Journal of Computational and Applied Mathematics

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1 Journal of Computational Applied Mathematics 236 (212) Contents lists available at SciVerse ScienceDirect Journal of Computational Applied Mathematics journal homepage: wwwelseviercom/locate/cam Semilocal convergence of a continuation method with Hölder continuous second derivative in Banach spaces DK Gupta, Prashanth M Department of Mathematics, Indian Institute of Technology, Kharagpur , India a r t i c l e i n f o a b s t r a c t Article history: Received 15 September 211 Received in revised form 9 February 212 Keywords: Continuation method Lipschitz continuous Hölder continuous Cubic convergence Recurrence relations Majorizing sequence In this paper, the semilocal convergence of a continuation method combining the Chebyshev method the convex acceleration of Newton s method used for solving nonlinear equations in Banach spaces is established by using recurrence relations under the assumption that the second Frëchet derivative satisfies the Hölder continuity condition This condition is mild works for problems in which the second Frëchet derivative fails to satisfy Lipschitz continuity condition A new family of recurrence relations are defined based on two constants which depend on the operator The existence uniqueness regions along with a closed form of the error bounds in terms of a real parameter α [, 1] for the solution x is given Two numerical examples are worked out to demonstrate the efficacy of our approach On comparing the existence uniqueness regions for the solution obtained by our analysis with those obtained by using majorizing sequences under Hölder continuity condition on F, it is found that our analysis gives improved results Further, we have observed that for particular values of the α, our analysis reduces to those for the Chebyshev method (α = ) the convex acceleration of Newton s method (α = 1) respectively with improved results 212 Elsevier BV All rights reserved 1 Introduction One of the most important challenging problems in scientific computing is that of finding efficiently the solutions of nonlinear equations by iterative methods in Banach spaces establish their convergence analysis There exists a large number of applications in chemical engineering, transportation, operation research etc that give rise to thouss of such equations depending on one or more parameters The boundary value problems appearing in kinetic theory of gases, elasticity other applied areas are reduced to solving nonlinear equations Dynamic systems are mathematically modeled by difference or differential equations their solutions usually represent the equilibrium states of the system obtained by solving nonlinear equations As a result, these problems are extensively studied many methods of various orders along with their local, semilocal global convergence analysis are developed in [1 4] Basically, two different kinds of approaches are used for convergence analysis The first one uses majorizing sequences obtained by the same iterative method applied to a scalar function to majorize the iterates The other one uses recurrence relations which has some advantages over majorizing sequences because, one can reduce the initial problem in a Banach space to a simpler problem with real