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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 743861, 14 pages
http://dx.doi.org/10.1155/2012/743861
Research Article

Generalized AOR Method for Solving Absolute Complementarity Problems

1Department of Mathematics, COMSATS Institute of Information Technology, Park Road, Chak Shahzad, Islamabad 44000, Pakistan
2Mathematics Department, College of Science, King Saud University, Riyadh 11451, Saudi Arabia

Received 25 January 2012; Accepted 9 February 2012

Academic Editor: Yonghong Yao

Copyright © 2012 Muhammad Aslam Noor et al. 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.

Abstract

We introduce and consider a new class of complementarity problems, which is called the absolute value complementarity problem. We establish the equivalence between the absolute complementarity problems and the fixed point problem using the projection operator. This alternative equivalent formulation is used to discuss the existence of a solution of the absolute value complementarity problem. A generalized AOR method is suggested and analyzed for solving the absolute the complementarity problems. We discuss the convergence of generalized AOR method for the L-matrix. Several examples are given to illustrate the implementation and efficiency of the method. Results are very encouraging and may stimulate further research in this direction.

1. Introduction

Complementarity theory introduced and studied by Lemke [1] and Cottle and Dantzig [2] has enjoyed a vigorous growth for the last fifty years. It is well known that both the linear and nonlinear programs can be characterized by a class of complementarity problems. The complementarity problems have been generalized and extended to study a wide class of problems, which arise in pure and applied sciences; see [124] and the references therein. Equally important is the variational inequality problem, which was introduced and studied in the early sixties. The theory of variational inequality has been developed not only to study the fundamental facts on the qualitative behavior of solutions but also to provide highly efficient new numerical methods for solving various nonlinear problems. For the recent applications, formulation, numerical results, and other aspects of the variational inequalities, see [1322].

Motivated and inspired by the research going on in these areas, we introduced and consider a new class of complmenetarity problems, which is called the absolute value complmenetarity problem. Related to the absolute value complementarity problem, we consider the problem of solving the absolute value variational inequality. We show that if the under lying set is a convex cone, then both these problems are equivalent. If the underlying set is the whole space, then the absolute value problem is equivalent to solving the absolute value equations, which have been studied extensively in recent years.

We use the projection technique to show that the absolute value complementarity problems are equivalent to the fixed point problem. This alternative equivalent form is used to study the existence of a unique solution of the absolute value complementarity problems under some suitable conditions. We again use the fixed point formulation to suggest and analyze a generalized AOR method for solving the absolute value complementarity problems. The convergence analysis of the proposed method is considered under some suitable conditions. Some examples are given to illustrate the efficiency and implementation of the proposed iterative methods. Results are very encouraging. The ideas and the technique of this paper may stimulate further research in these areas.

Let 𝑅𝑛 be an inner product space, whose inner product and norm are denoted by , and , respectively. For a given matrix 𝐴𝑅𝑛×𝑛, a vector 𝑏𝑅𝑛, we consider the problem of finding 𝑥𝑅𝑛 such that𝑥𝐾,𝐴𝑥|𝑥|𝑏𝐾,𝐴𝑥|𝑥|𝑏,𝑥=0,(1.1) where 𝐾={𝑥𝑅𝑛𝑥,𝑦0,𝑦𝐾} is the polar cone of a closed and convex cone 𝐾, 𝐴𝑅𝑛×𝑛,𝑏𝑅𝑛, and |𝑥| will denote the vector in 𝑅𝑛 with absolute values of components of 𝑥𝑅𝑛.

We remark that the absolute value complementarity problem (1.1) can be viewed as an extension of the complementarity problem considered by Lemke [1]. To solve the linear complementarity problems, several methods were proposed. These methods can be divided into two categories, the direct and indirect (iterative) methods. Lemke [1] and Cottle and Dantzig [2] developed the direct methods for solving linear complementarity problems based on the process of pivoting, whereas Mangasarian [10], Noor [14, 15], and Noor et al. [2022] considered the iterative methods. For recent applications, formulations, numerical methods, and other aspects of the complementarity problems and variational inequalities, see [124].

