• 沒有找到結果。

for the second-order cone complementarity problem

N/A
N/A
Protected

Academic year: 2022

Share "for the second-order cone complementarity problem"

Copied!
26
0
0

加載中.... (立即查看全文)

全文

(1)

DOI 10.1007/s10589-008-9182-9

A one-parametric class of merit functions

for the second-order cone complementarity problem

Jein-Shan Chen· Shaohua Pan

Received: 11 January 2007 / Revised: 21 April 2008 / Published online: 8 May 2008

© Springer Science+Business Media, LLC 2008

Abstract We investigate a one-parametric class of merit functions for the second- order cone complementarity problem (SOCCP) which is closely related to the popular Fischer–Burmeister (FB) merit function and natural residual merit function. In fact, it will reduce to the FB merit function if the involved parameter τ equals 2, whereas as τtends to zero, its limit will become a multiple of the natural residual merit function.

In this paper, we show that this class of merit functions enjoys several favorable prop- erties as the FB merit function holds, for example, the smoothness. These properties play an important role in the reformulation method of an unconstrained minimization or a nonsmooth system of equations for the SOCCP. Numerical results are reported for some convex second-order cone programs (SOCPs) by solving the unconstrained minimization reformulation of the KKT optimality conditions, which indicate that the FB merit function is not the best. For the sparse linear SOCPs, the merit function corresponding to τ= 2.5 or 3 works better than the FB merit function, whereas for the dense convex SOCPs, the merit function with τ= 0.1, 0.5 or 1.0 seems to have better numerical performance.

Keywords Second-order cone· Complementarity · Merit function · Jordan product

J.-S. Chen (



)

Department of Mathematics, National Taiwan Normal University, 88 Section 4, Ting-Chou Road, 11677 Taipei, Taiwan

e-mail:jschen@math.ntnu.edu.tw J.-S. Chen

Mathematics Division, National Center for Theoretical Sciences, Taipei Office, Taipei, Taiwain S. Pan

School of Mathematical Sciences, South China University of Technology, Guangzhou 510640, China e-mail:shhpan@scut.edu.cn

(2)

1 Introduction

We consider the conic complementarity problem of finding a vector ζ∈ Rnsuch that F (ζ )∈ K, G(ζ )∈ K, F (ζ ), G(ζ ) = 0, (1) where·, · is the Euclidean inner product, F : Rn→ Rn and G: Rn→ Rn are the mappings assumed to be continuously differentiable throughout this paper, andK is the Cartesian product of second-order cones (SOCs). In other words,

K = Kn1× Kn2× · · · × KnN, (2) where N, n1, . . . , nN≥ 1, n1+ · · · + nN= n, and

Kni:=

(x1, x2)∈ R × Rni−1| x2 ≤ x1

,

with ·  denoting the Euclidean norm and K1 denoting the set of nonnegative re- alsR+. We will refer to (1)–(2) as the second-order cone complementarity problem (SOCCP).

An important special case of the SOCCP corresponds to G(ζ )= ζ for all ζ ∈ Rn. Then (1) reduces to

F (ζ )∈ K, ζ ∈ K, F (ζ ), ζ = 0, (3) which is a natural extension of the nonlinear complementarity problem (NCP) [7,8]

withK = Rn+, the nonnegative orthant cone ofRn. Another important special case corresponds to the KKT optimality conditions of the convex second-order cone pro- gram (CSOCP):

minimize g(x)

subject to Ax= b, x ∈ K, (4)

where g: Rn→ R is a convex twice continuously differentiable function, A ∈ Rm×n has full row rank and b∈ Rm. From [4], we know that the KKT conditions of (4), which are sufficient but not necessary for optimality, can be reformulated as (1) with

F (ζ ):= ¯x +

I− AT(AAT)−1A

ζ, G(ζ ):= ∇g(F (ζ )) − AT(AAT)−1Aζ, (5) where ¯x ∈ Rn is any point such that A¯x = b. When g is linear, the CSOCP reduces to the linear SOCP which arises in numerous applications in engineering design, finance, robust optimization, and includes as special cases convex quadratically con- strained quadratic programs and linear programs; see [1,13] and references therein.