sequences vectors Besides, we find a symmetry between some special properties of the iteration method the corresponding ones in the system of recurrence relations Newton s method its variants are the quadratically convergent iterative methods used to solve these equations The well known Kantorovich theorem [5,6] gives sufficient conditions for the semilocal convergence of Newton s method as well as the error estimates existence uniqueness Corresponding author Tel: (o); fax: , addresses: dkg@mathsiitkgpernetin (DK Gupta), marojuprashanth@gmailcom (M Prashanth) /$ see front matter 212 Elsevier BV All rights reserved doi:1116/jcam212215

2 DK Gupta, M Prashanth / Journal of Computational Applied Mathematics 236 (212) regions of solutions Many researchers [7 12] have also studied the semilocal convergence of one point third-order iterative methods such as the Chebyshev method, the Halley method the Super-Halley method used for solving nonlinear equations under Kantorovich-type assumptions [5] on the involved operator in Banach spaces These methods are also used in many applications For example, they can be used in stiff systems [13], where a quick convergence is required Their main assumption for the convergence analysis was that the second Fréchet derivative satisfies Lipschitz continuity condition However, it is not always true as the following example illustrates Example Let us consider the integral equation of Fredholm type [14]: F(x)(s) = x(s) f (s) λ s s + t x(t)2+p dt, with s [, 1], x, f C[, 1], p (, 1] λ is a real number Under the sup-norm on the operator F, the second Fréchet derivative of F satisfies F (x) F (y) λ (1 + p)(2 + p) log 2 x y p, x, y Ω Hence, for p (, 1), F does not satisfy the Lipschitz continuity condition, but it satisfies the Hölder continuity condition Hernández Salanova [15] Xintao Ye Chong Li [16] studied the convergence of the Chebyshev method the Euler Halley method by using recurrence relations under the assumption that F satisfies the Hölder continuity condition The convergence of the Chebyshev method the convex acceleration of Newton s method using majorizing sequences under the Hölder continuity conditions are given in [17,18] Further, A family of Newton-type iterative processes solving nonlinear equations in Banach spaces, that generalizes the usually iterative methods of R-order at least three is considered by Hernández Romero in [19,2] They weaken the usual semilocal convergence conditions, for this type of iterative processes, assuming that the second Fréchet derivative is ω-conditioned This means that second Fréchet derivative satisfies F (x) F (y) ω( x y ), where ω is a nondecreasing continuous real function A continuation method is a parameter based method, giving a continuous connection between two functions f g Mathematically, a continuation method between two functions f, g : X Y, where X Y are Banach spaces, is defined as a continuous map h : [, 1] X Y such that h(α, x) = αf (x) + (1 α)g(x), α [, 1] h(, x) = g(x), h(1, x) = f (x) The continuation method was known as early as 193s It was used by Kinemitician in the 196s for solving mechanism synthesis problems It also gives a set of certain answers a simple iteration process to obtain solutions more exactly For further literature survey on it, one can refer to the works of [21 23] The