Convergence of an iterative method in Banach ...

Int. J. Applied Nonlinear Science, Vol. x, No. x, xxxx

Convergence of an iterative method in Banach spaces with Lipschitz continuous first derivative P. K. Parida Department of Mathematics, Center for Applied Mathematics, Central University of Jharkhand, Ranchi 835205, India E-mail: [email protected]

D.K. Gupta Department of Mathematics, Indian Institute of Technology, Kharagpur 721 302, West Bengal, India E-mail: [email protected] *Corresponding author Abstract: In this paper, the convergence of a third order Newton-like method used for solving F (x) = 0 in Banach spaces is established by using recurrence relations, when the first Fr´echet derivative of F satisfies Lipschitz condition. Here, we relaxed the necessary conditions on F in order to study the convergence. This work is useful when either second derivative of F may not exist or may not satisfy Lipschitz condition. An existence and uniqueness theorem is derived for the root x∗ of F (x) = 0. A numerical example is given to demonstrate the applicability of the method. Keywords: nonlinear operator equations; Lipschitz continuous; cubic convergence; recurrence relations; a priori error bounds. Reference to this paper should be made as follows: Parida, P.K. and Gupta, D.K. (xxxx) ‘Convergence of an iterative method in Banach spaces with Lipschitz continuous first derivative’, Int. J. Applied Nonlinear Science, Vol. x, No. x, pp. xxx–xxx. Biographical notes: P. K. Parida received his MSc degree from Utkal University, Orissa and PhD degree from IIT Kharagpur, India in 2007. At present he is working as an Assistant Professor at The Central University Jharkhand, Ranchi, India. He has published more than 14 research papers in international journal of repute. His research interests include numerical analysis computation. D.K. Gupta received his PhD in Science from IIT Kharagpur, India in 1985. Later, he joined there as a faculty member in the Department of Mathematics. Currently, he is working as a Senior Professor. His research interests focus on theoretical computer science, distributed databases, constraint satisfaction problems and numerical analysis.

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1

2

1

Introduction

Let F : Ω ⊆ X → Y be a nonlinear operator on an open convex subset Ω of a Banach space X with values in a Banach space Y and F (x) = 0,

(1)

be the given nonlinear operator equation. Iterative methods are used for solving these equations. Newton’s method and it’s variants are well known iterative methods. These methods are of second order and the convergence analysis given by Kantorovich theorem (Kantorovich and Akilov (1982); Rall (1979)) provides sufficient conditions to ensure convergence. In order to get faster convergence, many researchers (Candela and Marquina (1990a,b); Ezquerro and Hern´andez (1998); Hern´andez and Romero (2005); Hern´andez and Salanova (2000); Hern´andez (1998)) have studied some third order one point iterative methods (cf. Chebyshev’s method, Halley’s method and super-Halley method) to solve (1). The convergence of these methods are established by using recurrence relations under the assumption that the second Fr´echet derivative satisfies Lipschitz/H¨older condition. A very restrictive condition of one point iterative methods of order N is that they depend explicitly on the first (N − 1) derivatives of F . This implies that their informational efficiency is less than or equal to unity. Multi point iterative methods (Hern´andez (2001, 2000)) using information at a number of points are also studied to solve (1). These methods only use the values of F and it’s first derivative F ′ . Recently, Wu and Zhao (2006) have studied the convergence of a third order iterative method used for solving F (x) = 0. The iterative method used by them is defined as ) yn = xn − F ′ (xn )−1 F (xn )  ′ −1 ′ . (yn ) xn+1 = xn − F (xn )+F F (xn ) 2

(2)

This method being of third order is important because it does not contain higher order derivatives, but involves only the values of F and it’s first derivative F ′ . By using majorizing function, the convergence of this method is studied under the assumption that the second derivative of the operator F satisfies the Lipschitz condition. However, the method (2) is not efficient as its third order convergence require the satisfaction of the Lipchitz continuity condition on F ′′ . It can easily be seen that on some problems either F ′′ may not exist or may not satisfy Lipschitz continuity condition on F ′′ . To show this we consider the following example. Example. Consider F (x) = 0, where F : (0; A) → R; A > 0, and F (x) = ax1+q + bx, with a, b ∈ R and q ∈ [0, 1]. From this we have, when max-norm is used kF ′′ (x)k = max |a(1 + q)qxq−1 | = ∞. (0,A)

