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A parallel resolvent method for solving a system of nonlinear mixed variational inequalities
Journal of Inequalities and Applicationsvolume 2013, Article number: 509 (2013)
Abstract
In this paper, we introduce a system of generalized nonlinear mixed variational inequalities and obtain the approximate solvability by using the resolvent parallel technique. Our results may be viewed as an extension and improvement of the previously known results for variational inequalities.
1 Introduction and preliminaries
Variational inequality theory, which was introduced by Stampacchia [1] in 1964, has been witnessed as an interesting branch of mathematical and engineering sciences with a wide range of applications in industry, finance, economics and pure and applied sciences. In 2001, Verma [2] introduced a new system of strongly monotonic variational inequalities and studied the approximation solvability of the system based on the application of a projection method. The main and basic idea in this technique is to establish the equivalence between variational inequalities and fixed point problems. This alternative equivalence has been used to develop several projection iterative methods for solving variational inequalities and related optimization problems. Several extensions and generalizations of the system of strongly monotonic variational inequalities have been considered by many authors [3–12]. Inspired and motivated by research in this area, we introduce a system of generalized nonlinear mixed variational inequalities problem involving two different nonlinear operators. It is well known that if the nonlinear term in the mixed variational inequality is a proper, convex, and lower semicontinuous, then one can establish the equivalence between the mixed variational inequality and the fixed point problem. Using the parallel algorithm considered in [12], we suggest and analyze a parallel iterative method for solving this system. Our result may be viewed as an extension and improvement of the recent results.
Let ℋ be a real Hilbert space whose inner product and norm are denoted by $\u3008\cdot ,\cdot \u3009$ and $\parallel \cdot \parallel $, respectively. Let K be a nonempty closed convex subset of ℋ. Let ${T}_{1},{T}_{2}:K\times K\to \mathcal{H}$ be two nonlinear operators. Let ${\phi}_{1},{\phi}_{2}:\mathcal{H}\to \mathbb{R}\cup \{+\mathrm{\infty}\}$ be proper convex lower semicontinuous functions on ℋ. We consider a system of generalized nonlinear mixed variational inequalities (abbreviated as SMNVI) as follows: Find $({x}^{\ast},{y}^{\ast})\in K\times K$ such that
where $g:K\to K$ is a mapping and $\rho ,\eta >0$.
Note that if ${\phi}_{1}={\phi}_{2}={\delta}_{K}$, and $g=I$, where I is the identity operator, ${\delta}_{K}$ is the indicator function of K defined by
then problem (1.1) reduces to the following system of nonlinear variational inequalities (SNVI) considered in [3] of finding $({x}^{\ast},{y}^{\ast})\in K\times K$ such that
