Semilocal Convergence Domain of a Chandrasekhar Integral Equation

Abstract

In this study, we discuss the semilocal convergence analysis of a fourth-order iterative method in Banach spaces. We assume the Fréchet derivative satisfies the Lipschitz continuity condition, obtains suitable recurrence relations, and determines the domain of convergence under appropriate initial estimates. In addition, the uniqueness domain for the solution and the error bounds are obtained. Next, several numerical examples, one which includes a Chandrasekhar integral equation, are carried out to apply the theoretical findings for semilocal convergence. Then, a final overview is provided.

1. Introduction

Finding solutions to nonlinear equations of the form $\mathcal{P}\left(x\right)=0$ (where $\mathcal{P}:D\subseteq X\to Y$ is a differentiable nonlinear operator, $X,Y$ are complete normed linear spaces, and D is a non-empty, open, and convex set) is a fundamental and widely studied problem in applied mathematics. As it is often difficult to obtain an exact solution to the previous equation, we typically seek a numerical approximation instead. In such cases, we rely on approximation methods, which are typically iterative.

There are different types of convergence results used to approximate solutions to nonlinear equations. The first type, referred to as local convergence, depends on the existence of a presumed solution and requires the initial value to be sufficiently close to it. The second type, known as semilocal convergence, does not assume the existence of a specific solution, but the initial values must still satisfy certain conditions to guarantee convergence. Analyzing the semilocal convergence of iterative methods in Banach spaces is of great theoretical interest, as it enables the derivation of key results such as the existence and uniqueness of solutions, the convergence rate, a priori error estimates, and convergence regions, based solely on assumptions about the initial approximation $x_0$ rather than the solution itself. Such results have practical relevance in solving equations arising from differential and integral models. A semilocal convergence study can be developed through two distinct approaches, namely recurrent relations and recurrent functions.

For the classical Newton’s method [1], the semilocal convergence using recurrence relations was obtained by Kantorovich in [2]. Polyak [3] affirms that Kantorovich established in [4] a rigorous theoretical framework for the method of successive approximations applied to solving functional equations and introduced a semilocal analysis of Newton’s method by leveraging Banach’s contraction mapping principle. He focused on proving the conditions under which the method converges to a solution, thereby laying the groundwork for subsequent developments in numerical analysis and optimization. This was later refined to semilocal quadratic convergence in 1948, known as the Newton–Kantorovich theorem [5]. Kantorovich developed two distinct approaches to proving the aforementioned theorem, with one involving recurrence relations and the other employing majorant functions. The first proof [5] used recurrence relations, while the second [6] was founded on the majorant function framework. Mysovskikh [7] presented a more straightforward proof of semilocal quadratic convergence, based on slightly altered theoretical premises. Since the work carried out by Kantorovich, extensive research has explored the convergence and error bounds of Newton’s method based on the theorem’s assumptions or related conditions. Among these contributions, the convergence theorems by Ortega and Rheinboldt [8] stand out. Rall (see [9]) proposed an alternative approach for studying semilocal convergence, which relied on recurrence relations and was simpler to analyze. Furthermore, numerous variations of the Newton–Kantorovich theorem have been proposed, each with distinct assumptions and conclusions.

Several studies have explored the semilocal convergence of third-order methods, including the Chebyshev method, addressed in [10,11], and the Halley scheme, discussed in [12]. Many of these works build on Kantorovich’s framework for analyzing the Newton method (see [13,14]). Gutierrez and Hernández [15] provide a comprehensive theory that unifies these third-order methods. All of these papers established that the convergence of sequences in Banach spaces can be traced back to the convergence of a majorant sequence. Then, Hernández [16] introduced a novel form of recurrence relations for the Chebyshev method, composed of two sequences of positive real numbers.

Recently, scholars have carried out studies on semilocal convergence for higher-order iterative methods. For example, Chen et al. [17] studied the semilocal convergence for a modified Newton’s method using recurrence relations. Using that approach, Wang et al. [18] analyzed the semilocal convergence for a sixth-order variant of the Jarratt method. Hueso and Martínez [19] proved a convergence result for a third- and fourth-order family of methods. Jaiswal [20] examined the semilocal convergence of an existing eighth-order method for solving nonlinear equations in Banach spaces under relaxed conditions, utilizing Rall’s straightforward techniques. Later, Cordero et al. [21] used recurrence relations to prove the semilocal convergence in Banach spaces for the multidimensional extension of Chun’s scheme. Argyros et al. [22] provided the semilocal convergence of a two-step Jarratt-type method by using restricted convergence regions combined with majorizing scalar sequences.