aim of this paper is to study the semilocal convergence of a continuation method combining the Chebyshev method the convex acceleration of Newton s method for solving nonlinear equations F(x) =, where, F : Ω X Y be a nonlinear operator on an open convex domain Ω of a Banach space X with values in Banach space Y This is done by using recurrence relations under the assumption that the second Frëchet derivative satisfies the Hölder continuity condition The existence uniqueness theorem is given We have also derived a closed form of error bounds in terms of α [, 1] Two numerical examples are worked out to demonstrate the efficacy of our approach On comparing the existence uniqueness regions error bounds for the solution obtained by our analysis with those obtained by using majorizing sequences, it is found that our analysis gives improved results Further, we have observed that for particular values of the α, our analysis reduces to those for the Chebyshev method (α = ) the convex acceleration of Newton s method (α = 1) respectively This paper is organized in six sections Section 1 is the introduction In Section 2, some preliminary results are given Then, two real sequences are generated their properties are studied In Section 3, the recurrence relations are derived In Section 4, a convergence theorem is established for the existence uniqueness regions along with a priori error bounds for the solution In Section 5, two numerical examples are worked out to demonstrate, the efficacy of our convergence analysis a comparison of the existence uniqueness regions for the solution obtained by our analysis with those obtained by using majorizing sequences under the Hólder continuity condition are done Finally, conclusions form the Section 6 2 Preliminary results In this section, we shall derive a family of recurrence relations based on two constants in order to discuss the semilocal convergence of a continuation method combining two third order iterative methods namely, the Chebyshev method the convex acceleration of Newton s method used for solving (1) Let us assume that F (x ) 1 L(Y, X) exists at some point x Ω, where L(Y, X) is the set of bounded linear operators from Y into X The Chebyshev method the convex Acceleration of Newton s method are defined as follows For a suitable initial approximation x Ω, we define for n =, 1, 2,, the iterations x n+1 = J (x n ) = x n I L F (x n ) F (x n ) 1 F(x n ), (2) (1)

3 3176 DK Gupta, M Prashanth / Journal of Computational Applied Mathematics 236 (212) x n+1 = J 1 (x n ) = x n I + 12 L F (x n )(I L F (x n )) 1 F (x n ) 1 F(x n ) (3) where I is identity operator L F (x) is the linear operator given by L F (x) = F (x) 1 F (x)f (x) 1 F(x), x X (4) The continuation method between (2) (3) can now be defined for a suitable initial approximation x α, Ω, α [, 1] for n =, 1, 2,, by the iteration x α,n+1 = αj 1 (x α,n ) + (1 α)j (x α,n ) n (5) Replacing x n by x α,n in (2) (3) substituting the expressions for J (x α,n ) from (2) J 1 (x α,n ) from (3) in (5), we get y α,n = x α,n Γ α,n F(x α,n ) x α,n+1 = y α,n L (6) F (x α,n )G α (x α,n )(y α,n x α,n ) where, for Let G α (x α,n ) = I + αl F (x α,n )H α (x α,n ) H α (x α,n ) = (I L F (x α,n )) 1 L F (x α,n ) = F (x α,n ) 1 F (x α,n )F (x α,n ) 1 F(x α,n ) Let Γ α, = F (x α, ) 1 L(Y, X) exist at some x α,, let