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 Β© 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 π»1 (π
). 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]
π’π‘ β π’π‘π₯π₯ + 3π’π’π₯ = 2π’π₯ π’π₯π₯ + π’π’π₯π₯π₯
1 π’π‘ + (π’2 )π₯ + ππ₯ ππ’ (π‘, π₯) = 0, 2
(2)
π’ (0, π₯) = π’0 ,
(1)
β
has been investigated by many scholars. Equation (1) has peaked solitary wave solutions, which takes the form ππβ|π₯βππ‘| , π β π
. 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 π = π’0 β π’0π₯π₯ 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 [4β8]. For the initial value π’0 satisfying π’0 β π’0π₯π₯ β₯ 0 or π’0 β π’0π₯π₯ β€ 0, it is shown in [9] that the Camassa-Holm equation has unique global strong solutions in the Sobolev space π»π (π
) with π > 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 [12β16] and the references therein.
where ππ’ (π‘, π₯) = Ξβ2 (π’2 + (1/2)π’π₯2 ) = (1/2) β«ββ π|π₯βπ¦| (π’2 (π‘, π¦)+ (1/2)π’π¦2 (π‘, π¦))ππ¦. If (1) has a suitable smooth strong solution, we have the conservation law β«
β
ββ
(π’2 (π‘, π₯) + π’π₯2 (π‘, π₯)) ππ₯ β
=β«
ββ
(3) 2
(π’ (0, π₯) +
π’π₯2
(0, π₯)) ππ₯,
which derives σ΅© σ΅© βπ’βπΏβ (π
) β€ βπ’βπ»1 (π
) = σ΅©σ΅©σ΅©π’0 σ΅©σ΅©σ΅©π»1 (π
) .
(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: βπ’ (π‘, β
) β V (π‘, β
)βπΏ1 (π
) β€ ππππ‘ (β«
β
ββ
β
σ΅¨σ΅¨ 2 2 σ΅¨σ΅¨ σ΅¨σ΅¨π’0π₯ β V0π₯ σ΅¨σ΅¨ ππ₯) , σ΅¨ σ΅¨ ββ (5)
σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨π’0 (π₯) β V0 (π₯)σ΅¨σ΅¨σ΅¨ ππ₯ + β«
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where π‘ β [0, π0 ), functions π’ and V are two local or global strong solutions of problem (2) with initial data π’(0, β
) = π’0 β π»1 (π
) and V(0, β
) = V0 β π»1 (π
), respectively. Constant π depends on βπ’0 βπ»1 (π
) , βV0 βπ»1 (π
) , and the maximum existence time π0 . 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 Kruzkovβs 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 |ππΉ(π’)/ππ’| is bounded, then the function π»(π’, V) = sign(π’ β V)(πΉ(π’) β πΉ(V)) satisfies the Lipschitz condition in π’ and V, respectively. Lemma 3. Let π’0 β π»1 (π
). It holds that the function ππ’ (π‘, π₯) = ππ₯ ππ’ (π‘, π₯) = ππ₯ Ξβ2 (π’2 + (1/2)π’π₯2 ) satisfies σ΅© σ΅©σ΅© σ΅©σ΅©ππ’ (π‘, π₯)σ΅©σ΅©σ΅©πΏβ (π
) < β, σ΅©σ΅© σ΅© σ΅©σ΅©ππ’ (π‘, π₯)σ΅©σ΅©σ΅©πΏ1 (π
) < β, σ΅©σ΅©σ΅©ππ’ (π‘, π₯)σ΅©σ΅©σ΅© 2 < β, σ΅© σ΅©πΏ (π
) σ΅©σ΅© σ΅©σ΅© σ΅©σ΅©ππ’ (π‘, π₯)σ΅©σ΅©πΏβ (π
) < β, σ΅©σ΅© σ΅© σ΅©σ΅©ππ’ (π‘, π₯)σ΅©σ΅©σ΅©πΏ1 (π
) < β,
2. Notations and Several Lemmas Set ππ = [0, π] Γ π
for an arbitrary π > 0. The space of all infinitely differentiable functions π(π‘, π₯) with compact support in [0, π] Γ π
is denoted by πΆ0β (ππ ). We define π(π) as a function which is infinitely differentiable on (ββ, +β) such β that π(π) β₯ 0, π(π) = 0 for |π| β₯ 1 and β«ββ π(π)ππ = 1. For any number β > 0, we let πβ (π) = π(ββ1 π)/β. Then we have that πβ (π) is a function in πΆβ (ββ, β) and πβ (π) β₯ 0,
πβ (π) = 0 if |π| β₯ β,
σ΅¨ π σ΅¨σ΅¨ σ΅¨σ΅¨πβ (π)σ΅¨σ΅¨σ΅¨ β€ , β
β«
β
ββ
πβ (π) = 1.
