 (𝑡, 𝑥) − V (𝜏, 𝑦))

(24)

2 𝑢2 (𝑡, 𝑥) V (𝜏, 𝑦) ×( − ) 𝑔𝑥 2 2

We note that the first two terms in the integrand of (23) can be represented in the form

+ sign (𝑢 (𝑡, 𝑥) − V (𝜏, 𝑦))

𝑌ℎ = 𝑌 (𝑡, 𝑥, 𝜏, 𝑦, 𝑢 (𝑡, 𝑥) , V (𝜏, 𝑦)) 𝜆 ℎ (∗) .

× 𝑄𝑢 (𝑡, 𝑥) 𝑔} 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝜏 = 0. (21) Similarly, it has ∭∫

𝜉𝑇 ×𝜉𝑇

Since ‖𝑢‖𝐿∞ ≤ ‖𝑢0 ‖𝐻1 (𝑅) and ‖V‖𝐿∞ ≤ ‖V0 ‖𝐻1 (𝑅) , from Lemma 2, we know 𝑌ℎ satisfies the Lipschitz condition in 𝑢 and V, respectively. By the choice of 𝑔, we have 𝑌ℎ = 0 outside the region {(𝑡, 𝑥; 𝜏, 𝑦)} = {𝛿 ≤

󵄨 󵄨 { 󵄨󵄨󵄨V (𝜏, 𝑦) − 𝑢 (𝑡, 𝑥) 𝑔𝜏 󵄨󵄨󵄨 + sign (V (𝜏, 𝑦) − 𝑢 (𝑡, 𝑥)) ×(

V2 (𝜏, 𝑦) 𝑢2 (𝑡, 𝑥) − ) 𝑔𝑦 2 2

𝑡+𝜏 |𝑡 − 𝜏| ≤ 𝑇 − 2𝛿, ≤ ℎ, 2 2

󵄨 󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨𝑥 + 𝑦󵄨󵄨󵄨 󵄨𝑥 − 𝑦󵄨󵄨󵄨 ≤ 𝑟 − 2𝛿, 󵄨 ≤ ℎ} , 2 2 ∭∫

𝜉𝑇 ×𝜉𝑇

+ sign (V (𝜏, 𝑦) − 𝑢 (𝑡, 𝑥))

(25)

𝑌ℎ 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝜏

= ∭∫

[𝑌 (𝑡, 𝑥, 𝜏, 𝑦, 𝑢 (𝑡, 𝑥) , V (𝜏, 𝑦))

𝜉𝑇 ×𝜉𝑇

× 𝑄V (𝜏, 𝑦) 𝑔} 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝜏 = 0,

(26)

−𝑌 (𝑡, 𝑥, 𝑡, 𝑥, 𝑢 (𝑡, 𝑥) , V (𝑡, 𝑥)) ]

(22)

× 𝜆 ℎ (∗) 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝜏

from which we obtain 0 ≤ ∭∫

𝜉𝑇 ×𝜉𝑇

+ ∭∫

𝜉𝑇 ×𝜉𝑇

󵄨 󵄨 { 󵄨󵄨󵄨𝑢 (𝑡, 𝑥) − V (𝜏, 𝑦)󵄨󵄨󵄨 (𝑔𝑡 + 𝑔𝜏 )

× 𝜆 ℎ (∗) 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝜏

+ sign (𝑢 (𝑡, 𝑥) − V (𝜏, 𝑦)) ×(

= 𝐽1 (ℎ) + 𝐽2 .

2

Considering the estimate |𝜆(∗)| ≤ 𝑐/ℎ2 and the expression of function 𝑌ℎ , we have 󵄨󵄨󵄨𝐽1 (ℎ)󵄨󵄨󵄨 󵄨 󵄨

𝑢2 (𝑡, 𝑥) V (𝜏, 𝑦) − ) 2 2

× (𝑔𝑥 + 𝑔𝑦 ) } 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝜏

(23)

󵄨󵄨 󵄨 sign (𝑢 (𝑡, 𝑥) − V (𝑡, 𝑥)) + 󵄨󵄨󵄨󵄨∭∫ 󵄨󵄨 𝜉𝑇 ×𝜉𝑇

󵄨󵄨 󵄨 × (𝑄𝑢 (𝑡, 𝑥) − 𝑄V (𝜏, 𝑦)) 𝑔 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝜏󵄨󵄨󵄨󵄨 󵄨󵄨

= ∭∫

𝜉𝑇 ×𝜉𝑇

(𝐼1 + 𝐼2 + 𝐼3 ) 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝜏.

