Generalized AOR Method for Solving Absolute Complementarity Problems

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 [1–24] 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 [13–22].

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 𝑅^𝑛 be an inner product space, whose inner product and norm are denoted by ⟨⋅,⋅⟩ and ‖⋅‖, respectively. For a given matrix 𝐴∈𝑅^𝑛×𝑛, a vector 𝑏∈𝑅^𝑛, we consider the problem of finding 𝑥∈𝑅^𝑛 such that𝑥∈𝐾,𝐴𝑥−|𝑥|−𝑏∈𝐾^∗,⟨𝐴𝑥−|𝑥|−𝑏,𝑥⟩=0,(1.1) where 𝐾^∗={𝑥∈𝑅^𝑛∶⟨𝑥,𝑦⟩≥0,∀𝑦∈𝐾} is the polar cone of a closed and convex cone 𝐾, 𝐴∈𝑅^𝑛×𝑛,𝑏∈𝑅^𝑛, and |𝑥| will denote the vector in 𝑅^𝑛 with absolute values of components of 𝑥∈𝑅^𝑛.

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. [20–22] considered the iterative methods. For recent applications, formulations, numerical methods, and other aspects of the complementarity problems and variational inequalities, see [1–24].

Let 𝐾 be a closed and convex set in the inner product space 𝑅^𝑛. We consider the problem of finding 𝑥∈𝐾 such that⟨𝐴𝑥−|𝑥|−𝑏,𝑦−𝑥⟩≥0,∀𝑦∈𝐾.(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=𝑅^𝑛, then the problem (1.2) is equivalent to find 𝑥∈𝑅^𝑛 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 [7–12, 17–20, 23, 24]. We would like to emphasize that Mangasarian [1, 2, 4–11] has shown that the absolute value equations (1.3) are equivalent to the complementarity problems (1.1). Mangasarian [7–11] has used the complementarity approach to solve the absolute value equations (1.3). For other methods, see [7–12, 20–24] 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. 𝐵∈𝑅^𝑛×𝑛 is called an 𝐿-matrix if 𝑏_𝑖𝑖>0 for 𝑖=1,2,…,𝑛, and 𝑏_𝑖𝑗≤0 for 𝑖≠𝑗,𝑖,𝑗=1,2,…,𝑛.

Definition 1.2. A matrix 𝐴∈𝑅^𝑛×𝑛 is said to be positive definite matrix, if there exists a constant 𝛾>0, such that ⟨𝐴𝑥,𝑥⟩≥𝛾‖𝑥‖²,∀𝑥∈𝑅^𝑛,(1.4) and bounded if there exists a constant 𝛽>0 such that ‖𝐴𝑥‖≤𝛽‖𝑥‖,∀𝑥∈𝑅^𝑛.(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 𝑅^𝑛. For a given 𝑧∈𝑅^𝑛,𝑢∈𝐾 satisfies the inequality ⟨𝑢−𝑧,𝑢−𝑣⟩≥0,𝑣∈𝐾,(2.1) if and only if 𝑢=𝑃_𝐾𝑧,(2.2) where 𝑃_𝐾 is the projection of 𝑅^𝑛 onto the closed convex set 𝐾.

