Existence of Solutions and Algorithm for a System of Variational Inequalities

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Volume 2010, Article ID 182539,11pages doi:10.1155/2010/182539

Research Article

Existence of Solutions and Algorithm for

a System of Variational Inequalities

Yali Zhao,

1

_{Zunquan Xia,}

2

_{Liping Pang,}

2

_{and Liwei Zhang}

2

1_{Department of Mathematics, Bohai University, Jinzhou, Liaoning 121013, China}

2_{Department of Applied Mathematics, Dalian University of Technology, Dalian, Liaoning 116024, China}

Correspondence should be addressed to Yali Zhao,[email protected]

Received 13 May 2009; Revised 10 November 2009; Accepted 24 January 2010

Academic Editor: Nanjing Huang

Copyrightq2010 Yali Zhao et al. 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.

We obtain some existence results for a system of variational inequalitiesfor 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 thatXis a nonempty closed and convex subset ofRn;F:Rn → Rnis a vector-valued mapping. Variational inequality problemfor short, VIX, Fis to find anx∈X, such that

Fx, u−x ≥0, ∀u∈X. 1.1

We denote the solution set for VIX, Fby solX, F. In this paper, we suggest and study the following SVI: findx, y∈A×B, such that

Fx, y, u−x ≥0, ∀u∈A,

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whereF :A×B → Rn_,_G_:_A_×_B _→ _Rm_{are vector-valued mappings, and}_A_⊂_Rn_{, B} _⊂_Rm_.

The above SVI can be described as

VIA, F·, y,

VIB, Gx,·. 1.3

2. Existence and Uniqueness of Solutions for SVI

In this paper, otherwise specification,Rn_{is a}_n_{-dimensional Euclidean space, for all}_{x, y}_∈_Rn_, x, y denotes the inner product betweenx and y, x denotes norm ofx, that is, x

x, x.

In order to obtain our main results, we recall the following definitions and lemmas.

Definition 2.1. Let X ⊂ Rn _{be a nonempty subset, and let} _F _: _X _→ _Rn _{be a vector-valued}

mapping.

iFis said to be monotone if, for allx, y∈X,Fx−Fy, x−y ≥0.

iiFis said to be strictly monotone if, for allx, y∈X,x /y,Fx−Fy, x−y>0.

iiiFis said to be strongly monotone if there exists a constantα >0 such that

Fx−Fy, x−y≥αx−y2, ∀x, y∈X. 2.1

ivF is said to be coercive if there exists anx0 ∈ X and a constantR > 0 such that

x0 < Rand

Fx, x−x0 ≥0, ∀x∈X, x R. 2.2

vFis said to be Lipschitz continuous if there exists a constantL >0 such that

Fx−Fy≤Lx−y, ∀x, y∈X. 2.3

Remark 2.2. It is easy to see that

F is strongly monotone⇒

⎧ ⎨ ⎩

F is strictly monotone⇒F is monotone

F is coercive.

2.4

Based on the above all kinds of monotonicity, we have the following existence results for VIX, F.

Lemma 2.3see1. LetX ⊂ Rn_{be nonempty compact and convex set, and let} _F _: _X _→ _Rn _be

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Lemma 2.4see2. Let X ⊂ Rn _{be nonempty closed and convex set, and let}_F _: _X _→ _Rn _be

continuous mapping.

iIfFis strictly monotone, thenVIX, Fhas at most one solution,

iiIfFis coercive, thenVIX, Fmust has solution,

iiiIfFis strongly monotone, thenVIX, Fhas a unique solution.

In order to obtain the existence results for SVI, one needs to study parametric variational

inequalitiesVIA, F·, yandVIB, Gx,·in SVI.

