#### Abstract

The existence of the solution for a new system of generalized nonlinear mixed
quasi variational inclusions with *H*-monotone operators is proved by using implicit resolvent technique, and the sensitivity
analysis of solution in Hilbert spaces is given. Our results improve and generalize some results of the recent ones.

#### 1. Introduction

Sensitivity analysis of solution for variational inequalities and variational inclusions has been studied by many authors via quite different technique. (See [1–7] and the reference therein).

In 2004, Agarwal et al. [1] introduced and studied the following problem which is called the system of parametric generalized nonlinear mixed quasi variational inclusions.

Let be a real Hilbert space endowed with the product and norm , respectively. Let and be two nonempty open subsets of in which the parametric and take values. Let and be two maximal monotone mappings with respect to the first argument. and be nonlinear single-valued mappings. The system of parametric generalized nonlinear mixed quasi variational inclusions problem [1] is to find such that where and are two constants.

In this paper, we introduce a new system of parametric generalized nonlinear mixed quasi variational inclusions problem.

For each , , find , such that where , are nonlinear single-valued mappings, , are multivalued mappings, , are constants. By using implicit resolvent equations technique of -monotone operator, the existence of solution is proved and the sensitivity analysis of solution for the problem (1.2) is given. Our results improve and generalize the known results of [1, 2, 8, 9].

#### 2. Preliminaries

Let be a real Hilbert space, , be two nonempty open subsets of .

*Definition 2.1 (see [1]). *(i)A mapping is said to be monotone with respect to the first argument if
(ii) is said to be -strongly monotone with respect to the first argument if there exists a constant such that
(iii) is said to be -Lipschitz continuous if there exist such that

*Definition 2.2 (see [1]). *A mapping is said to be -Lipschitz continuous if there exist such that

*Definition 2.3 (see [1]). *A mapping is said to be -strongly monotone with respect to in the first argument if there exists such that
In a similar way, we can define the strong monotonicity of with respect to in the second argument.

*Definition 2.4 (see [9]). *Let be a single-valued mapping and be a multi-valued mapping. is said to be -monotone if is monotone with respect to the first argument and holds for all and .

*Definition 2.5 (see [9]). *Let be a strictly monotone mapping and be an -monotone mapping. The resolvent operator is defined by

Lemma 2.6 (see [9]). *Let be -strongly monotone with respect to the first argument, be -strongly monotone with respect to the first argument, and be -monotone, and be -monotone. Then for any fixed , the resolvent operator and are Lipschitz continuous:
**
where , are constants.*

Lemma 2.7. *Let be -monotone and be -monotone. For any fixed . is a solution of (1.2) if and only if ** here , .*

*Proof. *Assume that satisfies relations and , since , , then and holds if and only if
and hence is a solution of and .

#### 3. Main Results

Lemma 3.1. *Let , be two continuous mappings. If there exist , , , such that
**
then there exist such that
*

*Proof. *For any , let , , then by (3.1),,we have
where . Let , then
and hence,
Since , (3.5) implies that and are both cauchy sequences. Therefore, there exist such that . By continuity of and , .

Theorem 3.2. *Let be -Lipschitz continuous and be -Lipschitz continuous. Let be -Lipschitz continuous and -strongly monotone with respect to in the first argument, be -Lipschitz continuous, be -Lipschitz continuous and -strongly monotone with respect to in the second argument, be -Lipschitz continuous. Suppose that is -monotone and is -monotone If
**
then, for each , the problem (1.2) has an unique solution .*

*Proof. *Let . By defining of and Lemma 2.6, we have
Since is -Lipschitz continuous and -strongly monotone with respect to in the first argument, is -Lipschitz continuous, we have
Combining (3.7)–(3.9), we have
By defining of and Lemma 2.6, we have
Since is -Lipschitz continuous and -strongly monotone with respect to in the second argument, is -Lipschitz continuous. We have
Combining (3.11)–(3.13), we have
By (3.10) and (3.14), we have
By (3.6) and Lemma 3.1, there exist . such that . By Lemma 2.7, is a solution of (1.2). From (3.15), we easily see that the solution of (1.2) is unique.

*Assumption 1. *For implicit resolvent operator , there are two constants and such that

Theorem 3.3. *Suppose that the mappings , and are the same as in Theorem 3.2, and for any fixed , the mappings , are continuous (or Lipschitz continuous.) If Assumption 1 and condition (3.6) hold, then the solution for the problem (1.2) is continuous (or Lipschitz continuous).*

*Proof. *Suppose that , such that . From Theorem 3.2, we know that the problem (1.2) has solution ,, and .

Estimate and . By Lemma 2.7,we have
It follows from Assumption 1, Lemmas 2.6 and 2.7 that
Since is -Lipschitz continuous, -strongly monotone with respect to in the first argument and is -Lipschitz continuous, we conclude
It follows from (3.18) and (3.19) that
that is,
where
By Lemma 2.7, we have
It follows from Assumption 1, Lemmas 2.6 and 2.7 that
Since is -Lipschitz continuous and -strongly monotone with respect to in the second argument and is -Lipschitz continuous,we conclude
It follows from (3.23) and (3.24) that
That is,
where
By combining (3.21) and (3.26), we derive
That is,
where
By (3.6),
and hence

Estimate and .

By Lemma 2.7, we have
It follows from Assumption 1, Lemmas 2.6 and 2.7 that
Since is -Lipschitz continuous and -strongly monotone with respect to in the second argument and is -Lipschitz continuous,we conclude
It follows from (3.33) and (3.34) that
where defined by .

By Lemma 2.7, we have
As the proof of (3.25),we have
that is,
where is defined by .

Therefore
where is defined by .

(C) Prove that the conclusion of Theorem 3.3.

From (3.28) and (3.40), by the assumptions for , and in Theorem 3.3 and relation
we know that is continuous (or Lipschitz continuous).

From (3.29) and (3.39), by the assumptions for and in Theorem 3.3 and relation
we know that is continuous (or Lipschitz continuous).

#### Acknowledgment

The author would like to thank the anonymous referee for reading this paper carefully, providing valuable suggestions and comments. The work was supported by the National Science Foundation of China (no. 10561007) and Science and Technology Research Project of Education Department in Jiangxi province (GJJ10269).