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Journal of Applied Mathematics
Volume 2011, Article ID 921835, 15 pages
http://dx.doi.org/10.1155/2011/921835
Research Article

Sensitivity Analysis for a System of Generalized Nonlinear Mixed Quasi Variational Inclusions with H-Monotone Operators

Department of Science, Nanchang Institute of Technology, Nanchang 330099, China

Received 20 March 2011; Revised 2 June 2011; Accepted 13 June 2011

Academic Editor: Ya Ping Fang

Copyright © 2011 Han-Wen Cao. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 [17] 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 Ω and Λ be two nonempty open subsets of in which the parametric 𝜔 and 𝜆 take values. Let 𝑀×Ω2 and 𝑁×Λ2 be two maximal monotone mappings with respect to the first argument. 𝐻1,𝑆×Ω and 𝐻2,𝑇×Λ be nonlinear single-valued mappings. The system of parametric generalized nonlinear mixed quasi variational inclusions problem [1] is to find (𝑥,𝑦)× such that 𝐻0𝑥𝑦+𝜌1𝐻(𝑦,𝜔),𝑆(𝑦,𝜔)+𝜌𝑀(𝑥,𝜔),0𝑦𝑥+𝛾2(𝑥,𝜆)+𝑇(𝑥,𝜆)+𝛾𝑁(𝑦,𝜆),(1.1) where 𝜌>0 and 𝛾>0 are two constants.

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

For each 𝜔Ω, 𝜆Λ, find 𝑥=𝑥(𝜔,𝜆), 𝑦=𝑦(𝜔,𝜆) such that 0𝑓(𝑥,𝑦,𝜔)+𝜌1(𝐹(𝑥,𝑦,𝜔)+𝑀(𝑥,𝜔)),0𝑔(𝑥,𝑦,𝜆)+𝜌2(𝐺(𝑥,𝑦,𝜆)+𝑁(𝑦,𝜆)),(1.2) where 𝑓,𝐹××Ω, 𝑔,𝐺××Λ are nonlinear single-valued mappings, 𝑀×Ω2, 𝑁×Λ2 are multivalued mappings, 𝜌1>0, 𝜌2>0 are constants. By using implicit resolvent equations technique of 𝐻-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, Ω, Λ be two nonempty open subsets of .

Definition 2.1 (see [1]). (i)A mapping 𝑇×Ω is said to be monotone with respect to the first argument if 𝑇(𝑥,𝜔)𝑇(𝑦,𝜔),𝑥𝑦0,(𝑥,𝜔),(𝑦,𝜔)×Ω.(2.1)(ii)𝑇 is said to be 𝜅-strongly monotone with respect to the first argument if there exists a constant 𝜅>0 such that 𝑇(𝑥,𝜔)𝑇(𝑦,𝜔),𝑥𝑦𝜅𝑥𝑦2,(𝑥,𝜔),(𝑦,𝜔)×Ω.(2.2)(iii)𝑇 is said to be (𝜉,𝜂)-Lipschitz continuous if there exist 𝜉>0,𝜂>0 such that 𝑇𝑥,𝜔1𝑇𝑦,𝜔2𝜔𝜉𝑥𝑦+𝜂1𝜔2,𝑥,𝜔1,𝑦,𝜔2×Ω.(2.3)

Definition 2.2 (see [1]). A mapping 𝐹××Ω is said to be (𝜉,𝜂,𝜁)-Lipschitz continuous if there exist 𝜉>0,𝜂>0,𝜁>0 such that 𝐹𝑥1,𝑦1,𝜔1𝑥𝐹2,𝑦2,𝜔2𝑥𝜉1𝑥2𝑦+𝜂1𝑦2𝜔+𝜁1𝜔2𝑥,1,𝑦1,𝜔1,𝑥2,𝑦2,𝜔2××Ω.(2.4)

Definition 2.3 (see [1]). A mapping 𝐹××Ω is said to be 𝛼-strongly monotone with respect to 𝐻 in the first argument if there exists 𝛼>0 such that 𝐹𝑥1𝑥,𝑦,𝜔𝐹2,𝑦,𝜔,𝐻𝑥1𝐻𝑥2𝑥𝛼1𝑥22𝑥,1,𝑥,𝑦,𝜔2,𝑦,𝜔××Ω.(2.5) In a similar way, we can define the strong monotonicity of 𝐹 with respect to 𝐻 in the second argument.

Definition 2.4 (see [9]). Let 𝐻×Ω be a single-valued mapping and 𝑀×Ω2 be a multi-valued mapping. 𝑀 is said to be 𝐻-monotone if 𝑀 is monotone with respect to the first argument and (𝐻(,𝜔)+𝜌𝑀(,𝜔))()=holds for all 𝜌>0 and 𝜔Ω.