(6)
σ΅© σ΅©σ΅© σ΅©σ΅©ππ’ (π‘, π₯)σ΅©σ΅©σ΅©πΏ2 (π
) < β. The proof of Lemma 3 can be found in [13, 15] (see [13, Lemma 5.1]). Lemma 4. Let π’ be the strong solution of problem (2), π(π‘, π₯) β πΆ0β (ππ ), and π(0, π₯) = 0. Then β¬ {|π’ β π| ππ‘ + sign (π’ β π)
Assume that the function V(π₯) is locally integrable in (ββ, β). We define the approximation of function V(π₯) as Vβ (π₯) =
π₯βπ¦ 1 β ) V (π¦) ππ¦, β« π( β ββ β
β > 0.
lim
1
σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨V (π₯) β V (π₯0 )σ΅¨σ΅¨σ΅¨ ππ₯ = 0. |π₯βπ₯ |β€β
β«
(8)
0
Lemma 1 (see [17]). Let the function V(π‘, π₯) be bounded and measurable in cylinder Ξ© = [0, π] Γ π»π . If πΏ β (0, min[π, π]) and any number β β (0, πΏ), then the function 1 σ΅¨ σ΅¨ ββ« |(π‘βπ)/2|β€β, σ΅¨σ΅¨σ΅¨V (π‘, π₯) β V (π, π¦)σ΅¨σ΅¨σ΅¨ ππ₯ ππ‘ ππ¦ ππ β2 πΏβ€(π‘+π)/2β€πβπΏ, |(π₯βπ¦)/2|β€β, |(π₯+π¦)/2|β€πβπΏ (9)
satisfies limβ β 0 πβ = 0.
1 2 [π’ β π2 ] ππ₯ 2
(11)
β sign (π’ β π) ππ’ (π‘, π₯) π} ππ₯ ππ‘ = 0, where π is an arbitrary constant.
At any Lebesgue point π₯0 of the function V(π₯), we have limβ β 0 Vβ (π₯0 ) = V(π₯0 ). Since the set of points which are not Lebesgue points of V(π₯) has measured zero, we get Vβ (π₯) β V(π₯) as β β 0 almost everywhere. We introduce notation connected with the concept of a characteristic cone. For any π0 > 0, we define π > supπ‘β[0,β) βπ’βπΏβ (π
) < β. Let β§ designate the cone {(π‘, π₯) : |π₯| β€ π0 β ππ‘, 0 < π‘ < π0 = min(T, π0 πβ1 )}. We let ππ designate the cross section of the cone β§ by the plane π‘ = π, π β [0, π0 ]. Let π»π = {π₯ : |π₯| β€ π}, where π > 0.
