A superlinearly convergent hybrid algorithm for systems of nonlinear equations

Open Access
Research

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 Open image in new window be a continuous mapping and C R n Open image in new window be a nonempty, closed, and convex set. The inner product and norm are denoted by , Open image in new window and Open image in new window, respectively. Consider the problem of finding
x C such that F ( x ) = 0 . Open image in new window
(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 . Open image in new window
(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 [2, 3, 4, 5]. 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 [6, 7, 8, 9, 10, 11, 12, 13, 14].

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 Open image in new window and a vector x R n Open image in new window, the projection of x onto Ω is defined as
Π Ω ( x ) = arg min { y x | y Ω } . Open image in new window

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

Lemma 2.1Let Ω R n Open image in new windowbe 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 . Open image in new window

Algorithm 2.1 Choose x 0 C Open image in new window, parameters κ 0 [ 0 , 1 ) Open image in new window, λ, β ( 0 , 1 ) Open image in new window, γ 1 Open image in new window, γ 2 > 0 Open image in new window, a , b 0 Open image in new window, max { a , b } > 0 Open image in new window, and set k : = 0 Open image in new window.

Step 1. Compute F ( x k ) Open image in new window. If F ( x k ) = 0 Open image in new window, stop. Otherwise, let μ k = γ 1 F ( x k ) 1 / 2 Open image in new window, σ k = min { κ 0 , γ 2 F ( x k ) 1 / 2 } Open image in new window. Choose a positive semidefinite matrix G k R n × n Open image in new window. Compute d k R n Open image in new window such that
F ( x k ) + ( G k + μ k I ) d k = r k , Open image in new window
(2.1)
where
r k σ k μ k d k . Open image in new window
(2.2)

Stop if d k = 0 Open image in new window. Otherwise,

Step 2. Compute y k = x k + t k d k Open image in new window, where t k = β m k Open image in new window and m k Open image in new window 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 . Open image in new window
(2.3)
Step 3. Compute
x k + 1 = Π C k ( x k α k F ( y k ) ) , Open image in new window
where C k : = { x C : h k ( x ) 0 } Open image in new window and

Let k = k + 1 Open image in new window and return to Step 1.

Remark 2.1 When we take parameters a = 0 Open image in new window, b = 1 Open image in new window, and the search direction d k = x ¯ k x k Open image in new window, 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 Open image in new window satisfying conditions (2.1) and (2.2) exists. In fact, when we take d k = ( G k + μ k I ) 1 F ( x k ) Open image in new window, d k Open image in new window satisfies (2.1) and (2.2). Next, we need only to show the feasibility of (2.3).

Lemma 2.2For all nonnegative integerk, there exists a nonnegative integer m k Open image in new windowsatisfying (2.3).

Proof If d k = 0 Open image in new window, then it follows from (2.1) and (2.2) that F ( x k ) = 0 Open image in new window, which means Algorithm 2.1 terminates with x k Open image in new window being a solution of problem (1.1).

Now, we assume that d k 0 Open image in new window, for all k. By the definition of r k Open image in new window, the Cauchy-Schwarz inequality and the positive semidefiniteness of G k Open image in new window, 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 . Open image in new window
(2.5)
Suppose that the conclusion of Lemma 2.2 does not hold. Then there exists a nonnegative integer k 0 0 Open image in new window 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 . Open image in new window
(2.6)
Letting m Open image in new window 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 . Open image in new window

Which, together with (2.5), d k 0 0 Open image in new window, and σ k κ 0 < 1 Open image in new window, we conclude that λ 1 Open image in new window, which contradicts λ ( 0 , 1 ) Open image in new window. 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 3.1If the sequences { x k } Open image in new windowand { y k } Open image in new windoware generated by Algorithm  2.1, { x k } Open image in new windowis bounded andFis continuous, then { y k } Open image in new windowis 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 . Open image in new window
By the definition of μ k Open image in new window and σ k Open image in new window, it follows that
d k F ( x k ) μ k ( 1 σ k ) F ( x k ) 1 / 2 γ 1 ( 1 κ 0 ) . Open image in new window

From the boundedness of { x k } Open image in new window and the continuity of F, we conclude that { d k } Open image in new window is bounded, and hence so is { y k } Open image in new window. This completes the proof. □

Lemma 3.2Let x Open image in new windowbe a solution of problem (1.1) and the function h k Open image in new windowbe 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 . Open image in new window
(3.1)

In particular, if d k 0 Open image in new window, then h k ( x k ) > 0 Open image in new window.