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

Proof. Let 𝑥∈𝐾  be the solution of (1.2). Then ⟨𝐴𝑥−|𝑥|−𝑏,𝑦−𝑥⟩≥0,∀𝑦∈𝐾.(2.3) Since 𝐾  is a cone, taking 𝑦=0∈𝐾 and 𝑦=2𝑥∈𝐾, we have ⟨𝐴𝑥−|𝑥|−𝑏,𝑥⟩=0.(2.4) From inequality (1.2), we have 0≤⟨𝐴𝑥−|𝑥|−𝑏,𝑦−𝑥⟩==⟨⟨𝐴𝐴𝑥𝑥−−||𝑥𝑥||−−𝑏𝑏,,𝑦𝑦⟩⟩,−⟨𝐴∀𝑦𝑥∈−𝐾|𝑥,|−𝑏,𝑥⟩,(2.5) from which it follows that 𝐴𝑥−|𝑥|−𝑏∈𝐾^∗. Thus we conclude that 𝑥∈𝐾 is the solution of absolute complementarity problems (1.1).
Conversely, let 𝑥∈𝐾 be a solution of (1.1). Then 𝑥∈𝐾,𝐴𝑥−|𝑥|−𝑏∈𝐾^∗,⟨𝐴𝑥−|𝑥|−𝑏,𝑥⟩=0.(2.6) From (2.5) and (2.6), it follows that ⟨𝐴𝑥−|𝑥|−𝑏,𝑦−𝑥⟩≥0,∀𝑦∈𝐾.(2.7) Hence 𝑥∈𝐾 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 𝑅^𝑛. Then, for a constant 𝜌>0,𝑥∈𝐾 satisfies (1.2) if and only if 𝑥∈𝐾 satisfies the relation 𝑥=𝑃_𝐾(𝑥−𝜌[𝐴𝑥−|𝑥|−𝑏]),(2.8) where 𝑃_𝐾 is the projection of 𝑅^𝑛 onto the closed convex set 𝐾.

Proof. Let 𝑥∈𝐾  be the solution of (1.2). Then, for a positive constant 𝜌>0, ⟨𝑥−(𝑥−𝜌(𝐴𝑥−|𝑥|−𝑏)),𝑦−𝑥⟩≥0,∀𝑦∈𝐾.(2.9) Using Lemma 2.1, we have 𝑥=𝑃_𝐾(𝑥−𝜌[𝐴𝑥−|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=𝑃_𝐾(𝑥−𝜌[𝐴𝑥−|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 𝐴∈𝑅^𝑛×𝑛 be a positive definite matrix with constant 𝛼>0 and continuous with constant 𝛽>0. If 0<𝜌<2(𝛾−1)/(𝛽²−1),𝛾>1, then there exists a unique solution 𝑥∈𝐾 such that ⟨𝐴𝑥−|𝑥|−𝑏,𝑦−𝑥⟩≥0,∀𝑦∈𝐾,(2.12) where 𝐾 is a closed convex set in 𝑅^𝑛.

