Abstract

Under the new Hölder conditions, we consider the convergence analysis of the inverse-free Jarratt method in Banach space which is used to solve the nonlinear operator equation. We establish a new semilocal convergence theorem for the inverse-free Jarratt method and present an error estimate. Finally, three examples are provided to show the application of the theorem.

1. Introduction

We consider the following boundary value problem: Those are equivalent to the following nonlinear integral equation (see [1, 2]): where and is a twice Fréchet-differentiable operator. is the set of all continuous functions in ; is the Green function:

Instead of 2, we can try to solve a nonlinear operator equation , where Solving the nonlinear operator equation is an important issue in the engineering and technology field as these kinds of problems appear in many real-world applications. Economics [3], chemistry [4], and physics [58] are some of the examples of the scientific and engineering technology areas applied to solve these type of equations. In this study, we consider to establish a new semilocal convergence theorem of the Jarratt method in Banach space which is used to solve the nonlinear operator equation where is defined on an open convex of a Banach space with values in a Banach space .

There are a lot of methods of finding a solution of equation . Particularly iterative methods are often used to solve this problem (see [1, 2, 9, 10]). If we use the famous Newton method, we can proceed as Under a reasonable hypothesis, Newton’s method is the second-order convergence.

To improve the convergence order, many modified methods have been presented. The famous Halley’s method and the supper-Halley method are the third-order convergence. References [1122] give the convergence analysis for these methods. Now, we consider the following Jarratt method (see [2325]):

In this paper, we discuss the convergence of 7 for solving nonlinear operator equations in Banach spaces and establish a new semilocal convergence theorem under the following condition (see [20, 21]): where is a nondecreasing continuous function. Finally, the corresponding error estimate is also given.

2. Main Results

In the section, we establish a new semilocal convergence theorem and present the error estimate. Denote and . Suppose that and are the Banach spaces, is an open convex of the Banach space , and has continuous Fréchet derivative of the third-order. exists, for some , and satisfies (A5) is a nondecreasing continuous real function for such that , and there exists a positive real number such that for and .(A6)Denote , . Let , , , , , , where First, we get some lemmas.

Lemma 1. Suppose that , are given by 10. Then, is increasing and ;, , is increasing;, , , and .

Lemma 2. Suppose that , are given by 10. If then(i)the sequences , , are nonnegative and decreasing;(ii), .

Proof. (i) When , Suppose for . By Lemma 1, and are increasing; then (ii) By (i), is decreasing and . So, for all , This completes the proof of Lemma 2.

Lemma 3. Suppose that the conditions of Lemma 2 hold. Denote . Then (i), , ;(ii), .

Proof. First, by induction, we prove that (i) holds. Because and , we have Suppose that (i) holds for . Then we get and from (ii) we get This completes the proof of Lemma 3.

Lemma 4. Suppose that and are Banach spaces, is an open convex of the Banach space , has continuous Fréchet derivative of the second-order, and the sequences , are generated by 7. Then, for all natural numbers , the following formula holds:

Proof. Consider This completes the proof of Lemma 4.
By (A1)–(A6), 10, and 11, if , then where ; hence, . Consider By Banach lemma, exists, and

By Lemma 4, we have Hence, Hence, By hence exists, and . By induction, we can prove that the following Lemma 5 holds.

Lemma 5. Under the hypotheses of Lemma 2, the following items are true for all :(I) exists and   ;(II) ; (III); (IV) ; (V) ; (VI); (VII); (VIII) , where .

Theorem 6. Let and be two Banach spaces and has continuous Fréchet derivative of the third-order on a nonempty open convex . One supposes that exists for some and conditions (A1)–(A6) and 11 hold. If , then the sequence generated by 7 is well defined and converges to a unique solution of 2 in . Furthermore, the following error estimate is obtained: where and , .

Proof. Firstly, we prove that the sequence is a Cauchy one. From (II) and by Lemma 3, we have For , , By the Bernoulli inequality , so . Hence, we have Hence, is a Cauchy sequence and . Obviously, , for all , as if in 30; we obtain Following a similar procedure, we have , for all .
Now, let in 28. It follows that . Besides , since and is a bounded sequence, therefore by the continuity of in .
By letting in 30, we obtain error estimate 28.
To show uniqueness, let us assume that there exists a second solution of 2 in . Then By Banach lemma, we can obtain that the inverse of the linear operator exists and We get that .
This completes the proof of Theorem 6.

3. Application

In this section, we apply the convergence theorem and show three numerical examples.

Example 1. Consider the root of the equation on . Then, we easily get that does not satisfy () Hölder condition because, for all , Let then for and ; Let us consider a particular case of 2 from the operator given by the following nonlinear integral equation of mixed Hammerstein type (see [26]): where , , , for , are known functions and is a solution to be determined. If is () Hölder continuous in , for , the corresponding operator , does not satisfy () Hölder condition; for instance, the max-norm is considered. In this case, To solve this type of equations, we can consider where satisfy , where .

Remark 7. Observe that if is Lipschitz continuous in , we can choose , , so that Jarratt’s method is of -order, at least four order. If is Hölder continuous in , then we can choose , , , and Jarratt’s method is of -order, at least .

Example 2. Consider the case as follows: where the space is with the norm This equation arises in the theory of the radiative transfer and neutron transport and in the kinetic theory of gasses. Let us define the operator on by The first, the second, and the third derivatives of are defined by and we have To apply Theorem 6, we choose and we look for a domain in the form In this case, we have and from the Banach lemma, we obtain Then , , , , , and . This means that the hypothesis of Theorem 6 is satisfied. Then, the error bound becomes For , we get

Example 3. Let us consider the system of equations , where Then, we have Now, we choose and . We take the max-norm in and the norm max for . We define the norm of a bilinear operator on by where and .

Then, we get the following results: , , , and .

We get that the hypotheses of Theorem 6 are satisfied.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Authors’ Contribution

The authors have made the same contribution. All authors read and approved the final paper.

Acknowledgments

This work is supported by the National Basic Research 973 Program of China (no. 2011JB105001), National Natural Science Foundation of China (Grant no. 11371320), Zhejiang Natural Science Foundation (Grant no. LZ14A010002), the Foundation of Science and Technology Department (Grant no. 2013C31084) of Zhejiang Province, and Scientific Research Fund of Zhejiang Provincial Education Department (nos. Y201431077 and Y201329420) and by the Grant FEKT-S-11-2-921 of Faculty of Electrical Engineering and Communication, Brno University of Technology, Czech Republic.