A Third Order Newton-Like Method and Its Applications - MDPI

mathematics Article

A Third Order Newton-Like Method and Its Applications D. R. Sahu 1, *, Ravi P. Agarwal 2 and Vipin Kumar Singh 3 1 2 3

*

Department of Mathematics, Banaras Hindu University, Varanasi-221005, India Department of Mathematics, Texas A&M University-Kingsville, Kingsville, TX 78363-8202, USA; [email protected] Department of Mathematics, Banaras Hindu University, Varanasi-221005, India; [email protected] Correspondence: [email protected]

Received: 8 October 2018; Accepted: 7 December 2018; Published: 30 December 2018







Abstract: In this paper, we design a new third order Newton-like method and establish its convergence theory for finding the approximate solutions of nonlinear operator equations in the setting of Banach spaces. First, we discuss the convergence analysis of our third order Newton-like method under the ω-continuity condition. Then we apply our approach to solve nonlinear fixed point problems and Fredholm integral equations, where the first derivative of an involved operator does not necessarily satisfy the Hölder and Lipschitz continuity conditions. Several numerical examples are given, which compare the applicability of our convergence theory with the ones in the literature. Keywords: nonlinear operator equation; Fréchet derivative; ω-continuity condition; Newton-like method; Frédholm integral equation

1. Introduction Our purpose of this paper is to compute solution of nonlinear operator equation of the form F ( x ) = 0,

(1)

where F : D ⊂ X → Y is a nonlinear operator defined on an open convex subset D of a Banach space X with values into a Banach space Y. A lot of challenging problems in physics, numerical analysis, engineering, and applied mathematics are formulated in terms of finding roots of the equation of the form Equation (1). In order to solve such problems, we often use iterative methods. There are many iterative methods available in literature. One of the central method for solving such problems is the Newton method [1,2] defined by xn+1 = xn − ( Fx0 n )−1 F ( xn ) (2) for each n ≥ 0, where Fx0 denotes the Fréchet derivative of F at point x ∈ D. The Newton method and the Newton-like method are attractive because it converges rapidly from any sufficient initial guess. A number of researchers [3–20] have generalized and established local as well as semilocal convergence analysis of the Newton method Equation (2) under the following conditions: (a) Lipschitz condition: k Fx0 − Fy0 k ≤ K k x − yk for all x, y ∈ D and for some K > 0; (b) Hölder Lipschitz condition: k Fx0 − Fy0 k ≤ K k x − yk p for all x, y ∈ D and for some p ∈ (0, 1] and K > 0;

Mathematics 2019, 7, 31; doi:10.3390/math7010031

www.mdpi.com/journal/mathematics

Mathematics 2019, 7, 31

2 of 22

(c) ω-continuity condition: k Fx0 − Fy0 k ≤ ω (k x − yk) for all x, y ∈ D,where ω : [0, ∞) → [0, ∞) is a nondecreasing and continuous function. One can observe that the condition (c) is more general than the conditions (a) and (b). One can find numerical examples where the Lipschitz condition (a) and the Hölder continuity condition (b) on the first Fréchet derivative do not hold, but the ω-continuity condition (c) on first Fréchet derivative holds (see Example 1, [21]). On the other hand, many mathematical problems such as differential equations, integral equations, economics theory, game theory, variational inequalities, and optimization theory ([22,23]) can be formulated into the fixed point problem: Find x ∈ C such that x = G ( x ),

(3)

where G : C → X is an operator defined on a nonempty subset C of a metric space X. The easiest iterative method for constructing a sequence is Picard iterative method [24] which is given by x n +1 = G ( x n )

(4)

for each n ≥ 0. The Banach contraction principle (see [1,22,23,25]) provides sufficient conditions for the convergence of the iterative method Equation (4) to the fixed point of G. Banach spaces have more geometrical stricture with respect to metric spaces. For study fixed points of nonlinear smooth operators, Banach space structure is required. More details of Banach space theory and fixed point theory of nonlinear operators can be found in [1,22,23,26–28]. The Newton method and its variant [29,30] are also used to solve the fixed point problem of the form: ( I − G )( x ) = 0, (5) where I is the identity operator defined on X and G : D ⊂ X → X is a nonlinear Fréchet differentiable operator defined on an open convex subset D of a Banach space X. For finding approximate solution of the Equation (5), Bartle [31] used the Newton-like iterative method of the form xn+1 = xn − ( I − Gy0 n )

−1

( I − G ( xn ))

(6)

for each n ≥ 0, where Gx0 is Fréchet derivative of G at point x ∈ D and {yn } is the sequence of arbitrary points in D which are sufficiently closed to the desired solution of the Equation (5). Bartle [31] has discussed the convergence analysis of the form Equation (6) under the assumption that G is Fréchet differentiable at least at desired points and a modulus of continuity is known for G 0 as a function of x. The Newton method Equation (2) and the modified Newton method are the special cases of the form Equation (6). Following the idea of Bartle [31], Rall [32] introduced the following Stirling method for finding a solution of the fixed point problem Equation (5): (

y n = G ( x n ), xn+1 = xn − ( I − Gy0 n )−1 ( xn − G ( xn ))

(7)

for each n ≥ 0. Many researchers [33–35] have studied the Stirling method Equation (7) and established local as well as semilocal convergence analysis of the Stirling-like method. Recently, Parhi and Gupta [21,36] have discussed the semilocal convergence analysis of the following Stirling-like iterative method for computing a solution of operator Equation (5):    z n = G ( x n ), yn = xn − ( I − Gz0 n )−1 ( xn − G ( xn )),   x 0 −1 n+1 = yn − ( I − Gzn ) ( yn − G ( yn ))

(8)

Mathematics 2019, 7, 31

3 of 22

for each n ≥ 0. More precisely, Parhi and Gupta [21] have studied the semilocal convergence analysis of Equation (8) for computing a solution of the operator Equation (5), where G : D ⊂ X → X is a nonlinear Fréchet differentiable operator defined on an open convex subset D under the condition: (Ω) k Gx0 k ≤ k for all x ∈ D and for some k ∈ (0, 13 ]. There are some nonlinear Fréchet differentiable operators G : D ⊂ X → X defined on an open convex subset D which fail to satisfy the condition (Ω) (see Example 1). Therefore, ref. [21] (Theorem 1) is not applicable for such operators. So, there is the following natural question: Problem 1. Is it possible to develop the Stirling-like iterative method for computing a solution of the operator Equation (5), where the condition (Ω) does not hold? The main purpose of this paper is to design a new Newton-like method for solving the operator Equation (1) and provide an affirmative answer of the Problem 1. We prove our proposed Newton-like method has R-order of convergence at least 2p + 1 under the ω-continuity condition and it covers a wide variety of iterative methods. We derive the Stirling-like iterative method for computing a solution of the fixed point problem Equation (5), where (Ω) does not hold and hence it gives an affirmative answer to Question 1 and generalizes the results of Parhi and Gupta [21,36] in the context of the condition (Ω). In Section 2, we summarize some known concepts and results. In Section 3, we introduce a new Newton-like method for solving the operator Equation (1) and establish convergence theory of the proposed Newton-like method. In Section 4, we derive the Stirling-like iterative method from the proposed Newton-like method and establish a convergence theorem for computing a solution of the fixed point problem. Applications to Fredholm integral equations are also presented in Section 5, together with several numerical examples, which compare the applicability of our iterative technique with the ones in the literature. 2. Preliminary In this section, we discuss some technical results. Throughout the paper, we denote B( X, Y ) a collection of bounded linear operators from a Banach space X into a Banach space Y and B( X ) = B( X, X ). For some r > 0, Br [ x ] and Br ( x ) are the closed and open balls with center x and radius r, respectively, N0 = N ∪ {0} and Φ denote the collection of nonnegative, nondecreasing, continuous real valued functions defined on [0, ∞). Lemma 1. (Rall [37] (p. 50)) Let L be a bounded linear operator on a Banach space X. Then L−1 exists if and only if there is a bounded linear operator M in X such that M−1 exists and 1 . k M −1 k

k M − Lk < If L−1 exists, then we have

−1

L ≤ Lemma 2. Let 0 < k ≤ scalar equation

1 3

−1

M 1 − k1 − M−1 Lk

≤

−1

M 1 − k M−1 k k M − Lk

be a real number. Assume that q =

(1 − k p (1 + qt) p t) p+1 −



qptp + kp p+1

has a minimum positive root α. Then we have the following:

p

1 p +1

.