There have been various methods proposed for solving SOCPs and SOCCPs. They include the interior-point methods [2,3,15,16,19], the non-interior smoothing New- ton methods [6,9], and the smoothing-regularization method [11]. Recently, there was an alternative method [4] based on reformulating the SOCCP as an unconstrained minimization problem. In that approach, it aims to find a function ψ: Rn×Rn→ R+ satisfying

ψ (x, y)= 0 ⇐⇒ x ∈ K, y ∈ K, x, y = 0, (6)

(3)

so that the SOCCP can be reformulated as an unconstrained minimization problem min

ζ∈Rnf (ζ ):= ψ(F (ζ ), G(ζ )).

We call such ψ a merit function associated with the coneK.

A popular choice of ψ is the Fischer–Burmeister (FB) merit function ψFB(x, y):=1

2FB(x, y)2, (7)

where φFB: Rn× Rn→ Rnis the vector-valued FB function defined by

φFB(x, y):= (x2+ y2)1/2− (x + y), (8) with x2= x ◦ x denoting the Jordan product between x and itself, x1/2being a vector such that (x1/2)2= x, and x + y meaning the usual componentwise addition of vec- tors. The function ψFB was studied in [4] and particularly shown to be continuously differentiable (smooth). Another popular choice of ψ is the natural residual merit function

ψNR(x, y):=1

2NR(x, y)2,

where φNR: Rn× Rn→ Rnis the vector-valued natural residual function given by φNR(x, y):= x − (x − y)+

with (·)+meaning the projection in the Euclidean norm onto K. The function φNR

was studied in [9,11] which is involved in smoothing methods for the SOCCP. Com- pared with the FB merit function ψFB, the function ψNR has a drawback, i.e., its non-differentiability.

In this paper, we will investigate the following one-parametric class of functions ψτ(x, y) := 1

2τ(x, y)2, (9)

where τ is a fixed parameter from (0, 4) and φτ: Rn× Rn→ Rnis defined by φτ(x, y):=

(x− y)2+ τ(x ◦ y)1/2

− (x + y). (10)

Specifically, we prove that ψτ is a merit function associated withK which is continu- ously differentiable everywhere with computable gradient formulas (see Propositions 3.1–3.3), and hence the SOCCP can be reformulated as an unconstrained smooth minimization

ζmin∈Rnfτ(ζ ):= ψτ(F (ζ ), G(ζ )). (11) Also, we show that every stationary point of fτ solves the SOCCP under the condi- tion that∇F and −∇G are column monotone (see Proposition4.1). Observe that φτ

reduces to φFB when τ= 2, whereas its limit as τ → 0 becomes a multiple of φNR.

(4)

Thus, this class of merit functions has a close relation to two of the most important merit functions so that a closer look and study for it is worthwhile. In addition, this study is motivated by the work [12] where φτ was used to develop a nonsmooth Newton method for the NCP. This paper is mainly concerned with the merit function approach based on the unconstrained minimization problem (11). Numerical results are also reported for some convex SOCPs, which indicate that ψτ can be an alterna- tive for ψFBif a suitable τ is selected.

Throughout this paper,Rn denotes the space of n-dimensional real column vec- tors, andRn1× · · · × Rnm is identified withRn1+···+nm. Thus, (x1, . . . , xm)∈ Rn1×

· · · × Rnm is viewed as a column vector inRn1+···+nm. The notation I denotes an identity matrix of suitable dimension, and int(Kn)denotes the interior ofKn. For any differentiable mapping F: Rn→ Rm,∇F (x) ∈ Rn×mdenotes the transposed Jaco- bian of F at x. For a symmetric matrix A, we write A O (respectively, A  O) to mean A is positive semidefinite (respectively, positive definite). For nonnegative αand β, we write α= O(β) to mean α ≤ Cβ, with C > 0 independent of α and β.

Without loss of generality, in the rest of this paper we assume thatK = Kn(n >1).

All analysis can be carried over to the general case whereK has the structure as (2).

In addition, we always assume that τ satisfies 0 < τ < 4.

2 Preliminaries

It is known thatKn(n >1) is a closed convex self-dual cone with nonempty interior int(Kn):=

x= (x1, x2)∈ R × Rn−1| x1>x2 .

