Convergence of the GAOR Method for One Subclass of 𝐻-Matrix

Abstract

We discuss the convergence of the GAOR method to solve linear system which occurred in solving the weighted linear least squares problem. Moreover, we present one convergence theorem of the GAOR method when the coefficient matrix is a strictly doubly α diagonally dominant matrix which is a nonsingular 𝐻-matrix. Finally, we show that our results are better than previous ones by using four numerical examples.

1. Introduction

Consider the weighted linear least squares problem min_{𝑥∈𝑅𝑛}(𝐴𝑥−𝑏)^𝑇𝑊⁻¹(𝐴𝑥−𝑏), (1) where 𝑊 is the variance-covariance matrix. The problem has many scientific applications. A typical source is parameter estimation in mathematical modeling.

In order to solve it, we have to solve the following linear system: 𝐻𝑦=𝑓, (2) where 𝐻=𝐼−𝐵₁𝐷𝐶𝐼−𝐵₂ (3) is invertible. For example, in the variance-covariance matrix [1], a generalized SOR (GSOR) method to solve linear system (2) was proposed in [2], afterwards, a generalized AOR (GAOR) method to solve linear system (2) was established in [3] as follows:𝑦^(𝑘+1)=𝐿_𝜔,𝑟𝑦^(𝑘)+𝜔𝑘, (4) where 𝐿_𝜔,𝑟=()𝐵1−𝜔𝐼+𝜔𝐽+𝜔𝑟𝐾,𝑘=𝐼0−𝑟𝐶𝐼𝑓,𝐽=¹−𝐷−𝐶𝐵₂,𝐶𝐾=00𝐼−𝐵₁=0𝐶𝐷𝐶𝐼−𝐵₁𝐷. (5)

In [4–6], authors studied the convergence of the GAOR method for solving the linear system 𝐻𝑦=𝑓. In [4, 5], authors studied the convergence of the GAOR method for diagonally dominant coefficient matrices and gave the regions of convergence. In [6], authors studied the convergence of the GAOR method for strictly doubly diagonally dominant coefficient matrices and gave the regions of convergence. In [7], authors studied the preconditioned generalized AOR method for solving linear systems. They proposed two kinds of preconditioning that each one contains three preconditioners. They showed that the convergence rate of the preconditioned generalized AOR methods is better than that of the original method, whenever the original method is convergent. In [8], authors presented three kinds of preconditioners for preconditioned modified AOR method to solve systems of linear equations. They showed that the convergence rate of the preconditioned modified AOR method is better than that of the original method, whenever the original method is convergent.

Sometimes, the coefficient matrices of linear systems 𝐻𝑦=𝑓 are not strictly diagonally dominant or strictly doubly diagonally dominant. In this paper, we will discuss the convergence of the GAOR method when the coefficient matrices are strictly doubly 𝛼 diagonally dominant.

Throughout this paper, we denote the 𝑖-row sums of the modulus of the entries of 𝐽 and 𝐾 by 𝐽_𝑖 and 𝐾_𝑖, the 𝑖-column sums of the modulus of the entries of 𝐽 and 𝐾 by 𝐽^′_𝑖 and 𝐾^′_𝑖, respectively. And we denote the spectral radius of iterative matrix 𝐿_𝜔,𝑟 by 𝜌(𝐿_𝜔,𝑟): 𝑁={1,2,…,𝑛},𝑁₁=𝑖∣𝐾_𝑖=0,𝐾^′_𝑖,𝑁=0₂=𝑁−𝑁₁,𝑅_𝑖(𝐴)=_𝑘≠𝑖𝑎_𝑖𝑘,𝐶_𝑖(𝐴)=_𝑘≠𝑖𝑎_𝑘𝑖,𝑖,𝑘∈𝑁. (6)