3 Ezquerro and Hern´andez (1997, 2000, 2004) and Ezquerro et. al. (1998) relaxed this stronger condition and used Lipschitz condition on the first derivative of F to study the convergence of midpoint method, super-Halley-type method, Jarratt-type method and Halley-type method as all these are second derivative free methods. In this paper, the convergence of a third order Newton-like method used for solving F (x) = 0 in Banach spaces is established using recurrence relations, where the first Fr´echet derivative of F satisfies Lipschitz condition. This work is useful when either second derivative of F may not exist or may not satisfy Lipschitz condition. An existence and uniqueness theorem is derived for the root x∗ of F (x) = 0. A numerical example is given to demonstrate the applicability of the method. The paper is organized as follows. In section 2, two real scalar sequences are derived and their properties are studied. A recurrence relation for the iterative method (2) is derived in section 3. Based on these recurrence relations, an existence and uniqueness theorem for the root x∗ of F (x) = 0 is given in section 4. It establishes the convergence of the sequence {xn } and gives a priori error bounds for it. In section 5, a numerical example is worked out. Finally, conclusions form the section 6.

2

Construction of {an } & {cn } and their properties

In this section, we shall discuss the construction of two real sequences {an } & {cn } and study some of their properties for the iterative method (2). Let F be once Fr´echet differentiable operator in Ω and BL(Y, X) be the set of bounded linear operators from Y into X. Let Γ0 = F ′ (x0 )−1 ∈ BL(Y, X) exists at some point x0 ∈ Ω and the following conditions hold on F :  (C1) kΓ0 k ≤ β  (C2) kΓ0 F (x0 )k ≤ η  (C3) kF ′ (x) − F ′ (y)k ≤ M kx − yk, ∀ x, y ∈ Ω.

(3)

Let a0 = M βη and for n ∈ Z+ , define two real sequences cn = f (an )g(an ),

an+1 = an f (an )cn ,

(4)

where  f (x) = (2 − x)/(2 − 3x) . g(x) = x(4 − x)/(2 − x)2

(5)

Let r0 be the smallest positive zero of the polynomial h(x) = 5x2 − 8x + 2, then r0 = 0.310102. We need the following lemma in order to prove some properties of the sequences {an } and {cn } as we require. Lemma 1 Let f and g be the functions defined by the equation (5), then for x ∈ (0, r0 ], (i) f , g are increasing and f (x) > 1, (ii) f (δx) < f (x) and g(δx) < δg(x), for δ in (0, 1).

4 Proof. The proof is simple and hence omitted. The following lemmas establish the properties of the sequences {an } and {cn }. Lemma 2 Let a0 ∈ (0, r0 ] and h(a0 ) ≥ 0, then (i) cn f (an ) ≤ 1, (ii) {an }, {cn } are decreasing and an < 1 ∀n. Proof. Now from the definitions of f and g, we can easily conclude that cn f (an ) ≤ 1 iff, 0 ≤ 2(5a2n − 8an + 2) = 2h(an ). We will now prove the lemma by method of induction. For a0 ∈ (0, r0 ] and h(a0 ) ≥ 0, from above we can easily conclude that c0 f (a0 ) ≤ 1. Thus, (4) gives a1 = a0 f (a0 )c0 ≤ a0 < 1 and as f , g are increasing c1 = f (a1 )g(a1 ) ≤ f (a0 )g(a0 ) = c0 . Let the statements hold for n = k. Then, we get ak+1 = ak f (ak )ck ≤ ak < 1. Thus, ck+1 f (ak+1 ) = f (ak+1 )2 g(ak+1 ) ≤ f (ak )2 g(ak ) = ck f (ak ) ≤ 1, and ck+1 = f (ak+1 )g(ak+1 ) ≤ f (ak )g(ak ) = ck . Hence, by induction it holds for all n. 2 Lemma 3 Let us suppose that the hypothesis of Lemma 2 holds and define γ = a1 /a0 , then for n ≥ 1 we have, n−1

(i) an ≤ γ 2 (ii) cn < γ

2n

n

an−1 ≤ γ 2

−1

a0 , strictly hold for n ≥ 2,

/f (a0 ).

Proof. Since a1 = γa0 and from Lemma 2, we have a1 < a0 , so γ < 1. Thus, a2 = a1 f (a1 )c1 < γa0 f (γa0 )2 g(γa0 ) < γa0 f (a0 )2 γg(a0 ) = γ 2 a1 . Suppose (i) holds for n = k, then proceeding similarly as above, we have ak+1 < k γ 2 ak and k

k

k−1

ak+1 < γ 2 ak < · · · < γ 2 γ 2

k+1

0

· · · γ 2 a0 = γ 2

−1

a0 .