Definition 2.5 (see [9]). Let 𝐻×Ω be a strictly monotone mapping and 𝑀×Ω2 be an 𝐻-monotone mapping. The resolvent operator 𝐽𝐻𝑀(,𝜔),𝜌 is defined by 𝐽𝐻𝑀(,𝜔),𝜌(𝑢)=(𝐻(,𝜔)+𝜌𝑀(,𝜔))1(𝑢),𝑢.(2.6)

Lemma 2.6 (see [9]). Let 𝐻1×Ω be 𝛾1-strongly monotone with respect to the first argument, 𝐻2×Λ be 𝛾2-strongly monotone with respect to the first argument, and 𝑀×Ω2 be 𝐻1-monotone, and𝑁×Λ2 be 𝐻2-monotone. Then for any fixed 𝜔Ω,𝜆Λ, the resolvent operator 𝐽𝐻1𝑀(,𝜔),𝜌1 and 𝐽𝐻2𝑁(,𝜆),𝜌2 are Lipschitz continuous: 𝐽𝐻1𝑀(,𝜔),𝜌1(𝑢)𝐽𝐻1𝑀(,𝜔),𝜌11(𝑣)𝛾1𝐽𝑢𝑣,𝑢,𝑣,𝐻2𝑁(,𝜆),𝜌2(𝑢)𝐽𝐻2𝑁(,𝜆),𝜌21(𝑣)𝛾2𝑢𝑣,𝑢,𝑣,(2.7) where 𝛾1>0, 𝛾2>0 are constants.

Lemma 2.7. Let 𝑀×Ω2 be 𝐻1-monotone and 𝑁×Λ2 be 𝐻2-monotone. For any fixed (𝜔,𝜆)Ω×Λ. (𝑥(𝜔,𝜆),𝑦(𝜔,𝜆)) is a solution of (1.2) if and only if 𝑥(𝜔,𝜆)=𝐽𝐻1𝑀(,𝜔),𝜌1𝐻1(𝑥(𝜔,𝜆),𝜔)𝑓(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜔)𝜌1𝐹(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜔)𝑇(𝑥,𝑦,𝜔,𝜆),(2.81)𝑦(𝜔,𝜆)=𝐽𝐻2𝑁(,𝜆),𝜌2𝐻2(𝑦(𝜔,𝜆),𝜆)𝑔(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜆)𝜌2𝐺(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜆)𝑆(𝑥,𝑦,𝜔,𝜆),(2.82) here 𝑥=𝑥(𝜔,𝜆), 𝑦=𝑦(𝜔,𝜆).

Proof. Assume that (𝑥(𝜔,𝜆),𝑦(𝜔,𝜆)) satisfies relations 2.81 and 2.82, since 𝐽𝐻1𝑀(,𝜔),𝜌1=(𝐻1(,𝜔)+𝜌1𝑀(,𝜔))1, 𝐽𝐻2𝑀(,𝜔),𝜌2=(𝐻2(,𝜔)+𝜌2𝑁(,𝜔))1, then 2.81 and 2.82 holds if and only if 0𝑓(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜔)+𝜌1𝐹(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜔)+𝜌1𝑀(𝑥(𝜔,𝜆),𝜔),0𝑔(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜆)+𝜌2𝐺(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜆)+𝜌2𝑁(𝑦(𝜔,𝜆),𝜆),(2.9) and hence (𝑥(𝜔,𝜆)),𝑦(𝜔,𝜆) is a solution of 2.81 and 2.82.

3. Main Results

Lemma 3.1. Let 𝑇×, 𝑆× be two continuous mappings. If there exist Θ1, Θ2, 0<Θ1,Θ2<1, such that 𝑇𝑥1,𝑦1𝑥𝑇2,𝑦2+𝑆𝑥1,𝑦1𝑥𝑆2,𝑦2Θ1𝑥1𝑥2+Θ2𝑦1𝑦2,𝑥1,𝑥2,𝑦1,𝑦2.(3.1) then there exist 𝑥,𝑦 such that 𝑥𝑥=𝑇,𝑦,𝑦𝑥=𝑆,𝑦.(3.2)

Proof. For any 𝑥0,𝑦0, let 𝑥𝑛+1=𝑇(𝑥𝑥,𝑦𝑛), 𝑦𝑛+1=𝑆(𝑥𝑛,𝑦𝑛),𝑛=0,1,2,, then by (3.1),forall𝑥1,𝑥2,,𝑥𝑛,𝑥𝑛+1,𝑦1,𝑦2,,𝑦𝑛,𝑦𝑛+1,we have 𝑥𝑛+1𝑥𝑛+𝑦𝑛+1𝑦𝑛Θ1𝑥𝑛𝑥𝑛1+Θ2𝑦𝑛𝑦𝑛1𝑥Θ𝑛𝑥𝑛1+𝑦𝑛𝑦𝑛1,(3.3) where Θ=max{Θ1,Θ2}. Let 𝑎=𝑥1𝑥0+𝑦1𝑦0, then 𝑥𝑛+1𝑥𝑛+𝑦𝑛+1𝑦𝑛𝑥Θ𝑛𝑥𝑛1+𝑦𝑛𝑦𝑛1Θ2𝑥𝑛1𝑥𝑛2+𝑦𝑛1𝑦𝑛2Θ𝑛𝑥1𝑥0+𝑦1𝑦0=Θ𝑛𝑎,(3.4) and hence, 𝑥0𝑛+1𝑥𝑛Θ𝑛𝑦𝑎,0𝑛+1𝑦𝑛Θ𝑛𝑎.(3.5) Since 0<Θ=max{Θ1,Θ2}<1, (3.5) implies that {𝑥𝑛} and {𝑦𝑛} are both cauchy sequences. Therefore, there exist 𝑥,𝑦 such that 𝑥𝑛𝑥,𝑦𝑛𝑦(𝑛). By continuity of 𝑇 and 𝑆, 𝑥=𝑇(𝑥,𝑦),𝑦=𝑆(𝑥,𝑦).

