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Volume 2007, Article ID 76040, 15 pagesdoi:10.1155/2007/76040 Research Article An Algorithm Based on Resolvant Operators for Solving Positively Semidefinite Variational Inequalities Juhe

Volume 2007, Article ID 76040, 15 pages

doi:10.1155/2007/76040

Research Article

An Algorithm Based on Resolvant Operators for Solving

Positively Semidefinite Variational Inequalities

Juhe Sun, Shaowu Zhang, and Liwei Zhang

Received 16 June 2007; Accepted 19 September 2007

Recommended by Nan-Jing Huang

A new monotonicity,M-monotonicity, is introduced, and the resolvant operator of an M-monotone operator is proved to be single-valued and Lipschitz continuous With the

help of the resolvant operator, the positively semidefinite general variational inequality (VI) problem VI (S n

+,F + G) is transformed into a fixed point problem of a

nonexpan-sive mapping And a proximal point algorithm is constructed to solve the fixed point problem, which is proved to have a global convergence under the condition thatF in the

VI problem is strongly monotone and Lipschitz continuous Furthermore, a convergent path Newton method is given for calculating-solutions to the sequence of fixed point problems, enabling the proximal point algorithm to be implementable

Copyright © 2007 Juhe Sun et al This is an open access article distributed under the Cre-ative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

1 Introduction

In recent years, the variational inequality has been addressed in a large variety of prob-lems arising in elasticity, structural analysis, economics, transportation equilibrium, op-timization, oceanography, and engineering sciences [1,2] Inspired by its wide applica-tions, many researchers have studied the classical variational inequality and generalized it

in various directions Also, many computational methods for solving variational inequal-ities have been proposed (see [3–8] and the references therein) Among these methods, resolvant operator technique is an important one, which was studied in the 1990s by many researchers (such as [4,6,9]), and further studies developed recently [3,10,11]

As monotonicity plays an important role in the theory of variational inequality and its generalizations, in this paper, we introduce a new class of monotone operator:

M-monotone operator The resolvant operator associated with anM-monotone operator is

proved to be Lipschitz-continuous Applying the resolvant operator technique, we trans-form the positively semidefinite variational inequality (VI) problem VI(S n

+,F + G) into

a fixed point problem of a nonexpansive mapping and suggest a proximal point algo-rithm to solve the fixed point problem Under the condition thatF in the VI problem is

strongly monotone and Lipschitz-continuous, we prove that the algorithm has a global convergence To ensure the proposed proximal point algorithm is implementable, we in-troduce a path Newton algorithm whose step size is calculated by Armijo rule

In the next section, we recall some results and concepts that will be used in this paper

InSection 3, we introduce the definition of anM-monotone operator, and discuss

prop-erties of this kind of operators, especially the Lipschitz continuity of the resolvant opera-tor of anM-monotone operator InSection 4, we construct a proximal point algorithm, based on the results inSection 3, forVI(S n

+,F + G), and prove its global convergence To

ensure that the proposed proximal point algorithm inSection 4is implementable, we in-troduce a path Newton algorithm, inSection 5, in which the step size is calculated by Armijo rule

2 Preliminaries

Throughout this paper, we assume thatS ndenotes the space ofn × n symmetric matrices

andS n

+denote the cone ofn × n symmetric positive semidefinite matrices For A,B ∈ S n,

we define an inner product A,B  =tr(AB) which induces the norm  A  = A,A  Let

2S n denote the family of all the nonempty subsets ofS n We recall the following concepts, which will be used in the sequel

Definition 2.1 Let A,B,C : S n → S nbe single-valued operators and letM : S n × S n → S nbe mapping

(i)M(A, ·) is said to beα-strongly monotone with respect to A if there exists a

con-stantα > 0 satisfying



M(Ax,u) − M(Ay,u),x − y≥ α  x − y 2, ∀ x, y,u ∈ S n; (2.1)

(ii)M( ·,B) is said to be β-relaxed monotone with respect to B if there exists a constant

