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A regularization method for the second-order cone complementarity problem with the Cartesian P 0 -property
Shaohua Pan
a,∗, Jein-Shan Chen
b,1aSchool of Mathematical Sciences, South China University of Technology, Guangzhou 510641, China bDepartment of Mathematics, National Taiwan Normal University, Taipei 11677, Taiwan
Received 12 October 2007; accepted 14 February 2008
Abstract
We consider the Tikhonov regularization method for the second-order cone complementarity problem (SOCCP) with the Cartesian P0-property. We show that many results of the regularization method for the P0-nonlinear complementarity problem still hold for this important class of nonmonotone SOCCP. For example, under the more general setting, every regularized problem has the unique solution, and the solution trajectory generated is bounded if the original SOCCP has a nonempty and bounded solution set. We also propose an inexact regularization algorithm by solving the sequence of regularized problems approximately with the merit function approach based on Fischer–Burmeister merit function, and establish the convergence result of the algorithm.
Preliminary numerical results are also reported, which verify the favorable theoretical properties of the proposed method.
c
2008 Elsevier Ltd. All rights reserved.
Keywords:Second-order cone complementarity problem; Tikhonov regularization; Cartesian P0-property; Fischer–Burmeister merit function
1. Introduction
We consider the second-order cone complementarity problem (SOCCP) which is to find a point x ∈ Rnsuch that
x ∈K, F(x) ∈ K, hx, F(x)i = 0, (1)
where h·, ·i represents the Euclidean inner product, F : Rn → Rn is a mapping assumed to be continuously differentiable throughout this paper, and K is the Cartesian product of second-order cones (SOCs), also called Lorentz cones [9]. In other words,
K = Kn1×Kn2× · · · ×Knm, (2)
where m, n1, . . . , nm ≥1, n1+n2+ · · · +nm =n, and Kni :=n
x =(x1, x2) ∈ R × Rni−1|x1≥ kx2ko
∗Corresponding author. Tel.: +86 02087110153; fax: +86 224556567.
E-mail addresses:shhpan@scut.edu.cn(S. Pan),jschen@math.ntnu.edu.tw(J.-S. Chen).
1 Member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office.
0362-546X/$ - see front matter c 2008 Elsevier Ltd. All rights reserved.
doi:10.1016/j.na.2008.02.028
with k · k denoting the Euclidean norm and K1denoting the set of nonnegative reals R+. In what follows, we refer (1) and(2) to the SOCCP(F ). An important special case of(2)is K = Rn+, the nonnegative orthant in Rn, which corresponds to n1 = · · · = nm = 1 and m = n, and the SOCCP(F ) reduces to the nonlinear complementarity problem (NCP).
There exist various methods for solving the SOCCP(F ). They include the smoothing Newton method [1,12], the smoothing-regularization method [14], the merit function approaches [2,3], and the semismooth Newton method [17].
Most of these methods are proposed for the monotone SOCCP. In this paper, we will consider a particular method, i.e. the Tikhonov regularization method, for a class of nonmonotone SOCCP.
It is well known that the Tikhonov regularization method is designed to deal with the ill-posed problems which substitute the solution of the original problem with the solution of a sequence of well-posed problems whose solutions converge to a solution of the original problem; see [11,20] and the references therein. In the context of SOCCPs, the regularization scheme consists in solving a sequence of SOCCP(Fε):
x ∈K, Fε(x) ∈ K, hx, Fε(x)i = 0, (3)
whereε is a positive parameter tending to zero and Fε: Rn→ Rnis given by
Fε(x) := F(x) + εx. (4)
The regularization scheme was considered by [14], where it was used only to guarantee that the proposed smoothing algorithm could handle the monotone SOCCP. In this paper, we apply the regularization scheme for the SOCCP with the Cartesian P0-property.
Specifically, paralleling to the classical results of regularization methods for convex optimization problems [6], we try to generalize as much as possible the following results to the large class of SOCCP with F having the Cartesian
P0-property:
(a) The regularized problem SOCCP(Fε) has a unique solution x(ε) for every ε > 0.
(b) The trajectory x(ε) is continuous for ε > 0.
(c) For ε → 0, the trajectory x(ε) converges to the least l2-norm solution of SOCCP(F ) if the SOCCP(F ) has a nonempty solution set, and otherwise it diverges.
In Section 3, we generalize the result (a) to the more general setting, and concentrate on the partial extension of the results (b) and (c) in Section 4. Then, we propose an inexact regularization algorithm for the SOCCP(F ) in Section 5, and establish the corresponding convergence results. In Section 6, we report our numerical experience with the algorithm for solving some linear SOCPs from the DIMACS library and some SOCCPs generated randomly with the Cartesian P0-property, and make comparisons with the merit function approach [2] to verify the favorable theoretical properties of the proposed method. Finally, we conclude this paper with several open questions.
Throughout this paper, Rndenotes the space of n-dimensional real column vectors, and Rn1×· · ·×Rnmis identified with Rn1+···+nm. Thus,(x1, . . . , xm) ∈ Rn1×· · ·×Rnmis viewed as a column vector in Rn1+···+nm. For a differentiable mapping F : Rn → Rn, F0(x) ∈ Rn×n denotes the Jacobian matrix of F at x while ∇ F(x) ∈ Rn×n denotes the transpose Jacobian of F at x. If J and B are index sets such that J, B ⊆ {1, 2, . . . , m}, we denote MJ B by the block matrix consisting of the submatrices Mj k ∈ Rnj×nk of M with j ∈ J, k ∈ B, and xB by a vector consisting of subvectors xi ∈ Rni with i ∈ B. Given x ∈ Rn, [x]+and [x]−denote the minimum distance projection of x onto K and −K, respectively. For a set S, the notation int(S) denotes the interior of S. We write F = (F1, . . . , Fm) with Fi : Rn→ Rni and Fε=(Fε,1, . . . , Fε,m) with Fε,i : Rn→ Rni.
2. Preliminaries
We first review some basic concepts and properties related to the SOC Kl(l > 1), and then introduce the concepts of Cartesian P-properties and P-properties for a matrix M ∈ Rn×n and a nonlinear transformation F : Rn → Rn, respectively.
