A New Projection Algorithm for Generalized Variational Inequality

Abstract

We propose a new projection algorithm for generalized variational inequality with multivalued mapping. Our method is proven to be globally convergent to a solution of the variational inequality problem, provided that the multivalued mapping is continuous and pseudomonotone with nonempty compact convex values. Preliminary computational experience is also reported.

1. Introduction

We consider the following generalized variational inequality. To find and such that

(1.1)

where is a nonempty closed convex set in , is a multivalued mapping from into with nonempty values, and and denote the inner product and norm in , respectively.

Theory and algorithm of generalized variational inequality have been much studied in the literature [1–9]. Various algorithms for computing the solution of (1.1) are proposed. The well-known proximal point algorithm [10] requires the multivalued mapping to be monotone. Relaxing the monotonicity assumption, [1] proved if the set is a box and is order monotone, then the proximal point algorithm still applies for problem (1.1). Assume that is pseudomonotone, and [11] described a combined relaxation method for solving (1.1); see also [12, 13]. Projection-type algorithms have been extensively studied in the literature; see [14–17] and the references therein. Recently, [15] proposes a projection algorithm for generalized variational inequality with pseudomonotone mapping. In [15], choosing needs solving a single-valued variational inequality and hence is computationally expensive; see expression (2.1) in [15]. In this paper, we introduce a different projection algorithm for generalized variational inequality. In our method, can be taken arbitrarily. Moreover, the main difference of our method from that of [15] is the procedure of Armijo-type linesearch; see expression (2.2) in [15] and expression (2.2) in the next section.

Let be the solution set of (1.1), that is, those points satisfying (1.1). Throughout this paper, we assume that the solution set of problem (1.1) is nonempty and is continuous on with nonempty compact convex values satisfying the following property:

(1.2)

Property (1.2) holds if is pseudomonotone on in the sense of Karamardian [18]. In particular, if is monotone, then (1.2) holds.

The organization of this paper is as follows. In the next section, we recall the definition of continuous multivalued mapping, present the algorithm details, and prove the preliminary result for convergence analysis in Section 3. Numerical results are reported in the last section.

2. Algorithms

Let us recall the definition of continuous multivalued mapping. is said to be upper semicontinuous at if for every open set containing , there is an open set containing such that for all . F is said to be lower semicontinuous at , if we give any sequence converging to and any , there exists a sequence that converges to . is said to be continuous at if it is both upper semicontinuous and lower semicontinuous at . If is single valued, then both upper semicontinuity and lower semicontinuity reduce to the continuity of .

Let denote the projector onto and let be a parameter.

Proposition 2.1.

and solve problem (1.1) if and only if

(2.1)

Algorithm 2.2.

Choose and three parameters , and Set

Step 1.

If for some , stop; else take arbitrarily .

Step 2.

Let be the smallest nonnegative integer satisfying

(2.2)

where . Set .

Step 3.

Compute where , and

(2.3)

Let and go to Step 1.

Remark 2.3.

Since has compact convex values, has closed convex values. Therefore, in Step 2 is uniquely determined by .

Remark 2.4.

If is a single-valued mapping, the Armijo-type linesearch procedure (2.2) becomes that of Algorithm 2.2 in [14].

We show that Algorithm 2.2 is well defined and implementable.

Proposition 2.5.

If is not a solution of problem (1.1), then there exists a nonnegative integer satisfying (2.2).

Proof.

Suppose that for all , we have

(2.4)

where . Since is lower semicontinuous, , and as , for each , there is such that . Since ,

(2.5)

So . Let in (2.4), we have . This contradiction completes the proof.

Lemma 2.6.

For every and ,

(2.6)

Proof.

See [15, Lemma ].

Lemma 2.7.

Let be a closed convex set in , a real-valued function on , and the set . If is nonempty and is Lipschitz continuous on with modulus , then

(2.7)

where denotes the distance from to .

Proof.

See [14, Lemma ].

Lemma 2.8.

Let solve the variational inequality (1.1) and let the function be defined by (2.3). Then and . In particular, if then

Proof.

It follows from (2.3) that

(2.8)

where the first inequality follows from (2.2) and the last one follows from Lemma 2.6 and . If , then because . It remains to be proved that . Since , we have

(2.9)

on the other hand, assumption (1.2) implies that

(2.10)

Adding the last two expressions, we obtain that

(2.11)

It follows that

(2.12)

where the second inequality follows from assumption (1.2) and . Thus is verified.

3. Main Results

Theorem 3.1.

If is continuous with nonempty compact convex values on and condition (1.2) holds, then either Algorithm 2.2 terminates in a finite number of iterations or generates an infinite sequence converging to a solution of (1.1).

Proof.

Let be a solution of the variational inequality problem. By Lemma 2.8, . We assume that Algorithm 2.2 generates an infinite sequence . In particular, for every . By Step 3, it follows from Lemma in [14] that

(3.1)

where the last inequality is due to . It follows that the squence is nonincreasing, and hence is a convergent sequence. Therefore, is bounded and

(3.2)

By the boundedness of , there exists a convergent subsequence converging to .

If is a solution of problem (1.1), we show next that the whole sequence converges to . Replacing by in the preceding argument, we obtain that the sequence is nonincreasing and hence converges. Since is an accumulation point of , some subsequence of converges to zero. This shows that the whole sequence converges to zero, hence .

Suppose now that is not a solution of problem (1.1). We show first that in Algorithm 2.2 cannot tend to . Since is continuous with compact values, Proposition in [19] implies that is a bounded set, and so the sequence is bounded. Therefore, there exists a subsequence converging to . Since is upper semicontinuous with compact values, Proposition in [19] implies that is closed, and so . By the definition of , we have

(3.3)

If , then . The lower continuity of , in turn, implies the existence of such that converges to . Since , , and . Therefore and

(3.4)

Letting , we obtain the contradiction

(3.5)

with being continuous. Therefore, is bounded and so is .

It follows from (2.3) that

(3.6)

Since and are bounded, we have the sequence and hence the sequence is bounded. Thus, for some ,

(3.7)

Therefore, each function is Lipschitz continuous on with modulus . Noting that and applying Lemma 2.7, we obtain that

(3.8)

It follows from (3.8) and Lemma 2.8 that

(3.9)

Then (3.2) implies that

(3.10)

By the boundedness of , we obtain that Since is continuous and the sequences and are bounded, there exists an accumulation point of such that . This implies that solves the variational inequality (1.1). Similar to the preceding proof, we obtain that .

4. Numerical Experiments

In this section, we present some numerical experiments for the proposed algorithm. The MATLAB codes are run on a PC (with CPU Intel P-T2390) under MATLAB Version 7.0.1.24704(R14) Service Pack 1. We compare the performance of our Algorithm 2.2 and [15, Algorithm 1]. In the Tables 1 and 2, "It." denotes number of iteration, and "CPU" denotes the CPU time in seconds. The tolerance means when the procedure stops.

Table 1 Example 4.1.

Table 2 Example 4.2.

Example 4.1.

Let ,

(4.1)

and let be defined by

(4.2)

Then the set and the mapping satisfy the assumptions of Theorem 3.1 and (0,0,1) is a solution of the generalized variational inequality. Example 4.1 is tested in [15]. We choose , and for our algorithm; , and for Algorithm 1 in [15]. We use as the initial point.

Example 4.2.

Let ,

(4.3)

and be defined by

(4.4)

Then the set and the mapping satisfy the assumptions of Theorem 3.1 and (1,0,0,0) is a solution of the generalized variational inequality. We choose , and for the two algorithms.