πβ =
ππ
(7)
We call π₯0 a Lebesgue point of function V(π₯) if ββ0β
(10)
Proof. Let Ξ¦(π’) be an arbitrary twice smooth function on the line ββ < π’ < β. We multiply (2) by the function Ξ¦σΈ (π’)π(π‘, π₯), where π(π‘, π₯) β πΆ0β (ππ ). Integrating over ππ and transferring the derivatives with respect to π‘ and π₯ to the test function π, for any constant π, we obtain π’
β¬ {Ξ¦ (π’) ππ‘ + [β« Ξ¦σΈ (π§) π§ ππ§] ππ₯ π
ππ
(12)
σΈ
β Ξ¦ (π’) ππ’ (π‘, π₯) π} ππ₯ ππ‘ = 0, β
π’
in which we have used β«ββ [β«π Ξ¦σΈ (π§)π§ ππ§]ππ₯ ππ₯ β
=
σΈ
β β«ββ [πΞ¦ (π’)π’π’π₯ ]ππ₯. We have β«
β
ββ
π’
[β« Ξ¦σΈ (π§) π§ ππ§] ππ₯ ππ₯ π
β
1 1 [Ξ¦σΈ (π’) ( π’2 β π2 ) 2 2 ββ
=β«
β
1 π’ 2 β« (π§ β π2 ) Ξ¦σΈ σΈ (π§) ππ§] ππ₯ ππ₯. 2 π
(13)
Journal of Function Spaces and Applications
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Let Ξ¦β (π’) be an approximation of the function |π’ β π| and set Ξ¦(π’) = Ξ¦β (π’). Using the properties of the sign(π’ β π), (12), and (13) and sending β β 0, we have β¬ {|π’ β π| ππ‘ + sign (π’ β π) ππ
1 2 [π’ β π2 ] ππ₯ 2
β σ΅¨σ΅¨ 2 σ΅¨ σ΅¨ 2 2 σ΅¨σ΅¨ σ΅¨σ΅¨π’0 β V02 σ΅¨σ΅¨σ΅¨ ππ₯ + β« σ΅¨σ΅¨σ΅¨π’0π₯ σ΅¨σ΅¨ ππ₯ β V0π₯ σ΅¨ σ΅¨ σ΅¨ σ΅¨ ββ ββ
β€ π (β«
β
+β«
β
ββ
|π’ β V| ππ₯) , (16)
(14) in which we have used the Fubini theorem, βπ’βπΏβ β€ βπ’0 βπ»1 (π
) and βVβπΏβ β€ βV0 βπ»1 (π
) . The proof is completed.
β sign (π’ β π) ππ’ (π‘, π₯) π} ππ₯ ππ‘ = 0, which completes the proof.
3. Main Results In fact, the proof of (11) can also be found in [17]. Lemma 5. Assume π’ and V are two strong solutions of problem (2). It has σ΅¨σ΅¨ β σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨β« sign (π’ β V) [ππ’ (π‘, π₯) β πV (π‘, π₯)] π ππ₯σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ ββ σ΅¨σ΅¨ β€ πβ«
β
ββ
σ΅¨ σ΅¨ 2 σ΅¨ 2 σ΅¨σ΅¨ σ΅¨σ΅¨ + |π’ β V|) ππ₯. (σ΅¨σ΅¨σ΅¨σ΅¨π’02 β V02 σ΅¨σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨σ΅¨π’0π₯ β V0π₯ σ΅¨
(15)
β
ββ
(π’π₯2 β Vπ₯2 ) ππ₯ = β«
β
ββ
2 2 [(π’02 + π’0π₯ ) β (V02 + V0π₯ )] ππ₯ β
ββ«
ββ
(π’2 β V2 ) ππ₯,
σ΅¨σ΅¨ β 1 2 1 2 σ΅¨σ΅¨ β2 2 2 σ΅¨σ΅¨β« Ξ (π’ + π’π₯ β V β Vπ₯ ) σ΅¨σ΅¨ ββ 2 2
σ΅¨σ΅¨ σ΅¨ Γ ππ₯ [sign [π’ (π‘, π₯) β V (π‘, π₯)] π (π‘, π₯)] ππ₯σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨
σ΅¨σ΅¨ 1 β 1 1 σ΅¨ = σ΅¨σ΅¨σ΅¨ β¬ πβ|π₯βπ¦| (π’2 + π’π¦2 β V2 β Vπ¦2 ) ππ¦ σ΅¨σ΅¨ 2 ββ 2 2 σ΅¨σ΅¨ σ΅¨ Γ ππ₯ [sign [π’ (π‘, π₯) β V (π‘, π₯)] π (π‘, π₯)] ππ₯σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ β€
σ΅¨σ΅¨ 1 β σ΅¨σ΅¨σ΅¨σ΅¨ β 1 1 σ΅¨ β« σ΅¨σ΅¨β« (π’2 + π’π¦2 β V2 β Vπ¦2 ) ππ¦σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ 2 ββ σ΅¨σ΅¨ ββ 2 2 σ΅¨ σ΅¨ Γ πβ|π₯βπ¦| σ΅¨σ΅¨σ΅¨ππ₯ [sign [π’ (π‘, π₯) β V (π‘, π₯)] π (π‘, π₯)]σ΅¨σ΅¨σ΅¨ ππ₯
β€
σ΅¨σ΅¨ 1 σ΅¨σ΅¨σ΅¨σ΅¨ β 1 2 1 2 σ΅¨ 2 2 σ΅¨σ΅¨β« (π’ + π’π¦ β V β Vπ¦ ) ππ¦σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ 2 σ΅¨σ΅¨ ββ 2 2 Γβ«
β
ββ
βπ’ (π‘, β
) β V (π‘, β
)βπΏ1 (π
) β€ πππ‘ π (1 + π0 ) β
β σ΅¨ 2 σ΅¨ σ΅¨σ΅¨ 2 σ΅¨ σ΅¨σ΅¨π’0 (π₯) β V0 (π₯)σ΅¨σ΅¨σ΅¨ ππ₯ + β« σ΅¨σ΅¨σ΅¨σ΅¨π’0π₯ β V0π₯ σ΅¨σ΅¨σ΅¨σ΅¨ ππ₯) , ββ ββ (17)
Γ (β«
Proof. We have β«
Theorem 6. Let π’ and V be two local or global strong solutions of problem (2) with initial data π’(0, β
) = π’0 β π»1 (π
) and V(0, β
) = V0 β π»1 (π
), respectively. Let π0 be the maximum existence time of solutions π’ and V. For any π‘ β [0, π0 ), it holds that
σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ππ₯ [sign [π’ (π‘, π₯) β V (π‘, π₯)] π (π‘, π₯)]σ΅¨σ΅¨σ΅¨ ππ₯
σ΅¨σ΅¨ β σ΅¨σ΅¨ 1 1 σ΅¨ σ΅¨ β€ π σ΅¨σ΅¨σ΅¨β« (π’2 + π’π¦2 β V2 β Vπ¦2 ) ππ¦σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ ββ σ΅¨σ΅¨ 2 2
where π depends on βπ’0 βπ»1 (π
) and βV0 βπ»1 (π
) . From Theorem 6, we immediately obtain the uniqueness result. Theorem 7. Let π’(π‘, π₯) be a strong solution of (1) with π’0 β π»1 (π
), and let π0 be the maximum existence time of solution π’. Then any strong solution of (1) is unique. Proof of Theorem 6. For an arbitrary π > 0, set ππ = [0, π]Γπ
. Let π(π‘, π₯) β πΆ0β (ππ ). We assume that π(π‘, π₯) = 0 outside some cylinder z = {(π‘, π₯)} = [πΏ, π β 2πΏ] Γ π»πβ2πΏ ,
0 < 2πΏ β€ min (π, π) . (18)
We define π = π(
π₯βπ¦ π‘βπ π‘+π π₯+π¦ , ) πβ ( ) πβ ( ) 2 2 2 2
(19)
= π (β
β
β
) π β (β) , where (β
β
β
) = ((π‘ + π)/2, (π₯ + π¦)/2) and (β) = ((π‘ β π)/2, (π₯ β π¦)/2). The function π(π) is defined in (6). Note that ππ‘ + ππ = ππ‘ (β
β
β
) π β (β) , ππ₯ + ππ¦ = ππ₯ (β
β
β
) π β (β) .