For any x =(x1, x2), y = (y1, y2) ∈ R × Rl−1, define their Jordan product as
x ◦ y :=(hx, yi, x1y2+y1x2) . (5)
Write x + y to mean the usual componentwise addition of vectors and x2 to mean x ◦ x. Then ◦, + and e = (1, 0, . . . , 0)T ∈ Rl have the following basic properties [9,12]: (1) e ◦ x = x for all x ∈ Rl. (2) x ◦ y = y ◦ x for all x, y ∈ Rl. (3) x ◦(x2◦y) = x2◦(x ◦ y) for all x, y ∈ Rl. (4)(x + y) ◦ z = x ◦ z + y ◦ z for all x, y, z ∈ Rl. Notice that the Jordan product is not associative, but it is power associated, i.e.,x ◦(x ◦ x) = (x ◦ x) ◦ x for all x ∈ Rl. We stipulate x0=e. Besides, Klis not closed under Jordan product.
From [9,12], any vector x =(x1, x2) ∈ R × Rl−1has the spectral factorization:
x =λ1(x) · u(1)x +λ2(x) · u(2)x , (6)
whereλi(x) and u(i)x for i = 1, 2 are the spectral values and the associated spectral vectors given by λi(x) = x1+(−1)ikx2k, u(i)x = 1
2
1, (−1)ix¯2 , (7)
with ¯x2= x2
kx2kif x26=0 and otherwise ¯x2being any vector in Rl−1such that k ¯x2k =1. If x26=0, the factorization is unique. The spectral factorizations of x, x2and x1/2 have various interesting properties; see [9,12]. Here we list some that will be used later.
Property 2.1. For any x = (x1, x2) ∈ R × Rl−1, let λ1(x), λ2(x) and u(1)x , u(2)x be the spectral values and the associated spectral vectors. Then, the following results hold:
(a) For any x ∈ Rl, x2= [λ1(x)]2u(1)x + [λ2(x)]2u(2)x ∈Kl.
(b) x ∈ Kl ⇐⇒0 ≤λ1(x) ≤ λ2(x) and x ∈ int(Kl) ⇐⇒ 0 < λ1(x) ≤ λ2(x).
(c) For any x ∈ Kl, x1/2=
√λ1(x) u(1)x +
√λ2(x) u(2)x ∈Kl. (d) x ∈ Kl if and only if the symmetric matrix Lx :=hx
1 xT2 x2 x1I
i
is positive semidefinite, and x ∈int(Kl) if and only if Lxis positive definite.
Next we present the definitions of Cartesian P-properties for a matrix M ∈ Rn×n, which are special cases of those introduced by Chen and Qi [5] for a linear transformation.
Definition 2.1. A matrix M ∈ Rn×nis said to have
(a) the Cartesian P-property if for every nonzero z = (z1, . . . , zm) ∈ Rn with zi ∈ Rni, there exists an index ν ∈ {1, 2, . . . , m} such that hzν, (Mz)νi> 0;
(b) the Cartesian P0-property if for every nonzero z = (z1, . . . , zm) ∈ Rn with zi ∈ Rni, there exists a ν ∈ {1, 2, . . . , m} such that zν 6=0 and hzν, (Mz)νi ≥0.
Clearly, when m = n and n1= · · · =nm =1, M having the Cartesian P-property (or the Cartesian P0-property) coincides with M being a P-matrix (or P0-matrix) introduced in [4]. Let M be an n × n matrix with elements mi j. Then, M can be denoted by
M =
M11 M12 · · · M1m
M21 M22 · · · M2m
· · · · Mm1 Mm2 · · · Mmm
, (8)
where Mνl for eachν = 1, . . . , m and l = 1, . . . , m is an nν ×nl matrix consisting of those elements mk j with k = nν−1+1, . . . , nν, j = nj −1+1, . . . , nj and n0 =0. Let S be a proper subset of {1, 2, . . . , m} and denote by M(S) the matrix resulting from deleting the block matrix Mνl withν and l complementary to those indicated by S from M given as in(8). We call M(S) a principal block matrix of M. ByDefinition 2.1, it is not hard to verify that every principal block matrix M(S) must have the Cartesian P-property if the matrix M has the Cartesian P-property.
When m = n and n1 = · · · = nm = 1, this reduces to the well-known fact that every principal submatrix of a P-matrix is again a P-matrix. Particularly, assume that the matrix M, by rearrangement, is written as
M = MJ J MJ B
MBJ MBB
, (9)
where J and B are index sets such that J ∪ B = {1, 2, . . . , m} and J ∩ B = ∅. Then, when M has the Cartesian P-property and MJ J is nonsingular, we have the following result, which can be regarded as an extension of the fact that any Schur-complement of a P-matrix is also a P-matrix.
Proposition 2.1. Suppose that M defined as in(9)has the Cartesian P-property and the matrix MJ J is nonsingular.
Then its Schur-complement in the matrix M, i.e.
MbJ J =MBB−MBJ(MJ J)−1MJ B also has the Cartesian P-property.
Proof. Let yBbe an arbitrary nonzero vector with the dimension same as MBB. Let xJ be a vector with the dimension same as MJ J such that
MJ JxJ+MJ ByB=0, (10)
or equivalently,
xJ = −(MJ J)−1MJ ByB. (11)
Let z = (xJ, yB) ∈ Rn. Then, z 6= 0. From Definition 2.1(a) and the given assumption that M has the Cartesian P-property, there exists an index i ∈ {1, 2, . . . , m} such that
hzi, (Mz)ii> 0. (12)
Notice that the index i must belong to the set B. If not, i.e. i ∈ J , then from the definition of M we learn that inequality(12)is equivalent to
hxi, [MJ JxJ+MJ ByB]ii> 0,
which obviously contradicts equality(10). Now(12)is equivalent to hyi, [MBJxJ +MBByB]ii> 0.
Using the inequality and Eq.(11), we immediately have that
hyi, [MbJ JyB]ii = hyi, [MBByB−MBJ(MJ J)−1MJ ByB]ii
= hyi, [MBByB+MBJxJ]ii> 0.
