Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 810352, 8 pages http://dx.doi.org/10.1155/2014/810352
Research Article A Compact Difference Scheme for a Class of Variable Coefficient Quasilinear Parabolic Equations with Delay Wei Gu School of Statistics and Mathematics, Zhongnan University of Economics and Law, Wuhan, Hubei 430073, China Correspondence should be addressed to Wei Gu; wei
[email protected] Received 19 February 2014; Accepted 15 May 2014; Published 5 June 2014 Academic Editor: Zhongxiao Jia Copyright © 2014 Wei Gu. 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. A linearized compact difference scheme is provided for a class of variable coefficient parabolic systems with delay. The unique solvability, unconditional stability, and convergence of the difference scheme are proved, where the convergence order is four in space and two in time. A numerical test is presented to illustrate the theoretical results.
𝑢 (𝑥, 𝑡) = 𝜙 (𝑥, 𝑡) ,
1. Introduction From the twentieth century, more and more scholars have been attracted into the research on the theory of delay differential equations (DDEs) [1–4]. As we know, most DDEs have no analytical solutions; efficient numerical methods solving for DDEs and delay partial differential equations (DPDEs) need to be considered deeply. Recently, many scholars consider the numerical investigation on DPDEs. For instance, Marzban and Tabrizidooz [5] considered a hybrid approximation method for solving Hutchinson’s equation; Jackiewicz and Zubik-Kowal [6] considered Chebyshev spectral collocation and waveform relaxation methods for nonlinear DPDEs and finite difference methods were considered to solve delay parabolic partial differential equations in [7– 9]; Li et al. [10–12] constructed finite element methods to solve reaction-diffusion equations with delay. The numerical research of DPDEs focused on stability analysis can be referred to in [13]. The following variable coefficient parabolic systems with delay are considered in this paper:
𝑟 (𝑥, 𝑡) 𝑢𝑡 − 𝑑𝑢𝑥𝑥 = 𝑓 (𝑢 (𝑥, 𝑡) , 𝑢 (𝑥, 𝑡 − 𝑠) , 𝑥, 𝑡) , (𝑥, 𝑡) ∈ (0, 1) × (0, 𝑇] ,
(1)
𝑢 (0, 𝑡) = 𝛼 (𝑡) ,
𝑥 ∈ [0, 1] , 𝑡 ∈ [−𝑠, 0] ,
𝑢 (1, 𝑡) = 𝛽 (𝑡) ,
𝑡 ∈ (0, 𝑇] ,
(2) (3)
where 𝑑 > 0 is a constant and 𝑠 > 0 is the delay term, 𝑟(𝑥, 𝑡) ∈ 𝐶((0, 1) × (0, 𝑇]), 0 < 𝑐0 ≤ 𝑟(𝑥, 𝑡) ≤ 𝑐1 . In the special case of 𝑟(𝑥, 𝑡) = 1, numerical solutions of (1)–(3) have been considered in [14–17]. Ferreira and da Silva considered a backward Euler scheme and proved the stability and convergence by the energy method in [14]. A Crank-Nicolson scheme and a linearized compact difference scheme were proposed by Zhang and Sun in [15] and Sun and Zhang in [16], respectively. Q. Zhang and C. Zhang considered a new linearized compact multisplitting scheme in [17]. Gu and Wang constructed a Crank-Nicolson scheme in [18] to solve a special case of (1), where 𝑓 = 𝑓(𝑢(𝑥, 𝑡 − 𝑠)). In this paper, a linearized compact difference scheme solving for (1)–(3) will be constructed. The unique solvability, unconditional stability, and convergence of the difference scheme are proved, where the convergence order is four in space and two in time. A numerical test is presented to illustrate the theoretical results. The paper is organized as follows. In Section 2, a linearized compact difference scheme is constructed to solve (1)–(3). Section 3 considers the solvability, stability, and convergence of the provided difference scheme.
2
Abstract and Applied Analysis
In Section 4, a numerical test is presented to illustrate the theoretical results. Section 5 gives a brief discussion of this paper.
From Taylor expansion, we have 𝜏 2 𝜕3 𝑢 𝜕𝑢 (𝑥 , 𝜂𝑘 ) , (𝑥𝑖 , 𝑡𝑘+(1/2) ) = 𝛿𝑡 𝑈𝑖𝑘+(1/2) − 𝜕𝑡 24 𝜕𝑡3 𝑖 𝑖 𝜂𝑖𝑘 ∈ (𝑡𝑘 , 𝑡𝑘+1 ) ,
2. The Compact Difference Scheme and Local Truncation Error Throughout this paper, the following assumptions are assumed to be true.