Let 𝐾 be a closed and convex set in the inner product space 𝑅𝑛. We consider the problem of finding 𝑥𝐾 such that𝐴𝑥|𝑥|𝑏,𝑦𝑥0,𝑦𝐾.(1.2) The problem (1.2) is called the absolute value variational inequality, which is a special form of the mildly nonlinear variational inequalities, introduced and studied by Noor [13] in 1975.

If 𝐾=𝑅𝑛, then the problem (1.2) is equivalent to find 𝑥𝑅𝑛 such that𝐴𝑥|𝑥|𝑏=0,(1.3) which are known as the absolute value equations. These equations have been considered and studied extensively in recent years; see [712, 1720, 23, 24]. We would like to emphasize that Mangasarian [1, 2, 411] has shown that the absolute value equations (1.3) are equivalent to the complementarity problems (1.1). Mangasarian [711] has used the complementarity approach to solve the absolute value equations (1.3). For other methods, see [712, 2024] and the references therein.

In this paper, we suggest a generalized AOR method for solving absolute complementarity problem, which is easy to implement and gives almost exact solution of (1.3).

We also need the following definitions and concepts.

Definition 1.1. 𝐵𝑅𝑛×𝑛 is called an 𝐿-matrix if 𝑏𝑖𝑖>0 for 𝑖=1,2,,𝑛, and 𝑏𝑖𝑗0 for 𝑖𝑗,𝑖,𝑗=1,2,,𝑛.

Definition 1.2. A matrix 𝐴𝑅𝑛×𝑛 is said to be positive definite matrix, if there exists a constant 𝛾>0, such that 𝐴𝑥,𝑥𝛾𝑥2,𝑥𝑅𝑛,(1.4) and bounded if there exists a constant 𝛽>0 such that 𝐴𝑥𝛽𝑥,𝑥𝑅𝑛.(1.5)

2. Iterative Methods

To propose and analyze algorithm for absolute complementarity problems, we need the following well-known results.

Lemma 2.1 (see [16]). Let 𝐾 be a nonempty closed convex set in 𝑅𝑛. For a given 𝑧𝑅𝑛,𝑢𝐾 satisfies the inequality 𝑢𝑧,𝑢𝑣0,𝑣𝐾,(2.1) if and only if 𝑢=𝑃𝐾𝑧,(2.2) where 𝑃𝐾 is the projection of 𝑅𝑛 onto the closed convex set 𝐾.

Lemma 2.2. If 𝐾 is the positive cone in 𝑅𝑛, then 𝑥𝐾 is a solution of absolute value variational inequality (1.2) if and only if 𝑥𝐾 solves the absolute value complementarity problem (1.1).

Proof. Let 𝑥𝐾  be the solution of (1.2). Then 𝐴𝑥|𝑥|𝑏,𝑦𝑥0,𝑦𝐾.(2.3) Since 𝐾  is a cone, taking 𝑦=0𝐾 and 𝑦=2𝑥𝐾, we have 𝐴𝑥|𝑥|𝑏,𝑥=0.(2.4) From inequality (1.2), we have 0𝐴𝑥|𝑥|𝑏,𝑦𝑥=𝐴𝑥|𝑥|𝑏,𝑦𝐴𝑥|𝑥|𝑏,𝑥,=𝐴𝑥|𝑥|𝑏,𝑦,𝑦𝐾,(2.5) from which it follows that 𝐴𝑥|𝑥|𝑏𝐾. Thus we conclude that 𝑥𝐾 is the solution of absolute complementarity problems (1.1).
Conversely, let 𝑥𝐾 be a solution of (1.1). Then 𝑥𝐾,𝐴𝑥|𝑥|𝑏𝐾,𝐴𝑥|𝑥|𝑏,𝑥=0.(2.6) From (2.5) and (2.6), it follows that 𝐴𝑥|𝑥|𝑏,𝑦𝑥0,𝑦𝐾.(2.7) Hence 𝑥𝐾 is the solution of absolute variational inequality (1.2).