Thus, byDefinition 2.1(a), the matrix bMJ J has the Cartesian P-property. Definition 2.2 ([13]). A matrix M ∈ Rn×nis said to have
(a) the Jordan P-property (or the P1-property) if x ◦(Mx) ∈ −K ⇒ x = 0;
(b) the P-property if the condition that LxiL(Mx)i = L(Mx)iLxi, i = 1, 2, . . . , m and x ◦ (Mx) ∈ −K necessarily implies x = 0;
(c) the P0-property if M +εI for any ε > 0 has the P-property.
Proposition 2.2. (a) If a matrix M ∈ Rn×nhas the Cartesian P-property, then it also has the Jordan P-property and the P-property.
(b) If a matrix M ∈ Rn×nhas the Cartesian P0-property, then it has the P0-property.
Proof. (a) FromDefinition 2.2(a) and (b), it is not hard to see that the Jordan P-property implies the P-property.
Therefore, we only need to prove that the Cartesian P-property implies the Jordan P-property. Let x = (x1, . . . , xm) ∈ Rn with xi ∈ Rni be any vector such that x ◦(Mx) ∈ −K. From the Cartesian structure of K, we have
xi◦(Mx)i ∈ −Kni for i = 1, 2, . . . , m,
which, by the definition of Jordan product given by(5), means that
hxi, (Mx)ii ≤0 for all i = 1, 2, . . . , m. (13)
Now, suppose that x 6= 0. Then, fromDefinition 2.1(a), it follows that there exists an indexν ∈ {1, 2, . . . , m} such that hxν, (Mx)νi> 0, which clearly contradicts(13). Hence, M has the Jordan P-property.
(b) Observe that for anyε > 0, M + εI has the Cartesian P-property. By part (a) andDefinition 2.2(c), M has the P0-property.
The Cartesian P0-property may not imply the P-property. For example, let m = 2 and n1=n2=2, and consider
M =
1 1 0 0
1 1 0 0
0 0 2 2
0 0 2 2
and x =
−2 2
−1 1
.
It is easy to verify that M has the Cartesian P0-property, x ◦(Mx) = (0, 0, 0, 0) ∈ −K = −(K2×K2) and LxLM x =LM xLx =0, but x 6= 0, i.e., M has no P-property. Now, we are not clear whether the P-property implies the Cartesian P0-property.
Next we introduce definitions of Cartesian P-properties for a nonlinear mapping F : Rn → Rn in the setting of SOCs. The concepts of P-properties on Cartesian products in Rnwere first established by Facchinei and Pang [10].
Recently, Chen and Qi [5] and Kong et al. [15] extended the concepts of Cartesian P-properties to the setting of positive semidefinite cones and the general Euclidean Jordan algebra, respectively.
Definition 2.3. A nonlinear mapping F =(F1, . . . , Fm) with Fi : Rn→ Rni is said to
(a) have the uniform Cartesian P-property if there exists a constantρ > 0 such that, for any x, y ∈ Rn, there is an indexν ∈ {1, 2, . . . , m} such that
hxν−yν, Fν(x) − Fν(y)i ≥ ρkx − yk2;
(b) have the Cartesian P-property if for any x, y ∈ Rnwith x 6= y, there exists an indexν ∈ {1, 2, . . . , m} such that hxν−yν, Fν(x) − Fν(y)i > 0;
(c) have the Cartesian P0-property if for any x, y ∈ Rnwith x 6= y, there exists an indexν ∈ {1, 2, . . . , m} such that xν 6=yν and hxν−yν, Fν(x) − Fν(y)i ≥ 0.
(d) have the Cartesian R02-property if for any sequence {xk}satisfying the condition that kxkk → +∞, [−xk]+
kxkk →0, [−F(xk)]+
kxkk →0, (14)
there exists an indexν ∈ {1, 2, . . . , m} such that lim inf
k→+∞
λ2 Fν(xk) ◦ xνk kxνkk2 > 0.
ByDefinition 2.3, it is not difficult to verify the following one-way implications:
Uniform Cartesian P-property ⇒ Cartesian P-property ⇒ Cartesian P0-property, Uniform CartesianP-property ⇒ Cartesian R02-property.
Moreover, we see that, when m = 1, the Cartesian P-property (or the Cartesian P0-property) of F becomes the strict monotonicity (or monotonicity) of F . If the continuously differentiable mapping F has the Cartesian P-property (or P0-property), then its transposed Jacobian matrix ∇ F(x) at any x ∈ Rnhas the corresponding Cartesian P-properties.
When F degenerates into the affine function M x + q, F having the uniform Cartesian P-property coincides with M having the Cartesian P-property. In addition, byDefinition 2.3(b)–(c), we readily have the following result.
Proposition 2.3. For anyε > 0, let Fε : Rn → Rnbe given as in(4). If F has the Cartesian P0-property, then Fε will have the Cartesian P-property.
It should be pointed out that, when F has the Cartesian P-property, Fε must not have the uniform Cartesian P- property. A counterexample is given by [11] for the case m = 1.
Finally, paralleling toDefinition 2.2, we have the concepts of P-properties for a nonlinear mapping in the setting of SOCs, which are special cases of those given by [23].
Definition 2.4. A nonlinear mapping F =(F1, . . . , Fm) : Rn→ Rnis said to have (a) the Jordan P-property if(x − y) ◦ (F(x) − F(y)) ∈ −K ⇒ x = y;
(b) the P-property if the condition that Lxi−yiLFi(x)−Fi(y) = LFi(x)−Fi(y)Lxi−yi, i = 1, 2, . . . , m and (x − y) ◦ (F(x) − F(y)) ∈ −K implies x = y;
(c) the P0-property if F(x) + εx has the P-property for all ε > 0.
Proposition 2.4. (a) If a mapping F : Rn → Rn has the Cartesian P-property, then it must have the Jordan P- property and the P-property.
(b) If a mapping F : Rn→ Rnhas the Cartesian P0-property, then it must have the P0-property.