Wang et al. [23] established a semilocal convergence study of an iterative method using a recursive relation without requiring higher-order derivatives. They determined the domains of existence and uniqueness by appropriately choosing the initial point and applying the Lipschitz condition to the first-order Fréchet derivative across the region. In [24], a fourth-order family of methods for solving nonlinear equations was proposed and the stability of the class was analyzed using complex dynamical tools. Subsequently, ref. [25] provided local convergence results. Now, the purpose of this paper is to establish the semilocal convergence of this iterative scheme in Banach spaces and obtain error estimates through a system of recurrence relations.

This paper is organized as follows: Section 2 presents preliminary results and introduces the auxiliary functions. Section 3 outlines the development of the recurrence relations. In Section 4, the semilocal convergence is established. Section 5 provides numerical examples to support our theoretical findings. Section 6 concludes the paper and provides final remarks.

2. Preliminary Results

Let $X,Y$ be Banach spaces and $\mathcal{P}:D\subseteq X\to Y$ a nonlinear twice continuously Fréchet differentiable operator in an open convex domain of D. Consider the family of iterative schemes defined for $k=0,1,2,\dots$ by

$$\begin{array}{cc}\hfill & y_k=x_k-{\displaystyle \frac{2}{3}}{\mathcal{P}}^{\prime }{\left(x_k\right)}^{-1}\mathcal{P}\left(x_k\right)\hfill \\ \hfill & x_{k+1}=x_k-\left[\left({\displaystyle \frac{5}{8}}-\alpha \right)+\alpha {\mathcal{P}}^{\prime }{\left(y_k\right)}^{-1}{\mathcal{P}}^{\prime }\left(x_k\right)+{\displaystyle \frac{\alpha }{3}}{\mathcal{P}}^{\prime }{\left(x_k\right)}^{-1}{\mathcal{P}}^{\prime }\left(y_k\right)\right.\hfill \\ \hfill & \left.+\left({\displaystyle \frac{3}{8}}-{\displaystyle \frac{\alpha }{3}}\right){\left({\mathcal{P}}^{\prime }{\left(y_k\right)}^{-1}{\mathcal{P}}^{\prime }\left(x_k\right)\right)}^2\right]{\mathcal{P}}^{\prime }{\left(x_k\right)}^{-1}\mathcal{P}\left(x_k\right).\hfill \end{array}$$

(1)

where $\alpha \in \mathbb{C}$ is a parameter and $x_0$ is the initial estimate. This family was studied in [24], and its local convergence analysis was carried out in [25]. Let us denote ${\Gamma }_{x_k}={\left[{\mathcal{P}}^{\prime }\left(x_k\right)\right]}^{-1}$ and ${\Gamma }_{y_k}={\left[{\mathcal{P}}^{\prime }\left(y_k\right)\right]}^{-1}$.

Let $x_0\in D$ such that ${\Gamma }_{x_0}$ exists and

(C₁) 

$∥{\Gamma }_{x_0}∥\le \beta .$

(C₂) 

$∥{\Gamma }_{x_0}\mathcal{P}\left(x_0\right)∥\le \eta .$

(C₃) 

$∥{\mathcal{P}}^{″}\left(x\right)∥\le \rho ,\forall x\in D.$

(C₄) 

$∥{\mathcal{P}}^{″}\left(x\right)-{\mathcal{P}}^{″}\left(y\right)∥\le \vartheta ∥x-y∥,\forall x,y\in D.$

Let us define ${\beta }_0=∥{\Gamma }_{x_0}∥,{\eta }_0=∥{\Gamma }_{x_0}\mathcal{P}\left(x_0\right)∥,a_0=\rho {\beta }_0{\eta }_0$, and $b_0=\vartheta {\beta }_0{\eta }_0^2$, where ${\beta }_0,{\eta }_0\in {\mathbb{R}}^{+}$. From the first step in (1), we have

$$∥y_0-x_0∥=∥-{\displaystyle \frac{2}{3}}{\Gamma }_{x_0}\mathcal{P}\left(x_0\right)∥={\displaystyle \frac{2}{3}}{\eta }_0.$$