𝜕2 𝑢 (𝑥 , 𝑡 ) 𝜕𝑥2 𝑖 𝑘+(1/2)
(H1) Let 𝑚 be an integer satisfying 𝑚𝑠 ≤ 𝑇 < (𝑚 + 1)𝑠, denote 𝐼𝑙 = (𝑙𝑠, 𝑙𝑠 + 𝑠), 𝑙 = −1, 0, . . . , 𝑚 − 1, 𝐼𝑚 = (𝑚𝑠, 𝑇), and 𝐼 = ∪𝑚 𝑝=−1 𝐼𝑝 , assume that (1)–(3) has a unique solution 𝑢 ∈ 𝐶6,4 (𝐼 × (0, 𝑇]) and that 𝑢 and its partial derivatives are all bounded by a constant 𝑐2 ;
=
(H2) 𝑓(𝜇, ], 𝑥, 𝑡) has bounded first-order continuous partial derivatives, and we denote 𝑓 (𝑢 (𝑥, 𝑡) + 𝜀 , 𝑢 (𝑥, 𝑡 − 𝑠) + 𝜀 , 𝑥, 𝑡) , max 𝑐3 = 𝜇 1 2 |𝜀1 |≤𝜀0 , |𝜀2 |≤𝜀0 (4) 𝑐4 = max 𝑓] (𝑢 (𝑥, 𝑡) + 𝜀1 , 𝑢 (𝑥, 𝑡 − 𝑠) + 𝜀2 , 𝑥, 𝑡) , |𝜀1 |≤𝜀0 , |𝜀2 |≤𝜀0
=
where 𝜀0 > 0, 𝑐3 , and 𝑐4 are constants, (𝑥, 𝑡) ∈ (0, 1) × (0, 𝑇]. First let 𝑀 and 𝑗 be two positive integers; then, we take ℎ = 1/𝑀, 𝜏 = 𝑠/𝑗, 𝑥𝑖 = 𝑖ℎ, 𝑡𝑘 = 𝑘𝜏, 𝑡𝑘+1/2 = (𝑡𝑘 + 𝑡𝑘+1 )/2. Define Ωℎ𝜏 = Ωℎ × Ω𝜏 , where Ωℎ = {𝑥𝑖 | 0 ≤ 𝑖 ≤ 𝑀}, Ω𝜏 = {𝑡𝑘 | −𝑗 ≤ 𝑘 ≤ 𝑁}, 𝑁 = [𝑇/𝜏]. Denote 𝑈𝑖𝑘 = 𝑢(𝑥𝑖 , 𝑡𝑘 ), 0 ≤ 𝑖 ≤ 𝑀, −𝑗 ≤ 𝑘 ≤ 𝑁, throughout this paper. Let W = {V𝑖𝑘 | 0 ≤ 𝑖 ≤ 𝑀, −𝑗 ≤ 𝑘 ≤ 𝑁}
V𝑖𝑘+(1/2) =
V𝑖𝑘
𝛿𝑡 V𝑖𝑘+(1/2) = 𝑘 = 𝛿𝑥 V𝑖+(1/2)
V𝑖𝑘+1
+ 2
,
V𝑖𝑘+1 − V𝑖𝑘 , 𝜏 𝑘 V𝑖+1
− ℎ
V𝑖𝑘
,
(6)
ℎ2 𝜕4 𝑢 𝑘 𝜕4 𝑢 [ 4 (𝜉𝑖 , 𝑡𝑘 ) + 4 (𝜉𝑖𝑘+1 , 𝑡𝑘+1 )] 24 𝜕𝑥 𝜕𝑥
−
𝜏 2 𝜕4 𝑢 (𝑥 , 𝛾𝑘 ) , 8 𝜕𝑥2 𝜕𝑡2 𝑖 𝑖
𝜉𝑖𝑘 , 𝜉𝑖𝑘+1 ∈ (𝑥𝑖−1 , 𝑥𝑖+1 ) ,
(8)
𝛾𝑖𝑘 ∈ (𝑡𝑘 , 𝑡𝑘+1 ) ,
𝑓 (𝑢 (𝑥𝑖 , 𝑡𝑘+(1/2) ) , 𝑢 (𝑥𝑖 , 𝑡𝑘+(1/2)−𝑗 ) , 𝑥𝑖 , 𝑡𝑘+(1/2) ) 3 1 1 𝑘+1−𝑗 1 𝑘−𝑗 = 𝑓 ( 𝑈𝑖𝑘 − 𝑈𝑖𝑘−1 , 𝑈𝑖 + 𝑈𝑖 , 𝑥𝑖 , 𝑡𝑘+(1/2) ) 2 2 2 2 2 𝑘 3𝜏2 𝜕 𝑢 (𝑥𝑖 , 𝜌 ) + 𝑓𝜇 (𝜁𝑖𝑘 , 𝜍𝑖𝑘 , 𝑥𝑖 , 𝑡𝑘+(1/2) ) 8 𝜕𝑡2
−
2 𝑘 𝜏2 𝜕 𝑢 (𝑥𝑖 , ) 𝑓] (𝜁𝑖𝑘 , 𝜍𝑖𝑘 , 𝑥𝑖 , 𝑡𝑘+(1/2) ) , 8 𝜕𝑡2
where 𝜌𝑘 ∈ (𝑡𝑘−1 , 𝑡𝑘+1/2 ), 𝑘 ∈ (𝑡𝑘−𝑗 , 𝑡𝑘+1−𝑗 ), 𝜁𝑖𝑘 is between 𝑢(𝑥𝑖 , 𝑡𝑘+(1/2) ) and (3/2)𝑈𝑖𝑘 − (1/2)𝑈𝑖𝑘−1 , and 𝜍𝑖𝑘 is between 𝑘+1−𝑗 𝑘−𝑗 𝑢(𝑥𝑖 , 𝑡𝑘+(1/2)−𝑗 ) and (1/2)𝑈𝑖 + (1/2)𝑈𝑖 . Substituting (8) into (7), denote 𝑟𝑖𝑘+(1/2) = 𝑟(𝑥𝑖 , 𝑡𝑘+(1/2) ); we obtain 𝑟𝑖𝑘+(1/2) 𝛿𝑡 𝑈𝑖𝑘+(1/2) −
𝑑 𝜕2 𝑢 𝜕2 𝑢 [ 2 (𝑥𝑖 , 𝑡𝑘 ) + 2 (𝑥𝑖 , 𝑡𝑘+1 )] 2 𝜕𝑥 𝜕𝑥
1 ≤ 𝑖 ≤ 𝑀, 0 ≤ 𝑘 ≤ 𝑁 − 1,
where 𝑘
𝑅𝑖 =
𝜕𝑢 𝜕2 𝑢 (𝑥𝑖 , 𝑡𝑘+(1/2) ) − 𝑑 2 (𝑥𝑖 , 𝑡𝑘+(1/2) ) 𝜕𝑡 𝜕𝑥
0 ≤ 𝑖 ≤ 𝑀, 0 ≤ 𝑘 ≤ 𝑁 − 1.