Definition 1 (see [9]). We call matrix 𝐴∈𝐶^𝑛×𝑛 a strictly diagonally dominant matrix if 𝑎_𝑖𝑖>𝑅_𝑖(𝐴),∀𝑖∈𝑁 (7) and denote 𝐴∈𝑆𝐷.
We call matrix 𝐴∈𝐶^𝑛×𝑛 a strictly doubly diagonally dominant matrix if 𝑎_𝑖𝑖𝑎_𝑗𝑗>𝑅_𝑖(𝐴)𝑅_𝑗(𝐴),∀𝑖,𝑗∈𝑁 (8) and denote 𝐴∈𝑆𝐷𝐷.
We call matrix 𝐴∈𝐶^𝑛×𝑛 a strictly doubly 𝛼 diagonally dominant matrix if 𝑎_𝑖𝑖𝑎_𝑗𝑗>𝛼𝑅_𝑖(𝐴)+(1−𝛼)𝐶_𝑖(𝐴×)𝛼𝑅_𝑗(𝐴)+(1−𝛼)𝐶_𝑗(𝐴),∀𝑖,𝑗∈𝑁, (9) and denote 𝐴∈𝑆𝐷𝐷(𝛼), where 𝛼∈[0,1].

Obviously, a strictly doubly diagonally dominant matrix is a strictly doubly 𝛼 diagonally dominant matrix, but not vice versa.

For example, 11𝐴=43414134141411∈𝑆𝐷𝐷2, (10) but 𝐴∉𝑆𝐷𝐷.

Lemma 2 (see [9]). If 𝐴∈𝑆𝐷𝐷(𝛼), then 𝐴 is a nonsingular 𝐻-matrix.

In this paper, we study the convergence of the GAOR method for solving the linear system 𝐻𝑦=𝑓 for strictly doubly 𝛼 diagonally dominant coefficient matrices. Firstly, we obtain upper bound for the spectral radius of the matrix 𝐿_𝜔,𝑟 which is the iterative matrix of the GAOR iterative method. Moreover, we present one convergence theorem of the GAOR method. Finally, we present four numerical examples.

2. Upper Bound of the Spectral Radius of 𝐿_𝜔,𝑟

In this section, we obtain upper bound of the spectral radius of the iterative matrix 𝐿_𝜔,𝑟.

Theorem 3. Let 𝐻∈𝑆𝐷𝐷(𝛼), then 𝜌(𝐿_𝜔,𝑟) satisfies the following inequality: 𝜌𝐿_𝜔,𝑟≤max_{𝑖,𝑗𝑖≠𝑗}||+𝛼𝜔|𝐽1−𝜔|_𝑖+|𝜔𝑟|𝐾_𝑖+(1−𝛼)|𝜔|𝐽^′_𝑖+|𝜔𝑟|𝐾^′_𝑖×𝛼|𝜔|𝐽_𝑗+|𝜔𝑟|𝐾_𝑗+(1−𝛼)|𝜔|𝐽^′_𝑗+|𝜔𝑟|𝐾^′_𝑗^1/2. (11)

Proof. Let 𝜆 be an arbitrary eigenvalue of the iterative matrix 𝐿_𝜔,𝑟, then det𝜆𝐼−𝐿_𝜔,𝑟=0, (12) that is, det((𝜆+𝜔−1)𝐼−𝜔𝐽−𝜔𝑟𝐾)=0. (13) If (𝜆+𝜔−1)𝐼−𝜔𝐽−𝜔𝑟𝐾∈𝑆𝐷𝐷(𝛼), then |𝜆+𝜔−1|²>𝛼|𝜔|𝐽_𝑖+|𝜔𝑟|𝐾_𝑖+(1−𝛼)|𝜔|𝐽^′_𝑖+|𝜔𝑟|𝐾^′_𝑖×𝛼|𝜔|𝐽_𝑗+|𝜔𝑟|𝐾_𝑗+(1−𝛼)|𝜔|𝐽^′_𝑗+|𝜔𝑟|𝐾^′_𝑗,𝑖≠𝑗. (14) From Lemma 2, we know that (𝜆+𝜔−1)𝐼−𝜔𝐽−𝜔𝑟𝐾 (15) is nonsingular, that is det((𝜆+𝜔−1)𝐼−𝜔𝐽−𝜔𝑟𝐾)=det𝜆𝐼−𝐿_𝜔,𝑟≠0, (16) then 𝜆 is not an eigenvalue of the iterative matrix 𝐿_𝜔,𝑟.

