Báo cáo hóa học: " Research Article An Algorithm Based on Resolvant Operators for Solving Positively Semidefinite Variational Inequalities"

Hindawi Publishing Corporation Fixed Point Theory and Applications 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 (Sn +,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 that F in the VI problem is strongly monotone and Lipschitz continuous. Furthermore, a convergent path Newton method is given for calculating (cid:2)-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 an M-monotone operator is

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Fixed Point Theory and Applications

proved to be Lipschitz-continuous. Applying the resolvant operator technique, we trans- form the positively semidefinite variational inequality (V I) problem V I(Sn +,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 that F in the V I 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. In Section 3, we introduce the definition of an M-monotone operator, and discuss prop- erties of this kind of operators, especially the Lipschitz continuity of the resolvant opera- tor of an M-monotone operator. In Section 4, we construct a proximal point algorithm, based on the results in Section 3, for V I(Sn +,F + G), and prove its global convergence. To ensure that the proposed proximal point algorithm in Section 4 is implementable, we in- troduce a path Newton algorithm, in Section 5, in which the step size is calculated by Armijo rule.

2. Preliminaries

(cid:2)

Throughout this paper, we assume that Sn denotes the space of n × n symmetric matrices + denote the cone of n × n symmetric positive semidefinite matrices. For A,B ∈ Sn, and Sn we define an inner product (cid:3)A,B(cid:4) = tr(AB) which induces the norm (cid:5)A(cid:5) = (cid:3)A,A(cid:4). Let 2Sn denote the family of all the nonempty subsets of Sn. We recall the following concepts, which will be used in the sequel.

Definition 2.1. Let A,B,C : Sn → Sn be single-valued operators and let M : Sn × Sn → Sn be mapping.

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

stant α > 0 satisfying

(cid:4) (cid:3) M(Ax,u) − M(Ay,u),x − y ≥ α(cid:5)x − y(cid:5)2, ∀x, y,u ∈ Sn;

(2.1)

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

β > 0 satisfying

(cid:3) (cid:4)

M(u,Bx) − M(u,B y),x − y

≥ −β(cid:5)x − y(cid:5)2, ∀x, y,u ∈ Sn;

(2.2)

(iii) M(·, ·) is said to be αβ-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 to B with α ≥ β and α = β if and only if x = y, for all x, y,u ∈ Sn;

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

there exists a constant ξ > 0 satisfying

(cid:5) (cid:5)M(x,u) − M(y,u) (cid:5) (cid:5) ≤ ξ(cid:5)x − y(cid:5), ∀x, y,u ∈ Sn;

(2.3)

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

(cid:5)Ax − Ay(cid:5) ≤ t(cid:5)x − y(cid:5), ∀x, y ∈ Sn;

(2.4)

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

(cid:3)Bx − B y,x − y(cid:4) ≥ l(cid:5)Bx − B y(cid:5)2, ∀x, y ∈ Sn;

(2.5)

(vii) C is said to be r-strongly monotone with respect to M(A,B) if there exists a con-

stant r > 0 satisfying

(cid:3) (cid:4)

Cx − C y,M(Ax,Bx) − M(Ay,B y)

≥ r(cid:5)x − y(cid:5)2, ∀x, y ∈ Sn.

(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 : Sn → Sn, M : Sn × Sn → Sn be mappings. M is said to be coercive with respect to A and B if

(cid:3) (cid:4)

(2.7)

= +∞.

lim (cid:5)x(cid:5)→∞

M(Ax,Bx),x (cid:5)x(cid:5)

Definition 2.3. Let A,B : Sn → Sn, M : Sn × Sn → Sn be mappings. M is said to be bounded with respect to A and B if M(A(P),B(P)) is bounded for every bounded subset P of Sn. M is said to be semicontinuous with respect to A and B if for any fixed x, y,z ∈ Sn, the function t (cid:11)→ (cid:3)M(A(x + t y),B(x + t y)),z(cid:4) is continuous at 0+. Definition 2.4. T : Sn → 2Sn

is said to be monotone if

(cid:3)x − y,u − v(cid:4) ≥ 0, ∀u,v ∈ Sn, x ∈ Tu, y ∈ Tv;

(2.8)

and it is said to be maximal monotone if T is monotone and (I + cT)(Sn) = Sn for all c > 0, where I denotes the identity mapping on Sn.

3. M-Monotone operators

In this section, we introduce M-monotonicity of operators and discuss its properties. Definition 3.1. Let A,B : Sn → Sn be single-valued operators, M : Sn × Sn → Sn a mapping, and T : Sn → 2Sn a multivalue operator. T is said to be M-monotone with respect to A and B if T is monotone and (M(A,B) + cT)(Sn) = Sn holds for every c > 0. Remark 3.2. If M(A,B) = H, then the above definition reduces to H-monotonicity, which was studied in [5]. If M(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 : Sn → 2Sn is an M-monotone operator with respect to A and B, every matrix z ∈ Sn can be written in exactly one way as M(Ax,Bx) + cu, where u ∈ T(x).

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Fixed Point Theory and Applications

Proposition 3.4. Let M be αβ-symmetric monotone with respect to A and B and let T : Sn → 2Sn 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 (cid:3)x − y,u − v(cid:4) ≥ 0 for (v, y) ∈ Graph(T) implies that

x ∈ Tu.

(3.1)

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

(cid:3) (cid:4) ≥ 0, ∀(v, y) ∈ Graph(T).

(3.2)

x0 − y,u0 − v

(cid:7) (cid:7)

M

(3.3)

Since T is M-monotone with respect to A and B, (M(A,B) + cT)(Sn) = Sn holds for every c > 0, there exists (u1,x1) ∈ Graph(T) such that (cid:6) (cid:6) Au0,Bu0 Au1,Bu1

+ cx0 ∈ Sn.