−
𝑘
1 𝑘 𝑘 = ). (V + 10V𝑘𝑘 + V𝑖+1 12 𝑖−1
= 𝑓 (𝑢 (𝑥𝑖 , 𝑡𝑘+(1/2) ) , 𝑢 (𝑥𝑖 , 𝑡𝑘+(1/2)−𝑗 ) , 𝑥𝑖 , 𝑡𝑘+(1/2) ) ,
1 2 𝑘 (𝛿 𝑈 + 𝛿𝑥2 𝑈𝑖𝑘+1 ) 2 𝑥 𝑖
+ 𝜏2 𝑅𝑖 ,
Considering (1) at the point (𝑥𝑖 , 𝑡𝑘+(1/2) ), we have 𝑟 (𝑥𝑖 , 𝑡𝑘+(1/2) )
𝜏2 𝜕4 𝑢 (𝑥 , 𝛾𝑘 ) 8 (𝜕𝑥2 𝜕𝑡2 ) 𝑖 𝑖
1 1 𝑘+1−𝑗 1 𝑘−𝑗 3 (9) + 𝑈𝑖 , 𝑥𝑖 , 𝑡𝑘+(1/2) ) = 𝑓 ( 𝑈𝑖𝑘 − 𝑈𝑖𝑘−1 , 𝑈𝑖 2 2 2 2
𝑘 V𝑘 − 2V𝑖𝑘 + V𝑖−1 , 𝛿𝑥2 V𝑖𝑘 = 𝑖+1 ℎ2
AV𝑖𝑘
−
(5)
be the grid function space defined on Ωℎ𝜏 . The following notations are made:
1 𝜕2 𝑢 𝜕2 𝑢 [ 2 (𝑥𝑖 , 𝑡𝑘 ) + 2 (𝑥𝑖 , 𝑡𝑘+1 )] 2 𝜕𝑥 𝜕𝑥
𝑟𝑖𝑘+(1/2) 𝜕3 𝑢 𝑑 𝜕4 𝑢 𝑘 (𝑥 , 𝜂 ) − (𝑥 , 𝛾𝑘 ) 𝑖 𝑖 24 𝜕𝑡3 8 𝜕𝑥2 𝜕𝑡2 𝑖 𝑖 +
2 𝑘 3 𝜕 𝑢 (𝑥𝑖 , 𝜌 ) 𝑓𝜇 (𝜁𝑖𝑘 , 𝜍𝑖𝑘 , 𝑥𝑖 , 𝑡𝑘+(1/2) ) 8 𝜕𝑡2
−
2 𝑘 1 𝜕 𝑢 (𝑥𝑖 , ) 𝑓] (𝜁𝑖𝑘 , 𝜍𝑖𝑘 , 𝑥𝑖 , 𝑡𝑘+(1/2) ) . 8 𝜕𝑡2
(7)
(10)
Abstract and Applied Analysis
3
Acting operator A on both sides of (9), we have
Discretizing the initial and boundary conditions of (2) and (3), we obtain
A𝑟𝑖𝑘+(1/2) 𝛿𝑡 𝑈𝑖𝑘+(1/2) 2
𝑑 𝜕𝑢 𝜕𝑢 − [A 2 (𝑥𝑖 , 𝑡𝑘 ) + A 2 (𝑥𝑖 , 𝑡𝑘+1 )] 2 𝜕𝑥 𝜕𝑥 3 1 1 𝑘+1−𝑗 1 𝑘−𝑗 = A𝑓 ( 𝑈𝑖𝑘 − 𝑈𝑖𝑘−1 , 𝑈𝑖 + 𝑈𝑖 , 𝑥𝑖 , 𝑡𝑘+(1/2) ) 2 2 2 2 𝑘
+ 𝜏2 A𝑅𝑖 ,
𝑈𝑖𝑘 = 𝜙 (𝑥𝑖 , 𝑡𝑘 ) ,
2
(11)
𝑈0𝑘 = 𝛼 (𝑡𝑘 ) ,
𝑘 𝑈𝑀 = 𝛽 (𝑡𝑘 ) ,
3 1 1 𝑘+1−𝑗 1 𝑘−𝑗 = A𝑓 ( 𝑢𝑖𝑘 − 𝑢𝑖𝑘−1 , 𝑢𝑖 + 𝑢𝑖 , 𝑥𝑖 , 𝑡𝑘+(1/2) ) , 2 2 2 2 1 ≤ 𝑖 ≤ 𝑀 − 1, 0 ≤ 𝑘 ≤ 𝑁 − 1, (20)
Lemma 1 (see [19, 20]). Suppose that 𝑞(𝑥) ∈ 𝐶6 [𝑥𝑖−1 , 𝑥𝑖+1 ]; then, we have 1 [𝑞 (𝑥𝑖−1 ) + 10𝑞 (𝑥𝑖 ) + 𝑞 (𝑥𝑖+1 )] 12
𝑢𝑖𝑘 = 𝜙 (𝑥𝑖 , 𝑡𝑘 ) , 𝑢0𝑘 = 𝛼 (𝑡𝑘 ) ,
(12)
𝜃𝑖𝑘 ∈ (𝑥𝑖−1 , 𝑥𝑖+1 ) , ℎ4 𝜕6 𝑢 𝑘+1 𝜕2 𝑢 2 𝑘+1 (𝑥 , 𝑡 ) = 𝛿 𝑈 + (𝜃 , 𝑡𝑘+1 ) , 𝑥 𝑖 𝜕𝑥2 𝑖 𝑘+1 240 𝜕𝑥6 𝑖
𝑀−1 𝑖=1
(13)
𝑀
|V|1 = √ ℎ∑( 𝑖=1
V𝑖 − V𝑖−1 2 ), ℎ
(24)
‖V‖∞ = max V𝑖 . 0≤𝑖≤𝑀
Inserting (13) into (11), we have
By [16, 17, 19], we have the following two inequalities:
A𝑟𝑖𝑘+(1/2) 𝛿𝑡 𝑈𝑖𝑘+(1/2) − 𝑑𝛿𝑥2 𝑈𝑖𝑘+1/2 3 1 1 𝑘+1−𝑗 1 𝑘−𝑗 = A𝑓 ( 𝑈𝑖𝑘 − 𝑈𝑖𝑘−1 , 𝑈𝑖 + 𝑈𝑖 , 𝑥𝑖 , 𝑡𝑘+(1/2) ) (14) 2 2 2 2
‖V‖∞ ≤ ‖V‖ ≤
1 ≤ 𝑖 ≤ 𝑀 − 1, 0 ≤ 𝑘 ≤ 𝑁 − 1,
where 6
𝑑ℎ 𝜕 𝑢 𝑘 𝜕𝑢 (𝜃 , 𝑡 ) + 6 (𝜃𝑖𝑘+1 , 𝑡𝑘+1 )] . [ 480 𝜕𝑥6 𝑖 𝑘 𝜕𝑥 (15)
From 0 < 𝑐0 ≤ 𝑟𝑖𝑘+1/2 ≤ 𝑐1 and assumptions (H1) and (H2), we have 𝑘 𝑅 ≤ 𝑐5 , 1 ≤ 𝑖 ≤ 𝑀 − 1, 0 ≤ 𝑘 ≤ 𝑁 − 1, (16) 𝑖 such that 𝑘 𝑅𝑖 ≤ 𝑐6 (𝜏2 + ℎ4 ) ,
(23)
2 ‖V‖ = √ ℎ ∑ (V𝑖 ) ,
𝜃𝑖𝑘+1 ∈ (𝑥𝑖−1 , 𝑥𝑖+1 ) .