the following assumptions hold 1 Γ α, = F (x α, ) 1 β, 2 F (x α, ) 1 F(x α, ) η, 3 F (x) M, x Ω, (7) 4 F (x) F (y) N x y p, x, y Ω, p (, 1] a = Mβη b = Nβη 1+p f (x) = g(x, y) = 2(1 x) 2 4x + x 2 (α 1)x 3, αx2 2(1 x) Now, define the real sequences {a n } {b n }, where, y(1 + (α 1)x) + (1 x)(p + 1)(p + 2) + x 2 (1 + (α 1)x) + x3 (1 + (α 1)x) 2 (1) 2(1 x) 8(1 x) 2 a n+1 = a n f (a n ) 2 g(a n, b n ), b n+1 = b n f (a n ) 2+p g(a n, b n ) 1+p (11) Let r = be the smallest positive zero of the polynomial p(x) = (2α 2 4α + 2)x 6 (α 2 + 2α 3)x 5 + (18α 16)x 4 (8α + 1)x 3 (4α 36)x 2 32x + 8 = for α [, 1] We now describe the properties of the sequences {a n } {b n } through the following Lemmas Lemma 1 Let f g be the functions defined in (9) (1), respectively Then for x (, r ) (i) f is a increasing function f (x) > 1 in (, r ] for α [, 1] (ii) g is increasing in both arguments for y > α [, 1] (iii) f (γ x) < f (x) g(γ x, γ p+1 y) γ p+1 g(x, y) for γ (, 1), p (, 1], α [, 1] Proof The proof of Lemma 1 is trivial hence omitted here Lemma 2 For a fixed p (, 1] α [, 1], define the function (p + 1)(p + 2) (2α2 4α + 2)x 6 (α 2 + 2α 3)x 5 + (18α 16)x 4 Φ p (x) = 8 8(1 + (α 1)x)(1 x) (8α + 1)x3 (4α 36)x 2 32x + 8 8(1 + (α 1)x)(1 x) (12) (8) (9)

4 DK Gupta, M Prashanth / Journal of Computational Applied Mathematics 236 (212) If < a r b Φ p (a ), then (i) f (a n ) 2 g(a n, b n ) 1 for all n (ii) {a n } {b n } are decreasing a n r < 1, n Proof (i) can easily be proved for all n, as from the definitions of f g, we get if only if or, or, f (a n ) 2 g(a n, b n ) 1 4(1 a n ) 2 (2 4a n + a 2 (α n 1)a3 n )2 + a3 n (1 + (α 1)a n) 2 1 8(1 a n ) 2 (1 a n )(p + 1)(p + 2) + a 2 αa 2 n 2(1 a n ) + b n(1 + (α 1)a n ) (p + 1)(p + 2) (2α2 4α + 2)x 6 (α 2 + 2α 3)x 5 + (18α 16)x 4 b n 8 8(1 + (α 1)x)(1 x) (8α + 1)x3 (4α 36)x 2 32x + 8 8(1 + (α 1)x)(1 x) b n Φ p (a n ) n (1 + (α 1)a n) 2(1 a n ) To prove (ii), we shall use induction From (i) for < a r b Φ p (a ), we get f (a ) 2 g(a, b ) 1 Using (11), a 1 = a f (a ) 2 g(a, b ) a < 1 as f (x) > 1 in (, r ], we get b 1 = b f (a ) 2+p g(a, b ) 1+p < b f (a ) 2 g(a, b )[f (a ) 2 g(a, b )] p b Let us assume that (ii) holds for n = k Then, proceeding similarly as above, one can easily show that a k+1 a k r < 1 b k+1 b k Since, f g are increasing functions, this gives f (a k+1 ) 2 g(a k+1, b k+1 ) f (a k ) 2 g(a k, b k ) 1 Hence it is true for all n This proves the Lemma 2 Lemma 3 Let < a < r < b < Φ p (a ) Define γ = a 1 a, then for n 1 we have (i) a n γ (2+p)n 1 a n 1 γ ((2+p)n 1)/(1+p) a for n 1 (ii) b n (γ (2+p)n 1 ) (1+p) b n 1 γ (2+p)n 1 b (iii) f (a n )g(a n, b n ) γ (2+p)n f (a )g(a,b ) = γ (2+p)n, n γ f (a ) Proof Induction will be used to prove (i) (ii) Since a 1 = γ a a 1 < a from Lemma 2(i), we get γ < 1 Also by Lemma 1(i), we get +p b 1 = b f (a ) p+2 g(a, b ) p+1 < (f (a ) 2 g(a, b )) 1+p a1 b = b = γ 1+p b Suppose (i) (ii) hold for n = k, then Hence, a k+1 = a k f (a k ) 2 g(a k, b k ) γ (2+p)k 1 a k 1 f (a k 1 ) 2 g(γ (2+p)k 1 a k 1, (γ (2+p)k 1 ) 1+p b k 1 ) γ (2+p)k 1 a k 1 f (a k 1 ) 2 (γ (2+p)k 1 ) 1+p g(a k 1, b k 1 ) = γ (2+p)k a k a k+1 γ (2+p)k a k γ (2+p)k γ (2+p)k 1 a k 1 γ (2+p)k γ (2+p)k 1 γ (2+p) a γ ((2+p)k+1 1)/(1+p) a Again, from f (x) > 1 in (, r ], we get +p b k+1 = b k f (a k ) 2+p g(a k, b k ) 1+p b k [f (a k ) 2 g(a k, b k )] 1+p ak+1 b k (γ (2+p)k ) 1+p b k a a k