β > 0 satisfying



M(u,Bx) − M(u,By),x − y≥ − β  x − y 2, ∀ x, y,u ∈ S n; (2.2)

(iii)M( ·,·) is said to beαβ-symmetric monotone with respect to A and B if M(A, ·) is

α-strongly monotone with respect to A; and M( ·,B) is β-relaxed monotone with

respect toB with α ≥ β and α = β if and only if x = y, for all x, y,u ∈ S n;

(iv)M( ·,·) is said to beξ-Lipschitz-continuous with respect to the first argument if

there exists a constantξ > 0 satisfying

M(x,u) − M(y,u) ≤ ξ  x − y , ∀ x, y,u ∈ S n; (2.3)

(iv)A is said to be t-Lipschitz-continuous if there exists a constant t > 0 satisfying

(vi)B is said to be l-cocoercive if there exists a constant l > 0 satisfying

 Bx − By,x − y  ≥ l  Bx − By 2, ∀ x, y ∈ S n; (2.5) (vii)C is said to be r-strongly monotone with respect to M(A,B) if there exists a

con-stantr > 0 satisfying



Cx − Cy,M(Ax,Bx) − M(Ay,By)≥ r  x − y 2, ∀ x, y ∈ S n (2.6)

In a similar way to (v), we can define the Lipschitz continuity of the mapping M with

respect to the second argument

Definition 2.2 Let A,B : S n → S n,M : S n × S n → S nbe mappings.M is said to be coercive

with respect toA and B if

lim

 x →∞



M(Ax,Bx),x

Definition 2.3 Let A,B : S n → S n,M : S n × S n → S nbe mappings.M is said to be bounded

with respect toA and B if M(A(P),B(P)) is bounded for every bounded subset P of S n

M is said to be semicontinuous with respect to A and B if for any fixed x, y,z ∈ S n, the functiont →  M(A(x + ty),B(x + ty)),z is continuous at 0+

Definition 2.4 T : S n →2S nis said to be monotone if

 x − y,u − v  ≥0, ∀ u,v ∈ S n,x ∈ Tu, y ∈ Tv; (2.8) and it is said to be maximal monotone ifT is monotone and (I + cT)(S n)= S n for all

c > 0, where I denotes the identity mapping on S n

3.M-Monotone operators

In this section, we introduceM-monotonicity of operators and discuss its properties Definition 3.1 Let A,B : S n → S nbe single-valued operators,M : S n × S n → S na mapping, andT : S n →2S na multivalue operator.T is said to be M-monotone with respect to A and

B if T is monotone and (M(A,B) + cT)(S n)= S nholds for everyc > 0.

Remark 3.2 If M(A,B) = H, then the above definition reduces to H-monotonicity, which

was studied in [5] IfM(A,B) = I, then the definition of I-monotonicity is just the

maxi-mal monotonicity

Remark 3.3 Let T be a monotone operator and let c be a positive constant If T : S n →2S n

is anM-monotone operator with respect to A and B, every matrix z ∈ S ncan be written

in exactly one way asM(Ax,Bx) + cu, where u ∈ T(x).

Proposition 3.4 Let M be αβ-symmetric monotone with respect to A and B and let T :

S n →2S n be an M-monotone operator with respect to A and B, then T is maximal monotone Proof Since T is monotone, it is sufficient to prove the following property; inequality

 x − y,u − v  ≥0 for (v, y) ∈Graph(T) implies that

Suppose, by contradiction, that there exists some (u0,x0)∈Graph(T) such that



x0 − y,u0 − v≥0, ∀(v, y) ∈Graph(T). (3.2) SinceT is M-monotone with respect to A and B, (M(A,B) + cT)(S n)= S nholds for every

c > 0, there exists (u1,x1)∈Graph(T) such that

MAu1,Bu1+cx1 = MAu0,Bu0+cx0 ∈ S n (3.3)

It follows form (3.2) and (3.3) that

0≤ cx0 − x1,u0 − u1

= −MAu0,Bu0− MAu1,Bu1,u0 − u1

= −MAu0,Bu0− MAu1,Bu0,u0 − u1

−MAu1,Bu0− MAu1,Bu1,u0 − u1

≤ −(α − β)u0 − u1

≤0,

(3.4)

which yieldsu1 = u0 By (3.3), we have thatx1 = x0 Hence (u0,x0)∈Graph(T), which is

a contradiction Therefore (3.1) holds andT is maximal monotone This completes the

The following example shows that a maximal monotone operator may not be

M-monotone for someA and B.

