# The Penalty Method for a New System of Generalized Variational Inequalities

Authors: Yu-Chao Tang; Li-Wei Liu
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## Abstract

We consider a new system of generalized variational inequalities (SGVI). Using the penalty methods, we prove the existence of solution of SGVI in Hilbert spaces. Our results extend and improve some known results.


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## Body

## 1. Introduction

Throughout this work, letH be real Hilbert space with a norm ∥·∥ and inner product 〈·,·〉. Let K be a nonempty closed and convex subset of H. Given nonlinear mappings T1(x,y),T2(x,y):H×H→H, and ω1, ω2∈H, we consider the following problem:(1.1)〈rT1(x,y)-ω1,u-x〉≥0,∀u∈K,〈sT2(x,y)-ω2,u-y〉≥0,∀u∈K,
which is called the system of generalized variational inequality problem (SGVI), where r>0 and s>0 are constants. An element (x*,y*)∈K×K is called a solution of the problem (1.1) if(1.2)〈rT1(x*,y*)-ω1,u-x*〉≥0,∀u∈K,〈sT2(x*,y*)-ω2,u-y*〉≥0,∀u∈K.Special cases of problem (1.1) are as follows.(1) IfT1(x,y)=T2(x,y)=Tx, s=1, r=0, and ω2=ω, ω1=θ, then problem (1.1) reduces to the variational inequality(1.3)〈Tx-ω,u-x〉≥0,∀u∈K.
Problem (1.3) was introduced by Browder [1, 2] and studied by many authors (e.g., see [3–7]).(2) IfT1(x,y)=Tx+x-y, T2(x,y)=Ty+y-x, T:H→H, and ω1=ω2=θ, then problem (1.1) reduces to the system of variational inequality problem(1.4)〈Tx+x-y,u-x〉≥0,∀u∈K,〈Ty+y-x,u-y〉≥0,∀u∈K.
Problem (1.4) was introduced and studied by Verma [8].(3) IfT1(x,y)=T(x,y)+x-y, T2(x,y)=T(y,x)+y-x, and ω1=ω2=θ, then problem (1.1) becomes the following system of nonlinear variational inequalities(1.5)〈T(x,y)+x-y,u-x〉≥0,∀u∈K,〈T(y,x)+y-x,u-y〉≥0,∀u∈K,
which was considered by Chang et al. [9].Remark 1.1.
For a suitable choice ofT1 and T2, the problem (1.1) includes many kinds of known systems of variational inequalities as special case (see [4–10] and the references therein). In this work, by using the penalty method, we study the existence of solutions for SGVI.