We will show that 0 ≤ ∬ { |𝑢 (𝑡, 𝑥) − V (𝑡, 𝑥)| 𝑓𝑡 + sign (𝑢 (𝑡, 𝑥) − V (𝑡, 𝑥)) 𝜉𝑇

×(

𝑢2 (𝑡, 𝑥) V2 (𝑡, 𝑥) − ) 𝑓𝑥 } 𝑑𝑥 𝑑𝑡 2 2

𝑌 (𝑡, 𝑥, 𝑡, 𝑥, 𝑢 (𝑡, 𝑥) , V (𝑡, 𝑥))

[ [ 1 [ ≤ 𝑐 [ℎ + 2 [ ℎ [ [ × ∭∫

|(𝑡−𝜏)/2|≤ℎ, 𝛿≤(𝑡+𝜏)/2≤𝑇−𝛿, |(𝑥−𝑦)/2|≤ℎ, |(𝑥+𝑦)/2|≤𝑟−𝛿

󵄨󵄨 󵄨󵄨V (𝑡, 𝑥)

] ] ] 󵄨󵄨 −V (𝜏, 𝑦)󵄨󵄨 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝜏, ] , ] ] ] (27)

Journal of Function Spaces and Applications

5

where the constant 𝑐 does not depend on ℎ. Using Lemma 1, we obtain 𝐽1 (ℎ) → 0 as ℎ → 0. The integral 𝐽2 does not depend on ℎ. In fact, substituting 𝑡 = 𝛼, (𝑡 − 𝜏)/2 = 𝛽, 𝑥 = 𝜂, (𝑥 − 𝑦)/2 = 𝜉 and noting that ℎ

By Lemmas 1 and 3, we have 𝐾1 (ℎ) → 0 as ℎ → 0. Using (28), we have 𝐾2 = 22 ∬ 𝐼3 (𝛼, 𝜂, 𝛼, 𝜂, 𝑢 (𝛼, 𝜂) , V (𝛼, 𝜂)) 𝜉𝑇

∞

∫ ∫

−ℎ −∞

𝜆 ℎ (𝛽, 𝜉) 𝑑𝜉 𝑑𝛽 = 1,

(28)

ℎ

× {∫ ∫

∞

−ℎ −∞

we have 𝐽2 = 22 ∬ 𝑌ℎ (𝛼, 𝜂, 𝛼, 𝜂, 𝑢 (𝛼, 𝜂) , V (𝛼, 𝜂))

= 4∬ 𝐼3 (𝑡, 𝑥, 𝑡, 𝑥, 𝑢 (𝑡, 𝑥) , V (𝑡, 𝑥)) 𝑑𝑥 𝑑𝑡

𝜉𝑇

ℎ

× {∫ ∫

∞

−ℎ −∞

𝜆 ℎ (𝛽, 𝜉) 𝑑𝜉 𝑑𝛽} 𝑑𝜂 𝑑𝛼

= 4∬ sign (𝑢 (𝑡, 𝑥) − V (𝑡, 𝑥))

(29)

𝜉𝑇

× (𝑄𝑢 (𝑡, 𝑥) − 𝑄V (𝑡, 𝑥)) 𝑓 (𝑡, 𝑥) 𝑑𝑥 𝑑𝑡.

𝜉𝑇

From (30) and (33), we prove that inequality (24) holds. Let

Hence lim ∭∫

𝜉𝑇 ×𝜉𝑇

𝑌ℎ 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝜏

𝜔 (𝑡) = ∫

(30)

= 4∬ 𝑌ℎ (𝑡, 𝑥, 𝑡, 𝑥, 𝑢 (𝑡, 𝑥) , V (𝑡, 𝑥)) 𝑑𝑥 𝑑𝑡.

