Hindawi Publishing Corporation Advances in Numerical Analysis Volume 2013, Article ID 512084, 7 pages http://dx.doi.org/10.1155/2013/512084
Research Article Some Results on Preconditioned Mixed-Type Splitting Iterative Method Guangbin Wang1 and Fuping Tan2 1 2
Department of Mathematics, Qingdao University of Science and Technology, Qingdao 266061, China Department of Mathematics, Shanghai University, Shanghai 200444, China
Correspondence should be addressed to Guangbin Wang;
[email protected] Received 20 June 2013; Revised 14 September 2013; Accepted 27 September 2013 Academic Editor: Zhong-Zhi Bai Copyright © 2013 G. Wang and F. Tan. 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. We present a preconditioned mixed-type splitting iterative method for solving the linear system 𝐴𝑥 = 𝑏, where A is a Z-matrix. And we give some comparison theorems to show that the rate of convergence of the preconditioned mixed-type splitting iterative method is faster than that of the mixed-type splitting iterative method. Finally, we give one numerical example to illustrate our results.
where 𝑃𝐴 = 𝑀𝑝 − 𝑁𝑝 and 𝑀𝑝 is nonsingular. Thus, the equation above can also be written as
1. Introduction For solving linear system, 𝐴𝑥 = 𝑏,
(1)
where 𝐴 is an 𝑛 × 𝑛 square matrix and 𝑥 and 𝑏 are 𝑛-dimensional vectors, the basic iterative method is 𝑀𝑥𝑘+1 = 𝑁𝑥𝑘 + 𝑏,
𝑘 = 0, 1, . . . ,
𝑥𝑘+1 = 𝑇𝑥𝑘 + 𝑐,
𝑘 = 0, 1, . . . ,
where 𝑇 = 𝑀𝑝 −1 𝑁𝑝 and 𝑐 = 𝑀𝑝 −1 𝑃𝑏. In paper [1], Cheng et al. presented the mixed-type splitting iterative method as follows:
(2)
(𝐷 + 𝐷1 + 𝐿 1 − 𝐿) 𝑥𝑘+1 = (𝐷1 + 𝐿 1 + 𝑈) 𝑥𝑘 + 𝑏,
where 𝐴 = 𝑀 − 𝑁 and 𝑀 is nonsingular. Thus, (2) can be written as
𝑘 = 0, 1, 2 . . . ,
𝑥𝑘+1 = 𝑇𝑥𝑘 + 𝑐, −1
𝑘 = 0, 1, . . . ,
(3)
−1
where 𝑇 = 𝑀 𝑁 and 𝑐 = 𝑀 𝑏. Assuming that 𝐴 has unit diagonal entries, let 𝐴 = 𝐼 − 𝐿 − 𝑈, where 𝐼 is the identity matrix and −𝐿 and −𝑈 are strictly lower and strictly upper triangular parts of 𝐴, respectively. Transform the original system (1) into the preconditioned form as follows: 𝑃𝐴𝑥 = 𝑃𝑏.
(4)
Then, we can define the basic iterative scheme as follows: 𝑀𝑝 𝑥𝑘+1 = 𝑁𝑝 𝑥𝑘 + 𝑃𝑏,
𝑘 = 0, 1, . . . ,
(5)
(6)
(7)
with the following iterative matrix: −1
𝑇 = (𝐷 + 𝐷1 + 𝐿 1 − 𝐿) (𝐷1 + 𝐿 1 + 𝑈) ,
(8)
where 𝐷1 is an auxiliary nonnegative diagonal matrix, 𝐿 1 is an auxiliary strictly lower triangular matrix, and 0 ≤ 𝐿 1 ≤ 𝐿. In this paper, we will establish the preconditioned mixedtype splitting iterative method with the preconditioners 𝑃𝛼 = 𝐼 + 𝛼𝐿, 𝑃𝛽 = 𝐼 + 𝛽𝑈, and 𝑃𝛼𝛽 = 𝐼 + 𝛼𝐿 + 𝛽𝑈 for solving linear systems. And we obtain some comparison results which show that the rate of convergence of the preconditioned mixed-type splitting iterative method with 𝑃𝛼𝛽 is faster than that of the preconditioned mixed-type splitting iterative method with 𝑃𝛼 or 𝑃𝛽 . Finally, we give one numerical example to illustrate our results.
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Advances in Numerical Analysis
2. Preconditioned Mixed-Type Splitting Iterative Method
(2) The PAOR method is
For the linear system (1), we consider its preconditioned form as follows: 𝑃𝛼𝛽 𝐴𝑥 = 𝑃𝛼𝛽 𝑏,
(9)
with the preconditioner 𝑃𝛼𝛽 = 𝐼 + 𝛼𝐿 + 𝛽𝑈; that is, 𝐴 𝛼𝛽 𝑥 = 𝑏𝛼𝛽 .
𝐷1 =
1 (1 − 𝑤) 𝐷𝛼𝛽 , 𝑤
𝐿1 =
1 (𝑤 − 𝑟) 𝐿 𝛼𝛽 , 𝑤
̃ 𝑟𝑤 = (𝐷𝛼𝛽 − 𝑟𝐿 𝛼𝛽 )−1 [(1 − 𝑤) 𝐷𝛼𝛽 + (𝑤 − 𝑟) 𝐿 𝛼𝛽 + 𝑤𝑈𝛼𝛽 ] . 𝐿 (17) We need the following definitions and results.