Theorem 3.2. Let 𝐻1×Ω be (𝜉1,𝜂1)-Lipschitz continuous and 𝐻2×Λ be (𝜉2,𝜂2)-Lipschitz continuous. Let 𝑓××Ω be (𝜉𝑓,𝜂𝑓,𝜁𝑓)-Lipschitz continuous and 𝛾𝑓-strongly monotone with respect to 𝐻 in the first argument, 𝐹××Ω be (𝜉𝐹,𝜂𝐹,𝜁𝐹)-Lipschitz continuous,𝑔××Λ be (𝜉𝑔,𝜂𝑔,𝜁𝑔)-Lipschitz continuous and 𝛾𝑔-strongly monotone with respect to 𝐻2 in the second argument,𝐺××Λ be (𝜉𝐺,𝜂𝐺,𝜁𝐺)-Lipschitz continuous. Suppose that 𝑀×Ω2 is 𝐻1-monotone and 𝑁×Λ2 is 𝐻2-monotone If Θ1=1𝛾1𝜉212𝛾𝑓+𝜉2𝑓+𝜌1𝜉𝐹+𝜉𝑔+𝜌2𝜉𝐺𝛾2Θ<1,2=1𝛾2𝜉222𝛾𝑔+𝜂2𝑔+𝜌2𝜂𝐺+𝜂𝑓+𝜌1𝜂𝐹𝛾1<1,(3.6) then, for each 𝜔,𝜆Ω×Λ, the problem (1.2) has an unique solution (𝑥(𝜔,𝜆),𝑦(𝜔,𝜆)).

Proof. Let 𝑥1=𝑥1(𝜔,𝜆),𝑥2=𝑥2(𝜔,𝜆),𝑦1=𝑦1(𝜔,𝜆),𝑦2=𝑦2(𝜔,𝜆). By defining 2.81 of 𝑇 and Lemma 2.6, we have 𝑇𝑥1,𝑦1𝑥,𝜔,𝜆𝑇2,𝑦21,𝜔,𝜆𝛾1𝐻1𝑥1𝑥,𝜔𝑓1,𝑦1,𝜔𝜌1𝐹𝑥1,𝑦1,𝜔𝐻1𝑥2𝑥,𝜔+𝑓2,𝑦2,𝜔+𝜌1𝐹𝑥2,𝑦21,𝜔𝛾1𝐻1𝑥1,𝜔𝐻1𝑥2𝑥,𝜔𝑓1,𝑦1𝑥,𝜔+𝑓2,𝑦2+1,𝜔𝛾1𝜌1𝐹𝑥1,𝑦1𝑥,𝜔𝐹2,𝑦2.,𝜔(3.7) Since 𝑓 is (𝜉𝑓,𝜂𝑓,𝜁𝑓)-Lipschitz continuous and 𝛾𝑓-strongly monotone with respect to 𝐻1 in the first argument, 𝐹 is (𝜉𝐹,𝜂𝐹,𝜁𝐹)-Lipschitz continuous, we have 𝐻1𝑥1,𝜔𝐻1𝑥2𝑥,𝜔𝑓1,𝑦1𝑥,𝜔+𝑓2,𝑦2𝐻,𝜔1𝑥1,𝜔𝐻1𝑥2𝑥,𝜔𝑓1,𝑦1𝑥,𝜔+𝑓2,𝑦1+𝑓𝑥,𝜔2,𝑦1𝑥,𝜔𝑓2,𝑦2,𝜔𝜉212𝛾𝑓+𝜉2𝑓𝑥1𝑥2+𝜂𝑓𝑦1𝑦2,(3.8)𝐹𝑥1,𝑦1𝑥,𝜔𝐹2,𝑦2,𝜔𝜉𝐹𝑥1𝑥2+𝜂𝐹𝑦1𝑦2.