−∞

Since 𝐼3 = sign (𝑢 (𝑡, 𝑥) − V (𝜏, 𝑦)) (𝑄𝑢 (𝑡, 𝑥) − 𝑄V (𝜏, 𝑦)) 𝑓𝜆 ℎ (∗) , 𝜉𝑇 ×𝜉𝑇

𝜉𝑇 ×𝜉𝑇

𝜃ℎ = ∫

𝜉𝑇 ×𝜉𝑇

𝜎

−∞

𝜌ℎ (𝜎) 𝑑𝜎

(𝜃ℎ󸀠 (𝜎) = 𝜌ℎ (𝜎) ≥ 0)

𝑓 = [𝜃ℎ (𝑡 − 𝜏1 ) − 𝜃ℎ (𝑡 − 𝜏2 )] 𝜒 (𝑡, 𝑥) ,

[𝐼3 (𝑡, 𝑥, 𝜏, 𝑦) − 𝐼3 (𝑡, 𝑥, 𝑡, 𝑥)]

ℎ < min (𝜏1 , 𝑇0 − 𝜏2 ) ,

× 𝜆 ℎ (∗) 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝜏 + ∭∫

|𝑢 (𝑡, 𝑥) − V (𝑡, 𝑥)| 𝑑𝑥.

(35)

(36)

where

𝐼3 (𝑡, 𝑥, 𝑡, 𝑥) 𝜆 ℎ (∗) 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝜏

𝜒 (𝑡, 𝑥) = 𝜒𝜀 (𝑡, 𝑥) = 1 − 𝜃𝜀 (|𝑥| + 𝑁𝑡 − 𝑀0 + 𝜀) ,

= 𝐾1 (ℎ) + 𝐾2 , (31) we obtain 󵄨󵄨 󵄨 󵄨󵄨𝐾1 (ℎ)󵄨󵄨󵄨

𝜀 > 0.

1 ℎ2

(37)

We note that function 𝜒(𝑡, 𝑥) = 0 outside the cone ℧ and 𝑓(𝑡, 𝑥) = 0 outside the set z. For (𝑡, 𝑥) ∈ ℧, we have the relations 󵄨 󵄨 0 = 𝜒𝑡 + 𝑁 󵄨󵄨󵄨𝜒𝑥 󵄨󵄨󵄨 ≥ 𝜒𝑡 + 𝑁𝜒𝑥 .

≤ 𝑐 (ℎ +

(34)

and choose two numbers 𝜏1 and 𝜏2 ∈ (0, 𝑇0 ), 𝜏1 < 𝜏2 . In (24), we choose

𝐼3 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝜏

= ∭∫

∞

We define

𝜉𝑇

∭∫

(33)

𝜉𝑇

= 4∬ 𝑌ℎ (𝑡, 𝑥, 𝑡, 𝑥, 𝑢 (𝑡, 𝑥) , V (𝑡, 𝑥)) 𝑑𝑥 𝑑𝑡.

ℎ→0

𝜆 ℎ (𝛽, 𝜉) 𝑑𝜉 𝑑𝛽} 𝑑𝜂 𝑑𝛼

(38)

Applying (24) and (35)–(38), we have the inequality 0≤∬

× ∭∫

|(𝑡−𝜏)/2|≤ℎ, 𝛿≤(𝑡+𝜏)/2≤𝑇−𝛿, |(𝑥−𝑦)/2|≤ℎ, |(𝑥+𝑦)/2|≤𝑟−𝛿

󵄨󵄨 󵄨󵄨𝑄V (𝑡, 𝑥)

𝜉𝑇0

{[𝜌ℎ (𝑡 − 𝜏1 ) − 𝜌ℎ (𝑡 − 𝜏2 )] × 𝜒𝜀 |𝑢 (𝑡, 𝑥) − V (𝑡, 𝑥)|} 𝑑𝑥 𝑑𝑡

󵄨󵄨 󵄨 + 󵄨󵄨󵄨󵄨∬ [𝜃ℎ (𝑡 − 𝜏1 ) − 𝜃ℎ (𝑡 − 𝜏2 )] 󵄨󵄨 𝜉𝑇0 󵄨 −𝑄V (𝜏, 𝑦)󵄨󵄨󵄨 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝜏) . (32)

󵄨󵄨 󵄨 × [𝑄𝑢 (𝑡, 𝑥) − 𝑄V (𝑡, 𝑥)] 𝐵 (𝑡, 𝑥) 𝜒 (𝑡, 𝑥) 𝑑𝑥 𝑑𝑡󵄨󵄨󵄨󵄨 , 󵄨󵄨 (39) where 𝐵(𝑡, 𝑥) = sign[𝑢(𝑡, 𝑥) − V(𝑡, 𝑥)].