+ k p for any p ∈ (0, 1] and the

q p t2p = 0

Mathematics 2019, 7, 31

4 of 22

(1) q > k for all p ∈ (0, 1]. (2) α ∈ (0, 1). Proof. (1) This part is obvious. Indeed, we have 1 2− p 1 + kp − = + kp > 0 p+1 3 3(1 + p ) for all p ∈ (0, 1] and 0 < k ≤ 13 . (2) Set g(t) = (1 − k p (1 + qt) p t) p+1 −



qptp + kp p+1

p

q p t2p .

(9)

It is clear from the definition of g(t) that g(0) > 0, g(1) < 0 and g0 (t) < 0 in (0, 1). Therefore, g(t) is decreasing in (0, 1) and hence the Equation (9) has a minimum positive root α ∈ (0, 1). This completes the proof. Lemma 3. Let b0 ∈ (0, α) be a number such that k p (1 + qb0 ) p b0 < 1, where k, p, α and q are same as in Lemma 2. Define the real sequences {bn }, {θn } and {γn } by  bn +1 =

θn =

+ kp

p

2p

q p bn

bn ,

(10)

1 1 − k p (1 + qbn ) p bn

(11)

(1 − k p (1 + qbn ) p bn ) p+1

p

q p bn p p +1 + k 1 − k p (1 + qb



p

q p bn p +1



qbn2

n)

pb

, n

γn =

for each n ∈ N0 . Then we have the following: p q p b0 p +1

(1)

!p

+k p

2p

q p b0

(1−k p (1+qb0 ) p b0 ) p+1

< 1.

(2) The sequence {bn } is decreasing, that is bn+1 ≤ bn for all n ∈ N0 . (3) k p (1 + qbn ) p bn < 1 for all n ∈ N0 . n (4) bn+1 ≤ ξ (2p+1) bn for all n ∈ N0 . (5) θn ≤ ξ

(2p+1)n −1 p θ

for all n ∈ N0 , where θ0 = θ and ξ = γ0 θ p .

Proof. (1) Since the scalar equation g(t) = 0 defined by Equation (9) has a minimum positive root α ∈ (0, 1) and g(t) is decreasing in (0, 1) with g(0) > 0 and g(1) < 0. Therefore, g(t) > 0 in the interval (0, α) and hence   p

q p b0 p +1

p

+ kp

2p

q p b0

(1 − k p (1 + qb0 ) p b0 ) p+1

< 1.

(2) From (1) and Equation (10), we have b1 ≤ b0 . This shows that (2) is true for n = 0. Let j ≥ 0 be a fixed positive integer. Assume that (2) is true for n = 0, 1, 2, · · · , j. Now, using Equation (10), we have  p p p  p p p q b j +1 q bj 2p 2p p p p q b j +1 q p bj p +1 + k p +1 + k b j +2 =
p +1 b j +1 ≤
p +1 b j = b j +1 . 1 − k p (1 + qb j+1 ) p b j+1 1 − k p (1 + qb j ) p bn Thus (2) holds for n = j + 1. Therefore, by induction, (2) holds for all n ∈ N0 .

Mathematics 2019, 7, 31

5 of 22

(3) Since bn < bn−1 for each n = 1, 2, 3, · · · and k p (1 + qb0 ) p b0 < 1 for all p ∈ (0, 1], it follows that k p (1 + qbn ) p bn < k p (1 + qb0 ) p b0 < 1. (4) From (3), one can easily prove that the sequences {γn } and {θn } are well defined. Using Equations (10) and (11), one can easily observe that p

bn + 1 = γn θ n bn

(12)

for each n ∈ N0 . Put n = 0 and n = 1 in Equation (12), we have 0

b1 = γ0 θ p b0 = ξ (2p+1) b0 and 

=

b2

p

q p b1 p +1

+ kp

p

2p

q p b1

(1 − k p (1 + qb1 ) p b1 ) p+1  p p q p b0 p q p (ξb0 )2p p +1 + k

≤

b1

b1 (1 − k p (1 + qb0 ) p b0 ) p+1 p  p q p b0 2p p + k q p b0 p +1 2p b1 ≤ ξ (1 − k p (1 + qb0 ) p b0 ) p+1

= ξ 2p γ0 θ p b1 = ξ 2p+1 b1 . Hence (4) holds for n = 0 and n = 1. Let j > 1 be a fixed integer. Assume that (4) holds for each n = 0, 1, 2 · · · , j. From Equations (11) and (12), we have 

=

b j +2

≤

≤

p

q p b j +1 p +1

+ kp

p

2p

q p b j +1


p +1 b j +1 1 − k p (1 + qb j+1 ) p b j+1  p p p q b j +1 j p q p (ξ (2p+1) b j )2p p +1 + k
p +1 b j +1 1 − k p (1 + qb j+1 ) p b j+1  p p p q bj 2p p + k q p bj p +1 j (ξ 2p(2p+1) )
p +1 b j +1 1 − k p (1 + qb j ) p b j j

≤ ξ 2p(2p+1) ξ 2p(2p+1) = ξ (2p+1)

j +1

j −1

· · · ξ 2p(2p+1) ξ (2p+1) b j+1

b j +1 .

Thus (4) holds for n = j + 1. Therefore, by induction, (4) holds for all n ∈ N0 . (5) From Equation (11) and (4), one can easily observe that  θ1 =

p

q p b1 p +1

+ kp





qb12

1 − k p (1 + qb1 ) p b1

≤

p

q p b0 p +1

+ kp



q(ξb0 )2

1 − k p (1 + qb0 ) p b0

≤ξ

(2p+1)1 −1 p θ.

Mathematics 2019, 7, 31

6 of 22

Hence (5) holds for n = 1. Let j > 1 be a fixed integer. Assume that (5) holds for each n = 0, 1, 2 · · · , j. From Equation (11), we have  θ j +1

=

p

q p b j +1 p +1

+ kp



qb2j+1

1 − k p (1 + qb j+1 ) p b j+1 !2   p (2p+1) j+1 −1 q p b0 p 2p q ξ b0 p +1 + k

≤ = ξ

1 − k p (1 + qb0 ) p b0 (2p+1) j+1 −1 p

θ.

Thus (5) holds for n = j + 1. Therefore, by induction, (v) holds for all n ∈ N0 . This completes the proof. 3. Computation of a Solution of the Operator Equation (1) Let X and Y be Banach spaces and D be a nonempty open convex subset of X. Let F : D ⊂ X → Y be a nonlinear operator such that F is Fréchet differentiable at each point of D and let L ∈ B(Y, X ) such that ( I − LF )( D ) ⊆ D. To solve the operator Equation (1), we introduce the Newton-like algorithm as follows: Starting with x0 ∈ D and after xn ∈ D is defined, we define the next iterate xn+1 as follows:    zn = ( I − LF )( xn ), 1 F )( x ), yn = ( I − Fz0− n n   x 0−1 n+1 = ( I − Fzn F )( yn )

(13)

for each n ∈ N0 . If we take X = Y, F = I − G and L = I ∈ B( X ) in Equation (13), then the iteration process Equation (13) reduces to the Stirling-like iteration process Equation (8). Before proving the main result of the paper, we establish the following: Proposition 1. Let D be a nonempty open convex subset of a Banach space X, F : D ⊂ X → Y be a Fréchet differentiable at each point of D with values in a Banach space Y and L ∈ B(Y, X ) such that ( I − LF )( D ) ⊆ D. Let ω : [0, ∞) → [0, ∞) be a nondecreasing and continuous real-valued function. Assume that F satisfies the following conditions: (1) k Fx0 − Fy0 k ≤ ω (k x − yk) for all x, y ∈ D; (2) k I − LFx0 k ≤ c for all x ∈ D and for some c ∈ (0, ∞). Define a mapping T : D → D by T ( x ) = ( I − LF )( x ) for all x ∈ D. Then we have 0−1 0 0−1 k I − FTx FTy k ≤ k FTx kω (ck x − yk)

for all x, y ∈ D. Proof. For any x, y ∈ D, we have

(14)

Mathematics 2019, 7, 31

7 of 22

0−1 0 0−1 0 0 k I − FTx FTy k ≤ k FTx kk FTx − FTy k 0−1 ≤ k FTx kω (k Tx − Tyk) 0−1 = k FTx kω (k x − y − L( F ( x ) − F (y))k)   Z 1 0−1 = k FTx kω k x − y − L Fy0 +t( x−y) ( x − y)dtk 0 Z 1  0−1 ≤ k FTx kω k I − LFy0 +t(x−y) kdtk x − yk

≤

0 0−1 k FTx kω (ck x − yk).

