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Newton–Kantorovich theorem for a family of modified Halley’s method under Hölder continuity conditions in Banach space
Available online at www.sciencedirect.com
Applied Mathematics and Computation 202 (2008) 243–251 www.elsevier.com/locate/amc
Newton–Kantorovich theorem for a family of modified Halley’s method under Ho¨lder continuity conditions in Banach space Yueqing Zhao a,b, Qingbiao Wu a,* a
Department of Mathematics, Zhejiang University, Hangzhou, 310027 Zhejiang, PR China b Department of Mathematics, Taizhou University, Linhai, 317000 Zhejiang, PR China
Abstract In this study, under p-Ho¨lder continuity conditions a family of new modified Halley’s method with (2 + p)-order is proposed in Banach space which is used to solve the nonlinear operator equations. The Newton–Kantorovich convergence theorem for the family of new modified Halley’s method is established under Ho¨lder continuity conditions. The error estimate also is gotten. Finally, two examples are provided to show the application of our theorem. Ó 2008 Elsevier Inc. All rights reserved. Keywords: Nonlinear operator equation; Modified Halley’s method; Newton–Kantorovich theorem; Banach space; Ho¨lder continuity
1. Introduction In this study, under p-Ho¨lder continuity conditions we consider to establish the Newton–Kantorovich convergence theorem for a family new modified Halley’s method with (2 + p)-order in Banach space by using a technique based on constructing a system of real sequences which is used to solve the nonlinear operator equation F ðxÞ ¼ 0;
ð1Þ
where F is defined on an open convex X of a Banach space X with values in a Banach space Y. There are kinds of methods to find a solution of (1). Iterative methods are often used to solve this problem (see [1]). Since Kantorovich present the famous convergence result (see [2]), many Newton–Kantorovich type convergence theorems were gotten (see [3–10]). To improve the convergence order, many modified methods have been present (see [11–14]). The famous Halley’s method are third-order convergence which was widely discussed (see [15–22]). The famous Halley’s method is defined as *
Corresponding author. E-mail addresses: [email protected] (Y. Zhao), [email protected] (Q. Wu).
0096-3003/$ - see front matter Ó 2008 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2008.02.004
244
Y. Zhao, Q. Wu / Applied Mathematics and Computation 202 (2008) 243–251
8 1 y ¼ xn F 0 ðxn Þ F ðxn Þ; > > < n 1 1 Lðxn Þ ¼ F 0 ðxn Þ F 00 ðxn ÞF 0 ðxn Þ F ðxn Þ; > > : 1 xnþ1 ¼ y n þ 12 Lðxn ÞðI 12 Lðxn ÞÞ ðy n xn Þ:
ð2Þ
Now, we consider a new finite difference approximation of F 00 ðxÞ, kF 00 ðxÞðzn xn Þ ’ F 0 ðxn þ kðzn 1 xn ÞÞ F 0 ðxn Þ, where zn ¼ xn þ F 0 ðxn Þ F ðxn Þ; k > 0. A family of new modified Halley’s method with a parameter k is 8 1 > y n ¼ xn F 0 ðxn Þ F ðxn Þ; > > > > > < zn ¼ xn þ F 0 ðxn Þ1 F ðxn Þ;
ð3Þ
1 > H ðxn ; zn Þ ¼ 1k F 0 ðxn Þ ½F 0 ðxn þ kðzn xn ÞÞ F 0 ðxn Þ; > > > > > 1 : xnþ1 ¼ y n þ 12 H ðxn ; zn Þ I 12 H ðxn ; zn Þ ðy n xn Þ;
where k > 0. Convergence analysis under Ho¨lder continuity conditions for some methods were discussed (see [23–29]). In 1 p this study, under Ho¨lder continuity conditions kF 0 ðx0 Þ ðF 00 ðxÞ F 00 ðyÞÞk 6 N kx yk ; x; y 2 X; p 2 ð0; 1, we establish Newton–Kantorovich type convergence theorem for the family of new modified Halley’s method with (2 + p)-order in Banach space by using a technique based on constructing a system of real sequences under Ho¨lder continuity conditions in Banach space. The error estimate is obtained. Finally, two examples are provided to show the application of our theorem. 2. Convergence theorem In the section, we establish a Newton–Kantorovich convergence theorem and present the error estimate. Denote Bðx; rÞ ¼ fy 2 X jky xk < rg, Bðx; rÞ ¼ fy 2 X jky xk 6 rg. Suppose X and Y are the Banach spaces, X is an open convex of the Banach space X, F : X X ! Y has two orders continuous Fre´chet derivatives, 1 F 0 ðx0 Þ exists for some x0 2 X, and F satisfies ky 0 x0 k ¼ kF 0 ðx0 Þ1 F ðx0 Þk 6 g; 1
kF 0 ðx0 Þ F 00 ðxÞk 6 M
ð4Þ
x 2 X;
1
p
kF 0 ðx0 Þ ðF 00 ðxÞ F 00 ðyÞÞk 6 N kx yk ;
x; y 2 X; p 2 ð0; 1:
Denote a0 ¼ Mg;
b0 ¼ N g1þp ;
anþ1 ¼ an f 2 ðan Þgðan ; bn Þ;
bnþ1 ¼ bn f 2þp ðan Þg1þp ðan ; bn Þ;
ð5Þ
where f ðxÞ ¼
2x ; 2 3x
gðx; yÞ ¼
x2 ð2 xÞ
þ 2
2 þ ðp þ 2Þkp y; ð2 xÞðp þ 1Þðp þ 2Þ
k > 0; p 2 ð0; 1:
Firstly, we get some lemmas. We easily get Lemma 1. Lemma 1. Suppose f ðxÞ; gðx; yÞ are given by (6), then, 8 x 2 ð0; 12Þ, f ðxÞ increases and f ðxÞ > 1; 8 x 2 ð0; 12Þ, y > 0, gðx; yÞ increases; 8 c 2 ð0; 1Þ; x 2 ð0; 12Þ; f ðcxÞ < f ðxÞ. Moreover, gðcx; c1þp yÞ < c1þp gðx; yÞ:
ð6Þ
Y. Zhao, Q. Wu / Applied Mathematics and Computation 202 (2008) 243–251
245
Lemma 2. For p 2 ð0; 1, suppose f ðxÞ; gðx; yÞ are given by (6). Let hðxÞ ¼
4ðp þ 1Þðp þ 2Þ ð1 2xÞð1 xÞ: ð2 xÞð2 þ ðp þ 2Þkp
ð7Þ
If a0 2 ð0; 12Þ and b0 < hða0 Þ, then 2
(I) f ða0 Þ gða0 ; b0 Þ < 1; (II) The sequences fan g, fbn g are non-negative and decrease; 2an (III) 2a < 1 8 n P 0. n Proof. From the hypotheses, (I) follows immediately. (II): by (I), when n ¼ 1, 0 < a1 ¼ a0 f 2 ða0 Þgða0 ; b0 Þ < a0 ; 0 < b1 ¼ b0 f 2þp ða0 Þg1þp ða0 ; b0 Þ ¼ b0
f 2þ2p ða0 Þg1þp ða0 ; b0 Þ < b0 : f p ða0 Þ
Suppose when j ¼ 1; 2; . . . ; n, aj < aj1 , bj < bj1 . By the Lemma 1 and f, g is the increase, then anþ1 ¼ an f 2 ðan Þgðan ; bn Þ < an f 2 ða0 Þgða0 ; b0 Þ < an ; bnþ1 ¼ bn f 2þp ðan Þg1þp ðan ; bn Þ < bn f 2þ2p ðan Þg1þp ðan ; bn Þ < bn f 2þ2p ða0 Þg1þp ða0 ; b0 Þ < bn : (III): by (II), fan g decreases and a0 2 ð0; 12Þ: So, for all n P 0, 2an 2a0 6 < 1: 2 an 2 a0 This completes the proof of Lemma 2.
