Convergence of a Third-order family of methods in Banach spaces

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1 International Journal of Computational and Applied Mathematics. ISSN Volume 1, Number (17), pp Research India Publications Convergence of a Third-order family of methods in Banach spaces Gagandeep Department of Mathematics, Hans Raj Mahila Mahavidyalaya, Jalandhar-1448, India. R. Sharma 1 Department of Applied Sciences, DAV Institute of Engineering and Technology, Jalandhar-1448, India. Abstract In this paper, we study the semilocal convergence of a family of third-order methods for solving nonlinear equations in Banach spaces using recurrence relations. Recurrence relations for the family of methods are derived and finally an existenceuniqueness theorem is derived along with a priori error bounds. AMS subject classification: 65D99. Keywords: Nonlinear equations, Banach spaces, Recurrence relations, Semilocal convergence, Iterative methods. 1. Introduction The most well known Newton s method and its variants are used to solve nonlinear operator equations F(x) =. The convergence of these second order methods was established by Kantorovich Theorem (15, 16). The convergence of sequences obtained by these methods is derived from convergence of majorizing sequences 1. Rall in 18 established the convergence of Newton s method by using recurrence relations. With the same approach, various researchers established semilocal convergence of higher order 1 Corresponding author.

2 4 Gagandeep and R. Sharma methods in Banach spaces (see 3,, 13, 1, 7, 11, 17, 14, 1, 9,, 5,, 8, 1, 4, 6 and references there in). In this paper, we shall use recurrence relations to establish the semilocal convergence of a family of third-order methods 19 in Banach spaces. Based on these recurrence relations, an existence-uniqueness theorem is given and a priori error bounds are obtained for the method.. Recurrence relations In this section, we discuss a third order method 19 for solving nonlinear operator equations F(x) =, where F : X Y is a nonlinear operator on an open convex subset of a Banach space X with values in a Banach space Y. The third order method is defined as follows: y n = x n θf (x n ) 1 F(x n ), ( x n+1 = x n ) I 1 θ θ F (x n ) 1 F (y n ) F (x n ) 1 F (x n ). (.1) Let F be a twice Fréchet differentiable in and BL(Y, X) be the set of bounded linear operators from Y into X. Let us assume that Ɣ = F (x ) 1 BL(Y, X) exists at some x and the following conditions hold: (1) F (x) F (y) K x y,x,y, () F (x) M,x (3) Ɣ β, (4) Ɣ F(x ) η. Let us denote Now, we define the sequences a = Kβη. (.) a = b = 1,d = 1 + a, a n a n+1 =, 1 aa n d n b n+1 = a n+1 βηc n, d n+1 = (1 + 1 ) aa n+1b n+1 b n+1, (.3)

3 Convergence of a Third-order family of methods in Banach spaces 41 where with C n = M ( Mθ K n + K θ b + K nk n + K n = The polynomials C n and K n can be written as ) b n, (.4) ( 1 θ + aa ) nb n b n. (.5) C n = (P + P 1 a n b n + P a n b n )b n, K n = (Q + Q 1 a n b n )b n. Lemma.1. Under the previous assumptions, we prove the following: (I n ) Ɣ n = F (x n ) 1 a n β, (II n ) Ɣ n F(x n ) b n η, (IIIn) x n+1 x n d n η, (IV n ) x n+1 y n K n η. Proof. We use induction to prove the above claims. Notice that (I ) and (II ) follow immediately from the assumptions. To prove (III ), we consider (.1). Using the assumptions, it follows that x 1 x θ Kβ y x Ɣ F(x ) 1 + a η = d η, (.6) and (III ) holds. We have x 1 y = (1 θ)i + 1 θ F (x ) 1 (F (x ) F (y )) F (x ) 1 F (x ), (.7) so that x 1 y 1 θ θ + a θ Kβ y x Ɣ F(x ) η = K η, (.8) and (IV ) also holds. Following an inductive procedure and assuming that x n and aa n d n < 1, if x n+1, we have I Ɣ n F (x n+1 ) Ɣ n F (x n ) F (x n+1 ) aa n d n < 1. (.9)

4 4 Gagandeep and R. Sharma Then, from Banach lemma, Ɣ n+1 exists and Ɣ n+1 Ɣ n 1 Ɣ n F (x n ) F (x n+1 ) a n β = a n+1 β. (.1) 1 aa n d n Hence, by induction (.1) holds for all n. This proves condition (I n ). Using the first step of (1),wehave F(y n ) = F(y n ) θf(x n ) F (x n )(y n x n ) = (1 θ)f(x n ) + F(y n ) F(x n ) F (x n )(y n x n ) = (1 θ)f(x n ) + F (x n + t(y n x n ))(1 t)dt(y n x n ). (.11) Now subtract first step of (.1) from second, we get x n+1 y n = (1 θ)i + 1 θ F (x n ) 1 (F (x n ) F (y n )) F (x n ) 1 F (x n ). so that x n+1 y n 1 θ θ + aa nb n Using Taylor s formula, we have F(x n+1 ) = F(y n ) + F (y n )(x n+1 x n ) + θ a nβk y n x n F (y n + t(x n+1 y n ))(1 t)dt(x n+1 y n ) = (1 θ)f(x n ) + F (x n )(x n+1 x n ) + + F (x n + t(y n x n ))(1 t)dt(y n x n ) F (y n + t(x n+1 y n ))(1 t)dt(x n+1 y n ) (.1) Ɣ n F(x n ) b n η = K n η. (.13) + (F (y n ) F (x n ))(x n+1 y n ) = F (x n ) 1 θ F (x n ) 1 (F (y n ) F (x n ))Ɣ n F(x n ) + (F (y n ) F (x n ))(x n+1 y n ) + + F (y n + t(x n+1 y n ))(1 t)dt(x n+1 y n ) F (x n + t(y n x n ))(1 t)dt(y n x n ). (.14)