(3.9) Combining (3.7)–(3.9), we have 𝑇𝑥1,𝑦1𝑥,𝜔,𝜆𝑇2,𝑦21,𝜔,𝜆𝛾1𝜉212𝛾𝑓+𝜉2𝑓+𝜌1𝜉𝐹𝑥1𝑥2+𝜂𝑓+𝜌1𝜂𝐹𝛾1𝑦1𝑦2.(3.10) By defining 2.82 of 𝑆 and Lemma 2.6, we have 𝑆𝑥1,𝑦1𝑥,𝜔,𝜆𝑆2,𝑦21,𝜔,𝜆𝛾2𝐻2𝑦1𝑥,𝜆𝑔1,𝑦1,𝜆𝜌2𝐺𝑥1,𝑦1,𝜆𝐻2𝑦2𝑥,𝜆+𝑔2,𝑦2,𝜆+𝜌2𝐺𝑥2,𝑦21,𝜆𝛾2𝐻2𝑦1,𝜆𝐻2𝑦2𝑥,𝜆𝑔1,𝑦1𝑥,𝜆+𝑔2,𝑔2+𝜌,𝜆2𝛾2𝐺𝑥1,𝑦1𝑥,𝜆𝐺2,𝑦2.,𝜆(3.11) Since 𝑔 is (𝜉𝑔,𝜂𝑔,𝜁𝑔)-Lipschitz continuous and 𝛾𝑔-strongly monotone with respect to 𝐻2 in the second argument, 𝐺 is (𝜉𝐺,𝜂𝐺,𝜁𝐺)-Lipschitz continuous. We have 𝐻2𝑦1,𝜆𝐻2𝑦2𝑥,𝜆𝑔1,𝑦1𝑥,𝜆+𝑔2,𝑦2𝐻,𝜆2𝑦1,𝜆𝐻2𝑦2𝑥,𝜆𝑔1,𝑦1𝑥,𝜆+𝑔1,𝑦2+𝑔𝑥,𝜆1,𝑦2𝑥,𝜆𝑔2,𝑦2,𝜆𝜉222𝛾𝑔+𝜂2𝑔𝑦1𝑦2+𝜉𝑔𝑥1𝑥2,(3.12)𝐺𝑥1,𝑦1𝑥,𝜆𝐺2,𝑦2,𝜆𝜉𝐺𝑥1𝑥2+𝜂𝐺𝑦1𝑦2.(3.13) Combining (3.11)–(3.13), we have 𝑆𝑥1,𝑦1𝑥,𝜔,𝜆𝑆2,𝑦21,𝜔,𝜆𝛾2𝜉222𝛾𝑔+𝜂2𝑔+𝜌2𝜂𝐺𝑦1𝑦2+𝜉𝑔+𝜌2𝜉𝐺𝛾2𝑥1𝑥2.(3.14) By (3.10) and (3.14), we have 𝑇𝑥1,𝑦1𝑥,𝜔,𝜆𝑇2,𝑦2+𝑆𝑥,𝜔,𝜆1,𝑦1𝑥,𝜔,𝜆𝑆2,𝑦21,𝜔,𝜆𝛾1𝜉212𝛾𝑓+𝜉2𝑓+𝜌1𝜉𝐹+𝜉𝑔+𝜌2𝜉𝐺𝛾2𝑥1𝑥2+1𝛾2𝜉222𝛾𝑔+𝜂2𝑔+𝜌2𝜂𝐺+𝜂𝑓+𝜌1𝜂𝐹𝛾1𝑦1𝑦2=Θ1𝑥1𝑥2+Θ2𝑦1𝑦2.(3.15) By (3.6) and Lemma 3.1, there exist 𝑥=𝑥(𝜔,𝜆),𝑦=𝑦(𝜔,𝜆).(𝑥,𝑦) such that 𝑥=𝑇(𝑥,𝑦,𝜔,𝜆),𝑦=𝑆(𝑥,𝑦,𝜔,𝜆). By Lemma 2.7, (𝑥,𝑦) 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 𝐽𝐻1𝑀(,𝜔),𝜌1,𝐽𝐻2𝑁(,𝜆),𝜌2, there are two constants 𝜉>0 and 𝜂>0 such that 𝐽𝐻1𝑀(,𝜔),𝜌1(𝑢)𝐽𝐻1𝑀,𝜔,𝜌1(𝑢)𝜉𝜔𝜔,𝐽𝐻2𝑁(,𝜆),𝜌2(𝑣)𝐽𝐻2𝑁(,𝜆),𝜌2(𝑣)𝜂𝜆𝜆,𝑢,𝑣.(3.16)