For any x= (x1, x2), y= (y1, y2)∈ R × Rn−1, the Jordan product of x and y is defined by

x◦ y := (x, y, y1x2+ x1y2). (12) The Jordan product, unlike scalar or matrix multiplication, is not associative, which is a main source on complication in the analysis of SOCCP. The identity element under this product is e:= (1, 0, . . . , 0)T∈ Rn. For any x= (x1, x2)∈ R × Rn−1, the determinant of x is defined by det(x):= x12−x22. If det(x)= 0, then x is said to be invertible. If x is invertible, there exists a unique y= (y1, y2)∈ R × Rn−1satisfying x◦ y = y ◦ x = e. We call this y the inverse of x and denote it by x−1. For each x= (x1, x2)∈ R × Rn−1, let

Lx:=

x1 x2T x2 x1I

. (13)

It is easily verified that Lxy = x ◦ y and Lx+y= Lx+ Ly for any x, y∈ Rn, but generally L2x= LxLx= Lx2 and L−1x = Lx−1. If Lxis invertible, then the inverse of Lxis given by

L−1x = 1 det(x)

x1 −x2T

−x2

det(x) x1 I+ 1

x1x2x2T

⎦ . (14)

(5)

We next recall from [9] that each x= (x1, x2)∈ R × Rn−1admits a spectral fac- torization, associated withKn, of the form

x= λ1(x)· u(1)x + λ2(x)· u(2)x ,

where λ1(x), λ2(x)and u(1)x , u(2)x are the spectral values and the associated spectral vectors of x given by

λi(x)= x1+ (−1)ix2, u(i)x =1

2(1, (−1)i¯x2) for i= 1, 2,

with ¯x2= xx22 if x2= 0, and otherwise ¯x2 being any vector in Rn−1 such that

 ¯x2 = 1. If x2= 0, the factorization is unique. The spectral factorization of x has various interesting properties; see [9]. We list three properties that will be used later.

Property 2.1

(a) x2= λ21(x)· u(1)x + λ22(x)· u(2)x ∈ Knfor any x∈ Rn. (b) If x∈ Kn, then x1/2=√

λ1(x)· u(1)x +√

λ2(x)· u(2)x ∈ Kn.

(c) x∈ Kn⇐⇒ λ1(x)≥ 0 ⇐⇒ Lx O, x ∈ int(Kn)⇐⇒ λ1(x) >0⇐⇒ Lx O.

3 Smoothness of the function ψτ

In this section we will show that ψτ defined by (9) is a smooth merit function. First, by Properties2.1(a) and (b), φτ and ψτ are well-defined since for any x, y∈ Rn, we can verify that

(x− y)2+ τ(x ◦ y) =



x+τ− 2 2 y

2

+τ (4− τ) 4 y2

=



y+τ− 2 2 x

2

+τ (4− τ)

4 x2∈ Kn. (15) The following proposition shows that ψτ is indeed a merit function associated withKn.

Proposition 3.1 Let ψτ and φτbe given as in (9) and (10), respectively. Then,

ψτ(x, y)= 0 ⇐⇒ φτ(x, y)= 0 ⇐⇒ x ∈ Kn, y∈ Kn, x, y = 0.

Proof The first equivalence is clear by the definition of ψτ. We consider the second one.

⇐”. Since x ∈ K, y ∈ K and x, y = 0, we have x ◦ y = 0. Substituting it into the expression of φτ(x, y)then yields that φτ(x, y)= (x2+ y2)1/2− (x + y) = φFB(x, y). From Proposition 2.1 of [9], we immediately obtain φτ(x, y)= 0.

“⇒”. Suppose that φτ(x, y)= 0. Then, x + y =

(x− y)2+ τ(x ◦ y)1/2

. Squar- ing both sides yields x◦ y = 0. This implies that x + y = (x2+ y2)1/2, i.e.

(6)

φFB(x, y)= 0. From Proposition 2.1 of [9], it then follows that x∈ Kn, y∈ Knand

x, y = 0. 

In what follows, we focus on the proof of the smoothness of ψτ. We first introduce some notation that will be used in the sequel. For any x= (x1, x2), y= (y1, y2)∈ R × Rn−1, let

w= (w1, w2)= w(x, y) := (x − y)2+ τ(x ◦ y),

(16) z= (z1, z2)= z(x, y) :=

(x− y)2+ τ(x ◦ y)1/2

.