6

Journal of Function Spaces and Applications From (39), we obtain

0≤∬

𝜉𝑇0

Using the similar proof of (43), we get 𝑇0

{[𝜌ℎ (𝑡 − 𝜏1 ) − 𝜌ℎ (𝑡 − 𝜏2 )]

𝐿󸀠 (𝜏1 ) = − ∫ 𝜌ℎ (𝑡 − 𝜏1 ) 𝜔 (𝑡) 𝑑𝑡 󳨀→ −𝜔 (𝜏1 ) 0

× 𝜒𝜀 |𝑢 (𝑡, 𝑥) − V (𝑡, 𝑥)|} 𝑑𝑥 𝑑𝑡 + ∫ (𝜃ℎ (𝑡 − 𝜏1 ) − 𝜃ℎ (𝑡 − 𝜏2 ))

𝜏1

𝐿 (𝜏1 ) 󳨀→ 𝐿 (0) − ∫ 𝜔 (𝜎) 𝑑𝜎

0

󵄨󵄨 ∞ 󵄨󵄨 󵄨 󵄨 × 󵄨󵄨󵄨∫ [𝑄𝑢 (𝑡, 𝑥) − 𝑄V (𝑡, 𝑥)] 𝐵 (𝑡, 𝑥) 𝜒 (𝑡, 𝑥) 𝑑𝑥󵄨󵄨󵄨 𝑑𝑡. 󵄨󵄨 −∞ 󵄨󵄨 (40)

0

𝜏2

𝐿 (𝜏2 ) 󳨀→ 𝐿 (0) − ∫ 𝜔 (𝜎) 𝑑𝜎 0

𝜏2

𝐿 (𝜏1 ) − 𝐿 (𝜏2 ) 󳨀→ ∫ 𝜔 (𝜎) 𝑑𝜎 𝜏1

× 𝜒𝜀 |𝑢 (𝑡, 𝑥) − V (𝑡, 𝑥)|} 𝑑𝑥 𝑑𝑡 𝑇0

(41)

0

× (𝐺 + ∫

∞

−∞

∞

|𝑢 − V| 𝑑𝑥) 𝑑𝑡,

0

𝜏1

∞ 󵄨󵄨 2 󵄨 󵄨 2 2 󵄨󵄨 󵄨󵄨𝑢0 − V02 󵄨󵄨󵄨 𝑑𝑥 + ∫ 󵄨󵄨󵄨𝑢0𝑥 󵄨󵄨 𝑑𝑥) (𝜏2 − 𝜏1 ) . − V0𝑥 󵄨 󵄨 󵄨 󵄨 −∞ −∞ (49)

󵄨 󵄨󵄨 󵄨 󵄨 󵄨󵄨𝑢 (𝜏1 , 𝑥) − V (𝜏1 , 𝑥)󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨𝑢 (𝜏1 , 𝑥) − 𝑢0 (𝑥)󵄨󵄨󵄨 󵄨 󵄨 + 󵄨󵄨󵄨V (𝜏1 , 𝑥) − V0 (𝑥)󵄨󵄨󵄨 󵄨 󵄨 + 󵄨󵄨󵄨𝑢0 (𝑥) − V0 (𝑥)󵄨󵄨󵄨 .

0

|𝑢 (𝑡, 𝑥) − V (𝑡, 𝑥)| 𝑑𝑥} 𝑑𝑡 (42)

+ 𝑐 ∫ (𝜃ℎ (𝑡 − 𝜏1 ) − 𝜃ℎ (𝑡 − 𝜏2 )) 0

× (𝐺 + ∫

−∞

󵄨󵄨 󵄨󵄨 𝑇0 󵄨 󵄨󵄨 󵄨󵄨∫ 𝜌ℎ (𝑡 − 𝜏1 ) 𝜔 (𝑡) 𝑑𝑡 − 𝜔 (𝜏1 )󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 0 󵄨 󵄨 󵄨󵄨 𝑇0 󵄨󵄨 󵄨 󵄨 = 󵄨󵄨󵄨󵄨∫ 𝜌ℎ (𝑡 − 𝜏1 ) [𝜔 (𝑡) − 𝜔 (𝜏1 )] 𝑑𝑡󵄨󵄨󵄨󵄨 󵄨󵄨 0 󵄨󵄨

