Journal of Inequalities and Applications

, 2012:180

A superlinearly convergent hybrid algorithm for systems of nonlinear equations

Authors

    • Department of Mathematics and Computer ScienceYangtze Normal University

DOI: 10.1186/1029-242X-2012-180

Abstract

We propose a new algorithm for solving systems of nonlinear equations with convex constraints which combines elements of Newton, the proximal point, and the projection method. The convergence of the whole sequence is established under weaker conditions than the ones used in existing projection-type methods. We study the superlinear convergence rate of the new method if in addition a certain error bound condition holds. Preliminary numerical experiments show that our method is efficient.

MSC: 90C25, 90C30.

Keywords

nonlinear equations projection method global convergence superlinear convergence

1 Introduction

Let F : R n R n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq1_HTML.gif be a continuous mapping and C R n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq2_HTML.gif be a nonempty, closed, and convex set. The inner product and norm are denoted by , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq3_HTML.gif and https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq4_HTML.gif, respectively. Consider the problem of finding
x C such that F ( x ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ1_HTML.gif
(1.1)
Let S denote the solution set of (1.1). Throughout this paper, we assume that S is nonempty and F has the property that
F ( y ) , y x 0 , for all  y C  and all  x S . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ2_HTML.gif
(1.2)

The property (1.2) holds if F is monotone or more generally pseudomonotone on C in the sense of Karamardian [1].

Nonlinear equations have wide applications in reality. For example, many problems arising from chemical technology, economy, and communications can be transformed into nonlinear equations; see [25]. In recent years, many numerical methods for problem (1.1) with smooth mapping F have been proposed. These methods include the Newton method, quasi-Newton method, Levenberg-Marquardt method, trust region method, and their variants; see [614].

Recently, the literature [15] proposed a hybrid method for solving problem (1.1), which combines the Newton, proximal point, and projection methodologies. The method possesses a very nice globally convergent property if F is monotone and continuous. Under the assumptions of differentiability and nonsingularity, locally superlinear convergence of the method is proved. However, the condition of nonsingularity is too strong. Relaxing the nonsingularity assumption, the literature [16] proposed a modified version for the method by changing the projection way, and showed that under the local error bound condition which is weaker than nonsingularity, the proposed method converges superlinearly to the solution of problem (1.1). The numerical performances given in [16] show that the method is really efficient. However, the literatures [15, 16] need the mapping F to be monotone, which seems too stringent a requirement for the purpose of ensuring global convergence property and locally superlinear convergence of the hybrid method.

To further relax the assumption of monotonicity of F, in this paper, we propose a new hybrid algorithm for problem (1.1) which covers one in [16]. The global convergence of our method needs only to assume that F satisfies the property (1.2), which is much weaker than monotone or more generally pseudomonotone. We also discuss the superlinear convergence of our method under mild conditions. Preliminary numerical experiments show that our method is efficient.

2 Preliminaries and algorithms

For a nonempty, closed, and convex set Ω R n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq5_HTML.gif and a vector x R n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq6_HTML.gif, the projection of x onto Ω is defined as
Π Ω ( x ) = arg min { y x | y Ω } . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equa_HTML.gif

We have the following property on the projection operator; see [17].

Lemma 2.1 Let Ω R n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq7_HTML.gif be a closed convex set. Then it holds that
x Π Ω ( y ) 2 x y 2 y Π Ω ( y ) 2 , x Ω , y R n . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equb_HTML.gif

Algorithm 2.1 Choose x 0 C https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq8_HTML.gif, parameters κ 0 [ 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq9_HTML.gif, λ, β ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq10_HTML.gif, γ 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq11_HTML.gif, γ 2 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq12_HTML.gif, a , b 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq13_HTML.gif, max { a , b } > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq14_HTML.gif, and set k : = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq15_HTML.gif.