This completes the proof. Now, we are ready to prove our main results for solving the problem Equation (1) in the framework of Banach spaces. Theorem 1. Let D be a nonempty open convex subset of a Banach space X, F : D ⊂ X → Y a Fréchet differentiable at each point of D with values in a Banach space Y and L ∈ B(Y, X ) such that ( I − LF )( D ) ⊆ D. 1 ∈ B (Y, X ) exist. Let ω ∈ Φ and let α be the solution of Let x0 ∈ D be such that z0 = x0 − LF ( x0 ) and Fz0− 0 the Equation (9). Assume that the following conditions hold: (C1) k Fx0 − Fy0 k ≤ ω (k x − yk) for all x, y ∈ D; (C2) k I − LFx0 k ≤ k for all x ∈ D and for some k ∈ (0, 13 ]; 1 k ≤ β for some β > 0; (C3) k Fz0− 0 1 F ( x )k ≤ η for some η > 0; (C4) k Fz0− 0 0

(C5) ω (ts) ≤ t p ω (s), s ∈ [0, ∞), t ∈ [0, 1] and p ∈ (0, 1]; (C6) b0 = βω (η ) < α, q =

1 p +1

+ kp,

p q p b0 p +1

θ=

!

+k p

qb02

1−k p (1+qb0 ) p b0

and Br [ x0 ] ⊂ D, where r =

1+ q 1−θ η.

Then we have the following: (1) The sequence { xn } generated by Equation (13) is well defined, remains in Br [ x0 ] and satisfies the following estimates:  k y n −1 − z n −1 k ≤ k k y n −1 − x n −1 k ,      k xn − yn−1 k ≤ qbn−1 kyn−1 − xn−1 k,    kx − x n n−1 k ≤ (1 + qbn−1 )k yn−1 − xn−1 k, 0− 1 1k ≤ γ 0−1  Fzn exists and k Fz0− n−1 k Fzn−1 k,  n    k y n − x n k ≤ θ n −1 k y n −1 − x n −1 k ≤ θ n k y 0 − x 0 k ,    1 k ω (k y − x k) ≤ b k Fz0− n n n n

(15)

for all n ∈ N, where zn , yn ∈ Br [ x0 ], the sequences {bn }, {θn }, and {γn } are defined by Equations (10) and (11), respectively. (2) The sequence { xn } converges to the solution x ∗ ∈ Br [ x0 ] of the Equation (1). (3) The priory error bounds on x ∗ is given by:

(1 + qb0 )η

k xn − x ∗ k ≤ ξ 1/2p

2

 1−ξ

(2p+1)n p

  1

−p

γ0

n/p

γ0

for each n ∈ N0 . (4) The sequence { xn } has R-order of convergence at least 2p + 1.

ξ 1/2p

2

(2p+1)n

Mathematics 2019, 7, 31

8 of 22

1 F ( x ) is Proof. (1) First, we show that Equation (15) is true for n = 1. Since x0 ∈ D, y0 = x0 − Fz0− 0 0 well defined. Note that 1 ky0 − x0 k = k Fz0− F ( x0 )k ≤ η < r. 0

Hence y0 ∈ Br [ x0 ]. Using Equation (13), we have 1 ky0 − z0 k = k − Fz0− F ( x0 ) + LF ( x0 )k 0

= ky0 − x0 − LFz00 (y0 − x0 )k ≤ k I − LFz00 kky0 − x0 k ≤ k k y0 − x0 k. By Proposition 1 and (C2), we have 1 k x1 − y0 k = k Fz0− ( F (y0 ) − F ( x0 ) − Fz00 (y0 − x0 ))k 0

Z 1

1 0 k Fz0− ( Fx0 +t(y0 − x0 ) − Fy0 0 + Fy0 0 − Fz00 )kky0 − x0 kdt 0 Z 1  β k( Fx0 0 +t(y0 − x0 ) − Fy0 0 )kdt + k Fy0 0 − Fz00 k ky0 − x0 k 0 Z 1  β ω ((1 − t)ky0 − x0 k)dt + ω (ky0 − z0 k) ky0 − x0 k 0 Z 1  β (1 − t) p ω (ky0 − x0 k)dt + ω (kky0 − x0 k) ky0 − x0 k 0 Z 1  β (1 − t) p ω (ky0 − x0 k)dt + k p ω (ky0 − x0 k) ky0 − x0 k 0   1 β + k p ω (ky0 − x0 k)ky0 − x0 k p+1 qβω (η )ky0 − x0 k ≤ qb0 ky0 − x0 k.

≤

0

≤ = = = ≤ ≤ Thus we have

k x1 − x0 k ≤ k x1 − y0 k + ky0 − x0 k ≤ qb0 ky0 − x0 k + ky0 − x0 k ≤ (1 + qb0 )ky0 − x0 k < r,

(16)

which shows that x1 ∈ Br [ x0 ]. Note that z1 = ( I − LF )( x1 ) ∈ D. Using Proposition 1 and (C3)–(C5), we have 1 0 1 k I − Fz0− Fz1 k ≤ k Fz0− kω (kk x1 − x0 k) 0 0

≤

βω (k(1 + qb0 )ky0 − x0 k)

≤

βk p (1 + qb0 ) p ω (ky0 − x0 k)

≤ k p (1 + qb0 ) p βω (η ) ≤ (k(1 + qb0 )) p b0 < 1. 1 exists and Therefore, by Lemma 1, Fz0− 1 1 k Fz0− k≤ 1

Subsequently, we have

1k k Fz0− 1 0 = γ0 k Fz0− k. 0 1 − (k(1 + qb0 )) p b0

(17)

Mathematics 2019, 7, 31

9 of 22

1 F ( x )k ky1 − x1 k = k Fz0− 1 1 0− 1 = k Fz1 ( Fh( x1 ) − F (y0 ) − Fz00 ( x1 − y0 ))k i R1 1k 0 0 )k dt + k F 0 − F 0 k k x − y k ≤ k Fz0− k( F − F 0 1 y0 y0 y0 + t ( x1 − y0 ) 1 h 0 i z0 1 0− 1 ≤ k Fz1 k p+1 ω (k x1 − y0 k) + ω (kky0 − x0 k) k x1 − y0 k i h 1 p ω (k y − x k) qb k x − y k 1k ω ( qb k x − y k) + k ≤ k Fz0− 0 0 0 0 0 0 0 0 1  p+1p  p 1 k q b0 ω (k y − x k) + k p ω (k y − x k) qb k y − x k ≤ k Fz0− 0 0 0 0 0 0 0 p +1 1  p  p q b ≤ γ0 p+10 + k p βω (η )qb0 ky0 − x0 k p q p b0 p +1

≤

(18)

!

+k p qb02

1−(k (1+qb0 )) p b0

k y0 − x0 k

≤ θ k y0 − x0 k. From Equations (16) and (18), we have

k y1 − x0 k ≤ k y1 − x1 k + k x1 − x0 k ≤ θ ky0 − x0 k + (1 + qb0 )ky0 − x0 k ≤ (1 + qb0 )θ ky0 − x0 k + (1 + qb0 )ky0 − x0 k ≤ (1 + qb0 )(1 + θ )η < r and

k z1 − x0 | ≤ k z1 − y1 k + k y1 − x1 k + k x1 − x0 k ≤ (1 + k)ky1 − x1 k + (1 + qb0 )ky0 − x0 k ≤ (1 + q)θη + (1 + q)η = (1 + q)(1 + θ )η < r. This shows that z1 , y1 ∈ Br [ x0 ]. From Equations (17) and (18), we get 1 1 k Fz0− kω (ky1 − x1 k) ≤ γ0 k Fz0− kω (θ ky0 − x0 k) 0 1

≤ γ0 θ p βω (η ) ≤ γ0 θ p b0 = b1 . Thus we see that Equation (15) holds for n = 1. Let j > 1 be a fixed integer. Assume that Equation (15) is true for n = 1, 2, · · · , j. Since x j ∈ Br [ x0 ], it follows z j = ( I − LF )( x j ) ∈ D. Using (C3), (C4), Equations (13) and (15), we have

ky j − z j k = = ≤ ≤

1 F ( x )k = k( L − F 0−1 ) F ( x )k k LF ( x j ) − Fz0− j j zj j 1 ) F 0 ( x − y )k k( L − Fz0− j j zj j k I − LFz0 j kky j − x j k k k y j − x j k.