(2)

Prior to performing the semilocal convergence analysis, we apply the following lemmas:

Lemma 1

([26]). If operator $\mathcal{P}$ has continuous derivatives up to order $k+1$ in an open convex subset D and $∥{\mathcal{P}}^{\left(k+1\right)}\left(x\right)∥\le \rho$ for all $x\in D$, then if $x_0\in D$, the following is obtained:

$$\mathcal{P}\left(x\right)=\sum _{j=0}^k{\displaystyle \frac{1}{k!}}{\mathcal{P}}^{\left(k\right)}\left(x\right){\left(x-x_0\right)}^k+R_{x_0,k}\left(x-x_0\right),$$

where the remainder can be expressed as

$$R_{x_0,k}\left(x-x_0\right)={\displaystyle \frac{1}{k!}}{\int }_0^1{\left(1-t\right)}^k{\mathcal{P}}^{\left(k+1\right)}\left(x_0+t\left(x-x_0\right)\right){\left(x-x_0\right)}^{k+1}dt$$

and

$$∥R_{x_0,k}\left(x-x_0\right)∥\le {\displaystyle \frac{\rho }{\left(k+1\right)!}}{∥x-x_0∥}^{k+1}.$$

The previous lemma provides an expression and a bound for the remainder of Taylor’s expansion of the operator $\mathcal{P}$, and the next lemma (Banach Lemma) will allow us to guarantee the existence and boundedness of an inverse matrix. Both lemmas will be used in the upcoming proofs.

Lemma 2

(Banach Lemma [27]). Let $T\in \mathbb{L}\left(X,X\right)$, where $\mathbb{L}\left(X,X\right)$ is the space of all continuous linear operators on X into X. For such T, the norm

$$∥T∥=\underset{∥x∥\le 1}{sup}∥Tx∥$$

is defined. If $∥T∥<1$, then $I-T$ is an invertible matrix and

$$∥{\left(I-T\right)}^{-1}∥\le {\displaystyle \frac{1}{1-∥T∥}}.$$

Now, in order to set the semilocal study, we analyze the iterative method (1) step by step. Let us find a bound for $∥x_1-x_0∥$ and $∥x_1-y_0∥$. We have

$$x_1-x_0=\left(x_1-y_0\right)+\left(y_0-x_0\right),$$

and using the triangular inequality, we can write

$$∥x_1-x_0∥\le ∥x_1-y_0∥+∥y_0-x_0∥,$$

where from (1), we obtain the following:

$$x_1-x_0=-\left[\left({\displaystyle \frac{5}{8}}-\alpha \right)+\alpha {\Gamma }_{y_0}{\mathcal{P}}^{\prime }\left(x_0\right)+{\displaystyle \frac{\alpha }{3}}{\Gamma }_{x_0}{\mathcal{P}}^{\prime }\left(y_0\right)+\left({\displaystyle \frac{3}{8}}-{\displaystyle \frac{\alpha }{3}}\right){\left({\Gamma }_{y_0}{\mathcal{P}}^{\prime }\left(x_0\right)\right)}^2\right]{\Gamma }_{x_0}\mathcal{P}\left(x_0\right),$$

and

$$\begin{array}{cc}\hfill x_1-y_0& =\left[\left({\displaystyle \frac{1}{24}}+\alpha \right)-\alpha {\Gamma }_{y_0}{\mathcal{P}}^{\prime }\left(x_0\right)-{\displaystyle \frac{\alpha }{3}}{\Gamma }_{x_0}{\mathcal{P}}^{\prime }\left(y_0\right)-\left({\displaystyle \frac{3}{8}}-{\displaystyle \frac{\alpha }{3}}\right){\left({\Gamma }_{y_0}{\mathcal{P}}^{\prime }\left(x_0\right)\right)}^2\right]{\Gamma }_{x_0}\mathcal{P}\left(x_0\right).\hfill \end{array}$$

(3)