Step 1. Compute F ( x k ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq16_HTML.gif. If F ( x k ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq17_HTML.gif, stop. Otherwise, let μ k = γ 1 F ( x k ) 1 / 2 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq18_HTML.gif, σ k = min { κ 0 , γ 2 F ( x k ) 1 / 2 } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq19_HTML.gif. Choose a positive semidefinite matrix G k R n × n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq20_HTML.gif. Compute d k R n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq21_HTML.gif such that
F ( x k ) + ( G k + μ k I ) d k = r k , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ3_HTML.gif
(2.1)
where
r k σ k μ k d k . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ4_HTML.gif
(2.2)

Stop if d k = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq22_HTML.gif. Otherwise,

Step 2. Compute y k = x k + t k d k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq23_HTML.gif, where t k = β m k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq24_HTML.gif and m k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq25_HTML.gif is the smallest nonnegative integer m satisfying
F ( x k + β m d k ) , d k λ ( 1 σ k ) μ k d k 2 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ5_HTML.gif
(2.3)
Step 3. Compute
x k + 1 = Π C k ( x k α k F ( y k ) ) , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equc_HTML.gif
where C k : = { x C : h k ( x ) 0 } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq26_HTML.gif and
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ6_HTML.gif
(2.4)

Let k = k + 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq27_HTML.gif and return to Step 1.

Remark 2.1 When we take parameters a = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq28_HTML.gif, b = 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq29_HTML.gif, and the search direction d k = x ¯ k x k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq30_HTML.gif, our algorithm degrades into one in [16]. At this step of getting the next iterate, our projection way and projection region are also different from the one in [15].

Now we analyze the feasibility of Algorithm 2.1. It is obvious that d k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq31_HTML.gif satisfying conditions (2.1) and (2.2) exists. In fact, when we take d k = ( G k + μ k I ) 1 F ( x k ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq32_HTML.gif, d k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq31_HTML.gif satisfies (2.1) and (2.2). Next, we need only to show the feasibility of (2.3).

Lemma 2.2 For all nonnegative integer k, there exists a nonnegative integer m k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq25_HTML.gif satisfying (2.3).

Proof If d k = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq22_HTML.gif, then it follows from (2.1) and (2.2) that F ( x k ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq17_HTML.gif, which means Algorithm 2.1 terminates with x k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq33_HTML.gif being a solution of problem (1.1).

Now, we assume that d k 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq34_HTML.gif, for all k. By the definition of r k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq35_HTML.gif, the Cauchy-Schwarz inequality and the positive semidefiniteness of G k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq36_HTML.gif, we have
F ( x k ) , d k = ( d k ) T ( G k + μ k I ) ( d k ) ( d k ) T r k μ k d k 2 r k d k ( 1 σ k ) μ k d k 2 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ7_HTML.gif
(2.5)
Suppose that the conclusion of Lemma 2.2 does not hold. Then there exists a nonnegative integer k 0 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq37_HTML.gif such that (2.3) is not satisfied for any nonnegative integer m, i.e.,
F ( x k 0 + β m d k 0 ) , d k 0 < λ μ k 0 ( 1 σ k 0 ) d k 0 2 , m . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ8_HTML.gif
(2.6)
Letting m https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq38_HTML.gif and by the continuity of F, we have
F ( x k 0 ) , d k 0 λ μ k 0 ( 1 σ k 0 ) d k 0 2 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equd_HTML.gif

Which, together with (2.5), d k 0 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq39_HTML.gif, and σ k κ 0 < 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq40_HTML.gif, we conclude that λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq41_HTML.gif, which contradicts λ ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq42_HTML.gif. This completes the proof. □

3 Convergence analysis

In this section, we first prove two lemmas, and then analyze the global convergence of Algorithm 2.1.

Lemma 3.1 If the sequences { x k } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq43_HTML.gif and { y k } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq44_HTML.gif are generated by Algorithm  2.1, { x k } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq43_HTML.gif is bounded and F is continuous, then { y k } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq44_HTML.gif is also bounded.