+ cx1 = M

It follows form (3.2) and (3.3) that

(cid:3) (cid:4)

(cid:4) − M

0 ≤ c = −

(cid:7) (cid:4) (cid:3) = − − M (cid:7) (cid:4)

(3.4)

M (cid:3) M

− (cid:6) Au1,Bu1 (cid:6) Au1,Bu0 (cid:6) Au1,Bu1 (cid:7) ,u0 − u1 (cid:7) ,u0 − u1 (cid:7) ,u0 − u1 − M (cid:5) (cid:5)

x0 − x1,u0 − u1 (cid:3) (cid:7) (cid:6) Au0,Bu0 M (cid:6) Au0,Bu0 (cid:6) Au1,Bu0 (cid:5) (cid:5)u0 − u1

≤ −(α − β) ≤ 0,

which yields u1 = u0. By (3.3), we have that x1 = x0. Hence (u0,x0) ∈ Graph(T), which is a contradiction. Therefore (3.1) holds and T is maximal monotone. This completes the (cid:2) proof.

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

(cid:8)(cid:6) (cid:7)2(cid:9)

monotone for some A and B. Example 3.5. Let Sn = S2, T = I, and M(Ax,Bx) = x2 + 2E − x for all x ∈ S2, where E is an identity matrix. Then it is easy to see that I is maximal monotone. For all x ∈ S2, we have that (cid:5) (cid:5)

(cid:5) (cid:5)2 = (cid:5) (cid:5)2 = (cid:5) (cid:5)x2 + 2E − x + x (cid:5) (cid:5)x2 + 2E (cid:5) (cid:5)2 = tr (cid:6) M(A,B) + I (cid:7) (x)

x2 + 2E

(3.5)

≥ 8,

which means that 0 ¯∈(M(A,B) + I)(S2) and I is not M-monotone with respect to A and B. Proposition 3.6. Let T : Sn → 2Sn be a maximal monotone operator and let M : Sn × Sn → Sn 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, and αβ-symmetric monotone operator with respect

Juhe Sun et al.

5

to A and B, it follows from [9, Corollary 32.26] that M(A,B) + cT is surjective, that is, (M(A,B) + cT)(Sn) = Sn holds for every c > 0. Thus, T is an M-monotone operator with respect to A and B. The proof is complete. (cid:2) 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)−1 is single-valued. Proof. For any given u ∈ Sn, let x, y ∈ (M(A,B) + cT)−1(u). It follows that −M(Ax,Bx) + u ∈ Tx and −M(Ay,B y) + u ∈ T y. The monotonicity of T and M implies that

(cid:4) (cid:3) (cid:6)

0 ≤

− M(Ay,B y) + u (cid:7) ,x − y (cid:4) = −

(3.6)

− M(Ax,Bx) + u − (cid:3) M(Ay,B y) − M(Ax,Bx),x − y (cid:5) (cid:5) (cid:5) (cid:5)u0 − u1 ≤ −(α − β) ≤ 0.

From the symmetric monotonicity of M, we get that x = y. Thus (M(A,B) + cT)−1 is (cid:2) single-valued. This completes the proof. Definition 3.8. 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. The resolvant operator J M cT : Sn → Sn is defined by

J M cT (u) =

(cid:6) M(A,B) + cT (cid:7)−1(u), ∀u ∈ Sn.

(3.7)

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 : Sn → 2Sn is an M-monotone operator. Then the resolvant operator J M cT : Sn → Sn is Lipschitz-continuous with constant 1/(α − β), that is,

cT (u) − J M

cT(v)

(cid:5) (cid:5) ≤ 1 (cid:5) (cid:5)J M (cid:5)u − v(cid:5), ∀u,v ∈ Sn.

(3.8)

α − β

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

→ Sn be operators. Consider the general variational inequality problem

4. An algorithm for variational inequalities Let F,G : Sn + V I(Sn

+,F + G), defined by finding u ∈ Sn

+ such that (cid:4)

(cid:3)

.

F(u) + G(u),v − u

(4.1)

≥ 0, ∀v ∈ Sn +

We can rewrite it as the problem of finding u ∈ Sn

+ such that

0 ∈ G(u) + T(u),

(4.2)

where T ≡ F + (cid:2)(·;Sn

+). Let Sol(Sn

+,F + G) be the set of solutions of V I(Sn

+,F + G).

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Fixed Point Theory and Applications

Proposition 4.1. Let F,G : Sn → Sn be continuous and let M : Sn × Sn → Sn be a bounded, + coercive, semicontinuous, and αβ-symmetric monotone operator with respect to A : Sn → Sn and B : Sn → Sn. Then the following two properties hold for the map T ≡ F + (cid:2)(·;Sn

+):

(a) J M

+,Fcx), where Fcx(y) = M(Ay,B y) − M(Ax,Bx) +

cT (M(Ax,Bx) − cG(x)) = Sol(Sn 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

(cid:7) (cid:7) (cid:7)−1(cid:6) =

y ∈ J M cT

(cid:6) M(Ax,Bx) − cG(x) (cid:6) M(A,B) + cT

M(Ax,Bx) − cG(x)

(4.3)

is equivalent to

(cid:6) (cid:7)(cid:7) (cid:6) M(A,B) + cF + c(cid:2)

(y) + cG(x),

(4.4)

M(Ax,Bx) ∈

·;Sn +

or in other words,

(cid:6) (cid:7) . (cid:7) (cid:6) F(y) + G(x)

+ (cid:2)

(4.5)

0 ∈ M(Ay,B y) − M(Ax,Bx) + c

y;Sn +

This establishes (a).