𝑘
(22)
If V ∈ 𝑉, we introduce the following notations:
𝜕2 𝑢 ℎ4 𝜕6 𝑢 𝑘 A 2 (𝑥𝑖 , 𝑡𝑘 ) = 𝛿𝑥2 𝑈𝑖𝑘 + (𝜃 , 𝑡 ) , 𝜕𝑥 240 𝜕𝑥6 𝑖 𝑘
𝑅𝑖𝑘 = 𝜏2 A𝑅𝑖 +
1 ≤ 𝑘 ≤ 𝑁.
𝑉 = {V | V = (V0 , V1 , . . . , V𝑀) , V0 = V𝑀 = 0} .
From Lemma 1 and Taylor expansion, we obtain
6
𝑘 𝑢𝑀 = 𝛽 (𝑡𝑘 ) ,
(21)
Define the following grid function space on Ωℎ :
where 𝜔𝑖 ∈ (𝑥𝑖−1 , 𝑥𝑖+1 ).
4
0 ≤ 𝑖 ≤ 𝑀, −𝑗 ≤ 𝑘 ≤ 0,
3. The Solvability, Convergence, and Stability of the Compact Difference Scheme
ℎ4 (6) = 𝑞 (𝜔𝑖 ) , 240
+ 𝑅𝑖𝑘 ,
(19)
Replacing 𝑈𝑖𝑘 by 𝑢𝑖𝑘 in (14) and omitting 𝑅𝑖𝑘 , we obtain the following compact difference scheme:
Resorting to the following Lemma, we can obtain the estimation of the operator A.
A
1 ≤ 𝑘 ≤ 𝑁.
(18)
A𝑟𝑖𝑘+(1/2) 𝛿𝑡 𝑢𝑖𝑘+(1/2) − 𝑑𝛿𝑥2 𝑢𝑖𝑘+1/2
1 ≤ 𝑖 ≤ 𝑀 − 1, 0 ≤ 𝑘 ≤ 𝑁 − 1.
1 − 2 [𝑞 (𝑥𝑖−1 ) − 2𝑞 (𝑥𝑖 ) + 𝑞 (𝑥𝑖+1 )] ℎ
0 ≤ 𝑖 ≤ 𝑀, −𝑗 ≤ 𝑘 ≤ 0,
1 |V| , 2 1
(25)
1 |V| . √6 1
(26)
For the analysis of the difference scheme, the following Lemma is introduced. Lemma 2 (see [16, 17, 19]). Assume that {𝐹𝑘 | 𝑘 ≥ 0} to be nonnegative sequence and satisfies 𝑘
𝐹𝑘+1 ≤ 𝐴 + 𝐵𝜏∑𝐹𝑙 ,
(27)
𝑘 = 0, 1, 2, . . . ,
(28)
then 𝐹𝑘+1 ≤ 𝐴 exp (𝐵𝑘𝜏) ,
1 ≤ 𝑖 ≤ 𝑀 − 1, 0 ≤ 𝑘 ≤ 𝑁 − 1. (17)
𝑘 = 0, 1, . . . ;
𝑖=1
where 𝐴 and 𝐵 are nonnegative constants.
4
Abstract and Applied Analysis
Theorem 3. Under the condition that 5𝑐0 −𝑐1 > 0, the compact difference scheme (20)–(22) has a unique solution. Proof. Denote that 𝜆 = 𝑑𝜏/ℎ2 ; then, difference scheme (20)– (22) can be reformed as
Denote 𝑒𝑖𝑘 = 𝑈𝑖𝑘 − 𝑢𝑖𝑘 , 0 ≤ 𝑖 ≤ 𝑀, −𝑗 ≤ 𝑘 ≤ 𝑁; subtracting (20)–(22) from (14), (18), and (19), respectively, the following error equations can be obtained: A𝑟𝑖𝑘+(1/2) 𝛿𝑡 𝑒𝑖𝑘+(1/2) − 𝑑𝛿𝑥2 𝑒𝑖𝑘+1/2 3 1 1 𝑘+1−𝑗 1 𝑘−𝑗 = A [𝑓 ( 𝑈𝑖𝑘 − 𝑈𝑖𝑘−1 , 𝑈𝑖 + 𝑈𝑖 , 𝑥𝑖 , 𝑡𝑘+(1/2) ) 2 2 2 2
10 1 𝑘+1/2 𝜆 𝑘+1 − ) 𝑢𝑖−1 + ( 𝑟𝑖𝑘+1/2 + 𝜆) 𝑢𝑖𝑘+1 ( 𝑟𝑖−1 12 2 12
3 1 1 𝑘+1−𝑗 1 𝑘−𝑗 −𝑓 ( 𝑢𝑖𝑘 − 𝑢𝑖𝑘−1 , 𝑢𝑖 + 𝑢𝑖 , 𝑥𝑖 , 𝑡𝑘+(1/2) )] 2 2 2 2
1 𝑘+1/2 𝜆 𝑘+1 + ( 𝑟𝑖+1 − ) 𝑢𝑖+1 12 2
+ 𝑅𝑖𝑘 ,
1 𝑘+1/2 𝜆 𝑘 = ( 𝑟𝑖−1 + ) 𝑢𝑖−1 12 2
1 ≤ 𝑖 ≤ 𝑀 − 1, 0 ≤ 𝑘 ≤ 𝑁 − 1, (32) 𝑒𝑖𝑘 = 0,
10 1 𝑘+1/2 𝜆 𝑘 + ( 𝑟𝑖𝑘+1/2 − 𝜆) 𝑢𝑖𝑘 + ( 𝑟𝑖+1 + ) 𝑢𝑖+1 12 12 2 𝜏 1 𝑘−1 1 𝑘+1−𝑗 3 𝑘 + 𝑓 ( 𝑢𝑖−1 − 𝑢𝑖−1 , 𝑢𝑖−1 12 2 2 2
𝑒0𝑘 = 0,
10𝜏 1 1 𝑘+1−𝑗 3 𝑓 ( 𝑢𝑖𝑘 − 𝑢𝑖𝑘−1 , 𝑢𝑖 12 2 2 2
2 (5𝑐32 + 𝑐42 ) 3𝑇 𝐶 = 𝑐6 √ exp ( 𝑇) . 𝑑 (9𝑐0 − 𝑐1 ) 𝑑 (9𝑐0 − 𝑐1 )
𝜏≤(
𝜖0 1/2 ) , 4𝐶
(35)
𝜖0 1/4 ) , 4𝐶
(36)
0 ≤ 𝑘 ≤ 𝑁,
(37)
ℎ≤(
𝑘 𝑒 ≤ 𝐶 (𝜏2 + ℎ4 ) , ∞
1 𝑘−𝑗 + 𝑢𝑖+1 , 𝑥𝑖+1 , 𝑡𝑘+(1/2) ) . 2
where 𝜖0 > 0 is a constant.
