Applied Mathematics and Computation 188 (2007) 514–524 www.elsevier.com/locate/amc
Convergence analysis of the secant type methods Jinhai Chen a
a,*
, Zuhe Shen
b
Department of Mathematics, City University of Hong Kong, 83 Tat Chee Avenue, Kowloon, Hong Kong b Department of Mathematics, Nanjing University, Nanjing 210093, People’s Republic of China
Abstract In this paper, under the hypothesis that derivative satisfies some kinds of weak Lipschitz condition, the radius estimates of the convergence balls of secant type methods for operator equations are established in Banach space. Some applications and numerical experiments are also given. Ó 2006 Elsevier Inc. All rights reserved. Keywords: Secant type methods; Convergence ball; Uniqueness ball
1. Introduction Consider a Fre´chet differential nonlinear operator equation F ðxÞ ¼ 0;
ð1:1Þ
defined on a convex open subset D of a Banach space X with values in a Banach space Y. Solving the problem (1.1) is a basic and important problem in applied and computational mathematics. The most well-known method to solve (1.1) is the Newton’s method, which is quadratically convergent under the proper conditions showed by the well-known Kantorovich theorem [5,6]. Classical secant method can be used to solve the problem (1.1), instead of Newton’s method. Starting with two points, x1, x0 2 D, the secant method is described by the following procedure: 1
xnþ1 ¼ xn ½xn1 ; xn ; F F ðxn Þ;
n P 0;
ð1:2Þ
where the bound linear operator [x, y; F] is called a divided difference of first order for operator F on the points x and y and it satisfies the following equality: ½x; y; F ðx yÞ ¼ F ðxÞ F ðyÞ:
*
Corresponding author. E-mail address:
[email protected] (J. Chen).
0096-3003/$ - see front matter Ó 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2006.10.014
ð1:3Þ
J. Chen, Z. Shen / Applied Mathematics and Computation 188 (2007) 514–524
515
The convergence of the classical secant method (1.2) has been studied by many authors [1–3,6–8]. Since the finite difference approximation of the Jacobian is not very accurate, and the first iterations of this method are usually worse than Newton’s method, in [2,3], the following modification of secant method was discussed: xnþ1 ¼ xn ½xn ; xn þ an ðxn1 xn Þ; F 1 F ðxn Þ;
ð1:4Þ
n P 0;
where a 2 [0, 1]. This new iterative method is, in general, a good alternative to Newton’s method, since [xn, xn + an(xn1 xn); F] is always the good approximation to F 0 (xn). Moreover, it is more efficient than classical secant method (1.2), we refer the reader to [1,7] and the references therein. Now we suppose x* is a solution of (1.1), analyze the local convergence of secant type method (1.4) and in particular, estimate the radius of the convergence ball of the secant method. Let x* denote the solution of (1.1), B(x, r) denote an open ball with radius r and center x, and let Bðx; rÞ denote its closure. With the derivative of F satisfying the Lipschitz condition: 1
kF 0 ðx Þ ðF 0 ðxÞ F 0 ðyÞÞk 6 Lkx yk;
8x; y 2 Bðx ; rÞ; for some L > 0:
Traub and Wozniakowski [9] and Wang [10] independently gave an exact estimate of radius r ¼ vergence ball of Newton’s method: 1
xnþ1 ¼ xn F 0 ðxn Þ F ðxÞ;
ð1:5Þ 2 3L
for the con-
n ¼ 0; 1; 2; . . .