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 Open image in new window
(3.2)
λ b t k ( 1 σ k ) μ k d k 2 , Open image in new window
(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 , Open image in new window

where the inequality follows from condition (1.2) and the definition of  y k Open image in new window.

If d k 0 Open image in new window, then h k ( x k ) > 0 Open image in new window because σ k κ 0 < 1 Open image in new window. 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 } Open image in new window

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

Let x S Open image in new window and d k 0 Open image in new window. 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 , Open image in new window

where the first inequality follows from condition (1.2), the second one follows from (2.3), and the last one follows d k 0 Open image in new window, which shows that ( a F ( x k ) + b F ( y k ) ) Open image in new window is a descent direction of the function 1 2 x x 2 Open image in new window at the point x k Open image in new window.

We next prove our main result. Certainly, if Algorithm 2.1 terminates at Step k, then x k Open image in new window is a solution of problem (1.1). Therefore, in the following analysis, we assume that Algorithm 2.1 always generates an infinite sequence.

Theorem 3.1IfFis continuous onC, condition (1.2) holds and sup k G k < Open image in new window, then the sequence { x k } R n Open image in new windowgenerated by Algorithm  2.1 globally converges to a solution of (1.1).

Proof Let x Open image in new window be a solution of problem (1.1). Since x k + 1 = Π C k ( x k α k F ( y k ) ) Open image in new window, 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 , Open image in new window
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 , Open image in new window
which shows that the sequence { x k + 1 x } Open image in new window is nonincreasing, and hence is a convergent sequence. Therefore, { x k } Open image in new window is bounded and
lim k F ( y k ) , x k y k 2 F ( y k ) 2 = 0 . Open image in new window
(3.4)
From Lemma 3.1 and the continuity of F, we have that { F ( y k ) } Open image in new window is bounded; that is, there exists a positive constant M such that
F ( y k ) M , for all  k . Open image in new window
By (2.3) and the choices of σ k Open image in new window 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 . Open image in new window
This, together with inequality (3.4), we deduce that
lim k t k μ k d k = 0 . Open image in new window

Now, we consider the following two possible cases:

Suppose first that lim sup k t k > 0 Open image in new window. Then we must have
lim inf k μ k = 0 or lim inf k d k = 0 . Open image in new window
From the definition of μ k Open image in new window, the choice of d k Open image in new window and sup k G k < Open image in new window, each case of them follows that
lim inf k F ( x k ) = 0 . Open image in new window
Since F is continuous and { x k } Open image in new window is bounded, which implies that the sequence { x k } Open image in new window has some accumulation point x ˆ Open image in new window such that
F ( x ˆ ) = 0 . Open image in new window

This shows that x ˆ Open image in new window is a solution of problem (1.1). Replacing x Open image in new window by x ˆ Open image in new window in the preceding argument, we obtain that the sequence { x k x ˆ } Open image in new window is nonincreasing, and hence converges. Since x ˆ Open image in new window is an accumulation point of { x k } Open image in new window, some subsequence of { x k x ˆ } Open image in new window converges to zero, which implies that the whole sequence { x k x ˆ } Open image in new window converges to zero, and hence lim k x k = x ˆ Open image in new window.