The mathematical induction method will be used in the proof of this theorem. Denote (30)
Notice that 𝑢𝑘 (−𝑗 ≤ 𝑘 ≤ 0) is determined by the initial condition (21). Suppose that 𝑢𝑙 has been determined. Let 𝑘 = 𝑙 in (20); the linear algebraic equations with respect to 𝑢𝑙+1 can be obtained. Under the condition that 5𝑐0 − 𝑐1 > 0, we have
Proof. Acting ℎ𝛿𝑡 𝑒𝑖𝑘+(1/2) on (32) and summing up for 𝑖 from 1 to 𝑀 − 1, we obtain 𝑀−1
ℎ ∑ (A𝑟𝑖𝑘+(1/2) 𝛿𝑡 𝑒𝑖𝑘+(1/2) ) 𝛿𝑡 𝑒𝑖𝑘+(1/2) 𝑖=1
𝑀−1
− 𝑑ℎ ∑ (𝛿𝑥2 𝑒𝑖𝑘+1/2 ) 𝛿𝑡 𝑒𝑖𝑘+(1/2) 𝑖=1
𝑀−1 3 1 1 𝑘+1−𝑗 = ℎ ∑ {A [𝑓 ( 𝑈𝑖𝑘 − 𝑈𝑖𝑘−1 , 𝑈𝑖 2 2 2 𝑖=1
1 𝑘−𝑗 + 𝑈𝑖 , 𝑥𝑖 , 𝑡𝑘+(1/2) ) 2
10 𝑙+1/2 1 𝑙+1/2 𝜆 1 𝑙+1/2 𝜆 𝑟 + 𝜆 − 𝑟𝑖−1 − − 𝑟𝑖+1 − 𝑖 2 12 2 12 12 10 1 𝑙+1/2 1 𝑙+1/2 ≥ 𝑟𝑖𝑙+1/2 − 𝑟𝑖−1 − 𝑟𝑖+1 12 12 12
(34)
then we have
𝜏 1 𝑘−1 1 𝑘+1−𝑗 3 𝑘 − 𝑢𝑖+1 , 𝑢𝑖+1 𝑓 ( 𝑢𝑖+1 12 2 2 2
𝑘 ). 𝑢𝑘 = (𝑢0𝑘 , 𝑢1𝑘 , . . . , 𝑢𝑀
1 ≤ 𝑘 ≤ 𝑁.
If the following conditions are satisfied:
1 𝑘−𝑗 + 𝑢𝑖 , 𝑥𝑖 , 𝑡𝑘+(1/2) ) 2 +
𝑘 𝑒𝑀 = 0,
(33)
Theorem 4. Denote (29)
1 𝑘−𝑗 + 𝑢𝑖−1 , 𝑥𝑖−1 , 𝑡𝑘+(1/2) ) 2 +
0 ≤ 𝑖 ≤ 𝑀, −𝑗 ≤ 𝑘 ≤ 0,
3 1 1 𝑘+1−𝑗 − 𝑓 ( 𝑢𝑖𝑘 − 𝑢𝑖𝑘−1 , 𝑢𝑖 2 2 2 1 𝑘−𝑗 + 𝑢𝑖 , 𝑥𝑖 , 𝑡𝑘+(1/2) )]} 𝛿𝑡 𝑒𝑖𝑘+(1/2) 2
(31)
10𝑐0 − 2𝑐1 ≥ 12 > 0. Thus, the coefficient matrix of the linear algebraic system is strictly diagonally dominant and then there exists a unique solution 𝑢𝑙+1 . By the inductive principle, the proof ends.
𝑀−1
+ ℎ ∑ (𝑅𝑖𝑘 ) 𝛿𝑡 𝑒𝑖𝑘+(1/2) ,
1 ≤ 𝑖 ≤ 𝑀 − 1, 0 ≤ 𝑘 ≤ 𝑁 − 1.