SettingΩ {x, y ∈ Rn_×_Rm _| _x _∈ _{A, y} _∈ _B_}_/_∅_and _Ω_{is the feasible region of SVI,}

A⊂Rn_,_B_⊂_Rm_{are nonempty subset, and}_F_:_A_×_B _→ _Rn_,_G_:_A_×_B _→ _Rm_{are two continuous}

mappings. At first, one considersVIB, Gx,·, which is a parametric variational inequality with

respect toxin SVI.

Theorem 2.5. In VIB, Gx,·, assume thatA ⊂ Rn _and _B _⊂ _Rm _{are two compact and convex}

sets,G:A×B → Rm_{is continuous, and}_G_{is strictly monotone in}_y_{. Then, for any given}_x_∈_A_,

VIB, Gx,·has a unique solution and for allx ∈ A, there exists an implicit functiony ϕx

which is the unique solution toVIB, Gx,·. In addition, the implicit functionyϕxdetermined

byVIB, Gx,·is continuous onA.

Proof. iFor any givenx∈A, sinceB⊂Rm_{is compact and convex and}_G_{is continuous on}_A_×

B, then byLemma 2.3, parametric variational inequality VIB, Gx,·has solutions. In terms of strict monotonicity of the mappingGinyandLemma 2.4, we know that VIB, Gx,·has a unique solution. So, for allx∈A, the implicit functionyϕxdetermined by VIB, Gx,·

is well defined.

iiWe claim thatyϕxis continuous onA. In fact, for any givenx0∈A,{xn} ⊂A,

xn → x0asn → ∞, byi, we know that for alln, there existsyn ∈B, such thatynϕxn.

That is,

Gxn, yn

, y−yn ≥0, ∀y∈B. 2.5

Since{yn} ⊂Bis bounded, then there exists convergent subsequence{ynk}such thatynk →

y0asn → ∞, andy0 ∈B. In the following, we prove thaty0is a solution to VIB, Gx,·in

x0. For giveny∈B, there exists sequence{y_n_k}such thaty_n_k ∈B, k1,2, . . .andy_n_k → y

ask → ∞in view of the closedness ofB. Lettingyy_n

k in2.5, we have

Gxnk, ynk

, y_n_k −ynk ≥0, k1,2, . . . , 2.6

that is,

Gxnk, ynk

, y−ynk ≥ G

xnk, ynk

, y−y_n

k, k1,2, . . . , 2.7

observe thatGis continuous, and lettingk → ∞in2.7, we have

Gx0, y0

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Fory ∈ Bis arbitrary, theny0 is a solution to VIB, Gx,·, implyingy0 ϕx0. In order to explain thatϕis continuous atx0, we only need to know that the sequence{yn}satisfies

yn → y0asn → ∞. Let{ynl}be any subsequence of{yn}. Since{ynl}is bounded, there exists

a subsequence{yn_lk} ⊂ {ynl}such thatynlk → y0 ask → ∞. Using the method appeared in

Theorem 2.5, we can show that

Gx0, y0

, y−y₀ ≥0, ∀y∈B. 2.9

Thus, by the uniqueness of the solution to the problem VIB, Gx0,·, we conclude thaty₀

y0. Since,{ynl} ⊂ {yn}is arbitrary, we can conclude thatyn → y0asn → ∞, which means

that implicit functiony ϕxis continuous atx0. For x0 ∈ Ais arbitrary, we know that

yϕxis continuous onA.

FromTheorem 2.5, we see that in order to ensure thatyϕxx∈Ais well defined, the condition thatG is strictly monotone onA×B is necessary, but the boundedness ofB

is a strong condition. As usual,B is unbounded e.g., inequality constraint set B {y ∈ Rm _|_g_{x, y}_≥₀_}_{, where}_g _:_Rn_×_Rm _→ _Rl_{, is always unbounded}_{. So, we try to weaken the}

boundedness ofB. For this, we introduce the concept uniform coercivity ofGin VIB, Gx,·.

Definition 2.6. In VIB, Gx,·, letx0 ∈A;Gis said to be uniformly coercive nearx0, if there

exists some neighbourhoodV ofx0,y0∈BandR >0 such that y0 < Rand for allx∈V,

Gx, y, y−y0 ≥0, 2.10

wherey∈Band y R.