Under the hypothesis that F 0 (x) satisfies the following Lipschitz conditions: Z qðxÞ 0 1 0 0 s LðuÞ du; 8x 2 Bðx ; rÞ; kF ðx Þ ðF ðxÞ F ðx ÞÞk 6
ð1:6Þ
sqðxÞ
0
1
0
0
kF ðx Þ ðF ðxÞ F ðx ÞÞk 6
Z
qðxÞ
8x 2 Bðx ; rÞ;
LðuÞ du;
ð1:7Þ
0
where q(x) = kx x*k, xs = x* + s(x x*), and L is monotonic function, Wang [11,12] studied the convergence of the Newton’s method. In this paper, we study the convergence of the secant type method (1.4) with the following Lipschitz conditions [13–16] for operator F which are more general than (1.6), (1.7): 1
8x; y 2 Bðx ; rÞ;
ð1:8Þ
1
8x 2 Bðx ; rÞ;
ð1:9Þ
kF 0 ðx Þ ðF 0 ðxÞ F 0 ðyÞÞk 6 f0 ðkx ykÞ; kF 0 ðx Þ ðF 0 ðxÞ F 0 ðx ÞÞk 6 g0 ðkx x kÞ;
where f0, g0 are all monotonic functions. The radius estimate of the convergence ball is given, and the corresponding error estimate is also established. Especially, with the Lipschitz condition (1.5), we show the convergence of the sequence {xn}(n P 1) generated by the secant type method is quadratic at least. And a similar result under the Ho¨lder condition is also obtained. Finally, some applications and numerical experiments are provided to illuminate the convergence theories of secant type methods. The paper is organized as follows: in Sections 2–4, we give the local convergence theorems, the estimates of radii for the convergence ball of secant type method, the uniqueness of the solution and theirs applications. Section 5 shows the (1 + p)-order convergence of the secant type method. In Section 6, we present some numerical experiments. 2. Convergence analysis under weak Lipschitz conditions The condition on the function F kF ðxÞ F ðxs Þk 6 Lkx xs k;
8x 2 Bðx ; rÞ;
ð2:1Þ
where xs = x* + s(x x*), 0 6 s 6 1, is usually called radius Lipschitz condition in the ball B(x*, r) with constant L. Sometimes if it is only required to satisfy kF ðxÞ F ðx Þk 6 Lkx x k;
8x 2 Bðx ; rÞ;
ð2:2Þ
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J. Chen, Z. Shen / Applied Mathematics and Computation 188 (2007) 514–524
we call it the center Lipschitz condition in the ball B(x*, r) with constant L. Furthermore, the L in the Lipschitz condition need not be a constant, but a positive integrable function. If this is the case, then (2.1) or (2.2) is replaced by kF ðxÞ F ðxs Þk 6 f0 ðð1 sÞqðxÞÞ;
8x 2 Bðx ; rÞ; 0 6 s 6 1;
ð2:3Þ
or kF ðxÞ F ðx Þk 6 g0 ðqðxÞÞ;
8x 2 Bðx ; rÞ;
ð2:4Þ
x*k.