Proof. The proof is similar to that ofProposition 2.2, and we omit it. 3. Existence of regularized solutions
In this section, we show that the regularized problem SOCCP(Fε) has a unique solution x(ε) for every ε > 0 under the Cartesian P0-property of F and the following condition:
Condition A. For any sequence {xk} ⊆ Rn, when there exists i ∈ {1, 2, . . . , m} such that λ2(xik) → +∞, {Fε,i(xk)}
is bounded below, and
kFi(xk)k kxikk
is unbounded, there holds that
lim sup
k→+∞
* xik
kxikk, Fi(xk) kFi(xk)k
+
> 0.
The main tool to prove this result is the Fischer–Burmeister (FB) SOC complementarity function. The FB function was first introduced by Fischer [7,8], which plays a crucial role in the design of several nonsmooth Newton methods and merit function methods for the solution of NCPs. Recently, the function was extended to the setting of semidefinite complementarity problems [21,22] and SOCCPs [2], respectively.
Definition 3.1. A mappingφ : Rl × Rl → Rl is called an SOC complementarity function associated with the SOC Klif for any x, y ∈ Rl,
φ(x, y) = 0 ⇐⇒ x ∈Kl, y ∈ Kl, hx, yi = 0. (15)
The FB SOC complementarity function associated with Kl is defined as follows:
φFB(x, y) := (x2+y2)1/2−(x + y), ∀ (x, y) ∈ Rl× Rl. (16) ByProperty 2.1(a)–(c), clearly, the functionφFBis well defined in Rl× Rl. Moreover, it was shown in [12] thatφFB
satisfies the characterization (15). With the vector-valued function, Chen and Tseng [2] proposed a merit function approach for the SOCCP, and we recently developed a semismooth Newton method in [17]. In this section, we mainly employ the function as a theoretical tool. Define the operator ΦFB: Rn→ Rnby
ΦFB(x) :=
φFB(x1, F1(x)) φFB(xm, F... m(x))
, (17)
which induces a natural merit function ΨFB: Rn→ R+given by ΨFB(x) := 1
2kΦFB(x)k2=1 2
m
X
i =1
kφFB(xi, Fi(x))k2. (18)
The following proposition summarizes some important properties of ΨFB. Since their proofs are direct or can be found in [2,17], here we omit them.
Proposition 3.1. Let ΨFB: Rn→ R+be given as in(18). Then, the following results hold.
(a) x∗is a solution of the SOCCP(F ) if and only if x∗solves the systemΦFB(x) = 0.
(b) ΨFBis continuously differentiable everywhere on Rn.
(c) If F has the Cartesian P0-property, then every stationary point of ΨFBis a solution of the SOCCP(F ).
Analogously, for the SOCCP(Fε), we define the operator Φε: Rn→ Rnby
Φε(x) :=
φFB(x1, Fε,1(x)) φFB(xm, F... ε,m(x))
, (19)
where Fε,i : Rn→ Rni denotes the i th subvector of Fε. The natural merit function Ψε: Rn→ R+corresponding to Φεis then given by
Ψε(x) := 1
2kΦε(x)k2= 1 2
m
X
i =1
φFB(xi, Fε,i(x))
2. (20)
The following lemma plays a crucial role in proving the main result of this section. Since the proof can be found in Lemma 5.2 of [17], here we omit it.
Lemma 3.1. LetφFBbe defined as in(16). For any sequence {(xk, yk)} ⊆ Rl× Rl, letλk1≤λk2andµk1≤µk2denote the spectral values of xkand yk, respectively.
(a) Ifλk1→ −∞orµk1→ −∞, then kφFB(xk, yk)k → +∞.
(b) If {λk1}and {µk1}are bounded below, butλk2 → +∞,µk2 → +∞, and xk
kxkk◦ yk
kykk 9 0, then kφFB(xk, yk)k → +∞.
Proposition 3.2. Suppose that F : Rn → Rn has the Cartesian P0-property and satisfiesConditionA. Then the functionΨεgiven by(20)for anyε > 0 is coercive, i.e.,
lim
kxkk→∞
Ψε(xk) = +∞.
Proof. Suppose by contradiction that the conclusion does not hold. Then we can find an unbounded sequence {xk} ⊆ Rn with xk = (x1k, . . . , xmk) and xik ∈ Rni such that the sequence {Ψε(xk)} is bounded. Define the index set
J :=n
i ∈ {1, 2, . . . , m} | {kxikk}is unboundedo .
Since {xk}is unbounded, J 6= ∅. Subsequencing if necessary, we assume without loss of generality that {kxikk} → +∞for all i ∈ J . For each i ∈ J , we define
Ji :=nν ∈ {1, 2, . . . , ni} | {|xikν|}is unboundedo .
Let {yk}be a bounded sequence with yk=(y1k, . . . , ykm) and yik ∈ Rni defined as follows:
yiνk =0 if i ∈ J andν ∈ Ji; xikν otherwise.
From the definition of {yk}and the Cartesian P0-property of F , it follows that 0 ≤ max
1≤l≤m
D
xlk−ylk, Fl(xk) − Fl(yk)E
=D
xik−yik, Fi(xk) − Fi(yk)E
≤ nimax
ν∈Ji
xikνh
Fiν(xk) − Fiν(yk)i
=nixi jk h
Fi j(xk) − Fi j(yk)i , (21)
where i is an index from J for which the first maximum is attained, and j is an index from Ji for which the second maximum is attained. Without loss of generality, we assume that i and j are independent of k. Since i ∈ J and j ∈ Ji,
|xi jk| → + ∞. (22)
We now consider the two cases where xi jk → +∞and xi jk → −∞, respectively.