By applying Banach’s Lemma, we aim to prove the existence of the inverse of the operator ${\Gamma }_{x_0}{\mathcal{P}}^{\prime }\left(y_0\right)={\left[{\mathcal{P}}^{\prime }\left(x_0\right)\right]}^{-1}{\mathcal{P}}^{\prime }\left(y_0\right)$ and then the existence of the inverse of ${\Gamma }_{y_0}{\mathcal{P}}^{\prime }\left(x_0\right)={\left[{\mathcal{P}}^{\prime }\left(y_0\right)\right]}^{-1}{\mathcal{P}}^{\prime }\left(x_0\right)$. Thus, using (C₁), (C₃) and the definition of ${\beta }_0$, the following condition must be met:

$$\begin{array}{cc}\hfill ∥I-{\Gamma }_{x_0}{\mathcal{P}}^{\prime }\left(y_0\right)∥& \le ∥{\Gamma }_{x_0}∥∥{\mathcal{P}}^{\prime }\left(x_0\right)-{\mathcal{P}}^{\prime }\left(y_0\right)∥=∥{\Gamma }_{x_0}∥∥{\int }_{x_0}^{y_0}{\mathcal{P}}^{″}\left(x\right)dx∥\hfill \\ \hfill & \le ∥{\Gamma }_{x_0}∥{\int }_{x_0}^{y_0}∥{\mathcal{P}}^{″}\left(x\right)∥dx\le {\beta }_0\rho ∥y_0-x_0∥={\displaystyle \frac{2}{3}}{\beta }_0\rho {\eta }_0={\displaystyle \frac{2}{3}}{a}_0<1.\hfill \end{array}$$

(4)

By the Banach Lemma, ${\left[{\Gamma }_{x_0}{\mathcal{P}}^{\prime }\left(y_0\right)\right]}^{-1}$ exists and

$$∥{\Gamma }_{y_0}{\mathcal{P}}^{\prime }\left(x_0\right)∥\le {\displaystyle \frac{3}{3-2a_0}}.$$

(5)

Next, we prove the existence of the inverse of the operator ${\Gamma }_{y_0}{\mathcal{P}}^{\prime }\left(x_0\right)$. Hence, the next condition must be satisfied

$$∥I-{\Gamma }_{y_0}{\mathcal{P}}^{\prime }\left(x_0\right)∥=∥{\Gamma }_{y_0}{\mathcal{P}}^{\prime }\left(y_0\right)-{\Gamma }_{y_0}{\mathcal{P}}^{\prime }\left(x_0\right)∥=∥{\Gamma }_{y_0}{\mathcal{P}}^{\prime }\left(x_0\right)\left[I-{\Gamma }_{x_0}{\mathcal{P}}^{\prime }\left(y_0\right)\right]∥\le {\displaystyle \frac{2a_0}{3-2a_0}}<1,$$

if $2a_0<3-2a_0$, which leads to $a_0<\frac{3}{4}$ and by the Banach Lemma, ${\left[{\Gamma }_{y_0}{\mathcal{P}}^{\prime }\left(x_0\right)\right]}^{-1}$ exists and

$$∥{\Gamma }_{x_0}{\mathcal{P}}^{\prime }\left(y_0\right)∥\le {\displaystyle \frac{1}{1-∥I-{\Gamma }_{y_0}{\mathcal{P}}^{\prime }\left(x_0\right)∥}}\le {\displaystyle \frac{3-2a_0}{3-4a_0}}.$$

(6)

By taking the norm in (3) and using (4)–(6), the following is obtained:

$$∥x_1-y_0∥\le \left(\left|{\displaystyle \frac{1}{24}}+\alpha \right|+{\displaystyle \frac{3\left|\alpha \right|}{3-2a_0}}+{\displaystyle \frac{\left|\alpha \right|\left(3-2a_0\right)}{3\left(3-4a_0\right)}}+\left|{\displaystyle \frac{3}{8}}-{\displaystyle \frac{\alpha }{3}}\right|{\displaystyle \frac{9}{{\left(3-2a_0\right)}^2}}\right){\eta }_0$$

$$∥x_1-y_0∥\le p_{\alpha }\left(a_0\right){\eta }_0,$$

(7)

where

$$p_{\alpha }\left(t\right)=\left|{\displaystyle \frac{1}{24}}+\alpha \right|+{\displaystyle \frac{3\left|\alpha \right|}{3-2t}}+{\displaystyle \frac{\left|\alpha \right|\left(3-2t\right)}{3\left(3-4t\right)}}+\left|{\displaystyle \frac{3}{8}}-{\displaystyle \frac{\alpha }{3}}\right|{\displaystyle \frac{9}{{\left(3-2t\right)}^2}},$$