Proof. Uniqueness: Let 𝑥₁≠𝑥₂∈𝐾 be two solutions of (1.2). Then 𝐴𝐴𝑥𝑥₁₂−||𝑥₁||−𝑏,𝑦−𝑥₁≥0,∀𝑦∈𝐾,−||𝑥₂||−𝑏,𝑦−𝑥₂≥0,∀𝑦∈𝐾.(2.13) Taking 𝑦=𝑥₂∈𝐾 in (2.13) and 𝑦=𝑥₁∈𝐾 in (2.6), we have −𝐴𝐴𝑥₂𝑥₁+||𝑥₁||+𝑏,𝑥₁−𝑥₂≥0,−||𝑥₂||−𝑏,𝑥₁−𝑥₂≥0.(2.14) Adding the previous inequalities, we obtain 𝐴𝑥₁−𝑥₂−||𝑥₁||+||𝑥₂||,𝑥₁−𝑥₂≤0,(2.15) which implies that 𝐴𝑥₁−𝑥₂,𝑥₁−𝑥₂−‖‖𝑥₁−𝑥₂‖‖²≤0.(2.16) Since 𝐴 is positive definite, from (2.16), we have (𝛾−1)‖‖𝑥₁−𝑥₂‖‖²≤0.(2.17) As 𝛾>1, therefore from (2.17) we have ‖‖𝑥₁−𝑥₂‖‖²≤0,(2.18) which contradicts the fact that ‖𝑥₁−𝑥₂‖²≥0; hence 𝑥₁=𝑥₂.
Existence
Let 𝑥∈𝐾 be the solution of (1.2). Then ⟨𝐴𝑥−|𝑥|−𝑏,𝑦−𝑥⟩≥0,∀𝑦∈𝐾.(2.19) From Lemma 2.3, we have 𝑥=𝑃_𝐾(𝑥−𝜌[𝐴𝑥−|𝑥|−𝑏]).(2.20) Define a mapping 𝐹(𝑥)=𝑥=𝑃_𝐾(𝑥−𝜌[𝐴𝑥−|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 𝑥₁≠𝑥₂∈𝐾, consider ‖‖𝐹𝑥₁−𝐹𝑥₂‖‖=≤‖‖‖‖𝑃_𝐾𝑥𝑥₁−𝜌𝐴𝑥₁−||𝑥₁||−𝑏−𝑃_𝐾𝑥₂−𝜌𝐴𝑥₂−||𝑥₂||−𝑏₁‖‖≤‖‖𝑥−𝜌𝐴𝑥₁−||𝑥₁||−𝑏−𝑥₂−𝜌𝐴𝑥₂−||𝑥₂||−𝑏₁‖‖=‖‖𝑥₁−𝑥₂−𝜌𝐴𝑥₁−𝐴𝑥₂‖‖+𝜌‖‖||𝑥₁||−||𝑥₂||‖‖−𝑥₂−𝜌𝐴𝑥₁−𝑥₂‖‖+𝜌‖‖𝑥₁−𝑥₂‖‖,(2.22) where we have used the fact that 𝑃_𝐾 is nonexpansive. Now using positive definiteness of 𝐴, we have ‖‖𝑥₁−𝑥₂−𝜌𝐴𝑥₁−𝑥₂‖‖²≤=‖‖𝑥𝑥₁−𝑥₂−𝜌𝐴𝑥₁−𝑥₂,𝑥₁−𝑥₂−𝜌𝐴𝑥₁−𝑥₂₁−𝑥₂‖‖²−2𝜌𝛾𝑥₁−𝑥₂,𝐴𝑥₁−𝑥₂+𝜌²𝛽²‖‖𝑥₁−𝑥₂‖‖²=1−2𝛼𝜌+𝛽²𝜌²‖‖𝑥₁−𝑥₂‖‖².(2.23) From (2.22) and (2.23), we have ‖‖𝐹𝑥₁−𝐹𝑥₂‖‖≤𝜃‖‖𝑥₁−𝑥₂‖‖,(2.24) where 𝜃=(𝜌+√1−2𝛾𝜌+𝛽²𝜌²).Form 0<𝜌<2(𝛾−1)/(𝛽²−1) and 𝜌<1, we have 𝜃<1.
This shows that 𝐹(𝑥)  is a contraction mapping and has a fixed point 𝑥∈𝐾 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 𝑅^𝑛 and we define the projection 𝑃_𝐾𝑥 as𝑃_𝐾𝑥_𝑖=minmax0,𝑥_𝑖,𝑐_𝑖,𝑖=1,2,…,𝑛.(2.25) We recall the following well-known result.

Lemma 2.5 (see [3]). For any 𝑥  and 𝑦 in 𝑅^𝑛, the projection 𝑃_𝐾𝑥 has the following properties:(i)𝑃_𝐾(𝑥+𝑦)≤𝑃_𝐾𝑥+𝑃_𝐾𝑦, (ii)𝑃_𝐾𝑥−𝑃_𝐾𝑦≤𝑃_𝐾(𝑥−𝑦), (iii)𝑥≤𝑦⇒𝑃_𝐾𝑥≤𝑃_𝐾𝑦,(iv)𝑃_𝐾𝑥+𝑃_𝐾(−𝑥)≤|𝑥|,withequality,ifandonlyif−𝑐≤𝑥≤𝑐.

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,…,𝜔_𝑛) with 𝜔_𝑖∈𝑅₊ and let 𝛼 be a real number.