Example 3.5 Let S n = S2,T = I, and M(Ax,Bx) = x2+ 2E − x for all x ∈ S2, whereE is an

identity matrix Then it is easy to see thatI is maximal monotone For all x ∈ S2, we have that

M(A,B) + I

(x) 2

=x2+ 2E − x + x 2

=x2+ 2E 2

=tr

x2+ 2E2 ≥8, (3.5) which means that 0 ¯∈(M(A,B) + I)(S2) andI is not M-monotone with respect to A and B.

Proposition 3.6 Let T : S n →2S n be a maximal monotone operator and let M : S n × S n →

S n be a bounded, coercive, semicontinuous, and αβ-symmetric monotone operator with re-spect to A and B Then T is M-monotone with respect to A and B.

Proof For every c > 0, cT is maximal monotone since T is maximal monotone Since M is

bounded, coercive, semicontinuous, andαβ-symmetric monotone operator with respect

toA and B, it follows from [9, Corollary 32.26] thatM(A,B) + cT is surjective, that is,

(M(A,B) + cT)(S n)= S nholds for everyc > 0 Thus, T is an M-monotone operator with

Theorem 3.7 Let M be an αβ-symmetric monotone with respect to A and B and let T be

an M-monotone operator with respect to A and B Then the operator (M(A,B) + cT) −1is single-valued.

Proof For any given u ∈ S n, letx, y ∈(M(A,B) + cT) −1(u) It follows that − M(Ax,Bx) +

u ∈ Tx and − M(Ay,By) + u ∈ T y The monotonicity of T and M implies that

0≤− M(Ax,Bx) + u −− M(Ay,By) + u,x − y

= −M(Ay,By) − M(Ax,Bx),x − y

≤ −(α − β)u0 − u1

≤0.

(3.6)

From the symmetric monotonicity ofM, we get that x = y Thus (M(A,B) + cT) −1 is

Definition 3.8 Let M be an αβ-symmetric monotone with respect to A and B and let T be

anM-monotone operator with respect to A and B The resolvant operator J M

cT:S n → S nis defined by

J M

cT(u) =M(A,B) + cT−1

Theorem 3.9 Let M(A,B) be α-strongly monotone with respect to A and β-relaxed mono-tone with respect to B with α > β Suppose that T : S n →2S n is an M-monotone operator Then the resolvant operator J M

cT:S n → S n is Lipschitz-continuous with constant 1/(α − β), that is,

J M

cT(u) − J M

cT(v) ≤ 1

Since the proof ofTheorem 3.9is similar as that of [5, Theorem 2.2], we here omit it

4 An algorithm for variational inequalities

Let F,G : S n

+→ S n be operators Consider the general variational inequality problem

VI(S n

+,F + G), defined by finding u ∈ S n

+such that



F(u) + G(u),v − u≥0, ∀ v ∈ S n+. (4.1)

We can rewrite it as the problem of findingu ∈ S n

+such that

whereT ≡ F + ᏺ( ·;S n) Let Sol(S n,F + G) be the set of solutions of VI(S n,F + G).

Proposition 4.1 Let F,G : S n

+→ S n be continuous and let M : S n × S n → S n be a bounded, coercive, semicontinuous, and αβ-symmetric monotone operator with respect to A : S n → S n

and B : S n → S n Then the following two properties hold for the map T ≡ F + ᏺ( ·;S n

+):

(a)J M

cT(M(Ax,Bx) − cG(x)) =Sol(S n

+,F cx ), where F cx(y) = M(Ay,By) − M(Ax,Bx) + c(F(y) + G(x));

(b) If F is monotone, then T is M-monotone with respect to A and B.