By [10, Proposition 12.3.6], we can deduce that T is maximal monotone, it follows from Proposition 3.6, we get that T is M-monotone with respect to A and B. This com- (cid:2) pletes the proof. 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 ∈ Sn + is a solution of 0 ∈ G(u) + T(u) if and only if

u = J M cT

(cid:6) M(Au,Bu) − cG(u) (cid:7) ,

(4.6)

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

where J M

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. Let Assumption 4.3 hold and

(cid:11) (cid:8) (cid:9)

s2 − γ2

(cid:9)

(ξτ + ζt)2 − (α − β)2

s2 > γ2

.

(cid:10) (cid:10) (cid:10) (cid:10)c − (cid:10) (cid:10) (cid:10) (cid:10) ≤

,

(cid:8) (ξτ + ζt)2 − (α − β)2

(4.7)

s γ2

γ2

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(cid:7) (cid:7)(cid:5) (cid:5)

We can deduce that (cid:5) (cid:6) (cid:5)J M cT

− J M cT (cid:6) M(Ay,B y) − cG(y)

(cid:7)(cid:5) (cid:5) (cid:6) G(x) − G(y)

M(Ax,Bx) − cG(x) (cid:5) ≤ 1 (cid:5)M(Ax,Bx) − M(Ay,B y) − c α − β (cid:11)

(4.8)

(cid:5)x − y(cid:5) ≤

(ξτ + ζt)2 − 2cs + c2γ2 α − β

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

≤ (cid:5)x − y(cid:5),

which implies that J M recursion

(cid:7) (cid:6) (cid:6) M

xk

− cG

xk+1 ≡ J M cT

(4.9)

(cid:6) Axk,Bxk (cid:7)(cid:7) ,

which is desired to converge to a zero of G + T. Actually, this can be proved to be true. However, based on Lemma 4.2, we construct the following proximal point algorithm for V I(Sn

ckT (M(Axk,Bxk) − ckG(xk))(cid:5) ≤ εk.

+,F + G). Algorithm 4.5 Data. x0 ∈ Sn, c0 > 0, ε0 ≥ 0, and ρ0 > 0. Step 1. Set k = 0. Step 2. If xk ∈ Sol(Sn +,F + G), stop. Step 3. Find wk such that (cid:5)wk − J M Step 4. Set xk+1 ≡ (1 − ρk)xk + ρkwk and select ck+1, εk+1 and ρk+1. Set k ← k + 1 and go to Step 1.

The following theorem fully describes the convergence of Algorithm 4.5 for finding a

+,F + G).

k=1

(cid:12)∞

solution to V I(Sn 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 ∈ Sn be given, let {εk} ⊂ [0, ∞) satisfy E ≡ εk < ∞, {ck} ⊂ (cm, ∞), where cm > 0 and (cid:11)

(cid:8) (cid:9)

s2 − γ2

(cid:8) (cid:9)

(ξτ + ζt)2 − (α − β)2

s2 > γ2

(cid:10) (cid:10) (cid:10) (cid:10)ck − (cid:10) (cid:10) (cid:10) (cid:10) <

,

(ξτ + ζt)2 − (α − β)2

(4.10)

,

s γ2

γ2

which implies that

(cid:11) (cid:15) (cid:14) α − β −

(ξτ + ζt)2 − 2cks + c2

kγ2

(cid:13)L = (cid:11)

> 0.

(4.11)

(cid:15)2 (cid:14) α − β + 3

(ξτ + ζt)2 − 2cks + c2

kγ2

If {ρk} ⊆ [Rm,RM], where 0 < Rm ≤ RM ≤ p(cid:13)L, for all p ∈ [2,+∞), then the sequence {xk} generated by Algorithm 4.5 converges to a solution of V I(Sn

+,F + G).

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Fixed Point Theory and Applications

Proof. We introduce a new map

(cid:7) . (cid:6) M(A,B) − ckG

(4.12)

Qk ≡ I − J M ckT

ckT (M(A,B) − ckG), is also a

+), being a fixed point of J M

Clearly, any zero of G + F + (cid:2)(·;Sn zero of Qk. Now, let us prove that Qk is (cid:13)L-cocoercive.

(cid:4)

(cid:3) (cid:7) (cid:7)(cid:7) (cid:4) = − J M ckT (cid:7)

,x − y (cid:7)(cid:7)

(cid:4) (cid:6) M(Ay,B y) − ckG(y) (cid:6) G(x) − G(y) (cid:6) M(Ay,B y) − ck

,x − y

J M = (cid:5)x − y(cid:5)2 − ckT ≥ (cid:5)x − y(cid:5)2 − 1

(cid:7)(cid:5) (cid:5) (cid:5)x − y(cid:5) (cid:6) G(x) − G(y)

α − β

For x, y ∈ Sn we know that (cid:3) Qk(x) − Qk(y),x − y (cid:6) (cid:6) J M M(Ax,Bx) − ckG(x) x − y − ckT (cid:3) (cid:6) − J M M(Ax,Bx) ckT (cid:5) (cid:5)M(Ax,Bx) − M(Ay,B y) − ck (cid:11)

(ξτ + ζt)2 − 2cks + c2

kγ2(cid:5)x − y(cid:5)2

α − β

kγ2

≥ (cid:5)x − y(cid:5)2 − 1 (cid:11) (cid:16) (cid:17)

=

1 −

(cid:5)x − y(cid:5)2,

(ξτ + ζt)2 − 2cks + c2 α − β

(4.13)

(cid:5) (cid:5)2 (cid:7) (cid:7)(cid:7)(cid:5) (cid:5)2 (cid:5) (cid:5)Qk(x) − Qk(y) (cid:5) (cid:5)x − y − = − J M ckT (cid:7)(cid:4) (cid:6) M(Ay,B y) − ckG(y) (cid:6) (cid:7) M(Ay,B y) − ckG(y) − J M ckT (cid:7) (cid:7)(cid:5) (cid:5)2