From Lemma 2.1, it follows that both the problems (1.1) and (1.2) are equivalent.

In the next result, we prove the equivalence between the absolute value variational inequality (1.2) and the fixed point.

Lemma 2.3. Let 𝐾 be closed convex set in 𝑅𝑛. Then, for a constant 𝜌>0,𝑥𝐾 satisfies (1.2) if and only if 𝑥𝐾 satisfies the relation 𝑥=𝑃𝐾[](𝑥𝜌𝐴𝑥|𝑥|𝑏),(2.8) where 𝑃𝐾 is the projection of 𝑅𝑛 onto the closed convex set 𝐾.

Proof. Let 𝑥𝐾  be the solution of (1.2). Then, for a positive constant 𝜌>0, 𝑥(𝑥𝜌(𝐴𝑥|𝑥|𝑏)),𝑦𝑥0,𝑦𝐾.(2.9) Using Lemma 2.1, we have 𝑥=𝑃𝐾[](𝑥𝜌𝐴𝑥|𝑥|𝑏),(2.10) which is the required result.

Now using Lemmas 2.2 and 2.3, we see that the absolute value complementarity problem (1.1) is equivalent to the fixed point problem of the following type:𝑥=𝑃𝐾[](𝑥𝜌𝐴𝑥|𝑥|𝑏).(2.11) We use this alternative fixed point formulation to study the existence of a unique solution of the absolute value complementarity problem. Equation (1.1) and this is the main motivation of our next result.

Theorem 2.4. Let 𝐴𝑅𝑛×𝑛 be a positive definite matrix with constant 𝛼>0 and continuous with constant 𝛽>0. If 0<𝜌<2(𝛾1)/(𝛽21),𝛾>1, then there exists a unique solution 𝑥𝐾 such that 𝐴𝑥|𝑥|𝑏,𝑦𝑥0,𝑦𝐾,(2.12) where 𝐾 is a closed convex set in 𝑅𝑛.