Proof Combining inequality (2.5) with the Cauchy-Schwarz inequality, we obtain
μ k ( 1 σ k ) d k 2 F ( x k ) , d k F ( x k ) d k . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Eque_HTML.gif
By the definition of μ k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq45_HTML.gif and σ k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq46_HTML.gif, it follows that
d k F ( x k ) μ k ( 1 σ k ) F ( x k ) 1 / 2 γ 1 ( 1 κ 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equf_HTML.gif

From the boundedness of { x k } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq47_HTML.gif and the continuity of F, we conclude that { d k } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq48_HTML.gif is bounded, and hence so is { y k } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq44_HTML.gif. This completes the proof. □

Lemma 3.2 Let x https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq49_HTML.gif be a solution of problem (1.1) and the function h k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq50_HTML.gif be defined by (2.4). If condition (1.2) holds, then
h k ( x k ) λ b t k ( 1 σ k ) μ k d k 2 and h k ( x ) 0 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ9_HTML.gif
(3.1)

In particular, if d k 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq34_HTML.gif, then h k ( x k ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq51_HTML.gif.

Proof
h k ( x k ) = a F ( x k ) + b F ( y k ) , x k y k + a t k F ( x k ) , d k = a F ( x k ) , t k d k + b F ( y k ) , t k d k + a t k F ( x k ) , d k = b t k F ( y k ) , d k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ10_HTML.gif
(3.2)
λ b t k ( 1 σ k ) μ k d k 2 , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ11_HTML.gif
(3.3)
where the inequality follows from (2.3).
h k ( x ) = a F ( x k ) + b F ( y k ) , x y k + a t k F ( x k ) , d k = a F ( x k ) , x x k + a F ( x k ) , x k y k + b F ( y k ) , x y k + a t k F ( x k ) , d k 0 , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equg_HTML.gif

where the inequality follows from condition (1.2) and the definition of  y k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq52_HTML.gif.

If d k 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq34_HTML.gif, then h k ( x k ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq51_HTML.gif because σ k κ 0 < 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq40_HTML.gif. The proof is completed. □

Remark 3.1 Lemma 3.2 means that the hyperplane
H k : = { x R n | a F ( x k ) + b F ( y k ) , x y k + a t k F ( x k ) , d k = 0 } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equh_HTML.gif

strictly separates the current iterate from the solution set of problem (1.1).

Let x S https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq53_HTML.gif and d k 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq34_HTML.gif. Since
a F ( x k ) + b F ( y k ) , x k x = a F ( x k ) , x k x + b F ( y k ) , x k x = a F ( x k ) , x k x + b F ( y k ) , x k y k + b F ( y k ) , y k x b F ( y k ) , x k y k λ b t k μ k ( 1 σ k ) d k 2 > 0 , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equi_HTML.gif

where the first inequality follows from condition (1.2), the second one follows from (2.3), and the last one follows d k 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq34_HTML.gif, which shows that ( a F ( x k ) + b F ( y k ) ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq54_HTML.gif is a descent direction of the function 1 2 x x 2 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq55_HTML.gif at the point x k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq33_HTML.gif.

We next prove our main result. Certainly, if Algorithm 2.1 terminates at Step k, then x k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq33_HTML.gif is a solution of problem (1.1). Therefore, in the following analysis, we assume that Algorithm 2.1 always generates an infinite sequence.

Theorem 3.1 If F is continuous on C, condition (1.2) holds and sup k G k < https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq56_HTML.gif, then the sequence { x k } R n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq57_HTML.gif generated by Algorithm  2.1 globally converges to a solution of (1.1).