Algorithm 2.6. Step 1. Choose an initial vector 𝑥₀∈𝑅^𝑛 and a parameter Ω∈𝑅₊.Set𝑘=0.Step 2. Calculate 𝑥_𝑘+1=𝑃_𝐾𝑥_𝑘−𝐷⁻¹𝛼Ω𝐿𝑢_𝑘+1+(Ω𝐴−𝛼Ω𝐿)𝑢_𝑘−Ω||𝑥_𝑘||+𝑏.(2.27)Step 3. If 𝑥_𝑘+1=𝑥_𝑘, then stop. Else, set𝑘=𝑘+1 and go to Step 2.Now we define an operator 𝑔∶𝑅^𝑛→𝑅^𝑛 such that 𝑔(𝑥)=𝜉, where 𝜉 is the fixed point of the system 𝜉=𝑃_𝐾𝑥−𝐷⁻¹[𝛼Ω𝐿𝜉+(Ω𝐴−𝛼Ω𝐿)𝑥−Ω(|𝑥|+𝑏)].(2.28) We also assume that the set 𝜑={𝑥∈𝑅^𝑛∶𝑥≥0,𝐴𝑥−|𝑥|−𝑏≥0}(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 𝑔∶𝑅^𝑛→𝑅^𝑛 as defined in (2.28). Assume that𝐴∈𝑅^𝑛×𝑛 is an 𝐿-matrix. Also assume that 0<𝜔_𝑖≤1,0≤𝛼≤1. Then, for any 𝑥∈𝜑, the following holds:(i)𝑔(𝑥)≤𝑥, (ii)𝑥≤𝑦⇒𝑔(𝑥)≤𝑔(𝑦), (iii)𝜉=𝑔(𝑥)∈𝜑.