Then, w∈ Knand z∈ Kn. Moreover, by the definition of Jordan product, w1= w1(x, y)= x2+ y2+ (τ − 2)xTy,

(17) w2= w2(x, y)= 2(x1x2+ y1y2)+ (τ − 2)(x1y2+ y1x2).

Let λ1(w)and λ2(w)be the spectral values of w. By Property2.1(b), we have that z1= z1(x, y)=

λ2(w)+√ λ1(w)

2 ,

(18) z2= z2(x, y)=

λ2(w)−√ λ1(w)

2 ¯w2,

where ¯w2:= ww22 if w2= 0 and otherwise ¯w2 is any vector in Rn−1 satisfying

 ¯w2 = 1.

The following technical lemma describes the behavior of x, y when w= (x − y)2+ τ(x ◦ y) is on the boundary of Kn. In fact, it may be viewed as an extension of [4, Lemma 3.2].

Lemma 3.1 For any x= (x1, x2), y= (y1, y2)∈ R × Rn−1, if w /∈ int(Kn), then x12= x22, y12= y22, x1y1= x2Ty2, x1y2= y1x2; (19) x12+ y12+ (τ − 2)x1y1= x1x2+ y1y2+ (τ − 2)x1y2

= x22+ y22+ (τ − 2)x2Ty2. (20) If, in addition, (x, y)= (0, 0), then w2= 0, and furthermore,

xT2 w2

w2= x1, x1 w2

w2 = x2, y2T w2

w2= y1, y1 w2

w2 = y2. (21) Proof Since w= (x −y)2+τ(x ◦y) /∈ int(Kn), using (15) and [4, Lemma 3.2] yields



x1+τ− 2 2 y1

2

=

x2+τ− 2 2 y2

2, y12= y22,



x1+τ− 2 2 y1

 y2=



x2+τ− 2 2 y2

 y1,

(7)



x1+τ− 2 2 y1

 y1=



x2+τ− 2 2 y2

T

y2;



y1+τ− 2 2 x1

2

=

y2+τ− 2 2 x2

2, x12= x22,



y1+τ− 2 2 x1

 x2=



y2+τ− 2 2 x2

 x1,



y1+τ− 2 2 x1

 x1=



y2+τ− 2 2 x2

T

x2.

From these equalities, we readily get the results in (19). Since w ∈ Kn but w /int(Kn), we havex2+ y2+ (τ − 2)xTy= 2x1x2+ 2y1y2+ (τ − 2)(x1y2+ y1x2) by λ1(w)= 0. Applying the relations in (19) then gives the equalities in (20).

If, in addition, (x, y)= (0, 0), then it is clear that x1x2+ y1y2+ (τ − 2)x1y2 = x12+ y12+ (τ − 2)x1y1= 0. To prove the equalities in (21), it suffices to verify that x2Tww2

2= x1and x1 w2

w2= x2 by the symmetry of x and y in w. The verifications

are straightforward by (20) and x1y2= y1x2. 

By Lemma3.1, when w /∈ int(Kn), the spectral values of w are calculated as fol- lows:

λ1(w)= 0, λ2(w)= 4

x12+ y12+ (τ − 2)x1y1

. (22)

If (x, y)= (0, 0) also holds, then using (18), (20) and (22) yields that

z1(x, y)=

x12+ y12+ (τ − 2)x1y1, z2(x, y)=x1x2+ y1y2+ (τ − 2)x1y2



x12+ y12+ (τ − 2)x1y1

.

Thus, if (x, y)= (0, 0) and (x − y)2+ τ(x ◦ y) /∈ int(Kn), φτ(x, y)can be rewritten as

φτ(x, y)= z(x, y) − (x + y) =

⎜⎝



x12+ y12+ (τ − 2)x1y1− (x1+ y1)

x1x2+y1y2+(τ−2)x1y2

x12+y12+(τ−2)x1y1

− (x2+ y2)

⎠ . (23)

This specific expression will be employed in the proof of the following main result.