Proof Let x https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq49_HTML.gif be a solution of problem (1.1). Since x k + 1 = Π C k ( x k α k F ( y k ) ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq58_HTML.gif, it follows from Lemma 2.1 that
x k + 1 x 2 x k α k F ( y k ) x 2 x k + 1 x k + α k F ( y k ) 2 = x k x 2 2 α k F ( y k ) , x k x x k + 1 x k 2 2 α k F ( y k ) , x k + 1 x k , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equj_HTML.gif
i.e.,
x k x 2 x k + 1 x 2 2 α k F ( y k ) , x k x + x k + 1 x k 2 + 2 α k F ( y k ) , x k + 1 x k 2 α k F ( y k ) , x k y k + x k + 1 x k + α k F ( y k ) 2 α k 2 F ( y k ) 2 2 α k F ( y k ) , x k y k α k 2 F ( y k ) 2 = F ( y k ) , x k y k 2 F ( y k ) 2 , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equk_HTML.gif
which shows that the sequence { x k + 1 x } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq59_HTML.gif is nonincreasing, and hence is a convergent sequence. Therefore, { x k } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq43_HTML.gif is bounded and
lim k F ( y k ) , x k y k 2 F ( y k ) 2 = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ12_HTML.gif
(3.4)
From Lemma 3.1 and the continuity of F, we have that { F ( y k ) } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq60_HTML.gif is bounded; that is, there exists a positive constant M such that
F ( y k ) M , for all  k . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equl_HTML.gif
By (2.3) and the choices of σ k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq46_HTML.gif and λ, we have
F ( y k ) , x k y k 2 F ( y k ) 2 = t k 2 F ( y k ) , d k 2 F ( y k ) 2 t k 2 λ 2 ( 1 σ k ) 2 μ k 2 d k 4 M 2 λ 2 ( 1 κ 0 ) 2 t k 2 μ k 2 d k 4 M 2 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equm_HTML.gif
This, together with inequality (3.4), we deduce that
lim k t k μ k d k = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equn_HTML.gif

Now, we consider the following two possible cases:

Suppose first that lim sup k t k > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq61_HTML.gif. Then we must have
lim inf k μ k = 0 or lim inf k d k = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equo_HTML.gif
From the definition of μ k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq45_HTML.gif, the choice of d k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq31_HTML.gif and sup k G k < https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq56_HTML.gif, each case of them follows that
lim inf k F ( x k ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equp_HTML.gif
Since F is continuous and { x k } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq43_HTML.gif is bounded, which implies that the sequence { x k } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq43_HTML.gif has some accumulation point x ˆ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq62_HTML.gif such that
F ( x ˆ ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equq_HTML.gif

This shows that x ˆ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq62_HTML.gif is a solution of problem (1.1). Replacing x https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq49_HTML.gif by x ˆ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq62_HTML.gif in the preceding argument, we obtain that the sequence { x k x ˆ } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq63_HTML.gif is nonincreasing, and hence converges. Since x ˆ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq62_HTML.gif is an accumulation point of { x k } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq47_HTML.gif, some subsequence of { x k x ˆ } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq64_HTML.gif converges to zero, which implies that the whole sequence { x k x ˆ } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq64_HTML.gif converges to zero, and hence lim k x k = x ˆ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq65_HTML.gif.

Suppose now that lim k t k = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq66_HTML.gif. Let x ¯ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq67_HTML.gif be any accumulation point of { x k } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq43_HTML.gif and { x k j } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq68_HTML.gif be the corresponding subsequence converging to x ¯ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq67_HTML.gif. By the choice of t k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq69_HTML.gif, (2.3) implies that
F ( x k j + t k j β 1 d k j ) , d k j < λ ( 1 σ k j ) μ k j d k j 2 , for all  j . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equr_HTML.gif
Since F is continuous, we obtain by letting j https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq70_HTML.gif that
F ( x k j ) , d k j λ ( 1 σ k j ) μ k j d k j 2 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ13_HTML.gif
(3.5)

From (2.5) and (3.5), we conclude that λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq41_HTML.gif, which contradicts λ ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq42_HTML.gif. Hence, the case of lim k t k = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq66_HTML.gif is not possible. This completes the proof. □

Remark 3.2 Compared to the conditions of the global convergence used in literatures [15, 16], our conditions are weaker.

4 Convergence rate

In this section, we provide a result on the convergence rate of the iterative sequence generated by Algorithm 2.1. To establish this result, we need the following conditions (4.1) and (4.2).