(20)
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Journal of Function Spaces and Applications
Taking π = V(π, π¦) and assuming π(π‘, π₯) = 0 outside the cylinder z, from Lemma 4, we have ββ«
ππ Γππ
σ΅¨σ΅¨ σ΅¨ + σ΅¨σ΅¨σ΅¨σ΅¨β¬ sign (π’ (π‘, π₯) β V (π‘, π₯)) σ΅¨σ΅¨ ππ
σ΅¨σ΅¨ σ΅¨ Γ [ππ’ (π‘, π₯) β πV (π‘, π₯)] π ππ₯ ππ‘σ΅¨σ΅¨σ΅¨σ΅¨ . σ΅¨σ΅¨
σ΅¨ σ΅¨ { σ΅¨σ΅¨σ΅¨π’ (π‘, π₯) β V (π, π¦)σ΅¨σ΅¨σ΅¨ ππ‘ + sign (π’ (π‘, π₯) β V (π, π¦))
(24)
2 π’2 (π‘, π₯) V (π, π¦) Γ( β ) ππ₯ 2 2
We note that the first two terms in the integrand of (23) can be represented in the form
+ sign (π’ (π‘, π₯) β V (π, π¦))
πβ = π (π‘, π₯, π, π¦, π’ (π‘, π₯) , V (π, π¦)) π β (β) .
Γ ππ’ (π‘, π₯) π} ππ₯ ππ‘ ππ¦ ππ = 0. (21) Similarly, it has ββ«
ππ Γππ
Since βπ’βπΏβ β€ βπ’0 βπ»1 (π
) and βVβπΏβ β€ βV0 βπ»1 (π
) , from Lemma 2, we know πβ satisfies the Lipschitz condition in π’ and V, respectively. By the choice of π, we have πβ = 0 outside the region {(π‘, π₯; π, π¦)} = {πΏ β€
σ΅¨ σ΅¨ { σ΅¨σ΅¨σ΅¨V (π, π¦) β π’ (π‘, π₯) ππ σ΅¨σ΅¨σ΅¨ + sign (V (π, π¦) β π’ (π‘, π₯)) Γ(
V2 (π, π¦) π’2 (π‘, π₯) β ) ππ¦ 2 2
π‘+π |π‘ β π| β€ π β 2πΏ, β€ β, 2 2
σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨π₯ + π¦σ΅¨σ΅¨σ΅¨ σ΅¨π₯ β π¦σ΅¨σ΅¨σ΅¨ β€ π β 2πΏ, σ΅¨ β€ β} , 2 2 ββ«
ππ Γππ
+ sign (V (π, π¦) β π’ (π‘, π₯))
(25)
πβ ππ₯ ππ‘ ππ¦ ππ
= ββ«
[π (π‘, π₯, π, π¦, π’ (π‘, π₯) , V (π, π¦))
ππ Γππ
Γ πV (π, π¦) π} ππ₯ ππ‘ ππ¦ ππ = 0,
(26)
βπ (π‘, π₯, π‘, π₯, π’ (π‘, π₯) , V (π‘, π₯)) ]
(22)
Γ π β (β) ππ₯ ππ‘ ππ¦ ππ
from which we obtain 0 β€ ββ«
ππ Γππ
+ ββ«
ππ Γππ
σ΅¨ σ΅¨ { σ΅¨σ΅¨σ΅¨π’ (π‘, π₯) β V (π, π¦)σ΅¨σ΅¨σ΅¨ (ππ‘ + ππ )
Γ π β (β) ππ₯ ππ‘ ππ¦ ππ
+ sign (π’ (π‘, π₯) β V (π, π¦)) Γ(
= π½1 (β) + π½2 .
2
Considering the estimate |π(β)| β€ π/β2 and the expression of function πβ , we have σ΅¨σ΅¨σ΅¨π½1 (β)σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨
π’2 (π‘, π₯) V (π, π¦) β ) 2 2
Γ (ππ₯ + ππ¦ ) } ππ₯ ππ‘ ππ¦ ππ
(23)
σ΅¨σ΅¨ σ΅¨ sign (π’ (π‘, π₯) β V (π‘, π₯)) + σ΅¨σ΅¨σ΅¨σ΅¨ββ« σ΅¨σ΅¨ ππ Γππ
σ΅¨σ΅¨ σ΅¨ Γ (ππ’ (π‘, π₯) β πV (π, π¦)) π ππ₯ ππ‘ ππ¦ ππσ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨
= ββ«
ππ Γππ
(πΌ1 + πΌ2 + πΌ3 ) ππ₯ ππ‘ ππ¦ ππ.