At the same time, the corresponding ‘Lipschitz condition’ is referred to as having the L where q(x) = kx average. By Banach theorem [5,12], the following result can be obtained directly. Lemma 2.1. Suppose that F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 satisfies the Lipschitz condition: 1
kF 0 ðx Þ F 0 ðxÞ Ik 6 g0 ðqðxÞÞ;
8x 2 Bðx ; rÞ;
ð2:5Þ
where g0 is positive integrable function. Let r satisfy g0 ðrÞ 6 1;
ð2:6Þ
then F 0 (x) is invertible in this ball and kF 0 ðxÞ1 F 0 ðx Þk 6
1 : 1 g0 ðqðxÞÞ
ð2:7Þ
Theorem 2.1. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies the Lipschitz condition: kF 1 ðx ÞðF ðxÞ F ðxs ÞÞk 6 f0 ðð1 sÞqðxÞÞ;
8x 2 Bðx ; rÞ; 0 6 s 6 1;
ð2:8Þ
s
where x = x* + s(x x*), q(x) = kx x*k, and f0 is increasing. Let r > 0 satisfy R1 0
f0 ðð1 sÞrÞ ds 6 1: 1 f0 ðrÞ
ð2:9Þ
Then secant type method is convergent for all x0, x1 2 B(x*, r) and R1 f0 ðð1 sÞðð1 an Þqðxn Þ þ an qðxn1 ÞÞÞ ds kxnþ1 x k 6 kxn x k; R 10 1 0 f0 ðð1 an ð1 sÞÞqðxn Þ þ an ð1 sÞqðxn1 ÞÞ ds
n ¼ 0; 1; 2; . . . ;
ð2:10Þ
where R1 q¼
1
R 10 0
f0 ðð1 sÞðð1 a0 Þqðx0 Þ þ a0 qðx1 ÞÞÞ ds
f0 ðð1 a0 ð1 sÞÞqðx0 Þ þ a0 ð1 sÞqðx1 ÞÞ ds
ð2:11Þ
is less than 1. Proof. Arbitrarily choosing x0 2 B(x*, r), where r satisfies (2.9), then q determined by (2.11) is less than 1. In fact, by the monotonicity of f0, we have R1 R1 f ðð1 sÞðð1 a0 Þqðx0 Þ þ a0 qðx1 ÞÞÞ ds f0 ðð1 sÞrÞ ds 0 0 q¼ < 0 6 1: R1 R1 1 0 f0 ðð1 a0 ð1 sÞÞqðx0 Þ þ a0 ð1 sÞqðx1 ÞÞ ds 1 0 f0 ðrÞ ds Using the mean-value theorem of integration, the following expression can be obtained: Z 1 F ½x; yðx yÞ ¼ F ðxÞ F ðyÞ ¼ F 0 ðsx þ ð1 sÞyÞðx yÞ ds; 8x; y 2 Bðx ; rÞ; 0
J. Chen, Z. Shen / Applied Mathematics and Computation 188 (2007) 514–524
517
i.e., F ½x; y
Z
1
F 0 ðsx þ ð1 sÞyÞ ds;
8x; y 2 Bðx ; rÞ;
ð2:12Þ
0
we have Z 0 1 1 0 1 1 0 kI F 0 ðx Þ ½x; y; F k ¼ kF 0 ðx Þ ðF 0 ðx Þ ½x; y; F Þk ¼ F ðx Þ ðF ðx Þ F ðsx þ ð1 sÞyÞ dsÞ Z
0
1
6
kF 0 ðx Þ1 ðF 0 ðx Þ F 0 ðsx þ ð1 sÞyÞkÞ ds
0
Z
1
f0 ðksðx x Þ þ ð1 sÞðy x ÞkÞ ds
6 0
Z
1
f0 ðksðx x Þk þ kð1 sÞðy x ÞkÞ ds <
6 0
Z
1
f0 ðrÞ ds 6 1:
0
And then by Lemma 2.1, we know for all x,y 2 B(x*, r), [x, y; F] is invertible and k½x; y; F 1 F 0 ðx Þk 6
1
R1 0
1 f0 ðksðx x Þ þ ð1 sÞðy x ÞkÞ ds
;
8x; y 2 Bðx ; rÞ:
ð2:13Þ
Now if xn 2 B(x*, r), ~xn ¼ xn þ an ðxn1 xn Þ, we have by (1.4) 1
1
xnþ1 x ¼ xn x ½xn ; ~xn ; F F ðxn Þ ¼ ½xn ; ~xn ; F ðF ðx Þ F ðxn Þ þ ½xn ; ~xn ; F ðxn x ÞÞ Z 1 1 0 1 ¼ ½xn ; ~xn ; F F ðx Þ F 0 ðx Þ ð½xn ; ~xn ; F F 0 ðxs ÞÞðxn x Þ ds; 0