(10)
We apply the mixed-type splitting iterative method to it and have the corresponding preconditioned mixed-type splitting iterative method as follows: (𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 ) 𝑥𝑘+1 = (𝐷1 + 𝐿 1 + 𝑈𝛼𝛽 ) 𝑥𝑘 + 𝑏𝛼𝛽 , 𝑘 = 0, 1, 2, . . . , (11)
Definition 1 (see [2]). A matrix 𝐴 is a 𝑍-matrix if 𝑎𝑖𝑗 ≤ 0, for all 𝑖, 𝑗 = 1, 2, . . . 𝑛, such that 𝑖 ≠ 𝑗. A matrix 𝐴 is an 𝐿-matrix if 𝑎𝑖𝑖 > 0, 𝑖 = 1, 2, . . . 𝑛, and 𝑎𝑖𝑗 ≤ 0, for all 𝑖, 𝑗 = 1, 2, . . . 𝑛, such that 𝑖 ≠ 𝑗. Definition 2 (see [2]). A matrix 𝐴 is an 𝑀-matrix if 𝐴 is a nonsingular 𝑍-matrix, and 𝐴−1 ≥ 0. Definition 3 (see [2, 3]). Let 𝑀, 𝑁 ∈ 𝑅𝑛,𝑛 . Then, 𝐴 = 𝑀 − 𝑁 is called a regular splitting if 𝑀−1 ≥ 0 and 𝑁 ≥ 0; 𝐴 = 𝑀 − 𝑁 is called an 𝑀-splitting if 𝑀 is an 𝑀-matrix, 𝑁 ≥ 0. Lemma 4 (see [2]). Let 𝐴 ≥ 0 be an irreducible matrix. Then,
that is,
(1) 𝐴 has a positive real eigenvalue equal to its spectral radius;
−1
𝑥𝑘+1 = (𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 ) (𝐷1 + 𝐿 1 + 𝑈𝛼𝛽 ) 𝑥𝑘 −1
+ (𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 ) 𝑏𝛼𝛽 ,
𝑘 = 0, 1, 2 ⋅ ⋅ ⋅ , (12)
(2) to 𝜌(𝐴), there corresponds an eigenvector 𝑥 > 0; (3) 𝜌(𝐴) is a simple eigenvalue of 𝐴. Lemma 5 (see [4]). Let 𝐴 be a nonnegative matrix. Then,
So, the iterative matrix is ̃ = (𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 )−1 (𝐷1 + 𝐿 1 + 𝑈𝛼𝛽 ) , 𝑇
(13)
where 𝐷𝛼𝛽 , −𝐿 𝛼𝛽 , and −𝑈𝛼𝛽 are the diagonal, strictly lower, and strictly upper triangular matrices obtained from 𝐴 𝛼𝛽 , 𝐷1 is an auxiliary nonnegative diagonal matrix, 𝐿 1 is an auxiliary strictly lower triangular matrix, and 0 ≤ 𝐿 1 ≤ 𝐿 𝛼𝛽 . If we choose 𝛽 = 0, we have the following corresponding iterative matrix: −1
𝑇 = (𝐷𝛼 + 𝐷1 + 𝐿 1 − 𝐿 𝛼 ) (𝐷1 + 𝐿 1 + 𝑈𝛼 ) .
(14)
And if we choose 𝛼 = 0, we have the following corresponding iterative matrix: ̂ = (𝐷𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛽 )−1 (𝐷1 + 𝐿 1 + 𝑈𝛽 ) . 𝑇
(15)
If we choose certain auxiliary matrices, we can get the classical iterative methods as follows. (1) The PSOR method is 1 𝐷1 = (1 − 𝑟) 𝐷𝛼𝛽 , 𝑟
(2) if 𝐴𝑥 ≤ 𝛽𝑥 for some positive vector 𝑥, then 𝜌(𝐴) ≤ 𝛽. Moreover, if 𝐴 is irreducible and if 0 ≠ 𝛼𝑥 ≤ 𝐴𝑥 ≤ 𝛽𝑥 for some nonnegative vectors 𝑥, then 𝛼 ≤ 𝜌 (𝐴) ≤ 𝛽.
(18)
Lemma 6 (see [5]). Let 𝐴 = 𝑀 − 𝑁 be an 𝑀-splitting of 𝐴. Then, 𝜌(𝑀−1 𝑁) < 1 if and only if 𝐴 is a nonsingular 𝑀matrix. Lemma 7 (see [6, 7]). Let 𝐴 be a Z-matrix. Then, 𝐴 is a nonsingular 𝑀-matrix if and only if there is a positive vector 𝑥 such that 𝐴𝑥 ≥ 0. Lemma 8 (see [8]). Let 𝐴 = 𝑀 − 𝑁 be a regular splitting of 𝐴. Then, the splitting is convergent if and only if 𝐴−1 ≥ 0. Lemma 9 (see [9]). Let 𝐴 and 𝐵 be two 𝑛 × 𝑛 nonsingular lower triangular 𝐿-matrices. If 𝐴 ≥ 𝐵, then 𝐵−1 ≥ 𝐴−1 ≥ 0.
3. Convergence Analysis and Comparison Results
𝐿 1 = 0,
̃ 𝑟 = (𝐷𝛼𝛽 − 𝑟𝐿 𝛼𝛽 )−1 [(1 − 𝑟) 𝐷𝛼𝛽 + 𝑟𝑈𝛼𝛽 ] . 𝐿
(1) if 𝛼𝑥 ≤ 𝐴𝑥 for some nonnegative vector 𝑥, 𝑥 ≠ 0, then 𝛼 ≤ 𝜌(𝐴);
(16)
Theorem 10. Let 𝐴 be a nonsingular Z-matrix. Assume that 𝐷1 ≥ 0, 0 ≤ 𝐿 1 ≤ 𝐿 𝛼 , 𝛼 ∈ [0, 1], and 𝑇 and 𝑇 are the iterative
Advances in Numerical Analysis
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matrices given by (14) and (8), respectively. Consider the following.