We will show that 0 β€ β¬ { |π’ (π‘, π₯) β V (π‘, π₯)| ππ‘ + sign (π’ (π‘, π₯) β V (π‘, π₯)) ππ
Γ(
π’2 (π‘, π₯) V2 (π‘, π₯) β ) ππ₯ } ππ₯ ππ‘ 2 2
π (π‘, π₯, π‘, π₯, π’ (π‘, π₯) , V (π‘, π₯))
[ [ 1 [ β€ π [β + 2 [ β [ [ Γ ββ«
|(π‘βπ)/2|β€β, πΏβ€(π‘+π)/2β€πβπΏ, |(π₯βπ¦)/2|β€β, |(π₯+π¦)/2|β€πβπΏ
σ΅¨σ΅¨ σ΅¨σ΅¨V (π‘, π₯)
] ] ] σ΅¨σ΅¨ βV (π, π¦)σ΅¨σ΅¨ ππ₯ ππ‘ ππ¦ ππ, ] , ] ] ] (27)
Journal of Function Spaces and Applications
5
where the constant π does not depend on β. Using Lemma 1, we obtain π½1 (β) β 0 as β β 0. The integral π½2 does not depend on β. In fact, substituting π‘ = πΌ, (π‘ β π)/2 = π½, π₯ = π, (π₯ β π¦)/2 = π and noting that β
By Lemmas 1 and 3, we have πΎ1 (β) β 0 as β β 0. Using (28), we have πΎ2 = 22 β¬ πΌ3 (πΌ, π, πΌ, π, π’ (πΌ, π) , V (πΌ, π)) ππ
β
β« β«
ββ ββ
π β (π½, π) ππ ππ½ = 1,
(28)
β
Γ {β« β«
β
ββ ββ
we have π½2 = 22 β¬ πβ (πΌ, π, πΌ, π, π’ (πΌ, π) , V (πΌ, π))
= 4β¬ πΌ3 (π‘, π₯, π‘, π₯, π’ (π‘, π₯) , V (π‘, π₯)) ππ₯ ππ‘
ππ
β
Γ {β« β«
β
ββ ββ
π β (π½, π) ππ ππ½} ππ ππΌ
= 4β¬ sign (π’ (π‘, π₯) β V (π‘, π₯))
(29)
ππ
Γ (ππ’ (π‘, π₯) β πV (π‘, π₯)) π (π‘, π₯) ππ₯ ππ‘.
ππ
From (30) and (33), we prove that inequality (24) holds. Let
Hence lim ββ«
ππ Γππ
πβ ππ₯ ππ‘ ππ¦ ππ
π (π‘) = β«
(30)
= 4β¬ πβ (π‘, π₯, π‘, π₯, π’ (π‘, π₯) , V (π‘, π₯)) ππ₯ ππ‘.
ββ
Since πΌ3 = sign (π’ (π‘, π₯) β V (π, π¦)) (ππ’ (π‘, π₯) β πV (π, π¦)) ππ β (β) , ππ Γππ
ππ Γππ
πβ = β«
ππ Γππ
π
ββ
πβ (π) ππ
(πβσΈ (π) = πβ (π) β₯ 0)
π = [πβ (π‘ β π1 ) β πβ (π‘ β π2 )] π (π‘, π₯) ,
[πΌ3 (π‘, π₯, π, π¦) β πΌ3 (π‘, π₯, π‘, π₯)]
β < min (π1 , π0 β π2 ) ,
Γ π β (β) ππ₯ ππ‘ ππ¦ ππ + ββ«
|π’ (π‘, π₯) β V (π‘, π₯)| ππ₯.
(35)
(36)
where
πΌ3 (π‘, π₯, π‘, π₯) π β (β) ππ₯ ππ‘ ππ¦ ππ
π (π‘, π₯) = ππ (π‘, π₯) = 1 β ππ (|π₯| + ππ‘ β π0 + π) ,
= πΎ1 (β) + πΎ2 , (31) we obtain σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨πΎ1 (β)σ΅¨σ΅¨σ΅¨
π > 0.