5 Convergence of a Third-order family of methods in Banach spaces 43 Hence using (.13) in (.14), we have Therefore F(x n+1 ) K θ y n x n + K y n x n x n+1 y n + M y n x n + M x n+1 y n ( M Mθ ) K n + K θ b + K nk n + bn η = C n η. (.15) Ɣ n+1 F(x n+1 ) Ɣ n+1 F(x n+1 ) a n+1 βc n η = b n+1 η, (.16) So, by induction condition (II n ) holds for all n. Using (.16), we have x n+ x n θ F (x n+1 ) 1 F (x n+1 ) 1 F (y n+1 ) 1 Ɣ n+1 F(x n+1 ) a n+1βkb n+1 η b n+1 η = 1 + aa n+1b n+1 b n+1 η = d n+1 η. (.17) Hence, by induction, this inequality holds for all n. This proves condition (III n ). We have from (.13) and (.16) that x n+ y n+1 1 θ + 1 θ a n+1βk y n+1 x n+1 Ɣ n F(x n ) 1 θ + aa n+1b n+1 b n+1 η = K n+1 η. (.18) Hence, by induction, this inequality holds for all n. This proves condition (IV n ). 3. Convergence Analysis In this section, we establish the convergence of our third-order method (.1). To this end, we have to prove the convergence of the sequence x n defined in a Banach space or, which is same, to prove that d n is a Cauchy sequence and that the following assumptions hold: (1) x n, () aa n d n < 1,n N.

6 44 Gagandeep and R. Sharma The next two lemmas will show the Cauchy property for the sequence d n. Lemma 3.1. Assume that x is chosen so as to satisfy < d (, 3 1). Then, the sequence a n > is increasing, as n increases. < 1, that is, a a Proof. We show now that all the involved sequences are positive. Under the imposed conditions, we see that a,b,d,c,k are all positive, and also that 1 aa d >. Assume, now, that all a i,b i,d i,c i,k i, and 1 aa i d i are positive, for i =, 1,,...i,n Since C n > and b n+1 = a n+1 βηc n, it follows that a n+1,b n+1 have same sign, and so a n+1 b n+1 >. Further, from d n+1 = (1 + 1 aa n+1b n+1 )b n+1, we get that d n+1 has same sign as that of b n+1, and so, all the three terms a n+1,b n+1,d n+1 share the same sign. By absurd, we suppose that the implied sign is negative. Then d n +d n+1 <d n, and so, 1 aa n (d n +d n+1 )>1 aa n d n, which renders 1 aa n+1 d n+1 = 1 aa n(d n + d n+1 ) 1 aa n d n > 1, which implies aa n+1 d n+1 <, but that is impossible since a n+1,d n+1 have the same sign and a>. Next, since a n+1 = a n 1 aa n d n, then d n = 1 a ( 1 1 ), a n a n+1 n 1 and so, by telescoping, we get d i = 1 ( 1 1 ), where a = 1. a a a n This will render a n = i= i= 1. Certainly, since a>,d n 1 i >, for all i, 1 a d i i= n 1 n 1 then a d i increases as n increases and so, 1 a d i decreases as n increases, which implies that the reciprocal, a n is an increasing sequence, and consequently, a n a = 1. n=

7 Convergence of a Third-order family of methods in Banach spaces 45 We define the sequence c n = a n b n. Then the sequences {a n }, {b n }, {c n }, {d n } can be rewritten as a n+1 = a n 1 aa n d n = a n 1 a(c n + a c n ), b n+1 = a n+1 βηc n = βηb nc n (P + P 1 c n + P c n ) 1 a(c n + a c n ), c n+1 = a n+1 b n+1 = βηc n (P + P 1 c n + P cn ) 1 a(c n + a, c n ) d n+1 = (1 + 1 ) aa n+1b n+1 b n+1 = βηb nc n (P + P 1 c n + P cn ) ( 1 a(c n + a c n ) 1 + a ) c n+1. That the sequence {c n } is a decreasing sequence under the assumption that a 1 b 1 < 1 can be proved by using the mathematical induction. It is obvious that c 1 = a 1 b 1 < 1 = c. Assuming that c n <c n 1 for some n>, we have c n+1 = βηc n (P + P 1 c n + P c n ) 1 a(c n + a c n ) < βηc n 1 (P + P 1 c n 1 + P c n 1 ) 1 a(c n 1 + a c n 1 ) = c n. Therefore the sequence {c n } becomes a decreasing sequence with c n < 1 for all n. If <s<1 and c n sc n 1, then c n+1 = βηc n (P + P 1 c n + P c n ) 1 a(c n + a c n ) s βηc n 1 (P + P 1 sc n 1 + P s c n 1 ) 1 a(sc n 1 + a s c n 1 ) s βηc n 1 (P + P 1 c n 1 + P c n 1 ) 1 a(c n 1 + a c n 1 ) = s c n. Let ζ = c 1 c = c 1 = a 1 b 1, then we have <ζ <1 and c 1 ζc = ζ, so that c 1 ζc, c ζ c 1, c 3 ζ c, c n+1 ζ (n + n ) c = ζ n+1. 1 ζ.