On the other hand, since 𝐿 𝛼 = 𝐷𝛼 − 𝐼 + 𝛼𝐷 + 𝐿 − 𝛼𝐿 + 𝛼𝐿2 + 𝛼𝐸, we get 𝐸𝛼 − 𝑀𝛼 = (𝐼 + 𝛼𝐿) (𝐼 + 𝐷1 + 𝐿 1 − 𝐿)
(i) If 𝜌(𝑇) < 1, then 𝜌(𝑇) < 𝜌(𝑇) < 1. (ii) Let 𝐴 be irreducible. Assume that 1 − 𝛼 ∑𝑖−1 𝑙=1 𝑎𝑖𝑙 𝑎𝑙𝑗 > 0 and 𝑎𝑖𝑗 + ∑𝑛𝑘=𝑖+1 𝑎𝑖𝑘 𝑎𝑘𝑗 ≤ 0;
− (𝐷𝛼 + 𝐷1 + 𝐿 1 − 𝐿 𝛼 ) = (𝐼 + 𝐷1 + 𝐿 1 − 𝐿) + 𝛼𝐿 (𝐼 + 𝐷1 + 𝐿 1 − 𝐿) − (𝐷𝛼 + 𝐷1 + 𝐿 1 − 𝐿 𝛼 )
then,one has
= 𝐼 − 𝐿 + 𝛼𝐿 + 𝛼𝐿𝐷1 + 𝛼𝐿𝐿 1 − 𝛼𝐿2 − 𝐷𝛼 + 𝐿 𝛼
(1) 𝜌(𝑇) ≥ 𝜌(𝑇), if 𝜌(𝑇) ≥ 1,
= 𝐼 − 𝐿 + 𝛼𝐿 + 𝛼𝐿𝐷1 + 𝛼𝐿𝐿 1 − 𝛼𝐿2 − 𝐷𝛼 + 𝐷𝛼
(2) 𝜌(𝑇) ≤ 𝜌(𝑇), if 𝜌(𝑇) < 1.
− 𝐼 + 𝛼𝐷 + 𝐿 − 𝛼𝐿 + 𝛼𝐿2 + 𝛼𝐸 = 𝛼 (𝐿𝐷1 + 𝐿𝐿 1 + 𝐷 + 𝐸) ≥ 0,
Proof. Let
(22) 𝑀𝛼 = 𝐷𝛼 + 𝐷1 + 𝐿 1 − 𝐿 𝛼 ,
which implies that
𝑁𝛼 = 𝐷1 + 𝐿 1 + 𝑈𝛼 , 𝑀 = 𝐼 + 𝐷1 + 𝐿 1 − 𝐿, 𝑁 = 𝐷1 + 𝐿 1 + 𝑈,
−1 −1 𝐴−1 𝛼 𝐸𝛼 − 𝐴 𝛼 𝑀𝛼 = 𝐴 𝛼 (𝐸𝛼 − 𝑀𝛼 ) ≥ 0.
(19)
𝐸𝛼 = (𝐼 + 𝛼𝐿) (𝐼 + 𝐷1 + 𝐿 1 − 𝐿) ,
Then, we have 𝐴 = 𝑀 − 𝑁,
𝐴 𝛼 = 𝑀𝛼 − 𝑁𝛼 = 𝐸𝛼 − 𝐹𝛼 .
(20)
(i) Since 𝐴 is a nonsingular Z-matrix and 𝐷1 ≥ 0, 0 ≤ 𝐿 1 ≤ 𝐿 𝛼 , it is clear that 𝑀 = 𝐼 + 𝐷1 + 𝐿 1 − 𝐿 is a nonsingular 𝑀matrix and the splitting 𝐴 = 𝑀 − 𝑁 = (𝐼 + 𝐷1 + 𝐿 1 − 𝐿) − (𝐷1 + 𝐿 1 + 𝑈)
−1 −1 Therefore, 𝐴−1 𝛼 𝐸𝛼 ≥ 𝐴 𝛼 𝑀𝛼 ≥ 0. So, we have 𝜌(𝑀𝛼 𝑁𝛼 ) ≤ 𝜌(𝐸𝛼−1 𝐹𝛼 ); that is,
𝜌 (𝑇) ≤ 𝜌 (𝑇) < 1.
𝐹𝛼 = (𝐼 + 𝛼𝐿) (𝐷1 + 𝐿 1 + 𝑈) .
(21)
is an 𝑀-splitting. Since 𝜌(𝑇) < 1, it follows from Lemma 6 that 𝐴 is a nonsingular 𝑀-matrix. Then, by Lemma 7, there is a positive vector 𝑥 such that 𝐴𝑥 ≥ 0, so 𝐴 𝛼 𝑥 = (𝐼+𝛼𝐿)𝐴𝑥 ≥ 0. By Lemma 7, 𝐴 𝛼 is also a nonsingular 𝑀-matrix. Obviously, we can get that 𝐷𝛼 is a positive diagonal matrix. And from 𝐿 𝛼 is nonnegative, we know that 𝑀𝛼 being a 𝑍-matrix. Since 𝐷𝛼−1 𝐿 𝛼 ≥ 0 is a strictly lower triangular matrix, so that 𝜌(𝐷𝛼−1 𝐿 𝛼 ) = 0 < 1. So, we have (𝐼 + 𝐷𝛼−1 𝐷1 + 𝐷𝛼−1 𝐿 1 − 𝐷𝛼−1 𝐿 𝛼 )−1 ≥ 0. Then, 𝑀𝛼−1 = (𝐼 + 𝐷𝛼−1 𝐷1 + 𝐷𝛼−1 𝐿 1 − 𝐷𝛼−1 𝐿 𝛼 )−1 𝐷𝛼−1 ≥ 0; hence, 𝑀𝛼 is a nonsingular 𝑀-matrix. For 𝑗 ≠ 𝑖+1, it is obvious that (𝑈𝛼 )𝑖𝑗 = −𝑎𝑖𝑗 +𝛼𝑎𝑖𝑖+1 𝑎𝑖+1𝑗 ≥ 0. And for 𝑗 = 𝑖+1, we have (𝑈𝛼 )𝑖𝑗 = (𝛼−1)𝑎𝑖𝑗 ≥ 0. Thus, 𝑈𝛼 ≥ 0 and 𝑁𝛼 ≥ 0. We have proven that 𝐴 𝛼 = 𝑀𝛼 − 𝑁𝛼 and 𝐴 = 𝑀 − 𝑁 are both 𝑀-splittings and 𝐸𝛼−1 𝐹𝛼 = 𝑀−1 𝑁, two splittings 𝐴 𝛼 = 𝑀𝛼 − 𝑁𝛼 = 𝐸𝛼 − 𝐹𝛼 are nonnegative.