Theorem 3.3. Suppose that the mappings 𝐻1,𝑓,𝐹,𝑀,𝐻2,𝑔,𝐺, and 𝑁 are the same as in Theorem 3.2, and for any fixed 𝑥,𝑦, the mappings 𝜔𝐻1(𝑥,𝜔),𝜔𝑓(𝑥,𝑦,𝜔),𝜔𝐹(𝑥,𝑦,𝜔),𝜆𝐻2(𝑦,𝜆),𝜆𝑔(𝑥,𝑦,𝜆),𝜆𝐺(𝑥,𝑦,𝜆) are continuous (or Lipschitz continuous.) If Assumption 1 and condition (3.6) hold, then the solution (𝑥(𝜔,𝜆),𝑦(𝜔,𝜆)) for the problem (1.2) is continuous (or Lipschitz continuous).

Proof. Suppose that 𝜔,𝜔Ω, 𝜆,𝜆Λ such that 𝜔𝜔,𝜆𝜆. From Theorem 3.2, we know that the problem (1.2) has solution (𝑥(𝜔,𝜆),𝑦(𝜔,𝜆)),𝑥(𝜔,𝜆),𝑦(𝜔,𝜆)), (𝑥(𝜔,𝜆),𝑦(𝜔,𝜆)) and (𝑥(𝜔,𝜆),𝑦(𝜔,𝜆)).
(A) Estimate 𝑥(𝜔,𝜆)𝑥(𝜔,𝜆) and 𝑦(𝜔,𝜆)𝑦(𝜔,𝜆). By Lemma 2.7,we have 𝑥(𝜔,𝜆)=𝐽𝐻1𝑀(,𝜔),𝜌1𝐻1(𝑥(𝜔,𝜆),𝜔)𝑓(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜔)𝜌1𝐹(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜔)𝐽𝐻1𝑀(,𝜔),𝜌1𝑥(𝑝),𝜔,𝜆=𝐽𝐻1𝑀,𝜔,𝜌1𝐻1𝑥,𝜔,𝜆𝜔𝑥𝑓𝜔,𝜆,𝑦,𝜔,𝜆𝜔𝜌1𝐹𝑥𝜔,𝜆,𝑦,𝜔,𝜆𝜔𝐽𝐻1𝑀,𝜔,𝜌1(𝑞).(3.17) It follows from Assumption 1, Lemmas 2.6 and 2.7 that 𝑥(𝜔,𝜆)𝑥=𝐽𝜔,𝜆𝐻1𝑀(,𝜔),𝜌1(𝑝)𝐽𝐻1𝑀,𝜔,𝜌1𝐽(𝑞)𝐻1𝑀(,𝜔),𝜌1(𝑝)𝐽𝐻1𝑀(,𝜔),𝜌1+𝐽(𝑞)𝐻1𝑀(,𝜔),𝜌1(𝑞)𝐽𝐻1𝑀,𝜔,𝜌11(𝑞)𝛾1𝑝𝑞+𝜉𝜔𝜔.(3.18) Since 𝑓 is (𝜉𝑓,𝜂𝑓,𝜁𝑓)-Lipschitz continuous, 𝛾𝑓 -strongly monotone with respect to 𝐻1 in the first argument and 𝐹 is (𝜉𝐹,𝜂𝐹,𝜁𝐹)-Lipschitz continuous, we conclude 𝐻𝑝𝑞=1(𝑥(𝜔,𝜆),𝜔)𝑓(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜔)𝜌1𝐹(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜔)𝐻1𝑥,𝜔,𝜆𝜔𝑥+𝑓𝜔,𝜆,𝑦,𝜔,𝜆𝜔+𝜌1𝐹𝑥𝜔,𝜆,𝑦,𝜔,𝜆𝜔𝐻1(𝑥(𝜔,𝜆),𝜔)𝐻1𝑥𝑥𝜔,𝜆,𝜔𝑓(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜔)+𝑓+𝐻𝜔,𝜆,𝑦(𝜔,𝜆),𝜔1𝑥𝜔,𝜆,𝜔𝐻1𝑥,𝜔,𝜆𝜔+𝑓𝑥𝑥𝜔,𝜆,𝑦(𝜔,𝜆),