If for eachx∈A,Gis uniformly coercive nearx, thenGis said to be uniformly coercive onA.

Lemma 2.7. In VIB, Gx,·, let B ⊂ Rm _{be nonempty closed and convex set, and let} _G _be

uniformly coercive near x0 ∈ A, then there exists some neighbourhood V ⊂ A ofx0, such that

x∈VSolB, Gx,·is bounded set.

Proof. For givenx0 ∈V, by the definition of the uniform coercivity ofGnearx0, there exists

some neighbourhoodV ⊂Aofx0,y0∈B,andR >0 such that y0 < Rand for allx∈V,

Gx, y, y−y0 ≥0, ∀y∈B, yR. 2.11

LetBR :{y∈Rm| y ≤R} ∩B.It is obvious thatBRis a nonempty bounded closed convex

subset ofRm_._{In view of}_{Lemma 2.3}_{, we know that VI}_B

R, Gx0, ymust have solution. That is, there exists any₀∈BRsuch that

Gx0, y₀

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Now, we state thaty₀∈SolB, Gx0,·and y₀ ≤R.In fact, if y₀ < R, for ally∈B, y /∈BR,

joiny₀ andy intoλy₀ 1−λy with 0 ≤ λ ≤ 1. Then take small enoughλ > 0 such that

yλy 1−λy₀∈BR. Substitutingywithyin2.12, we have

Gx0, y₀

, y−y₀ ≥0, 2.13

implying thaty₀∈SolB, Gx0,·and y0 < R. On the other hand, if y0 R, substituting

ywithy₀in2.11, we have

Gx0, y₀

, y₀−y0 ≥0, 2.14

which, by plus2.12, we get

Gx0, y₀

, y−y0 ≥0, ∀y∈BR, y0< R. 2.15

For all y ∈ B,y /∈BR, consider the connection of yand y0; following the same argument, we have thaty₀ ∈ SolB, Gx0,·and y0 R.Therefore,y0 ∈SolB, Gx0,·and y0 ≤

R. That is, SolB, Gx0,·is bounded. For x0 ∈ V is arbitrary, the conclusion holds. This completes the proof.

If the boundedness ofBis replaced by the uniform coercivity ofGinTheorem 2.5, then we have the following result.

Theorem 2.8. In VIB, Gx,·, let B ⊂ Rm be nonempty closed and convex set, and let G be

uniformly coercive on A with respect to B and strict monotone in y. Then for each x ∈ A,

VIB, Gx,·has a unique solution, and for allx∈A, the implicit functionyϕxdetermined by

VIB, Gx,·is continuous onA.

Proof. i For given x ∈ A, by Lemma 2.4 and the coercivity of G on B, we know that

VIB, Gx,·has solution. Noting thatGis strictly monotone iny, VIB, Gx,·has a unique solution, and so the implicit functionyϕxis well defined.

ii For given x0 ∈ A, {xn} ⊂ A, satisfying xn → x0 as n → ∞. By Lemma 2.7, there exists some neighbourhood U ⊂ A of x0 and bounded open set V, such that

x∈USolB, Gx,· ⊂ V, that is, the solution set of VIB, Gx,·denoted by{y ∈ B | y

ϕx, x∈U} ⊂V.{xn} ⊂Asuch thatxn → x0asn → ∞. Let{xn} ⊂Uwithout generality, then{yn} ⊂V is bounded; the following argument is similar toTheorem 2.5, so it is omitted, and this completes the proof.

SetΛ {y ∈ B | y ϕx, x ∈ A}; it is to see that Λ ⊂ B. We will investigate the parametric variational inequality VIA, F·, ywith respect toyin SVI.