For x S https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq71_HTML.gif, there are positive constants δ, c 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq72_HTML.gif, and c 2 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq73_HTML.gif such that
c 1 dist ( x , S ) F ( x ) , x N ( x , δ ) , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ14_HTML.gif
(4.1)
and
F ( x ) F ( y ) G k ( x y ) c 2 x y 2 , x , y N ( x , δ ) , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ15_HTML.gif
(4.2)
where dist ( x , S ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq74_HTML.gif denotes the distance from x to solution set S, and
N ( x , δ ) = { x R n | x x δ } . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equs_HTML.gif
If F is differentiable and F ( ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq75_HTML.gif is locally Lipschitz continuous with modulus θ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq76_HTML.gif, then there exists a constant L 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq77_HTML.gif such that
F ( y ) F ( x ) F ( x ) ( y x ) L 1 y x 2 , x , y N ( x , δ ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ16_HTML.gif
(4.3)
In fact, by the mean value theorem of vector valued function, we have
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equt_HTML.gif
where L 1 = θ / 2 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq78_HTML.gif. Under assumptions (4.2) or (4.3), it is readily shown that there exists a constant L 2 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq79_HTML.gif such that
F ( y ) F ( x ) L 2 y x , x , y N ( x , δ ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ17_HTML.gif
(4.4)

In 1998, the literature [15] showed that their proposed method converged superlinearly when the underlying function F is monotone, differentiable with F ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq80_HTML.gif being nonsingular, and ∇F is locally Lipschitz continuous. It is known that the local error bound condition given in (4.1) is weaker than the nonsingular. Recently, under conditions (4.1), (4.2), and the underlying function F being monotone and continuous, the literature [16] obtained the locally superlinear rate of convergence of the proposed method.

Next, we analyze the superlinear convergence rate of the iterative sequence under a weaker condition. In the rest of section, we assume that x k x https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq81_HTML.gif, k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq82_HTML.gif, where x S https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq71_HTML.gif.

Lemma 4.1 Let G R n × n https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq83_HTML.gif be a positive semidefinite matrix and μ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq84_HTML.gif. Then

(1) ( G + μ I ) 1 1 μ https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq85_HTML.gif;

(2) ( G + μ I ) 1 G 2 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq86_HTML.gif.

Proof See [18]. □

Lemma 4.2 Suppose that F is continuous and satisfies conditions (1.2), (4.1), and (4.2). If there exists a positive constant N such that G k N https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq87_HTML.gif for all k, then for all k sufficiently large,

(1) c 3 d k F ( x k ) c 4 d k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq88_HTML.gif;

(2) F ( x k ) + G k d k c 5 d k 3 / 2 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq89_HTML.gif, where c 3 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq90_HTML.gif, c 4 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq91_HTML.gif and c 5 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq92_HTML.gif are all positive constants.

Proof For (1), let x k N ( x , 1 2 δ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq93_HTML.gif and x ˆ k S https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq94_HTML.gif be the closest solution to x k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq33_HTML.gif. We have
x ˆ k x x ˆ k x k + x k x δ , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equu_HTML.gif
i.e., x ˆ k N ( x , δ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq95_HTML.gif. Thus, by (2.1), (2.2), (4.2), and Lemma 4.1, we have
d k ( G k + μ k I ) 1 F ( x k ) + ( G k + μ k I ) 1 r k ( G k + μ k I ) 1 [ F ( x ˆ k ) F ( x k ) G k ( x ˆ k x k ) ] + ( G k + μ k I ) 1 G k ( x ˆ k x k ) + 1 μ k r k c 2 μ k x ˆ k x k 2 + 2 x ˆ k x k + σ k d k . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equv_HTML.gif
By x k x ˆ k = dist ( x k , S ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq96_HTML.gif and σ k κ 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq97_HTML.gif, it follows that
( 1 κ 0 ) d k ( c 2 μ k dist ( x k , S ) + 2 ) dist ( x k , S ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equw_HTML.gif
From (4.1) and the choice of μ k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq45_HTML.gif, it holds that
c 2 μ k dist ( x k , S ) c 1 1 c 2 F ( x k ) γ 1 F ( x k ) 1 / 2 = c 2 γ 1 c 1 F ( x k ) 1 / 2 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equx_HTML.gif
From the boundedness of { F ( x k ) } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq98_HTML.gif, there exists a positive constant M 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq99_HTML.gif such that
F ( x k ) 1 / 2 M 1 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equy_HTML.gif
Therefore,
d k c 2 M 1 + 2 γ 1 c 1 c 1 γ 1 ( 1 κ 0 ) dist ( x k , S ) c 2 M 1 + 2 γ 1 c 1 c 1 2 γ 1 ( 1 κ 0 ) F ( x k ) . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ18_HTML.gif
(4.5)