Using Equations (13) and (19), we have

(19)

Mathematics 2019, 7, 31

10 of 22

1 F ( y )k k x j+1 − y j k = k Fz0− j j 0− 1 ≤ k Fz j kk F (y j ) − F ( x j ) − Fz0 j (y j − x j )k hR i 1 1k 0 0 k dt k y − x k ≤ k Fz0− k F − F j j zj j h R0 x j + t ( y j − x j ) i 1 1k 0 0 k dt + k F 0 − F 0 k k y − x k ≤ k Fz0− k F − F j j y y z j j j j h R0 x j + t ( y j − x j ) i 1 0− 1 = k Fz j k 0 ω ( x j + t(y j − x j ) − y j )dt + ω (ky j − z j k) ky j − x j k hR i 1 1k ≤ k Fz0− ω (( 1 − t )k y − x k) dt + ω ( k k y − x k) ky j − x j k j j j j j h R0 i 1 0− 1 p p ≤ k Fz j k 0 ((1 − t) + k )ω (ky j − x j k)dt ky j − x j k i h 1 p ω (k y − x k)k y − x k 1k + k ≤ k Fz0− j j j j p +1 j

(20)

1 k ω (k y − x k)k y − x k = qk Fz0− j j j j j = qb j ky j − x j k.

From Equation (20), we have

k x j +1 − x j k ≤ k x j +1 − y j k + k y j − x j k ≤ qb j ky j − x j k + ky j − x j k ≤ (1 + qb j )ky j − x j k.

(21)

Using Equations (20) and (21) and the triangular inequality, we have j

k x j +1 − x 0 k ≤

∑ k x s +1 − x s k

s =0 j

≤

∑ (1 + qbs )kys − xs k

s =0 j

≤

∑ (1 + qb0 )θ s ky0 − x0 k

s =0

≤ (1 + qb0 ) ≤

1 − θ j +1 η 1−θ

(1 + q ) η = r, 1−θ

which implies that xk+1 ∈ Br [ x0 ]. Again, by using Proposition 1, (C2), (C5), and Equation (21), we have 1 0 1 k I − Fz0− Fz j+1 k ≤ k Fz0− kω (kk x j+1 − x j k) j j 1 p ≤ k Fz0− kk (1 + qb j ) p ω (ky j − x j k) j

≤ k p (1 + qb j ) p b j < 1. 1 exists and Therefore, by Lemma 1, Fz0− j +1

1 k Fz0− k≤ j +1

1k k Fz0− j

1 − k p (1 + qb j ) p b j

Using Equations (13), (C2), and (21), we have

1 = γ j k Fz0− k. j

Mathematics 2019, 7, 31

11 of 22

1 ky j+1 − x j+1 k = k Fz0− F ( x j+1 )k j +1 1 = k Fz0− ( F ( x j+1 ) − F (y j ) − Fz0 j ( x j+1 − y j ))k j +1 Z 1  0−1 0 0 0 0 k Fy j +t(x j+1 −y j ) − Fy j kdt + k Fy j − Fz j k k x j+1 − y j k ≤ k Fz j+1 k 0 Z 1  0−1 ≤ k Fz j+1 k ω (tk x j+1 − y j k)dt + ω (ky j − z j k) k x j+1 − y j k 0 Z 1  1 ≤ k Fz0− k ω ( tqb k y − x k) dt + ω ( k k y − x k) qb j ky j − x j k j j j j j j +1 0 " p p # q bj p 0−1 ≤ γ j k Fz j k ω (ky j − x j k) + k ω (ky j − x j k) qb j ky j − x j k p+1 # " p p q bj p 1 + k k Fz0− kω (ky j − x j k)qb j ky j − x j k = γj j p+1 # " p p q bj ≤ γj + k p qb2j ky j − x j k p+1

≤ θ j k y j − x j k ≤ θ j +1 k y 0 − x 0 k , k y j +1 − x 0 k ≤ k y j +1 − x j +1 k + k x j +1 − x 0 k j

≤ θ j +1 k y 0 − x 0 k +

∑ k x s +1 − x s k

s =0 j

≤ θ j +1 k y 0 − x 0 k +

∑ (1 + qb0 )θ s ky0 − x0 k

s =0 j +1

≤ (1 + qb0 ) ∑ θ s η s =0

≤

(1 + q ) η =r 1−θ

and

k z j +1 − x 0 k ≤ k z j +1 − y j +1 k + k y j +1 − x j +1 k + k x j +1 − x 0 k j

≤ (1 + k)ky j+1 − x j+1 k +

∑ (1 + qb0 )θ s η

s =0 j

≤ (1 + q ) θ j +1 η +

∑ (1 + q ) θ s η

s =0 j +1

≤

∑ (1 + q ) θ s η < r

s =0

which implies that z j+1 , y j+1 ∈ Br ( x0 ). Also, we have 1 1 k Fz0− kω (ky j+1 − x j+1 k) ≤ γ j k Fz0− kω (θ j ky j − x j k) j +1 j p

1 ≤ γ j θ j k Fz0− kω (ky j − x j k) j p

≤ γ j θ j b j = b j +1 . Hence we conclude that Equation (15) is true for n = j + 1. Therefore, by induction, Equation (15) is true for all n ∈ N0 .

Mathematics 2019, 7, 31

12 of 22

(2) First, we show that the sequence { xn } is a Cauchy sequence. For this, letting m, n ∈ N0 and using Lemma 3, we have m + n −1

∑

k xm+n − xn k ≤

k x j +1 − x j k

j=n m + n −1

∑

≤

(1 + qb j )ky j − x j k

j=n m + n −1 j −1

∑ ∏ θ i k y0 − x0 k

≤ (1 + qb0 )

j=n

i =0

m + n −1 j −1

∑

≤ (1 + qb0 )

∏ξ

j=n

k y0 − x0 k

i =0

m + n −1 j −1

∑

≤ (1 + qb0 )

(2p+1)i −1 p θ

∏ξ

j=n

(2p+1)i p

i =0

− 1p

γ0 ky0 − x0 k

j −1

∑

= (1 + qb0 )

(2p + 1)i p i =0

∑

m + n −1

ξ

j=n m + n −1

∑

≤ (1 + qb0 )

ξ

− 1p

γ0 ky0 − x0 k

j (2p+1) j −1 −p 2p2 γ

!

k y0 − x0 k.

0

j=n

By Bernoulli’s inequality, for each j ≥ 0 and y > −1, we have (1 + y) j ≥ 1 + jy. Hence we have k xm+n − xn k ≤ (1 + qb0 )ξ ≤ (1 + qb0 )ξ = (1 + qb0 )ξ = (1 + qb0 )ξ = (1 + qb0 )ξ

−

−

−

1 2p2

1 2p2

1 2p2

− np γ0 − np

γ0

− np γ0

(2p+1)n −1 2p2 (2p+1)n −1 2p2

ξ ξ  ξ

− np γ0 − np

γ0

(2p+1)n 2p2 (2p+1)n 2p2

(2p+1)n 2p2



+ξ +ξ +ξ

(2p+1)n (2p+1) 2p2 (2p+1)n (1+2p) 2p2

(2p+1)n



− 1p γ0

1 +1 p 2p2

1



n p 1 + ξ (2p+1) γ0−1   m  n p 1− ξ (2p+1) γ0−1



n

1

1−(ξ (2p+1) γ0−1 ) p

− 1p

γ0



+···+ξ +···+ξ − 1p γ0

(2p+1)n (2p+1)m−1 2p2

(2p+1)n (1+2(m−1) p) 2p2

+···+ξ

+···+



1) − (m− p γ0

(2p+1)n

n ξ (2p+1) γ0−1



!

1) − (m− p

γ0

1 + m −1 p 2p2



η ! η 

1) − (m− p η γ0

(22)

 m −1  p

η

 η.

Since the sequence { xn } is a Cauchy sequence and hence it converges to some point x ∗ ∈ Br [ x0 ]. From Equations (13), (C2), and (15), we have

k LF ( xn )k ≤ kzn − yn k + kyn − xn k ≤ k kyn − xn k + kyn − xn k ≤ (1 + k)θ n η.