1
¼ ½xn ; ~xn ; F F 0 ðx Þ
Z
1
1
F 0 ðx Þ ðF 0 ðsxn þ ð1 sÞ~xn Þ F 0 ðxs ÞÞðxn x Þ ds;
0
where xs = x* + s(xn x*). Hence, by inequality (2.13) we obtain Z 1 1 1 kF 0 ðx Þ ðF 0 ðsxn þ ð1 sÞ~xn Þ F 0 ðxs ÞÞk qðxn Þ ds kxnþ1 x k 6 k½xn ; ~xn ; F F 0 ðx Þk R1 6 1
R1 0
0
f0 ðksðxn x Þ þ ð1 sÞð~xn x ÞkÞ ds
R1 6 1
R 10 0
1
R 10 0
qðxn Þ
f0 ðð1 sÞðð1 an Þqðxn Þ þ an qðxn1 ÞÞÞ ds
f0 ðð1 an ð1 sÞÞqðxn Þ þ an ð1 sÞqðxn1 ÞÞ ds
R1 6
0
f0 ðkð1 sÞð~xn x ÞkÞ ds
f0 ðð1 sÞðð1 an Þqðxn Þ þ an qðxn1 ÞÞÞ ds
f0 ðð1 an ð1 sÞÞqðxn Þ þ an ð1 sÞqðxn1 ÞÞ ds
qðxn Þ qðxn Þ:
Taking n = 0 above, we obtain kx1 x*k 6 qkx0 x*k < kx0 x*k. Hence, i.e. x1 2 B(x*, r), this shows that (1.4) can be continued an infinite number of times. By mathematical induction, all xn belong to B(x*, r) and q(xn) = kxn x*k decreases monotonically. Thus (2.10) follows. h Theorem 2.2. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies the Lipschitz condition: kF 1 ðx ÞðF ðxÞ F ðx ÞÞk 6 g0 ðqðxÞÞ;
8x 2 Bðx ; rÞ;
where q(x) = kx x*k, and g0 is increasing. Let r > 0 satisfy R1 g0 ðrÞ þ 0 g0 ðsrÞ ds 6 1: 1 g0 ðrÞ
ð2:14Þ
ð2:15Þ
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J. Chen, Z. Shen / Applied Mathematics and Computation 188 (2007) 514–524
Then secant type method is convergent for all x0, x1 2 B(x*, r), n = 0, 1, 2, . . ., and R1
kxnþ1 x k 6
g0 ðð1 an ð1 sÞÞqðxn Þ þ an ð1 sÞqðxn1 ÞÞ ds kxn x k R1 1 0 g0 ðð1 an ð1 sÞÞqðxn Þ þ an ð1 sÞqðxn1 ÞÞ ds R1 g ðsqðxn ÞÞ ds 0 0 þ kxn x k; R1 1 0 g0 ðð1 an ð1 sÞÞqðxn Þ þ an ð1 sÞqðxn1 ÞÞ ds 0
ð2:16Þ
where R1 g0 ðð1 an ð1 sÞÞqðx0 Þ þ a0 ð1 sÞqðx1 ÞÞ ds þ 0 g0 ðsqðx0 ÞÞ ds R1 1 0 g0 ðð1 a0 ð1 sÞÞqðx0 Þ þ a0 ð1 sÞqðx1 ÞÞ ds
R1 q¼
0
ð2:17Þ
is less than 1. Proof. Arbitrarily choosing x0 2 B(x*, r), where r satisfies (2.15), then q determined by (2.17) is less than 1. In fact, by the monotonicity of f0, g0, we have R1 R1 g0 ðð1 an ð1 sÞÞqðx0 Þ þ a0 ð1 sÞqðx1 ÞÞ ds þ 0 g0 ðsqðx0 ÞÞ ds g0 ðrÞ þ 0 g0 ðsrÞ ds 6 1: < R1 1 g0 ðrÞ 1 0 g0 ðð1 a0 ð1 sÞÞqðx0 Þ þ a0 ð1 sÞqðx1 ÞÞ ds
R1 q¼
0
By the expression (2.12), we have Z 0 1 1 0 0 kI F ðx Þ ½x; y; F k ¼ kF ðx Þ ðF ðx Þ ½x; y; F Þk ¼ F ðx Þ ðF ðx Þ F ðsx þ ð1 sÞyÞ dsÞ 0
1
1
0
Z
0
0
1
1
kF 0 ðx Þ ðF 0 ðx Þ F 0 ðsx þ ð1 sÞyÞk dsÞ
6 0
Z
1
g0 ðksðx x Þ þ ð1 sÞðy x ÞkÞ ds
6 0
Z 6
1
g0 ðksðx x Þk þ kð1 sÞðy x ÞkÞ ds <
0
Z
1
g0 ðrÞ ds 6 1:
0
And then by Lemma 2.1, we know for all x 2 B(x*, r), [x, y; F] is invertible and 1
k½x; y; F F 0 ðx Þk 6
1
R1 0
1 g0 ðksðx
x Þ
þ ð1 sÞðy
x ÞkÞ ds
;
8x; y 2 Bðx ; rÞ:
ð2:18Þ
Now if xn 2 B(x*, r), ~xn ¼ xn þ an ðxn1 xn Þ, we have by (1.4) 1
1
xnþ1 x ¼ xn x ½xn ; ~xn ; F F ðxn Þ ¼ ½xn ; ~xn ; F ðF ðx Þ F ðxn Þ þ ½xn ; ~xn ; F ðxn x ÞÞ Z 1 1 1 ¼ ½xn ; ~xn ; F F 0 ðx Þ F 0 ðx Þ ð½xn ; ~xn ; F F 0 ðxs ÞÞðxn x Þ ds; 1
¼ ½xn ; ~xn ; F F 0 ðx Þ 1
¼ ½xn ; ~xn ; F F 0 ðx Þ
Z Z
0 1
1
F 0 ðx Þ ðF 0 ðsxn þ ð1 sÞ~xn Þ F 0 ðxs ÞÞðxn x Þ ds 0 1
1
F 0 ðx Þ ðF 0 ðsxn þ ð1 sÞ~xn Þ F 0 ðx Þ þ F 0 ðx Þ F 0 ðxs ÞÞðxn x Þ ds; 0
J. Chen, Z. Shen / Applied Mathematics and Computation 188 (2007) 514–524
519
where xs = x* + s(xn x*). Hence, by inequality (2.18) we obtain 1
kxnþ1 x k 6 k½xn ; ~xn ; F F 0 ðx Þk
Z
1
1
ðkF 0 ðx Þ ðF 0 ðsxn þ ð1 sÞ~xn Þ F 0 ðx ÞÞk
0 1
þ kF 0 ðx Þ ðF ðx Þ F 0 ðxs ÞÞkÞ qðxn Þ ds R1 ðg0 ðksðxn x Þ þ ð1 sÞð~xn x ÞkÞ þ g0 ðsðxn x ÞÞÞ ds 6 0 qðxn Þ R1 1 0 g0 ðksðxn x Þ þ ð1 sÞð~xn x ÞkÞ ds R1 R1 g0 ðð1 an ð1 sÞÞqðxn Þ þ an ð1 sÞqðxn1 ÞÞ ds þ 0 g0 ðsqðxn ÞÞ ds 6 0 qðxn Þ: R1 1 0 g0 ðð1 an ð1 sÞÞqðxn Þ þ an ð1 sÞqðxn1 ÞÞ ds Taking n = 0 above, we obtain kx1 x*k 6 qkx0 x*k < kx0 x*k. Hence, i.e. x1 2 B(x*, r), this shows that (1.4) can be continued an infinite number of times. By mathematical induction, all xn belong to B(x*, r) and q(xn) = kxn x*k decreases monotonically. Thus (2.16) follows. h Similar to the proof procedures of Theorems 2.1 and 2.2, the following conclusion can be established: Theorem 2.3. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies Lipschitz conditions (2.8) and (2.14). Let r > 0 satisfy R1 0
f0 ðð1 sÞrÞ ds 6 1: 1 g0 ðrÞ
ð2:19Þ
Then secant type method is convergent for all x0, x1 2 B(x*, r) and R1
kxnþ1 x k 6
1
R 10 0
f0 ðð1 sÞðð1 an Þqðxn Þ þ an qðxn1 ÞÞÞ ds
g0 ðð1 an ð1 sÞÞqðxn Þ þ an ð1 sÞqðxn1 ÞÞ ds
kxn x k;
n ¼ 0; 1; 2; . . . ;
ð2:20Þ
where R1 q¼
1
R 10 0
f0 ðð1 sÞðð1 a0 Þqðx0 Þ þ a0 qðx1 ÞÞÞ ds
g0 ðð1 a0 ð1 sÞÞqðx0 Þ þ a0 ð1 sÞqðx1 ÞÞ ds
ð2:21Þ
is less than 1. 3. The uniqueness ball for the solution Theorem 3.1. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies the Lipschitz condition: kF 1 ðx ÞðF ðxÞ F ðx ÞÞk 6 g0 ðqðxÞÞ;
8x 2 Bðx ; rÞ;
where xs = x* + s(x x*), q(x) = kx x*k, and g0 is increasing. Let r > 0 satisfy Rr g ðtÞ dt 0 0 6 1: r Then Eq. (1.1) has a unique solution x* in B(x*, r).