Proof. Uniqueness: Let 𝑥1𝑥2𝐾 be two solutions of (1.2). Then 𝐴𝑥1||𝑥1||𝑏,𝑦𝑥10,𝑦𝐾,𝐴𝑥2||𝑥2||𝑏,𝑦𝑥20,𝑦𝐾.(2.13) Taking 𝑦=𝑥2𝐾 in (2.13) and 𝑦=𝑥1𝐾 in (2.6), we have 𝐴𝑥1+||𝑥1||+𝑏,𝑥1𝑥20,𝐴𝑥2||𝑥2||𝑏,𝑥1𝑥20.(2.14) Adding the previous inequalities, we obtain 𝐴𝑥1𝑥2||𝑥1||+||𝑥2||,𝑥1𝑥20,(2.15) which implies that 𝐴𝑥1𝑥2,𝑥1𝑥2𝑥1𝑥220.(2.16) Since 𝐴 is positive definite, from (2.16), we have 𝑥(𝛾1)1𝑥220.(2.17) As 𝛾>1, therefore from (2.17) we have 𝑥1𝑥220,(2.18) which contradicts the fact that 𝑥1𝑥220; hence 𝑥1=𝑥2.
Existence
Let 𝑥𝐾 be the solution of (1.2). Then 𝐴𝑥|𝑥|𝑏,𝑦𝑥0,𝑦𝐾.(2.19) From Lemma 2.3, we have 𝑥=𝑃𝐾[](𝑥𝜌𝐴𝑥|𝑥|𝑏).(2.20) Define a mapping 𝐹(𝑥)=𝑥=𝑃𝐾[](𝑥𝜌𝐴𝑥|𝑥|𝑏).(2.21) To show that the mapping 𝐹(𝑥) defined by (2.21) has a fixed point, it is enough to prove that 𝐹(𝑥) is a contraction mapping. For 𝑥1𝑥2𝐾, consider 𝐹𝑥1𝑥𝐹2=𝑃𝐾𝑥1𝜌𝐴𝑥1||𝑥1||𝑏𝑃𝐾𝑥2𝜌𝐴𝑥2||𝑥2||𝑥𝑏1𝜌𝐴𝑥1||𝑥1||𝑥𝑏2𝜌𝐴𝑥2||𝑥2||𝑥𝑏1𝑥2𝜌𝐴𝑥1𝐴𝑥2||𝑥+𝜌1||||𝑥2||=𝑥1𝑥2𝑥𝜌𝐴1𝑥2𝑥+𝜌1𝑥2,(2.22) where we have used the fact that 𝑃𝐾 is nonexpansive. Now using positive definiteness of 𝐴, we have 𝑥1𝑥2𝑥𝜌𝐴1𝑥22=𝑥1𝑥2𝑥𝜌𝐴1𝑥2,𝑥1𝑥2𝑥𝜌𝐴1𝑥2𝑥1𝑥22𝑥2𝜌𝛾1𝑥2𝑥,𝐴1𝑥2+𝜌2𝛽2𝑥1𝑥22=12𝛼𝜌+𝛽2𝜌2𝑥1𝑥22.(2.23) From (2.22) and (2.23), we have 𝐹𝑥1𝑥𝐹2𝑥𝜃1𝑥2,(2.24) where 𝜃=(𝜌+12𝛾𝜌+𝛽2𝜌2).Form 0<𝜌<2(𝛾1)/(𝛽21) and 𝜌<1, we have 𝜃<1.
This shows that 𝐹(𝑥)  is a contraction mapping and has a fixed point 𝑥𝐾 satisfying the absolute value variational inequality (1.2).

For the sake of simplicity, we consider the special case, when 𝐾=[0,𝑐] is a closed convex set in 𝑅𝑛 and we define the projection 𝑃𝐾𝑥 as𝑃𝐾𝑥𝑖=minmax0,𝑥𝑖,𝑐𝑖,𝑖=1,2,,𝑛.(2.25) We recall the following well-known result.

Lemma 2.5 (see [3]). For any 𝑥  and 𝑦 in 𝑅𝑛, the projection 𝑃𝐾𝑥 has the following properties:(i)𝑃𝐾(𝑥+𝑦)𝑃𝐾𝑥+𝑃𝐾𝑦, (ii)𝑃𝐾𝑥𝑃𝐾𝑦𝑃𝐾(𝑥𝑦), (iii)𝑥𝑦𝑃𝐾𝑥𝑃𝐾𝑦,(iv)𝑃𝐾𝑥+𝑃𝐾(𝑥)|𝑥|,withequality,ifandonlyif𝑐𝑥𝑐.

We now suggest the iterative methods for solving the absolute value complmentarity problems (1.1). For this purpose, we decompose the matrix 𝐴  as,𝐴=𝐷+𝐿+𝑈,(2.26) where 𝐷 is the diagonal matrix, and 𝐿  and 𝑈 are strictly lower and strictly upper triangular matrices, respectively. Let Ω=diag(𝜔1,𝜔2,,𝜔𝑛) with 𝜔𝑖𝑅+ and let 𝛼 be a real number.