(23)
(24)
(ii) Let 𝐴 = 𝐼 − 𝐿 − 𝑈 be irreducible. Since 𝐿 + 𝑈 is a nonnegative and irreducible matrix, and according to the proof of Lemma 4 in paper [9], we can obtain that 𝑇 and 𝑇 are nonnegative and irreducible matrices. Thus, from Lemma 4, we know that there exists a positive vector 𝑥 = (𝑥1 , 𝑥2 ⋅ ⋅ ⋅ 𝑥𝑛 )𝑇 such that 𝑇𝑥 = 𝜆𝑥, where we denote 𝜆 = 𝜌(𝑇), which is equivalent to (𝐷1 + 𝐿 1 + 𝑈) 𝑥 = 𝜆 (𝐼 + 𝐷1 + 𝐿 1 − 𝐿) 𝑥, (𝑈 − 𝜆𝐼 + 𝜆𝐿) 𝑥 = [(𝜆 − 1) 𝐷1 + (𝜆 − 1) 𝐿 1 ] 𝑥.
(25)
Let 𝐿𝑈 = 𝐷 + 𝐸 + 𝐹, where 𝐷, 𝐸, and 𝐹 are the diagonal, lower triangular, and upper triangular parts of 𝐿𝑈, respectively. So, 𝐴 𝛼 = 𝐷𝛼 − 𝐿 𝛼 − 𝑈𝛼 = (𝐼 − 𝛼𝐷) − (𝐿 − 𝛼𝐿 + 𝛼𝐿2 + 𝛼𝐸) − (𝑈 + 𝛼𝐹) ,
(26)
where 𝐷𝛼 = 𝐼 − 𝛼𝐷, 𝐿 𝛼 = 𝐿 − 𝛼𝐿 + 𝛼𝐿2 + 𝛼𝐸, 𝑈𝛼 = 𝑈 + 𝛼𝐹. Now, we consider 𝑇𝑥 − 𝑇𝑥 = (𝐷𝛼 + 𝐷1 + 𝐿 1 − 𝐿 𝛼 ) −1 × (𝐷1 + 𝐿 1 + 𝑈𝛼 ) 𝑥 − 𝜆𝑥 −1
= (𝐷𝛼 + 𝐷1 + 𝐿 1 − 𝐿 𝛼 )
× [(𝐷1 + 𝐿 1 + 𝑈𝛼 ) − 𝜆 (𝐷𝛼 + 𝐷1 + 𝐿 1 − 𝐿 𝛼 )] 𝑥 −1
= (𝐷𝛼 + 𝐷1 + 𝐿 1 − 𝐿 𝛼 )
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Advances in Numerical Analysis × [(𝐷1 + 𝐿 1 + 𝑈 + 𝛼𝐹)
Proof. Let 𝑀𝛼𝛽 = 𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 ,
− 𝜆 (𝐼 − 𝛼𝐷 + 𝐷1 + 𝐿 1
𝑁𝛼𝛽 = 𝐷1 + 𝐿 1 + 𝑈𝛼𝛽 ,
−𝐿 + 𝛼𝐿 − 𝛼𝐿2 − 𝛼𝐸)] 𝑥
𝑀 = 𝐼 + 𝐷1 + 𝐿 1 − 𝐿,
−1
= (𝐷𝛼 + 𝐷1 + 𝐿 1 − 𝐿 𝛼 )
𝑁 = 𝐷1 + 𝐿 1 + 𝑈,
× { [(𝐷1 + 𝐿 1 + 𝑈) − 𝜆 (𝐼 + 𝐷1 + 𝐿 1 − 𝐿)] 𝑥 + [𝛼𝐹 + 𝜆 (𝛼𝐷 − 𝛼𝐿 + 𝛼𝐿2 + 𝛼𝐸)] 𝑥}
(30)
𝐸𝛼 = (𝐼 + 𝛼𝐿 + 𝛽𝑈) (𝐼 + 𝐷1 + 𝐿 1 − 𝐿) , 𝐹𝛼 = (𝐼 + 𝛼𝐿 + 𝛽𝑈) (𝐷1 + 𝐿 1 + 𝑈) .
−1
Then, we have
= (𝐷𝛼 + 𝐷1 + 𝐿 1 − 𝐿 𝛼 )
𝐴 = 𝑀 − 𝑁,
× [𝛼𝐹 + 𝜆 (𝛼𝐷 − 𝛼𝐿 + 𝛼𝐿2 + 𝛼𝐸)] 𝑥 −1
= (𝐷𝛼 + 𝐷1 + 𝐿 1 − 𝐿 𝛼 )
× [(𝜆 − 1) 𝛼𝐿 (𝐷1 + 𝐿 1 ) + (𝜆 − 1) 𝛼 (𝐷 + 𝐸)] 𝑥 −1
= (𝜆 − 1) (𝐷𝛼 + 𝐷1 + 𝐿 1 − 𝐿 𝛼 ) × 𝛼 [𝐿𝐷1 + 𝐿𝐿 1 + 𝐷 + 𝐸] 𝑥.