+

(cid:6) (cid:6) J M M(Ax,Bx) − ckG(x) ckT (cid:6) (cid:3) x − y,J M M(Ax,Bx) − ckG(x) = (cid:5)x − y(cid:5)2 − 2 ckT (cid:5) (cid:6) (cid:6) (cid:5)J M M(Ay,B y) − ckG(y) M(Ax,Bx) − ckG(x) ckT (cid:11)

kγ2

(cid:5)x − y(cid:5)2 ≤ (cid:5)x − y(cid:5)2 + 2 − J M ckT kγ2 (ξτ + ζt)2 − 2cks + c2 α − β (cid:11)

(cid:5)x − y(cid:5)2

(ξτ + ζt)2 − 2cks + c2 α − β

kγ2

(cid:11) ⎞

= ⎠ (cid:5)x − y(cid:5)2.

+ ⎛ ⎝1 + 3

(ξτ + ζt)2 − 2cks + c2 α − β

(4.14)

Inequalities (4.13) and (4.14) imply that (cid:3)

(cid:4)

Qk(x) − Qk(y),x − y (cid:11)

−1

kγ2

kγ2

(cid:11) ⎤ ⎤ (cid:5) (cid:5)2 ⎦ ⎦ ≥ ⎡ ⎣1+3 (cid:5) (cid:5)Qk(x)−Qk(y)

(ξτ + ζt)2 − 2cks + c2 α − β

(ξτ + ζt)2 − 2cks + c2 α − β (cid:5) (cid:5)2.

⎡ ⎣1− (cid:5) (cid:5)Qk(x) − Qk(y) = (cid:13)L

(4.15)

Juhe Sun et al.

9

For all k, we denote by xk the point computed exactly by the resolvent. That is,

(cid:7)

xk+1 ≡

(cid:6) M (cid:6) xk (cid:7)(cid:7) . − ckG (cid:6) 1 − ρk (cid:6) Axk,Bxk

(4.16)

(cid:7) xk + ρkJ M ckT

For every zero x∗ of T, we obtain (cid:7)

(cid:5) (cid:5)2 (cid:5) (cid:5)xk+1 − x∗ (cid:5) (cid:5)2 = − x∗ (cid:3) (cid:7) (cid:4) (cid:7)(cid:5) (cid:5)2 = (cid:5) (cid:5)Qk (cid:6) xk

+ ρ2 k

Qk (cid:5) (cid:5)Qk

≤ (cid:7) ,xk − x∗ (cid:7)(cid:5) (cid:6) (cid:5)2 xk

(4.17)

≤

(cid:6) (cid:6) xk − Qk x∗ (cid:5) (cid:7)(cid:5) (cid:6) (cid:5)Qk (cid:5)2 + ρ2 xk k (cid:7)(cid:5) (cid:7)(cid:5) (cid:6) (cid:5)2 (cid:5)Qk xk (cid:7)(cid:5) (cid:7)(cid:5) (cid:6) (cid:5)2 (cid:5)Qk xk ≤

(cid:5) (cid:6) (cid:5)xk − ρkQk xk (cid:5) (cid:5) (cid:5)xk − x∗ (cid:5)2 − 2ρk (cid:5) (cid:5) (cid:5)xk − x∗ (cid:5)2 − 2ρk(cid:13)L (cid:5) (cid:5) (cid:6) (cid:5)2 − ρk (cid:5)xk − x∗ 2(cid:13)L − ρk (cid:5) (cid:5) (cid:6) (cid:5)xk − x∗ (cid:5)2 − Rm 2(cid:13)L − RM (cid:5) (cid:5) (cid:5)2. (cid:5)xk − x∗ ≤

(cid:5) (cid:5) (cid:5) (cid:5)xk+1 − x∗ (cid:5) (cid:5)xk+1 − xk+1

Since (cid:5)xk − xk(cid:5) ≤ ρkεk, we get that (cid:5) (cid:5) ≤

≤

(4.18)

(cid:5) (cid:5)x0 − x∗ ≤

ρiεi

(cid:5) (cid:5)x0 − x∗ ≤ (cid:5) (cid:5) (cid:5)xk+1 − x∗ (cid:5) + (cid:5) (cid:5) (cid:5)xk − x∗ (cid:5) + ρkεk k(cid:26) (cid:5) (cid:5) + i=0 (cid:5) (cid:5) + p(cid:13)LE.

Therefore, the sequence {xk} is bounded. On the other hand, we have that

(cid:7)(cid:5) (cid:5)2 (cid:5) (cid:5)xk+1 − x∗ (cid:5) (cid:5)2 = (cid:6) xk+1 − xk+1 (cid:4) (cid:5) (cid:5)2 =

(cid:5) (cid:5)2 ≤ (cid:5) (cid:5)xk+1 − xk+1

(4.19)

(cid:7) ≤

+ ρ2

(cid:5) (cid:3) (cid:5)xk+1 − xk+1 xk+1 − x∗,xk+1 − xk+1 + (cid:5) (cid:5) (cid:5) (cid:5) (cid:5) + (cid:5)xk+1 − xk+1 (cid:5) (cid:5)xk+1 − x∗ (cid:5) (cid:5) + p(cid:13)LE kε2 k

(cid:5) (cid:5)xk+1 − x∗ + (cid:5) (cid:5) (cid:5)xk+1 − x∗ (cid:5)2 + 2 (cid:5) (cid:5) (cid:5)2 + 2 (cid:5)xk+1 − x∗ (cid:5) (cid:5) (cid:5)2 + 2ρkεk (cid:5)xk − x∗ (cid:7)(cid:5) (cid:5)Qk (cid:6)(cid:5) (cid:5)x0 − x∗ (cid:7)(cid:5) (cid:6) (cid:5)2. xk − Rm (cid:6) 2(cid:13)L − RM