For (ii), we have n

cn = f (an )g(an ) < f (γ 2

−1

as γ = c0 f (a0 ) = f (a0 )2 g(a0 ).

n

a0 )g(γ 2 2

−1

n

a0 ) < γ 2

−1

n

f (a0 )g(a0 ) = γ 2 /f (a0 )

5

3 Recurrence Relations In this section, we shall derive the recurrence relations for the iterative method given by (2). First of all we give some preliminary results. Now y0 exists as Γ0 = F ′ (x0 )−1 exists. So its easy to see that M kΓ0 kky0 − x0 k ≤ M βη = a0 . Also we have

F ′ (y0 ) + F ′ (x0 ) a0

F ′ (x0 ) − F ′ (y0 ) 1 < 1.

I − Γ0

= Γ0

≤ M kΓ0 kkx0 − y0 k ≤ 2 2 2 2 Hence  ′

by

F (y0 )+F ′ (x0 ) 2

Banach’s theorem −1 ′ F (x0 ) exists and

(Kantorovich

and

Akilov

(1982)),

 F ′ (y ) + F ′ (x ) −1

2

0 0 F ′ (x0 ) ≤ .

2 2 − a0

Thus, we have

 F ′ (y ) + F ′ (x ) −1 2

0 0 F ′ (x0 ) kF ′ (x0 )−1 F (x0 )k ≤ ky0 − x0 k kx1 − x0 k ≤ 2 2 − a0 and

4 − a0 ky0 − x0 k. 2 − a0 On the basis of above mentioned results, we can now prove the following inequalities for n ≥ 1: kx1 − y0 k ≤ kx1 − x0 k + kx0 − y0 k ≤

(i) (ii) (iii) (iv) (v)

′ −1 ′ kΓn Γ−1 F (xn−1 )k ≤ f (an−1 ), n−1 k = kF (xn ) kyn − xn k = kΓn F (xn )k ≤ cn−1 kyn−1 − xn−1 k, M kΓn kkyn − xn k ≤ an , kxn+1 − xn k ≤ 2/(2 − an )kyn − xn k, kxn+1 − yn k ≤ (4 − an )/(2 − an )kyn − xn k.

The following lemma is used to establish the above inequalities.

     

(6)

    

Lemma 4 Let F be the given operator and the sequences {xn } and {yn } are generated by the method (2). Then ∀n ∈ Z+ , we have Γn F (xn+1 ) =

 F ′ (x ) + F ′ (y ) −1 h F ′ (y ) − F ′ (x ) i n n n n Γn F (xn ) 2 2 Z 1 + Γn [F ′ (xn + t(xn+1 − xn )) − F ′ (xn )](xn+1 − xn )dt. 0

Proof. From the method (2), we obtain Z

xn+1

Γn F ′ (x)dx xn Z xn+1 = Γn F (xn ) + (xn+1 − xn ) + [Γn F ′ (x) − I]dx

Γn F (xn+1 ) = Γn F (xn ) +

xn

6 Z xn+1  F ′ (x ) + F ′ (y ) −1 n n = Γn F (xn ) − F (xn ) + Γn [F ′ (x) − F ′ (xn )]dx 2 xn Z xn+1 h  F ′ (x ) + F ′ (y ) −1 i n n ′ = I− F (xn ) Γn F (xn ) + Γn [F ′ (x) − F ′ (xn )]dx 2 xn  F ′ (x ) + F ′ (y ) −1 h F ′ (y ) − F ′ (x ) i n n n n Γn F (xn ) = 2 2 Z 1 + Γn [F ′ (xn + t(xn+1 − xn )) − F ′ (xn )](xn+1 − xn )dt.  0

The conditions (I) − (V ) will be proved by induction. Assume that x1 ∈ Ω. We now have kI − Γ0 F ′ (x1 )k ≤ M kΓ0 kkx0 − x1 k ≤

2 2a0 M βky0 − x0 k ≤ < 1. 2 − a0 2 − a0

Hence by Banach’s theorem, Γ1 = F ′ (x1 )−1 exists and kΓ1 Γ−1 0 k≤

2 − a0 1 = = f (a0 ). 1 − 2a0 /(2 − a0 ) 2 − 3a0

(7)

Now, according to Lemma 4,

 F ′ (x ) + F ′ (y ) −1 h F ′ (y ) − F ′ (x ) i

0 0 0 0 kΓ0 F (x1 )k = Γ0 F (x0 ) 2 2 Z 1

+ Γ0 [F ′ (x0 + t(x1 − x0 )) − F ′ (x0 )](x1 − x0 )dt 0

M M 2 ≤ kΓ0 k ky0 − x0 k2 + kΓ0 k kx1 − x0 k2 2 − a0 2 2 M 2M ≤ kΓ0 kky0 − x0 k2 + kΓ0 kky0 − x0 k2 2 − a0 (2 − a0 )2 h a 2a0 i 0 ≤ + ky0 − x0 k 2 − a0 (2 − a0 )2 a0 (4 − a0 ) ≤ ky0 − x0 k = g(a0 )ky0 − x0 k. (2 − a0 )2

and so ky1 − x1 k = kΓ1 F (x1 )k ≤ kΓ1 Γ−1 0 kkΓ0 F (x1 )k ≤ f (a0 )g(a0 )ky0 − x0 k = c0 ky0 − x0 k.(8) Also M kΓ1 kky1 − x1 k ≤ M kΓ0 kf (a0 )2 g(a0 )ky0 − x0 k ≤ a0 f (a0 )c0 = a1 .