Proof. To prove (i), we need to verify that 𝜉_𝑖≤𝑥_𝑖,𝑖=1,2,…,𝑛 hold with 𝜉_𝑖 satisfying 𝜉_𝑖=𝑃_𝐾𝑥^𝑖−𝑎^−1𝑖𝑖𝛼𝜔^𝑖𝑖−1_𝑗=1𝐿_𝑖𝑗𝜉_𝑗−𝑥_𝑗+𝜔_𝑖(𝐴𝑥−|𝑥|−𝑏)_𝑖.(2.30) To prove the required result, we use mathematical induction. For this let 𝑖=1, 𝜉₁=𝑃_𝐾𝑥₁−𝑎⁻¹¹¹𝜔₁(𝐴𝑥−|𝑥|−𝑏)₁.(2.31) Since 𝐴𝑥−|𝑥|−𝑏≥0,𝜔_𝑖>0, therefore 𝜉₁≤𝑥₁. For 𝑖=2, we have 𝜉₂=𝑃_𝐾𝑥₂−𝑎⁻¹²²𝛼𝜔₂𝐿₂₁𝜉₁−𝑥₁+𝜔₂(𝐴𝑥−|𝑥|−𝑏)₂.(2.32) Here 𝐴𝑥−|𝑥|−𝑏≥0,𝜔_𝑖>0,𝐿₂₁≤0, and 𝜉₁−𝑥₁≤0. This implies that 𝜉₂≤𝑥₂.
Suppose that 𝜉_𝑖≤𝑥_𝑖for𝑖=1,2,…,𝑘−1.(2.33) We have to prove that the statement is true, for 𝑖=𝑘, that is, 𝜉_𝑘≤𝑥_𝑘.(2.34) Consider 𝜉_𝑘=𝑃_𝐾𝑥^𝑘−𝑎^−1𝑘𝑘𝛼𝜔^𝑘𝑘−1_𝑗=1𝐿_𝑘𝑗𝜉_𝑗−𝑥_𝑗+𝜔_𝑘(𝐴𝑥−|𝑥|−𝑏)_𝑘=𝑃_𝐾,𝑥_𝑘−𝑎+^−1𝑘𝑘𝜔𝛼_𝑘𝜔_𝑘𝐿_𝑘1𝜉₁−𝑥₁+𝐿_𝑘2𝜉₂−𝑥₂+⋯+𝐿_𝑘𝑘−1𝜉_𝑘−1−𝑥_𝑘−1(𝐴𝑥−|𝑥|−𝑏)_𝑘.(2.35) Since 𝐴𝑥−|𝑥|−𝑏≥0,𝜔_𝑘>0,𝐿_𝑘1,𝐿_𝑘2,…,𝐿_𝑘𝑘−1≤0, and 𝜉_𝑖≤𝑥_𝑖 for 𝑖=1,2,…,𝑘−1, from (2.35) we can write 𝜉_𝑘≤𝑥_𝑘. Hence (i) is proved.
Now we prove (ii). For this let us suppose that 𝜉=𝑔(𝑥) and 𝜙=𝑔(𝑦). We will prove 𝑥≤𝑦⟹𝜉≤𝜙.(2.36) As 𝜉=𝑃_𝐾𝑥−𝐷⁻¹[𝛼Ω𝐿𝜉+(Ω𝐴−𝛼Ω𝐿)𝑥−Ω(|𝑥|+𝑏)],(2.37) so 𝜉_𝑖  can be written as 𝜉_𝑖=𝑃_𝐾⎛⎜⎜⎜⎝𝑥_𝑖−𝑎^−1𝑖𝑖⎡⎢⎢⎢⎣𝛼𝜔_𝑖𝑖−1_𝑗=1𝐿_𝑖𝑗𝜉_𝑗+𝜔_𝑖𝑎_𝑖𝑖𝑥_𝑖+(1−𝛼)𝜔_𝑖𝑖−1_𝑗=1𝐿_𝑖𝑗𝑥_𝑗+𝜔_𝑖𝑛_{𝑗=1𝑗≠𝑖}𝑈_𝑖𝑗𝑥_𝑗−𝜔_𝑖||𝑥_𝑖||−𝜔_𝑖𝑏_𝑖=𝑃_𝐾⎤⎥⎥⎥⎦⎞⎟⎟⎟⎠⎛⎜⎜⎜⎝1−𝜔_𝑖𝑥_𝑖−𝑎^−1𝑖𝑖⎡⎢⎢⎢⎣𝛼𝜔_𝑖𝑖−1_𝑗=1𝐿_𝑖𝑗𝜉_𝑗+(1−𝛼)𝜔_𝑖𝑖−1_𝑗=1𝐿_𝑖𝑗𝑥_𝑗+𝜔_𝑖𝑛_{𝑗=1𝑗≠𝑖}𝑈_𝑖𝑗𝑥_𝑗−𝜔_𝑖||𝑥_𝑖||−𝜔_𝑖𝑏_𝑖⎤⎥⎥⎥⎦⎞⎟⎟⎟⎠.(2.38) Similarly, for 𝜙_𝑖, we have 𝜙_𝑖=𝑃_𝐾⎛⎜⎜⎜⎝1−𝜔_𝑖𝑦_𝑖−𝑎^−1𝑖𝑖⎡⎢⎢⎢⎣𝛼𝜔_𝑖𝑖−1_𝑗=1𝐿_𝑖𝑗𝜙_𝑗+(1−𝛼)𝜔_𝑖𝑖−1_𝑗=1𝐿_𝑖𝑗𝑦_𝑗+𝜔_𝑖𝑛_{𝑗=1𝑗≠𝑖}𝑈_𝑖𝑗𝑦_𝑗−𝜔_𝑖||𝑦_𝑖||−𝜔_𝑖𝑏_𝑖⎤⎥⎥⎥⎦⎞⎟⎟⎟⎠,(2.39) and for 𝑖=1, 𝜙₁=𝑃_𝐾⎛⎜⎜⎜⎝1−𝜔₁𝑦₁−𝑎⁻¹¹¹𝜔₁⎡⎢⎢⎢⎣_𝑛_{𝑗=1𝑗≠𝑖}𝑈_1𝑗𝑦_𝑗−||𝑦₁||−𝑏₁≥𝑃_𝐾⎤⎥⎥⎥⎦⎞⎟⎟⎟⎠⎛⎜⎜⎜⎝1−𝜔₁𝑥₁−𝑎⁻¹¹¹𝜔₁⎡⎢⎢⎢⎣_𝑛_{𝑗=1𝑗≠𝑖}𝑈_1𝑗𝑥_𝑗−||𝑥₁||−𝑏₁=𝜉₁⎤⎥⎥⎥⎦⎞⎟⎟⎟⎠.