Letting E2 =

k < ∞, we have for every k, ε2 (cid:5) (cid:5)x0 − x∗

(cid:7) (cid:12)∞ i=0 (cid:5) (cid:5)xk+1 − x∗ (cid:5) (cid:5)2 ≤ (cid:6)(cid:5) (cid:5)x0 − x∗

(cid:7) k(cid:26)

(4.20)

(cid:5) (cid:5) + p(cid:13)LE (cid:5) (cid:6) (cid:5)Qk xk (cid:7)(cid:5) (cid:5)2. (cid:5) (cid:5)2 + 2p(cid:13)LE (cid:6) 2(cid:13)L − RM

+ p2(cid:13)L2E2 − Rm

i=0

Passing to the limit k → ∞, one has that

(cid:5)Qk(xk)(cid:5)2 < ∞, implying that (cid:7) (cid:12)∞ i=0 (cid:6) xk

Qk

= 0.

(4.21)

lim k→∞

10

Fixed Point Theory and Applications

According to Remark 3.3, for every k, there exists a unique pair (yk,vk) in gphT ckT (M(Axk,Bxk) −

such that zk = M(Axk,Bxk) − ckG(xk) = M(Ayk,B yk) + ckvk. Then J M ckG(xk)) = yk. So that Qk(xk) → 0 implies that (xk − yk) → 0, vk → 0.

Since ck is bounded away from zero, it follows that c−1

k Qk(xk) → 0. Since xk is bounded, it has at least a limit point. Let x∞ be such a limit point and assume that the subse- quence {xki : ki ∈ k} converges to x∞. It follows that {yki : ki ∈ k} also converges to x∞. For every (y,v) in gphT by the monotonicity of T, we have that (cid:3)y − yk,v − vk(cid:4) ≥ 0. Let- ting ki(∈ k) → ∞, we get that (cid:3)y − y∞,v − vk(cid:4) ≥ 0. We see that T is M-monotone due to Proposition 4.1, this implies that (x∞, −G(x∞)) ∈ gphT, that is, −G(x∞) ∈ T(x∞). This (cid:2) completes the proof.

5. Solving an approximate fixed point to J M ckT How to calculate wk at Step 3 is the key in Algorithm 4.5. If εk = 0, this amounts to the exact solution of V I(Sn

+,Fk), where Fk(x) = M(Ax,Bx) − M

(cid:7) (cid:7)(cid:7) . (cid:6) Axk,Bxk (cid:6) F(x) + G(xk

+ ck

(5.1)

ckT (M(Axk,Bxk) − ckG(xk)) is +,Fk)

+,Fk). Hence, wk is an inexact solution of the V I(Sn

+,Fk)) ≤ εk.

Now, we consider the case of εk > 0. We can prove that J M the unique solution of the V I(Sn satisfying dist(wk,Sol(Sn 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

(cid:7)(cid:7) (cid:5) (cid:5) (cid:6) wk (cid:7)(cid:5) (cid:5). (cid:6) Fk

dist

(cid:6) wk,Sol ≤ c(k)

(5.2)

(cid:6) Sn +,Fk (cid:7)nat Sn +

(cid:5) (cid:5)x − y

Proof. By Assumption 4.3, we can easily get that Fk is L(cid:17)(k)-Lipschitz-continuous and η(k)-strongly monotone, where L(cid:17)(k) = ξτ + ζt + ckL and η(k) = α − β + ckm, that is, (cid:5) (cid:5) ≤ L(cid:17)(k) (cid:4)

(cid:3)

(5.3)

.

+,Fk). We

(cid:5) (cid:5)Fk(x) − Fk(y) Fk(x) − Fk(y),x − y (cid:5) (cid:5), ≥ η(k)(cid:5)x − y(cid:5)2, ∀x, y ∈ Sn +

(cid:3) (cid:4) (cid:6) wk − r

.

+ is the natural map associated with the V I(Sn + (wk), where (Fk)nat Let r = (Fk)nat Sn Sn +(wk − Fk(wk)), that is, have that wk − r = ΠSn (cid:7) y − wk + r,Fk

(5.4)

≥ 0, ∀y ∈ Sn +

For all x∗ ∈ Sol(Sn

+,Fk) and wk − r ∈ Sn

+, we also have that (cid:6) x∗

(cid:7)(cid:4) (cid:3) wk − r − x∗,Fk ≥ 0.

(5.5)

From (5.4) and (5.5), we get that (cid:4)

(cid:7) (cid:7) (cid:4) (cid:3) (cid:6) wk (cid:6) wk − r (cid:3) x∗ − wk,Fk

r,Fk (cid:3) (cid:7)(cid:4)

(cid:4) (cid:3) (cid:7)(cid:4) (cid:3)

(5.6)

(cid:6) wk (cid:5) (cid:5)r

+

r,Fk

(cid:5) (cid:5)2 ≥ 0,

+

(cid:7)(cid:4)

x∗ − wk,Fk (cid:6) x∗ −

− (cid:6) x∗ = (cid:3) x∗ − wk,Fk − r + (cid:6) wk (cid:3) r,Fk

x∗ − wk,r (cid:7)(cid:4) ≥ 0.

(5.7)

Juhe Sun et al.