𝐴 𝛼𝛽 = 𝑀𝛼𝛽 − 𝑁𝛼𝛽 = 𝐸𝛼𝛽 − 𝐹𝛼𝛽 .
(31)
(i) By a similar proof of Theorem 10, we can prove that 𝐴 𝛼𝛽 = 𝑀𝛼𝛽 − 𝑁𝛼𝛽 and 𝐴 = 𝑀 − 𝑁 are both 𝑀-splitting and −1 𝐸𝛼𝛽 𝐹𝛼𝛽 = 𝑀−1 𝑁, two splittings 𝐴 𝛼𝛽 = 𝑀𝛼𝛽 −𝑁𝛼𝛽 = 𝐸𝛼𝛽 −𝐹𝛼𝛽 , are nonnegative. ̂+𝐿− On the other hand, since 𝐿 𝛼𝛽 = 𝐷𝛼𝛽 − 𝐼 + 𝛼𝐷 + 𝛽𝐷 2 ̂ 𝛼𝐿 + 𝛼𝐿 + 𝛼𝐸 + 𝛽𝐸, we get 𝐸𝛼𝛽 − 𝑀𝛼𝛽 = (𝐼 + 𝛼𝐿 + 𝛽𝑈) (𝐼 + 𝐷1 + 𝐿 1 − 𝐿)
(27) Since 𝐷𝛼 +𝐷1 +𝐿 1 −𝐿 𝛼 is an 𝑀-matrix, and 𝐿𝐷1 +𝐿𝐿 1 +𝐷+𝐸 ≥ 0, we have the following. (1) If 𝜆 ≥ 1, then 𝑇𝑥 ≥ 𝑇𝑥 = 𝜆𝑥. By Lemma 5, we get 𝜌(𝑇) ≥ 𝜌(𝑇). (2) If 𝜆 < 1, then 𝑇𝑥 ≤ 𝑇𝑥 = 𝜆𝑥. By Lemma 5, we get 𝜌(𝑇) ≤ 𝜌(𝑇). Theorem 11. Let 𝐴 be a nonsingular 𝑍-matrix. Assume that ̃ and 𝑇 are the iter𝐷1 ≥ 0, 0 ≤ 𝐿 1 ≤ 𝐿 𝛼𝛽 , 𝛼, 𝛽 ∈ [0, 1], and 𝑇 ative matrices given by (13) and (8), respectively. Consider the following.
− (𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 ) = (𝐼 + 𝐷1 + 𝐿 1 − 𝐿) + 𝛼𝐿 (𝐼 + 𝐷1 + 𝐿 1 − 𝐿) + 𝛽𝑈 (𝐼 + 𝐷1 + 𝐿 1 − 𝐿) − (𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 ) = 𝐼 − 𝐿 + 𝛼𝐿 (𝐼 + 𝐷1 + 𝐿 1 − 𝐿) + 𝛽𝑈 (𝐼 + 𝐷1 + 𝐿 1 − 𝐿) − 𝐷𝛼𝛽 + 𝐿 𝛼𝛽 = 𝐼 − 𝐿 + 𝛼𝐿 (𝐼 + 𝐷1 + 𝐿 1 − 𝐿) + 𝛽𝑈 (𝐼 + 𝐷1 + 𝐿 1 − 𝐿) − 𝐷𝛼𝛽 + 𝐷𝛼𝛽 ̂ + 𝐿 − 𝛼𝐿 − 𝐼 + 𝛼𝐷 + 𝛽𝐷
(i) If 𝜌(𝑇) < 1, then ̃ < 𝜌 (𝑇) < 1. 𝜌 (𝑇)
(28)
̂ + 𝛽𝑈 (𝐷1 + 𝐿 1 ) = 𝛽 (𝑈 − 𝐹)
(ii) Let 𝐴 be irreducible. Assume that 𝑖−1
𝑛
𝑙=1
𝑘=𝑖+1
+ 𝛼 (𝐷 + 𝐸) + 𝛼𝐿 (𝐷1 + 𝐿 1 ) ≥ 0,
1 − 𝛼 ∑ 𝑎𝑖𝑙 𝑎𝑙𝑗 − 𝛽 ∑ 𝑎𝑖𝑘 𝑎𝑘𝑗 > 0, 𝑛
𝑎𝑖𝑗 + ∑ 𝑎𝑖𝑘 𝑎𝑘𝑗 ≤ 0; 𝑘=𝑖+1
Then, one has ̃ ≥ 𝜌(𝑇) if 𝜌(𝑇) ≥ 1, (1) 𝜌(𝑇) ̃ ≤ 𝜌(𝑇) if 𝜌(𝑇) < 1. (2) 𝜌(𝑇)
+ 𝛼𝐿2 + 𝛼𝐸 + 𝛽𝐸̂
(32) (29)
which implies that −1 −1 𝐴−1 𝛼𝛽 𝐸𝛼𝛽 − 𝐴 𝛼𝛽 𝑀𝛼𝛽 = 𝐴 𝛼𝛽 (𝐸𝛼𝛽 − 𝑀𝛼𝛽 ) ≥ 0. −1 Therefore, 𝐴−1 𝛼𝛽 𝐸𝛼𝛽 ≥ 𝐴 𝛼𝛽 𝑀𝛼𝛽 −1 −1 𝜌(𝑀𝛼𝛽 𝑁𝛼𝛽 ) ≤ 𝜌(𝐸𝛼𝛽 𝐹𝛼𝛽 ); that is,
𝜌 (𝑇) ≤ 𝜌 (𝑇) < 1.