Proof We have that the inclusion

y ∈ J M

cT

M(Ax,Bx) − cG(x)=M(A,B) + cT−1M(Ax,Bx) − cG(x) (4.3)

is equivalent to

M(Ax,Bx) ∈M(A,B) + cF + cᏺ·;S n

+



(y) + cG(x), (4.4)

or in other words,

0∈ M(Ay,By) − M(Ax,Bx) + cF(y) + G(x)+ᏺy;S n

+



This establishes (a).

By [10, Proposition 12.3.6], we can deduce thatT is maximal monotone, it follows

fromProposition 3.6, we get thatT is M-monotone with respect to A and B This

Lemma 4.2 Let M be an αβ-symmetric monotone with respect to A and B and let T be an M-monotone operator with respect to A and B Then u ∈ S n

+is a solution of 0 ∈ G(u) + T(u)

if and only if

u = J M

cT

where J M

cT =(M(A,B) + cT) −1and c > 0 is a constant.

In order to obtain our results, we need the following assumption

Assumption 4.3 The mappings F, G, M, A, B satisfy the following conditions.

(1)F is L-Lipschitz-continuous and m-strongly monotone.

(2)M(A, · ) is α-strongly monotone with respect to A; and M( ·,B) is β-relaxed monotone with respect to B with α > β.

(3)M( ·,· ) is ξ-Lipschitz-continuous with respect to the first argument and ζ-Lipschitz-continuous with respect to the second argument.

(4)A is τ-Lipschitz-continuous and B is t-Lipschitz-continuous.

(5)G is γ-Lipschitz-continuous and s-strongly monotone with respect to M(A,B) Remark 4.4 LetAssumption 4.3hold and

c − s

γ2

s2− γ2 

(ξτ + ζt)2−(α − β)2

γ2 , s2> γ2 

(ξτ + ζt)2−(α − β)2

We can deduce that

J M

cT

M(Ax,Bx) − cG(x)− J M

cT

M(Ay,By) − cG(y)

≤ α −1βM(Ax,Bx) − M(Ay,By) − c

G(x) − G(y)

≤

(ξτ + ζt)2−2cs + c2γ2

α − β  x − y 

≤  x − y ,

(4.8)

which implies thatJ M

cT(M(A,B) − cG) is nonexpansive Then, it is natural to consider the

recursion

x k+1 ≡ J M

cT

MAx k,Bx k

− cGx k

which is desired to converge to a zero ofG + T Actually, this can be proved to be true.

However, based onLemma 4.2, we construct the following proximal point algorithm for

VI(S n

+,F + G).

Algorithm 4.5

Data x0∈ S n,c0 > 0, ε0 ≥ 0, and ρ0 > 0.

Step 1 Set k = 0.

Step 2 If x k ∈Sol(S n

+,F + G), stop.

Step 3 Find w k such that  w k − J M

ckT(M(Ax k,Bx k)− c k G(x k)) ≤ ε k Step 4 Set x k+1 ≡(1− ρ k)x k+ρ k w k and select c k+1,ε k+1 and ρ k+1 Set k ← k + 1 and go to Step 1.

The following theorem fully describes the convergence ofAlgorithm 4.5for finding a solution toVI(S n

+,F + G).

Theorem 4.6 Suppose that Algorithm 4.5 holds Let M be bounded, coercive, semicon-tinuous, and αβ-symmetric monotone with respect to A and B; and let F be monotone and Lipschitz-continuous Let x0∈ S n be given, let { ε k } ⊂[0,∞ ) satisfy E ≡∞ k =1ε k < ∞ ,

{ c k } ⊂(c m,∞ ), where c m > 0 and

c k − s

γ2 <

s2− γ2 

(ξτ + ζt)2−(α − β)2

γ2 , s2> γ2 

(ξτ + ζt)2−(α − β)2

which implies that

L =



α − β −(ξτ + ζt)2−2c k s + c2

k γ2 



α − β + 3(ξτ + ζt)2−2c k s + c2

k γ2 2 > 0. (4.11)

If { ρ k } ⊆[R m,R M ], where 0 < R m ≤ R M ≤ pL, for all p ∈[2, +∞ ), then the sequence { x k }

generated by Algorithm 4.5 converges to a solution of VI(S n,F + G).