1 β2
(37)
We note that function π(π‘, π₯) = 0 outside the cone β§ and π(π‘, π₯) = 0 outside the set z. For (π‘, π₯) β β§, we have the relations σ΅¨ σ΅¨ 0 = ππ‘ + π σ΅¨σ΅¨σ΅¨ππ₯ σ΅¨σ΅¨σ΅¨ β₯ ππ‘ + πππ₯ .
β€ π (β +
(34)
and choose two numbers π1 and π2 β (0, π0 ), π1 < π2 . In (24), we choose
πΌ3 ππ₯ ππ‘ ππ¦ ππ
= ββ«
β
We define
ππ
ββ«
(33)
ππ
= 4β¬ πβ (π‘, π₯, π‘, π₯, π’ (π‘, π₯) , V (π‘, π₯)) ππ₯ ππ‘.
ββ0
π β (π½, π) ππ ππ½} ππ ππΌ
(38)
Applying (24) and (35)β(38), we have the inequality 0β€β¬
Γ ββ«
|(π‘βπ)/2|β€β, πΏβ€(π‘+π)/2β€πβπΏ, |(π₯βπ¦)/2|β€β, |(π₯+π¦)/2|β€πβπΏ
σ΅¨σ΅¨ σ΅¨σ΅¨πV (π‘, π₯)
ππ0
{[πβ (π‘ β π1 ) β πβ (π‘ β π2 )] Γ ππ |π’ (π‘, π₯) β V (π‘, π₯)|} ππ₯ ππ‘
σ΅¨σ΅¨ σ΅¨ + σ΅¨σ΅¨σ΅¨σ΅¨β¬ [πβ (π‘ β π1 ) β πβ (π‘ β π2 )] σ΅¨σ΅¨ ππ0 σ΅¨ βπV (π, π¦)σ΅¨σ΅¨σ΅¨ ππ₯ ππ‘ ππ¦ ππ) . (32)
σ΅¨σ΅¨ σ΅¨ Γ [ππ’ (π‘, π₯) β πV (π‘, π₯)] π΅ (π‘, π₯) π (π‘, π₯) ππ₯ ππ‘σ΅¨σ΅¨σ΅¨σ΅¨ , σ΅¨σ΅¨ (39) where π΅(π‘, π₯) = sign[π’(π‘, π₯) β V(π‘, π₯)].
6
Journal of Function Spaces and Applications From (39), we obtain
0β€β¬
ππ0
Using the similar proof of (43), we get π0
{[πβ (π‘ β π1 ) β πβ (π‘ β π2 )]
πΏσΈ (π1 ) = β β« πβ (π‘ β π1 ) π (π‘) ππ‘ σ³¨β βπ (π1 ) 0
Γ ππ |π’ (π‘, π₯) β V (π‘, π₯)|} ππ₯ ππ‘ + β« (πβ (π‘ β π1 ) β πβ (π‘ β π2 ))
π1
πΏ (π1 ) σ³¨β πΏ (0) β β« π (π) ππ
0
σ΅¨σ΅¨ β σ΅¨σ΅¨ σ΅¨ σ΅¨ Γ σ΅¨σ΅¨σ΅¨β« [ππ’ (π‘, π₯) β πV (π‘, π₯)] π΅ (π‘, π₯) π (π‘, π₯) ππ₯σ΅¨σ΅¨σ΅¨ ππ‘. σ΅¨σ΅¨ ββ σ΅¨σ΅¨ (40)
0
π2
πΏ (π2 ) σ³¨β πΏ (0) β β« π (π) ππ 0
π2
πΏ (π1 ) β πΏ (π2 ) σ³¨β β« π (π) ππ π1
Γ ππ |π’ (π‘, π₯) β V (π‘, π₯)|} ππ₯ ππ‘ π0
(41)
0
Γ (πΊ + β«
β
ββ
β
|π’ β V| ππ₯) ππ‘,
0
π1
β σ΅¨σ΅¨ 2 σ΅¨ σ΅¨ 2 2 σ΅¨σ΅¨ σ΅¨σ΅¨π’0 β V02 σ΅¨σ΅¨σ΅¨ ππ₯ + β« σ΅¨σ΅¨σ΅¨π’0π₯ σ΅¨σ΅¨ ππ₯) (π2 β π1 ) . β V0π₯ σ΅¨ σ΅¨ σ΅¨ σ΅¨ ββ ββ (49)
σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨π’ (π1 , π₯) β V (π1 , π₯)σ΅¨σ΅¨σ΅¨ β€ σ΅¨σ΅¨σ΅¨π’ (π1 , π₯) β π’0 (π₯)σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨ + σ΅¨σ΅¨σ΅¨V (π1 , π₯) β V0 (π₯)σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨ + σ΅¨σ΅¨σ΅¨π’0 (π₯) β V0 (π₯)σ΅¨σ΅¨σ΅¨ .