(33)
≥ 0. So, we have
(34)
Advances in Numerical Analysis
5
(ii) Let 𝐴 𝛼𝛽 = 𝑃𝛼𝛽 𝐴 = (𝐼 + 𝛼𝐿 + 𝛽𝑈)𝐴 = 𝐼 − 𝐿 − 𝑈 + 𝛼𝐿(𝐼 − 𝐿 − 𝑈) + 𝛽𝑈(𝐼 − 𝐿 − 𝑈) ̂ − (𝐿 − 𝛼𝐿 + 𝛼𝐿2 + 𝛼𝐸 + 𝛽𝐸) ̂ = 𝐼 − 𝛼𝐷 − 𝛽𝐷 − (𝑈 + 𝛼𝐹 − 𝛽𝑈 + 𝛽𝐹̂ + 𝛽𝑈2 ) ,
̂ = 𝐼 − 𝛼𝐷 − 𝛽𝐷,
̂ 𝐿 𝛼 = 𝐿 − 𝛼𝐿 + 𝛼𝐿2 + 𝛼𝐸 + 𝛽𝐸, ̂ 𝑈𝛼𝛽 = 𝑈 − 𝛽𝑈 + 𝛽𝑈2 + 𝛼𝐹 + 𝛽𝐹.
(36)
By (25), we have ̃ − 𝜆𝑥 = (𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 )−1 𝑇𝑥 × (𝐷1 + 𝐿 1 + 𝑈𝛼𝛽 ) 𝑥 − 𝜆𝑥 −1
= (𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 ) × [(𝐷1 + 𝐿 1 + 𝑈𝛼𝛽 )
−𝜆 (𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 )] 𝑥 −1
= (𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 )
× [(𝐷1 + 𝐿 1 + 𝑈 + 𝛼𝐹 − 𝛽𝑈 + 𝛽𝐹̂ + 𝛽𝑈2 )
̂ 𝑥 +𝛽𝑈 (𝐷1 + 𝐿 1 ) + 𝛽 (𝑈 − 𝐹)] = (𝜆 − 1) (𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 )
−1
× [ (𝛼𝐿 + 𝛽𝑈) (𝐷1 + 𝐿 1 ) ̂ 𝑥. +𝛼 (𝐷 + 𝐸) + 𝛽 (𝑈 − 𝐹)] (37) If 𝑎𝑖𝑗 + ∑𝑛𝑘=𝑖+1 𝑎𝑖𝑘 𝑎𝑘𝑗 ≤ 0, then by the proof of Theorem 10, we have 𝑈 − 𝐹̂ ≥ 0. Therefore, one has the following. ̃ − 𝜆𝑥 ≥ 0 but not equal to 0. There(1) If 𝜆 ≥ 1, then 𝑇𝑥 ̃ ̃ ≥ 𝜆 = 𝜌(𝑇). fore, 𝑇𝑥 ≥ 𝜆𝑥. By Lemma 5, we get 𝜌(𝑇) ̃ − 𝜆𝑥 ≤ 0 but not equal to 0. There(2) If 𝜆 < 1, then 𝑇𝑥 ̃ ≤ 𝜆𝑥. By Lemma 5, we get 𝜌(𝑇) ̃ ≤ 𝜆 = 𝜌(𝑇). fore, 𝑇𝑥 Remark. If we choose 𝛼 = 0 in Theorem 11, we have a similar result which is showed by the following corollary. Corollary 12. Let 𝐴 be a nonsingular Z-matrix. Assume that ̂ and 𝑇 are the 𝐷1 ≥ 0, 0 ≤ 𝐿 1 ≤ 𝐿 𝛽 , 𝛽 ∈ [0, 1], and 𝑇 iterative matrices given by (15) and (8), respectively. Consider the following. (i) If 𝜌(𝑇) < 1, then
̂ + 𝐷1 + 𝐿 1 − 𝐿 − 𝜆 (𝐼 − 𝛼𝐷 − 𝛽𝐷 ̂ 𝑥 +𝛼𝐿 − 𝛼𝐿 − 𝛼𝐸 − 𝛽𝐸)]
−1
× [𝛼𝐿 (𝐷1 + 𝐿 1 ) + 𝛼 (𝐷 + 𝐸)
(35)
̂ + 𝐸̂ + 𝐹, ̂ and 𝐷, 𝐸, 𝐹, 𝐷, ̂ 𝐸, ̂ and where 𝐿𝑈 = 𝐷 + 𝐸 + 𝐹, 𝑈𝐿 = 𝐷 𝐹̂ are the diagonal, strictly lower, and strictly upper triangular matrices of 𝐿𝑈 and 𝑈𝐿, respectively. And denote 𝐴 𝛼𝛽 = 𝐷𝛼𝛽 − 𝐿 𝛼𝛽 − 𝑈𝛼𝛽 ; then according to (35), we have 𝐷𝛼𝛽
= (𝜆 − 1) (𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 )
̂ < 𝜌 (𝑇) < 1. 𝜌 (𝑇)
(38)
2
−1
= (𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 )
× { [(𝐷1 + 𝐿 1 + 𝑈) − 𝜆 (𝐼 + 𝐷1 + 𝐿 1 − 𝐿)] 𝑥 + [𝛼𝐹 − 𝛽𝑈 + 𝛽𝐹̂ + 𝛽𝑈2 ̂ + 𝛼𝐿 − 𝜆 (−𝛼𝐷 − 𝛽𝐷 ̂ 𝑥} −𝛼𝐿2 − 𝛼𝐸 − 𝛼𝐸)] −1
= (𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 )
× [𝛼𝐹 − 𝛽𝑈 + 𝛽𝐹̂ + 𝛽𝑈2 ̂ + 𝛼𝐿 − 𝛼𝐿2 − 𝛼𝐸 − 𝛼𝐸)] ̂ 𝑥 −𝜆 (−𝛼𝐷 − 𝛽𝐷 −1
= (𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 )
× [𝛼𝐹 + 𝜆 (𝛼𝐷 − 𝛼𝐿 + 𝛼𝐿2 + 𝛼𝐸) ̂ + 𝛽𝐸)] ̂ 𝑥 −𝛽𝑈 + 𝛽𝑈2 + 𝛽𝐹̂ + 𝜆 (𝛽𝐷
(ii) Let 𝐴 be irreducible. Assume that 𝑛
1 − 𝛽 ∑ 𝑎𝑖𝑘 𝑎𝑘𝑗 > 0, 𝑘=𝑖+1
𝑛
𝑎𝑖𝑗 + ∑ 𝑎𝑖𝑘 𝑎𝑘𝑗 ≤ 0;
(39)
𝑘=𝑖+1