5 3178 DK Gupta, M Prashanth / Journal of Computational Applied Mathematics 236 (212) Hence, b k+1 = (γ (2+p)k ) 1+p b k (γ (2+p)k ) 1+p (γ (2+p)k 1 ) 1+p (γ (2+p) ) 1+p b γ (2+p)k+1 1 b (iii) follows from f (a n )g(a n, b n ) f (γ (2+p)n 1/(1+p) a )g(γ (2+p)n 1/(1+p) a, γ (2+p)n 1 b ) γ f (a )g(a (2+p)n, b ) = γ (2+p)n /f (a ) γ as γ = a 1 /a = f (a ) 2 g(a, b ) Thus, the Lemma 3 is proved 3 Recurrence relations In this section, the recurrence relations will be derived for the method (6) under the assumptions of the previous section Now y α, exists as Γ α, = F (x α, ) 1 exists Thus, we get L F (x α, ) M Γ α, Γ α, F(x α, ) a Using Banach Lemma, this gives Thus, Also, (I L F (x α, )) a G α (x α, ) 1 + αa 1 a x α,1 y α, a αa y α, x α, 1 a x α,1 x α, x α,1 y α, + y α, x α, (2 a + (α 1)a 2) y α, x α, 2(1 a ) N Γ α, y α, x α, 1+p Nβη 1+p = b (13) We shall now prove the following inequalities for n 1, α [, 1] (I) Γ α,n = F (x α,n ) 1 f (a n 1 ) Γ α,n 1, (II) Γ α,n F(x α,n ) f (a n 1 )g(a n 1, b n 1 ) Γ α,n 1 F(x α,n 1 ), (III) L F (x α,n ) M Γ α,n Γ α,n F(x α,n ) a n, (IV) N Γ α,n Γ α,n F(x α,n ) 1+p b n, (2 an + (α 1)a 2 n (V) x α,n+1 x α,n ) (14) Γ α,n F(x α,n ), 2(1 a n ) (VI) y α,n, x α,n+1 B(x α,, Rη), for R = 2 a + (α 1)a 2 2(1 a )(1 γ ) = 1 f (a ) This will require the following Lemma Lemma 4 Let the sequence {x α,n } {y α,n } be generated by (6) Then, for all n Z +, using Taylor s theorem, we get F(x α,n+1 ) = F (x α,n + t(y α,n x α,n ))I G α (x α,n )(y α,n x α,n ) 2 (1 t)dt [F (x α,n + t(y α,n x α,n )) F (x α,n )]G α (x α,n )(y α,n x α,n ) 2 (1 t)dt F (x α,n + t(y α,n x α,n ))(y α,n x α,n )(x α,n+1 y α,n )dt F (x α,n + t(x α,n+1 y α,n ))(1 t)dt(x α,n+1 y α,n ) 2

6 DK Gupta, M Prashanth / Journal of Computational Applied Mathematics 236 (212) Now, conditions (I) (VI) can be proved by using induction For x α,1 Ω, we get I Γ α, F (x α,1 ) M Γ α, x α,1 x α, (2 a + (α 1)a 2) M Γ α, y α, x α, 2(1 a ) a (2 a + (α 1)a 2) < 1 2(1 a ) Hence, Γ α,1 = F (x α,1 ) 1 exists By using Banach Lemma, we get Γ α,1 Γ α, 1 M Γ α, x α, x α,1 2(1 a ) 2 4a + a 2 (α 1)a3 Thus, y α,1 also exists Using Lemma 4, we get From F(x α,1 ) M 2 I G α(x α, ) y α, x α, 2 + N Γ α, = f (a ) Γ α, (15) + M y α, x α, x α,1 y α, + M 2 x α,1 y α, 2 I G α (x α, ) αa 1 a, t p (1 t)dt y α, x α, 2+p G α (x α, ) we get F(x α,1 ) M αa y α, x α, 2 N (1 + (α a (p + 1)(p + 2) y α, x α, p+2 1)a ) (1 a ) + M y α, x α, 2 a (1 + (α 1)a ) + M a 2 2(1 a ) 8 y α, x α, 2 (1 + (α 1)a ) 2 (1 a ) 2 M αa η 2 N (1 + (α 1)a ) + η p a (p + 1)(p + 2) (1 a ) + M a (1 + (α 1)a ) η 2 + M a 2 2(1 a ) 8 η2 (1 + (α 1)a ) 2 (1 a ) 2 This gives Γ α,1 F(x α,1 ) Γ α,1 F(x α,1 ) f (a ) Also, we have f (a )g(a, b )η (p + 1)(p + 2)(1 a ) + a 2 αa 2 2(1 a ) + b (1 + (α 1)a ) (1 + (α 1)a ) 2(1 a ) L F (x α,1 ) M Γ α,1 Γ α,1 F(x α,1 ) M Γ α, f (a ) 2 g(a, b ) y α, x α, + a3 (1 + (α 1)a ) 2 η 8(1 a ) 2 (16) a f (a ) 2 g(a, b ) = a 1 (17) N Γ α,1 Γ α,1 F(x α,1 ) 1+p N Γ α, f (a )f (a ) 1+p g(a, b ) 1+p y α, x α, By using Banach Lemma, this gives (I L F (x α,1 )) 1 1 a 1 as L F (x α,1 ) 1 Thus, G α (x α,1 ) 1 + αa 1 1 a 1 b f (a ) 2+p g(a, b ) 1+p = b 1 (18)