Corollary 2.9. InVIA, F·, y, letA⊂Rn_{, B}_⊂_Rm_{be two nonempty compact and convex subsets,}

F:A×B → Rnbe continuous and strict monotone inx. Then for each giveny∈Λ,VIA, F·, y

has a unique solution, and for ally∈Λ, the implicit functionxψydetermined byVIA, F·, y

is continuous onΛ.

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Lemma 2.10see3,Brouwer fixed point theorem. LetX ⊂ Rn _{be nonempty compact and}

convex set, and letT :X → Xbe continuous. Then there exists anx0, such thatx0Tx0.

Theorem 2.11. In SVI, letA⊂Rn_{, B}_⊂_Rm_{be two compact and convex subset, and let}_F_:_A_×_B _→

Rn_and_G_:_A_×_B _→ _Rm_{be two continuous mappings and strict monotone in}_x_and_y_{, respectively.}

Then SVI has solution.

Proof. By the given conditions of Theorems2.11and2.5, we know that there exists continuous

implicit function y ϕx x ∈ A determined by parametric variational inequality VIB, Gx,· with respect to x in SVI. Also denoted the range of y ϕx by Λ. By

Corollary 2.9, there exists continuous implicit functionx ψydetermined by parametric variational inequality VIA, F·, ywith respect toyin SVI such that for ally∈Λ,xψy

is the unique solution to VIA, F·, y. Let Φx ψϕx,for allx ∈ A.Making use of Brouwer fixed point theorem Lemma 2.10, we have that there exists x0 ∈ A, such that

x0 Φx0 ψϕx0. Settingy0 ϕx0, by the definitions ofϕ and ψ, we know that

x0, y0is a solution of SVI.

Corollary 2.12. In SVI, letA⊂Rn_{be nonempty compact and convex subset, let}_B_⊂_Rm_{be nonempty}

closed and convex subset, letF :A×B → Rn_and_G_:_A_×_B _→ _Rm_{be two continuous mappings,}

andF, Gbe strict monotone inxandy, respectively. LetGbe uniformly coercive onAwith respect

toB. Then SVI has solution.

Proof. ByTheorem 2.8and similar argument inTheorem 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. LetF :Rn_×_Rm _→ _Rn _{be vector-valued mapping; if there exists}_{α >}_{0, such}

that for ally∈Rm_,

Fx1, y

−Fx2, y

, x1−x2

≥α x1−x2 2, ∀x1, x2∈Rn, 2.16

thenFis said to be uniformly strongly monotone inx.

Lemma 2.14. InVIA, F·, y, letFbe uniformly strongly monotone, thenFis uniformly coercive.

Proof. For giveny0 ∈ B, we only need to prove that F is uniformly coercive at y0. Let us

consider VIA, F·, y. Assume that x0 ∈ Ais a solution to VIA, F·, y, and x0 < R,

R >0. SinceFis strongly monotone, then there existsα >0, such that for ally∈B,

Fx, z−Fz, y, x−z≥α x−z 2, ∀x, z∈A. 2.17

Lettingzx0, yy0in2.17, we have

Fx, y0

, x−x0

≥α x−x0 2

Fx0, y0

, x−x0

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Noting thatx0∈Ais a solution to VIA, F·, y, we have

Fx0, y0

, x−x0

≥0, ∀x∈A. 2.19

Combining2.18, we obtain

Fx, y0

, x−x0 ≥0, ∀x∈A, x R. 2.20

SinceFis continuous, then there exists some neighbourhoodUofy0, such that for ally∈U,

Fx, y, x−x0

≥0, ∀x∈A, x R, 2.21

which implies thatFis uniform coercive aty0.

Using the uniformly strong monotonicity ofF, we can obtain the following existence result for SVI under the condition thatAis a nonempty closed and convex subset ofRn_.

Corollary 2.15. In SVI, letA⊂Rnbe a nonempty closed and convex set, letB⊂ Rmbe a compact

and convex set, letF, Gbe two continuous mapping, letFbe uniformly strongly monotone inx, and

letGbe strictly monotone iny. Then SVI has solution.