However, 𝜆 is an eigenvalue of the iterative matrix 𝐿_𝜔,𝑟, then there exists at least a couple of 𝑖,𝑗∈𝑁(𝑖≠𝑗), such that |𝜆+𝜔−1|²≤𝛼|𝜔|𝐽_𝑖+|𝜔𝑟|𝐾_𝑖+(1−𝛼)|𝜔|𝐽^′_𝑖+|𝜔𝑟|𝐾^′_𝑖×𝛼|𝜔|𝐽_𝑗+|𝜔𝑟|𝐾_𝑗+(1−𝛼)|𝜔|𝐽^′_𝑗+|𝜔𝑟|𝐾^′_𝑗. (17) That is, |𝜆|²−2|1−𝜔||𝜆|+(1−𝜔)²≤𝛼|𝜔|𝐽_𝑖+|𝜔𝑟|𝐾_𝑖+(1−𝛼)|𝜔|𝐽^′_𝑖+|𝜔𝑟|𝐾^′_𝑖×𝛼|𝜔|𝐽_𝑗+|𝜔𝑟|𝐾_𝑗+(1−𝛼)|𝜔|𝐽^′_𝑗+|𝜔𝑟|𝐾^′_𝑗. (18) It is easy to find that the discriminant of a curve of second order Δ=4𝛼|𝜔|𝐽_𝑖+|𝜔𝑟|𝐾_𝑖+(1−𝛼)|𝜔|𝐽^′_𝑖+|𝜔𝑟|𝐾^′_𝑖×𝛼|𝜔|𝐽_𝑗+|𝜔𝑟|𝐾_𝑗+(1−𝛼)|𝜔|𝐽^′_𝑗+|𝜔𝑟|𝐾^′_𝑗≥0, (19) then the solution of (18) satisfies |𝜆|≤||+𝛼|𝜔|𝐽1−𝜔_𝑖+|𝜔𝑟|𝐾_𝑖+(1−𝛼)|𝜔|𝐽^′_𝑖+|𝜔𝑟|𝐾^′_𝑖×𝛼|𝜔|𝐽_𝑗+|𝜔𝑟|𝐾_𝑗+(1−𝛼)|𝜔|𝐽^′_𝑗+|𝜔𝑟|𝐾^′_𝑗^1/2,𝑖≠𝑗. (20) So, 𝜌(𝐿_𝜔,𝑟) satisfies the following inequality: 𝜌𝐿_𝜔,𝑟≤max_{𝑖,𝑗𝑖≠𝑗}||+𝛼𝜔|𝐽1−𝜔|_𝑖+|𝜔𝑟|𝐾_𝑖+(1−𝛼)|𝜔|𝐽^′_𝑖+|𝜔𝑟|𝐾^′_𝑖×𝛼|𝜔|𝐽_𝑗+|𝜔𝑟|𝐾_𝑗+(1−𝛼)|𝜔|𝐽^′_𝑗+|𝜔𝑟|𝐾^′_𝑗^1/2. (21)

3. Convergence of the GAOR Method

In this section, we investigate the convergence of the GAOR method to solve linear system (2). We assume that 𝐻 is a strictly doubly 𝛼 diagonally dominant coefficient matrix and obtain the regions of convergence of the GAOR method.

Theorem 4. If 𝐻∈𝑆𝐷𝐷(𝛼), 𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗<1,∀𝑖,𝑗∈𝑁,𝑖≠𝑗, (22) then the GAOR method converge if 𝜔,𝑟 satisfy either(Ι) 0<𝜔≤1, |𝑟|<min_{𝑖.𝑗𝑖≠𝑗}1−𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗×𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖+𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗⁻¹,−𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗−𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖+Δ₁×2𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗⁻¹ (23)or(ΙΙ)  1<𝜔<min2,min_{𝑖,𝑗𝑖≠𝑗}2−2𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗1−𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗,|𝑟|<min_{𝑖.𝑗𝑖≠𝑗}(2−𝜔)²−𝜔²𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖×𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗×𝜔²𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗+𝜔²𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗×𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖⁻¹,−𝜔²𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗−𝜔²𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖+Δ₂×𝜔²𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖×𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗⁻¹, (24)where Δ₁=𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗+𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖²+4𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗×1−𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗,Δ₂=𝜔²𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗+𝜔²𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖²+4𝜔²𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗×𝜔²−𝜔²𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖×𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗−4𝜔+4. (25)