7 318 DK Gupta, M Prashanth / Journal of Computational Applied Mathematics 236 (212) Hence, x α,2 y α,1 a 1(1 + (α 1)a 1 ) y α,1 x α,1 2(1 a 1 ) x α,2 x α,1 x α,2 y α,1 + y α,1 x α,1 Now, for = 1, we have f (a ) (2 a 1 + (α 1)a 2) 1 y α,1 x α,1 (19) 2(1 a 1 ) y α,1 x α, (2 a + (α 1)a 2) η = Rη (2) 2(1 a )(1 γ ) x α,2 x α, 2 a + (α 1)a 2 η = Rη (21) 2(1 a )(1 γ ) Using (15) (21), the conditions (I) (VI) hold for n = 1 Now let us assume that the conditions (I) (VI) hold for n = k x α,k Ω for all α [, 1] Proceeding similarly as above, we can prove that these conditions also hold for n = k + 1 Hence, by induction they hold for all n 4 Convergence theorem In this section we shall establish the convergence theorem a closed form of the error bounds based on α [, 1] for the method (6) Let us denote γ = a 1 /a, = 1/f (a ) R = 2 a +(α 1)a 2 2(1 a )(1 γ ) for all α [, 1] Let B(x α,, Rη) = {x X : x x α, < Rη} B(x α,, Rη) = {x X : x x α, Rη} represent the open closed balls around x α, Also, let us assume that < a r b Φ p (a ) hold, where r be the smallest positive zero of the polynomial (2α 2 4α + 2)x 6 (α 2 + 2α 3)x 5 + (18α 16)x 4 (8α + 1)x 3 (4α 36)x 2 32x + 8 = for all α [, 1] Theorem 1 Under the assumptions given in (7) B(x α,, Rη) Ω, the method (6) starting from x α, generates a sequence of iterates {x α,n } converging to the root x of F(x) = with R-order at least (2 + p) In this case x α,n, y α,n x lie in B(x α,, Rη) x is unique in B(x α,, 2 Mβ Rη) Ω Further the error bounds on x is given by x x α,n (2 γ ((2+p)n 1)/(1+p) a + (α 1)γ ((2+p)n 1)/(1+p) a 2) γ ((2+p)n 1)/(1+p) n η 2(1 γ ((2+p)n 1)/(1+p) a ) 1 γ (2+p)n (22) Proof It is sufficient to show that {x α,n } is a Cauchy sequence in order to establish the convergence of {x α,n } Using (14), we can give y α,n x α,n f (a n 1 )g(a n 1, b n 1 ) y α,n 1 x α,n 1 n 1 f (a j )g(a j, b j ) y α, x α, n 1 j= f (a j )g(a j, b j ) η j= (23) x α,m+n x α,m x α,m+n x α,m+n 1 + x α,m+1 x α,m 2 a m+n 1 + (α 1)a 2 m+n 1 y α,m+n 1 x α,m+n 1 + 2(1 a m+n 1 ) + 2 a m + (α 1)a 2 m y α,m x α,m 2(1 a m ) m+n 2 2 a m + (α 1)a 2 m 2(1 a m ) j= m 1 f (a j )g(a j, b j ) + + j= f (a j )g(a j, b j ) η (24)

8 DK Gupta, M Prashanth / Journal of Computational Applied Mathematics 236 (212) Now, for a = r, we have b = Φ p (a ) = Hence, from Lemma 2, we obtain f (a ) 2 g(a, b ) = 1, a n = a n 1 = = a b n = b n 1 = = b = This gives y α,n x α,n (f (a )g(a, b )) n y α, x α, = n η x α,m+n x α,m (2 a + (α 1)a 2 ) m 2(1 a ) n 1 η (25) Hence, if we take m =, we get x α,n x α, (2 a + (α 1)a 2 ) 2(1 a ) n 1 η (26) Thus, x α,n B(x α,, Rη) Similarly we can prove that y α,n B(x α,, Rη) Also, we can conclude that {x α,n } is a Cauchy sequence For < a < r < b < Φ p (a ) from (14), we get on using Lemma 3(iii), for n 1, n 1 n 1 y α,n x α,n f (a j )g(a j, b j ) η (γ (2+p)j )η = γ ((2+p)n 1)/(1+p) n Also from (24), we get j= x α,m+n x α,m 2 a m + (α 1)a 2 m 2(1 a m ) By Bernoulli s inequality, we get < 2 a m + (α 1)a 2 m 2(1 a m ) j= m+n 2 j= < (2 a m + (α 1)a 2 m ) m 2(1 a m ) m 1 f (a j )g(a j, b j ) + + j= f (a j )g(a j, b j ) γ ((2+p)m+n 1 1)/(1+p) m+n γ ((2+p)m 1)/(1+p) m γ ((2+p)m+n 1 1)/(1+p) n 1 + +γ ((2+p)m 1)/(1+p) η < (2 a mγ ((2+p)m 1)/(1+p) + (α 1)a 2 γ ((2+p)m 1)/(1+p) m ) m 2(1 a m γ ((2+p)m 1)/(1+p) ) γ ((2+p)m 1)/(1+p) γ (2+p)m [(2+p) n 1 1]/(1+p) n γ (2+p)m[(2+p) 1]/(1+p) + 1 η x α,m+n x α,m < (2 a mγ ((2+p)m 1)/(1+p) + (α 1)a 2 m γ ((2+p)m 1)/(1+p) ) m 2(1 a m γ ((2+p)m 1)/(1+p) ) Thus, for m =, we get x α,n x α, < 2 a + (α 1)a 2 