Case (1): xki j → +∞. In this case, since Fi j(yk) is bounded by the continuity of Fi j(y), inequality(21)implies that Fi j(xk) does not tend to −∞. This in turn implies that
n
Fi j(xk) + εxi jko
→ + ∞. (23)
Case (2): xi jk → −∞. Now, using inequality(21)and the boundedness of Fi j(yk) immediately yields that Fi j(xk) does not tend to +∞. This in turn implies that
n
Fi j(xk) + εxi jko
→ − ∞. (24)
From Eq.(22)–(24)and the definition of Fε,i(x), we thus obtain that
kxikk → +∞, kFε,i(xk)k → +∞. (25)
Ifλ1(xik) → −∞ or λ1[Fε,i(xk)] → −∞, then fromLemma 3.1(a) we readily obtain that kφFB(xik, Fε,i(xk))k → +∞. Otherwise, Eq.(25)implies that {xik}and {Fε,i(xk)} are bounded below, but λ2(xik) → +∞ and λ2[Fε,i(xk)] → +∞. We next prove that
lim
k→+∞
xik kxikk
◦ Fε,i(xk)
kFε,i(xk)k 9 0, (26)
and consequently fromLemma 3.1(b) it follows that kφFB(xik, Fε,i(xk))k → +∞. From the first two equations of (21)and the boundedness of {yk}and {Fi(yk)}, it is not hard to verify that hkxxikk
ik,kFFi(xk)
ε,i(xk)ki ≥ 0 for all sufficiently large k. Notice that
* xik
kxikk, Fε,i(xk) kFε,i(xk)k
+
=
* xik
kxikk, Fi(xk) kFε,i(xk)k
+
+ εkxikk
kFε,i(xk)k, ∀ k. (27)
Therefore, if the sequence
kFi(xk)k kxikk
is bounded, then equality(27)implies that
lim sup
k→+∞
* xik
kxikk, Fε,i(xk) kFε,i(xk)k
+
> 0. (28)
If the sequence
kFi(xk)k kxkik
is unbounded, then usingCondition Aand equality(27), it is easy to verify that(28)also holds. Clearly, Eq.(28)implies(26), and we thus get kφFB(xik, Fε,i(xk))k → +∞. This contradicts the boundedness of {Ψε(xk)}.
Proposition 3.2states that underCondition Aand the Cartesian P0-property of F the level set
Lγ(x) := {x ∈ Rn|Ψε(x) ≤ γ } (29)
is bounded for every γ ≥ 0. Now we are in a position to prove the following main result. Notice that, when m = n and n1 = · · · = nm = 1, the Cartesian P0-property of F is equivalent to requiring that F is P0-function;
whereasCondition Aautomatically holds since the assumption thatλ2(xik) → +∞, {Fε,i(xk)} is bounded below, and
kFi(xk)k kxikk
is unbounded implies that there exists a subsequence {xik}k∈K satisfying xik → +∞and Fi(xk) → +∞
for k ∈ K , and consequently lim supk→∞
xik
kxikk,kFFi(xk)
i(xk)k
> 0. Thus, the assertion ofProposition 3.2reduces to that of [11, Proposition 3.4].
Theorem 3.1. Assume that the mapping F : Rn→ Rnhas the Cartesian P0-property and satisfiesConditionA. Then for everyε > 0 the problem SOCCP(Fε) has a unique bounded solution x(ε).
Proof. Let ε > 0. Then the mapping Fε has the Cartesian P-property by Proposition 2.3. This means that the regularized problem SOCCP(Fε) has at most one solution. In fact, suppose that x(ε) and ˆx(ε) are two different solutions of the SOCCP(Fε). From the Cartesian P-property of Fε, it then follows that there exists an index i ∈ {1, 2, . . . , m} such that
0< hxi(ε) − ˆxi(ε), Fε,i(x(ε)) − Fε,i( ˆx(ε))i
= hxi(ε), Fε,i(x(ε))i − hxi(ε), Fε,i( ˆx(ε))i
− h ˆxi(ε), Fε,i(x(ε))i + h ˆxi(ε), Fε,i( ˆx(ε))i
= −hxi(ε), Fε,i( ˆx(ε))i − h ˆxi(ε), Fε,i(x(ε))i, (30)
where the last equality is due to hxi(ε), Fε,i(x(ε))i = 0 and h ˆxi(ε), Fε,i( ˆx(ε))i = 0. Note that the two terms on the right-hand side of(30)are nonpositive since xi(ε), ˆxi(ε) ∈ Kni and Fε,i(x(ε)), Fε,i( ˆx(ε)) ∈ Kni. Thus, we obtain a contradiction with inequality(30).
To prove the existence of a solution, let x0∈ Rnbe an arbitrary point and defineγ := Ψε(x0). ByProposition 3.2, the corresponding level set Lγ(x) is nonempty and compact. Therefore, the continuous function Ψε(x) attains a global minimum x(ε) on Lγ(x) which, by the definition of level sets, is also a global minimum of Ψε(x) on Rn. Therefore, x(ε) is a stationary point of Ψε(x). Since the mapping Fε has the Cartesian P-property, we have fromProposition 3.1(c) that x(ε) is a solution of the regularized problem SOCCP(Fε). Furthermore, this solution is bounded. Combining with the discussions above, we complete the proof.
4. Behaviour of the solution path
FromTheorem 3.1, we learn that the regularized problem SOCCP(Fε) for everyε > 0 has a unique solution x(ε) when the mapping F has the Cartesian P0-property and satisfies Condition A. Thus, as the parameterε tends to 0, the solution of the regularized problem SOCCP(Fε) generates a solution path P := {x(ε) | ε > 0}. The aim of this section is to study the properties of the trajectory P. Specifically, we prove that, if F has the uniform Cartesian P-property, the path P is bounded asε → 0 and the bound is dependent on the constant ρ involved in the uniform Cartesian P-property. We also illustrate that in this case the path P is not locally Lipschitz continuous asε → 0. Then, for the case that F has the Cartesian P0-property and satisfiesCondition A, we provide the condition to guarantee that x(ε) remains bounded asε → 0. The reason why we are interested in the boundedness of x(ε) is due to the following evident result.
Theorem 4.1. Let {εk}be a sequence of positive values converging to0. If {x(εk)} converges to a point ¯x, then ¯x solves the SOCCP(F ).
The following proposition states that the solution x(ε) of SOCCP(Fε) is bounded for anyε ≥ 0 if F has the uniform Cartesian P-property, but the bound is dependent on the constantρ involved in the uniform Cartesian P-property.