𝜔𝑓𝜔,𝜆,𝑦+𝑓𝑥𝜔,𝜆,𝜔𝜔,𝜆,𝑦𝑥𝜔,𝜆,𝜔𝑓𝜔,𝜆,𝑦,𝜔,𝜆𝜔+𝜌1𝑥𝐹(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜔)𝐹𝜔,𝜆,𝑦𝜔,𝜆,𝜔+𝜌1𝐹𝑥𝜔,𝜆,𝑦𝑥𝜔,𝜆,𝜔𝐹𝜔,𝜆,𝑦,𝜔,𝜆𝜔𝜉212𝛾𝑓+𝜉2𝑓𝑥(𝜔,𝜆)𝑥+𝐻𝜔,𝜆1𝑥𝜔,𝜆,𝜔𝐻1𝑥,𝜔,𝜆𝜔+𝜂𝑓𝑦(𝜔,𝜆)𝑦+𝑓𝑥𝜔,𝜆𝜔,𝜆,𝑦𝑥𝜔,𝜆,𝜔𝑓𝜔,𝜆,𝑦,𝜔,𝜆𝜔+𝜌1𝜉𝐹𝑥(𝜔,𝜆)𝑥𝜔,𝜆+𝜌1𝜂𝐹𝑦(𝜔,𝜆)𝑦𝜔,𝜆+𝜌1𝐹𝑥𝜔,𝜆,𝑦𝑥𝜔,𝜆,𝜔𝐹𝜔,𝜆,𝑦,𝜔,𝜆𝜔.(3.19) It follows from (3.18) and (3.19) that 𝑥(𝜔,𝜆)𝑥𝜔,𝜆𝜃1𝑥(𝜔,𝜆)𝑥+1𝜔,𝜆𝛾1𝜂𝑓+𝜌1𝜂𝐹𝑦(𝜔,𝜆)𝑦+1𝜔,𝜆𝛾1𝐻1𝑥𝜔,𝜆,𝜔𝐻1𝑥,𝜔,𝜆𝜔+1𝛾1𝑓𝑥𝜔,𝜆,𝑦𝑥𝜔,𝜆,𝜔𝑓𝜔,𝜆,𝑦,𝜔,𝜆𝜔+𝜌1𝛾1𝐹𝑥𝜔,𝜆,𝑦𝑥𝜔,𝜆,𝜔𝐹𝜔,𝜆,𝑦,𝜔,𝜆𝜔+𝜉𝜔𝜔,(3.20) that is, 𝑥(𝜔,𝜆)𝑥𝜂𝜔,𝜆𝑓+𝜌1𝜂𝐹1𝜃1𝛾1𝑦(𝜔,𝜆)𝑦+1𝜔,𝜆1𝜃11𝛾1𝐻1𝑥𝜔,𝜆,𝜔𝐻1𝑥,𝜔,𝜆𝜔+1𝛾1𝑓𝑥𝜔,𝜆,𝑦𝑥𝜔,𝜆,𝜔𝑓𝜔,𝜆,𝑦,𝜔,𝜆𝜔+𝜌1𝛾1𝐹𝑥𝜔,𝜆,𝑦𝑥𝜔,𝜆,𝜔𝐹𝜔,𝜆,𝑦,𝜔,𝜆𝜔+𝜉𝜔𝜔,(3.21) where 𝜃1=1𝛾1𝜉212𝛾𝑓+𝜉2𝑓+𝜌1𝜉𝐹,() By Lemma 2.7, we have 𝑦(𝜔,𝜆)=𝐽𝐻2𝑁(,𝜆)𝜌2𝐻2(𝑦(𝜔,𝜆),𝜆)𝑔(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜆)𝜌2𝐺(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜆)𝐽𝐻2𝑁(,𝜆),𝜌2𝑦(𝑠),𝜔,𝜆=𝐽𝐻2𝑁(,𝜆),𝜌2𝐻2𝑦𝑥𝜔,𝜆,𝜆𝑔𝜔,𝜆,𝑦𝜔,𝜆,𝜆𝜌2𝐺𝑥𝜔,𝜆,𝑦𝜔,𝜆,𝜆𝐽𝐻2𝑁(,𝜆),𝜌2(𝑡).(3.22) It follows from Assumption 1, Lemmas 2.6 and 2.7 that 𝑦(𝜔,𝜆)𝑦=𝐽𝜔,𝜆𝐻2𝑁(,𝜆),𝜌2(𝑠)𝐽𝐻2𝑁(,𝜆),𝜌21(𝑡)𝛾2𝑠𝑡.(3.23) Since 𝑔 is (𝜉𝑔,𝜂𝑔,𝜁𝑔)-Lipschitz continuous and 𝛾𝑔-strongly monotone with respect to 𝐻2 in the second argument and 𝐺 is (𝜉𝐺,𝜂𝐺,𝜁𝐺)-Lipschitz continuous,we conclude 𝐻𝑠𝑡=2(𝑦(𝜔,𝜆),𝜆)𝑔(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜆)𝜌2𝐺(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜆)𝐻2𝑦𝑥𝜔,𝜆,𝜆+𝑔𝜔,𝜆,𝑦𝜔,𝜆,𝜆+𝜌2𝐺𝑥𝜔,𝜆,𝑦𝐻𝜔,𝜆,𝜆2(𝑦(𝜔,𝜆),𝜆)𝐻2𝑦𝜔,𝜆,𝜆𝑔(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜆)+𝑔𝑥(𝜔,𝜆),𝑦+𝑔𝜔,𝜆,𝜆𝑥(𝜔,𝜆),𝑦𝑥𝜔,𝜆,𝜆𝑔𝜔,𝜆,𝑦𝜔,𝜆,𝜆+𝜌2𝑥𝐺(𝑥(𝜔,𝜆),𝑦(𝜔,𝜆),𝜆)𝐺𝜔,𝜆,𝑦𝜔,𝜆,𝜆𝜉222𝛾𝑔+𝜂2𝑔𝑦(𝜔,𝜆)𝑦𝜔,𝜆+𝜉𝑔𝑥(𝜔,𝜆)𝑥𝜔,𝜆+𝜌2𝜉𝐺𝑥(𝜔,𝜆)𝑥𝜔,𝜆+𝜌2𝜂𝐺𝑦(𝜔,𝜆)𝑦.