Lemma 3. Under the conditions of Lemma 2, denote c ¼ aa10 ¼ f 2 ða0 Þgða0 ; b0 Þ < 1, then n1
(i) an < cð2þpÞ an1 < c
ð2þpÞn 1 1þp
(ii) f ðan Þgðan ; bn Þ < cð2þpÞ
n
1
n1
1þp
bn < ðcð2þpÞ Þ
a0 ;
f ða0 Þgða0 ; b0 Þ ¼
ð2þpÞn
c f ða0 Þ
bn1 < cð2þpÞ
n
1
b0 8 n P 2;
8 n P 1.
Proof. Firstly, by induction, we prove the (i) holds. Because a1 ¼ ca0 and f ða0 Þ > 1, we have b1 ¼ b0 f 2þp ða0 Þg1þp ða0 ; b0 Þ < b0 f 2þ2p ða0 Þg1þp ða0 ; b0 Þ ¼ c1þp b0 : Suppose the (i) holds for n. Then, we get n1
n1
n1
n1
anþ1 ¼ an f 2 ðan Þgðan ; bn Þ < cð2þpÞ an1 f 2 ðcð2þpÞ an1 Þ gðcð2þpÞ an1 ; ðcð2þpÞ Þ ð2þpÞn1
1þp
bn1 Þ
ð2þpÞn1 1þp
2
n
n1
n
anþ1 < cð2þpÞ an < cð2þpÞ cð2þpÞ an1 < < cð2þpÞ cð2þpÞ
n1
0
cð2þpÞ a0 ¼ c
ð2þpÞnþ1 1 1þp
a0
and bnþ1 ¼ bn f
2þp
ðan Þg n
1þp
1þp
ðan ; bn Þ < bn f n1
1þp
2þ2p
ðan Þg
1þp
ðan ; bn Þ ¼ bn n
< ðcð2þpÞ Þ ðcð2þpÞ Þ bn1 < < ðcð2þpÞ Þ 1þp ð2þpÞnþ1 1 nþ1 1þp ¼ c b0 ¼ cð2þpÞ 1 b0 :
1þp
anþ1 an n1
ðcð2þpÞ Þ
1þp 1þp
n
< ðcð2þpÞ Þ 0
ðcð2þpÞ Þ
1þp
1þp
bn
b0
246
Y. Zhao, Q. Wu / Applied Mathematics and Computation 202 (2008) 243–251
(ii): ð2þpÞn 1 ð2þpÞn 1 ð2þpÞn 1 1þp f ðan Þgðan ; bn Þ < f c 1þp a0 g c 1þp a0 ; ðc 1þp Þ b0 ð2þpÞn 1 1þp < c 1þp f ða0 Þgða0 ; b0 Þ n
¼ cð2þpÞ
n
1
f ða0 Þgða0 ; b0 Þ ¼
cð2þpÞ ; f ða0 Þ
n P 1:
This completes the proof of Lemma 3. Lemma 4. Suppose X and Y are the Banach spaces, X is an open convex of the Banach space X, F : X X ! Y has two orders continuous Fre´chet derivatives. The sequences fxn g; fy n g are generated by (3). Then, for all natural number n, the following formula holds: Z 1 2 F ðxnþ1 Þ ¼ F 00 ðy n þ tðxnþ1 y n ÞÞð1 tÞ dtðxnþ1 y n Þ 0 Z 1 Z 1 1 00 00 þ F ðxn þ tðy n xn ÞÞð1 tÞ dt F ðxn ktðy n xn ÞÞ ðy n xn ÞP ðxn ; zn Þðy n xn Þ 2 0 0 Z 1 1 00 F ðxn þ tðy n xn ÞÞt dtðy n xn ÞH ðxn ; zn ÞP ðxn ; zn Þðy n xn Þ; ð8Þ þ 2 0 where P ðxn ; zn Þ ¼ ðI 12 H ðxn ; zn ÞÞ1 . Proof. By (3), we get zn xn ¼ ðy n xn Þ and F ðxnþ1 Þ ¼ F ðxnþ1 Þ F ðy n Þ F 0 ðy n Þðxnþ1 y n Þ þ F ðy n Þ þ F 0 ðy n Þðxnþ1 y n Þ Z 1 2 ¼ F 00 ðy n þ tðxnþ1 y n ÞÞð1 tÞ dtðxnþ1 y n Þ þ F ðy n Þ þ F 0 ðy n Þðxnþ1 y n Þ; 0
1 F ðxn ÞH ðxn ; zn Þ ¼ ½F 0 ðxn þ kðzn xn ÞÞ F 0 ðxn Þ ¼ k 0
Z
1
F 00 ðxn ktðy n xn ÞÞ dtðy n xn Þ; 0
1 1 F 0 ðy n Þðxnþ1 y n Þ ¼ F 0 ðxn ÞH ðxn ; zn ÞP ðxn ; zn Þðy n xn Þ þ ½F 0 ðy n Þ F 0 ðxn ÞH ðxn ; zn ÞP ðxn ; zn Þðy n xn Þ 2 2 Z 1 1 F 00 ðxn ktðy n xn ÞÞ dtðy n xn ÞP ðxn ; zn Þðy n xn Þ ¼ 2 0 Z 1 1 00 þ F ðxn þ tðy n xn ÞÞ dtðy n xn ÞH ðxn ; zn ÞP ðxn ; zn Þðy n xn Þ; 2 0 Z 1 2 0 F ðy n Þ ¼ F ðy n Þ F ðxn Þ F ðxn Þðy n xn Þ ¼ F 00 ðxn þ tðy n xn ÞÞð1 tÞ dtðy n xn Þ ¼
Z 0
0
1
1 F 00 ðxn þ tðy n xn ÞÞð1 tÞ dtðy n xn Þ I H ðxn ; zn Þ P ðxn ; zn Þðy n xn Þ: 2
Hence the (8) holds. This completes the proof of Lemma 4.