We obtain that the left-hand side of (1) by setting c 3 : = c 1 2 γ 1 ( 1 κ 0 ) c 2 M 1 + 2 γ 1 c 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq100_HTML.gif.

For the right-hand side part, it follows from (2.1) and (2.2) that
F ( x k ) G k + μ k I d k + r k ( G k + μ k I + σ k μ k ) d k ( N + γ 1 M 1 + κ 0 γ 1 M 1 ) d k . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equz_HTML.gif

We obtain the right-hand side part by setting c 4 : = N + γ 1 M 1 + κ 0 γ 1 M 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq101_HTML.gif.

For (2), using (2.1) and (2.2), we have
F ( x k ) + G k d k μ k d k + r k ( 1 + σ k ) μ k d k ( 1 + κ 0 ) γ 1 F ( x k ) 1 / 2 d k ( 1 + κ 0 ) γ 1 c 4 1 / 2 d k 3 / 2 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equaa_HTML.gif

By setting c 5 : = ( 1 + κ 0 ) γ 1 c 4 1 / 2 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq102_HTML.gif, we obtain the desired result. □

Lemma 4.3 Suppose that the assumptions in Lemma  4.2 hold. Then for all k sufficiently large, it holds that
y k = x k + d k . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equab_HTML.gif
Proof By lim k x k = x https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq103_HTML.gif and the continuity of F, we have
lim k F ( x k ) = F ( x ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equac_HTML.gif
By Lemma 4.2(1), we obtain that
lim k d k = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equad_HTML.gif
which means that x k + d k N ( x , δ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq104_HTML.gif for all k sufficiently large. Hence, it follows from (4.2) that
F ( x k + d k ) = F ( x k ) + G k d k + R k , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ19_HTML.gif
(4.6)
where R k c 2 d k 2 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq105_HTML.gif. Using (2.1) and (2.2), (4.6) can be written as
F ( x k + d k ) = μ k d k + r k + R k . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ20_HTML.gif
(4.7)
Hence,
F ( x k + d k ) , d k = μ k d k , d k r k d k R k d k μ k d k 2 σ k μ k d k 2 c 2 d k 3 = ( 1 c 2 d k μ k ( 1 σ k ) ) μ k ( 1 σ k ) d k 2 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equae_HTML.gif
By Lemma 4.2(1) and the choices of μ k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq45_HTML.gif and σ k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq46_HTML.gif, for k sufficiently large, we obtain
1 1 c 2 d k μ k ( 1 σ k ) 1 c 2 c 3 1 F ( x k ) ( 1 κ 0 ) γ 1 F ( x k ) 1 / 2 = 1 c 2 c 3 1 F ( x k ) 1 / 2 ( 1 κ 0 ) γ 1 λ , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equaf_HTML.gif

where the last inequality follows from lim k F ( x k ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq106_HTML.gif.

Therefore,
F ( x k + d k ) , d k λ μ k ( 1 σ k ) d k 2 , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equag_HTML.gif

which implies that (2.3) holds with t k = 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq107_HTML.gif for all k sufficiently large, i.e., y k = x k + d k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq108_HTML.gif. This completes the proof. □

From now on, we assume that k is large enough so that y k = x k + d k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq108_HTML.gif.