Proof. ByLemma 2.14andCorollary 2.12, it is easy to see that the conclusion holds.

Corollary 2.16. In SVI, Let A and B be nonempty closed and convex subsets, let F, G be two

continuous mappings, and letFbe uniformly strongly monotone inx, and letGbe uniformly coercive

and strict monotone iny. Then SVI has solution.

Furthermore, IfFandGare Lipschitz continuous inxandy, respectively, one can obtain the

following existence and uniqueness result for SVI.

Theorem 2.17. In SVI, letA, Bbe nonempty compact and convex subsets, letFbe uniformly strongly

monotone with constantα1 > 0and Lipschitz continuous with Lipschitz constantβ1 > 0inx, and

Lipschitz continuous with constantγ1 > 0 in y, and let G be uniformly strongly monotone with

constantα2 >0and Lipschitz continuous with constantβ2 >0iny, and Lipschitz continuous with

constantγ2>0inx. If there exists constantsρ1, ρ2>0such that

max

1−2ρ1α1ρ₁2β2₁ρ2γ2,

1−2ρ2α2ρ2₂β2₂ρ1γ1

<1, 2.22

then SVI has a unique solution.

In order to proveTheorem 2.17, we need the following lemma.

Lemma 2.18see4. In SVI, letA, Bbe nonempty closed and convex subsets, and letF, Gbe two

continuous mappings. SVI has solutionx, yif and only ifx, ysatisfies

xPA

x−ρ1F

x, y,

yPB

y−ρ2G

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where PA·, PB· denote the projection from Rn and Rm to A and B, respectively; furthermore,

projection operator is nonexpansive andρ1, ρ2>0are constants.

The Proof of

Theorem 2.17

For arbitrary given constantρ1, ρ2>0, defineTρ1:A×B → AandTρ2:A×B → Bby

Tρ1

x, yPA

x−ρ1F

x, y,

Tρ2

x, yPB

y−ρ2G

x, y, ∀x, y∈A×B. 2.24

For anyx1, y1,x2, y2∈A×B, it follows from2.24andLemma 2.18that

_T_ρ 1

x1, y1

−Tρ1

x2, y2

≤x1−x2−ρ1

Fx1, y1

−Fx2, y2

≤x1−x2−ρ1

Fx1, y1

−Fx2, y1ρ1F

x2, y1

−Fx2, y2

≤1−2ρ1α1ρ2₁β2₁ x1−x2 ρ1γ1y1−y2.

2.25

We have used the strong monotonicity and Lipschitz continuity of F in x and Lipschitz continuity ofFiny. Similarly, we have

_T_ρ 2

x1, y1

−Tρ2

x2, y2

≤y1−y2−ρ2

Gx1, y1

−Gx2, y2

≤y1−y2−ρ2

Gx1, y1

−Gx2, y1ρ2G

x2, y1

−Gx2, y2

≤1−2ρ2α2ρ22β22y1−y2ρ2γ2 x1−x2 .

2.26

It follows from2.25and2.26that

Tρ1

x1, y1

−Tρ1

x2, y2Tρ2

x1, y1

−Tρ2

x2, y2

≤1−2ρ1α1ρ2₁β₁2ρ2γ2

x1−x2

1−2ρ2α2ρ22β22ρ1γ1

y1−y2

≤k x1−x2 y1−y2,

2.27

where

kmax

1−2ρ1α1ρ₁2β2₁ρ2γ2,

1−2ρ2α2ρ₂2β2₂ρ1γ1

. 2.28

Define · 1onA×Bby

_{x, y}

1 x y, ∀

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It is easy to see thatA×B, · 1is a Banach space. For any givenρ1, ρ2 > 0, defineSρ1,ρ2 : A×B → A×Bby