Proposition 3.2 The function ψτgiven by (9) is differentiable at every (x, y)∈ Rn× Rn. Moreover,xψτ(0, 0)= ∇yψτ(0, 0)= 0; if (x − y)2+ τ(x ◦ y) ∈ int(Kn), then

xψτ(x, y)= Lx+τ−2

2 yL−1z − I

φτ(x, y),

yψτ(x, y)= (24) Ly+τ−2

2 xL−1z − I

φτ(x, y);

(8)

if (x, y)= (0, 0) and (x − y)2+ τ(x ◦ y) ∈ int(Kn), then x12+ y12+ (τ − 2)x1y1= 0 and

xψτ(x, y)=

 x1+τ−22 y1



x12+ y12+ (τ − 2)x1y1

− 1



φτ(x, y),

(25)

yψτ(x, y)=

 y1+τ−22 x1



x12+ y12+ (τ − 2)x1y1

− 1



φτ(x, y).

Proof Case (1): (x, y)= (0, 0). For any u = (u1, u2), v= (v1, v2)∈ R × Rn−1, let μ1, μ2 be the spectral values of (u− v)2+ τ(u ◦ v) and ξ(1), ξ(2) be the spectral vectors. Then,

2

ψτ(u, v)− ψτ(0, 0)

=u2+ v2+ (τ − 2)(u ◦ v)1/2

− u − v2

=√μ1ξ(1)+√

μ2ξ(2)− u − v2

≤

2+ u + v2

.

In addition, from the definition of spectral value, it follows that

μ2= u2+ v2+ (τ − 2)uTv+ 2(u1u2+ v1v2)+ (τ − 2)(u1v2+ v1u2)

≤ 2u2+ 2v2+ 3|τ − 2|uv ≤ 5(u2+ v2).

Now combining the last two equations, we have ψτ(u, v)− ψτ(0, 0)= O(u2+

v2). This shows that ψτ is differentiable at (0, 0) withxψτ(0, 0) =

yψτ(0, 0)= 0.

Case (2): (x− y)2+ τ(x ◦ y) ∈ int(Kn). By [5, Proposition 5], z(x, y) defined by (18) is continuously differentiable at such (x, y), and consequently φτ(x, y)is also continuously differentiable at such (x, y) since φτ(x, y)= z(x, y) − (x + y). Notice that

z2(x, y)=



x+τ − 2 2 y

2

+τ (4− τ) 4 y2, which leads to ∇xz(x, y)Lz = Lx+τ−2

2 y by taking differentiation on both sides about x. Since Lz O by Property2.1(c), it follows that∇xz(x, y)= Lx+τ−2

2 yL−1z . Consequently,

xφτ(x, y)= ∇xz(x, y)− I = Lx+τ−2

2 yL−1z − I.

This together with∇xψτ(x, y)= ∇xφτ(x, y)φτ(x, y)proves the first formula of (24).

For the symmetry of x and y in ψτ, the second formula also holds.

Case (3): (x, y)= (0, 0) and (x − y)2+ τ(x ◦ y) /∈ int(Kn). For any x = (x1, x2), y= (y1, y2)∈ R × Rn−1, it is easy to verify that

τ(x, y)=x2+ y2+ (τ − 2)(x◦ y)1/22+ x+ y2

(9)

− 2

x2+ y2+ (τ − 2)(x◦ y)1/2

, x+ y

= x2+ y2+ (τ − 2)x, y + x+ y2

− 2

x2+ y2+ (τ − 2)(x◦ y)1/2

, x+ y ,

where the second equality uses the fact that z2= z2, e for any z ∈ Rn. Since

x2+ y2+ (τ − 2)x, y + x + y2 is clearly differentiable in (x, y), it suffices to show that [x2+ y2 + (τ − 2)(x ◦ y)]1/2, x + y is differen- tiable at (x, y)= (x, y). By Lemma3.1, w2= w2(x, y)= 0, which implies w2 = w2(x, y)= 2x1x2+2y1y2+(τ −2)(x1y2+y1x2)= 0 for all (x, y)∈ Rn×Rnsuf- ficiently near to (x, y). Let μ1, μ2be the spectral values of x2+y2+(τ −2)(x◦y).