𝜔,𝜆(3.24) It follows from (3.23) and (3.24) that 𝑦(𝜔,𝜆)𝑦1𝜔,𝜆𝛾2𝜉222𝛾𝑔+𝜂2𝑔+𝜌2𝜂𝐺𝑦(𝜔,𝜆)𝑦+1𝜔,𝜆𝛾2𝜉𝑔+𝜌2𝜉𝐺𝑥(𝜔,𝜆)𝑥.𝜔,𝜆(3.25) That is, 𝑦(𝜔,𝜆)𝑦1𝜔,𝜆1𝜃21𝛾2𝜉𝑔+𝜌2𝜉𝐺𝑥(𝜔,𝜆)𝑥,𝜔,𝜆(3.26) where 𝜃2=1𝛾2𝜉222𝛾𝑔+𝜂2𝑔+𝜌2𝜂𝐺.() By combining (3.21) and (3.26), we derive 𝑥(𝜔,𝜆)𝑥𝜂𝜔,𝜆𝑓+𝜌1𝜂𝐹𝜉𝑔+𝜌2𝜉𝐺1𝜃11𝜃2𝛾1𝛾2𝑥(𝜔,𝜆)𝑥+1𝜔,𝜆1𝜃11𝛾1𝐻1𝑥𝜔,𝜆,𝜔𝐻1𝑥,𝜔,𝜆𝜔+1𝛾1𝑓𝑥𝜔,𝜆,𝑦𝑥𝜔,𝜆,𝜔𝑓𝜔,𝜆,𝑦,𝜔,𝜆𝜔+𝜌1𝛾1𝐹𝑥𝜔,𝜆,𝑦𝑥𝜔,𝜆,𝜔𝐹𝜔,𝜆,𝑦,𝜔,𝜆𝜔+𝜉𝜔𝜔𝑠.(3.27) That is, 𝑥(𝜔,𝜆)𝑥1𝜔,𝜆11𝜅1𝜃11𝛾1𝐻1𝑥𝜔,𝜆,𝜔𝐻1𝑥,𝜔,𝜆𝜔+1𝛾1𝑓𝑥𝜔,𝜆,𝑦𝑥𝜔,𝜆,𝜔𝑓𝜔,𝜆,𝑦,𝜔,𝜆𝜔+𝜌1𝛾1𝐹𝑥𝜔,𝜆,𝑦𝑥𝜔,𝜆,𝜔𝐹𝜔,𝜆,𝑦,𝜔,𝜆𝜔+𝜉𝜔𝜔,(3.28)𝑦(𝜔,𝜆)𝑦𝜉𝜔,𝜆𝑔+𝜌2𝜉𝐺1𝜃2𝛾2111𝜅1𝜃11𝛾1𝐻1𝑥𝜔,𝜆,𝜔𝐻1𝑥,𝜔,𝜆𝜔+1𝛾1𝑓𝑥𝜔,𝜆,𝑦𝑥𝜔,𝜆,𝜔𝑓𝜔,𝜆,𝑦,𝜔,𝜆𝜔+𝜌1𝛾1𝐹𝑥𝜔,𝜆,𝑦𝑥𝜔,𝜆,𝜔𝐹𝜔,𝜆,𝑦,𝜔,𝜆𝜔+𝜉𝜔𝜔,(3.29) where 𝜂𝜅=𝑓+𝜌1𝜂𝐹𝜉𝑔+𝜌2𝜉𝐺1𝜃11𝜃2𝛾1𝛾2.() By (3.6), 1𝜃1>1𝛾2𝜉𝑔+𝜌2𝜉𝐺,1𝜃2>1𝛾1𝜂𝑓+𝜌1𝜂𝐹,(3.30) and hence 𝜉0<𝑔+𝜌2𝜉𝐺𝜂𝑓+𝜌1𝜂𝐹1𝜃11𝜃2𝛾1𝛾2<1,thatis,0<𝜅<1.(3.31)
(B) Estimate 𝑦(𝜔,𝜆)𝑦(𝜔,𝜆) and 𝑥(𝜔,𝜆)𝑥(𝜔,𝜆).
By Lemma 2.7, we have 𝑦𝜔,𝜆=𝐽𝐻2𝑁(,𝜆),𝜌2𝐻2𝑦𝜔,𝜆,𝜆𝑥𝑔𝜔,𝜆,𝑦𝜔,𝜆,𝜆𝜌2𝐺𝑥𝜔,𝜆,𝑦𝜔,𝜆,𝜆=𝐽𝐻2𝑁(,𝜆),𝜌2𝑦(𝑚),𝜔,𝜆=𝐽𝐻2𝑁(,𝜆),𝜌2𝐻2𝑦𝑥𝜔,𝜆,𝜆𝑔𝜔,𝜆,𝑦𝜔,𝜆,𝜆𝜌2𝐺𝑥𝜔,𝜆,𝑦𝜔,𝜆,𝜆=𝐽𝐻2𝑁(,𝜆),𝜌2(𝑛).(3.32) It follows from Assumption 1, Lemmas 2.6 and 2.7 that 𝑦𝜔,𝜆𝑦=𝐽𝜔,𝜆𝐻2𝑁(,𝜆),𝜌2(𝑚)𝐽𝐻2𝑁(,𝜆),𝜌2𝐽(𝑛)𝐻2𝑁(,𝜆),𝜌2(𝑚)𝐽𝐻2𝑁(,𝜆),𝜌2+𝐽(𝑛)𝐻2𝑁(,𝜆),𝜌2(𝑛)𝐽𝐻2𝑁(,𝜆),𝜌21(𝑛)𝛾2𝑚𝑛+𝜂𝜆𝜆.(3.33) Since 𝑔 is (𝜉𝑔,𝜂𝑔,𝜁𝑔)-Lipschitz continuous and 𝛾𝑔-strongly monotone with respect to 𝐻2 in the second argument and 𝐺 is (𝜉𝐺,𝜂𝐺,𝜁𝐺)-Lipschitz continuous,we conclude 