∞

−∞

By the properties of the function 𝜌ℎ (𝜎) for ℎ ≤ min(𝜏1 , 𝑇0 − 𝜏1 ), we have

(43)

(50)

Thus, from (42), (43), (48), (49), and (50), for any 𝑡 ∈ [0, 𝑇0 ], we have ∫

|𝑢 − V| 𝑑𝑥) 𝑑𝑡.

∞

Let 𝜏1 → 0 and 𝜏2 → 𝑡, and note that

𝑇0

∞

(48)

𝜏2

= (∫

𝑇0

as ℎ 󳨀→ 0.

𝑇0

0 ≤ ∫ { [𝜌ℎ (𝑡 − 𝜏1 ) − 𝜌ℎ (𝑡 − 𝜏2 )]

−∞

(47)

Furthermore, if ℎ → 0, we have

∞

×∫

as ℎ 󳨀→ 0.

∫ (𝜃ℎ (𝑡 − 𝜏1 ) − 𝜃ℎ (𝑡 − 𝜏2 )) 𝐺 𝑑𝑡 󳨀→ 𝐺 ∫ 𝑑𝑡

2 2 where 𝐺 = ∫−∞ |𝑢02 −V02 |𝑑𝑥+∫−∞ |𝑢0𝑥 −V0𝑥 |𝑑𝑥 and 𝑐 is defined in Lemma 5. Letting 𝜀 → 0 in (41) and sending 𝑀0 → ∞, we have

∞

(46)

Then, we get

{[𝜌ℎ (𝑡 − 𝜏1 ) − 𝜌ℎ (𝑡 − 𝜏2 )]

+ 𝑐 ∫ (𝜃ℎ (𝑡 − 𝜏1 ) − 𝜃ℎ (𝑡 − 𝜏2 ))

as ℎ 󳨀→ 0.

Similarly, we have

Using Lemma 5, we have

𝜉𝑇0

(45)

from which we obtain

𝑇0

0≤∬

as ℎ 󳨀→ 0,

|𝑢 (𝑡, 𝑥) − V (𝑡, 𝑥)| 𝑑𝑥 ∞

≤∫

−∞

|𝑢 (0, 𝑥) − V (0, 𝑥)| 𝑑𝑥 ∞

∞ 󵄨 󵄨 󵄨 2 2 󵄨󵄨 󵄨󵄨 𝑑𝑥) + 𝑐𝑇0 (∫ 󵄨󵄨󵄨󵄨𝑢02 − V02 󵄨󵄨󵄨󵄨 𝑑𝑥 + ∫ 󵄨󵄨󵄨󵄨𝑢0𝑥 − V0𝑥 󵄨 −∞ −∞ 𝑡

∞

0

−∞

+∫ ∫

(51)

|𝑢 − V| 𝑑𝑥,

from which we complete the proof of Theorem 6 by using the Gronwall inequality.

1 𝜏1 +ℎ 󵄨󵄨 󵄨 ≤𝑐 ∫ 󵄨𝜔 (𝑡) − 𝜔 (𝜏1 )󵄨󵄨󵄨 𝑑𝑡 󳨀→ 0 as ℎ 󳨀→ 0, ℎ 𝜏1 −ℎ 󵄨

Acknowledgments

where 𝑐 is independent of ℎ. Set 𝑇0

𝑇0

𝑡−𝜏1

0

0

−∞

𝐿 (𝜏1 ) = ∫ 𝜃ℎ (𝑡 − 𝜏1 ) 𝜇 (𝑡) 𝑑𝑡 = ∫ ∫

𝜌ℎ (𝜎) 𝑑𝜎𝜔 (𝑡) 𝑑𝑡. (44)

Thanks are given to referees whose suggestions are very helpful to the paper. This work is supported by both the Fundamental Research Funds for the Central Universities (JBK120504) and the Applied and Basic Project of Sichuan Province (2012JY0020).

Journal of Function Spaces and Applications

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