𝑖=1
(38) Mathematical induction will be used to prove this theorem. Notice that ‖𝑒𝑘 ‖∞ = 0 (−𝑗 ≤ 𝑘 ≤ 0) and suppose that
Abstract and Applied Analysis
5
(37) is true for 0 ≤ 𝑘 ≤ 𝑙; we will show that (37) is also true for 𝑘 = 𝑙 + 1. In the following, each term of (38) will be estimated: 𝑀−1
ℎ ∑ (A𝑟𝑖𝑘+(1/2) 𝛿𝑡 𝑒𝑖𝑘+(1/2) ) 𝛿𝑡 𝑒𝑖𝑘+(1/2)
From the inequality above, we obtain 𝑀−1 3 1 1 𝑘+1−𝑗 1 𝑘−𝑗 + 𝑈𝑖 , 𝑥𝑖 , 𝑡𝑘+(1/2) ) ℎ ∑ {A [𝑓 ( 𝑈𝑖𝑘 − 𝑈𝑖𝑘−1 , 𝑈𝑖 2 2 2 2 𝑖=1
3 1 1 𝑘+1−𝑗 1 𝑘−𝑗 −𝑓 ( 𝑢𝑖𝑘 − 𝑢𝑖𝑘−1 , 𝑢𝑖 + 𝑢𝑖 , 𝑥𝑖 , 𝑡𝑘+(1/2) )]} 2 2 2 2
𝑖=1
𝑘+(1/2) 𝑘+(1/2) (𝑟𝑖−1 + 𝑟𝑖+1 ) 2 ℎ 𝑀−1 𝑘+(1/2) ≥ − ] (𝛿𝑡 𝑒𝑖𝑘+(1/2) ) ∑ [9𝑟𝑖 12 𝑖=1 2
≥
𝑀−1 3 1 𝑘+1−𝑗 1 𝑘−𝑗 1 ≤ ℎ ∑ {A (𝑐3 𝑒𝑖𝑘 − 𝑒𝑖𝑘−1 + 𝑐4 𝑒𝑖 + 𝑒𝑖 )} 2 2 2 2 𝑖=1
9𝑐0 − 𝑐1 𝑘+(1/2) 2 𝛿 𝑒 , 12 𝑡 𝑀−1
−𝑑ℎ ∑ (𝛿𝑥2 𝑒𝑖𝑘+(1/2) ) 𝛿𝑡 𝑒𝑖𝑘+(1/2) = 𝑖=1
× 𝛿𝑡 𝑒𝑖𝑘+(1/2)
𝑑 𝑘+1 2 𝑘 2 (𝑒 1 − 𝑒 1 ) , 2𝜏
× 𝛿𝑡 𝑒𝑖𝑘+(1/2) ≤
𝑀−1
ℎ ∑ (𝑅𝑖𝑘 ) 𝛿𝑡 𝑒𝑖𝑘+(1/2)
𝑀−1 3 1 𝑘+1−𝑗 1 𝑘−𝑗 2 1 × ∑ {A (𝑐3 𝑒𝑖𝑘 − 𝑒𝑖𝑘−1 + 𝑐4 𝑒𝑖 + 𝑒𝑖 )} 2 2 2 2 𝑖=1
𝑖=1
≤ ≤
𝑀−1 2 2 9𝑐 − 𝑐 𝑀−1 6 ℎ ∑ (𝑅𝑖𝑘 ) + 0 1 ℎ ∑ (𝛿𝑡 𝑒𝑖𝑘+(1/2) ) 9𝑐0 − 𝑐1 𝑖=1 24 𝑖=1 2 9𝑐 − 𝑐 6 2 𝑐2 (𝜏2 + ℎ4 ) + 0 1 𝛿𝑡 𝑒𝑘+(1/2) . 9𝑐0 − 𝑐1 6 24
+
(39)
≤
0 ≤ 𝑘 ≤ 𝑙.
6 ℎ 9𝑐0 − 𝑐1
+
(40) ≤
From (H2), we have 3 𝑘 1 𝑘−1 1 𝑘+1−𝑗 1 𝑘−𝑗 𝑓 ( 𝑈 − 𝑈 , 𝑈 + 𝑈𝑖 , 𝑥𝑖 , 𝑡𝑘+(1/2) ) 2 𝑖 2 2 𝑖 2 𝑖
3 1 1 𝑘+1−𝑗 1 𝑘−𝑗 −𝑓 ( 𝑢𝑖𝑘 − 𝑢𝑖𝑘−1 , 𝑢𝑖 + 𝑢𝑖 , 𝑥𝑖 , 𝑡𝑘+(1/2) ) 2 2 2 2 3 𝑘 1 𝑘−1 1 𝑘+1−𝑗 1 𝑘−𝑗 ≤ 𝑐3 𝑒𝑖 − 𝑒𝑖 + 𝑐4 𝑒𝑖 + 𝑒𝑖 , 2 2 2 2
9𝑐0 − 𝑐1 𝑘+(1/2) 2 𝛿 𝑒 24 𝑡 𝑖
𝑀−1 3 1 1 𝑘+1−𝑗 1 𝑘−𝑗 2 + 𝑒𝑖 ) × ∑ (𝑐3 𝑒𝑖𝑘 − 𝑒𝑖𝑘−1 + 𝑐4 𝑒𝑖 2 2 2 2 𝑖=1
From the inductive assumption and (36), we have 𝜖 𝑘 𝑒 ≤ 𝐶 (𝜏2 + ℎ4 ) ≤ 0 , ∞ 2
6 ℎ 9𝑐0 − 𝑐1
9𝑐0 − 𝑐1 𝑘+(1/2) 2 𝛿 𝑒 24 𝑡
12 9𝑐0 − 𝑐1 𝑀−1 2 3 1 × [𝑐32 ℎ ∑ ( 𝑒𝑖𝑘 − 𝑒𝑖𝑘−1 ) 2 𝑖=1 2 𝑀−1 1 𝑘+1−𝑗 1 𝑘−𝑗 2 +𝑐42 ℎ ∑ ( 𝑒𝑖 + 𝑒𝑖 ) ] 2 𝑖=1 2
(41) +
1 ≤ 𝑖 ≤ 𝑀, 0 ≤ 𝑘 ≤ 𝑙. ≤ It then follows that
9𝑐0 − 𝑐1 𝑘+(1/2) 2 𝛿 𝑒 24 𝑡
12 9𝑐0 − 𝑐1 2 2 5 𝑀−1 × [ 𝑐32 ℎ ∑ ((𝑒𝑖𝑘 ) + (𝑒𝑖𝑘−1 ) ) 2 𝑖=1
A [𝑓 ( 3 𝑈𝑘 − 1 𝑈𝑘−1 , 1 𝑈𝑘+1−𝑗 + 1 𝑈𝑘−𝑗 , 𝑥 , 𝑡 𝑖 𝑘+(1/2) ) 2 𝑖 2 𝑖 2 𝑖 2 𝑖
3 1 1 𝑘+1−𝑗 1 𝑘−𝑗 −𝑓 ( 𝑢𝑖𝑘 − 𝑢𝑖𝑘−1 , 𝑢𝑖 + 𝑢𝑖 , 𝑥𝑖 , 𝑡𝑘+(1/2) )] 2 2 2 2 3 𝑘 1 𝑘−1 1 𝑘+1−𝑗 1 𝑘−𝑗 ≤ A (𝑐3 𝑒𝑖 − 𝑒𝑖 + 𝑐4 𝑒𝑖 + 𝑒𝑖 ) , 2 2 2 2 1 ≤ 𝑖 ≤ 𝑀 − 1, 0 ≤ 𝑘 ≤ 𝑙. (42)
1 𝑀−1 𝑘+1−𝑗 2 𝑘−𝑗 2 ) + (𝑒𝑖 ) )] + 𝑐42 ℎ ∑ ((𝑒𝑖 2 𝑖=1 + =
9𝑐0 − 𝑐1 𝑘+(1/2) 2 𝛿 𝑒 24 𝑡
30 2 2 𝑐32 (𝑒𝑘 + 𝑒𝑘−1 ) 9𝑐0 − 𝑐1
6
Abstract and Applied Analysis +
6 2 2 𝑐2 (𝑒𝑘+1−𝑗 + 𝑒𝑘−𝑗 ) 9𝑐0 − 𝑐1 4
9𝑐 − 𝑐 2 + 0 1 𝛿𝑡 𝑒𝑘+(1/2) , 24
Table 1: Numerical results of (54) when ℎ = 1/10, 𝜏 = 1/100.