Theorem 1. Suppose X and Y are the Banach spaces, X is an open convex of the Banach space X, F : X X ! Y 1 has two orders continuous Fre´chet derivatives, F 0 ðx0 Þ exists for some x0 2 X, and F satisfies (4). Denote a0 ¼ Mg; b0 ¼ N g1þp , under conditions of the Lemma 2, if Bðx0 ; ðj1 kj þ kÞRgÞ X, then, the sequence fxn g generated by (3) from the initial point x0 is well defined, xn 2 Bðx0 ; RgÞ and converges to the unique solution x in Bðx0 ; M2 RgÞ \ X and
Y. Zhao, Q. Wu / Applied Mathematics and Computation 202 (2008) 243–251
2
kx xn k 6 2c
ð2þpÞn 1 1þp
a0
1 1
n cð2þpÞ D
c
ð2þpÞn 1 1þp
Dn g;
247
ð9Þ
2 1 where R ¼ 2a 1cD ; c ¼ aa10 ; D ¼ f ða1 0 Þ. 0
Proof. ky 0 x0 k ¼ g < Rg; ðR > 1Þ; kx0 þ kðz0 x0 Þ x0 k 6 kg, hence, we get y 0 2 X; x0 þ kðz0 x0 Þ 2 X. By (4), we have 1 1 kH ðx0 ; z0 Þk 6 Mkkz0 x0 k ¼ MkF 0 ðx0 Þ F 0 ðx0 Þkky 0 x0 k 6 Mg ¼ a0 < 1: k 1 2 ¼ Hence P ðxn ; zn Þ; exists and kP ðxn ; zn Þk 6 ; 1 a0 =2 2 a0 N kF 0 ðx0 Þ1 F 0 ðx0 Þkky 0 x0 k1þp 6 N g1þp ¼ b0 ; 1 2 2 kx1 x0 k 6 kx1 y 0 k þ ky 0 x0 k 6 1 þ a0 ky x0 k; ky 0 x0 k ¼ 2 2 a0 2 a0 0 2a0 1 1 < 1: kF 0 ðx0 Þ F 0 ðx1 Þ Ik ¼ kF 0 ðx0 Þ ðF 0 ðx1 Þ F 0 ðx0 ÞÞk 6 Mkx1 x0 k 6 2 a0 By Banach Lemma, F 0 ðx1 Þ1 exists, and 1 1 1 ¼ f ða0 ÞkF 0 ðx0 Þ F 0 ðx0 Þk: kF 0 ðx1 Þ F 0 ðx0 Þk 6 2a0 1 2a0 By conditions of Theorem 1 and Lemma 4, we obtain Z 1 Z 1 1 kF 0 ðx0 Þ1 F 00 ðx0 þ tðy 0 x0 ÞÞð1 tÞ dt F 0 ðx0 Þ1 F 00 ðx0 ktðy 0 x0 ÞÞ dtk 2 0 0 Z 1 1 6 kF 0 ðx0 Þ ½F 00 ðx0 þ tðy 0 x0 ÞÞ F 00 ðx0 Þð1 tÞ dt 0
Z 1 1 F 00 ðx0 ktðy 0 x0 ÞÞ F 00 ðx0 Þ dtk þ kF 0 ðx0 Þ1 2 0 Z Z 1 1 1 pp 2 þ ðp þ 2Þkp p p p 6 N ky 0 x0 k ; Ntp ð1 tÞ dtky 0 x0 k þ N k t dtky 0 x0 k ¼ 2 2ðp þ 1Þðp þ 2Þ 0 0 M 2Ma0 2 þ ðp þ 2Þkp 1 2 2 2þp 0 ky 0 x0 k þ N ky 0 x0 k kF ðx0 Þ F ðx1 Þk 6 kx1 y 0 k þ 2 4ð2 a0 Þ ð2 a0 Þðp þ 1Þðp þ 2Þ Ma0 2 þ ðp þ 2Þkp 2 2þp N ky 0 x0 k : 6 ky x k þ 0 0 2 ð2 a0 Þðp þ 1Þðp þ 2Þ ð2 a0 Þ Hence, 1
1
ky 1 x1 k ¼ kF 0 ðx1 Þ F 0 ðx0 ÞF 0 ðx0 Þ F ðx1 Þk 6 f ða0 Þgða0 ; b0 Þky 0 x0 k and 1
H ðx1 ; z1 Þ 6 MkF 0 ðx1 Þ F 0 ðx0 Þkky 1 x1 k 6 a0 f 2 ða0 Þgða0 ; b0 Þ ¼ a1 ; N kF 0 ðx1 Þ1 F 0 ðx0 Þk ky 1 x1 k1þp 6 N ky 0 x0 k1þp f ða0 Þ2þp gða0 ; b0 Þ1þp ¼ b1 ; 2 ky x1 k: kx2 x1 k 6 kx2 y 1 k þ ky 1 x1 k 6 2 a1 1 By the induction, we can proved the following formulae hold: (i) kF 0 ðxn Þ1 F 0 ðx0 Þk 6 f ðan1 ÞkF 0 ðxn1 Þ1 F 0 ðx0 Þk; (ii) ky n xn k 6 f ðan1 Þgðan1 ; bn1 Þky n1 xn1 k; (iii) H ðxn ; zn Þ 6 MkF 0 ðxn Þ1 F 0 ðx0 Þk ky n xn k 6 an ;
248
Y. Zhao, Q. Wu / Applied Mathematics and Computation 202 (2008) 243–251 1
(iv) N kF 0 ðxn Þ F 0 ðx0 Þk ky n xn k 2 (v) kxnþ1 xn k 6 2a ky n xn k: n
1þp
6 bn ;
Now, we prove the fxn g is a Cauchy sequence. For n P 1, " # n1 Y f ðaj Þgðaj ; bj Þ ky 0 x0 k: ky n xn k 6 f ðan1 Þgðan1 ; bn1 Þky n1 xn1 k 6 6 j¼0 n1 Y
n1 ð2þpÞj Y ð2þpÞn 1 c ¼ c 1þp Dn : f ðaj Þgðaj ; bj Þ 6 f ða0 Þ j¼0 j¼0
So, kxnþm xn k 6 kxnþm xnþm1 k þ kxnþm1 xnþm2 k þ þ kxnþ1 xn k ! nþm2 n1 Y Y 2 6 f ðaj Þgðaj ; bj Þ þ þ f ðaj Þgðaj ; bj Þ g 2 an j¼0 j¼0 ð2þpÞnþm1 1 ð2þpÞn 1 2 c ð1þpÞ Dnþm1 þ þ c ð1þpÞ Dn g 2 an ð2þpÞn 1 ð2þpÞn ðð2þpÞm1 1Þ 2 ð1þpÞ ¼ c ð1þpÞ Dn g c Dm1 þ þ 1 : 2 an 6
By the Bernouilli inequality ð1 þ xÞk 1 > kx, so ð2 þ pÞk 1 > kð1 þ pÞ, we have 2
kxnþm xn k 6 2c
ð2þpÞn 1 1þp
a0
1 n
1 cð2þpÞ D
c
ð2þpÞn 1 1þp
Dn g:
ð10Þ
Therefore, fxn g is a Cauchy sequence. By (10), let n ¼ 0, we can get kxm x0 k 6
2 1 g ¼ Rg: 2 a0 1 cD
ð11Þ
Moreover, from a similar reasoning, we easily know zn ; y n 2 Bðx0 ; RgÞ; n P 0 and xn þ kðzn xn Þ 2 Bðx0 ; ðj1 kj þ kÞRgÞ. By letting m ! 1 in (10), we obtain (9). To see that x is a solutio of (1), we have kF 0 ðxn Þ1 F ðxn Þk ¼ ky n xn k ! 0 as n ! 1. Taking into account that kF ðxn Þk 6 kF 0 ðxn ÞkkF 0 ðxn Þ1 F ðxn Þk, and the sequence kF 0 ðxn Þk is bounded. We deduce that F ðx Þ ¼ 0 from the continuity of F. By (11), we obtain x 2 Sðx0 ; RgÞ: Now we prove the unique. Suppose y also is the solution of the Eq. (1) in Bðx0 ; M2 RgÞ \ X. Then, we have Z 1 Z 1 kF 0 ðx0 Þ1 ½F 0 ðx þ tðy x ÞÞ F 0 ðx0 Þk dt 6 M kx þ tðy x Þ x0 k dt 0
Z
0 1
½ð1 tÞkx x0 k þ tky x0 k dt
6M 0
M 2 < ðRg þ RgÞ ¼ 1: 2 M R1 By Banach Lemma, ½ 0 F 0 ðx0 Þ1 F 0 ðx þ tðy x ÞÞ dt1 exists. Because of Z 1 1 1 0 ¼ F 0 ðx0 Þ ½F ðy Þ F ðx Þ ¼ F 0 ðx0 Þ F 0 ðx þ tðy x ÞÞ dtðy x Þ; 0
we know y ¼ x . This completes the proof of Theorem 1.