Sρ1,ρ2

x, yTρ1

x, y, Tρ2

x, y, ∀x, y∈A×B. 2.30

By assumption, we know that 0< k <1. It follows from2.27that

_S_ρ

1,ρ2x1, y1−Sρ1,ρ2x2, y2₁ ≤kx1, y1

−x2, y2, 2.31

which implies thatSρ1,ρ2 : A×B → A×Bis a contraction operator. Hence, there exists a

uniquex∗, y∗∈A×B, such that

Sρ1,ρ2

x∗, y∗x∗, y∗. 2.32

That is,

x∗PA

x∗−ρ1F

x∗, y∗,

y∗PB

y∗−ρ2G

x∗, y∗. 2.33

ByLemma 2.18,x∗, y∗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 3.1see,5. Let{cn}and{kn}be two real sequence of nonnegative numbers that satisfy the following conditions.

i0≤kn<1, n0,1,2, . . .andlim sup_nkn <1,

iicn1≤kncn, n0,1,2, . . . .

Then,cnconverges to 0 asn → ∞.

Algorithm 3.2. Let A, B, F, G, ρ1, and ρ2 be the same as in Theorem 2.17. For any given

x0, y0∈A×B, define iterative sequence{xn, yn}by

xn1anxn 1−anPA

xn−ρ1F

xn, yn

, n0,1,2, . . . ,

yn1anyn 1−anPB

xn−ρ2G

xn, yn

, n0,1,2, . . . , 3.1

where

0≤an <1, lim sup n

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Theorem 3.3. Let A, B, F, G, ρ1, and ρ2 be the same as in Theorem 2.17. Assume that all the

conditions ofTheorem 2.17hold. Then,xn, yngenerated byAlgorithm 3.2converges to the unique

solutionx∗, y∗of SVI and there existsd∈0,1, such that

xn−x∗ yn−y∗≤dn

x0−x∗ y0−y∗, ∀n≥0. 3.3

Proof. By Theorem 2.17, SVI admits a unique solutionx∗, y∗. It follows from Lemma 2.18

that

x∗anx∗ 1−anPA

x∗−ρ1F

x∗, y∗,

y∗any∗ 1−anPB

y∗−ρ2G

x∗, y∗. 3.4

It follows from3.1and3.4that

xn1−x∗ ≤an xn−x∗ 1−anPA

xn−ρ1F

xn, yn

−PA

x∗−ρ1F

x∗, y∗

≤an xn−x∗ 1−anxn−x∗ρ1

Fxn, yn

−Fx∗, y∗

≤an xn−x∗ 1−an

1−2ρ1α1ρ21β12 xn−x∗ 1−anρ1γ1yn−y∗, _y_n₁₋_y∗_≤_a

nyn−y∗ 1−anPB

yn−ρ2G

xn, yn

−PB

y∗−ρ2G

x∗, y∗

≤anyn−y∗ 1−anyn−y∗ρ2

Gxn, yn

−Gx∗, y∗

≤anyn−y∗ 1−an

1−2ρ2α2ρ₂2β₂2yn−y∗ 1−anρ2γ2 xn−x∗ .

3.5

By3.5, we get

xn1−x∗ yn1−y∗≤an

xn−x∗ yn−y∗ 1−ank

xn−x∗ yn−y∗

k 1−kan

xn−x∗ yn−y∗,

3.6

where 0≤k <1 is defined by

kmax

1−2ρ1α1ρ₁2β2₁ρ2γ2,

1−2ρ2α2ρ₂2β2₂ρ1γ1

. 3.7

Set

cn xn−x∗ yn−y∗, knk 1−kan. 3.8

Then3.6can be rewritten as

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By3.2, we know that lim sup_nkn<1. It follows fromLemma 3.1that 0≤kn≤d <1 and that

xn−x∗ yn−y∗≤dn

x0−x∗ y0−y∗, ∀n≥0 3.10

for all n ≥ 0. Therefore,xn, ynconverges geometrically to the unique solutionx∗, y∗ of

SVI. This completes the proof.

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 Province20071097.

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