0 ≤ 𝑘 ≤ 𝑙. (43)
Inserting (39)–(43) into (38), we obtain 𝑑 𝑘+1 2 𝑘 2 (𝑒 1 − 𝑒 1 ) 2𝜏 30 2 2 𝑐32 (𝑒𝑘 + 𝑒𝑘−1 ) 9𝑐0 − 𝑐1
≤
+
6 2 2 𝑐42 (𝑒𝑘+1−𝑗 + 𝑒𝑘−𝑗 ) 9𝑐0 − 𝑐1
+
2 6 𝑐62 (𝜏2 + ℎ4 ) , 9𝑐0 − 𝑐1
(44)
0 ≤ 𝑘 ≤ 𝑙.
2 2𝜏 6 + 𝑐62 (𝜏2 + ℎ4 ) , 𝑑 9𝑐0 − 𝑐1
(45)
0 ≤ 𝑘 ≤ 𝑙.
2𝜏 6 ) + 𝑐2 𝑑 9𝑐0 − 𝑐1 4
𝑚−1 2
2 × ∑ (𝑒𝑚 + 𝑒 𝑚=0
𝑙 2 2𝜏 6 𝑐62 ∑ (𝜏2 + ℎ4 ) 𝑑 9𝑐0 − 𝑐1 𝑚=0
+
𝑘
≤
𝑑 (9𝑐0 − 𝑐1 )
12𝑐62
By the inductive principle, this completes the proof.
𝑟 (𝑥, 𝑡) V𝑡 − 𝑑V𝑥𝑥 = 𝑓 (V (𝑥, 𝑡) , V (𝑥, 𝑡 − 𝑠) , 𝑥, 𝑡) , (𝑥, 𝑡) ∈ (0, 1) × (0, 𝑇] , V (𝑥, 𝑡) = 𝜙 (𝑥, 𝑡) + 𝜓 (𝑥, 𝑡) , V (0, 𝑡) = 𝛼 (𝑡) ,
𝑥 ∈ [0, 1] , 𝑡 ∈ [−𝑠, 0] ,
V (1, 𝑡) = 𝛽 (𝑡) ,
(49)
𝑡 ∈ (0, 𝑇] ,
V𝑖𝑘 = 𝜙 (𝑥𝑖 , 𝑡𝑘 ) + 𝜓𝑖𝑘 , V0𝑘 = 𝛼 (𝑡𝑘 ) ,
0 ≤ 𝑖 ≤ 𝑀, −𝑗 ≤ 𝑘 ≤ 0,
𝑘 V𝑀 = 𝛽 (𝑡𝑘 ) ,
1 ≤ 𝑘 ≤ 𝑁.
Similar to the proof of Theorem 4, the following stability result can be obtained
𝜏
12𝑐62 𝑇 2 2 × ∑ 𝑒𝑚 + (𝜏2 + ℎ4 ) , 𝑑 (9𝑐 − 𝑐 ) 0 1 𝑚=1 𝑘
Theorem 5. Denote 0 ≤ 𝑘 ≤ 𝑙. (46)
By Lemma 2, we have 12𝑐62 𝑇
8.652𝑒 − 009 1.623𝑒 − 008 2.258𝑒 − 008 2.805𝑒 − 008 3.288𝑒 − 008 3.725𝑒 − 008 4.129𝑒 − 008 4.509𝑒 − 008 4.870𝑒 − 008 5.218𝑒 − 008
1 ≤ 𝑖 ≤ 𝑀 − 1, 0 ≤ 𝑘 ≤ 𝑁 − 1, (50)
(𝑙 + 1) 𝜏 2 2 2 ≤ 𝜏 ∑ 𝑒𝑚 + (𝜏 + ℎ4 ) 𝑑 (9𝑐0 − 𝑐1 ) 𝑚=1 𝑑 (9𝑐0 − 𝑐1 ) 4 (5𝑐32 + 𝑐42 )
0.667184 0.727837 0.788490 0.849143 0.909796 0.970449 1.031102 1.091755 1.152408 1.213061
3 1 1 𝑘+1−𝑗 1 𝑘−𝑗 = A𝑓 ( V𝑖𝑘 − V𝑖𝑘−1 , V𝑖 + V𝑖 , 𝑥𝑖 , 𝑡𝑘+(1/2) ) , 2 2 2 2
𝑚=0
𝑐42 )
0.667184 0.727837 0.788490 0.849143 0.909796 0.970449 1.031102 1.091755 1.152408 1.213061
A𝑟𝑖𝑘+(1/2) 𝛿𝑡 V𝑖𝑘+(1/2) − 𝑑𝛿𝑥2 V𝑖𝑘+1/2
2 2 × ∑ (𝑒𝑚+1−𝑗 + 𝑒𝑚−𝑗 )
24 (5𝑐32
(0.5, 0.1) (0.5, 0.2) (0.5, 0.3) (0.5, 0.4) (0.5, 0.5) (0.5, 0.6) (0.5, 0.7) (0.5, 0.8) (0.5, 0.9) (0.5, 1.0)
where 𝜓(𝑥, 𝑡) is the perturbation caused by 𝜙(𝑥, 𝑡). The following difference scheme solving for (49) can be obtained:
𝑘
+
|𝑢(𝑥𝑖 , 𝑡𝑘 ) − 𝑢𝑖𝑘 |
To discuss the stability of the difference scheme (20)–(22), we consider the following problem:
Summing up (45) for 𝑘, noticing (33), and exploiting (26), we have 𝑒𝑘+1 2 ≤ 𝑒0 2 + 2𝜏 30 𝑐2 1 1 𝑑 9𝑐0 − 𝑐1 3 𝑘
Exact solution
2 (5𝑐32 + 𝑐42 ) 3𝑇 𝑙+1 𝑒 ≤ 𝑐6 √ exp ( 𝑇) (𝜏2 + ℎ4 ) . ∞ 𝑑 (9𝑐0 − 𝑐1 ) 𝑑 (9𝑐0 − 𝑐1 ) (48)
𝑘+1 2 𝑘 2 2𝜏 30 2 𝑘 2 𝑘−1 2 𝑒 ≤ 𝑒 + 1 1 𝑑 9𝑐0 − 𝑐1 𝑐3 (𝑒 + 𝑒 ) 2𝜏 6 2 2 𝑐42 (𝑒𝑘+1−𝑗 + 𝑒𝑘−𝑗 ) 𝑑 9𝑐0 − 𝑐1
Numerical solution
From (25), we obtain
The above inequality has the following form:
+
(𝑥, 𝑡)
4 (5𝑐32 + 𝑐42 )
𝑙+1 2 2 4 2 𝑒 ≤ 1 𝑑 (9𝑐 − 𝑐 ) exp ( 𝑑 (9𝑐 − 𝑐 ) 𝑇) (𝜏 + ℎ ) . 0 1 0 1 (47)
𝜂𝑖𝑘 = V𝑖𝑘 − 𝑢𝑖𝑘 ,
0 ≤ 𝑖 ≤ 𝑀, −𝑗 ≤ 𝑘 ≤ 𝑁.