Y. Zhao, Q. Wu / Applied Mathematics and Computation 202 (2008) 243–251
249
3. Applications Example 1. Now we employ iterative method (3) to solve the follow equation and compare these methods with Newton’s method and Halley’s method. We defined as f ðxÞ ¼ x8=3 256:
ð12Þ
The first and the second derivative of F is defined by 8 f 0 ðxÞ ¼ x5=3 ; 3
f 00 ðxÞ ¼
40 2=3 x 9
and we have jf 00 ðxÞ f 00 ðyÞj ¼
40 2=3 40 jx y 2=3 j 6 jx yj2=3 : 9 9
To apply Theorem 1, we choose x0 ¼ 9 and we look for a domain in the form X ¼ Bðx0 ; 9Þ In this case, we obtain 1
1
jf 0 ðx0 Þ j ¼ 0:00963001 . . . ; M ¼ 0:29396 . . . ;
g ¼ jf 0 ðx0 Þ f ðx0 Þj ¼ 0:90971 . . . 2 N ¼ 0:042800 . . . ; p ¼ : 3
3:49651 Then a0 ¼ 0:267423 < 1=2, and b0 ¼ 0:0365556 < hða0 Þ ¼ 2þ8k if and only if k 2 ð0; 208:115Þ and 2=3 =3 Bðx0 ; ðj1 kj þ kÞRgÞ X if and only if k 2 ð0; 4:36894Þ. Hence, when k 2 ð0; 4:36894Þ, the conditions of Theorem 1 are fulfilled. ffiffiffiffiffiffiffiffiffi p 0 0 ðxÞÞf 0 ðxÞ 8 1 Denote x ¼ 2563 ¼ 8, by (3) xnþ1 ¼ vðxn Þ, where vðxÞ ¼ x 1 þ 2k1 f ðxþkf ðxÞ=f 0 0 0 1 f ðxþkf ðxÞ=f ðxÞÞf ðxÞ f 0 ðxÞ 12k f ðxÞ f 0 ðxÞ . We have v000 ðx Þ ¼ 0 if f 0 ðxÞ
2 f 00 ðx Þ2 11 k ¼ þ 0 000 ¼ : 3 f ðx Þf ðx Þ 6 So, we get the convergence of the sequence fxn g generated by modified Halley’s method (3) with k ¼ 116 is four orders. Now, we compare some of these methods. We analyze the error xk x for these methods with Newton’s method, Halley’s method and the family of new modified Halley’s method (3). In these cases, we have taken x0 ¼ 9 (see Table 1). Example 2. Consider the case as follows: Z 1 1 s 15=7 xðtÞ dt; xðsÞ ¼ 1 þ 4 0 sþt
ð13Þ
where the space is X ¼ C½0; 1 with norm kxk ¼ max jxðsÞj:
ð14Þ
06s61
Table 1 Error computing results Step k k k k
¼1 ¼2 ¼3 ¼4
Newton’s
Halley’s 2
9.028453 10 8.375337 104 7.305968 108 5.560122 1016
3
6.608249 10 2.293402 109 9.598411 1029 7.036505 1087
Methods (3) with a ¼ 11 6
Methods (3) with a ¼ 1
1.042665 103 1.806115 1015 1.626801 1062 1.070757 10250
3.553708 103 1.623402 1010 1.547442 1032 1.340229 1098
250
Y. Zhao, Q. Wu / Applied Mathematics and Computation 202 (2008) 243–251
This equation arises in the theory of the radiative transfer, neutron transport and in the kinetic theory of gasses. Let us define the operator F on X by 1 F ðxÞ ¼ xðsÞ 4
Z
1 0
s xðtÞ15=7 dt 1: sþt
ð15Þ
The first and the second derivative of F is defined by Z 15 1 s 8=7 0 xðtÞ uðtÞ dt; u 2 X; F ðxÞuðsÞ ¼ uðsÞ 28 0 s þ t Z 30 1 s 1=7 xðtÞ uðtÞvðtÞ dt F 00 ðxÞðuvÞðsÞ ¼ 49 0 s þ t and we have k½F 00 ðxÞ F 00 ðyÞuvk 6
30 max 49 s2½0;1
Z 0
1
s 66 1=7 1=7 1=7 jðxðtÞ yðtÞ ÞuðtÞvðtÞj dt 6 ln 2kx yk kuvk: sþt 100
To apply Theorem 1, we choose x0 ¼ x0 ðsÞ ¼ 1 and we look for a domain in the form X ¼ Bðx0 ; 2Þ Cð½0; 1Þ: In this case, we have kF 0 ðx0 Þ Ik 6
15 ln 2 < 1; 28
from the Banach Lemma, we obtain 28 7 ln 2 ; kF 0 ðx0 Þ1 F ðx0 Þk 6 ¼ g ¼ 0:275639 . . . ; 28 15 ln 2 28 15 ln 2 28 30 1=7 28 30 2 ln 2 ¼ 0:745300 . . . ; N ¼ ln 2 ¼ 0:675036 . . . ; M¼ 28 15 ln 2 49 28 15 ln 2 49
kF 0 ðx0 Þ1 k 6
1 p¼ : 7
2:555215 Then a0 ¼ 0:205434 < 1=2, and b0 ¼ 0:154781 < hða0 Þ ¼ 2þ15=7k 1=7 if and only if k 2 ð0; 652259:9 Þ and Bðx0 ; ðj1 kj þ kÞRgÞ X if and only if k 2 ð0; 3:02928 . . .Þ. Hence, when k 2 ð0; 3:02928 . . .Þ the conditions of Theorem 1 are fulfilled. Now we choose a ¼ 0:1 then the error bound becomes
1
kxn x k 6
1 0:102717 0:231879 0:231879
ð15=7Þn 1 8=7
ð15=7Þn 1 8=7
1 1 0:231879
ð15=7Þn
0:771048
0:771048n 0:2756398:
For n ¼ 1; 2; 3; 4, we get kx1 x k 6 5:22418 102 ;
kx2 x k 6 1:6614 103 ;
kx3 x k 6 1:55627 106 ;
kx4 x k 6 6:81657 1013 :
Acknowledgements The authors thank editors and anonymous referee for his or her careful reading and valuable comments on improving the original manuscript. This work was supported by Natural Science Foundation of Zhejiang Province (Grant No. Y606154). References [1] J.M. Ortega, W.C. Rheinbolt, Iterative Solution of Nonlinear Equations in Servral Variables, Academic Press, New York, 1970. [2] L. Kantorovich, On Newton method (Russian), Trudy Mat. Inst. Steklov 28 (1949) 104–144.