Mathematics 2019, 7, 31

13 of 22

Taking the limit as n → ∞ and using the continuity of F and the linearity of L, we have F ( x ∗ ) = 0. (3) Taking the limit as m → ∞ in Equation (22), we have

(1 + qb0 )η

k x ∗ − xn k ≤ ξ 1/2p

2

for each n ∈ N0 . (4) Here we prove

 1−ξ

(2p+1)n p

  1

−p

γ0

ξ 1/2p

2

(2p+1)n

n/p

(23)

γ0

k x n +1 − x ∗ k ≤K k xn − x ∗ k2p+1

for all n ∈ N0 and for some K > 0. One can easily observe that there exists n0 > 0 such that

k xn − x ∗ k < 1

(24)

whenever n ≥ n0 . Using Equations (13) and (24), we have

kzn − x ∗ k = k xn − x ∗ − LF ( xn )k = k xn − x ∗ − L( F ( xn ) − F ( x ∗ ))k = k xn − x ∗ − L ≤

Z 1 0

Z 1 0

Fx0 ∗ +t( xn − x∗ ) ( xn − x ∗ )dtk

k I − LFx0 ∗ +t(xn − x∗ ) kk xn − x ∗ kdt

≤ k k xn − x ∗ k and

1 F ( x )k kyn − x ∗ k = k xn − x ∗ − Fz0− n n 1 [ F 0 ( x − x ∗ ) − F ( x )]k = k Fz0− n zn n n 1 kk F ( x ) − x ∗ − F 0 ( x − x ∗ )k ≤ k Fz0− n zn n n R1 0 0− 1 = k Fzn kk 0 ( Fx∗ +t(xn − x∗ ) − Fz0n )( xn − x ∗ )kdt R 1 k 1 k F0 ≤ k Fz0− − Fz0n kk xn − x ∗ kdt n 0 x ∗ +t( xn − x ∗ ) R 1 k 1 (k F 0 − Fx0 ∗ k + k Fx0 ∗ − Fz0n k)k xn − x ∗ kdt ≤ k Fz0− n 0 x ∗ +t( xn − x ∗ ) R 1 k 1 ( ω ( t k x − x ∗ k) + ω (k z − x ∗ k))k x − x ∗ k dt ≤ k Fz0− n n n n 0 R1 p ∗ p ∗ 0− 1 ≤ k Fzn k 0 (t k xn − x k ω (1) + ω (kk xn − x k))k xn − x ∗ kdt   1 p ω (1)k x − x ∗ k p+1 1k ≤ k Fz0− n p +1 + k n 1 k qω (1)k x − x ∗ k p+1 . = k Fz0− n n

Using Equations (13), (24) and (25), we have

(25)

Mathematics 2019, 7, 31

14 of 22

k x n +1 − x ∗ k 1 = kyn − Fz0− F (yn ) − x ∗ k n 1 ≤ k Fz0− kk F (yn ) − F ( x ∗ ) − Fz0n (yn − x ∗ )k n 1 = k Fz0− kk n 1 ≤ k Fz0− k n 1 = k Fz0− k n 1 ≤ k Fz0− k n

Z 1

0 Z 1 0

Z 1 0

( Fx0 ∗ +t(yn − x∗ ) − Fz0n )(yn − x ∗ )kdt

k Fx0 ∗ +t(yn − x∗ ) − Fz0n kkyn − x ∗ kdt

(k Fx0 ∗ +t(yn − x∗ ) − Fx0 ∗ k + k Fx0 ∗ − Fz0n k)kyn − x ∗ kdt

Z 1

(ω (tkyn − x ∗ k) + ω (kk xn − x ∗ k))kyn − x ∗ kdt  Z 1   0−1 p 0−1 ∗ p +1 ∗ = k Fzn k t ω k Fzn kqω (1)k xn − x k + ω (kk xn − x k) dt 0

0

1 ×k Fz0− kqω (1)k xn − x ∗ k p+1 n  Z 1   p ∗ p ( p +1) 0−1 p ∗ p 1 2 t k x − x k ω k F k qω ( 1 ) + k k x − x k ω ( 1 ) dt ≤ k Fz0− k n n zn n 0

×qω (1)k xn − x ∗ k p+1
  ∗ p2 0−1 0−1 2 k xn − x k ω k Fzn k qω (1) = k Fzn k + k p qω (1)k xn − x ∗ k2p+1 p+1 = Kn k xn − x ∗ k2p+1 , where Kn =

1 2 k Fz0− k n




 2 1 k qω (1) k xn − x ∗ k p ω k Fz0− p n + k qω (1). p+1

1 −1 , where σ > 0. Then, for all x ∈ B ( x ∗ ), we have Let k Fx0− ∗ k ≤ d and 0 < d < ω ( σ ) σ 1 0 0−1 0 0 k I − Fx0− ∗ Fx k ≤ k Fx ∗ kk Fx ∗ − Fx k ≤ dω ( σ ) < 1

and so, by Lemma 1, we have

k Fx0−1 k ≤

d := λ. 1 − dω (σ )

Since xn → x ∗ and zn → x ∗ as n → ∞, there exists a positive integer N0 such that 1 k Fz0− k≤ n

d 1 − dω (σ )

for all n ≥ N0 . Thus, for all n ≥ N0 , one can easily observe that K n ≤ λ2



 2 σ p ω (λqω (1)) + k p qω (1) = K. p+1

This shows that the R-order of convergence at least (2p + 1). This completes the proof. 4. Applications 4.1. Fixed Points of Smooth Operators Let X be a Banach spaces and D be a nonempty open convex subset of X. Let G : D ⊂ X → X be a nonlinear operator such that D is Fréchet differentiable at each point of D and let L ∈ B( X, X ) such that ( I − L( I − G ))( D ) ⊆ D. For F = I − G, the Newton-like algorithm Equation (13) reduces to the following Stirling-like method for computing fixed point of the operator G:

Mathematics 2019, 7, 31

15 of 22

Starting with x0 ∈ D and after xn ∈ D is defined, we define the next iterate xn+1 as follows:    zn = ( I − L( I − G ))( xn ), yn = ( I − ( I − Gz0 n )−1 ( I − G ))( xn ),   x 0 −1 n+1 = ( I − ( I − Gzn ) ( I − G ))( yn )

(26)

for each n ∈ N0 . For the choice of X = Y and F = I − G, Theorem 1 reduces to the following: Theorem 2. Let D be a nonempty open convex subset of a Banach space X, G : D → X be a Fréchet differentiable at each point of D with values into itself. Let L ∈ B( X ) be such that ( I − L( I − G ))( D ) ⊆ D. Let x0 ∈ D be such that z0 = x0 − L( x0 − G ( x0 )) and let ( I − Gz0 0 )−1 ∈ B( X ) exist. Let ω ∈ Φ and α be a solution of the Equation (9). Assume that the conditions (C5)–(C6) and the following conditions hold: (C7) k( I − Gz0 0 )−1 k ≤ β for some β > 0; (C8) k( I − Gz0 0 )−1 ( x0 − G ( x0 ))k ≤ η for some η > 0; (C9) k Gx0 − Gy0 k ≤ ω (k x − yk) for all x, y ∈ D; (C10) k I − L( I − Gx0 )k ≤ k for all x ∈ D and for some k ∈ (0, 31 ]. Then the sequence { xn } generated by Equation (26) is well defined, remains in Br [ x0 ] and converges to the fixed point x ∗ ∈ Br [ x0 ] of the operator G and the sequence { xn } has R-order of convergence at least 2p + 1. We give an example to illustrate Theorem 2. Example 1. Let X = Y = R and D = (−1, 1) ⊂ X. Define a mapping G : D → R by G(x) = for all x ∈ D. Define L : R → R by L( x ) =

7.9 7 x

1.1x3 − x 6

(27)

for all x ∈ R. One can easily observe that

( I − L( I − G ))( x ) ∈ D for all x ∈ D. Clearly, G is differentiable on D and its derivative at x ∈ D is Gx0 = with k Gx0 k ≤ 0.3833 for all x ∈ D and G 0 satisfies the Lipschitz condition

3.3x2 −1 6

and Gx0 is bounded

k Gx0 − Gy0 k ≤ K k x − yk for all x, y ∈ D, where K = 1.1. For x0 = 0.3, we have z0 = ( I − L( I − G ))( x0 ) = −0.0894135714,

k( I − Gz0 0 )−1 k ≤ 0.860385626 = β,

k( I − Gz0 0 )−1 ( x0 − G ( x0 ))k ≤ 0.29687606 = η. For p = 1, q =

5 6

and ω (t) = Kt for all t ≥ 0, we have b0 = βω (η ) = 0.280970684 < 1,  θ=

p

q p b0 p +1

+ kp



qb02

1 − k(1 + qb0 )b0

= 0.0335033167 < 1

and r = 0.563139828. Hence all the conditions of Theorem 2 are satisfied. Therefore, the sequence { xn } generated by Equation (26) is in Br [ x0 ] and it converges to the fixed point x ∗ = 0 ∈ Br [ x0 ] of G.