Then we can compute that 2

x2+ y2+ (τ − 2)(x◦ y)1/2

, x+ y

=√ μ2



x1+ y1 +[2(x1x2 + y1y2)+ (τ − 2)(x1y2 + y1x2)]T(x2 + y2)

2(x1x2 + y1y2)+ (τ − 2)(x1y2 + y1x2)



+√ μ1



x1+ y1[2(x1x2+ y1y2)+ (τ − 2)(x1y2+ y1x2)]T(x2+ y2)

2(x1x2+ y1y2)+ (τ − 2)(x1y2+ y1x2)

 .

(26) Since λ2(w) >0 and w2(x, y)= 0, the first term on the right-hand side of (26) is differentiable at (x, y)= (x, y). Now, we claim that the second term is o(xx + y− y), i.e., it is differentiable at (x, y) with zero gradient. To see this, no- tice that w2(x, y)= 0, and hence μ1= x2+ y2+ (τ − 2)x, y − 2(x1x2+ y1y2)+ (τ − 2)(x1y2+ y1x2), viewed as a function of (x, y), is differentiable at (x, y)= (x, y). Moreover, μ1= λ1(w)= 0 when (x, y)= (x, y). Thus, the first- order Taylor’s expansion of μ1at (x, y) yields

μ1 = O(x− x + y− y).

Also, since w2(x, y)= 0, by the product and quotient rules for differentiation, the function

x1+ y1[2(x1x2+ y1y2)+ (τ − 2)(x1y2+ y1x2)]T(x2+ y2)

2(x1x2 + y1y2)+ (τ − 2)(x1y2 + y1x2) (27) is differentiable at (x, y)= (x, y), and it has value 0 at (x, y)= (x, y) due to

x1+ y1[x1x2+ y1y2+ (τ − 2)x1y2]T(x2+ y2)

x1x2+ y1y2+ (τ − 2)x1y2

= x1− x2T w2

w2+ y1− y2T w2

w2= 0.

Hence, the function in (27) is O(x− x + y− y) in magnitude, which together with μ1= O(x− x + y− y) shows that the second term on the right-hand side

(10)

of (26) is O

(x− x + y− y)3/2

= o

x− x + y− y .

Thus, we have shown that ψτ is differentiable at (x, y). Moreover, we see that 2∇ψτ(x, y)is the sum of the gradient ofx2+ y2+ (τ − 2)x, y + x+ y2 and the gradient of the first term on the right-hand side of (26), evaluated at (x, y)= (x, y).

The gradient ofx2+ y2+ (τ − 2)x, y + x+ y2with respect to x, evaluated at (x, y)= (x, y), is 2x + (τ − 2)y + 2(x + y). The derivative of the first term on the right-hand side of (26) with respect to x1, evaluated at (x, y)= (x, y), works out to be

√ 1 λ2(w)



x1+τ− 2 2 y1

 +



x2+τ− 2 2 y2

T

w2

w2



×



x1+ y1+ (x2+ y2)T w2

w2



+ λ2(w)



1+ (x2+τ−22 y2)T(x2+ y2)

x1x2+ y1y2+ (τ − 2)x1y2

wT2(x2+ y2)· w2T(x2+τ−22 y2)

x1x2+ y1y2+ (τ − 2)x1y2 · w22



= 2(x1+τ−22 y1)(x1+ y1)



x21+ y12+ (τ − 2)x1y1 + 2

x12+ y12+ (τ − 2)x1y1,

where the equality follows from Lemma3.1. Similarly, the gradient of the first term on the right of (26) with respect to x2, evaluated at (x, y)= (x, y), works out to be

√ 1 λ2(w)



x2+τ− 2 2 y2

 +



x1+τ− 2 2 y1

 w2

w2

×



x1+ y1+ (x2+ y2)T w2

w2



+ λ2(w)

(2x1+ (τ − 2)y1)x2+τ2(x1+ y1)y2

x1x2+ y1y2+ (τ − 2)x1y2

.wT2(x2+ y2)· (x1+τ−22 y1)w2

x1x2+ y1y2+ (τ − 2)x1y2 · w22

= 2(2x1+ (τ − 2)y1)x2+τ2(x1+ y1)y2



x12+ y12+ (τ − 2)x1y1

.