## 2. Preliminaries

In the sequel, we give some definitions and lemmas. In what follows,→ and ⇀ stand for strong and weak convergence, respectively.Definition 2.1 (see [11, pages 96–105]).
LetT:H→H be a mapping.(i)
The mappingT is said to be pseudo-monotone if D(T) is closed convex set and its restrictions to finite-dimensional subspaces are demicontinuous, and for every sequence {xn}⊂D(T), xn⇀x in H, the inequality lim supn→∞〈Txn,xn-x〉≤0 implies that
(2.1)〈Tx,x-y〉≤lim infn→∞〈Txn,xn-y〉,∀y∈D(T).(ii)
The mappingT is said to have the generalized pseudo-monotone property if for any sequence {[xn,Txn]} with xn⇀x in X and Txn⇀f in H such that
(2.2)lim supn→∞〈Txn,xn-x〉≤0,
we have f=Tx and 〈Txn,xn〉→〈f,x〉 as n→∞.(iii)
The mappingT is said to be monotone if for any x,y∈D(T), the inequality 〈Tx-Ty,x-y〉≥0 holds.(iv)
The monotone mappingT is said to be maximal if the inequality 〈Tx-g,x-y〉≥0 for all [x,Tx]∈G(T)(graph of T) implies [y,g]∈G(T).(v)
The mappingT is said to be coercive if there exists a continuous increasing function C:R+→R+ with C(r)→∞ as r→∞ such that for x∈D(T)(2.3)〈Tx,x〉≥C(∥x∥)∥x∥.Definition 2.2.
LetT:H×H→H be a mapping.(i)
The mappingT is said to be (ξ,η)-Lipschitz continuous if there exist constants ξ>0,η>0 such that
(2.4)∥T(x1,y1)-T(x2,y2)∥≤ξ∥x1-x2∥+η∥y1-y2∥,∀x1,x2,y1,y2∈H.(ii)
The mappingT is said to be α-strongly monotone in the first argument if there exists α>0 such that for each fixed y∈H, we have
(2.5)〈T(x1,y)-T(x2,y),x1-x2〉≥α∥x1-x2∥2,∀x1,x2∈H.(iii)
The mappingT is said to be β-strongly monotone in the second argument if there exists β>0 such that for each fixed x∈H, we have
(2.6)〈T(x,y1)-T(x,y2),y1-y2〉≥β∥y1-y2∥2,∀y1,y2∈H.(iv)
The mappingT is said to be coercive in the first (second) argument if there exists a continuous increasing function C1:R+→R+(C2:R+→R+) with C1(r)→∞(C2(r)→∞) as r→∞ such that for each fixed y∈H(x∈H), we have
(2.7)〈T(x,y),x〉≥C1(∥x∥)∥x∥,∀x∈H,〈T(x,y),y〉≥C2(∥y∥)∥y∥,∀y∈H.Definition 2.3 (see [11]).
(i) A single-valued bounded demicontinuous monotone operatorP:H→H with the property that K=kerP=N(P)={x∈H:Px=0} is said to be the penalty operator of K⊂H.
(ii)P is demicontinuous if xn→x implies that Pxn⇀Px.
(iii)P is hemicontinuous if h∈H, tn>0, x+tnh∈H, tn→0 implies that P(x+tnh)⇁Px, where ⇁ stands for weak* convergence in H.Lemma 2.4 (see [11, page 267]).
LetPK:H→K be the projection on K, then the mapping P=I-PK is a penalty operator of K.Lemma 2.5 (see [11, page 98]).
(i) Any maximal monotone mappingT:H→H with D(T)=H is a pseudo-monotone one.
(ii) Any pseudo-monotone mapping has the generalized pseudo-monotone property.
(iii) The monotone mappingT is maximal if and only if R(I+λT)=H,forallλ>0.