(8)

replacing (2) and (7) in (2), we have

$$∥x_1-x_0∥\le \left(p_{\alpha }\left(a_0\right)+{\displaystyle \frac{2}{3}}\right){\eta }_0=g_{\alpha }\left(a_0\right){\eta }_0,$$

(9)

where by definition

$$g_{\alpha }\left(t\right)=p_{\alpha }\left(t\right)+{\displaystyle \frac{2}{3}}.$$

(10)

The result obtained from Lemma 2 is applied to $I-{\Gamma }_{x_0}{\mathcal{P}}^{\prime }\left(x_1\right)$ in order to prove the existence of the inverse matrix of ${\Gamma }_{x_0}{\mathcal{P}}^{\prime }\left(x_1\right)$ and the inverse of ${\mathcal{P}}^{\prime }\left(x_1\right)$. Therefore, the following condition must hold:

$$\begin{array}{cc}\hfill ∥I-{\Gamma }_{x_0}{\mathcal{P}}^{\prime }\left(x_1\right)∥& \le ∥{\Gamma }_{x_0}∥∥{\mathcal{P}}^{\prime }\left(x_0\right)-{\mathcal{P}}^{\prime }\left(x_1\right)∥=∥{\Gamma }_{x_0}∥∥{\int }_{x_0}^{x_1}{\mathcal{P}}^{″}\left(x\right)dx∥\hfill \\ \hfill & \le ∥{\Gamma }_{x_0}∥{\int }_{x_0}^{x_1}∥{\mathcal{P}}^{″}\left(x\right)∥dx\le {\beta }_0\rho ∥x_1-x_0∥\hfill \\ \hfill & \le {\beta }_0\rho g_{\alpha }\left(a_0\right){\eta }_0=a_0{g}_{\alpha }\left(a_0\right)<1,\hfill \end{array}$$

by the Banach Lemma, ${\Gamma }_{x_1}{\mathcal{P}}^{\prime }\left(x_0\right)$ exists and

$$∥{\Gamma }_{x_1}{\mathcal{P}}^{\prime }\left(x_0\right)∥\le {\displaystyle \frac{1}{1-∥I-{\Gamma }_{x_0}{\mathcal{P}}^{\prime }\left(x_1\right)∥}}\le {\displaystyle \frac{1}{1-a_0{g}_{\alpha }\left(a_0\right)}}.$$

(11)

Therefore,

$$∥{\Gamma }_{x_1}∥\le {\displaystyle \frac{∥{\mathcal{P}}^{\prime }\left(x_0\right)∥}{1-∥I-{\Gamma }_{x_0}{\mathcal{P}}^{\prime }\left(x_1\right)∥}}\le {\displaystyle \frac{1}{1-a_0{g}_{\alpha }\left(a_0\right)}}=∥{\mathcal{P}}^{\prime }\left(x_0\right)∥f_{\alpha }\left(a_0\right),$$

(12)

where by definition

$$f_{\alpha }\left(t\right)={\displaystyle \frac{1}{1-t{g}_{\alpha }\left(t\right)}}.$$

(13)

Taylor’s expansion of $\mathcal{P}\left(x_1\right)$ around $x_0$ can be described as follows:

$$\begin{array}{cc}\hfill \mathcal{P}\left(x_1\right)& =\mathcal{P}\left(x_0\right)+{\mathcal{P}}^{\prime }\left(x_0\right)\left(x_1-x_0\right)+{\displaystyle \frac{1}{2}}{\mathcal{P}}^{″}\left(x_0\right){\left(x_1-x_0\right)}^2\hfill \\ \hfill & +{\int }_0^1\left[{\mathcal{P}}^{″}\left(x_0+t\left(x_1-x_0\right)\right)-{\mathcal{P}}^{″}\left(x_0\right)\right]{\left(x_1-x_0\right)}^2\left(1-t\right)dt,\hfill \end{array}$$

(14)

Multiplying by ${\mathcal{P}}^{\prime }\left(x_0\right)$, the first step in (1), we have

$$-{\displaystyle \frac{3}{2}}{\mathcal{P}}^{\prime }\left(x_0\right)\left(y_0-x_0\right)=\mathcal{P}\left(x_0\right),$$

(15)