Proof. Because 𝐻∈𝑆𝐷𝐷(𝛼), then the GAOR method converge if max_{𝑖,𝑗𝑖≠𝑗}|+𝛼|𝜔|𝐽|1−𝜔_𝑖+|𝜔𝑟|𝐾_𝑖+(1−𝛼)|𝜔|𝐽^′_𝑖+|𝜔𝑟|𝐾^′_𝑖×𝛼|𝜔|𝐽_𝑗+|𝜔𝑟|𝐾_𝑗+(1−𝛼)|𝜔|𝐽^′_𝑗+|𝜔𝑟|𝐾^′_𝑗^1/2<1. (26) That is, |1−𝜔|+max_{𝑖,𝑗𝑖≠𝑗}𝛼|𝜔|𝐽_𝑖+|𝜔𝑟|𝐾_𝑖+(1−𝛼)|𝜔|𝐽^′_𝑖+|𝜔𝑟|𝐾^′_𝑖×𝛼|𝜔|𝐽_𝑗+|𝜔𝑟|𝐾_𝑗+(1−𝛼)|𝜔|𝐽^′_𝑗+|𝜔𝑟|𝐾^′_𝑗^1/2<1. (27) Firstly, when 0<𝜔≤1, we have 𝜔²>𝛼|𝜔|𝐽_𝑖+|𝜔𝑟|𝐾_𝑖+(1−𝛼)|𝜔|𝐽^′_𝑖+|𝜔𝑟|𝐾^′_𝑖×𝛼|𝜔|𝐽_𝑗+|𝜔𝑟|𝐾_𝑗+(1−𝛼)|𝜔|𝐽^′_𝑗+|𝜔𝑟|𝐾^′_𝑗. (28) That is, |𝑟|²𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗+|𝑟|𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗+𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖+𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗−1<0. (29) Then, we have the following conditions.(1) When 𝑖,𝑗∈𝑁₁, then 𝐾_𝑖=𝐾_𝑗=0, 𝐾^′_𝑖=𝐾^′_𝑗=0. From 𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗<1, (30)we have |𝑟|<+∞. (31)(2) When 𝑖∈𝑁₁, 𝑗∈𝑁₂ or 𝑖∈𝑁₂, 𝑗∈𝑁₁, then |𝑟|𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖+𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗+𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗−1<0. (32)From [𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖][𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗]<1, we have |𝑟|<1−𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗×𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖+𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗⁻¹. (33)(3) When 𝑖,𝑗∈𝑁₂, it is easy to prove that the discriminant of a curve of second order of (29) Δ₁>0, and then |𝑟|<−𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗−𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖+Δ₁×2𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗⁻¹. (34)So, when 0<𝜔≤1, we have |𝑟|<min_{𝑖,𝑗𝑖≠𝑗}1−𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗×𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖+𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗⁻¹,−𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗−𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖+Δ₁×2𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗⁻¹. (35) Secondly, when 1<𝜔<2, we have 4−4𝜔+𝜔²>𝛼|𝜔|𝐽_𝑖+|𝜔𝑟|𝐾_𝑖+(1−𝛼)|𝜔|𝐽^′_𝑖+|𝜔𝑟|𝐾^′_𝑖×𝛼|𝜔|𝐽_𝑗+|𝜔𝑟|𝐾_𝑗+(1−𝛼)|𝜔|𝐽^′_𝑗+|𝜔𝑟|𝐾^′_𝑗. (36) That is, |𝑟|²𝜔²𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗+|𝑟|𝜔²𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗+𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖+𝜔²𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗−(𝜔−2)²<0. (37) Then we have the following conditions.(1) When 𝑖,𝑗∈𝑁₁, then 𝐾_𝑖=𝐾_𝑗=0, 𝐾^′_𝑖=𝐾^′_𝑗=0. From inequality (37), we have 𝜔²𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗−(𝜔−2)²<0. (38)From [𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖][𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗]<1, we have 1<𝜔<min_{𝑖,𝑗𝑖≠𝑗}2−2𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗1−𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗,|𝑟|<+∞. (39)(2) When 𝑖∈𝑁₁, 𝑗∈𝑁₂ or 𝑖∈𝑁₂, 𝑗∈𝑁₁, then 𝐾_𝑖=𝐾^′_𝑖=0 or 𝐾_𝑗=𝐾^′_𝑗=0. From inequality (37), we have |𝑟|𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗+𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝜔²+𝜔²𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗−(𝜔−2)²<0, (40)so, |𝑟|<(2−𝜔)²−𝜔²𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖×𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗×𝜔²𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗+𝜔²𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖⁻¹. (41)(3) When 𝑖,𝑗∈𝑁₂, then 𝐾_𝑖>0, 𝐾_𝑗>0, 𝐾^′_𝑖>0, 𝐾^′_𝑗>0.
It is easy to find that the discriminant of a curve of second order of (37) Δ₂>0. Then we have |𝑟|<−𝜔²𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗−𝜔²𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖+Δ₂×𝜔²𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗⁻¹. (42) So, when 1<𝜔<min{2,min_{𝑖,𝑗,𝑖≠𝑗}((2−2[𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖][𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗])/(1−[𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖][𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗]))}, we have |𝑟|<min_{𝑖,𝑗𝑖≠𝑗}(2−𝜔)²−𝜔²𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗×𝜔²𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗+𝜔²𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖⁻¹,−𝜔²𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐽_𝑗+(1−𝛼)𝐽^′_𝑗−𝜔²𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗𝛼𝐽_𝑖+(1−𝛼)𝐽^′_𝑖+Δ₂×𝜔²𝛼𝐾_𝑖+(1−𝛼)𝐾^′_𝑖𝛼𝐾_𝑗+(1−𝛼)𝐾^′_𝑗⁻¹. (43)