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Applied Mathematics and Computation 202 (2008) 243–251 www.elsevier.com/locate/amc
Newton–Kantorovich theorem for a family of modified Halley’s method under Ho¨lder continuity conditions in Banach space Yueqing Zhao a,b, Qingbiao Wu a,* a
Department of Mathematics, Zhejiang University, Hangzhou, 310027 Zhejiang, PR China b Department of Mathematics, Taizhou University, Linhai, 317000 Zhejiang, PR China
Abstract In this study, under p-Ho¨lder continuity conditions a family of new modified Halley’s method with (2 + p)-order is proposed in Banach space which is used to solve the nonlinear operator equations. The Newton–Kantorovich convergence theorem for the family of new modified Halley’s method is established under Ho¨lder continuity conditions. The error estimate also is gotten. Finally, two examples are provided to show the application of our theorem. Ó 2008 Elsevier Inc. All rights reserved. Keywords: Nonlinear operator equation; Modified Halley’s method; Newton–Kantorovich theorem; Banach space; Ho¨lder continuity
1. Introduction In this study, under p-Ho¨lder continuity conditions we consider to establish the Newton–Kantorovich convergence theorem for a family new modified Halley’s method with (2 + p)-order in Banach space by using a technique based on constructing a system of real sequences which is used to solve the nonlinear operator equation F ðxÞ ¼ 0;
ð1Þ
where F is defined on an open convex X of a Banach space X with values in a Banach space Y. There are kinds of methods to find a solution of (1). Iterative methods are often used to solve this problem (see [1]). Since Kantorovich present the famous convergence result (see [2]), many Newton–Kantorovich type convergence theorems were gotten (see [3–10]). To improve the convergence order, many modified methods have been present (see [11–14]). The famous Halley’s method are third-order convergence which was widely discussed (see [15–22]). The famous Halley’s method is defined as *
Corresponding author. E-mail addresses: [email protected] (Y. Zhao), [email protected] (Q. Wu).
0096-3003/$ - see front matter Ó 2008 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2008.02.004
244
Y. Zhao, Q. Wu / Applied Mathematics and Computation 202 (2008) 243–251
8 1 y ¼ xn F 0 ðxn Þ F ðxn Þ; > > < n 1 1 Lðxn Þ ¼ F 0 ðxn Þ F 00 ðxn ÞF 0 ðxn Þ F ðxn Þ; > > : 1 xnþ1 ¼ y n þ 12 Lðxn ÞðI 12 Lðxn ÞÞ ðy n xn Þ:
ð2Þ
Now, we consider a new finite difference approximation of F 00 ðxÞ, kF 00 ðxÞðzn xn Þ ’ F 0 ðxn þ kðzn 1 xn ÞÞ F 0 ðxn Þ, where zn ¼ xn þ F 0 ðxn Þ F ðxn Þ; k > 0. A family of new modified Halley’s method with a parameter k is 8 1 > y n ¼ xn F 0 ðxn Þ F ðxn Þ; > > > > > < zn ¼ xn þ F 0 ðxn Þ1 F ðxn Þ;
ð3Þ
1 > H ðxn ; zn Þ ¼ 1k F 0 ðxn Þ ½F 0 ðxn þ kðzn xn ÞÞ F 0 ðxn Þ; > > > > > 1 : xnþ1 ¼ y n þ 12 H ðxn ; zn Þ I 12 H ðxn ; zn Þ ðy n xn Þ;
where k > 0. Convergence analysis under Ho¨lder continuity conditions for some methods were discussed (see [23–29]). In 1 p this study, under Ho¨lder continuity conditions kF 0 ðx0 Þ ðF 00 ðxÞ F 00 ðyÞÞk 6 N kx yk ; x; y 2 X; p 2 ð0; 1, we establish Newton–Kantorovich type convergence theorem for the family of new modified Halley’s method with (2 + p)-order in Banach space by using a technique based on constructing a system of real sequences under Ho¨lder continuity conditions in Banach space. The error estimate is obtained. Finally, two examples are provided to show the application of our theorem. 2. Convergence theorem In the section, we establish a Newton–Kantorovich convergence theorem and present the error estimate. Denote Bðx; rÞ ¼ fy 2 X jky xk < rg, Bðx; rÞ ¼ fy 2 X jky xk 6 rg. Suppose X and Y are the Banach spaces, X is an open convex of the Banach space X, F : X X ! Y has two orders continuous Fre´chet derivatives, 1 F 0 ðx0 Þ exists for some x0 2 X, and F satisfies ky 0 x0 k ¼ kF 0 ðx0 Þ1 F ðx0 Þk 6 g; 1
kF 0 ðx0 Þ F 00 ðxÞk 6 M
ð4Þ
x 2 X;
1
p
kF 0 ðx0 Þ ðF 00 ðxÞ F 00 ðyÞÞk 6 N kx yk ;
x; y 2 X; p 2 ð0; 1:
Denote a0 ¼ Mg;
b0 ¼ N g1þp ;
anþ1 ¼ an f 2 ðan Þgðan ; bn Þ;
bnþ1 ¼ bn f 2þp ðan Þg1þp ðan ; bn Þ;
ð5Þ
where f ðxÞ ¼
2x ; 2 3x
gðx; yÞ ¼
x2 ð2 xÞ
þ 2
2 þ ðp þ 2Þkp y; ð2 xÞðp þ 1Þðp þ 2Þ
k > 0; p 2 ð0; 1:
Firstly, we get some lemmas. We easily get Lemma 1. Lemma 1. Suppose f ðxÞ; gðx; yÞ are given by (6), then, 8 x 2 ð0; 12Þ, f ðxÞ increases and f ðxÞ > 1; 8 x 2 ð0; 12Þ, y > 0, gðx; yÞ increases; 8 c 2 ð0; 1Þ; x 2 ð0; 12Þ; f ðcxÞ < f ðxÞ. Moreover, gðcx; c1þp yÞ < c1þp gðx; yÞ:
ð6Þ
Y. Zhao, Q. Wu / Applied Mathematics and Computation 202 (2008) 243–251
245
Lemma 2. For p 2 ð0; 1, suppose f ðxÞ; gðx; yÞ are given by (6). Let hðxÞ ¼
4ðp þ 1Þðp þ 2Þ ð1 2xÞð1 xÞ: ð2 xÞð2 þ ðp þ 2Þkp
ð7Þ
If a0 2 ð0; 12Þ and b0 < hða0 Þ, then 2
(I) f ða0 Þ gða0 ; b0 Þ < 1; (II) The sequences fan g, fbn g are non-negative and decrease; 2an (III) 2a < 1 8 n P 0. n Proof. From the hypotheses, (I) follows immediately. (II): by (I), when n ¼ 1, 0 < a1 ¼ a0 f 2 ða0 Þgða0 ; b0 Þ < a0 ; 0 < b1 ¼ b0 f 2þp ða0 Þg1þp ða0 ; b0 Þ ¼ b0
f 2þ2p ða0 Þg1þp ða0 ; b0 Þ < b0 : f p ða0 Þ
Suppose when j ¼ 1; 2; . . . ; n, aj < aj1 , bj < bj1 . By the Lemma 1 and f, g is the increase, then anþ1 ¼ an f 2 ðan Þgðan ; bn Þ < an f 2 ða0 Þgða0 ; b0 Þ < an ; bnþ1 ¼ bn f 2þp ðan Þg1þp ðan ; bn Þ < bn f 2þ2p ðan Þg1þp ðan ; bn Þ < bn f 2þ2p ða0 Þg1þp ða0 ; b0 Þ < bn : (III): by (II), fan g decreases and a0 2 ð0; 12Þ: So, for all n P 0, 2an 2a0 6 < 1: 2 an 2 a0 This completes the proof of Lemma 2.
