Existence of Solutions and Algorithm for a System of Variational Inequalities

Abstract

We obtain some existence results for a system of variational inequalities (for short, denoted by SVI) by Brouwer fixed point theorem. We also establish the existence and uniqueness theorem using the projection technique for the SVI and suggest an iterative algorithm and analysis convergence of the algorithm.

1. Questions under Consideration in This Paper

Suppose that is a nonempty closed and convex subset of ; is a vector-valued mapping. Variational inequality problem (for short, VI) is to find an , such that

(1.1)

We denote the solution set for VI by . In this paper, we suggest and study the following SVI: find , such that

(1.2)

where , are vector-valued mappings, and The above SVI can be described as

(1.3)

2. Existence and Uniqueness of Solutions for SVI

In this paper, otherwise specification, is a n-dimensional Euclidean space, for all , denotes the inner product between and , denotes norm of , that is, .

In order to obtain our main results, we recall the following definitions and lemmas.

Definition 2.1.

Let be a nonempty subset, and let be a vector-valued mapping.

(i) is said to be monotone if, for all ,

(ii) is said to be strictly monotone if, for all , ,

(iii) is said to be strongly monotone if there exists a constant such that

(2.1)

(iv) is said to be coercive if there exists an and a constant such that and

(2.2)

(v) is said to be Lipschitz continuous if there exists a constant such that

(2.3)

Remark 2.2.

It is easy to see that

(2.4)

Based on the above all kinds of monotonicity, we have the following existence results for .

Lemma 2.3 (see [1]).

Let be nonempty compact and convex set, and let be continuous mapping. Then must has solution.

Lemma 2.4 (see [2]).

Let be nonempty closed and convex set, and let be continuous mapping.

(i)If is strictly monotone, then has at most one solution,

(ii)If is coercive, then must has solution,

(iii)If is strongly monotone, then has a unique solution.

In order to obtain the existence results for SVI, one needs to study parametric variational inequalities and in SVI.

Setting and is the feasible region of SVI, , are nonempty subset, and , are two continuous mappings. At first, one considers , which is a parametric variational inequality with respect to in SVI.

Theorem 2.5.

In , assume that and are two compact and convex sets, is continuous, and is strictly monotone in . Then, for any given , has a unique solution and for all , there exists an implicit function which is the unique solution to . In addition, the implicit function determined by is continuous on .

Proof.

• (i) For any given , since is compact and convex and is continuous on , then by Lemma 2.3, parametric variational inequality has solutions. In terms of strict monotonicity of the mapping in and Lemma 2.4, we know that has a unique solution. So, for all , the implicit function determined by is well defined.

• (ii) We claim that is continuous on . In fact, for any given , , as , by (i), we know that for all , there exists , such that . That is,

(2.5)

Since is bounded, then there exists convergent subsequence such that as , and . In the following, we prove that is a solution to in . For given , there exists sequence such that and as in view of the closedness of . Letting in (2.5), we have

(2.6)

that is,

(2.7)

observe that is continuous, and letting in (2.7), we have

(2.8)

For is arbitrary, then is a solution to , implying . In order to explain that is continuous at , we only need to know that the sequence satisfies as . Let be any subsequence of . Since is bounded, there exists a subsequence such that as . Using the method appeared in Theorem 2.5, we can show that

(2.9)

Thus, by the uniqueness of the solution to the problem , we conclude that . Since, is arbitrary, we can conclude that as , which means that implicit function is continuous at . For is arbitrary, we know that is continuous on .

From Theorem 2.5, we see that in order to ensure that is well defined, the condition that is strictly monotone on is necessary, but the boundedness of is a strong condition. As usual, is unbounded (e.g., inequality constraint set , where , is always unbounded). So, we try to weaken the boundedness of . For this, we introduce the concept uniform coercivity of in .

Definition 2.6.

In , let ; is said to be uniformly coercive near , if there exists some neighbourhood of , and such that and for all ,

(2.10)

where and .

If for each , is uniformly coercive near , then is said to be uniformly coercive on .

Lemma 2.7.

In , let be nonempty closed and convex set, and let be uniformly coercive near , then there exists some neighbourhood of , such that is bounded set.

Proof.

For given , by the definition of the uniform coercivity of near , there exists some neighbourhood of , and such that and for all ,

(2.11)

Let It is obvious that is a nonempty bounded closed convex subset of In view of Lemma 2.3, we know that must have solution. That is, there exists an such that

(2.12)

Now, we state that and In fact, if , for all , join and into with . Then take small enough such that . Substituting with in (2.12), we have

(2.13)

implying that and . On the other hand, if , substituting with in (2.11), we have

(2.14)

which, by plus (2.12), we get

(2.15)

For all , , consider the connection of and ; following the same argument, we have that and Therefore, and . That is, is bounded. For is arbitrary, the conclusion holds. This completes the proof.

If the boundedness of is replaced by the uniform coercivity of in Theorem 2.5, then we have the following result.

Theorem 2.8.

In , let be nonempty closed and convex set, and let be uniformly coercive on with respect to and strict monotone in . Then for each , has a unique solution, and for all , the implicit function determined by is continuous on .

Proof.

• (i) For given , by Lemma 2.4 and the coercivity of on , we know that has solution. Noting that is strictly monotone in , has a unique solution, and so the implicit function is well defined.

• (ii) For given , , satisfying as . By Lemma 2.7, there exists some neighbourhood of and bounded open set , such that , that is, the solution set of denoted by . such that as . Let without generality, then is bounded; the following argument is similar to Theorem 2.5, so it is omitted, and this completes the proof.