Suppose now that lim k t k = 0 Open image in new window. Let x ¯ Open image in new window be any accumulation point of { x k } Open image in new window and { x k j } Open image in new window be the corresponding subsequence converging to x ¯ Open image in new window. By the choice of t k Open image in new window, (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 . Open image in new window
Since F is continuous, we obtain by letting j Open image in new window that
F ( x k j ) , d k j λ ( 1 σ k j ) μ k j d k j 2 . Open image in new window
(3.5)

From (2.5) and (3.5), we conclude that λ 1 Open image in new window, which contradicts λ ( 0 , 1 ) Open image in new window. Hence, the case of lim k t k = 0 Open image in new window 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 Open image in new window, there are positive constants δ, c 1 Open image in new window, and c 2 Open image in new window such that
c 1 dist ( x , S ) F ( x ) , x N ( x , δ ) , Open image in new window
(4.1)
and
F ( x ) F ( y ) G k ( x y ) c 2 x y 2 , x , y N ( x , δ ) , Open image in new window
(4.2)
where dist ( x , S ) Open image in new window denotes the distance from x to solution set S, and
N ( x , δ ) = { x R n | x x δ } . Open image in new window
If F is differentiable and F ( ) Open image in new window is locally Lipschitz continuous with modulus θ > 0 Open image in new window, then there exists a constant L 1 > 0 Open image in new window such that
F ( y ) F ( x ) F ( x ) ( y x ) L 1 y x 2 , x , y N ( x , δ ) . Open image in new window
(4.3)
In fact, by the mean value theorem of vector valued function, we have
where L 1 = θ / 2 Open image in new window. Under assumptions (4.2) or (4.3), it is readily shown that there exists a constant L 2 > 0 Open image in new window such that
F ( y ) F ( x ) L 2 y x , x , y N ( x , δ ) . Open image in new window
(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 ) Open image in new window 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 Open image in new window, k Open image in new window, where x S Open image in new window.

Lemma 4.1Let G R n × n Open image in new windowbe a positive semidefinite matrix and μ > 0 Open image in new window. Then

(1) ( G + μ I ) 1 1 μ Open image in new window;

(2) ( G + μ I ) 1 G 2 Open image in new window.

Proof See [18]. □

Lemma 4.2Suppose thatFis continuous and satisfies conditions (1.2), (4.1), and (4.2). If there exists a positive constantNsuch that G k N Open image in new windowfor allk, then for allksufficiently large,

(1) c 3 d k F ( x k ) c 4 d k Open image in new window;

(2) F ( x k ) + G k d k c 5 d k 3 / 2 Open image in new window, where c 3 Open image in new window, c 4 Open image in new windowand c 5 Open image in new windoware all positive constants.

Proof For (1), let x k N ( x , 1 2 δ ) Open image in new window and x ˆ k S Open image in new window be the closest solution to x k Open image in new window. We have
x ˆ k x x ˆ k x k + x k x δ , Open image in new window
i.e., x ˆ k N ( x , δ ) Open image in new window. 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 . Open image in new window
By x k x ˆ k = dist ( x k , S ) Open image in new window and σ k κ 0 Open image in new window, it follows that
( 1 κ 0 ) d k ( c 2 μ k dist ( x k , S ) + 2 ) dist ( x k , S ) . Open image in new window
From (4.1) and the choice of μ k Open image in new window, 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 . Open image in new window
From the boundedness of { F ( x k ) } Open image in new window, there exists a positive constant M 1 Open image in new window such that
F ( x k ) 1 / 2 M 1 . Open image in new window
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 ) . Open image in new window
(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 Open image in new window.

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 . Open image in new window

We obtain the right-hand side part by setting c 4 : = N + γ 1 M 1 + κ 0 γ 1 M 1 Open image in new window.

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 . Open image in new window

By setting c 5 : = ( 1 + κ 0 ) γ 1 c 4 1 / 2 Open image in new window, we obtain the desired result. □

Lemma 4.3Suppose that the assumptions in Lemma  4.2 hold. Then for allksufficiently large, it holds that
y k = x k + d k . Open image in new window
Proof By lim k x k = x Open image in new window and the continuity of F, we have
lim k F ( x k ) = F ( x ) = 0 . Open image in new window
By Lemma 4.2(1), we obtain that
lim k d k = 0 , Open image in new window
which means that x k + d k N ( x , δ ) Open image in new window 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 , Open image in new window
(4.6)
where R k c 2 d k 2 Open image in new window. 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 . Open image in new window
(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 . Open image in new window
By Lemma 4.2(1) and the choices of μ k Open image in new window and σ k Open image in new window, 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 , Open image in new window

where the last inequality follows from lim k F ( x k ) = 0 Open image in new window.