8 46 Gagandeep and R. Sharma On other hand the sequence {d n } under the assumption that a 1 b 1 < 1wehave d n = aa nd n = (c n + a aa n c n ) 1 a n ( c n + a 1 n) c = c n + a a c n ( 1 + a ) c n ( 1 + a ) ζ n. 1 ζ Since {a n } is an increasing sequence, and a 1. We thus have proved the following estimates. Lemma 3.. We assume that a 1 b 1 < 1. Then the sequence {c n } is a decreasing sequence and for all n N we have the following estimates where <ζ = a 1 b 1 < 1. c n+1 ζ n+1. 1 ζ, ( d n 1 + a ) ζ n. 1 ζ Lemma 3.3. The sequence d n > is convergent sequence and its limit is. Proof. Since a n 1 is an increasing, then 1/a n 1 is a decreasing sequence and further 1/a n 1. Therefore 1/a n is convergent to a limit L. Since d n = 1 ( 1 1 ), a a n a n+1 then d n is convergent to the limit 1 (L L) =. a n 1 ) 1 Remark 3.4. Clearly d i <, since d i = lim d i = lim (1 1an = i n a i= i= i=1 1 a (1 L), where L would be the finite limit of 1/a n. Theorem 3.5. Let X, Y be Banach spaces and F be a twice Fréchet differentiable in an open convex domain of Banach space X and BL(Y, X) be set of bounded linear operators from Y into X. Let us assume that Ɣ = F (x ) 1 BL(Y, X) exists at some x and the following conditions hold: (1) F (x) F (y) K x y,x,y, () F (x) M, (3) Ɣ β,

9 Convergence of a Third-order family of methods in Banach spaces 47 (4) Ɣ F(x ) η. Let us denote a = Kβη. Suppose that x is chosen so as to satisfy a (, 3 1) and a 1 b 1 < 1. Then, if B(x,rη), where r = d n, then the sequence {x n } defined by (1)and starting at x converges to a solution x of the solution F(x) =. In this case, the solution x and the iterates x n belong to B(x,rη), and x is the only solution of F(x) = inb(x, rη). Kβ Furthermore, the error bound on x depends on the sequence {d n } given by x n+1 x k=n+1 d k η n= (1 + a/)η ζ k=n+1 ζ k,ζ = a 1 b 1. (3.1) Proof. It is easy to see that the sequence {x n } is convergent. Hence, there exists a limit x such that lim n x n = x. The sequence {a n } is bounded above since a n = 1 n 1 1 a d i i= 1. 1 a d i Since lim d n =, so we have lim b n =. This indicates that lim C n =. Thus by n n n and by the continuity of F, we proved that Also, where r = F(x ) =. x n+1 x x n+1 x n + x n x n x 1 x n d k η rη, (3.) k= d n. We conclude that {x n } lies in B(x,rη)and taking limit as n n= we have x B(x,rη).To show the uniqueness of the solution, suppose that y B(x, is another solution of F(x) =. Then, = F(y ) F(x ) = i= Kβ Rη) F (x + t(y x ))dt(y x ). (3.3)

10 48 Gagandeep and R. Sharma To show that y = x, we have to show that the operator invertible. Now, for Ɣ Kβ Kβ < Kβ (rη + Kβ F (x + t(y x )) F (x ) dt x + t(y x ) x dt ((1 t) x x +t y x )dt it follows from Banach s Theorem 15 that the operator (F (x + t(y x ))dt is rη) = 1, (3.4) an inverse, and consequently, y = x. For every m n + 1, we have (F (x + t(y x ))dt has x m x n+1 x m x m 1 + x m 1 x m + + x n+ x n+1 m 1 k=n+1 By taking m,weget and from Lemma 3.1 d k η rη. (3.5) x n+1 x k=n+1 d k η rη. (3.6) x n+1 x d k η k=n+1 (1 + a/)η ζ k=n+1 ζ k, <ζ <1. (3.7) which shows that {x n } converges and completes the proof. 4. Conclusion In this paper, the recurrence relations are developed for establishing the convergence of a family of third-order methods for solving F(x) = in Banach spaces. Based on recurrence relations, we prove a semilocal convergence, which shows the existenceuniqueness theorem for this family of methods and a priori error bounds.

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