4. Examples

In this section, we give four numerical examples to show that our results are better than previous ones.

Example 1. Let 11𝐻=3131311313131=𝐼−𝐵₁𝐷𝐶𝐼−𝐵₂, (44) where 𝐼−𝐵₁=113131,𝐼−𝐵₂=(1). (45)

It is easy to know that 𝐻∈𝑆𝐷𝐷(1/2) and 1𝐽=0−3−13−1310−3−131304,𝐾=00000094929,𝐽₁=𝐽₂=𝐽₃=23,𝐾₁=𝐾₂=0,𝐾₃=109,𝐽^′₁=𝐽^′₂=𝐽^′₃=23,𝐾^′₁=𝐾^′₂=49,𝐾^′₃=29. (46) So 𝑁₁={1,2},𝑁₂={3},12𝐽_𝑖𝐽_𝑗+12𝐽^′_𝑖𝐽^′_𝑗=49<1,𝑖,𝑗∈𝑁,𝑖≠𝑗. (47)

By Theorem 4, we obtain the following regions of convergence: (Ι) 0<𝜔≤1 and |𝑟|<(√31−4)/2, or(ΙΙ) 1<𝜔<6/5 and |𝑟|<(27/4)(1/𝜔−1/2)²−3/4.

In addition, 𝐻∈𝑆𝐷𝐷.

By Theorem 2 of paper [6], we obtain the following regions of convergence:(Ι) 0<𝜔≤1 and |𝑟|<3/4, or(ΙΙ) 1<𝜔<6/5 and |𝑟|<(27/20)((2−𝜔)/𝜔)²−3/5.

From Figure 1, we know that the regions of convergence got by Theorem 4 in this paper (bounded by blue lines) are larger than ones got by Theorem 2 of paper [6] (bounded by green lines). From Example 1 of paper [6], we know that the regions of convergence got by Theorem 2 of paper [6] are larger than ones of paper [4] and paper [5]. So, the regions of convergence got by Theorem 4 in this paper are larger than ones of paper [4–6].