(2.40) Since 𝑦₁≥𝑥₁, therefore −|𝑦₁|≤−|𝑥₁|. Hence it is true for 𝑖=1. Suppose it is true for 𝑖=1,2,…𝑘−1; we will prove it for 𝑖=𝑘; for this consider 𝜙_𝑘=𝑃_𝐾⎛⎜⎜⎜⎝1−𝜔_𝑘𝑦_𝑘−𝑎^−1𝑘𝑘⎡⎢⎢⎢⎣𝛼𝜔_𝑘𝑘−1_𝑗=1𝐿_𝑘𝑗𝜙_𝑗+(1−𝛼)𝜔_𝑘𝑘−1_𝑗=1𝐿_𝑘𝑗𝑦_𝑗+𝜔_𝑘𝑛_{𝑗=1𝑗≠𝑖}𝑈_𝑘𝑗𝑦_𝑗−𝜔_𝑘||𝑦_𝑘||−𝜔_𝑘𝑏_𝑘≥𝑃_𝐾⎤⎥⎥⎥⎦⎞⎟⎟⎟⎠⎛⎜⎜⎜⎝1−𝜔_𝑘𝑥_𝑘−𝑎^−1𝑘𝑘⎡⎢⎢⎢⎣𝛼𝜔_𝑘𝑘−1_𝑗=1𝐿_𝑘𝑗𝜉_𝑗+(1−𝛼)𝜔_𝑘𝑘−1_𝑗=1𝐿_𝑘𝑗𝑥_𝑗+𝜔_𝑘𝑛_{𝑗=1𝑗≠𝑖}𝑈_𝑘𝑗𝑥_𝑗−𝜔_𝑘||𝑥_𝑘||−𝜔_𝑘𝑏_𝑘=𝜉_𝑘⎤⎥⎥⎥⎦⎞⎟⎟⎟⎠.(2.41) Since 𝑥≤𝑦, and 𝜉_𝑖≤𝜙_𝑖 for 𝑖=1,2,…𝑘−1, 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)=𝑃_𝐾(𝜉−𝐷⁻¹Ω[𝛼𝐿(𝜆−𝜉)+𝐴𝜉−|𝜉|−𝑏]) from (i) 𝑔(𝜉)=𝜆≤𝜉. Also by definition of 𝑔,𝜉=𝑔(𝑥)≥0 and 𝜆=𝑔(𝜉)≥0.
Now 𝜆_𝑖=𝑃_𝐾𝜉^𝑖−𝑎^−1𝑖𝑖𝛼𝜔^𝑖𝑖−1_𝑗=1𝐿_𝑖𝑗𝜆_𝑗−𝜉_𝑗+𝜔_𝑖𝐴𝜉−||𝜉||−𝑏_𝑖.(2.43) For 𝑖=1,𝜉₁≥0 by definition of 𝑔. Suppose that (𝐴𝜉−|𝜉|−𝑏)_𝑖<0, so 𝜆₁=>𝑃𝑃_𝐾𝐾𝜉₁−𝑎⁻¹¹¹𝜔₁𝐴𝜉−||𝜉||−𝑏₁𝜉₁=𝜉₁,(2.44) which contradicts the fact that 𝜆≤𝜉. Therefore, (𝐴𝜉−|𝜉|−𝑏)_𝑖≥0.
Now we prove it for any 𝑘 in 𝑖=1,2,…,𝑛. Suppose the contrary (𝐴𝜉−|𝜉|−𝑏)_𝑖<0; then 𝜆_𝑘=𝑃_𝐾𝜉^𝑘−𝑎^−1𝑘𝑘𝛼𝜔^𝑘𝑘−1_𝑗=1𝐿_𝑘𝑗𝜆_𝑗−𝜉_𝑗+𝜔_𝑘𝐴𝜉−||𝜉||−𝑏_𝑘.(2.45) As it is true for all 𝛼∈[0,1], it should be true for 𝛼=0. That is, 𝜆_𝑘=>𝑃𝑃_𝐾𝐾𝜉_𝑘−𝑎^−1𝑘𝑘𝜔_𝑘𝐴𝜉−||𝜉||−𝑏_𝑘𝜉_𝑘=𝜉_𝑘,(2.46) which contradicts the fact that 𝜆≤𝜉. So (𝐴𝜉−|𝜉|−𝑏)_𝑘≥0, 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 𝐴∈𝑅^𝑛×𝑛 is an 𝐿-matrix. Also assume that 0<𝜔_𝑖≤1,0≤𝛼≤1. Then for any initial vector 𝑥₀∈𝜑, the sequence {𝑥_𝑘},𝑘=0,1,2,…, defined by Algorithm 2.6 has the following properties:(i)0≤𝑥_𝑘+1≤𝑥_𝑘≤𝑥₀;𝑘=0,1,2,…, (ii)lim_𝑘→∞𝑥_𝑘=𝑥^∗ is a unique solution of the absolute value complementarity problem (1.1).