2(1 a ) Hence, x α,n B(x α,, Rη) Also y α,n B(x α,, Rη) follows from η 1 γ (2+p)mn n η (27) 1 γ (2+p)m (1 γ n n ) η (28) (1 γ ) y α,n x α, y α,n+1 x α,n+1 + x α,n+1 x α,n + + x α,1 x α, y α,n+1 x α,n a n + (α 1)a 2 n y α,n x α,n + 2(1 a n ) + 2 a + (α 1)a 2 y α, x α, 2(1 a ) < < 2 a + (α 1)a 2 2(1 a ) 1 γ n+2 n+2 1 γ η < Rη On taking the limit as n in (25) (27), we get x B(x α,, Rη) To show that x is a solution of F(x) = We have that F(x α,n ) F (x α,n ) Γ α,n F(x α,n ) the sequence { F (x α,n ) } is bounded as F (x α,n ) F (x α, ) + M x α,n x α, < F (x α, ) + MRη

9 3182 DK Gupta, M Prashanth / Journal of Computational Applied Mathematics 236 (212) Since F is continuous, by taking limit as n α [, 1] we get F(x ) = To prove the uniqueness of the solution, if y be the another solution of (1) in B(x α,, 2 Rη) Ω then we have Mβ = F(y ) F(x ) = F (x + t(y x ))dt(y x ) Clearly, y = x 1, if F (x + t(y x ))dt is invertible This follows from Γ [F (x + t(y x )) F (x α, )]dt Mβ x + t(y x ) x α, dt by Banach Lemma Thus, y = x 5 Numerical examples Mβ Mβ 2 (1 t) x x α, + t y x α, dt Rη + 2 Mβ Rη = 1 Example 1 Let X be the space of all continuous functions on [, 1] consider the integral equation F(x) = where F(x)(s) = x(s) f (s) λ s s + t x(t)2+p dt (29) with s [, 1],x, f Ω X, p (, 1] λ is a real number The given integral equation is called Fredholm-type integral equations Here norm is taken as the sup-norm Now, it is easy to find that first second Frëchet derivatives of F as F (x)u(s) = u(s) λ(2 + p) F (x)uv(s) = λ(2 + p)(1 + p) s s + t x(t)1+p u(t)dt, s u Ω s + t x(t)p uv(t)dt, u, v Ω Clearly, F does not satisfy the Lipschitz continuity condition as for p (, 1) for all x, y Ω F (x) F s (y) = λ(2 + p)(1 + p) s + t [x(t)p y(t) p ]dt s λ (2 + p)(1 + p) max s [,1] s + t dt x(t)p y(t) p λ (2 + p)(1 + p) log 2 x y p However, it satisfies the Hölder continuity condition for p (, 1] N = λ (2 + p)(1 + p) log 2 To obtain a bound for Γ α,, we find F(x α, ) x α, f + λ log 2 x α, p I F (x α, ) λ (2 + p) log 2 x α, 1+p Now, if λ (2 + p) log 2 x α, 1+p < 1, then by Banach Lemma, we get Also, Hence, Γ α, = F (x α, ) 1 1 = β 1 λ (2 + p) log 2 x α, 1+p F (x) λ (2 + p)(1 + p) log 2 x p Γ α, F(x α, ) x α, f + λ log 2 x α, p 1 λ (2 + p) log 2 x α, 1+p Now, for λ = 1/4, p = 1/5, f (s) = 1 x α, = x (s) = 1, we get Γ α, β = , Γ α, F(x α, ) η = 2851, N = , b = Nβη 1+p = The conditions of Theorem 1 requires to find the values of a parameter

10 DK Gupta, M Prashanth / Journal of Computational Applied Mathematics 236 (212) Fig 1 Conditions on the parameters S Fig 2 Conditions of the parameter S S such that S B(x α,, Rη) Ω The values of S (125574, 11458) are obtained from Fig 1 for α = such that S (R(S)η + 1) > Φ p (a (S)) b Also, a (S) r = if only if S < From M = M(S) = S p, a = a (S) = M(S)βη = 2751S p for S = 9, we get Ω = B(1, 9), M = 79934, a = b = < Φ p (a ) Thus, the conditions of Theorem 1 are satisfied Hence, a solution of (29) exists in B(1, ) Ω the solution is unique in the ball B(1, 174) Ω Now, for α = 1,we get S (12117, ) from Fig 2 Again taking S = 9 then a = < b < Φ p (a ),we find that a solution of (29) exists in B(1, ) Ω the solution is unique in the ball B(1, ) Ω On the other h, working with majorizing sequences for α [, 1], we get the solution exists in the ball B(1, ) Ω unique in the ball B(1, ) On