(9)

Again as Γ1 = F ′ (x1 )−1 exists, so y1 exists. Hence,

F ′ (y1 ) + F ′ (x1 ) a1

F ′ (x1 ) − F ′ (y1 ) 1 < 1.

= Γ1

≤ M kΓ1 kkx1 − y1 k =

I − Γ1 2 2 2 2

7  ′ −1 ′ (x1 ) So by Banach’s theorem (Kantorovich and Akilov (1982)), F (y1 )+F F ′ (x1 ) 2 exists and

 F ′ (y ) + F ′ (x ) −1

2

1 1 F ′ (x1 ) ≤ .

2 2 − a1 Hence,

kx2 − x1 k ≤

2 ky1 − x1 k 2 − a1

(10)

kx2 − y1 k ≤

4 − a1 ky1 − x1 k. 2 − a1

(11)

and

Hence, for n = 1, equations (7)–(11) proved the conditions (I) − −(V ) respectively. Now consider that these inequalities hold for n = k and xk ∈ Ω. Proceeding similarly as above, one can easily prove these conditions for n = k + 1. Hence, by induction it holds for all n.

4 Convergence Analysis The existence and uniqueness theorem for the root x∗ of F (x) = 0 is given below. It establishes the convergence of the sequence {xn } and gives a priori error bounds 2 for it. Let us denote γ = a1 /a0 , ∆ = 1/f (a0 ), R = (2−a0 )(1−γ∆) , B(x0 , Rη) = {x ∈ ¯ 0 , Rη) = {x ∈ X : kx − x0 k ≤ Rη}. X : kx − x0 k < Rη} and B(x Theorem 1 Let the conditions 0 < a0 ≤ r0 and h(a0 ) ≥ 0 hold, where r0 = 0.310102 is the smallest positive zero of the polynomial h(x) = 5x2 − 8x + 2. Let ¯ 0 , Rη) ⊆ Ω hold. Then starting from F satisfies the conditions given by (3) and B(x x0 , the sequence {xn } generated by the method (2) converges R-quadratically to a ¯ 0 , Rη) of the equation (1) with both xn , yn lie in B(x ¯ 0 , Rη) and solution x∗ ∈ B(x x∗ is unique in B(x0 , 2/(M β) − Rη) ∩ Ω. Furthermore, the error bounds on x∗ is given by n

kx∗ − xn k ≤

∆n 2γ 2 −1 η. n −1 2 (2 − γ a0 ) (1 − γ 2n ∆)

(12)

Proof. Here in order to prove {xn } is convergent, it is sufficient to show that {xn } is a Cauchy sequence. Now for a0 = r0 , h(a0 ) = 0 and f (a0 )c0 = 1. Hence from (4), an = an−1 = · · · = a0 and cn = cn−1 = · · · = c0 . So from (6), we have kyn − xn k ≤ cn−1 kyn−1 − xn−1 k = c0 kyn−1 − xn−1 k ≤ · · · ≤ cn0 ky0 − x0 k = ∆n η and kxn+1 − xn k ≤

2 2 kyn − xn k ≤ ∆n η. 2 − an 2 − a0

8 Thus kxm+n − xm k ≤ kxm+n − xm+n−1 k + · · · + kxm+1 − xm k i 2 h m+n−1 2∆m  1 − ∆n  ≤ ∆ + · · · + ∆m η = η. 2 − a0 2 − a0 1 − ∆

(13)

¯ 0 , Rη). Similarly one can prove that yn ∈ Hence, if we take m = 0, xn ∈ B(x ¯ B(x0 , Rη). Also from the above equation and Lemma 1, we can conclude that, {xn } is a Cauchy sequence. Let 0 < a0 < r0 and h(a0 ) > 0. Now from (6) and Lemma 3(ii), for n ≥ 1, we have kyn − xn k ≤ cn−1 kyn−1 − xn−1 k ≤ · · · ≤

n−1 Y

(cj )ky0 − x0 k
−1 and every integer k ≥ 0, we have (1 + x)k − 1 ≥ kx. Hence we get m

m

2γ 2 −1 ∆m 1 − γ 2 n ∆n kxm+n − xm k < η. (2 − γ 2m −1 a0 ) (1 − γ 2m ∆)

(14)

For m = 0, we obtain kxn − x0 k