Algorithm 2.6. Step 1. Choose an initial vector 𝑥0𝑅𝑛 and a parameter Ω𝑅+.Set𝑘=0.Step 2. Calculate 𝑥𝑘+1=𝑃𝐾𝑥𝑘𝐷1𝛼Ω𝐿𝑢𝑘+1+(Ω𝐴𝛼Ω𝐿)𝑢𝑘||𝑥Ω𝑘||+𝑏.(2.27)Step 3. If 𝑥𝑘+1=𝑥𝑘, then stop. Else, set𝑘=𝑘+1 and go to Step 2.Now we define an operator 𝑔𝑅𝑛𝑅𝑛 such that 𝑔(𝑥)=𝜉, where 𝜉 is the fixed point of the system 𝜉=𝑃𝐾𝑥𝐷1[]𝛼Ω𝐿𝜉+(Ω𝐴𝛼Ω𝐿)𝑥Ω(|𝑥|+𝑏).(2.28) We also assume that the set 𝜑={𝑥𝑅𝑛𝑥0,𝐴𝑥|𝑥|𝑏0}(2.29) of the absolute value complementarity problem is nonempty.

To prove the convergence of Algorithm 2.6, we need the following result.

Theorem 2.7. Consider the operator 𝑔𝑅𝑛𝑅𝑛 as defined in (2.28). Assume that𝐴𝑅𝑛×𝑛 is an 𝐿-matrix. Also assume that 0<𝜔𝑖1,0𝛼1. Then, for any 𝑥𝜑, the following holds:(i)𝑔(𝑥)𝑥, (ii)𝑥𝑦𝑔(𝑥)𝑔(𝑦), (iii)𝜉=𝑔(𝑥)𝜑.