## 3. Main Results

We consider first the problem of finding(x,y)∈H×H such that(3.1)ω1=rT1(x,y)+1ɛPx,ω2=sT2(x,y)+1δPy,
which is called the system of nonlinear variational equation, where r>0,s>0, ɛ>0,δ>0 and ω1,ω2∈H, P=I-PK is the penalty operator of K (see Lemma 2.4). We will prove the existence of solutions for problem (3.1).Lemma 3.1.
LetH be real Hilbert space and K⊂H be a nonempty closed convex subset of H. Let mapping T1:H×H→H be (ξ1,η1)-Lipschitz continuous and α1-strongly monotone in the first argument and mapping T2:H×H→H be (ξ2,η2)-Lipschitz continuous and α2-strongly monotone in the second argument. If
(3.2)γ1=1-2α1r+ξ12r2+sξ2<1,γ2=1-2α2s+η22s2+rη1<1,
then for each ɛ>0,δ>0, problem (3.1) admits a solution (xɛ,yδ).Proof.
SinceP is continuous monotone operator, P is hemicontinuous monotone operator. By Corollary 2.3 in [11], P is maximal monotone. So, problem (3.1) is equivalent to the following problem
(3.3)x=JPɛ[x-rT1(x,y)+ω1],y=JPδ[y-sT2(x,y)+ω2],
where JPɛ=(I+(1/ε)P)-1, JPδ=(I+(1/δ)P)-1, and I is the identity mapping on H.
For any givenx0∈H,y0∈H, we can compute the sequences {xn} and {yn} by Picard iterative schemes:
(3.4)xn+1=JPɛ(xn-rT1(xn,yn)+ω1),n≥0,(3.5)yn=JPδ(xn-sT2(xn,yn)+ω2),n≥0.
Since JPɛ is nonexpansive [11] and T1 is (ξ1,η1)-Lipschitz continuous and α1-strongly monotone in the first argument, then by (3.4), we have
(3.6)∥xn+1-xn∥=∥JPɛ(xn-rT1(xn,yn)+ω1)-JPɛ(xn-1-rT1(xn-1,yn-1)+ω1)∥≤∥xn-xn-1-r(T1(xn,yn)-T1(xn-1,yn-1))∥≤∥xn-xn-1-r(T1(xn,yn)-T1(xn-1,yn))∥+r∥T1(xn-1,yn)-T1(xn-1,yn-1)∥≤1-2α1r+ξ12r2∥xn-xn-1∥+rη1∥yn-yn-1∥.
Similarly, we get from (3.5)
(3.7)∥yn+1-yn∥≤1-2α2s+η22s2∥yn-yn-1∥+sξ2∥xn-xn-1∥.
By (3.6) and (3.7), we have
(3.8)∥xn+1-xn∥+∥yn+1-yn∥≤(1-2α1r+ξ12r2+sξ2)∥xn-xn-1∥+(1-2α2s+η22s2+rη1)∥yn-yn-1∥=γ1∥xn-xn-1∥+γ2∥yn-yn-1∥≤γ(∥xn-xn-1∥+∥yn-yn-1∥),
where γ=max(γ1,γ2). By (3.2), we know that 0<γ<1, and (3.8) implies that {xn} and {yn} are both Cauchy sequences. Thus, there exist xɛ∈H, yδ∈H such that xn→xɛ,yn→yδ(as n→∞). By continuity of JPɛ, JPδ, T1, and T2 and algorithm (3.4) and (3.5), we know that (xɛ,yδ) satisfies the following relation:
(3.9)xɛ=JPɛ(xɛ-rT1(xɛ,yδ)+ω1),yδ=JPδ(yδ-sT2(xɛ,yδ)+ω2).
Therefore, (xɛ,yδ) is a solution of problem (3.1). This completes the proof of Lemma 3.1.Remark 3.2.
IfT is (s,t)-Lipschitz continuous, then T is bounded, that is, T map bounded set to bounded set.Now, we give our main results.Theorem 3.3.
LetH be a real Hilbert space and K⊂H be a nonempty closed convex subset with θ∈K. Let T1,T2:H×H→H be the same as in Lemma 3.1. If the mapping T1 is coercive with respect to the first argument and the mapping T2 is coercive with respect to the second argument, then for any ω1∈H,ω2∈H, there exists one solution (x*,y*) of the SGVI (1.1).Proof.