11

Adding (5.6) and (5.7), we deduce that

(cid:3) (cid:7)(cid:4) (cid:7) (cid:5) (cid:5)wk − x∗ (cid:5) (cid:5)2 ≤

η(k)

(cid:6) x∗ (cid:3) (cid:6) wk (cid:4) (cid:7) (cid:7)(cid:4) (cid:6) wk ≤ −

(5.8)

− Fk (cid:5) (cid:5) (cid:6) x∗ r,Fk (cid:5) (cid:5)wk − x∗

x∗ − wk,Fk (cid:3) wk − x∗,r (cid:5) (cid:5)wk − x∗ (cid:7)(cid:5) (cid:5)wk − x∗

− Fk − (cid:5)r(cid:5)2 + (cid:5) (cid:5) + (cid:5)r(cid:5)L(cid:17)(k) (cid:5) (cid:5) (cid:5)r(cid:5). = ≤ (cid:5)r(cid:5) (cid:6) 1 + L(cid:17)(k)

(cid:7)(cid:7) (cid:5) (cid:5) (cid:6) Fk

dist

(cid:6) wk,Sol ≤ c(k) (cid:7)(cid:5) (cid:5),

(5.9)

(cid:6) Sn +,Fk

Hence, (cid:5)wk − x∗(cid:5) ≤ η(k)−1(1 + L(cid:17)(k))(cid:5)r(cid:5). This implies that (cid:6) (cid:7)nat wk Sn +

(cid:2)

where c(k) = η(k)−1(1 + L(cid:17)(k)). This completes the proof.

Consequently, the computation of wk can be accomplished by obtaining an inexact

solution of V I(Sn

+,Fk) satisfying the residual condition (cid:6) Fk

(cid:5) (cid:5) (cid:6) wk

εk.

(5.10)

+(·) is directionally differentiable and strongly semismooth Sn

(cid:7)nat Sn + (cid:7)(cid:5) (cid:5) ≤ 1 c(k) (cid:27)

We note that the operator everywhere (see, e.g., [12]). If Fk(·) is continuously differentiable, then we get that

(cid:28) (cid:7) (cid:6) Fk

(w) = w −

(cid:6) w − Fk(w)

(5.11)

Sn +

(cid:7)nat Sn +

is directionally differentiable.

In what follows, we present the following path Newton method for solving the equa-

+ (wk) = 0. tion (Fk)nat Sn

Algorithm 5.2 Data. w0 ∈ Sn, γ ∈ (0,1), and ρ ∈ (0,1). Step 1. Set j = 0. + (w j) = 0, stop. Step 2. If (Fk)nat Sn

(cid:7)(cid:7)(cid:5) (cid:6) (cid:5) (cid:5) (cid:7)(cid:5) (cid:5) (cid:6) w j (cid:7)(cid:5) (cid:5).

p j

(cid:5) ≤ (cid:6) Fk (cid:6) Fk (cid:6) ρi ¯τ j (cid:6) 1 − γρi ¯τ j

(5.12)

+ (w j)] and consider the corresponding path p j(·) = Step 3. Select an element V j ∈ ∂[(Fk)nat Sn w j − (·)V −1 + (w j) with domain I j = [0, ¯τ j) for some ¯τ j ∈ (0,1]. Find the smallest non- j (Fk)nat Sn negative integer i j such that with i = i j, ρi ¯τ j ∈ I j and (cid:7)nat Sn +

(cid:7)nat Sn +

Step 4. Set τ j = ρi j ¯τ j, w j+1 = p j(τ j), and j ← j + 1; go to Step 2. Theorem 5.3. Let F, G, M, A, and B satisfy all the conditions of Assumption 4.3. If for all w ∈ Sn + (w)] is nonsingular, then the sequence {w j } generated + every matrix in ∂[(Fk)nat Sn by Algorithm 5.2 has at least one accumulation point and every accumulation point of the sequence {w j } is the zero point of (Fk)nat + . Sn

12

Fixed Point Theory and Applications

+ (w j)(cid:5)} is monotonically decreasing; thus it is Proof. The nonnegative sequence {(cid:5)(Fk)nat Sn bounded. It follows from Lemma 5.1 that the sequence {w j } is bounded. That implies that {w j } has at least one accumulation point.

(cid:7)

Assume that there exists a subsequence {w jm } of {w j } converging to w∗ such that (cid:6) w∗

(cid:6) Fk (cid:18)= 0,

(5.13)

(cid:7)nat Sn +

(cid:5) (cid:5) (cid:6) w jm (cid:7) .

that is, there exist positive constants δ and η satisfying (cid:6) Fk

(cid:7)(cid:5) (cid:5) ≥ η, ∀w jm ∈ B (cid:6) w∗,δ

(5.14)

(cid:7)nat Sn +

(cid:6) (cid:5)h(cid:5)2

By the strong semismoothness of (Fk)nat + , we have that Sn (cid:6) Fk

(cid:6) Fk

(w) − V (h) = O

(w + h) −

(cid:7) , ∀w ∈ B (cid:6) w∗,δ

(5.15)

(cid:7) , h ∈ Sn +, (cid:7)nat Sn + (cid:7)nat Sn +

(cid:31) (cid:5) (cid:5)

where V ∈ ∂(Fk)nat + (w) is nonsingular and there is a positive constant (cid:29)c such that Sn (cid:5) (cid:5)V −1

≤ (cid:29)c. (cid:30) (cid:5)V (cid:5),

max

(5.16)

sup w∈B(w∗,δ)

(w)

V ∈∂(Fk)nat Sn +

(cid:6) (cid:7) (cid:6) (cid:7) (cid:6) w jm − O − = (cid:6) Fk (cid:6) Fk

V jm

p jm(τ) − w jm

p jm(τ)

(5.17)

+ (w jm)], we have that Letting τ ∈ (0, ¯τ jm), h = p jm(τ) − w jm, and V jm ∈ ∂[(Fk)nat Sn (cid:5) (cid:6)(cid:5) (cid:5)2(cid:7) (cid:7) (cid:7)nat (cid:5)p jm(τ) − w jm . Sn +

(cid:7)nat Sn +

It follows that (cid:6)

(cid:7) (cid:6) (cid:7) (cid:6) (cid:5) (cid:5) (cid:7)(cid:5) (cid:5)

w jm

−

O

(cid:6) Fk (cid:6) Fk

p jm(τ)

p jm(τ) − w jm

(cid:7)nat Sn + (cid:7)nat Sn +

.