If ${T}_{1}={T}_{2}=T$ and $g=I$, where I is the identity operator, then problem (1.1) is equivalent to the following system of nonlinear mixed variational inequalities (SNVI) considered in [7, 8] of finding $({x}^{\ast},{y}^{\ast})\in K\times K$ such that
If ${\phi}_{1}={\phi}_{2}={\delta}_{K}$ and ${T}_{1},{T}_{2}:K\to \mathcal{H}$ are univariate mappings, then problem (1.1) is reduced to the following system of nonlinear variational inequalities (SNVI) considered in [12] of finding $({x}^{\ast},{y}^{\ast})\in K\times K$ such that
where $g:K\to K$ is a mapping.
If ${T}_{1}={T}_{2}=T$, $g=I$ and ${\phi}_{1}={\phi}_{2}={\delta}_{K}$, where T is a univariate mapping defined by $T:K\to \mathcal{H}$, then problem (1.1) reduces to the following system of variational inequalities (SVI) considered in [2] of finding $({x}^{\ast},{y}^{\ast})\in K\times K$ such that
We also need the following wellknown results.
Definition 1.1 Define the norm $\parallel \cdot \parallel $ on $\mathcal{H}\times \mathcal{H}$ by
Definition 1.2 For any maximal monotone operator T, the resolvent operator associated with T, for any $\lambda >0$, is defined by
Remark 1.1 It is well known that the subdifferential ∂φ of a proper convex lower semicontinuous function $\phi :\mathcal{H}\to \mathbb{R}\cup \{+\mathrm{\infty}\}$ is a maximal monotone operator. We can define its resolvent operator by
where $\lambda >0$ and ${J}_{\phi}$ is defined everywhere.
Lemma 1.1 [13]
For a given $u,z\in \mathcal{H}$ satisfies the inequality
if and only if $u={J}_{\phi}^{\lambda}(z)$, where ${J}_{\phi}^{\lambda}(u)={(I+\lambda \partial \phi )}^{1}(u)$ is the resolvent operator and $\lambda >0$.
If φ is the indicator function of a closed convex set $K\subseteq \mathcal{H}$, then the resolvent operator ${J}_{\phi}^{\lambda}(\cdot )$ reduces to the projection operator ${P}_{K}(\cdot )$. It is well known that ${J}_{\phi}^{\lambda}$ is nonexpansive, i.e.,
Based on Lemma 1.1, similar to that in [8] and [7], the following statement gives equivalent characterization of problem (1.1).
Lemma 1.2 Problem (1.1) is equivalent to finding $({x}^{\ast},{y}^{\ast})\in K\times K$ such that
where ${J}_{{\phi}_{i}}^{1}={(I+\partial {\phi}_{i})}^{1}$, $i=1,2$.
Proof Suppose that $({x}^{\ast},{y}^{\ast})\in K\times K$ is a solution of the following generalized nonlinear mixed variational inequalities (abbreviated as SNMVI):
where $g:K\to K$ is a mapping and $\rho >0$, ${\rho}^{\prime}>0$, $\eta >0$, ${\eta}^{\prime}>0$. Using Lemma 1.1, we can easily show that problem (1.7) is equivalent to
where ${J}_{{\phi}_{1}}^{{\rho}^{\prime}}={(I+{\rho}^{\prime}\partial {\phi}_{1})}^{1}$, ${J}_{{\phi}_{2}}^{{\eta}^{\prime}}={(I+{\eta}^{\prime}\partial {\phi}_{2})}^{1}$. Let ${\rho}^{\prime}={\eta}^{\prime}=1$. Then problem (1.7) reduces to problem (1.1) and ${J}_{{\phi}_{i}}^{1}={(I+\partial {\phi}_{i})}^{1}$, $i=1,2$. This completes the proof. □
Remark 1.2 If ${T}_{1}={T}_{2}=T$ and $g=I$, where I is the identity operator, then Lemma 1.2 reduces to Lemma 1.2 in [7].
Definition 1.3 A mapping $T:K\times K\to \mathcal{H}$ is said to be