Mathematics 2019, 7, 31

16 of 22

Remark 1. In Example 1, k Gx0 k ≤ 0.38333 > 13 . Thus the condition (Ω) does not hold and so we can not apply Parhi and Gupta [21] (Theorem 1) for finding fixed points of the operators like G defined by (27). Thus the Stirling-like method defined by Equation (26) provides an affirmative answer of the Problem 1. If the condition (Ω) holds, then Theorem 2 with L = I reduces to the main result of Parhi and Gupta [21] as follows: Corollary 1. [21] (Theorem 1) Let D be a nonempty open convex subset of a Banach space X and G : D → D be a Fréchet differentiable operator and let x0 ∈ D with z0 = G ( x0 ). Let ( I − Gz0 0 )−1 ∈ B( X ) exists and ω ∈ Φ. Assume that the conditions (C5)–(C9) and the following condition holds: (C11) k Gx0 k ≤ k for all x ∈ D and for some k ∈ (0, 31 ]. Then the sequence { xn } generated by Equation (8) is well defined, remains in Br [ x0 ] and converges to the fixed point x ∗ ∈ Br [ x0 ] of the operator G with R−order of the convergence at least 2p + 1. Example 2. Let X = Y = R and D = (−6, 6) ⊂ X. Define a mapping G : D → R by G(x) = 2 + e

sin x 5

for all x ∈ D. It is obvious that G is Fréchet differentiable on D and its Fréchet derivative at x ∈ D is sin x Gx0 = cos5 x e 5 . Clearly, Gx0 is bounded with k Gx0 k ≤ 0.22 < 13 = k and

k Gx0 − Gy0 k ≤ K k x − yk for all x, y ∈ D, where K = 0.245. For x0 = 0, we have z0 = G ( x0 ) = 3,

k( I − Gz0 0 )−1 k ≤ 0.834725586524139 = β

and

k( I − Gz0 0 )−1 ( x0 − G ( x0 ))k ≤ 2.504176759572418 = η. For p = 1, q =

5 6

and ω (t) = Kt for all t ≥ 0, we have b0 = βKη = 0.512123601526580 < 1,  θ=

p

q p b0 p +1

 + k p qb02

1 − k(1 + qb0 )b0

= 0.073280601270728 < 1

and r = 5.147038576039456. Hence all the conditions of Theorem 2 with L = I are satisfied. Therefore, the sequence { xn } generated by Equation (26) is in Br [ x0 ] and it converges to the fixed point x ∗ = 3.023785446275295 ∈ Br [ x0 ] of G (Table 1). Table 1. A priori error bounds. n

k xn − x∗ k

0 1 2 3 4 5

3.0237854462752 1.7795738211156 × 10−2 6.216484249588206 × 10−6 2.335501569916687 × 10−9 8.775202786637237 × 10−13 4.440892098500626 × 10−16

Mathematics 2019, 7, 31

17 of 22

4.2. Fredholm Integral Equations Let X be a Banach space over the field F (R or C) with the norm k · k and D be an open convex subset of X. Further, let B( X ) be the Banach space of bounded linear operators from X into itself. Let S ∈ B( X ), u ∈ X and λ ∈ F. We investigate a solution x ∈ X of the nonlinear Fredholm-type operator equation: x − λSQ( x ) = u, (28) where Q : D → X is continuously Fréchet differentiable on D. The operator Equation (28) has been discussed in [10,38,39]. Define an operator F : D → X by F ( x ) = x − λSQ( x ) − u

(29)

for all x ∈ D. Then solving the operator Equation (29) is equivalent to solving the operator Equation (1). From Equation (29), we have Fx0 (h) = h − λSQ0x (h) (30) for all h ∈ X. Now, we apply Theorem 1 to solve the operator Equation (28). Theorem 3. Let X be a Banach space and D an open convex subset of X. Let Q : D → X be a continuously Fréchet differentiable mapping at each point of D. Let L, S ∈ B( X ) and u ∈ X. Assume that, for any x0 ∈ D, z0 = x0 − L( x0 − λSQ( x0 ) − u) and ( I − λSQ0z0 )−1 exist. Assume that the condition (C6) and the following conditions hold: (C12) (C13) (C14) (C15) (C16) (C17)

( I − L( I − λSQ))( x ) − u ∈ D for all x ∈ D; k( I − λSQ0z0 )−1 k ≤ β for some β > 0; k( I − λSQ0z0 )−1 ( x0 − λSQ( x0 ) − u)k ≤ η for some η > 0; k Q0x − Q0y k ≤ ω0 (k x − yk) for all x, y ∈ D, where ω0 ∈ Φ; ω0 (st) ≤ s p ω0 (t), s ∈ [0, 1] and t ∈ [0, ∞); k I − L( I − λSQ0x )k ≤ k, k ≤ 31 for all x ∈ D.

Then we have the following: (1) The sequence { xn } generated by    zn = xn − L( xn − λSQ( xn ) − u), yn = xn − ( I − λSQ0zn )−1 ( xn − λSQ( xn ) − u),   x 0 −1 n+1 = yn − ( I − λSQzn ) ( yn − λSQ ( yn ) − u )

(31)

for each n ∈ N0 is well defined, remains in Br [ x0 ] and converges to a solution x ∗ of the Equation (28). (2) The R-order convergence of sequence { xn } is at least 2p + 1. Proof. Let F : D → X be an operator defined by Equation (29). Clearly, F is Fréchet differentiable at each point of D and its Fréchet derivative at x ∈ D is given by Equation (30). Now, from (C13) 1 k ≤ β and so it follows that (C3) holds. From (C14), Equations (29) and Equation (30), we have k Fz0− 0 0− 1 and (30), we have k Fz0 ( F ( x0 ))k ≤ η. Hence (C4) is satisfied. For all x, y ∈ D, using (C15), we have

k Fx0 − Fy0 k = sup{k( Fx0 − Fy0 )zk : z ∈ X, kzk = 1} ≤ |λ|kSk sup{k Q0x − Q0y kkzk : z ∈ X, kzk = 1} ≤ |λ|kSkω0 (k x − yk) = ω (k x − yk),

Mathematics 2019, 7, 31

18 of 22

where ω (t) = |λ|kSkω0 (t). Clearly, ω ∈ Φ and, from (C16), we have ω (st) ≤ s p ω (t) for all s ∈ [0, 1] and t ∈ (0, ∞]. Thus (C1) and (C5) hold. (C2) follows from (C17) for c = k ∈ (0, 13 ]. Hence all the conditions of Theorem 1 are satisfied. Therefore, Theorem 3 follows from Theorem 1. This completes the proof. Let D = X = Y = C [ a, b] be the space of all continuous real valued functions defined on [ a, b] ⊂ R with the norm k x k = sup | x (t)|. Consider, the following nonlinear integral equation: t∈[ a,b]

x (s) = g(s) + λ

Z b a

K (s, t)(µ( x (t))1+ p + ν( x (t))2 )dt

(32)

for all s ∈ [ a, b] and p ∈ (0, 1], where g, x ∈ C [ a, b] with g(s) ≥ 0 for all s ∈ [ a, b], K : [ a, b] × [ a, b] → R is a continuous nonnegative real-valued function and µ, ν, λ ∈ R. Define two mappings S, Q : D → X by Sx (s) =

Z b a

K (s, t) x (t)dt

(33)

for all s ∈ [ a, b] and Qx (s) = µ( x (s))1+ p + ν( x (s))2

(34)

for all µ, ν ∈ R and s ∈ [ a, b]. One can easily observe that K is bounded on [ a, b] × [ a, b], that is, there exists a number M ≥ 0 such that |K (s, t)| ≤ M for all s, t ∈ [ a, b]. Clearly, S is bounded linear operator with kSk ≤ M(b − a) and Q is Fréchet differentiable and its Fréchet derivative at x ∈ D is given by Q0x h(s) = (µ(1 + p) x p + 2νx )h(s)