0
|π’ (π‘, π₯) β V (π‘, π₯)| ππ₯} ππ‘ (42)
+ π β« (πβ (π‘ β π1 ) β πβ (π‘ β π2 )) 0
Γ (πΊ + β«
ββ
σ΅¨σ΅¨ σ΅¨σ΅¨ π0 σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨β« πβ (π‘ β π1 ) π (π‘) ππ‘ β π (π1 )σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ 0 σ΅¨ σ΅¨ σ΅¨σ΅¨ π0 σ΅¨σ΅¨ σ΅¨ σ΅¨ = σ΅¨σ΅¨σ΅¨σ΅¨β« πβ (π‘ β π1 ) [π (π‘) β π (π1 )] ππ‘σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ 0 σ΅¨σ΅¨
β
ββ
By the properties of the function πβ (π) for β β€ min(π1 , π0 β π1 ), we have
(43)
(50)
Thus, from (42), (43), (48), (49), and (50), for any π‘ β [0, π0 ], we have β«
|π’ β V| ππ₯) ππ‘.
β
Let π1 β 0 and π2 β π‘, and note that
π0
β
(48)
π2
= (β«
π0
as β σ³¨β 0.
π0
0 β€ β« { [πβ (π‘ β π1 ) β πβ (π‘ β π2 )]
ββ
(47)
Furthermore, if β β 0, we have
β
Γβ«
as β σ³¨β 0.
β« (πβ (π‘ β π1 ) β πβ (π‘ β π2 )) πΊ ππ‘ σ³¨β πΊ β« ππ‘
2 2 where πΊ = β«ββ |π’02 βV02 |ππ₯+β«ββ |π’0π₯ βV0π₯ |ππ₯ and π is defined in Lemma 5. Letting π β 0 in (41) and sending π0 β β, we have
β
(46)
Then, we get
{[πβ (π‘ β π1 ) β πβ (π‘ β π2 )]
+ π β« (πβ (π‘ β π1 ) β πβ (π‘ β π2 ))
as β σ³¨β 0.
Similarly, we have
Using Lemma 5, we have
ππ0
(45)
from which we obtain
π0
0β€β¬
as β σ³¨β 0,
|π’ (π‘, π₯) β V (π‘, π₯)| ππ₯ β
β€β«
ββ
|π’ (0, π₯) β V (0, π₯)| ππ₯ β
β σ΅¨ σ΅¨ σ΅¨ 2 2 σ΅¨σ΅¨ σ΅¨σ΅¨ ππ₯) + ππ0 (β« σ΅¨σ΅¨σ΅¨σ΅¨π’02 β V02 σ΅¨σ΅¨σ΅¨σ΅¨ ππ₯ + β« σ΅¨σ΅¨σ΅¨σ΅¨π’0π₯ β V0π₯ σ΅¨ ββ ββ π‘
β
0
ββ
+β« β«
(51)
|π’ β V| ππ₯,
from which we complete the proof of Theorem 6 by using the Gronwall inequality.
1 π1 +β σ΅¨σ΅¨ σ΅¨ β€π β« σ΅¨π (π‘) β π (π1 )σ΅¨σ΅¨σ΅¨ ππ‘ σ³¨β 0 as β σ³¨β 0, β π1 ββ σ΅¨
Acknowledgments
where π is independent of β. Set π0
π0
π‘βπ1
0
0
ββ
πΏ (π1 ) = β« πβ (π‘ β π1 ) π (π‘) ππ‘ = β« β«
πβ (π) πππ (π‘) ππ‘. (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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