Proposition 4.1. Suppose that F has the uniform Cartesian P-property. Then, for anyε ≥ 0, we have
kx(ε)k ≤ ρ−1k[−F(0)]+k, (31)
whereρ > 0 is the constant involved in the uniform Cartesian P-property.
Proof. Since the uniform Cartesian P-property implies the Cartesian R02-property and the P-property, from [23, Theorem 3.1] and the proof ofProposition 4.3(b) below, it follows that x(ε) exists for any ε ≥ 0. If x(ε) ≡ 0 for anyε ≥ 0, then inequality(31)is direct. Suppose that x(ε) 6= 0 for some ε ≥ 0. Since x(ε) is the solution of the SOCCP(Fε), it follows that
xi(ε) ∈ Kni, Fε,i(x(ε)) ∈ Kni and xi(ε), Fε,i(x(ε)) = 0, i = 1, 2, . . . , m.
By this and the uniform Cartesian P-property of F , we have that ρkx(ε)k2≤ max
1≤i ≤m
hxi(ε), Fi(x(ε)) − Fi(0)i
= max
1≤i ≤m
hxi(ε), − εxi(ε) − Fi(0)i
≤ max
1≤i ≤m
hxi(ε), −Fi(0)i
≤ max
1≤i ≤m
hxi(ε), [−Fi(0)]+i
≤ kx(ε)kk[−Fi(0)]+k,
where the third inequality is since xi(ε) ∈ Kni, −Fi(0) = [−Fi(0)]++ [−Fi(0)]−and [−Fi(0)]− ∈ −Kni. This leads to the desired result.
Remark 4.1. (a) FromProposition 4.1, when F has the uniform Cartesian P-property, the SOCCP(F ) has a unique bounded solution. Furthermore, if F(0) ∈ K, the regularized problem SOCCP(Fε) for everyε ≥ 0 has the unique solution x(ε) = 0.
(b) When F is an affine function M x + q with M ∈ Rn×nand q ∈ Rn, the assumption ofProposition 4.1is equivalent to requiring that M has the Cartesian P-property.
Proposition 4.2. Suppose that F has the uniform Cartesian P-property. Then, for anyε1, ε2≥0, there holds that
kx(ε1) − x(ε2)k ≤ ρ−1kε1x(ε1) − ε2x(ε2)k, (32)
whereρ > 0 is the constant same asProposition4.1.
Proof. Without loss of generality, we assume thatε16=ε2. Let y(ε1) := Fε1(x(ε1)), y(ε2) := Fε2(x(ε2)).
Since x(ε1) and x(ε2) are the solutions of the problem SOCCP(Fε1) and SOCCP(Fε2), respectively, we have xi(ε1), yi(ε1) ∈ Kni with hxi(ε1), yi(ε1)i = 0 and xi(ε2), yi(ε2) ∈ Kni with hxi(ε2), yi(ε2)i = 0 for all i =1, 2, . . . , m. From this, it then follows that
hxi(ε1) − xi(ε2), Fi(x(ε1)) − Fi(x(ε2))i = hxi(ε1) − xi(ε2), yi(ε1) − ε1xi(ε1) − yi(ε2) + ε2xi(ε2)i
= −hxi(ε1), yi(ε2)i − hxi(ε2), yi(ε1)i + hxi(ε1) − xi(ε2), ε2xi(ε2) − ε1xi(ε1)i
≤ hxi(ε1) − xi(ε2), ε2xi(ε2) − ε1xi(ε1)i,
where the inequality holds since −hxi(ε1), yi(ε2)i ≤ 0 and −hxi(ε2), yi(ε1)i ≤ 0. Using the last inequality and the uniform Cartesian P-property of F , we have that
ρkx(ε1) − x(ε2)k2≤ max
1≤i ≤mhxi(ε1) − xi(ε2), Fi(x(ε1)) − Fi(x(ε2))i
≤ max
1≤i ≤mhxi(ε1) − xi(ε2), ε2xi(ε2) − ε1xi(ε1)i ,
≤ max
1≤i ≤mkxi(ε1) − xi(ε2)k kε2xi(ε2) − ε1xi(ε1)k
≤ kx(ε1) − x(ε2)k kε2x(ε2) − ε1x(ε1)k ,
which immediately implies the desired result. Thus, we complete the proof.
Propositions 4.1 and4.2characterize some properties of the path P as ε → 0 under the uniform Cartesian P- property of F . However, these results cannot imply the locally Lipschitz continuity of P asε → 0. The following counterexample illustrates the fact.
Example 4.1. Let m = 2 and n1=n2=2. Let F be given by F(x) = Mx + q, where
M =
1 0 0 0
0 1 0 0
0 0 2 0
0 0 0 2
, q =
−1 +ε 0ε 0 0
for any givenε > 0.
Since the matrix M has the Cartesian P-property, the mapping F has the uniform Cartesian P-property. For the SOCCP(Fε), i.e., to find x such that
x ∈K2×K2, Fε(x) ∈ K2×K2, hx, Fε(x)i = 0,
we can verify that x(ε) = (1/ε, 0, 0, 0)Tis the unique solution. Obviously, x(ε) is not locally Lipschitz continuous as ε → 0, and furthermore, x(ε) even has no bound.
Next, we concentrate on the case where F has the Cartesian P0-property and satisfiesCondition A. Under this case, we cannot prove the continuity of the mappingε → x(ε) at any ε > 0 like the NCP case. The main reason is that we cannot obtain the result corresponding to Theorem 3.1 of [16] under the Cartesian P-property of F , although when ∇ F(x) has the Cartesian P-property, its every principal block matrix has the Cartesian P-property, and the Schur-complement of a matrix with the Cartesian P-property also has the Cartesian P-property byProposition 2.1.
For this case, we can state the following result, whose proof will be postponed until the next section.
Theorem 4.2. Suppose that F has the Cartesian P0-property and satisfiesConditionA. If the solution setS∗of the SOCCP(F ) is nonempty and bounded, then the pathPε¯= {x(ε) | ε ∈ (0, ¯ε ]} is bounded for any ¯ε > 0 and
limε↓0dist x(ε) | S∗ = 0.