= (cid:5) (cid:5) − V jm (cid:5) (cid:5)p jm(τ) − w jm (cid:5) (cid:6)(cid:5) (cid:5)2(cid:7) (cid:5)p jm(τ) − w jm (cid:5) (cid:5) (cid:5) (cid:5)p jm(τ) − w jm

(5.18)

So we can choose a positive t∗ small enough so that

(cid:9)

.

(5.19)

(cid:6) 0,t∗

, ∀t ∈

o(t) t

≤ 1 − γ (cid:29)c

(cid:7) (cid:9)

.

(cid:6) 0,τ ∗

(5.20)

, ∀τ ∈

(cid:5) ≤ 1 − γ (cid:29)c

From the definition of p jm, we know that there exists a constant τ ∗ ∈ (0, ¯τ jm] small enough so that (cid:5)p jm(τ) − w jm (cid:5) ≤ t∗, for all τ ∈ (0,τ ∗], which implies that (cid:5) (cid:6)(cid:5) (cid:5) (cid:5)p jm(τ) − w jm o (cid:5) (cid:5) (cid:5)p jm(τ) − w jm

(cid:7) (cid:6) (cid:6) (cid:5) (cid:5) (cid:7)(cid:5) (cid:5)≤ (cid:6) Fk (cid:6) Fk

p jm(τ)

(cid:7)nat Sn +

p jm(τ) − w jm (cid:5) (cid:5) o

(5.21)

(cid:6) w jm (cid:6) Fk ≤(1 − τ) (cid:7)(cid:5) (cid:5) 1 − γ (cid:29)c (cid:6) w jm (cid:6) w jm

It follows from (5.16), (5.18), (5.19), and (5.20) that (cid:7)(cid:5) (cid:5) (cid:6) (cid:7)nat (cid:5) (cid:5)V jm w jm + Sn (cid:5) (cid:6)(cid:5) (cid:7) + (cid:5) (cid:5)p jm(τ) − w jm (cid:5) (cid:5)p jm(τ) − w jm (cid:5) (cid:5) + (cid:5)p jm(τ) − w jm (cid:5) (cid:5) (cid:5) (cid:5) (cid:5) (cid:7)(cid:5) (cid:6) (cid:7)nat (cid:5) (cid:5) (cid:5)V −1 (cid:5) (cid:5) + τ Fk Sn jm + (cid:5) (cid:7)(cid:5) (cid:6) (cid:5) (cid:5), ∀τ ∈ (0,τ ∗]. Fk

≤ (1 − τγ) (cid:7)nat Sn + (cid:7)nat Sn +

Juhe Sun et al.

13

(cid:6)

Consequently, (cid:5) (cid:5)

(cid:6) w jm (cid:6) Fk (cid:5) (cid:6) (cid:5) Fk

p jm(τ)

(cid:7)(cid:5) (cid:5) ≤ (1 − τγ) (cid:7)(cid:5) (cid:5), ∀τ ∈ (0,τ ∗].

(5.22)

(cid:7)nat Sn + (cid:7)nat Sn +

By the definition of the step-size τ jm, it follows that there exists ξ ∈ (0,τ ∗) such that τ jm ≥ ξ for all jm. Indeed, if no such ξ exists, then {τ jm } converges to zero. This implies that the sequence of integers {i jm } is unbounded. Consequently, by the definition of i jm, we have, for all jm sufficiently large, (cid:6)

(cid:7)(cid:7)(cid:5) (cid:5) (cid:5) (cid:7)(cid:5) (cid:5)

pk

(cid:5) > (cid:6) wk (cid:6) Fk (cid:6) Fk (cid:6) ρi jm −1 ¯τk (cid:6) 1 − γρi jm −1 ¯τk (cid:7)(cid:5) (cid:5);

(5.23)

(cid:7)nat Sn + (cid:7)nat Sn +

but this contradicts (5.22) with τ ≡ ρi jm −1 ¯τ jm. Consequently, the desired ξ exists. The in- equality (5.22) implies that

(cid:5) (cid:5) (cid:7)(cid:5) (cid:5) (cid:6) w jm+1 (cid:6) w jm (cid:7)(cid:5) (cid:5). (cid:7)(cid:5) (cid:5) ≤ (cid:6) Fk (cid:6) Fk (cid:6) 1 − τ jmγ

(5.24)

+ (w jm)(cid:5) ≥ Passing to the limit m → ∞, we deduce a contradiction because limm→∞ (cid:5)(Fk)nat Sn η > 0 and the sequence {τ jm } is bounded away from zero. This yields that (Fk)nat + (w∗) = 0. Sn (cid:2) This completes the proof.