Lemma 4.4 Suppose that the assumptions in Lemma  4.2 hold. Set x ˜ k : = x k α k F ( y k ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq109_HTML.gif. Then for all k sufficiently large, there exists a positive constant c 6 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq110_HTML.gif such that
https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equah_HTML.gif
Proof Set
H k 1 = { x R n | F ( y k ) , x y k = 0 } . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equai_HTML.gif
Then x ˜ k = Π H k 1 ( x k ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq111_HTML.gif and y k H k 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq112_HTML.gif. Hence, the vectors x k x ˜ k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq113_HTML.gif and y k x ˜ k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq114_HTML.gif are orthogonal. That is,
y k x ˜ k = y k x k sin θ k = d k sin θ k , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ21_HTML.gif
(4.8)
where θ k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq115_HTML.gif is the angle between x ˜ k x k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq116_HTML.gif and y k x k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq117_HTML.gif. Because x ˜ k x k = α k F ( y k ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq118_HTML.gif and y k x k = d k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq119_HTML.gif, the angle between F ( y k ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq120_HTML.gif and μ k d k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq121_HTML.gif is also θ k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq115_HTML.gif. By (4.7), we obtain
F ( y k ) ( μ k d k ) = R k + r k , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equaj_HTML.gif
which implies that the vectors F ( y k ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq120_HTML.gif, μ k d k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq121_HTML.gif and R k + r k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq122_HTML.gif constitute a triangle. Since lim k μ k = lim k γ 1 F ( x k ) 1 / 2 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq123_HTML.gif and lim k α k = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq124_HTML.gif. So for all k sufficiently large, we have
sin θ k r k + R k μ k d k σ k + c 2 d k μ k γ 2 F ( x k ) 1 / 2 + c 2 F ( x k ) c 3 γ 1 F ( x k ) 1 / 2 = ( γ 2 + c 2 c 3 γ 1 ) F ( x k ) 1 / 2 , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equak_HTML.gif
which, together with (4.8) and Lemma 4.2(1), we obtain
y k x ˜ k ( γ 2 + c 2 c 3 γ 1 ) F ( x k ) 1 / 2 d k c 6 d k 3 / 2 , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equal_HTML.gif

where c 6 = c 4 1 / 2 ( γ 2 + c 2 c 3 γ 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq125_HTML.gif. This completes the proof. □

Now, we turn our attention to local rate of convergence analysis.

Theorem 4.1 Suppose that the assumptions in Lemma  4.2 hold. Then the sequence { dist ( x k , S ) } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq126_HTML.gif Q-superlinearly converges to 0.

Proof By the definition of x ˜ k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq127_HTML.gif, Lemma 4.2(1) and (4.4), for sufficiently large k, we have
x ˜ k x x k α k F ( y k ) x x k x + F ( y k ) , x k y k F ( y k ) 2 F ( y k ) x k x + d k x k x + c 3 1 F ( x k ) = x k x + c 3 1 F ( x k ) F ( x ) ( 1 + L 2 c 3 1 ) x k x , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equam_HTML.gif
which implies that lim k x ˜ k x = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq128_HTML.gif because lim k x k x = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq129_HTML.gif. Thus, x ˜ k N ( x , δ ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq130_HTML.gif for k sufficiently large, which, together with (4.2), Lemma 4.2, Lemma 4.4, and the definition of x ˜ k https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq127_HTML.gif, we obtain
F ( x ˜ k ) F ( x k ) + G k ( x ˜ k x k ) + c 2 x ˜ k x k 2 F ( x k ) + G k ( y k x k ) + G k x ˜ k y k + c 2 x ˜ k x k 2 c 5 d k 3 / 2 + N c 6 d k 3 / 2 + c 2 α k F ( y k ) 2 ( c 5 + N c 6 ) d k 3 / 2 + c 2 d k 2 = ( c 5 + N c 6 + c 2 d k 1 / 2 ) d k 3 / 2 ( c 5 + N c 6 + c 2 c 3 1 / 2 F ( x k ) 1 / 2 ) d k 3 / 2 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equan_HTML.gif
Because { F ( x k ) } https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq98_HTML.gif is bounded, there exists a positive constant c 7 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq131_HTML.gif such that
F ( x ˜ k ) c 7 d k 3 / 2 . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equ22_HTML.gif
(4.9)
On the other hand, from Lemma 3.2, we know that
S C H k , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equao_HTML.gif
where S is the solution set of problem (1.1). Since x k + 1 = Π C H k ( x ˜ k ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq132_HTML.gif, it follows from Lemma 2.1 that
x k + 1 x 2 x ˜ k x 2 x k + 1 x ˜ k 2 , x S , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equap_HTML.gif
which implies that
x k + 1 x x ˜ k x . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equaq_HTML.gif
Therefore, together with inequalities (4.1), (4.5), and (4.9), we have
dist ( x k + 1 , S ) dist ( x ˜ k , S ) 1 c 1 F ( x ˜ k ) c 7 c 1 d k 3 / 2 c 7 c 1 ( c 2 M 1 + 2 γ 1 c 1 c 1 γ 1 ( 1 κ 0 ) ) 3 / 2 dist 3 / 2 ( x k , S ) , https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equar_HTML.gif