Then, replacing (15) in (14), we obtain

$$\begin{array}{cc}\hfill \mathcal{P}\left(x_1\right)& =-{\displaystyle \frac{3}{2}}{\mathcal{P}}^{\prime }\left(x_0\right)\left(y_0-x_0\right)+{\mathcal{P}}^{\prime }\left(x_0\right)\left(x_1-x_0\right)+{\displaystyle \frac{1}{2}}{\mathcal{P}}^{″}\left(x_0\right){\left(x_1-x_0\right)}^2\hfill \\ \hfill & +{\int }_0^1\left[{\mathcal{P}}^{″}\left(x_0+t\left(x_1-x_0\right)\right)-{\mathcal{P}}^{″}\left(x_0\right)\right]{\left(x_1-x_0\right)}^2\left(1-t\right)dt,\hfill \\ \hfill \mathcal{P}\left(x_1\right)& ={\mathcal{P}}^{\prime }\left(x_0\right)\left[\left(x_1-y_0\right)-{\displaystyle \frac{1}{2}}\left(y_0-x_0\right)\right]+{\displaystyle \frac{1}{2}}{\mathcal{P}}^{″}\left(x_0\right){\left(x_1-x_0\right)}^2\hfill \\ \hfill & +{\int }_0^1\left[{\mathcal{P}}^{″}\left(x_0+t\left(x_1-x_0\right)\right)-{\mathcal{P}}^{″}\left(x_0\right)\right]{\left(x_1-x_0\right)}^2\left(1-t\right)dt,\hfill \\ \hfill \mathcal{P}\left(x_1\right)& =\left[{\mathcal{P}}^{\prime }\left(y_0\right)-{\int }_{x_0}^{y_0}{\mathcal{P}}^{″}\left(t\right)dt\right]\left[\left(x_1-y_0\right)-{\displaystyle \frac{1}{2}}\left(y_0-x_0\right)\right]+{\displaystyle \frac{1}{2}}{\mathcal{P}}^{″}\left(x_0\right){\left(x_1-x_0\right)}^2\hfill \\ \hfill & +{\int }_0^1\left[{\mathcal{P}}^{″}\left(x_0+t\left(x_1-x_0\right)\right)-{\mathcal{P}}^{″}\left(x_0\right)\right]{\left(x_1-x_0\right)}^2\left(1-t\right)dt.\hfill \end{array}$$

Now, Taylor’s expansion of ${\mathcal{P}}^{\prime }\left(y_0\right)$ around $x_0$ is

$${\mathcal{P}}^{\prime }\left(y_0\right)={\mathcal{P}}^{\prime }\left(x_0\right)+{\mathcal{P}}^{″}\left(x_0\right)\left(y_0-x_0\right)+{\int }_0^1\left[{\mathcal{P}}^{″}\left(x_0+t\left(y_0-x_0\right)\right)-{\mathcal{P}}^{″}\left(x_0\right)\right]\left(y_0-x_0\right)dt.$$

From the first step in (1), we have $y_0-x_0=-{\displaystyle \frac{2}{3}}\left(u_0-x_0\right)$, where

$$u_0=x_0-{\Gamma }_{x_0}\mathcal{P}\left(x_0\right).$$

Therefore,

$$\begin{array}{cc}\hfill {\mathcal{P}}^{\prime }\left(y_0\right)& ={\mathcal{P}}^{\prime }\left(x_0\right)+{\mathcal{P}}^{″}\left(x_0\right)\left[-{\displaystyle \frac{2}{3}}\left(u_0-x_0\right)\right]+{\int }_0^1\left[{\mathcal{P}}^{″}\left(x_0+t\left[-{\displaystyle \frac{2}{3}}\left(u_0-x_0\right)\right]\right)-{\mathcal{P}}^{″}\left(x_0\right)\right]\hfill \\ \hfill & \left[-{\displaystyle \frac{2}{3}}\left(u_0-x_0\right)\right]dt,\hfill \\ \hfill {\mathcal{P}}^{\prime }\left(y_0\right)& ={\mathcal{P}}^{\prime }\left(x_0\right)-{\displaystyle \frac{2}{3}}{\mathcal{P}}^{″}\left(x_0\right)\left(u_0-x_0\right)-{\displaystyle \frac{2}{3}}{\int }_0^1\left[{\mathcal{P}}^{″}\left(x_0+t\left[-{\displaystyle \frac{2}{3}}\left(u_0-x_0\right)\right]\right)-{\mathcal{P}}^{″}\left(x_0\right)\right]\hfill \\ \hfill & \left(u_0-x_0\right)dt.\hfill \end{array}$$