As an immediate consequence ofTheorem 4.2, we have the following conclusion.
Corollary 4.1. Suppose that F has the Cartesian P0-property and satisfies Condition A. If the SOCCP(F ) has a unique solution ¯x , thenlimε↓0x(ε) = ¯x.
As illustrated by Example 4.6 of [11], it is impossible to remove the boundedness assumption of S∗ without destroying the boundedness of the path Pε¯. To this end, we next provide some conditions to guarantee the nonemptyness and boundedness of S∗.
Proposition 4.3. The SOCCP(F ) has a nonempty and bounded solution set S∗under one of the following conditions:
(a) F is monotone, and the SOCCP(F ) is strictly feasible, i.e. there is ¯x ∈ Rnsatisfying ¯x, F( ¯x) ∈ int(K).
(b) The mapping F has the P0-property and the Cartesian R02-property.
Proof. (a) Since F(x) is monotone and ∇ F(x) is positive semidefinite, the result is direct by Proposition 6 of [2].
(b) We prove that in this case a stronger result holds, that is, the following SOCCP(F, q)
x ∈K, F(x) + q ∈ K, hx, F(x) + qi = 0 (33)
has a nonempty and bounded solution set for all q ∈ Rn. By Theorem 3.1 of [23], we only need to prove that for any 4> 0, the following set
{x : x solves(33)with kqk ≤ 4} (34)
is bounded. Suppose that the set is not bounded. Then there exists a sequence {qk}with kqkk ≤ 4and a sequence {xk}with kxkk → +∞such that for any k,
xk∈K, yk =F(xk) + qk∈K and xk◦yk =0. (35)
Without loss of generality, we assume that kxikk → +∞. This is equivalent to saying that for any k, 1
2
m
X
i =1
kφFB(xik, yik)k2=0. (36)
Using Lemma 8 of [2] and the boundedness of qk, we then obtain that kxkk → +∞, lim
k→+∞
[−xk]+
kxkk →0, lim
k→+∞
[−yk]+
kxkk →0, and lim
k→+∞
k[qk]+k
kxkk →0. (37) Noting that
k[qk]+k = k[yk−F(xk)]+k ≥ k[−F(xk)]+k,
where the inequality is due to [2, Lemma 7 (c)], we have from the last term in(37)that lim
k→+∞
k[−F(xk)]+k kxkk →0.
This, together with the first two terms in(37), shows that {xk}satisfies condition (14). By the Cartesian R02- property of F , there existsν ∈ {1, 2, . . . , m} such that
lim inf
k→+∞
λ2[xνk◦Fν(xk)]
kxkk2 > 0.
However, from Eq.(35)and the boundedness of qk, we have λ2[xkν◦Fν(xk)]
kxkk2 = λ2[−xkν◦qνk] kxkk2 → 0.
This leads to a contradiction. Consequently, the set defined by(34)is bounded.
Notice that the Cartesian R02-property is implied by the R0-property in [23]. Hence,Proposition 4.3(b) provides a weaker condition for S∗being nonempty and bounded. ByTheorem 3.1andPropositions 4.3(b) and2.4(b), we have the following result.
Corollary 4.2. Suppose that F has the Cartesian P0-property and the Cartesian R02-property and satisfies ConditionA. Then the pathPε¯ = {x(ε) | ε ∈ (0, ¯ε ]} is bounded for any ¯ε > 0 and limε↓0dist(x(ε) | S∗) = 0.
5. Inexact regularization method
The discussions from the last two sections show that the original SOCCP(F ) can be solved by calculating the exact solutions of a sequence of regularized problems SOCCP(Fε). However, in practice, it is usually not possible to solve the SOCCP(Fε) exactly for eachε > 0. In this section, we propose an inexact regularization algorithm which only requires inexact solutions of these subproblems, but preserves all convergence properties of its exact counterpart. First, let us describe the specific algorithm.
Algorithm 5.1 (Inexact Regularization Method).
(S.0) Chooseε0> 0 and τ0> 0, and set k := 0.
(S.1) Compute an approximate solution xkof SOCCP (Fε) such that
Ψε(xk) ≤ τk. (38)
(S.2) Terminate the iteration if a suitable criterion is satisfied.
(S.3) Chooseεk+1> 0 and τk+1> 0, set k := k + 1, and go to (S.1).
Clearly, if we takeτk =0 at each iteration, then xk =x(εk). In addition, the point xk can be easily obtained by applying any effective gradient-type unconstrained optimization algorithm to the minimization problem
x ∈RminnΨε(x), (39)
becauseΨε(x) is continuously differentiable everywhere and has bounded level sets for those SOCCPs with F having the Cartesian P0-property and satisfyingConditionA. In our numerical experiments, we adopt the BFGS algorithm to compute xk.
The following well-known Mountain Pass Theorem [18] will be used in the convergence analysis ofAlgorithm 5.1.
Lemma 5.1. Suppose that f : Rn→ R is smooth and coercive. Let C ⊆ Rnbe a nonempty compact set and denote c by the least value of f on the boundary of C , i.e. ¯¯ c :=minx ∈∂C f(x). If there are two points a ∈ C and b 6∈ C such that f(a) < ¯c and f (b) < ¯c, then there exists a point z ∈ Rnsuch that ∇ f(z) = 0 and f (z) ≥ ¯c.
Now we establish the convergence results ofAlgorithm 5.1. To this end, assume thatAlgorithm 5.1generates an infinite sequence so that the termination criterion in (S.2) is never active. The analysis technique adopted is similar to that of [11].
Theorem 5.1. Let F be the mapping having the Cartesian P0-property and satisfyingConditionA. Assume that the solution setS∗of the SOCCP(F ) is nonempty and bounded. Ifεk →0 andτk→0, then the sequence {xk}generated byAlgorithm5.1remains bounded, and every accumulation point of {xk}is a solution of the SOCCP(F ).