which implies that the order of superlinear convergence is at least 1.5. This completes the proof. □

Remark 4.1 Compared with the proof of the locally superlinear convergence in literatures [15, 16], our conditions are weaker.

5 Numerical experiments

In this section, we present some numerical experiments results to show the efficiency of our method. The MATLAB codes are run on a notebook computer with CPU2.10GHZ under MATLAB Version 7.0. Just as done in [16], we take G k = F ( x k ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq133_HTML.gif and use the left division operation in MATLAB to solve the system of linear equations (2.1) at each iteration. We choose b = 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq29_HTML.gif, λ = 0.96 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq134_HTML.gif, κ 0 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq135_HTML.gif, β = 0.7 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq136_HTML.gif, and γ 1 = 1 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq137_HTML.gif. ‘Iter.’ denotes the number of iteration and ‘CPU’ denotes the CPU time in seconds. We choose F ( x k ) 10 6 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq138_HTML.gif as the stop criterion. The example is tested in [16].

Example Let
F ( x ) = ( 1 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 ) ( x 1 x 2 x 3 x 4 ) + ( x 1 3 x 2 3 2 x 3 3 2 x 4 3 ) + ( 10 1 3 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equas_HTML.gif
and the constraint set C be taken as
C = { x R 4 | i = 1 4 x i 3 , x i 0 , i = 1 , 2 , 3 , 4 } . https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_Equat_HTML.gif
From Tables 1-2, we can see that our algorithm is efficient if parameters are chosen properly. We can also observe that the algorithm’s operation results change with the value of a. When we take a = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq28_HTML.gif, the operation results are not best, that is to say, the direction F ( y k ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq120_HTML.gif is not an optimal one.
Table 1

Numerical results of Example with a = 10 15 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq139_HTML.gif

Initial point

Iter.

CPU

F ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq140_HTML.gif

(3,0,0,0)

11

0.10

1.07 × 10−8

(1,1,0,0)

13

0.09

1.62 × 10−9

(0,1,0,1)

15

0.04

2.46 × 10−9

(0,0,0,1)

21

0.18

9.92 × 10−10

(1,0,0,2)

16

0.54

5.66 × 10−10

Table 2

Numerical results of Example with a = 0 https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq28_HTML.gif

Initial point

Iter.

CPU

F ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-180/MediaObjects/13660_2012_Article_303_IEq140_HTML.gif

(3,0,0,0)

11

0.10

1.07 × 10−8

(1,1,0,0)

13

0.12

1.62 × 10−9

(0,1,0,1)

19

0.14

1.17 × 10−9

(0,0,0,1)

18

0.18

1.44 × 10−9

(1,0,0,2)

15

0.21

7.88 × 10−9

Acknowledgements

The author wish to thank the anonymous referees for their suggestions and comments. This work is also supported by the Educational Science Foundation of Chongqing, Chongqing of China (Grant No. KJ111309).

Copyright information

© Zheng; licensee Springer 2012

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.