Lemma 3. Under the conditions of Lemma 2, denote c ¼ aa10 ¼ f 2 ða0 Þgða0 ; b0 Þ < 1, then n1
(i) an < cð2þpÞ an1 < c
ð2þpÞn 1 1þp
(ii) f ðan Þgðan ; bn Þ < cð2þpÞ
n
1
n1
1þp
bn < ðcð2þpÞ Þ
a0 ;
f ða0 Þgða0 ; b0 Þ ¼
ð2þpÞn
c f ða0 Þ
bn1 < cð2þpÞ
n
1
b0 8 n P 2;
8 n P 1.
Proof. Firstly, by induction, we prove the (i) holds. Because a1 ¼ ca0 and f ða0 Þ > 1, we have b1 ¼ b0 f 2þp ða0 Þg1þp ða0 ; b0 Þ < b0 f 2þ2p ða0 Þg1þp ða0 ; b0 Þ ¼ c1þp b0 : Suppose the (i) holds for n. Then, we get n1
n1
n1
n1
anþ1 ¼ an f 2 ðan Þgðan ; bn Þ < cð2þpÞ an1 f 2 ðcð2þpÞ an1 Þ gðcð2þpÞ an1 ; ðcð2þpÞ Þ ð2þpÞn1
1þp
bn1 Þ
ð2þpÞn1 1þp
2
n
n1
n
anþ1 < cð2þpÞ an < cð2þpÞ cð2þpÞ an1 < < cð2þpÞ cð2þpÞ
n1
0
cð2þpÞ a0 ¼ c
ð2þpÞnþ1 1 1þp
a0
and bnþ1 ¼ bn f
2þp
ðan Þg n
1þp
1þp
ðan ; bn Þ < bn f n1
1þp
2þ2p
ðan Þg
1þp
ðan ; bn Þ ¼ bn n
< ðcð2þpÞ Þ ðcð2þpÞ Þ bn1 < < ðcð2þpÞ Þ 1þp ð2þpÞnþ1 1 nþ1 1þp ¼ c b0 ¼ cð2þpÞ 1 b0 :
1þp
anþ1 an n1
ðcð2þpÞ Þ
1þp 1þp
n
< ðcð2þpÞ Þ 0
ðcð2þpÞ Þ
1þp
1þp
bn
b0
246
Y. Zhao, Q. Wu / Applied Mathematics and Computation 202 (2008) 243–251
(ii): ð2þpÞn 1 ð2þpÞn 1 ð2þpÞn 1 1þp f ðan Þgðan ; bn Þ < f c 1þp a0 g c 1þp a0 ; ðc 1þp Þ b0 ð2þpÞn 1 1þp < c 1þp f ða0 Þgða0 ; b0 Þ n
¼ cð2þpÞ
n
1
f ða0 Þgða0 ; b0 Þ ¼
cð2þpÞ ; f ða0 Þ
n P 1:
This completes the proof of Lemma 3. Lemma 4. Suppose X and Y are the Banach spaces, X is an open convex of the Banach space X, F : X X ! Y has two orders continuous Fre´chet derivatives. The sequences fxn g; fy n g are generated by (3). Then, for all natural number n, the following formula holds: Z 1 2 F ðxnþ1 Þ ¼ F 00 ðy n þ tðxnþ1 y n ÞÞð1 tÞ dtðxnþ1 y n Þ 0 Z 1 Z 1 1 00 00 þ F ðxn þ tðy n xn ÞÞð1 tÞ dt F ðxn ktðy n xn ÞÞ ðy n xn ÞP ðxn ; zn Þðy n xn Þ 2 0 0 Z 1 1 00 F ðxn þ tðy n xn ÞÞt dtðy n xn ÞH ðxn ; zn ÞP ðxn ; zn Þðy n xn Þ; ð8Þ þ 2 0 where P ðxn ; zn Þ ¼ ðI 12 H ðxn ; zn ÞÞ1 . Proof. By (3), we get zn xn ¼ ðy n xn Þ and F ðxnþ1 Þ ¼ F ðxnþ1 Þ F ðy n Þ F 0 ðy n Þðxnþ1 y n Þ þ F ðy n Þ þ F 0 ðy n Þðxnþ1 y n Þ Z 1 2 ¼ F 00 ðy n þ tðxnþ1 y n ÞÞð1 tÞ dtðxnþ1 y n Þ þ F ðy n Þ þ F 0 ðy n Þðxnþ1 y n Þ; 0
1 F ðxn ÞH ðxn ; zn Þ ¼ ½F 0 ðxn þ kðzn xn ÞÞ F 0 ðxn Þ ¼ k 0
Z
1
F 00 ðxn ktðy n xn ÞÞ dtðy n xn Þ; 0
1 1 F 0 ðy n Þðxnþ1 y n Þ ¼ F 0 ðxn ÞH ðxn ; zn ÞP ðxn ; zn Þðy n xn Þ þ ½F 0 ðy n Þ F 0 ðxn ÞH ðxn ; zn ÞP ðxn ; zn Þðy n xn Þ 2 2 Z 1 1 F 00 ðxn ktðy n xn ÞÞ dtðy n xn ÞP ðxn ; zn Þðy n xn Þ ¼ 2 0 Z 1 1 00 þ F ðxn þ tðy n xn ÞÞ dtðy n xn ÞH ðxn ; zn ÞP ðxn ; zn Þðy n xn Þ; 2 0 Z 1 2 0 F ðy n Þ ¼ F ðy n Þ F ðxn Þ F ðxn Þðy n xn Þ ¼ F 00 ðxn þ tðy n xn ÞÞð1 tÞ dtðy n xn Þ ¼
Z 0
0
1
1 F 00 ðxn þ tðy n xn ÞÞð1 tÞ dtðy n xn Þ I H ðxn ; zn Þ P ðxn ; zn Þðy n xn Þ: 2
Hence the (8) holds. This completes the proof of Lemma 4.