Then,

$$\begin{array}{cc}\hfill {\Gamma }_{x_0}\mathcal{P}\left(x_1\right)& ={\Gamma }_{x_0}\left[{\mathcal{P}}^{\prime }\left(x_0\right)-{\displaystyle \frac{2}{3}}{\mathcal{P}}^{″}\left(x_0\right)\left(u_0-x_0\right)-{\displaystyle \frac{2}{3}}{\int }_0^1\left[{\mathcal{P}}^{″}\left(x_0+t\left[-{\displaystyle \frac{2}{3}}\left(u_0-x_0\right)\right]\right)-{\mathcal{P}}^{″}\left(x_0\right)\right]\right.\hfill \\ \hfill & \left.\left(u_0-x_0\right)dt-{\int }_{x_0}^{y_0}{\mathcal{P}}^{″}\left(t\right)dt\right]\left[\left(x_1-y_0\right)-{\displaystyle \frac{1}{2}}\left(y_0-x_0\right)\right]+{\displaystyle \frac{1}{2}}{\Gamma }_{x_0}{\mathcal{P}}^{″}\left(x_0\right){\left(x_1-x_0\right)}^2\hfill \\ \hfill & +{\Gamma }_{x_0}{\int }_0^1\left[{\mathcal{P}}^{″}\left(x_0+t\left(x_1-x_0\right)\right)-{\mathcal{P}}^{″}\left(x_0\right)\right]{\left(x_1-x_0\right)}^2\left(1-t\right)dt\hfill \end{array}$$

and taking the norm

$$\begin{array}{cc}\hfill ∥{\Gamma }_{x_0}\mathcal{P}\left(x_1\right)∥& \le ∥{\Gamma }_{x_0}\left[{\mathcal{P}}^{\prime }\left(x_0\right)-{\displaystyle \frac{2}{3}}{\mathcal{P}}^{″}\left(x_0\right)\left(u_0-x_0\right)-{\displaystyle \frac{2}{3}}{\int }_0^1\left[{\mathcal{P}}^{″}\left(x_0+t\left[-{\displaystyle \frac{2}{3}}\left(u_0-x_0\right)\right]\right)-{\mathcal{P}}^{″}\left(x_0\right)\right]-{\int }_{x_0}^{y_0}{\mathcal{P}}^{″}\left(t\right)dt\right]\left[\left(x_1-y_0\right)\right.\hfill \\ \hfill & \left.-\frac{1}{2}\left(y_0-x_0\right)\right]+{\displaystyle \frac{1}{2}}{\Gamma }_{x_0}{\mathcal{P}}^{″}\left(x_0\right){\left(x_1-x_0\right)}^2+{\Gamma }_{x_0}{\int }_0^1\left[{\mathcal{P}}^{″}\left(x_0+t\left(x_1-x_0\right)\right)-{\mathcal{P}}^{″}\left(x_0\right)\right]{\left(x_1-x_0\right)}^2\left(1-t\right)dt∥,\hfill \\ \hfill ∥{\Gamma }_{x_0}\mathcal{P}\left(x_1\right)∥& \le ∥{\Gamma }_{x_0}{\mathcal{P}}^{\prime }\left(x_0\right)∥∥x_1-y_0∥+{\displaystyle \frac{2}{3}}∥{\Gamma }_{x_0}∥∥{\mathcal{P}}^{″}\left(x_0\right)∥∥u_0-x_0∥∥x_1-y_0∥+{\displaystyle \frac{2}{9}}∥{\Gamma }_{x_0}∥\vartheta {∥u_0-x_0∥}^2∥x_1-y_0∥\hfill \\ \hfill & +∥{\Gamma }_{x_0}∥\rho ∥y_0-x_0∥∥x_1-y_0∥+{\displaystyle \frac{1}{2}}∥{\Gamma }_{x_0}{\mathcal{P}}^{\prime }\left(x_0\right)∥∥y_0-x_0∥+{\displaystyle \frac{1}{2}}{\displaystyle \frac{2}{3}}∥{\Gamma }_{x_0}∥∥{\mathcal{P}}^{″}\left(x_0\right)∥∥u_0-x_0∥∥y_0-x_0∥\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}{\displaystyle \frac{2}{9}}∥{\Gamma }_{x_0}∥\vartheta {∥u_0-x_0∥}^2∥y_0-x_0∥+{\displaystyle \frac{1}{2}}∥{\Gamma }_{x_0}∥\rho {∥y_0-x_0∥}^2+{\displaystyle \frac{1}{2}}{\beta }_0\rho {∥x_1-x_0∥}^2+∥{\Gamma }_{x_0}∥{\displaystyle \frac{\vartheta }{6}}{∥x_1-x_0∥}^3,\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill & {\int }_0^1∥{\mathcal{P}}^{″}\left(x_0+t\left(x_1-x_0\right)\right)-{\mathcal{P}}^{″}\left(x_0\right)∥{∥x_1-x_0∥}^2∥\left(1-t\right)dt∥\hfill \\ \hfill & \le {\int }_0^1\vartheta ∥x_0+t{x}_1-t{x}_0-x_0∥{∥x_1-x_0∥}^2∥\left(1-t\right)dt∥\le {\int }_0^1\vartheta ∥t∥{∥x_1-x_0∥}^3∥\left(1-t\right)dt∥\hfill \\ \hfill & =\vartheta {∥x_1-x_0∥}^3{\int }_0^1∥\left(t-t^2\right)dt∥=\vartheta {∥x_1-x_0∥}^3∥{\left.{\displaystyle \frac{t^2}{2}}-{\displaystyle \frac{t^3}{3}}\right|}_0^1∥={\displaystyle \frac{\vartheta }{6}}{∥x_1-x_0∥}^3\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & {\displaystyle \frac{2}{3}}{\int }_0^1∥\left[{\mathcal{P}}^{″}\left(x_0+t\left[-{\displaystyle \frac{2}{3}}\left(u_0-x_0\right)\right]\right)-{\mathcal{P}}^{″}\left(x_0\right)\right]\left(u_0-x_0\right)dt∥\hfill \\ \hfill & \le {\displaystyle \frac{2}{3}}{\int }_0^1\vartheta ∥x_0-{\displaystyle \frac{2}{3}}\left(t{u}_0-t{x}_0\right)-x_0∥∥u_0-x_0∥∥dt∥\le {\displaystyle \frac{4}{9}}{\int }_0^1\vartheta {∥u_0-x_0∥}^2∥tdt∥\hfill \\ \hfill & ={\displaystyle \frac{4}{9}}\vartheta {∥u_0-x_0∥}^2{\int }_0^1∥tdt∥={\displaystyle \frac{4}{9}}\vartheta {∥u_0-x_0∥}^2∥{\left.{\displaystyle \frac{t^2}{2}}\right|}_0^1∥={\displaystyle \frac{2}{9}}\vartheta {∥u_0-x_0∥}^2.\hfill \end{array}$$