Then, combining the last two gradient expressions yields that

(11)

2∇xψτ(x, y)

= 2x + (τ − 2)y + 2(x + y) −

2



x12+ y12+ (τ − 2)x1y1

0

− 2



x12+ y12+ (τ − 2)x1y1

 (x1+τ−22 y1)(x1+ y1) (2x1+ (τ − 2)y1)x2+τ2(x1+ y1)y2

.

Using the fact that x1y2= y1x2and noting that φτ can be simplified as the one in (23) under this case, we readily rewrite the above expression for∇xψτ(x, y)in the form of (25). By symmetry,∇yψτ(x, y)also holds as the form of (25).  Proposition3.2shows that ψτ is differentiable with a computable gradient. To establish the continuity of the gradient of ψτ or the smoothness of ψτ, we need the following two crucial technical lemmas whose proofs are provided in theAppendix.

Lemma 3.2 For any x= (x1, x2), y= (y1, y2)∈ R × Rn−1, if w2= 0, then



x1+τ − 2 2 y1



+ (−1)i



x2+τ − 2 2 y2

T

w2

w2

2

≤



x2+τ− 2 2 y2



+ (−1)i



x1+τ − 2 2 y1

 w2

w2

2 ≤ λi(w)

for i= 1, 2. Furthermore, these relations also hold when interchanging x and y.

Lemma 3.3 For all (x, y) satisfying (x− y)2+ τ(x ◦ y) ∈ int(Kn), we have that

Lx+τ−22 yL−1z 

F ≤ C, L

y+τ−22 xL−1z 

F≤ C, (28)

where C > 0 is a constant independent of x, y and τ , and·F denotes the Frobenius norm.

Proposition 3.3 The function ψτ defined by (9) is smooth everywhere onRn× Rn. Proof By Proposition3.2and the symmetry of x and y in∇ψτ, it suffices to show that∇xψτ is continuous at every (a, b)∈ Rn× Rn. If (a− b)2+ τ(a ◦ b) ∈ int(Kn), the conclusion has been shown in Proposition3.2. We next consider the other two cases.

Case (1): (a, b)= (0, 0). By Proposition3.2, we need to show that∇xψτ(x, y)

→ 0 as (x, y) → (0, 0). If (x − y)2+ τ(x ◦ y) ∈ int(Kn), thenxψτ(x, y)is given by (24), whereas if (x, y)= (0, 0) and (x −y)2+τ(x ◦y) /∈ int(Kn), thenxψτ(x, y) is given by (25). Notice that

Lx+τ−2

2 yL−1z and x1+τ−22 y1



x12+ y12+ (τ − 2)x1y1

(12)

are bounded with bound independent of x, y and τ , using the continuity of φτ(x, y) immediately yields the desired result.

Case (2): (a, b)= (0, 0) and (a − b)2+ τ(a ◦ b) /∈ int(Kn). We will show that

xψτ(x, y)→ ∇xψτ(a, b)by the two subcases: (2a) (x, y)= (0, 0) and (x − y)2+ τ (x◦y) /∈ int(Kn)and (2b) (x−y)2+τ(x ◦y) ∈ int(Kn). In subcase (2a),xψτ(x, y) is given by (25). Noting that the right-hand side of (25) is continuous at (a, b), the desired result follows.

Next, we prove that∇xψτ(x, y)→ ∇xψτ(a, b) in subcase (2b). From (24), we have that

xψτ(x, y)=



x+τ− 2 2 y



− Lx+τ−2

2 yL−1z (x+ y) − φτ(x, y). (29) On the other hand, since (a, b)= (0, 0) and (a − b)2+ τ(a ◦ b) /∈ int(Kn),

a2+ b2+ (τ − 2)aTb= 2(a1a2+ b1b2)+ (τ − 2)(a1b2+ b1a2) = 0, (30) and moreover from (20) it follows that

a2+ b2+ (τ − 2)aTb= 2

a21+ b21+ (τ − 2)a1b1



= 2

a22+ b22+ (τ − 2)a2Tb2

= 2(a1a2+ b1b2)+ (τ − 2)a1b2. (31)

Using the equalities in (31), it is not hard to verify that a1+τ−22 b1



a12+ b21+ (τ − 2)a1b1

(a− b)2+ τ(a ◦ b)1/2

= a +τ− 2 2 b.