Figure 1 

Example 2. Let 11𝐻=4341413414141=𝐼−𝐵₁𝐷𝐶𝐼−𝐵₂, (48) where 𝐼−𝐵₁=114141,𝐼−𝐵₂=(1). (49)

It is easy to know that 𝐻∈𝑆𝐷𝐷(1/2) and 1𝐽=0−4−34−1430−4−14−1405,𝐾=000000516616,𝐽16₁=𝐽₂=1,𝐽₃=12,𝐽^′₁=𝐽^′₂=12,𝐽^′₃=32,𝐾₁=𝐾₂=0,𝐾₃=1,𝐾^′₁=𝐾^′₂=516,𝐾^′₃=38. (50) So, 𝑁₁={1,2},𝑁₂={3},12𝐽_𝑖𝐽_𝑗+12𝐽^′_𝑖𝐽^′_𝑗=58<1,𝑖,𝑗∈𝑁,𝑖≠𝑗. (51)

By Theorem 4, we obtain the following regions of convergence:(Ι) 0<𝜔≤1 and |𝑟|<(4/55)(√2289−43), or(ΙΙ) 1<𝜔<4(2−√3) and |𝑟|<(8/55)7040(1/𝜔−1/2)²+529−344/55.But 𝐻∉𝑆𝐷𝐷, so we cannot use Theorem 2 of paper [6]. And we cannot use the results of paper [4, 5].

Example 3. Let 11𝐻=3160701613131016016131=𝐼−𝐵₁𝐷𝐶𝐼−𝐵₂, (52) where 𝐼−𝐵₁=11301,𝐼−𝐵₂=116131. (53)

Obviously, 𝐻∈𝑆𝐷𝐷(1) and 1𝐽=0−3−160700−6−13−13100−610−6−130,1𝐾=0000000031910011867136.18 (54)

By Theorem 4, we obtain the following regions of convergence: (Ι) 0<𝜔≤1 and |𝑟|<1/3, or (ΙΙ) 1<𝜔<8−4√3 and |𝑟|<(4/3)((2−𝜔)/𝜔)²−1.

Consider the following linear system: 𝐻𝑦=𝑓, (55) where 11𝐻=31607016131310160161313,𝑓=2523232, (56) which has the exact solution 𝑦=(1,1,1,1)^𝑇. (57)

We choose the initial solution 𝑦₀=(0,0,0,0)^𝑇. If 𝜔=101/100,𝑟=−1/5, we need 28 iterations to achieve four decimal digits of accuracy in the solution components. If 𝜔=101/100, 𝑟=1/5, we need 18 iterations to achieve four decimal digits of accuracy in the solution components. If 𝜔=1/2,𝑟=1/4, we need 22 iterations to achieve four decimal digits of accuracy in the solution components.

Example 4. Consider 𝐻=^{111960⋯⋯⋯⋯0−1962111960⋯⋯⋯0−196−196211196196⋯⋯196−1961960−211196⋯196196}₋₁^{96⋮⋮⋮⋮⋮⋮⋮⋮⋮}_{1960⋯0−211196196−1961960⋯⋯0−211196−1961960⋯⋯⋯0−211−1961960⋯⋯⋯⋯0−2151×51}, (58)
where 𝐼−𝐵₁=11−196211,𝐷=0⋯⋯01961960⋯0.96 (59) Obviously, 𝐻∈𝑆𝐷𝐷(1/2), and 𝐽=^{10−1960⋯⋯⋯⋯0−196210−1960⋯⋯⋯0−196196210−−196196⋯⋯−−1961960196210−−196196⋯−−196}₁^{96⋮⋮⋱⋱⋱⋱⋱⋮⋮}_{1960⋯0210−−196−1961961960⋯⋯0210−−1961961960⋯⋯⋯0210−1961960⋯⋯⋯⋯02051×51},𝐾=^{0000⋯⋯⋯000000⋯⋯⋯0023−964609−192161920⋯000−49−19216−1961921600⋯000−}₋₁^{9216⋮⋮⋮⋮⋮⋮⋮⋮⋮}_{−1961921600⋯000−−19216−1961921600⋯000−−19216−1961921600⋯000−−19216−1961921600⋯000−921651×51}. (60)

By Theorem 4, we obtain the following regions of convergence: (Ι) 0<𝜔≤1 and |𝑟|<2208/3481, or (ΙΙ) 1<𝜔<192/169 and |𝑟|<(18432−16224𝜔)/3481𝜔.

Acknowledgments

The authors would like to thank the referees for their valuable comments and suggestions, which greatly improved the original version of this paper. This work was supported by the National Natural Science Foundation of China (No. 11001144) and the Natural Science Foundation of Shandong Province (No. ZR2012AL09).