Proof. Suppose that the sequence {xk}is unbounded. Then, passing to a subsequence if necessary, we assume that {kxkk} → +∞. This, together with the boundedness of S∗, means that there exists a compact set C ⊆ Rn with S∗⊂intC and xk 6∈Cfor sufficiently large k. Let x∗∈S∗be a solution of the SOCCP(F ). Then we have
ΨFB(x∗) = 0 and ¯c := min
x ∈∂CΨFB(x) > 0. (40)
Letδ := ¯c/4. Notice that Ψε(x) viewed as the function of x and ε is continuous on the compact set C × [0, ˜ε], and so is uniformly continuous on C × [0, ˜ε]. Hence, there exists an ˜ε > 0 such that for all x ∈ C and ε ∈ [0, ˜ε]
|Ψε(x) − ΨFB(x)| ≤ δ. (41)
Combining(41)with(40), we have that for all sufficiently large k, Ψεk(x∗) ≤ 1
4c¯ (42)
and
c := min
x ∈∂CΨεk(x) ≥ 3
4c.¯ (43)
On the other hand, Ψεk(xk) ≤ τk byAlgorithm 5.1andτk →0, which means that Ψεk(xk) ≤ 1
4c¯ (44)
for all k large enough. Now using(42)–(44)and setting a = x∗and b = xkinLemma 5.1, there exists a vector ˆx ∈ Rn such that
∇Ψεk( ˆx) = 0 and Ψεk( ˆx) ≥ c ≥ 3 4c¯> 0.
This says that ˆxis a stationary point of Ψεk(x), but not a solution of the SOCCP(Fεk). However, byProposition 3.1(c), we know that any stationary point of Ψεk(x) is a solution of the SOCCP(Fεk). Thus, we obtain a contradiction.
Obviously,Theorem 4.2follows fromTheorem 5.1by settingτk =0 for all k. AlsoCorollaries 4.1and4.2can be easily extended to the inexact framework.
Corollary 5.1. Suppose that F has the Cartesian P0-property and satisfies ConditionA. Let {xk}be the sequence generated byAlgorithm5.1. Ifεk →0 andτk →0, and the SOCCP(F ) has a unique solution ¯x , thenlimεk→0xk= ¯x . Corollary 5.2. Suppose that F has the Cartesian P0-property and the Cartesian R02-property and satisfies ConditionA. Let {xk}be the sequence generated byAlgorithm5.1. If εk →0 andτk →0, then {xk}is bounded and its every accumulation point is a solution of the SOCCP(F ).
In addition, byProposition 4.3(a), we also have the following corollary.
Corollary 5.3. Suppose that F is monotone and satisfiesConditionAand the SOCCP(F ) is strictly feasible. Let {xk} be the sequence generated byAlgorithm5.1. If εk → 0 andτk → 0, then {xk}is bounded and every accumulation point is a solution of the SOCCP(F ).
Finally, we stress that, as far as we know, the inexact regularizationAlgorithm 5.1studied in this section is currently the only algorithm to guarantee the SOCCP(F ) with the Cartesian P0-property and a nonempty bounded solution set can actually be solved.
6. Numerical experiments
In this section, we report our preliminary numerical experience with the inexact regularization method for solving some SOCPs and SOCCPs, and make numerical comparisons with the merit function approach [2] which reformulates the SOCCP(F ) as:
x ∈RminnΨFB(x). (45)
All experiments were done at a PC with 2.8GHz CPU and 512MB memory. The computer codes were all written in Matlab 6.5. The subproblem(39)inAlgorithm 5.1and the minimization problem(45)were both solved by the limited-memory BFGS method with 5 limited-memory vector-updates. To improve the numerical performance of the BFGS method, we replaced the monotone Armijo line search by a nonmonotone line search as described by Zhang and Hager [24]. In other words, in the BFGS method, we computed the smallest nonnegative integer m such that
f(xk+βmdk) ≤ Wk+σβm∇f(xk)Tdk
where f(x) = Ψε(x) or ΨFB(x), dk was the direction of the kth iterate, and Wk :=(ηk−1Qk−1Wk−1+ f(xk))/Qk with Qk =ηk−1Qk−1+1.
Throughout the experiments, we adoptedβ = 0.5, σ = 10−4, W0 = f(x0), Q0 =1 andηk ≡ 0.85 for all k. In addition, we updatedεkandτkinAlgorithm 5.1by the formula:
εk =0.1εk−1 and τk =εk for all k,
where the initial regularization parameterε0was given in the corresponding examples. We terminatedAlgorithm 5.1 and the merit function approach [2] whenever one of the following conditions was satisfied: (1) ΨFB(xk) ≤ 10−6and
|hxk, F(xk)i| ≤ 10−5; (2) the steplength was less than 10−10.
To verify the efficiency of the regularization method, we first applied the inexact regularization method for solving a class of monotone SOCCPs, which correspond to the KKT optimality conditions of the linear SOCPs from the DIMACS Implementation Challenge library [19]. The standard linear SOCPs can be described as follows:
min cTx
s.t. Ax = b, x ∈ K, (46)
where A ∈ Rm×n has full row rank, b ∈ Rm and c ∈ Rn. From [2], we know that the KKT conditions of(46)are equivalent to finding a pointζ ∈ Rnsuch that
F(ζ ) ∈ K, G(ζ ) ∈ K, hF(ζ ), G(ζ)i = 0 (47)
with
F(ζ ) := d + (I − AT(AAT)−1A)ζ, G(ζ) := c − AT(AAT)−1Aζ, where d satisfies Ax = b. Hence, the corresponding regularized SOCCP problem is
Fε(ζ ) ∈ K, Gε(ζ ) ∈ K, hFε(ζ), Gε(ζ )i = 0 (48)
with
Fε(ζ ) := d + (I − AT(AAT)−1A)ζ + εζ, Gε(ζ ) := c − AT(AAT)−1Aζ + εζ, and the merit functions Ψεand ΨFBare specialized as
Ψε(ζ ) = 1 2
m
X
i =1
kφFB(Fε(ζ), Gε(ζ ))k2 and ΨFB(ζ ) = 1 2
m
X
i =1
kφFB(F(ζ ), G(ζ ))k2.