This together with the expression of∇xψτ(a, b)given by (25) yields

xψτ(a, b)=



a+τ− 2 2 b



a1+τ−22 b1



a12+ b21+ (τ − 2)a1b1

(a+ b) − φτ(a, b). (32)

Comparing (29) with (32), we see that if we wish to prove∇xψτ(x, y)→ ∇xψτ(a, b) as (x, y)→ (a, b), it suffices to show that

Lx+τ−2

2 yL−1z (x+ y) → a1+τ−22 b1



a21+ b21+ (τ − 2)a1b1

(a+ b), (33)

which is also equivalent to proving the following three relations

Lx+τ−2 2 yL−1z



x+τ− 2 2 y



a1+τ−22 b1



a12+ b21+ (τ − 2)a1b1



a+τ− 2 2 b

 , (34)

(13)

Ly+τ−2 2 xL−1z



y+τ− 2 2 x



b1+τ−22 a1



a12+ b21+ (τ − 2)a1b1



b+τ− 2 2 a

 , (35)

4− τ

2 Lx−yL−1z



y+τ− 2 2 x



4−τ

2 (a1− b1)



a12+ b21+ (τ − 2)a1b1



b+τ− 2 2 a

 . (36)

By the symmetry of x and y in (34) and (35), we only prove (34) and (36). Let

1, ζ2):= Lx+τ−2 2 yL−1z



x+τ− 2 2 y

 ,

(37) 1, ξ2):= Lx−yL−1z



y+τ − 2 2 x

 .

Then showing (34) and (36) reduces to proving the following relations hold as (x, y)→ (a, b):

ζ1(a1+τ−22 b1)2



a21+ b21+ (τ − 2)a1b1

,

(38) ζ2a1+τ−22 b1



a21+ b21+ (τ − 2)a1b1



a2+τ− 2 2 b2

 ,

ξ1(a1− b1)(b1+τ−22 a1)



a21+ b21+ (τ − 2)a1b1

,

(39) ξ2(a1− b1)



a21+ b21+ (τ − 2)a1b1



b2+τ− 2 2 a2

 .

To verify (38), we take (x, y) sufficiently near to (a, b). By (30), we may assume that w2= w2(x, y)= 0. Let s = (s1, s2)= x +τ−22 y. Using (14) and (37), we can calculate that

ζ1= 1 det(z)



s12z1− 2s1s2Tz2+det(z)

z1 s22+(z2Ts2)2 z1

 ,

=s22

z1 +(s1z1− s2Tz2)2

z1det(z) , (40)

ζ2= 1 det(z)



s1z1s2− zT2s2s2− s21z2+s1det(z) z1 s2+s1

z1zT2s2z2



= s1

z1

s2+(s1z1− s2Tz2) det(z)

 s2s1

z1

z2



. (41)

參考文獻

相關文件

Although we have obtained the global and superlinear convergence properties of Algorithm 3.1 under mild conditions, this does not mean that Algorithm 3.1 is practi- cally efficient,

Because both sets R m  and L h i ði ¼ 1; 2; :::; JÞÞ are second-order regular, similar to [19, Theorem 3.86], we state in the following theorem that there is no gap between

Abstract In this paper, we consider the smoothing Newton method for solving a type of absolute value equations associated with second order cone (SOCAVE for short), which.. 1

Chen, Conditions for error bounds and bounded level sets of some merit func- tions for the second-order cone complementarity problem, Journal of Optimization Theory and

We propose a primal-dual continuation approach for the capacitated multi- facility Weber problem (CMFWP) based on its nonlinear second-order cone program (SOCP) reformulation.. The

Chen, “Alternative proofs for some results of vector- valued functions associated with second-order cone,” Journal of Nonlinear and Convex Analysis, vol.. Chen, “The convex and

By this, the second-order cone complementarity problem (SOCCP) in H can be converted into an unconstrained smooth minimization problem involving this class of merit functions,

Abstract We investigate some properties related to the generalized Newton method for the Fischer-Burmeister (FB) function over second-order cones, which allows us to reformulate