ð3:1Þ
ð3:2Þ
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J. Chen, Z. Shen / Applied Mathematics and Computation 188 (2007) 514–524
Proof. Suppose x0 2 B(x*, r), x0 5 x* is also a solution of (1.1). Using the integral expressions (2.12), we have 1
1
kI F 0 ðx Þ ½x0 ; x ; F k ¼ kF 0 ðx Þ ðF 0 ðx Þ ½x0 ; x ; F Þk Z 0 1 1 0 0 ¼ F ðx Þ ðF ðx Þ F ðsx0 þ ð1 sÞx Þ dsÞ Z
0
1
1
kF 0 ðx Þ ðF 0 ðx Þ F 0 ðsx0 þ ð1 sÞx Þk dsÞ 6 0 Rr Z 1 g0 ðtÞ dt ¼ 1; g0 ðsrÞ ds ¼ 0 < r 0
6
Z
1
g0 ðksðx0 x ÞkÞ ds 0
and then by Lemma 2.1, we know [x0, x*; F] is invertible and 1
x0 x ¼ ½x0 ; x ; F ðF ðx0 Þ F ðx ÞÞ ¼ 0:
ð3:3Þ
This is in contradiction with the assumption. Thus, it follows that Eq. (1.1) has a unique solution x* in B(x*, r). h 4. Applications In the study of the secant type methods, the assumption that the derivative is Lipschitz continuous is considered traditional. In this section, we will apply the obtained results to some concrete cases. By taking f0(s) = L1sp, g0(s) = L2sp, L1, L2, p are constants, the following corollaries are obtained directly. Corollary 4.1. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies the Lipschitz condition: p
kF 1 ðx ÞðF ðxÞ F ðxs ÞÞk 6 L1 ðð1 sÞqðxÞÞ ; s
x*
x*),
where x = + s(x q(x) = kx sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pþ1 p r¼ 6 1: L1 ðp þ 2Þ
x*k.
8x 2 Bðx ; rÞ; 0 6 s 6 1;
ð4:1Þ
Let r > 0 satisfy ð4:2Þ
Then secant type method is convergent for all x0, x1 2 B(x*, r), q¼
L1 ðð1 a0 Þqðx0 Þ þ a0 qðx1 ÞÞp <1 R1 p ðp þ 1Þð1 L1 0 ðð1 a0 ð1 sÞÞqðx0 Þ þ a0 ð1 sÞqðx1 ÞÞ dsÞ
ð4:3Þ
and the inequality (2.10) holds. Corollary 4.2. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies the Lipschitz condition: p
kF 1 ðx ÞðF ðxÞ F ðx ÞÞk 6 L2 ðqðxÞÞ ;
8x 2 Bðx ; rÞ;
where q(x) = kx x*k. Let r > 0 satisfy sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pþ1 p 6 1: r¼ L2 ð2p þ 3Þ Then secant type method is convergent for all x0, x1 2 B(x*, r), R1 p p 0Þ L2 0 ðð1 an ð1 sÞÞqðx0 Þ þ a0 ð1 sÞqðx1 ÞÞ ds þ L2 qðx pþ1 q¼ <1 R1 1 L2 0 ðð1 a0 ð1 sÞÞqðx0 Þ þ a0 ð1 sÞqðx1 ÞÞp ds and the inequality (2.16) holds.
ð4:4Þ
ð4:5Þ
ð4:6Þ
J. Chen, Z. Shen / Applied Mathematics and Computation 188 (2007) 514–524
521
Corollary 4.3. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies Lipschitz conditions (4.1) and (4.4). Let r > 0 satisfy sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pþ1 p r¼ 6 1: ð4:7Þ L1 þ L2 ðp þ 1Þ Then secant type method is convergent for all x0, x1 2 B(x*, r), p
q¼
L1 ðð1 a0 Þqðx0 Þ þ a0 qðx1 ÞÞ <1 R1 ðp þ 1Þð1 L2 0 ðð1 a0 ð1 sÞÞqðx0 Þ þ a0 ð1 sÞqðx1 ÞÞp dsÞ
ð4:8Þ
and the inequality (2.20) holds. Corollary 4.4. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies the Lipschitz conditions: p
kF 1 ðx ÞðF ðxÞ F ðx ÞÞk 6 L2 ðqðxÞÞ ;
8x 2 Bðx ; rÞ;
ð4:9Þ
s
where x = x* + s(x x*), q(x) = kx x*k, and g0 are increasing. Let r > 0 satisfy sffiffiffiffiffiffiffiffiffiffiffi p p þ 1 r¼ 6 1: L2
ð4:10Þ
Then Eq. (1.1) has a unique solution x* in B(x*, r). It follows from Corollaries 4.1, 4.2 and 4.3, 4.4 that the radius estimates of convergence balls ffi of secant type qffiffiffiffiffiffiffiffiffiffiffi
methods (1.4) agrees with the results of method in [9,10,12,4] r ¼ 3L2 1 ; r ¼ Newton’s 1 larger than the known results in [3,8] r ¼ 3L1 .