then, one has ̂ ≥ 𝜌(𝑇) if 𝜌(𝑇) ≥ 1, (1) 𝜌(𝑇) ̂ ≤ 𝜌(𝑇) if 𝜌(𝑇) < 1. (2) 𝜌(𝑇) Now, one will provide some results to show the relations ̃ 𝜌(𝑇), and 𝜌(𝑇). ̂ among 𝜌(𝑇), Theorem 13. Let 𝐴 = (𝑎𝑖𝑗 ) ∈ 𝑅𝑛×𝑛 be a nonsingular 𝑍-matrix. ̃ and 𝑇 be iterative matrices given by (13) and (14), respecLet 𝑇 tively. Assume that 𝛼, 𝛽 ∈ [0, 1], 𝐷1 ≥ 0, 0 ≤ 𝐿 1 ≤ 𝐿 𝛼𝛽 . If 𝑛 𝑛 1−𝛼 ∑𝑖−1 𝑙=1 𝑎𝑖𝑙 𝑎𝑙𝑗 −𝛽 ∑𝑘=𝑖+1 𝑎𝑖𝑘 𝑎𝑘𝑗 > 0 and 𝑎𝑖𝑗 +∑𝑘=𝑖+1 𝑎𝑖𝑘 𝑎𝑘𝑗 ≤ 0, then ̃ ≥ 𝜌(𝑇) if 𝜌(𝑇) ≥ 1; (1) 𝜌(𝑇) ̃ ≤ 𝜌(𝑇) if 𝜌(𝑇) < 1. (2) 𝜌(𝑇)
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Proof. Since 𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 and 𝐷𝛼 + 𝐷1 + 𝐿 1 − 𝐿 𝛼 are two lower triangular 𝐿-matrices with 𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 ≤ 𝐷𝛼 + 𝐷1 + 𝐿 1 − 𝐿 𝛼 , by Lemma 9, we have
By the proof of Corollary 12 and Theorem 11, we consider ̃ − 𝑇𝑥 ̂ = 𝑇𝑥 ̃ − 𝜆𝑥 − (𝑇𝑥 ̂ − 𝜆𝑥) 𝑇𝑥 −1
−1
(𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 )
≥ (𝐷𝛼 + 𝐷1 + 𝐿 1 − 𝐿 𝛼 )
−1
≥ 0. (40)
= (𝜆 − 1) (𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 ) × [(𝛼𝐿 + 𝛽𝑈) (𝐷1 + 𝐿 1 ) + 𝛼𝐷 ̂ 𝑥 +𝛼𝐸 + 𝛽 (𝑈 − 𝐹)]
By the proof of Theorems 10 and 11, we consider
− (𝐷𝛼 + 𝐷1 + 𝐿 1 − 𝐿 𝛼 ) ̃ − 𝑇𝑥 = 𝑇𝑥 ̃ − 𝜆𝑥 − (𝑇𝑥 − 𝜆𝑥) 𝑇𝑥
̂ 𝑥 × [𝛽𝑈 (𝐷1 + 𝐿 1 ) + 𝛽 (𝑈 − 𝐹)] −1
−1
= (𝜆 − 1) (𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 )
≥ (𝜆 − 1) (𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 )
× [(𝛼𝐿 + 𝛽𝑈) (𝐷1 + 𝐿 1 ) + 𝛼𝐷
× (𝛼𝐿 (𝐷1 + 𝐿 1 ) + 𝛼𝐷 + 𝛼𝐸) 𝑥.
̂ 𝑥 +𝛼𝐸 + 𝛽 (𝑈 − 𝐹)] − (𝐷𝛼 + 𝐷1 + 𝐿 1 − 𝐿 𝛼 )
(41)
−1
× [𝛼𝐿 (𝐷1 + 𝐿 1 ) + 𝛼𝐷 + 𝛼𝐸] 𝑥 −1
̂ 𝑥. × (𝛽𝑈 (𝐷1 + 𝐿 1 ) + 𝛽 (𝑈 − 𝐹)) In view of the proof of Theorem 11, we have 𝛽𝑈(𝐷1 +𝐿 1 )+ ̂ ≥ 0. 𝛽(𝑈 − 𝐹) Therefore, one has the following. (1) If 𝜆 ≥ 1, the right-hand side of the above inequality is ̃ ≥ 𝜌(𝑇). more than zero. By Lemma 8, 𝜌(𝑇) (2) If 𝜆 < 1, the right-hand side of the above inequality ̃ ≤ 𝜌(𝑇). is more than zero. By Lemma 8, 𝜌(𝑇) Theorem 14. Let 𝐴 = (𝑎𝑖𝑗 ) ∈ 𝑅𝑛×𝑛 be a nonsingular Z-matrix. ̃ and 𝑇 ̂ be iterative matrices given by (13) and (15), respecLet 𝑇 tively. Assume that 𝛼, 𝛽 ∈ [0, 1], 𝐷1 ≥ 0, 0 ≤ 𝐿 1 ≤ 𝐿 𝛼𝛽 . If 𝑛 𝑛 1−𝛼 ∑𝑖−1 𝑙=1 𝑎𝑖𝑙 𝑎𝑙𝑗 −𝛽 ∑𝑘=𝑖+1 𝑎𝑖𝑘 𝑎𝑘𝑗 > 0 and 𝑎𝑖𝑗 +∑𝑘=𝑖+1 𝑎𝑖𝑘 𝑎𝑘𝑗 ≤ 0, then ̃ ≥ 𝜌(𝑇) ̂ if 𝜌(𝑇) ≥ 1, (1) 𝜌(𝑇) ̃ ≤ 𝜌(𝑇) ̂ if 𝜌(𝑇) < 1. (2) 𝜌(𝑇)
≥ (𝐷𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛽 )
(1) If 𝜆 ≥ 1, the right-hand side of the above inequality is ̃ ≥ 𝜌(𝑇). ̂ more than zero. By Lemma 8, 𝜌(𝑇)
Remark. The results (theorems and corollaries) in Section 3 are in some sense the generalized Stein-Rosenberg-type theorems like those in the papers [10–13]. The results (theorems and corollaries) in Section 3 are the comparisons of spectral radius of iterative matrices between the mixed-type splitting method and the preconditioned mixed-type splitting method, while the results in the papers [10–13] are the comparisons of spectral radius of iterative matrices between the parallel decomposition-type relaxation method and its special case.