Proof We introduce a new map

Q k ≡ I − J M

ckT

Clearly, any zero ofG + F + ᏺ( ·;S n

+), being a fixed point ofJ M

ckT(M(A,B) − c k G), is also a

zero ofQ k Now, let us prove thatQ kisL-cocoercive.

Forx, y ∈ S nwe know that



Q k(x) − Q k(y),x − y

=x − y −J M

ckT

M(Ax,Bx) − c k G(x)− J M

ckT

M(Ay,By) − c k G(y),x − y

=  x − y 2−J M

ckT

M(Ax,Bx)− J M

ckT

M(Ay,By) − c k

G(x) − G(y),x − y

≥  x − y 2− 1

α − βM(Ax,Bx) − M(Ay,By) − c k

G(x) − G(y)x − y 

≥  x − y 2− α −1β

(ξτ + ζt)2−2c k s + c2

k γ2 x − y 2

=



1−

(ξτ + ζt)2−2c k s + c2

k γ2

α − β



 x − y 2,

(4.13)

Q k(x) − Q k(y) 2

=x − y −

J M ckT

M(Ax,Bx) − c k G(x)− J M

ckT

M(Ay,By) − c k G(y) 2

=  x − y 2−2

x − y,J M ckT

M(Ax,Bx) − c k G(x)− J M

ckT

M(Ay,By) − c k G(y)

+J M

ckT

M(Ax,Bx) − c k G(x)− J M

ckT

M(Ay,By) − c k G(y) 2

≤  x − y 2+ 2

(ξτ + ζt)2−2c k s + c2

k γ2

α − β  x − y 2

+

(ξτ + ζt)2−2c k s + c2

k γ2

α − β  x − y 2

=

⎛

⎝1 + 3

(ξτ + ζt)2−2c k s + c2

k γ2

α − β

⎞

⎠x − y 2.

(4.14) Inequalities (4.13) and (4.14) imply that



Q k(x) − Q k(y),x − y

≥

⎡

⎣1−

(ξτ + ζt)2−2c k s + c2

k γ2

α − β

⎤

⎦

⎡

⎣1+3

(ξτ + ζt)2−2c k s + c2

k γ2

α − β

⎤

⎦

−1

Q k(x) − Q k(y) 2

LQ k(x) − Q k(y) 2

.

(4.15)

For allk, we denote by x kthe point computed exactly by the resolvent That is,

x k+1 ≡1− ρ k

x k+ρ k J M ckT

MAx k,Bx k

− c k Gx k

For every zerox ∗ofT, we obtain

x k+1 − x ∗ 2

=x k − ρ k Q k

x k

− x ∗ 2

=x k − x ∗ 2

−2ρ k

Q k

x k

− Q k

x ∗

,x k − x ∗

+ρ2

kQ k

x k 2

≤x k − x ∗ 2

−2ρ k LQ k

x k 2

+ρ2

kQ k

x k 2

≤x k − x ∗ 2

− ρ k

2L − ρ kQ k

x k 2

≤x k − x ∗ 2

− R m

2L − R MQ k

x k 2

≤x k − x ∗ 2

.

(4.17)

Since x k − x k  ≤ ρ k ε k, we get that

x k+1 − x ∗  ≤ x k+1 − x ∗+x k+1 − x k+1

≤x k − x ∗+ρ k ε k

≤x0− x ∗+k

i =0

ρ i ε i

≤x0− x ∗+pLE.