Proof. To prove (i), we need to verify that 𝜉𝑖𝑥𝑖,𝑖=1,2,,𝑛 hold with 𝜉𝑖 satisfying 𝜉𝑖=𝑃𝐾𝑥𝑖𝑎1𝑖𝑖𝛼𝜔𝑖𝑖1𝑗=1𝐿𝑖𝑗𝜉𝑗𝑥𝑗+𝜔𝑖(𝐴𝑥|𝑥|𝑏)𝑖.(2.30) To prove the required result, we use mathematical induction. For this let 𝑖=1, 𝜉1=𝑃𝐾𝑥1𝑎111𝜔1(𝐴𝑥|𝑥|𝑏)1.(2.31) Since 𝐴𝑥|𝑥|𝑏0,𝜔𝑖>0, therefore 𝜉1𝑥1. For 𝑖=2, we have 𝜉2=𝑃𝐾𝑥2𝑎122𝛼𝜔2𝐿21𝜉1𝑥1+𝜔2(𝐴𝑥|𝑥|𝑏)2.(2.32) Here 𝐴𝑥|𝑥|𝑏0,𝜔𝑖>0,𝐿210, and 𝜉1𝑥10. This implies that 𝜉2𝑥2.
Suppose that 𝜉𝑖𝑥𝑖for𝑖=1,2,,𝑘1.(2.33) We have to prove that the statement is true, for 𝑖=𝑘, that is, 𝜉𝑘𝑥𝑘.(2.34) Consider 𝜉𝑘=𝑃𝐾𝑥𝑘𝑎1𝑘𝑘𝛼𝜔𝑘𝑘1𝑗=1𝐿𝑘𝑗𝜉𝑗𝑥𝑗+𝜔𝑘(𝐴𝑥|𝑥|𝑏)𝑘,=𝑃𝐾𝑥𝑘𝑎1𝑘𝑘𝛼𝜔𝑘𝐿𝑘1𝜉1𝑥1+𝐿𝑘2𝜉2𝑥2++𝐿𝑘𝑘1𝜉𝑘1𝑥𝑘1+𝜔𝑘(𝐴𝑥|𝑥|𝑏)𝑘.(2.35) Since 𝐴𝑥|𝑥|𝑏0,𝜔𝑘>0,𝐿𝑘1,𝐿𝑘2,,𝐿𝑘𝑘10, and 𝜉𝑖𝑥𝑖 for 𝑖=1,2,,𝑘1, from (2.35) we can write 𝜉𝑘𝑥𝑘. Hence (i) is proved.
Now we prove (ii). For this let us suppose that 𝜉=𝑔(𝑥) and 𝜙=𝑔(𝑦). We will prove 𝑥𝑦𝜉𝜙.(2.36) As 𝜉=𝑃𝐾𝑥𝐷1[]𝛼Ω𝐿𝜉+(Ω𝐴𝛼Ω𝐿)𝑥Ω(|𝑥|+𝑏),(2.37) so 𝜉𝑖  can be written as 𝜉𝑖=𝑃𝐾𝑥𝑖𝑎1𝑖𝑖𝛼𝜔𝑖𝑖1𝑗=1𝐿𝑖𝑗𝜉𝑗+𝜔𝑖𝑎𝑖𝑖𝑥𝑖+(1𝛼)𝜔𝑖𝑖1𝑗=1𝐿𝑖𝑗𝑥𝑗+𝜔𝑖𝑛𝑗=1𝑗𝑖𝑈𝑖𝑗𝑥𝑗𝜔𝑖||𝑥𝑖||𝜔𝑖𝑏𝑖=𝑃𝐾1𝜔𝑖𝑥𝑖𝑎1𝑖𝑖𝛼𝜔𝑖𝑖1𝑗=1𝐿𝑖𝑗𝜉𝑗+(1𝛼)𝜔𝑖𝑖1𝑗=1𝐿𝑖𝑗𝑥𝑗+𝜔𝑖𝑛𝑗=1𝑗𝑖𝑈𝑖𝑗𝑥𝑗𝜔𝑖||𝑥𝑖||𝜔𝑖𝑏𝑖.(2.38) Similarly, for 𝜙𝑖, we have 𝜙𝑖=𝑃𝐾1𝜔𝑖𝑦𝑖𝑎1𝑖𝑖𝛼𝜔𝑖𝑖1𝑗=1𝐿𝑖𝑗𝜙𝑗+(1𝛼)𝜔𝑖𝑖1𝑗=1𝐿𝑖𝑗𝑦𝑗+𝜔𝑖𝑛𝑗=1𝑗𝑖𝑈𝑖𝑗𝑦𝑗𝜔𝑖||𝑦𝑖||𝜔𝑖𝑏𝑖,(2.39) and for 𝑖=1, 𝜙1=𝑃𝐾1𝜔1𝑦1𝑎111𝜔1𝑛𝑗=1𝑗𝑖𝑈1𝑗𝑦𝑗||𝑦1||𝑏1𝑃𝐾1𝜔1𝑥1𝑎111𝜔1𝑛𝑗=1𝑗𝑖𝑈1𝑗𝑥𝑗||𝑥1||𝑏1=𝜉1.(2.40) Since 𝑦1𝑥1, therefore |𝑦1||𝑥1|. Hence it is true for 𝑖=1. Suppose it is true for 𝑖=1,2,𝑘1; we will prove it for 𝑖=𝑘; for this consider 𝜙𝑘=𝑃𝐾1𝜔𝑘𝑦𝑘𝑎1𝑘𝑘𝛼𝜔𝑘𝑘1𝑗=1𝐿𝑘𝑗𝜙𝑗+(1𝛼)𝜔𝑘𝑘1𝑗=1𝐿𝑘𝑗𝑦𝑗+𝜔𝑘𝑛𝑗=1𝑗𝑖𝑈𝑘𝑗𝑦𝑗𝜔𝑘||𝑦𝑘||𝜔𝑘𝑏𝑘𝑃𝐾1𝜔𝑘𝑥𝑘𝑎1𝑘𝑘𝛼𝜔𝑘𝑘1𝑗=1𝐿𝑘𝑗𝜉𝑗+(1𝛼)𝜔𝑘𝑘1𝑗=1𝐿𝑘𝑗𝑥𝑗+𝜔𝑘𝑛𝑗=1𝑗𝑖𝑈𝑘𝑗𝑥𝑗𝜔𝑘||𝑥𝑘||𝜔𝑘𝑏𝑘=𝜉𝑘.(2.41) Since 𝑥𝑦, and 𝜉𝑖𝜙𝑖 for 𝑖=1,2,𝑘1, hence it is true for 𝑘 and (ii) is verified.
Next we prove (iii), that is, 𝜉=𝑔(𝑥)𝜑.(2.42) Let 𝜆=𝑔(𝜉)=𝑃𝐾(𝜉𝐷1Ω[𝛼𝐿(𝜆𝜉)+𝐴𝜉|𝜉|𝑏]) from (i) 𝑔(𝜉)=𝜆𝜉. Also by definition of 𝑔,𝜉=𝑔(𝑥)0 and 𝜆=𝑔(𝜉)0.
Now 𝜆𝑖=𝑃𝐾𝜉𝑖𝑎1𝑖𝑖𝛼𝜔𝑖𝑖1𝑗=1𝐿𝑖𝑗𝜆𝑗𝜉𝑗+𝜔𝑖||𝜉||𝐴𝜉𝑏𝑖.(2.43) For 𝑖=1,𝜉10 by definition of 𝑔. Suppose that (𝐴𝜉|𝜉|𝑏)𝑖<0, so 𝜆1=𝑃𝐾𝜉1𝑎111𝜔1||𝜉||𝐴𝜉𝑏1>𝑃𝐾𝜉1=𝜉1,(2.44) which contradicts the fact that 𝜆𝜉. Therefore, (𝐴𝜉|𝜉|𝑏)𝑖0.
Now we prove it for any 𝑘 in 𝑖=1,2,,𝑛. Suppose the contrary (𝐴𝜉|𝜉|𝑏)𝑖<0; then 𝜆𝑘=𝑃𝐾𝜉𝑘𝑎1𝑘𝑘𝛼𝜔𝑘𝑘1𝑗=1𝐿𝑘𝑗𝜆𝑗𝜉𝑗+𝜔𝑘||𝜉||𝐴𝜉𝑏𝑘.(2.45) As it is true for all 𝛼[0,1], it should be true for 𝛼=0. That is, 𝜆𝑘=𝑃𝐾𝜉𝑘𝑎1𝑘𝑘𝜔𝑘||𝜉||𝐴𝜉𝑏𝑘>𝑃𝐾𝜉𝑘=𝜉𝑘,(2.46) which contradicts the fact that 𝜆𝜉. So (𝐴𝜉|𝜉|𝑏)𝑘0, for any 𝑘 in 𝑖=1,2,,𝑛. Hence 𝜉=𝑓(𝑥)𝜑.