Proof. Since 𝑥₀∈𝜑, by (i) of Theorem 2.7, we have 𝑥₁≤𝑥₀ and 𝑥₁∈𝜑. Recursively using Theorem 2.7 we obtain 0≤𝑥_𝑘+1≤𝑥_𝑘≤𝑥₀;𝑘=0,1,2,….(2.47) From (i) we observe that the sequence {𝑥_𝑘}is monotone bounded; therefore, it converges to some 𝑥^∗∈𝑅^𝑛+ satisfying 𝑥^∗==𝑃𝑃_𝐾𝐾𝑥^∗−𝐷⁻¹𝛼Ω𝐿𝑥^∗+(Ω𝐴−𝛼Ω𝐿)𝑥^∗−Ω||𝑥^∗||+𝑏𝑥^∗−𝐷⁻¹Ω𝐴𝑥^∗−Ω||𝑥^∗||−Ω𝑏.(2.48) Hence 𝑥^∗is 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 that𝜔_𝑖=𝑎+(𝑏−𝑎)𝑖𝑛,𝑖=1,2,…,𝑛,(3.1) where 𝜔_𝑖  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 𝐴∈𝑅^𝑛×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.

Table 1 

Computational results.

Example 3.2. Consider the ordinary differential equation: 𝑑²𝑥𝑑𝑡²−|𝑥|=1−𝑥²,0≤𝑥≤1,𝑥(0)=0,𝑥(1)=1.(3.2) The exact solution is 𝑥(𝑡)=0.7378827425sin(𝑡)−3cos(𝑡)+3−𝑥²−0.7310585786𝑒_−𝑥,𝑥<0,−0.2689414214𝑒^𝑥+1+𝑥²,𝑥>0.(3.3) We take 𝑛=10; the matrix 𝐴 is given by 𝑎_𝑖,𝑗=⎧⎪⎨⎪⎩−102,214,2,ffooothrre𝑗rw=𝑗𝑗i==s𝑖,e𝑖𝑖.+−11,,𝑖𝑖==12,,23,,……,,𝑛𝑛,−1,(3.4) The constant vector 𝑏 is given by 𝑏=120121,111271,112112,110251,19261,18251,17221,15271,14201,−11426120^𝑇.(3.5) Here 𝐴 is not 𝐿-matrix. The comparison between the exact solution and the approximate solutions is given in Figure 1.

Figure 1 

Example 3.3. Let the matrix 𝐴 be given by 𝑎_𝑖,𝑗=⎧⎪⎨⎪⎩8−0,,1,ffooothrre𝑗rw=𝑗𝑗i==s𝑖,e𝑖𝑖.+−11,,𝑖𝑖==12,,23,,……,,𝑛𝑛,−1,(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‖<10⁻⁶. 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].

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.