Theorem 1. Suppose X and Y are the Banach spaces, X is an open convex of the Banach space X, F : X X ! Y 1 has two orders continuous Fre´chet derivatives, F 0 ðx0 Þ exists for some x0 2 X, and F satisfies (4). Denote a0 ¼ Mg; b0 ¼ N g1þp , under conditions of the Lemma 2, if Bðx0 ; ðj1 kj þ kÞRgÞ X, then, the sequence fxn g generated by (3) from the initial point x0 is well defined, xn 2 Bðx0 ; RgÞ and converges to the unique solution x in Bðx0 ; M2 RgÞ \ X and
Y. Zhao, Q. Wu / Applied Mathematics and Computation 202 (2008) 243–251
2
kx xn k 6 2c
ð2þpÞn 1 1þp
a0
1 1
n cð2þpÞ D
c
ð2þpÞn 1 1þp
Dn g;
247
ð9Þ
2 1 where R ¼ 2a 1cD ; c ¼ aa10 ; D ¼ f ða1 0 Þ. 0
Proof. ky 0 x0 k ¼ g < Rg; ðR > 1Þ; kx0 þ kðz0 x0 Þ x0 k 6 kg, hence, we get y 0 2 X; x0 þ kðz0 x0 Þ 2 X. By (4), we have 1 1 kH ðx0 ; z0 Þk 6 Mkkz0 x0 k ¼ MkF 0 ðx0 Þ F 0 ðx0 Þkky 0 x0 k 6 Mg ¼ a0 < 1: k 1 2 ¼ Hence P ðxn ; zn Þ; exists and kP ðxn ; zn Þk 6 ; 1 a0 =2 2 a0 N kF 0 ðx0 Þ1 F 0 ðx0 Þkky 0 x0 k1þp 6 N g1þp ¼ b0 ; 1 2 2 kx1 x0 k 6 kx1 y 0 k þ ky 0 x0 k 6 1 þ a0 ky x0 k; ky 0 x0 k ¼ 2 2 a0 2 a0 0 2a0 1 1 < 1: kF 0 ðx0 Þ F 0 ðx1 Þ Ik ¼ kF 0 ðx0 Þ ðF 0 ðx1 Þ F 0 ðx0 ÞÞk 6 Mkx1 x0 k 6 2 a0 By Banach Lemma, F 0 ðx1 Þ1 exists, and 1 1 1 ¼ f ða0 ÞkF 0 ðx0 Þ F 0 ðx0 Þk: kF 0 ðx1 Þ F 0 ðx0 Þk 6 2a0 1 2a0 By conditions of Theorem 1 and Lemma 4, we obtain Z 1 Z 1 1 kF 0 ðx0 Þ1 F 00 ðx0 þ tðy 0 x0 ÞÞð1 tÞ dt F 0 ðx0 Þ1 F 00 ðx0 ktðy 0 x0 ÞÞ dtk 2 0 0 Z 1 1 6 kF 0 ðx0 Þ ½F 00 ðx0 þ tðy 0 x0 ÞÞ F 00 ðx0 Þð1 tÞ dt 0
Z 1 1 F 00 ðx0 ktðy 0 x0 ÞÞ F 00 ðx0 Þ dtk þ kF 0 ðx0 Þ1 2 0 Z Z 1 1 1 pp 2 þ ðp þ 2Þkp p p p 6 N ky 0 x0 k ; Ntp ð1 tÞ dtky 0 x0 k þ N k t dtky 0 x0 k ¼ 2 2ðp þ 1Þðp þ 2Þ 0 0 M 2Ma0 2 þ ðp þ 2Þkp 1 2 2 2þp 0 ky 0 x0 k þ N ky 0 x0 k kF ðx0 Þ F ðx1 Þk 6 kx1 y 0 k þ 2 4ð2 a0 Þ ð2 a0 Þðp þ 1Þðp þ 2Þ Ma0 2 þ ðp þ 2Þkp 2 2þp N ky 0 x0 k : 6 ky x k þ 0 0 2 ð2 a0 Þðp þ 1Þðp þ 2Þ ð2 a0 Þ Hence, 1
1
ky 1 x1 k ¼ kF 0 ðx1 Þ F 0 ðx0 ÞF 0 ðx0 Þ F ðx1 Þk 6 f ða0 Þgða0 ; b0 Þky 0 x0 k and 1
H ðx1 ; z1 Þ 6 MkF 0 ðx1 Þ F 0 ðx0 Þkky 1 x1 k 6 a0 f 2 ða0 Þgða0 ; b0 Þ ¼ a1 ; N kF 0 ðx1 Þ1 F 0 ðx0 Þk ky 1 x1 k1þp 6 N ky 0 x0 k1þp f ða0 Þ2þp gða0 ; b0 Þ1þp ¼ b1 ; 2 ky x1 k: kx2 x1 k 6 kx2 y 1 k þ ky 1 x1 k 6 2 a1 1 By the induction, we can proved the following formulae hold: (i) kF 0 ðxn Þ1 F 0 ðx0 Þk 6 f ðan1 ÞkF 0 ðxn1 Þ1 F 0 ðx0 Þk; (ii) ky n xn k 6 f ðan1 Þgðan1 ; bn1 Þky n1 xn1 k; (iii) H ðxn ; zn Þ 6 MkF 0 ðxn Þ1 F 0 ðx0 Þk ky n xn k 6 an ;
248
Y. Zhao, Q. Wu / Applied Mathematics and Computation 202 (2008) 243–251 1
(iv) N kF 0 ðxn Þ F 0 ðx0 Þk ky n xn k 2 (v) kxnþ1 xn k 6 2a ky n xn k: n
1þp
6 bn ;
Now, we prove the fxn g is a Cauchy sequence. For n P 1, " # n1 Y f ðaj Þgðaj ; bj Þ ky 0 x0 k: ky n xn k 6 f ðan1 Þgðan1 ; bn1 Þky n1 xn1 k 6 6 j¼0 n1 Y
n1 ð2þpÞj Y ð2þpÞn 1 c ¼ c 1þp Dn : f ðaj Þgðaj ; bj Þ 6 f ða0 Þ j¼0 j¼0
So, kxnþm xn k 6 kxnþm xnþm1 k þ kxnþm1 xnþm2 k þ þ kxnþ1 xn k ! nþm2 n1 Y Y 2 6 f ðaj Þgðaj ; bj Þ þ þ f ðaj Þgðaj ; bj Þ g 2 an j¼0 j¼0 ð2þpÞnþm1 1 ð2þpÞn 1 2 c ð1þpÞ Dnþm1 þ þ c ð1þpÞ Dn g 2 an ð2þpÞn 1 ð2þpÞn ðð2þpÞm1 1Þ 2 ð1þpÞ ¼ c ð1þpÞ Dn g c Dm1 þ þ 1 : 2 an 6
By the Bernouilli inequality ð1 þ xÞk 1 > kx, so ð2 þ pÞk 1 > kð1 þ pÞ, we have 2
kxnþm xn k 6 2c
ð2þpÞn 1 1þp
a0
1 n
1 cð2þpÞ D
c
ð2þpÞn 1 1þp
Dn g:
ð10Þ
Therefore, fxn g is a Cauchy sequence. By (10), let n ¼ 0, we can get kxm x0 k 6
2 1 g ¼ Rg: 2 a0 1 cD
ð11Þ
Moreover, from a similar reasoning, we easily know zn ; y n 2 Bðx0 ; RgÞ; n P 0 and xn þ kðzn xn Þ 2 Bðx0 ; ðj1 kj þ kÞRgÞ. By letting m ! 1 in (10), we obtain (9). To see that x is a solutio of (1), we have kF 0 ðxn Þ1 F ðxn Þk ¼ ky n xn k ! 0 as n ! 1. Taking into account that kF ðxn Þk 6 kF 0 ðxn ÞkkF 0 ðxn Þ1 F ðxn Þk, and the sequence kF 0 ðxn Þk is bounded. We deduce that F ðx Þ ¼ 0 from the continuity of F. By (11), we obtain x 2 Sðx0 ; RgÞ: Now we prove the unique. Suppose y also is the solution of the Eq. (1) in Bðx0 ; M2 RgÞ \ X. Then, we have Z 1 Z 1 kF 0 ðx0 Þ1 ½F 0 ðx þ tðy x ÞÞ F 0 ðx0 Þk dt 6 M kx þ tðy x Þ x0 k dt 0
Z
0 1
½ð1 tÞkx x0 k þ tky x0 k dt
6M 0
M 2 < ðRg þ RgÞ ¼ 1: 2 M R1 By Banach Lemma, ½ 0 F 0 ðx0 Þ1 F 0 ðx þ tðy x ÞÞ dt1 exists. Because of Z 1 1 1 0 ¼ F 0 ðx0 Þ ½F ðy Þ F ðx Þ ¼ F 0 ðx0 Þ F 0 ðx þ tðy x ÞÞ dtðy x Þ; 0
we know y ¼ x . This completes the proof of Theorem 1.