(1)
relaxed g$(\gamma ,r)$cocoercive if there exist constants $\gamma >0$ and $r>0$ such that for all $x,y\in K$,
$$\u3008T(x,u)T(y,v),g(x)g(y)\u3009\ge (\gamma ){\parallel T(x,u)T(y,v)\parallel}^{2}+r{\parallel g(x)g(y)\parallel}^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall}u,v\in K;$$ 
(2)
gμLipschitz continuous in the first variable if there exists a constant $\mu >0$ such that for all $x,y\in K$,
$$\parallel T(x,u)T(y,v)\parallel \le \mu \parallel g(x)g(y)\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall}u,v\in K.$$
Remark 1.3 If T is a univariate mapping and $g=I$, where I is the identity operator, then Definition 1.3 reduces to the standard definition of relaxed $(\gamma ,r)$cocoercive and Lipschitz continuous, respectively.
Definition 1.4 A mapping $g:K\to \mathcal{H}$ is said to be αexpansive if there exists a constant $\alpha >0$ such that for all $x,y\in \mathcal{H}$,
Lemma 1.3 [14]
Suppose that $\{{\delta}_{n}\}$ is a nonnegative sequence satisfying the following inequality:
where ${n}_{0}$ is a nonnegative number, ${\lambda}_{n}\in [0,1]$ with ${\sum}_{n=0}^{\mathrm{\infty}}{\lambda}_{n}=\mathrm{\infty}$, and ${\sigma}_{n}=o({\lambda}_{n})$. Then ${lim}_{n\to \mathrm{\infty}}{\delta}_{n}=0$.
2 Algorithms
In this section, we suggest a parallel algorithm associated with the resolvent operator for solving the system of SNMVI. Our results extend and improve the corresponding results in [2, 3, 7, 11, 12]. In fact, using Lemma 1.2, we suggest the following iterative method for solving problem (1.1).
Algorithm 2.1 For arbitrarily chosen initial points ${x}_{0},{y}_{0}\in K$ (and $g({x}_{0}),g({y}_{0})\in K$), compute the sequences $\{{x}_{n}\}$ and $\{{y}_{n}\}$ such that
where ${J}_{{\phi}_{i}}^{1}={(I+\partial {\phi}_{i})}^{1}$, $i=1,2$, is the resolvent operator, $\rho ,\eta >0$, ${\alpha}_{n}\in [0,1]$ and ${\beta}_{n}\in [0,1]$ for all $n\ge 0$.
As reported in [12], one of the attractive features of Algorithm 2.1 is that it is suitable for implementing on two different processor computers. In other words, ${x}_{n+1}$ and ${y}_{n+1}$ are solved in parallel, and Algorithm 2.1 is the socalled parallel resolvent method. We refer the interested reader to papers [15–17] and references therein for more examples and ideas of parallel iterative methods.
If ${\phi}_{1}={\phi}_{2}={\delta}_{K}$, and $g=I$, ${\delta}_{K}$ is the indicator function of K, then Algorithm 2.1 reduces to the following algorithm.
Algorithm 2.2 For arbitrarily chosen initial points ${x}_{0},{y}_{0}\in K$, compute the sequences $\{{x}_{n}\}$ and $\{{y}_{n}\}$ such that
where $\rho ,\eta >0$, ${\alpha}_{n}\in [0,1]$ and ${\beta}_{n}\in [0,1]$ for all $n\ge 0$.
If ${T}_{1}={T}_{2}=T$ and $g=I$, then Algorithm 2.1 reduces to the following algorithm.
Algorithm 2.3 For arbitrarily chosen initial points ${x}_{0},{y}_{0}\in K$, compute the sequences $\{{x}_{n}\}$ and $\{{y}_{n}\}$ such that
where ${J}_{{\phi}_{i}}^{1}={(I+\partial {\phi}_{i})}^{1}$, $i=1,2$, is the resolvent operator, $\rho ,\eta >0$, ${\alpha}_{n}\in [0,1]$ and ${\beta}_{n}\in [0,1]$ for all $n\ge 0$.
If ${\phi}_{1}={\phi}_{2}={\delta}_{K}$ and ${T}_{1},{T}_{2}:K\to \mathcal{H}$ are univariate mappings, then Algorithm 2.1 reduces to the following algorithm.
Algorithm 2.4 For arbitrarily chosen initial points ${x}_{0},{y}_{0}\in K$ (and $g({x}_{0}),g({y}_{0})\in K$), compute the sequences $\{{x}_{n}\}$ and $\{{y}_{n}\}$ such that
where $\rho ,\eta >0$, ${\alpha}_{n}\in [0,1]$ and ${\beta}_{n}\in [0,1]$ for all $n\ge 0$.
If ${T}_{1}={T}_{2}=T$, $g=I$ and ${\phi}_{1}={\phi}_{2}={\delta}_{K}$, where T is a univariate mapping defined by $T:K\to \mathcal{H}$, then Algorithm 2.1 reduces to the following algorithm.
Algorithm 2.5 For arbitrarily chosen initial points ${x}_{0},{y}_{0}\in K$ (and $g({x}_{0}),g({y}_{0})\in K$), compute the sequences $\{{x}_{n}\}$ and $\{{y}_{n}\}$ such that
where $\rho ,\eta >0$, ${\alpha}_{n}\in [0,1]$ and ${\beta}_{n}\in [0,1]$ for all $n\ge 0$.
3 Main results
In this section, based on Algorithm 2.1, we now present the approximation solvability of problem (1.1) involving relaxed g$(\gamma ,r)$cocoercive and gμLipschitz continuous in the first variable mappings in Hilbert settings.
Theorem 3.1 Let ℋ be a real Hilbert space. Let K be a nonempty closed convex subset of ℋ, and let ${T}_{i}:K\times K\to \mathcal{H}$ be relaxed g$({\gamma}_{i},{r}_{i})$cocoercive and g${\mu}_{i}$Lipschitz continuous in the first variable for $i=1,2$. Let $g:K\to K$ be an αexpansive mapping. Suppose that $({x}^{\ast},{y}^{\ast})\in K\times K$ is the unique solution to problem (1.1) and $\{{x}_{n}\}$, $\{{y}_{n}\}$ are generated by Algorithm 2.1. If $\{{\alpha}_{n}\}$ and $\{{\beta}_{n}\}$ are two sequences in $[0,1]$ satisfying the following conditions:

(i)
${\alpha}_{n}{\theta}_{2}{\beta}_{n}\ge 0$ and ${\beta}_{n}{\theta}_{1}{\alpha}_{n}\ge 0$ such that ${\sum}_{n=0}^{\mathrm{\infty}}{\alpha}_{n}{\theta}_{2}{\beta}_{n}=\mathrm{\infty}$, ${\sum}_{n=0}^{\mathrm{\infty}}{\beta}_{n}{\theta}_{1}{\alpha}_{n}=\mathrm{\infty}$,

(ii)
${\theta}_{1}=\sqrt{1+2\rho {\gamma}_{1}{{\mu}_{1}}^{2}2\rho {r}_{1}+{\rho}^{2}{\mu}_{1}}$ such that $0<{\theta}_{1}<1$,

(iii)
${\theta}_{2}=\sqrt{1+2\eta {\gamma}_{2}{{\mu}_{2}}^{2}2\eta {r}_{2}+{\eta}^{2}{\mu}_{2}}$ such that $0<{\theta}_{2}<1$,
then the sequences $\{{x}_{n}\}$ and $\{{y}_{n}\}$ converge to ${x}^{\ast}$ and ${y}^{\ast}$, respectively.
Proof Since $({x}^{\ast},{y}^{\ast})\in K\times K$ is the unique solution to problem (1.1), from Lemma 1.2 it follows that
We first evaluate $\parallel g({x}_{n+1})g({x}^{\ast})\parallel $ for all $n\ge 0$. From (2.1) and the nonexpansive property of the resolvent operator, we have
Notice that ${T}_{1}$ is relaxed g$({\gamma}_{1},{r}_{1})$cocoercive and g${\mu}_{1}$Lipschitz continuous in the first variable. Then we have
where ${\theta}_{1}=\sqrt{1+2\rho {\gamma}_{1}{{\mu}_{1}}^{2}2\rho {r}_{1}+{\rho}^{2}{\mu}_{1}}<1$ in view of assumption (ii). Substituting (3.3) into (3.2), we have
Similarly, since ${T}_{2}$ is relaxed g$({\gamma}_{2},{r}_{2})$cocoercive and g${\mu}_{2}$Lipschitz continuous in the first variable, we have
where ${\theta}_{2}=\sqrt{1+2\eta {\gamma}_{2}{{\mu}_{2}}^{2}2\eta {r}_{2}+{\eta}^{2}{\mu}_{2}}<1$ in view of assumption (iii). It follows from (3.4) and (3.5) that
where ${w}_{1n}=1({\alpha}_{n}{\theta}_{2}{\beta}_{n})$ and ${w}_{2n}=1({\beta}_{n}{\theta}_{1}{\alpha}_{n})$.
From assumption (i) and Lemma 1.3, we can obtain
and so
which implies that sequences $\{g({x}_{n})\}$ and $\{g({y}_{n})\}$ converge to $g({x}^{\ast})$ and $g({y}^{\ast})$, respectively. Since g is αexpansive, it follows that $\{{x}_{n}\}$ and $\{{y}_{n}\}$ converge to ${x}^{\ast}$ and ${y}^{\ast}$, respectively. This completes the proof. □
The following theorems can be obtained from Theorem 3.1 immediately.
Theorem 3.2 [3]
Let ℋ be a real Hilbert space. Let K be a nonempty closed convex subset of ℋ, and let ${T}_{i}:K\times K\to \mathcal{H}$ be relaxed $({\gamma}_{i},{r}_{i})$cocoercive and ${\mu}_{i}$Lipschitz continuous in the first variable for $i=1,2$. Suppose that $({x}^{\ast},{y}^{\ast})\in K\times K$ is the unique solution to problem (1.2) and $\{{x}_{n}\}$, $\{{y}_{n}\}$ are generated by Algorithm 2.2. If $\{{\alpha}_{n}\}$ and $\{{\beta}_{n}\}$ are two sequences in $[0,1]$ satisfying the following conditions:

(1)
${\alpha}_{n}{\theta}_{2}{\beta}_{n}\ge 0$ and ${\beta}_{n}{\theta}_{1}{\alpha}_{n}\ge 0$ such that ${\sum}_{n=0}^{\mathrm{\infty}}{\alpha}_{n}{\theta}_{2}{\beta}_{n}=\mathrm{\infty}$, ${\sum}_{n=0}^{\mathrm{\infty}}{\beta}_{n}{\theta}_{1}{\alpha}_{n}=\mathrm{\infty}$,

(2)
${\theta}_{1}=\sqrt{1+2\rho {\gamma}_{1}{{\mu}_{1}}^{2}2\rho {r}_{1}+{\rho}^{2}{\mu}_{1}}$ such that $0<{\theta}_{1}<1$,

(3)
${\theta}_{2}=\sqrt{1+2\eta {\gamma}_{2}{{\mu}_{2}}^{2}2\eta {r}_{2}+{\eta}^{2}{\mu}_{2}}$ such that $0<{\theta}_{2}<1$,
then the sequences $\{{x}_{n}\}$ and $\{{y}_{n}\}$ converge to ${x}^{\ast}$ and ${y}^{\ast}$, respectively.
Theorem 3.3 [12]
Let ℋ be a real Hilbert space. Let K be a nonempty closed convex subset of ℋ, and let ${T}_{i}:K\to \mathcal{H}$ be relaxed g$({\gamma}_{i},{r}_{i})$cocoercive and g${\mu}_{i}$Lipschitz continuous for $i=1,2$. Let $g:K\to K$ be an αexpansive mapping. Suppose that $({x}^{\ast},{y}^{\ast})\in K\times K$ is the unique solution to problem (1.4) and $\{{x}_{n}\}$, $\{{y}_{n}\}$ are generated by Algorithm 2.4. If $\{{\alpha}_{n}\}$ and $\{{\beta}_{n}\}$ are two sequences in $[0,1]$ satisfying the following conditions:

(1)
$0\le {\alpha}_{n},{\beta}_{n}\le 1$, ${\alpha}_{n}{\theta}_{2}{\beta}_{n}\ge 0$ and ${\beta}_{n}{\theta}_{1}{\alpha}_{n}\ge 0$ such that ${\sum}_{n=0}^{\mathrm{\infty}}{\alpha}_{n}{\theta}_{2}{\beta}_{n}=\mathrm{\infty}$, ${\sum}_{n=0}^{\mathrm{\infty}}{\beta}_{n}{\theta}_{1}{\alpha}_{n}=\mathrm{\infty}$,

(2)
${\theta}_{1}=\sqrt{1+2\rho {\gamma}_{1}{{\mu}_{1}}^{2}2\rho {r}_{1}+{\rho}^{2}{\mu}_{1}}$ such that $0<{\theta}_{1}<1$,

(3)
${\theta}_{2}=\sqrt{1+2\eta {\gamma}_{2}{{\mu}_{2}}^{2}2\eta {r}_{2}+{\eta}^{2}{\mu}_{2}}$ such that $0<{\theta}_{2}<1$,
then the sequences $\{{x}_{n}\}$ and $\{{y}_{n}\}$ converge to ${x}^{\ast}$ and ${y}^{\ast}$, respectively.
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Acknowledgements
This work was supported by the Natural Science Foundation of China (60804065, 11371015), the Key Project of Chinese Ministry of Education (211163), Sichuan Youth Science and Technology Foundation (2012JQ0032).
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Keywords
 resolvent operator
 parallel projection
 relaxed cocoercive
 generalized nonlinear mixed variational inequalities