Set ; it is to see that . We will investigate the parametric variational inequality with respect to in SVI.

Corollary 2.9.

In , let be two nonempty compact and convex subsets, be continuous and strict monotone in . Then for each given , has a unique solution, and for all , the implicit function determined by is continuous on .

Proof.

The conclusion holds directly from Theorem 2.5.

Lemma 2.10 (see [3, (Brouwer fixed point theorem)]).

Let be nonempty compact and convex set, and let be continuous. Then there exists an , such that .

Theorem 2.11.

In SVI, let be two compact and convex subset, and let and be two continuous mappings and strict monotone in and , respectively. Then SVI has solution.

Proof.

By the given conditions of Theorems 2.11 and 2.5, we know that there exists continuous implicit function determined by parametric variational inequality with respect to in SVI. Also denoted the range of by . By Corollary 2.9, there exists continuous implicit function determined by parametric variational inequality with respect to in SVI such that for all , is the unique solution to . Let for all Making use of Brouwer fixed point theorem (Lemma 2.10), we have that there exists , such that . Setting , by the definitions of and , we know that is a solution of SVI.

Corollary 2.12.

In SVI, let be nonempty compact and convex subset, let be nonempty closed and convex subset, let and be two continuous mappings, and be strict monotone in and , respectively. Let be uniformly coercive on with respect to . Then SVI has solution.

Proof.

By Theorem 2.8 and similar argument in Theorem 2.11, our conclusion holds.

Now, we give the definition of uniformly strong monotonicity, which is stronger condition than the uniformly coercivity.

Definition 2.13.

Let be vector-valued mapping; if there exists , such that for all ,

(2.16)

then is said to be uniformly strongly monotone in .

Lemma 2.14.

In , let be uniformly strongly monotone, then is uniformly coercive.

Proof.

For given , we only need to prove that is uniformly coercive at . Let us consider . Assume that is a solution to , and , . Since is strongly monotone, then there exists , such that for all ,

(2.17)

Letting in (2.17), we have

(2.18)

Noting that is a solution to , we have

(2.19)

Combining (2.18), we obtain

(2.20)

Since is continuous, then there exists some neighbourhood of , such that for all ,

(2.21)

which implies that is uniform coercive at .

Using the uniformly strong monotonicity of , we can obtain the following existence result for SVI under the condition that is a nonempty closed and convex subset of .

Corollary 2.15.

In SVI, let be a nonempty closed and convex set, let be a compact and convex set, let be two continuous mapping, let be uniformly strongly monotone in , and let be strictly monotone in . Then SVI has solution.

Proof.

By Lemma 2.14 and Corollary 2.12, it is easy to see that the conclusion holds.

Corollary 2.16.

In SVI, Let and be nonempty closed and convex subsets, let be two continuous mappings, and let be uniformly strongly monotone in , and let be uniformly coercive and strict monotone in . Then SVI has solution.

Furthermore, If and are Lipschitz continuous in and , respectively, one can obtain the following existence and uniqueness result for SVI.

Theorem 2.17.

In SVI, let be nonempty compact and convex subsets, let be uniformly strongly monotone with constant and Lipschitz continuous with Lipschitz constant in , and Lipschitz continuous with constant in , and let be uniformly strongly monotone with constant and Lipschitz continuous with constant in , and Lipschitz continuous with constant in . If there exists constants such that

(2.22)

then SVI has a unique solution.

In order to prove Theorem 2.17, we need the following lemma.

Lemma 2.18 (see [4]).

In SVI, let be nonempty closed and convex subsets, and let be two continuous mappings. SVI has solution if and only if satisfies

(2.23)

where denote the projection from and to and , respectively; furthermore, projection operator is nonexpansive and are constants.

The Proof of Theorem 2.17

For arbitrary given constant , define and by

(2.24)

For any , it follows from (2.24) and Lemma 2.18 that

(2.25)

We have used the strong monotonicity and Lipschitz continuity of in and Lipschitz continuity of in . Similarly, we have

(2.26)

It follows from (2.25) and (2.26) that

(2.27)

where

(2.28)

Define on by

(2.29)

It is easy to see that is a Banach space. For any given , define by

(2.30)

By assumption, we know that . It follows from (2.27) that

(2.31)

which implies that is a contraction operator. Hence, there exists a unique , such that

(2.32)

That is,

(2.33)

By Lemma 2.18, is the unique solution of SVI.

3. Iterative Algorithm and Convergence

In this section, we will construct an iterative algorithm for approximating the unique solution of SVI and discuss the convergence analysis of the algorithm.

Lemma 3.1 (see, [5]).

Let and be two real sequence of nonnegative numbers that satisfy the following conditions.

(i) and ,

(ii)

Then, converges to 0 as .

Algorithm 3.2.

Let and be the same as in Theorem 2.17. For any given , define iterative sequence by

(3.1)

where

(3.2)

Theorem 3.3.

Let and be the same as in Theorem 2.17. Assume that all the conditions of Theorem 2.17 hold. Then, generated by Algorithm 3.2 converges to the unique solution of SVI and there exists , such that

(3.3)

Proof.

By Theorem 2.17, SVI admits a unique solution . It follows from Lemma 2.18 that

(3.4)

It follows from (3.1) and (3.4) that

(3.5)

By (3.5), we get

(3.6)

where is defined by

(3.7)

Set

(3.8)

Then (3.6) can be rewritten as

(3.9)

By (3.2), we know that . It follows from Lemma 3.1 that and that

(3.10)

for all . Therefore, converges geometrically to the unique solution of SVI. This completes the proof.