p
pþ1 L1 ðpþ2Þ
, and are also
5. Convergence analysis under Ho¨lder condition In Sections 2–4, we established the local convergence theorems for the secant type method (1.4), and under the weak Lipschitz conditions, we also proved that the sequence {xn} generated by the secant type method (1.4) converges to a solution x* of (1.1) linearly at least. In this section, under the Ho¨lder condition (0 < p 6 1) and taking an = O(kxn x*k), we will show that the convergence order can be improved to 1 + p at least. Specially, if the Lipschitz condition (1.5) holds, the convergence order is quadratic at least. Theorem 5.1. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies the Ho¨lder condition (4.1) with 0 < p 6 1. Let r > 0 satisfy (4.2). Then secant type method is convergent for all x0, x1 2 B(x*, r), n = 0, 1, 2, . . ., kxnþ1 x k 6
ðp þ 1Þð1 L1
R1 0
L1 ð1 an þ Oð1Þqðxn1 ÞÞp
pþ1
p
kxn x k
p
kxn x k
p
kxn x kpþ1 :
ðð1 a0 ð1 sÞÞqðxn Þ þ an ð1 sÞqðxn1 ÞÞ dsÞ
:
ð5:1Þ
Proof. By Theorem 2.1, it is easy to see kxnþ1 x k 6 ¼
L1 ðð1 an Þqðxn Þ þ an qðxn1 ÞÞ
ðp þ 1Þð1 L1 ðp þ 1Þð1 L1
R1 0
R1 0
p
ðð1 an ð1 sÞÞqðxn Þ þ an ð1 sÞqðxn1 ÞÞ dsÞ L1 ð1 an þ Oð1Þqðxn1 ÞÞ
p
ðð1 a0 ð1 sÞÞqðxn Þ þ an ð1 sÞqðxn1 ÞÞ dsÞ
So (5.1) holds, and that shows the convergence order of the sequence {xn}(n P 0) generated by the secant type method (1.4) is at least 1 + p. Specially, if the Lipschitz condition (1.5) holds, the convergence order is quadratic at least. The proof of the theorem is complete. h
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Similar to the proof procedure of Theorem 5.1, we can obtain the following (1 + p)-order convergence results of secant type method under other Ho¨lder conditions. Theorem 5.2. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies the Ho¨lder condition (4.4) with 0 < p 6 1. Let r > 0 satisfy (4.5). Then secant type method is convergent for all x0, x1 2 B(x*, r), n = 0, 1, 2, . . ., R1 p L2 L ð1 an ð1 sÞ þ Oð1Þð1 sÞqðxn1 ÞÞ ds þ pþ1 0 1 1þp kxnþ1 x k 6 ð5:2Þ kxn x k : R1 p 1 L2 0 ðð1 an ð1 sÞÞqðxn Þ þ an ð1 sÞqðxn1 ÞÞ ds Theorem 5.3. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies the Ho¨lder conditions (4.1) and (4.4) with 0 < p 6 1. Let r > 0 satisfy (4.7). Then secant type method is convergent for all x0, x1 2 B(x*, r), n = 0, 1, 2, . . . , p
kxnþ1 x k 6
ðp þ 1Þð1 L2
R1
L1 ðð1 an Þ þ an qðxn1 ÞÞ
ðð1 an ð1 sÞÞqðxn Þ þ an ð1 sÞqðxn1 ÞÞp dsÞ 0
kxn x k
1þp
:
ð5:3Þ
6. Numerical experiments In this section, we apply the convergence ball result given in Section 2 to solve the following nonlinear equations [13,14]. Example 6.1. Let us consider F ðxÞ ¼ ex 1;
x 2 ð1; 1Þ:
In this case, we have x* = 1 and F 0 (x*) = 1, and by simple computations, we can obtain kF 0 ðx ÞðF 0 ðxÞ F 0 ðyÞÞk ¼ kex ey k 6 ekx yk; 0
0
0
8x; y 2 ð1; 1Þ;
kF ðx ÞðF ðxÞ F ðx ÞÞk ¼ ke e k 6 ðe 1Þkx x k; x
0
8x; y 2 ð1; 1Þ:
2 0:324947. Then the radius of convergence ball for secant type method from Theorem 2.3: r ¼ 3e2 102 In Table 1, we consider the secant type method (1.4) with an ¼ kxn1 xn k. In order to obtain a good approximation, secant method (1.2) needs more than 42 function evaluations and 41 inversions. Whereas, secant type method needs only 16 function evaluations and 8 inversions to arrive at the exact solution.