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3. Applications Example 1. Now we employ iterative method (3) to solve the follow equation and compare these methods with Newton’s method and Halley’s method. We defined as f ðxÞ ¼ x8=3 256:
ð12Þ
The first and the second derivative of F is defined by 8 f 0 ðxÞ ¼ x5=3 ; 3
f 00 ðxÞ ¼
40 2=3 x 9
and we have jf 00 ðxÞ f 00 ðyÞj ¼
40 2=3 40 jx y 2=3 j 6 jx yj2=3 : 9 9
To apply Theorem 1, we choose x0 ¼ 9 and we look for a domain in the form X ¼ Bðx0 ; 9Þ In this case, we obtain 1
1
jf 0 ðx0 Þ j ¼ 0:00963001 . . . ; M ¼ 0:29396 . . . ;
g ¼ jf 0 ðx0 Þ f ðx0 Þj ¼ 0:90971 . . . 2 N ¼ 0:042800 . . . ; p ¼ : 3
3:49651 Then a0 ¼ 0:267423 < 1=2, and b0 ¼ 0:0365556 < hða0 Þ ¼ 2þ8k if and only if k 2 ð0; 208:115Þ and 2=3 =3 Bðx0 ; ðj1 kj þ kÞRgÞ X if and only if k 2 ð0; 4:36894Þ. Hence, when k 2 ð0; 4:36894Þ, the conditions of Theorem 1 are fulfilled. ffiffiffiffiffiffiffiffiffi p 0 0 ðxÞÞf 0 ðxÞ 8 1 Denote x ¼ 2563 ¼ 8, by (3) xnþ1 ¼ vðxn Þ, where vðxÞ ¼ x 1 þ 2k1 f ðxþkf ðxÞ=f 0 0 0 1 f ðxþkf ðxÞ=f ðxÞÞf ðxÞ f 0 ðxÞ 12k f ðxÞ f 0 ðxÞ . We have v000 ðx Þ ¼ 0 if f 0 ðxÞ
2 f 00 ðx Þ2 11 k ¼ þ 0 000 ¼ : 3 f ðx Þf ðx Þ 6 So, we get the convergence of the sequence fxn g generated by modified Halley’s method (3) with k ¼ 116 is four orders. Now, we compare some of these methods. We analyze the error xk x for these methods with Newton’s method, Halley’s method and the family of new modified Halley’s method (3). In these cases, we have taken x0 ¼ 9 (see Table 1). Example 2. Consider the case as follows: Z 1 1 s 15=7 xðtÞ dt; xðsÞ ¼ 1 þ 4 0 sþt
ð13Þ
where the space is X ¼ C½0; 1 with norm kxk ¼ max jxðsÞj:
ð14Þ
06s61
Table 1 Error computing results Step k k k k
¼1 ¼2 ¼3 ¼4
Newton’s
Halley’s 2
9.028453 10 8.375337 104 7.305968 108 5.560122 1016
3
6.608249 10 2.293402 109 9.598411 1029 7.036505 1087
Methods (3) with a ¼ 11 6
Methods (3) with a ¼ 1
1.042665 103 1.806115 1015 1.626801 1062 1.070757 10250
3.553708 103 1.623402 1010 1.547442 1032 1.340229 1098
250
Y. Zhao, Q. Wu / Applied Mathematics and Computation 202 (2008) 243–251
This equation arises in the theory of the radiative transfer, neutron transport and in the kinetic theory of gasses. Let us define the operator F on X by 1 F ðxÞ ¼ xðsÞ 4
Z
1 0
s xðtÞ15=7 dt 1: sþt
ð15Þ
The first and the second derivative of F is defined by Z 15 1 s 8=7 0 xðtÞ uðtÞ dt; u 2 X; F ðxÞuðsÞ ¼ uðsÞ 28 0 s þ t Z 30 1 s 1=7 xðtÞ uðtÞvðtÞ dt F 00 ðxÞðuvÞðsÞ ¼ 49 0 s þ t and we have k½F 00 ðxÞ F 00 ðyÞuvk 6
30 max 49 s2½0;1
Z 0
1
s 66 1=7 1=7 1=7 jðxðtÞ yðtÞ ÞuðtÞvðtÞj dt 6 ln 2kx yk kuvk: sþt 100
To apply Theorem 1, we choose x0 ¼ x0 ðsÞ ¼ 1 and we look for a domain in the form X ¼ Bðx0 ; 2Þ Cð½0; 1Þ: In this case, we have kF 0 ðx0 Þ Ik 6
15 ln 2 < 1; 28
from the Banach Lemma, we obtain 28 7 ln 2 ; kF 0 ðx0 Þ1 F ðx0 Þk 6 ¼ g ¼ 0:275639 . . . ; 28 15 ln 2 28 15 ln 2 28 30 1=7 28 30 2 ln 2 ¼ 0:745300 . . . ; N ¼ ln 2 ¼ 0:675036 . . . ; M¼ 28 15 ln 2 49 28 15 ln 2 49
kF 0 ðx0 Þ1 k 6
1 p¼ : 7
2:555215 Then a0 ¼ 0:205434 < 1=2, and b0 ¼ 0:154781 < hða0 Þ ¼ 2þ15=7k 1=7 if and only if k 2 ð0; 652259:9 Þ and Bðx0 ; ðj1 kj þ kÞRgÞ X if and only if k 2 ð0; 3:02928 . . .Þ. Hence, when k 2 ð0; 3:02928 . . .Þ the conditions of Theorem 1 are fulfilled. Now we choose a ¼ 0:1 then the error bound becomes
1
kxn x k 6
1 0:102717 0:231879 0:231879
ð15=7Þn 1 8=7
ð15=7Þn 1 8=7
1 1 0:231879
ð15=7Þn
0:771048
0:771048n 0:2756398:
For n ¼ 1; 2; 3; 4, we get kx1 x k 6 5:22418 102 ;
kx2 x k 6 1:6614 103 ;
kx3 x k 6 1:55627 106 ;
kx4 x k 6 6:81657 1013 :
Acknowledgements The authors thank editors and anonymous referee for his or her careful reading and valuable comments on improving the original manuscript. This work was supported by Natural Science Foundation of Zhejiang Province (Grant No. Y606154). References [1] J.M. Ortega, W.C. Rheinbolt, Iterative Solution of Nonlinear Equations in Servral Variables, Academic Press, New York, 1970. [2] L. Kantorovich, On Newton method (Russian), Trudy Mat. Inst. Steklov 28 (1949) 104–144.
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