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Existence of Solutions and Algorithm for a System of Variational Inequalities
Fixed Point Theory and Applications volume 2010, Article number: 182539 (2010)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ1_HTML.gif)
We denote the solution set for VI by
. In this paper, we suggest and study the following SVI: find
, such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ2_HTML.gif)
where ,
are vector-valued mappings, and
The above SVI can be described as
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ3_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ4_HTML.gif)
(iv) is said to be coercive if there exists an
and a constant
such that
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ5_HTML.gif)
(v) is said to be Lipschitz continuous if there exists a constant
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ6_HTML.gif)
Remark 2.2.
It is easy to see that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ7_HTML.gif)
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,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ8_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ9_HTML.gif)
that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ10_HTML.gif)
observe that is continuous, and letting
in (2.7), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ11_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ12_HTML.gif)
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
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ13_HTML.gif)
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
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ14_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ15_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ16_HTML.gif)
implying that and
. On the other hand, if
, substituting
with
in (2.11), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ17_HTML.gif)
which, by plus (2.12), we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ18_HTML.gif)
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
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ19_HTML.gif)
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
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ20_HTML.gif)
Letting in (2.17), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ21_HTML.gif)
Noting that is a solution to
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ22_HTML.gif)
Combining (2.18), we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ23_HTML.gif)
Since is continuous, then there exists some neighbourhood
of
, such that for all
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ24_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ25_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ26_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ27_HTML.gif)
For any , it follows from (2.24) and Lemma 2.18 that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ28_HTML.gif)
We have used the strong monotonicity and Lipschitz continuity of in
and Lipschitz continuity of
in
. Similarly, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ29_HTML.gif)
It follows from (2.25) and (2.26) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ30_HTML.gif)
where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ31_HTML.gif)
Define on
by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ32_HTML.gif)
It is easy to see that is a Banach space. For any given
, define
by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ33_HTML.gif)
By assumption, we know that . It follows from (2.27) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ34_HTML.gif)
which implies that is a contraction operator. Hence, there exists a unique
, such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ35_HTML.gif)
That is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ36_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ37_HTML.gif)
where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ38_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ39_HTML.gif)
Proof.
By Theorem 2.17, SVI admits a unique solution . It follows from Lemma 2.18 that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ40_HTML.gif)
It follows from (3.1) and (3.4) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ41_HTML.gif)
By (3.5), we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ42_HTML.gif)
where is defined by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ43_HTML.gif)
Set
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ44_HTML.gif)
Then (3.6) can be rewritten as
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ45_HTML.gif)
By (3.2), we know that . It follows from Lemma 3.1 that
and that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F182539/MediaObjects/13663_2009_Article_1216_Equ46_HTML.gif)
for all . Therefore,
converges geometrically to the unique solution
of SVI. This completes the proof.
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Acknowledgment
The authors are grateful to the referee for his/her valuable suggestions and comments that help to present the paper in the present form. This work was supported by the Doctoral Initiating Foundation of Liaoning Province (20071097).
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Zhao, Y., Xia, Z., Pang, L. et al. Existence of Solutions and Algorithm for a System of Variational Inequalities. Fixed Point Theory Appl 2010, 182539 (2010). https://doi.org/10.1155/2010/182539
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DOI: https://doi.org/10.1155/2010/182539