(cid:7)nat Sn + (cid:7)nat Sn +

Remark 5.4. As stated above, Algorithm 5.2 generates a sequence converging to the zero point of (Fk)nat + , Step 3 in Algorithm 4.5 is implementable. Obviously, Algorithm 5.2 stops Sn within a finite number of iterations at a wk such that (5.10) holds. Example 5.5. Assume that there exists a positive constant ¯c such that

(cid:5) (cid:5) ≤ ¯c.

sup

(cid:5) (cid:5)V JF(w)

(5.25)

(cid:27)

V ∈∂

(xk −ck(F(w)+G(xk)))

sup w∈Sn +

Sn +

(cid:27)

Let ck ∈ (0,1/ ¯c). Suppose that M(Aw,Bw) = w, for all w ∈ Sn + and F is Lipschitz-continu- +(xk − ck(F(w) + G(xk))), for + (w) = w − ous and strongly monotone. We have (Fk)nat Sn Sn (cid:27) all w ∈ Sn +(xk − ck(F(w) + G(xk)))}, + (w)] ⊂ {I − ckV JF(w) | V ∈ ∂ +. Then ∂[(Fk)nat Sn Sn for all w ∈ Sn + (w)] is nonsingular for all +. We easily get that every matrix in ∂[(Fk)nat Sn w ∈ Sn +. It follows from Theorem 5.3 that every accumulation point of {wk} generated by Algorithm 5.2 is the zero point of (Fk)nat + . Sn

At first sight, the M-monotonicity of T = F + (cid:2)(·,Sn

+) seems having little use be- cause the algorithm based on maximal monotonicity can also solve the V I(Sn +,F + G) directly. However, we will see that in some practical cases the variational inequality us- ing Algorithm 4.5, which is based on M-monotone operator, is actually much simpler to solve and easier to analyze than using algorithm based on maximal monotone map. We illustrate this by the following example. Example 5.6. Let F : Sn +

→ Sn be defined by

(5.26)

F(x) = S(x) +

x ∀x ∈ Sn +,

1 16

x, G(x) = 1 8

→ Sn is s-Lipschitz-continuous and monotone with (cid:3)S(x),x(cid:4) ≥ −∞.

where S : Sn +

14

Fixed Point Theory and Applications

We have F is (s + (1/16))-Lipschitz-continuous, (1/16)-strongly monotone, and G is

(1/8)-Lipschitz-continuous.

Now, we take M(Ax,Bx) = Ax + Bx, where Ax = (1 + (ck/16))x and Bx = −ckS(x) − (ck/8)x for all x ∈ Sn and 0 < ck < 16. Then, we can easily prove that M(·, ·) is Lipschitz- continuous with first and second arguments, M(A,B) is bounded and semicontinuous; and A and B are both Lipschitz-continuous. It is also easy to see that (cid:4)

(cid:7)(cid:7) (cid:3) (cid:4) − ckS(x) + (cid:6) ck/16

x,x

= +∞,

(5.27)

lim (cid:5)x(cid:5)→∞

= lim (cid:5)x(cid:5)→∞ (cid:3) M(Ax,Bx),x (cid:5)x(cid:5) (cid:6) 1 − (cid:5)x(cid:5)

which implies that M(A,B) is coercive. Also, we can deduce that M(A,B) is (1 + (ck/16))- strongly monotone with respect to A and ck(s + (1/8))-relaxed monotone with respect to B and (1 + (ck/16)) > ck(s + (1/8)), if we let s < (1/ck) − (1/16). Also, we can prove that G is strongly monotone with respect to M(A,B).

We choose x0 ∈ Sn

+, {εk}, {ck}, and {ρk} satisfying Theorem 4.6 and compute {wk} by

the residual rule

(cid:28) (cid:7) (cid:7) (cid:7) (cid:7)(cid:7)(cid:7) (cid:5) (cid:5) (cid:6) wk (cid:6) wk − M (cid:6) F (cid:6) wk (cid:7)(cid:5) (cid:5) = (cid:6) Fk − ck (cid:6) xk (cid:5) (cid:5) (cid:5) (cid:5)

+ M

+ G

Sn + (cid:28)

(cid:7)nat S+ (cid:6) Awk,Bwk ! (cid:7) = (cid:6) − ckS

xk

xk

(cid:5) (cid:5) (cid:5) (cid:5)wk − (cid:5) (cid:5) (cid:5) (cid:5)wk − (cid:6) Axk,Bxk !(cid:5) (cid:5) (cid:5) (cid:5)

+

1 − 3

ck 16

Sn +

≤

εk,

η(k) 1 + L(cid:17)(k)

(5.28)

! (cid:28) (cid:7)

xk

εk.

that is, wk can be computed as follows: (cid:5) (cid:5) (cid:5) (cid:5)wk −

!(cid:5) (cid:5) (cid:5) (cid:5) ≤ − ckS(xk

+

1 − 3

(5.29)

ck 16

η(k) 1 + L(cid:17)(k)

Sn +

It follows from Theorem 4.6 that the sequence {xk} generated by Algorithm 4.5 converges to a solution of

" #

.

S(x) +

(5.30)

x +

x, y − x

≥ 0, ∀y ∈ Sn +

1 16

1 8

Note that the core of proximal point algorithm is the calculation of wk. As we have seen, if we use [10, Algorithm 12.3.8], which is based on the maximal monotonicity of T = F + (cid:2)(·,Sn

+), wk will be computed as

(cid:28) (cid:7)

xk

− (cid:6) wk xk − ckS (cid:5) (cid:5) (cid:5) (cid:5)wk − !(cid:5) (cid:5) (cid:5) (cid:5) ≤

εk,

(5.31)

ck 8

η(k) 1 + L(cid:17)(k)

Sn +

which is more complicated to solve than (5.29). This example verifies the above com- ments.

Juhe Sun et al.

15

Acknowledgments

This work was supported by the National Natural Science Foundation of China under project Grant no. 10771026 and by the Scientific Research Foundation for the Returned Overseas Chinese Scholars, State Education Ministry.

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Juhe Sun: Department of Applied Mathematics, Dalian University of Technology, Dalian, Liaoning 116024, China Email address: juhesun@163.com

Shaowu Zhang: Department of Applied Mathematics, Dalian University of Technology, Dalian, Liaoning 116024, China Email address: zhangsw@dlut.edu.cn

Liwei Zhang: Department of Applied Mathematics, Dalian University of Technology, Dalian, Liaoning 116024, China Email address: lwzhang@dlut.edu.cn