We now consider the convergence criteria of Algorithm 2.6 and this is the main motivation of our next result.

Theorem 2.8. Assume that 𝐴𝑅𝑛×𝑛 is an 𝐿-matrix. Also assume that 0<𝜔𝑖1,0𝛼1. Then for any initial vector 𝑥0𝜑, the sequence {𝑥𝑘},𝑘=0,1,2,, defined by Algorithm 2.6 has the following properties:(i)0𝑥𝑘+1𝑥𝑘𝑥0;𝑘=0,1,2,, (ii)lim𝑘𝑥𝑘=𝑥 is a unique solution of the absolute value complementarity problem (1.1).

Proof. Since 𝑥0𝜑, by (i) of Theorem 2.7, we have 𝑥1𝑥0 and 𝑥1𝜑. Recursively using Theorem 2.7 we obtain 0𝑥𝑘+1𝑥𝑘𝑥0;𝑘=0,1,2,.(2.47) From (i) we observe that the sequence {𝑥𝑘}is monotone bounded; therefore, it converges to some 𝑥𝑅𝑛+ satisfying 𝑥=𝑃𝐾𝑥𝐷1𝛼Ω𝐿𝑥+(Ω𝐴𝛼Ω𝐿)𝑥||𝑥Ω||+𝑏=𝑃𝐾𝑥𝐷1Ω𝐴𝑥||𝑥Ω||.Ω𝑏(2.48) Hence 𝑥is the solution of the absolute value complementarity problem (1.1).