𝐻𝑚𝑛=2𝑦𝜔,𝜆,𝜆𝑥𝑔𝜔,𝜆,𝑦𝜔,𝜆,𝜆𝜌2𝐺𝑥𝜔,𝜆,𝑦𝜔,𝜆,𝜆𝐻2𝑦𝑥𝜔,𝜆,𝜆+𝑔𝜔,𝜆,𝑦𝜔,𝜆,𝜆+𝜌2𝐺𝑥𝜔,𝜆,𝑦𝐻𝜔,𝜆,𝜆2𝑦𝜔,𝜆,𝜆𝐻2𝑦,𝜔,𝜆𝜆𝑥𝑔𝜔,𝜆,𝑦𝜔,𝜆,𝜆𝑥𝑔𝜔,𝜆,𝑦,𝜔,𝜆𝜆+𝐻2𝑦,𝜔,𝜆𝜆𝐻2𝑦+𝑔𝑥𝜔,𝜆,𝜆𝜔,𝜆,𝑦,𝜔,𝜆𝜆𝑥𝑔𝜔,𝜆,𝑦,𝜔,𝜆𝜆+𝑔𝑥𝜔,𝜆,𝑦,𝜔,𝜆𝜆𝑥𝑔𝜔,𝜆,𝑦𝜔,𝜆,𝜆+𝜌2𝐺𝑥𝜔,𝜆,𝑦𝜔,𝜆,𝜆𝑥𝐺𝜔,𝜆,𝑦,𝜔,𝜆𝜆+𝜌2𝐺𝑥𝜔,𝜆,𝑦,𝜔,𝜆𝜆𝑥𝐺𝜔,𝜆,𝑦𝜔,𝜆,𝜆𝜉222𝛾𝑔+𝜂2𝑔𝑦𝜔,𝜆𝑦𝜔,𝜆+𝜉𝑔𝑥𝜔,𝜆𝑥𝜔,𝜆+𝜌2𝜉𝐺𝑥𝜔,𝜆𝑥𝜔,𝜆+𝜂𝐺𝑦𝜔,𝜆𝑦+𝑔𝑥𝜔,𝜆𝜔,𝜆,𝑦,𝜔,𝜆𝜆𝑥𝑔𝜔,𝜆,𝑦+𝐻𝜔,𝜆,𝜆2𝑦,𝜔,𝜆𝜆𝐻2𝑦𝜔,𝜆,𝜆+𝜌2𝐺𝑥𝜔,𝜆,𝑦,𝜔,𝜆𝜆𝑥𝐺𝜔,𝜆,𝑦.𝜔,𝜆,𝜆(3.34) It follows from (3.33) and (3.34) that 𝑦𝜔,𝜆𝑦𝜉𝜔,𝜆𝑔+𝜌2𝜉𝐺1𝜃2𝛾2𝑥𝜔,𝜆𝑥+1𝜔,𝜆1𝜃21𝛾2𝐻2𝑦,𝜔,𝜆𝜆𝐻2𝑦+1𝜔,𝜆,𝜆𝛾2𝑔𝑥𝜔,𝜆,𝑦,𝜔,𝜆𝜆𝑥𝑔𝜔,𝜆,𝑦+𝜌𝜔,𝜆,𝜆2𝛾2𝐺𝑥𝜔,𝜆,𝑦,𝜔,𝜆𝜆𝑥𝐺𝜔,𝜆,𝑦𝜔,𝜆,𝜆+𝜂𝜆𝜆,(3.35) where 𝜃2 defined by .
By Lemma 2.7, we have 𝑥𝜔,𝜆=𝐽𝐻1𝑀,𝜔,𝜌1𝐻1𝑥𝜔,𝜆,𝜔𝑥𝑓𝜔,𝜆,𝑦𝜔,𝜆,𝜔𝜌1𝐹𝑥𝜔,𝜆,𝑦𝜔,𝜆,𝜔,𝑥𝜔,𝜆=𝐽𝐻1𝑀,𝜔,𝜌1𝐻1𝑥,𝜔,𝜆𝜔𝑥𝑓𝜔,𝜆,𝑦,𝜔,𝜆𝜔𝜌1𝐹𝑥𝜔,𝜆,𝑦,𝜔,𝜆𝜔.(3.36) As the proof of (3.25),we have 𝑥𝜔,𝜆𝑥1𝜔,𝜆𝛾1𝜉212𝛾𝑓+𝜉2𝑓+𝜌1𝜉𝐹𝑥𝜔,𝜆𝑥+1𝜔,𝜆𝛾1𝜂𝑓+𝜌1𝜂𝐹𝑦𝜔,𝜆𝑦,𝜔,𝜆(3.37) that is, 𝑥𝜔,𝜆𝑥𝜂𝜔,𝜆𝑓+𝜌1𝜂𝐹1𝜃1𝛾1𝑦𝜔,𝜆𝑦,𝜔,𝜆(3.38) where 𝜃1 is defined by .
Therefore 𝑦𝜔,𝜆𝑦1𝜔,𝜆11𝜅1𝜃21𝛾2𝐻2𝑦,𝜔,𝜆𝜆𝐻2𝑦+1𝜔,𝜆,𝜆𝛾2𝑔𝑥𝜔,𝜆,𝑦,𝜔,𝜆𝜆𝑥𝑔𝜔,𝜆,𝑦+𝜌𝜔,𝜆,𝜆2𝛾2𝐺𝑥𝜔,𝜆,𝑦,𝜔,𝜆𝜆𝑥𝐺𝜔,𝜆,𝑦𝜔,𝜆,𝜆+𝜂𝜆𝜆,𝑥(3.39)𝜔,𝜆𝑥𝜂𝜔,𝜆𝑓+𝜌1𝜂𝐹1𝜃1𝛾1111𝜅1𝜃21𝛾2𝐻2𝑦,𝜔,𝜆𝜆𝐻2𝑦+1𝜔,𝜆,𝜆𝛾2𝑔𝑥𝜔,𝜆,𝑦,𝜔,𝜆𝜆𝑥𝑔𝜔,𝜆,𝑦+𝜌𝜔,𝜆,𝜆2𝛾2𝐺𝑥𝜔,𝜆,𝑦,𝜔,𝜆𝜆𝑥𝐺𝜔,𝜆,𝑦𝜔,𝜆,𝜆+𝜂𝜆𝜆,(3.40) where 𝜅 is defined by .
(C) Prove that the conclusion of Theorem 3.3.
From (3.28) and (3.40), by the assumptions for 𝐻1,𝑓,𝐹,𝐻2,𝑔, and 𝐺 in Theorem 3.3 and relation 𝑥(𝜔,𝜆)𝑥𝜔,𝜆𝑥(𝜔,𝜆)𝑥+𝑥𝜔,𝜆𝜔,𝜆𝑥𝜔,𝜆,(3.41) we know that 𝑥(𝜔,𝜆) is continuous (or Lipschitz continuous).
From (3.29) and (3.39), by the assumptions for 𝐻1,𝑓,𝐹,𝐻2,𝑔 and 𝐺 in Theorem 3.3 and relation 𝑦(𝜔,𝜆)𝑦𝜔,𝜆𝑦(𝜔,𝜆)𝑦+𝑦𝜔,𝜆𝜔,𝜆𝑦𝜔,𝜆,(3.42) we know that 𝑦(𝜔,𝜆) 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).

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