comparing these results, we see that for α =, our existence region for the solution is improved but not the uniqueness region However, for α = 1, both the existence uniqueness regions are improved Example 2 Consider the boundary value problem given by y + y y 3 =, y() = y(1) = (3) To find the solution, we divided the interval [, 1] into n subintervals by taking stepsize h = 1 n Let {z k} be the points of the subdivision with = z < z 1 < z 2 < < z n = 1 corresponding values of the function y = y(z ), y 1 = y(z 1 ),, y n = y(z n ) Using approximations for the first second derivatives given by y i = (y i+1 y i )/h, y i = (y i 1 2y i + y i+1 )/h 2, i = 1, 2,, n 1 noting that y = = y n, we define the operator F : R n 1 R n 1 by F(y) = G(y) + hj(y) 2h 2 g(y),

11 3184 DK Gupta, M Prashanth / Journal of Computational Applied Mathematics 236 (212) where, G = 2 4, J = 1 4 y 3 1 y 1 y 3 2 g(y) =, y = y 2 y n 1 y 3 n 1 Then, we get y 2 1 y 2 2 F (y) = G + hj 6h 2 y 2 3, y 2 n 1 y 1 y 2 F (y) = 12h 2 y 3 y n 1 Let x R n 1, A R n 1 R n 1, define the norms of x A by x = Now, we get max x i, A = max a ik 1 i n 1 1 i n 1 F (x) F (y) 12 x y This gives Γ α, β = , Γ α, F(x α, ) η = , F (x) M = 22824, N = 12, a = Mβη = , b = Also, for α =, we get a < r = 38778, b < Φ p (a ) Hence, all the conditions of Theorem 1 are satisfied Thus, the solution of Eq (3) exists in the ball B(1, ) unique in the ball B(1, ) Ω Now for α = 1, a solution of Eq (3) exists in the ball B(1, 17145) unique in the ball B(1, 15941) Ω But, by using Majorizing sequences, we get the solution exists in B(1, 1717) for α [, 1] unique in B(1, 26736) Ω Comparing these results, one can easily conclude that the existence uniqueness regions of solution for α = α = 1 are improved by our approach 6 Conclusions The semilocal convergence of a continuation method combining the Chebyshev method the convex acceleration of Newton s method used for solving nonlinear equations in Banach spaces is established by using recurrence relations under the assumption that the second Frëchet derivative satisfies the Hölder continuity condition This condition is mild works for problems in which the second Frëchet derivative is either difficult to compute or fails to satisfy Lipschitz continuity condition A new family of recurrence relations are defined based on the two constants, which depend on the operator An existence uniqueness regions along with a priori error bounds for the solution x is given A closed form of the error bounds is also derived in terms of a real parameter α [, 1] Two numerical examples are worked out to demonstrate the efficacy of our approach On comparing the existence uniqueness regions for the solution obtained by our analysis with those obtained by using majorizing sequences under Hölder continuity condition, it is found that our analysis gives improved results Further, we have observed that for particular values of α, our analysis reduces to those for the Chebyshev method (α = ) the convex acceleration of Newton s method (α = 1) respectively, with improved results References [1] JM Ortega, WC Rheinboldt, Iterative Solution of Nonlinear Equations in Several Variables, Academic Press, 197 [2] AM Ostrowski, Solution of Equations Systems of Equations, Academic Press Inc, 1966

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