Letɛ>0,δ>0. Then, by Lemma 3.1, the problem (3.1) has a solution (xɛ,yδ).
By the monotonicity ofP, we have
(3.10)〈ω1,xɛ〉=〈T1(xɛ,yδ),xɛ〉+1ɛ〈Pxɛ,xɛ〉=〈T1(xɛ,yδ),xɛ〉+1ɛ〈Pxɛ-Pθ,xɛ-θ〉≥〈T1(xɛ,yδ),xɛ〉≥C1(∥xɛ∥)∥xɛ∥,
since T1 is coercive with respect to the first argument. Hence {xε∣ɛ>0} remains bounded as ɛ→0. Similarly, we have
(3.11)〈ω2,yδ〉=〈T2(xɛ,yδ),yδ〉+1δ〈Pyδ,yδ〉≥〈T2(xɛ,yδ),yδ〉≥C2(∥yδ∥)∥yδ∥.
Hence {yδ∣δ>0} remains also bounded as δ→0. The boundedness of T1 and T2 and the fact that
(3.12)Pxɛ=ɛ(ω1-rT1(xɛ,yδ))Pyδ=δ(ω2-sT2(xɛ,yδ))
implies that ∥Pxɛ∥≤c and ∥Pyδ∥≤c for some constant c>0. By reflexivity of H, we can choose sequences {xn} and {yn}, xn=xɛn, yn=yδn such that xn⇀x*∈H,yn⇀y*∈H, and T1(xn,yn)⇀u,T2(xn,yn)⇀v as ɛn→0 and δn→0. Using the fact that ∥Pxn∥→0 and ∥Pyn∥→0 as n→∞ and taking the limit in the monotonicity relation, we get that
(3.13)0≤limn→∞〈Px-Pxn,x-xn〉=〈Px,x-x*〉,∀x∈H,0≤limn→∞〈Px-Pyn,x-yn〉=〈Px,x-y*〉,∀x∈H.
Set x=x*+tz with t>0 and let z be arbitrarily chosen in H. Then 〈P(x*+tz),z〉≥0 and by the hemicontinuity of P, we get that 〈Px*,z〉≥0 for all z∈H. Hence Px*=0, that is, x*∈K. Similarly, we have Py*=0, that is, y*∈K.
By (3.1) and monotonicity of P, we have
(3.14)〈rT1(xn,yn)-ω1,x*-xn〉=1ɛn〈Px*-Pxn,x*-xn〉≥0,〈sT2(xn,yn)-ω2,y*-yn〉=1δn〈Py*-Pyn,y*-yn〉≥0.
Therefore
(3.15)lim supn→∞〈rT1(xn,yn),xn-x*〉≤lim supn→∞〈ω1,xn-x*〉=0,lim supn→∞〈sT2(xn,yn),yn-y*〉≤lim supn→∞〈ω2,yn-y*〉=0,
that is,
(3.16)lim supn→∞〈T1(xn,yn),xn-x*〉≤0,lim supn→∞〈T2(xn,yn),yn-y*〉≤0.
By Lemma 2.5(ii) and Definition 2.1, we deduce that
(3.17)u=T1(x*,y*),〈T1(xn,yn),xn〉→〈u,x*〉asn→∞,v=T2(x*,y*),〈T2(xn,yn),yn〉→〈v,y*〉asn→∞.
By (3.1), (3.17), and monotonicity of P, we have
(3.18)〈rT1(x*,y*),x*-u〉=limn→∞〈rT1(xn,yn),x*-u〉=lim infn→∞〈rT1(xn,yn),x*-u〉=lim infn→∞〈ω1-1ɛnPxn,xn-u〉=lim infn→∞〈ω1,xn-u〉-lim supn→∞1ɛn〈Pxn-Px,xn-u〉≤〈ω1,x*-u〉∀u∈K,〈sT2(x*,y*),y*-u〉=lim infn→∞〈sT2(xn,yn),y*-u〉=lim infn→∞〈ω2-1δnPyn,yn-u〉=lim infn→∞〈ω2,yn-u〉-lim supn→∞1δn〈Pyn-Px,yn-u〉≤〈ω2,y*-u〉∀u∈K.
Hence
(3.19)〈rT1(x*,y*)-ω1,u-x*〉≥0,∀u∈K,〈sT2(x*,y*)-ω2,u-y*〉≥0,∀u∈K.
This completes the proof of Theorem 3.3.IfT1(x,y)=Tx, r=1, ω1=ω, ω2=θ, and s=0, then we have the following theorem.Theorem 3.4.
LetH be a real Hilbert space and K⊂H be a nonempty closed convex subset with 0∈K. Let T:H→H be α-strongly monotone, ξ-Lipschitz continuous, and coercive operator. If
(3.20)1-2rα+r2ξ2<1,
then for each ω∈H, there exists at least one solution x* of problem (1.3).



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