Table 1 Error estimates in max-norm, x1 = 0.4, x0 = 0.3 Iteration
Secant (1.2)
Secant type (1.4)
2 4 6 8 10 14 18 22 26 30 34 38 40 41
4.29150e002 7.67916e003 1.34777e003 2.35696e004 4.11919e005 1.25784e006 3.84086e008 1.17282e009 3.58122e011 1.09357e012 1.39888e014 4.44089e016 2.2204e016 0.00000e000
2.88623e004 7.39838e009 1.84297e013 0.00000e000
J. Chen, Z. Shen / Applied Mathematics and Computation 188 (2007) 514–524
523
Example 6.2. Consider the system of nonlinear equations ( 2x1 19 x21 x2 ¼ 0; x1 þ 2x2 19 x22 ¼ 0: The associated nonlinear operator F : R2 ! R2 is given by F 1 ðx1 ; x2 Þ; F ðx1 ; x2 Þ ¼ ; F 2 ðx1 ; x2 Þ where F 1 ðx1 ; x2 Þ ¼ 2x1 19 x21 x2 and F 2 ðx1 ; x2 Þ ¼ x1 þ 2x2 19 x22 . We use the infinity norm kxk = kxk1 = max(kx1k, kx2k) and the induced matrix norms, and take the divided difference of F, [u, v;F] for u,v 2 R2 as: F i ðu1 ; v1 Þ F i ðv1 ; v2 Þ ; u1 v 1 F i ðu1 ; v2 Þ F i ðu1 ; v2 Þ ½u; v; F i2 ¼ u1 v 1 ½u; v; F i1 ¼
where i = 1, 2. Therefore 0 ½u; v; F ¼ @
2 19
u21 w21 v1 w1
1
1
1 2 19
u22 w22 v2 w2
A;
Now taking x* = (9, 9), then we can easily verify that x* is a solution of the problem. Furthermore, ! 2 x 1 2 0 1 1 9 0 0 F ðxÞ ¼ : ; F ðx Þ ¼ 1 2 29 x2 1 0 By simple computations, we have 2 kF 0 ðx ÞðF 0 ðxÞ F 0 ðyÞÞk ¼ kx yk; 8x; y 2 R2 ; 9 2 0 0 0 kF ðx ÞðF ðxÞ F ðx ÞÞk ¼ kx x k; 8x; y 2 R2 : 9 Then the radius of convergence ball for secant type method from Theorem 2.3: r = 3.00.
Table 2 Error estimates in max-norm, x1 = (11.5, 11.5), x0 = (11.0, 11.0) Iteration
Secant (1.2)
Secant type (1.4)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
4.48980e001 5.67443e002 5.89451e002 5.91877e004 5.92122e005 5.92147e006 5.92150e007 5.92150e008 5.92150e009 5.92150e010 5.92149e011 5.92237e012 6.03961e014 5.32907e015 0.00000e000
3.19602e001 1.02492e002 2.29426e005 2.55217e008 2.83276e011 3.19744e014 0.00000e000
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We consider the same choice for parameters an with the previous example. As shown in Table 2, the secant type method (1.4) has better numerical behaviors than classical secant method (1.2). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16]
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