(35)

for all h ∈ C [ a, b]. For all x, y ∈ D, we have

k Q0x − Q0y k = ≤ ≤ ≤ =

sup{k( Q0x − Q0y )hk : h ∈ C [ a, b], khk = 1} sup{k(µ(1 + p)( x p − y p ) + 2ν( x − y))hk : h ∈ C [ a, b], k hk = 1} sup{(|µ|(1 + p)k x p − y p k + 2|ν|k x − yk)khk : h ∈ C [ a, b], k hk = 1} |µ|(1 + p)k x − yk p + 2|ν|k x − yk ω0 (k x − yk),

(36)

where ω0 (t) = |µ|(1 + p)t p + 2|ν|t, t ≥ 0 with ω0 (st) ≤ s p ω0 (t)

(37)

for all s ∈ [0, 1] and t ∈ [0, ∞). For any x ∈ D, using Equations (33) and (35), we have

kSQ0x k 0 = sup{k n SQ x hk : h R∈ X, khk = 1} o b = sup sups∈[a,b] a K (s, t)(µ(1 + p)( x (t)) p + 2νx (t))h(t)dt : h ∈ X, khk = 1 nR o b ≤ sup a |K (s, t)|(|µ|(1 + p)| x (t)| p + 2|ν|| x (t)|)|h(t)|dt : h ∈ X, khk = 1 ≤ (|µ|(1 + p)k x k p + 2|ν|k x k) M(b − a) < 1. We now apply Theorem 3 to solve the Fredholm integral Equation (32).

(38)

Mathematics 2019, 7, 31

19 of 22

Theorem 4. Let D = X = Y = C [ a, b] and µ, ν, λ, M ∈ R. Let S, Q : D → X be operators defined by Equations (33) and (34), respectively. Let L ∈ B( X ) and x0 ∈ D be such that z0 = x0 − L( x0 − λSQ( x0 ) − g) ∈ D. Assume that the condition (C6) and the following conditions hold: (C18)

1 1−|λ|(|µ|(1+ p)kz0 k p +2|ν|kz0 k) M (b− a)

(C19)

k x0 − gk+|λ|(|µ|k x0 k p+1 +2|ν|k x0 k2 ) M(b− a) 1−|λ|(|µ|(1+ p)kz0 k p +2|ν|kz0 k) M(b− a)

= β for some β > 0; = η for some η > 0;

(C20) k I − Lk + |λ|k Lk(|µ|(1 + p)k x k p + 2|ν|k x k) M (b − a) ≤

1 for all x ∈ D. 3

Then the sequence generated by Equation (31) with u = g ∈ X is well defined, remains in Br [ x0 ] and converges to the solution x ∗ ∈ Br [ x0 ] of the Equation (32) with R-order convergence at least (2p + 1). Proof. Note that D = X = Y = C [ a, b]. Obviously, (C12) holds. Using Equations (C20), (33), (35) and (38), we have

k I − ( I − λSQ0z0 )k ≤ |λ|(|µ|(1 + p)kz0 k p + 2|ν|kz0 k) M(b − a) < 1. Therefore, by Lemma 1, ( I − λSQ0z0 )−1 exists and 1 . 1 − |λ|(|µ|(1 + p)kz0 k p + 2|ν|kz0 k) M (b − a)

k( I − λSQ0z0 )−1 k ≤

(39)

Hence Equations (C18) and (39) implies (C13) holds. Using Equations (C19), (38) and (39), we have

k( I − λSQ0z0 )−1 ( x0 − λSQ( x0 ) − g)k ≤ k( I − λSQ0z0 )−1 k(k x0 − gk + kλSQ( x0 )k) k x0 − gk + |λ|(|µ|k x0 k p+1 + 2|ν|k x0 k2 ) M(b − a) 1 − |λ|(|µ|(1 + p)kz0 k p + 2|ν|kz0 k) M(b − a) ≤ η. ≤

Thus the condition (C14) is satisfied. The conditions (C15) and (C16) follow from Equations (36) and (37), respectively. Now, from Equation (C20) and (38), we have

k I − L( I − λSQ0x )k ≤ k I − Lk + k LkkλSQ0x k ≤ k I − Lk + k Lk|λ|(|µ|(1 + p)k x k p + 2|ν|k x k) M(b − a) 1 ≤ . 3 This implies that (C17) holds. Hence all the conditions of Theorem 3 are satisfied. Therefore, Theorem 4 follows from Theorem 3. This completes the proof. Now, we give one example to illustrate Theorem 3. Example 3. Let X = Y = C [0, 1] be the space of all continuous real valued functions defined on [0, 1]. Let D = { x : x ∈ C [0, 1], k x k < 23 } ⊂ C [0, 1]. Consider the following nonlinear integral equation: x (s) = sin(πs) +

1 10

Z 1 0

cos(πs) sin(πt)( x (t)) p+1 dt, p ∈ (0, 1].

Define two mappings S : X → X and Q : D → Y by S( x )(s) =

Z 1 0

K (s, t) x (t)dt,

Q( x )(s) = ( x (s)) p+1 ,

(40)

Mathematics 2019, 7, 31

20 of 22

where K (s, t) = cos(πs) sin(πt). For u = sin(πs), the problem Equation (40) is equivalent to the problem Equation (28). Here, one can easily observe that S is bounded linear operator with kSk ≤ 1 and Q is Fréchet differentiable with Q0x h(s) = ( p + 1)( x (s)) p h(s) for all h ∈ X and s ∈ [0, 1]. For all x, y ∈ D, we have

k Q0x − Q0y k ≤ ( p + 1)k x − yk p = ω0 (k x − yk), where ω0 (t) = ( p + 1)t p for any t ≥ 0. Clearly, ω0 ∈ Φ. Define a mapping F : D → X by F ( x )(s) = x (s) −

1 SQ( x )(s) − sin(πs). 10

Clearly, F is Fréchet differentiable on D. We now show that (C12) holds for L = I ∈ B( X ). Note that

 

1 1 3 p +1 3

+1 < k( I − L( I − λSQ))( x ) − uk = SQ( x )(s) + sin(πs) ≤ 10 10 2 2 for all x ∈ D. Thus ( I − L( I − λSQ))( x ) − u ∈ D for all x ∈ D. For all x ∈ D, we have

kI −

Fx0 k

p+1 p+1 ≤ kxk p ≤ 10 10

 p 3 = k. 2

Therefore, by Lemma 1, Fx0−1 exists and Fx0−1 U (s)

R1 ( p + 1) cos(πs) 0 sin(πt)( x (t)) p U (t)dt = U (s) + R1 10 − ( p + 1) 0 sin(πt) cos(πt)( x (t)) p dt

(41)

for all U ∈ Y. Let x0 (s) = sin(πs), ω (t) =

ω0 ( t ) 10

=

p +1 p 10 t

. Then we have the following:

cos(πs) R 1

(sin(πt)) p+2 dt; cos(πs) R 1 p+2 dt; (b) z0 (s) = x0 (s) − F ( x0 (s)) = sin(πs) + 10 0 (sin( πt )) (a) x0 ∈ X, F ( x0 (s)) = −

10

10 p+1 = β; 10 p+1 −( p+1)11 p 10 p 0− 1 k Fz0 F ( x0 )k ≤ 10 p+1 −( p+1)11 p

0

1k ≤ (c) k Fz0− 0

(d)

(e) b0 = βω (η ) =

= η;

( p+1)10 p( p+1) (10 p+1 −( p+1)11 p ) p+1

and q = 

qb0 2 +k



1 p +1

+



p +1 10

p


2 3 p . 2

qb0 2

(1+qb )η

One can easily observe that θ = 1−k(1+q)b < 1 for all p ∈ (0, 1] and r = 1−θ0 . Hence all the 0 conditions of Theorem 3 are satisfied. Therefore, the sequence { xn } generated by Equation (31) is well defined, remains in Br [ x0 ] and converges to a solution of the integral Equation (40). For p = 1, the convergence behavior of Newton-like method Equation (31) is given in Table 2. Table 2. Iterates of Newton-like method Equation (31). n

xn (s)

zn (s)

yn (s)

0 1 2 3

sin(πs) sin(πs) + 0.0424794035 cos(πs) sin(πs) + 0.0424795616 cos(πs) sin(πs) + 0.0424796111 cos(πs)

sin(πs) + 0.0424413182 cos(πs) sin(πs) + 0.0424791962 cos(πs) sin(πs) + 0.042479611 cos(πs) sin(πs) + 0.0424796109 cos(πs)

sin(πs) + 0.0425947671 cos(πs) sin(πs) + 0.0424796116 cos(πs) sin(πs) + 0.0424796112 cos(πs) sin(πs) + 0.0424796113 cos(πs)

5. Conclusions The semilocal convergence of the third order Newton-like method for finding zeros of an operator from a Banach space to another Banach space and the corresponding Stirling-like method for finding

Mathematics 2019, 7, 31

21 of 22

fixed points of an operator on a Banach space are established under the ω-continuity condition. Our iterative technique is applied to nonlinear Fredholm-type operator equations. The R-order of our methods are clearly shown to be equal to at least 2p + 1 for any p ∈ (0, 1]. Some numerical examples are given in support of our work, where earlier work cannot apply. In future, our iterative techniques can be applied in optimization problems. Author Contributions: The authors have contributed equally to this paper. Funding: This research has received no exteranl funding. Acknowledgments: We would like to thank the reviewers for their valuable suggestions for improvement of this paper. Conflicts of Interest: The authors declare no conflict of interest.