4. Numerical Example Consider the following equation: −Δ𝑢 +
𝜕𝑢 𝜕𝑢 + = 𝑓, 𝜕𝑥 𝜕𝑦
−1
≥ 0. (42)
(44)
in the unit square Ω with Dirichlet boundary conditions. If we apply the central difference scheme on a uniform grid with 𝑁 × 𝑁 interior nodes (𝑁2 = 𝑛) to the discretization of the above equation, we can get a system of linear equations with the coefficient matrix 𝐴 = 𝐼 ⊗ 𝑃 + 𝑄 ⊗ 𝐼,
Proof. Since 𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 and 𝐷𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛽 are two lower triangular 𝐿-matrices with 𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 ≤ 𝐷𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛽 , by Lemma 9, we have −1
Since 𝛼𝐿(𝐷1 + 𝐿 1 ) + 𝛼𝐷 + 𝛼𝐸 ≥ 0, we get the following.
(2) If 𝜆 < 1, the right-hand side of the above inequality is ̃ ≤ 𝜌(𝑇). ̂ more than zero. By Lemma 8, 𝜌(𝑇)
≥ (𝜆 − 1) (𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 )
(𝐷𝛼𝛽 + 𝐷1 + 𝐿 1 − 𝐿 𝛼𝛽 )
(43)
−1
(45)
where ⊗ denotes the Kronecker product, 2+ℎ 2−ℎ , 1, − ), 8 8 1−ℎ 1+ℎ , 1, − ) 𝑄 = tridiag (− 4 4
𝑃 = tridiag (−
(46)
are 𝑁 × 𝑁 tridiagonal matrices, and the step size is ℎ = 1/𝑁.
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Table 1 𝛼(𝛽) 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
𝜌(𝑇) 0.563691 0.563691 0.563691 0.563691 0.563691 0.563691 0.563691 0.563691 0.563691 0.563691 0.563691
𝜌(𝑇) 0.563691 0.554915 0.545958 0.536827 0.527531 0.518081 0.508493 0.498782 0.488967 0.479065 0.469097
̂ 𝜌(𝑇) 0.563691 0.554027 0.544027 0.533673 0.522949 0.511842 0.500331 0.488412 0.476073 0.463309 0.450121
̃ 𝜌(𝑇) 0.563691 0.545537 0.527441 0.509401 0.491418 0.473488 0.455608 0.437775 0.419983 0.402227 0.384541
We choose 𝑁 = 5; then 𝐴 ∈ 𝑅25×25 . If we choose (1) 𝐷1 = 0.45𝐷𝛼 , 𝐿 1 = 0.4𝐿 𝛼 , 𝛼 ∈ [0, 0.5], (2) 𝐷1 = 0.45𝐷𝛽 , 𝐿 1 = 0.4𝐿 𝛽 , 𝛽 ∈ [0, 0.5], (3) 𝐷1 = 0.45𝐷𝛼𝛽 , 𝐿 1 = 0.4𝐿 𝛼𝛽 , 𝛼 = 𝛽 ∈ [0, 0.5], then we can obtain the following results by Theorems 10–14. Table 1 shows that that the rate of convergence of the preconditioned mixed-type splitting method is faster than that of the mixed-type splitting method. And it shows that the rate of convergence of the preconditioned mixed-type splitting method with 𝑃𝛼𝛽 is faster than that of the preconditioned mixed-type splitting method with 𝑃𝛼 or 𝑃𝛽 .
Acknowledgments This work was supported by the National Natural Science Foundation of China (Grant no. 11001144) and the Natural Science Foundation of Shandong Province of China (ZR2012AL09).
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[8] O. Axelsson, Iterative Solution Methods, Cambridge University Press, Cambridge, UK, 1994. [9] T. Z. Huang, G. H. Cheng, and X. Y. Cheng, “Modified SORtype iterative method for Z-matrices,” Applied Mathematics and Computation, vol. 175, no. 1, pp. 258–268, 2006. [10] Z.-Z. Bai, “On the comparisons of the multisplitting unsymmetric aor methods for M-matrices,” Calcolo, vol. 32, no. 3-4, pp. 207–220, 1995. [11] Z.-Z. Bai, “The generalized Stein-Rosenberg type theorem for the PDAOR-method,” Bulletin of the Institute of Mathematics Academia Sinica, vol. 19, pp. 329–335, 1997 (Chinese). [12] Z.-Z. Bai, “A class of parallel decomposition-type relaxation methods for large sparse systems of linear equations,” Linear Algebra and Its Applications, vol. 282, no. 1–3, pp. 1–24, 1998. [13] Z.-Z. Bai and R. Nabben, “Some properties of the block matrices in the parallel decomposition-type relaxation methods,” Applied Numerical Mathematics, vol. 29, no. 2, pp. 167–170, 1999.
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