3. Numerical Results

In this section, we consider several examples to show the efficiency of the proposed method. The convergence of the generalized AOR method is guaranteed for L-matrices only but it is also possible to solve different types of systems. The elements of the diagonal matrix  Ω are chosen from the interval [𝑎,𝑏] such that𝜔𝑖=𝑎+(𝑏𝑎)𝑖𝑛,𝑖=1,2,,𝑛,(3.1) where 𝜔𝑖  is the 𝑖th diagonal element of Ω. All the experiments are performed with Intel(R) Core(TM) 2 × 2.1 GHz, 1 GB RAM, and the codes are written in Matlab 7.

Example 3.1. We test Algorithm 2.6 on 𝑚 consecutively generated solvable random problems 𝐴𝑅𝑛×𝑛, and 𝑛 ranging from 10 to 1000. We chose a random matrix 𝐴 from a uniform distribution on [0,1], such that whose all diagonal elements are equal to 1000 and 𝑥  is chosen randomly from a uniform distribution on [0,1]. The constant vector is computed as 𝑏=𝐴𝑥|𝑥|. The computational results are shown in Table 1.

tab1
Table 1: Computational results.

Example 3.2. Consider the ordinary differential equation: 𝑑2𝑥𝑑𝑡2|𝑥|=1𝑥2,0𝑥1,𝑥(0)=0,𝑥(1)=1.(3.2) The exact solution is 𝑥(𝑡)=0.7378827425sin(𝑡)3cos(𝑡)+3𝑥2,𝑥<0,0.7310585786𝑒𝑥0.2689414214𝑒𝑥+1+𝑥2,𝑥>0.(3.3) We take 𝑛=10; the matrix 𝐴 is given by 𝑎𝑖,𝑗=242,for𝑗=𝑖,121,for𝑗=𝑖+1,𝑖=1,2,,𝑛1,𝑗=𝑖1,𝑖=2,3,,𝑛,0,otherwise.(3.4) The constant vector 𝑏 is given by 𝑏=120,121117,121112,121105,12196,12185,12172,12157,12140,12114620121𝑇.(3.5) Here 𝐴 is not 𝐿-matrix. The comparison between the exact solution and the approximate solutions is given in Figure 1.

743861.fig.001
Figure 1

Example 3.3. Let the matrix 𝐴 be given by 𝑎𝑖,𝑗=8,for𝑗=𝑖,1,for𝑗=𝑖+1,𝑖=1,2,,𝑛1,𝑗=𝑖1,𝑖=2,3,,𝑛,0,otherwise.(3.6) Let 𝑏=(6,5,5,,5,6)𝑇, the problem size 𝑛 ranging from 4 to 1024. The stopping criteria are 𝐴𝑥|𝑥|𝑏<106. We choose initial guess 𝑥0 as 𝑥0=(0,0,,0)𝑇. The computational results are shown in Table 2.
In Table 2 TOC denotes total time taken by CPU. The rate of convergence of AOR method is better than iterative method [21].

tab2
Table 2: Computational results.

4. Conclusion

In this paper, we have introduced a new class of complementarity problems, known as the absolute value complementarity problems. We have used the projection technique to establish the equivalence between the absolute value variational inequalities, fixed point problems, and the absolute value complementarity problems. This equivalence is used to discuss the existence of a unique solution of the absolute value problems under some suitable conditions. We have also used this alternative equivalent formulation to suggest and analyze an iterative method for solving the absolute value complementarity problems. Some special cases are also discussed. The results and ideas of this paper may be used to solve the variational inequalities and related optimization problems. This is another direction for future research.

Acknowledgments

This paper is supported by the Visiting Professor Program of King Saud University, Riyadh, Saudi Arabia. The authors are also grateful to Dr. S. M. Junaid Zaidi, Rector, COMSATS Institute of Information Technology, Pakistan, for providing the excellent research facilities.

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