Reference 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

Kantorovich, L.V.; Akilov, G.P. Functional Analysis; Aregamon Press: Oxford, UK, 1982. Kantorovich, L.V. On Newton’s method for functional equations. Dokl. Akad. Nauk. SSSR 1948, 59, 1237–1240. (In Russian) Rheinbolt, W.C. A unified Convergence theory for a class of iterative processes. SIAM J. Numer. Anal. 1968, 5, 42–63. [CrossRef] Argyros, I.K.; Cho, Y.J.; Hilout, S. Numerical Methods for Equations and its Applications; CRC Press/Taylor & Francis Group Publ. Comp.: New York, NY, USA, 2012. Argyros, I.K.; Hilout, S. Improved generaliged differentiability conditions for Newton-like methods. J. Complex. 2010, 26, 316–333. [CrossRef] Argyros, I.K.; Hilout, S. Majorizing sequences for iterative methods. J. Comput. Appl. Math. 2012, 236, 1947–1960. [CrossRef] Argyros, I.K. An improved error analysis for Newton-like methods under generalized conditions. J. Comput. Appl. Math. 2003, 157, 169–185. [CrossRef] Argyros, I.K.; Hilout, S. On the convergence of Newton-type methods under mild differentiability conditions. Number. Algorithms 2009, 52, 701–726. [CrossRef] Sahu, D.R.; Singh, K.K.; Singh, V.K. Some Newton-like methods with sharper error estimates for solving operator equations in Banach spaces. Fixed Point Theory Appl. 2012, 78, 1–20. [CrossRef] Sahu, D.R.; Singh, K.K.; Singh, V.K. A Newton-like method for generalized operator equations in Banach spaces. Numer. Algorithms 2014, 67, 289–303. [CrossRef] Sahu, D.R.; Cho, Y.J.; Agarwal, R.P.; Argyros, I.K. Accessibility of solutions of operator equations by Newton-like Methods. J. Comlex. 2015, 31, 637–657. [CrossRef] Argyros, I.K. A unifying local-semilocal convergence analysis and applications for two-point Newton-like methods in Banach space. J. Math. Anal. Appl. 2004, 298, 374–397. [CrossRef] Argyros, I.K.; Cho, Y.J.; George, S. On the “Terra incognita” for the Newton-Kantrovich method. J. Korean Math. Soc. 2014, 51, 251–266. [CrossRef] Argyros, I.K.; Cho, Y.J.; George, S. Local convergence for some third-order iterative methods under weak conditions. J. Korean Math. Soc. 2016, 53, 781–793. [CrossRef] Ren, H.; Argyros, I.K.; Cho, Y.J. Semi-local convergence of Steffensen-type algorithms for solving nonlinear equations. Numer. Funct. Anal. Optim. 2014, 35, 1476–1499. [CrossRef] Ezquerro, J.A.; Hernández, M.A.; Salanova, M.A. A discretization scheme for some conservative problems. J. Comput. Appl. Math. 2000, 115, 181–192. [CrossRef] Ezquerro, J.A.; Hernández, M.A.; Salanova, M.A. A newton like method for solving some boundery value problems. Numer. Funct. Anal. Optim. 2002, 23, 791–805. [CrossRef] Ezquerro, J.A.; Hernández, M.A. Generalized differentiability conditions for Newton’s method. IMA J. Numer. Anal. 2002, 22, 187–205. [CrossRef] Proinov, P.D. New general convergence theory for iterative process and its applications to NewtonKantorovich type theores. J. Complex. 2010, 26, 3–42. [CrossRef] Sahu, D.R.; Yao, J.C.; Singh, V.K.; Kumar, S. Semilocal Convergence Analysis of S-iteration Process of Newton-Kantorovich Like in Banach Spaces. J. Optim. Theory Appl. 2017, 172, 102–127. [CrossRef]

Mathematics 2019, 7, 31

21. 22. 23. 24. 25.

26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39.

22 of 22

Parhi, S.K.; Gupta, D.K. Convergence of a third order method for fixed points in Banach spaces. Numer. Algorithms 2012, 60, 419–434. [CrossRef] Agarwal, R.P.; O’Regan, D.; Sahu, D.R. Fixed Point Theory for Lipschitzian-Type Mappings with Applications; Topological Fixed Point Theory and its Applications; Springer: New York, NY, USA, 2009; p. 6. Agarwal, R.P.; Meehan, M.; O’Regan, D. Fixed Point Theory and Applications; Cambridge University Press: Cambridge, UK, 2004. Picard, E. Memorire sur la theorie des equations aux derivees partielles et la methode aes approximations successive. J. Math. Pures Appl. 1980, 6, 145–210. Cho, Y.J. Survey on metric fixed point theory and applications. In Advances on Real and Complex Analysis with Applications; Trends in Mathematics; Ruzhahsky, M., Cho, Y.J., Agarwal, P., Area, I., Eds.; Birkhäuser, Springer: Basel, Switzerland, 2017. Granas, A.; Dugundji, J. Fixed Point Theory; Springer: New York, NY, USA, 2003. Zeidler, E. Nonlinear Functional Analysis and Its Applications I: Fixed-Point Theorems; Springer: New York, NY, USA, 1986. Zeidler, E. Nonlinear Functional Analysis and Its Applications III: Variational Methods and Applications; Springer: New York, NY, USA, 1985. Argyros, I.K. On Newton’s method under mild differentiability conditions and applications. Appl. Math. Comput. 1999, 102, 177–183. [CrossRef] Ortega, J.M.; Rheinboldt, W.C. Iterative Solution of Nonlinear Equations in Several Variables; Academic Press: New York, NY, USA; London, UK, 1970. Bartle, R.G. Newton’s method in Banach spaces. Proc. Am. Math. Soc. 1955, 6, 827–831. Rall, L.B. Convergence of Stirling’s method in Banaeh spaces. Aequa. Math. 1975, 12, 12–20. [CrossRef] Parhi, S.K.; Gupta, D.K. Semilocal convergence of a Stirling-like method in Banach spaces. Int. J. Comput. Methods 2010, 7, 215–228. [CrossRef] Argyros, I.K.; Muruster, S.; George, S. On the Convergence of Stirling’s Method for Fixed Points Under Not Necessarily Contractive Hypotheses. Int. J. Appl. Comput. Math. 2017, 3, 1071–1081. [CrossRef] Alshomrani, A.S.; Maroju, P.; Behl, R. Convergence of a Stirling-like method for fixed points in Banach spaces. J. Comput. Appl. Math. 2018. [CrossRef] Parhi, S.K.; Gupta, D.K. A third order method for fixed points in Banach spaces. J. Math. Anal. Appl. 2009, 359, 642–652. [CrossRef] Rall, L.B. Computational Solution of Nonlinear Operator Equations; John Wiley and Sons: New York, NY, USA, 1969. Hernández, M.A.; Salanova, M.A. A Newton-like iterative process for the numerical solution of Fredholm nonlinear integral equations. J. Integr. Equ. Appl. 2005, 17, 1–17. [CrossRef] Kohaupt, L. A Newton-like method for the numerical solution of nonlinear Fredholm-type operator equations. Appl. Math. Comput. 2012, 218, 10129–10148. [CrossRef] c 2018 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access

article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).