Therefore,
F ( x k + d k ) , d k λ μ k ( 1 σ k ) d k 2 , Open image in new window

which implies that (2.3) holds with t k = 1 Open image in new window for all k sufficiently large, i.e., y k = x k + d k Open image in new window. This completes the proof. □

From now on, we assume that k is large enough so that y k = x k + d k Open image in new window.

Lemma 4.4Suppose that the assumptions in Lemma  4.2 hold. Set x ˜ k : = x k α k F ( y k ) Open image in new window. Then for allksufficiently large, there exists a positive constant c 6 Open image in new windowsuch that
Proof Set
H k 1 = { x R n | F ( y k ) , x y k = 0 } . Open image in new window
Then x ˜ k = Π H k 1 ( x k ) Open image in new window and y k H k 1 Open image in new window. Hence, the vectors x k x ˜ k Open image in new window and y k x ˜ k Open image in new window are orthogonal. That is,
y k x ˜ k = y k x k sin θ k = d k sin θ k , Open image in new window
(4.8)
where θ k Open image in new window is the angle between x ˜ k x k Open image in new window and y k x k Open image in new window. Because x ˜ k x k = α k F ( y k ) Open image in new window and y k x k = d k Open image in new window, the angle between F ( y k ) Open image in new window and μ k d k Open image in new window is also θ k Open image in new window. By (4.7), we obtain
F ( y k ) ( μ k d k ) = R k + r k , Open image in new window
which implies that the vectors F ( y k ) Open image in new window, μ k d k Open image in new window and R k + r k Open image in new window constitute a triangle. Since lim k μ k = lim k γ 1 F ( x k ) 1 / 2 = 0 Open image in new window and lim k α k = 0 Open image in new window. 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 , Open image in new window
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 , Open image in new window

where c 6 = c 4 1 / 2 ( γ 2 + c 2 c 3 γ 1 ) Open image in new window. This completes the proof. □

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

Theorem 4.1Suppose that the assumptions in Lemma  4.2 hold. Then the sequence { dist ( x k , S ) } Open image in new windowQ-superlinearly converges to 0.

Proof By the definition of x ˜ k Open image in new window, 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 , Open image in new window
which implies that lim k x ˜ k x = 0 Open image in new window because lim k x k x = 0 Open image in new window. Thus, x ˜ k N ( x , δ ) Open image in new window for k sufficiently large, which, together with (4.2), Lemma 4.2, Lemma 4.4, and the definition of x ˜ k Open image in new window, 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 . Open image in new window
Because { F ( x k ) } Open image in new window is bounded, there exists a positive constant c 7 Open image in new window such that
F ( x ˜ k ) c 7 d k 3 / 2 . Open image in new window
(4.9)
On the other hand, from Lemma 3.2, we know that
S C H k , Open image in new window
where S is the solution set of problem (1.1). Since x k + 1 = Π C H k ( x ˜ k ) Open image in new window, it follows from Lemma 2.1 that
x k + 1 x 2 x ˜ k x 2 x k + 1 x ˜ k 2 , x S , Open image in new window
which implies that
x k + 1 x x ˜ k x . Open image in new window
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 ) , Open image in new window

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 ) Open image in new window and use the left division operation in MATLAB to solve the system of linear equations (2.1) at each iteration. We choose b = 1 Open image in new window, λ = 0.96 Open image in new window, κ 0 = 0 Open image in new window, β = 0.7 Open image in new window, and γ 1 = 1 Open image in new window. ‘Iter.’ denotes the number of iteration and ‘CPU’ denotes the CPU time in seconds. We choose F ( x k ) 10 6 Open image in new window 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 ) Open image in new window
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 } . Open image in new window
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 Open image in new window, the operation results are not best, that is to say, the direction F ( y k ) Open image in new window is not an optimal one.
Table 1

Numerical results of Example with a = 10 15 Open image in new window

Initial point

Iter.

CPU

F ( x ) Open image in new window

(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 Open image in new window

Initial point

Iter.

CPU

F ( x ) Open image in new window

(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

Notes

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).

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© 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.

Authors and Affiliations

  1. 1.Department of Mathematics and Computer ScienceYangtze Normal UniversityFulingChina

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