(4.18)

Therefore, the sequence{ x k }is bounded On the other hand, we have that

x k+1 − x ∗ 2

=x k+1 − x ∗+

x k+1 − x k+1 2

=x k+1 − x ∗ 2

+ 2

x k+1 − x ∗,x k+1 − x k+1

+x k+1 − x k+1 2

≤x k+1 − x ∗ 2

+ 2x k+1 − x ∗x k+1 − x k+1+x k+1 − x k+1 2

≤x k − x ∗ 2

+ 2ρ k ε kx0− x ∗+pLE

+ρ2

k ε2

k

− R m

2L − R MQ k

x k 2

.

(4.19)

LettingE2 =∞ i =0ε2

k < ∞, we have for everyk,

x k+1 − x ∗ 2

≤x0− x ∗ 2

+ 2pLEx0− x ∗+pLE

+p2L2E2 − R m

2L − R M k

i =0

Q k

x k 2

Passing to the limitk → ∞, one has that∞

i =0 Q k(x k)2< ∞, implying that lim

k →∞ Q k

x k

According to Remark 3.3, for every k, there exists a unique pair (y k,v k) in gphT

such thatz k = M(Ax k,Bx k)− c k G(x k)= M(Ay k,By k) +c k v k ThenJ M

ckT(M(Ax k,Bx k)−

c k G(x k))= y k So thatQ k(x k)→0 implies that (x k − y k)→0,v k →0

Sincec kis bounded away from zero, it follows thatc −1

k Q k(x k)→0 Sincex kis bounded,

it has at least a limit point Letx ∞ be such a limit point and assume that the subse-quence{ x ki:k i ∈ k }converges tox ∞ It follows that{ y ki:k i ∈ k }also converges tox ∞ For every (y,v) in gphT by the monotonicity of T, we have that  y − y k,v − v k  ≥0 Let-tingk i(∈ k) → ∞, we get that y − y ∞,v − v k  ≥0 We see thatT is M-monotone due to

Proposition 4.1, this implies that (x ∞,− G(x ∞))∈gphT, that is, − G(x ∞)∈ T(x ∞) This

5 Solving an approximate fixed point toJ M

ckT

How to calculatew kat Step 3 is the key inAlgorithm 4.5 Ifε k =0, this amounts to the exact solution ofVI(S n

+,F k), where

F k(x) = M(Ax,Bx) − MAx k,Bx k

+c k

F(x) + G(x k

Now, we consider the case ofε k > 0 We can prove that J M

ckT(M(Ax k,Bx k)− c k G(x k)) is the unique solution of theVI(S n

+,F k) Hence,w k is an inexact solution of theVI(S n

+,F k) satisfying dist(w k, Sol(S n

+,F k))≤ ε k

Lemma 5.1 Let F, G, M, A, B satisfy all the conditions of Assumption 4.3 Then a constant c(k) > 0 exists such that

dist

w k, Sol

S n

+,F k

≤ c(k)F knat

S n

+



Proof By Assumption 4.3, we can easily get that F k isL (k)-Lipschitz-continuous and η(k)-strongly monotone, where L (k) = ξτ + ζt + c k L and η(k) = α − β + c k m, that is,

F k(x) − F k(y) ≤ L (k)x − y,



F k(x) − F k(y),x − y≥ η(k)  x − y 2, ∀ x, y ∈ S n+. (5.3)

Letr =(F k)nat

S n+(w k), where (F k)nat

S n+ is the natural map associated with theVI(S n

+,F k) We have thatw k − r =ΠS n+(w k − F k(w k)), that is,



y − w k+r,F k

w k

For allx ∗ ∈Sol(S n

+,F k) andw k − r ∈ S n

+, we also have that



w k − r − x ∗,F k

x ∗

From (5.4) and (5.5), we get that



x ∗ − w k,F k

w k

− r+

r,F k

w k

− r

=x ∗ − w k,F k

w k

−x ∗ − w k,r+

r,F k

w k

+r 2



x ∗ − w k,F k

x ∗

−r,F k

x ∗

.

(4.15)

For allk, we denote by x kthe... k

According to Remark 3.3, for every k, there exists a unique... VI(S n,F + G).

Proof We introduce a new map

Q k