(51)
Then, there exist constants 𝑐7 and 𝑐8 such that 0 𝑀−1 𝑘 2 𝜂 ≤ 𝑐7 √𝜏ℎ ∑ ∑ (𝜓𝑖𝑘 ) ∞
(52)
𝑚=−𝑗 𝑖=𝑙
under the condition that h and 𝜏 are small enough and max−𝑗≤𝑘≤0, 0≤𝑖≤𝑀|𝜓𝑖𝑘 | ≤ 𝑐8 .
Abstract and Applied Analysis
7
Table 2: Maximum norm errors of (54) with different step-sizes. 𝜏 1/100 1/400 1/1600 1/6400 1/25600 ×10 7
𝐸∞ (ℎ, 𝜏) 5.220𝑒 − 008 3.295𝑒 − 009 2.059𝑒 − 010 1.283𝑒 − 011 6.566𝑒 − 013
𝐸∞ (ℎ, 𝜏)/𝐸∞ (ℎ/2, 𝜏/2) ∗ 15.842 15.999 16.046 19.546
×10−8 5
|U(x, t) − Uht (x, t)|
ℎ 1/10 1/20 1/40 1/80 1/160
−8
4 3 2 1 0 1 0.5
|U(x, 1) − Uht (x, 1)|
6
x
0
5
0
0.2
0.4
0.6 t
0.8
1
4 h = 1/10, 𝜏 = 1/100
3
Figure 2: Error surface maps of difference scheme (20)–(22) solving for problem (54) with step-size ℎ = 1/10; 𝜏 = 1/100.
2 1 0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x h = 1/40, 𝜏 = 1/1600
h = 1/20, 𝜏 = 1/400
h = 1/80, 𝜏 = 1/6400
×10−9 4
Figure 1: Error curves of difference scheme (20)–(22) solving for problem (54) with different step-sizes when 𝑡 = 1.
|U(x, t) − Uht (x, t)|
h = 1/10, 𝜏 = 1/100
3 2 1 0 1
4. Numerical Test In this section, a numerical test is considered to validate the algorithms provided in this paper, and the numerical solutions 𝑢𝑖𝑘 of the example are obtained by exploiting scheme (20)–(22). Define 𝑢 (𝑥 , 𝑡 ) − 𝑢𝑘 . 𝐸∞ (ℎ, 𝜏) = max (53) 𝑖 𝑘 𝑖 0≤𝑖≤𝑀, 0≤𝑘≤𝑁 Example 1. Consider the following problem:
x
0.5 0
0
0.2
0.4
0.6
0.8
1
t
h = 1/20, 𝜏 = 1/400
Figure 3: Error surface maps of difference scheme (20)–(22) solving for problem (54) with step-size ℎ = 1/20; 𝜏 = 1/400.
𝑟 (𝑥, 𝑡) 𝑢𝑡 − 2𝑢𝑥𝑥 = 𝑢 (𝑥, 𝑡) + 𝑢 (𝑥, 𝑡 − 0.1) + 2𝑒−𝑥 , 𝑥 ∈ (0, 1) , 𝑡 ∈ (0, 1] , −𝑥
𝑢 (𝑥, 𝑡) = 𝑒
𝑢 (0, 𝑡) = 1 + 𝑡,
(1 + 𝑡) ,
𝑥 ∈ (0, 1) , 𝑡 ∈ [−0.1, 0] ,
𝑢 (1, 𝑡) = 𝑒−1 (1 + 𝑡) ,
(54)
𝑡 ∈ (0, 1] ,
where 𝑟(𝑥, 𝑡) = 4𝑡 + 5.9. The exact solution of (54) is 𝑢(𝑥, 𝑡) = 𝑒−𝑥 (1 + 𝑡). Table 1 provides some numerical results of difference scheme (20)–(22) solving for (54) with step-sizes ℎ = 1/10 and 𝜏 = 1/100. Table 2 gives the maximum absolute errors between numerical solutions and exact solutions with
different step-sizes. From Table 2, we can see that when the space step-size and the time step-size are reduced by a factor of 1/2 and 1/4, respectively, then the maximum absolute errors are reduced by a factor of approximately 1/16. Figure 1 provides us the error curves of numerical solutions for (54) at 𝑡 = 1 by using scheme (20)–(22). Figures 2 and 3 give the error surface of the numerical solutions with step-sizes ℎ = 1/10, 𝜏 = 1/100, and ℎ = 1/20, 𝜏 = 1/400, respectively. Generally speaking, from the results of the tables and the figures provided, we can see that the numerical results are coincident with the theoretical results.
8
5. Conclusion In this paper, a compact difference scheme is constructed to solve a type of variable coefficient delay partial differential equations, and the difference scheme is proved to be unconditionally stable and convergent. Finally, a numerical test is presented to illustrate the theoretical results.
Conflict of Interests The author declares that there is no conflict of interests regarding the publication of this paper.
Acknowledgments This work is supported by the Project Sponsored by the Scientific Research Foundation for the Returned Overseas Chinese Scholars, State Education Ministry (no. 2013693) and the National Natural Science Foundation of China (nos. 71301166, 11301544, and 11201487).
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