Using (2) and conditions (C₁)–(C₄), we have

$$\begin{array}{cc}\hfill ∥{\Gamma }_{x_0}\mathcal{P}\left(x_1\right)∥& \le p_{\alpha }\left(a_0\right){\eta }_0+{\displaystyle \frac{2}{3}}{\beta }_0\rho p_{\alpha }\left(a_0\right){\eta }_0^2+{\displaystyle \frac{2}{9}}{\beta }_0\vartheta p_{\alpha }\left(a_0\right){\eta }_0^3+{\displaystyle \frac{2}{3}}{\beta }_0\rho p_{\alpha }\left(a_0\right){\eta }_0^2\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}{\displaystyle \frac{2}{3}}{\eta }_0+{\displaystyle \frac{1}{3}}{\displaystyle \frac{2}{3}}{\beta }_0\rho {\eta }_0^2+{\displaystyle \frac{1}{9}}{\displaystyle \frac{2}{3}}{\beta }_0\vartheta {\eta }_0^3+{\displaystyle \frac{1}{2}}{\displaystyle \frac{2}{3}}{\displaystyle \frac{2}{3}}{\beta }_0\rho {\eta }_0^2+{\displaystyle \frac{1}{2}}{\beta }_0\rho {\left[g_{\alpha }\left(a_0\right){\eta }_0\right]}^2\hfill \\ \hfill & +{\beta }_0{\displaystyle \frac{\vartheta }{6}}{\left[g_{\alpha }\left(a_0\right){\eta }_0\right]}^3.\hfill \end{array}$$

Now, using (8), (9), and the definition for $a_0,b_0$, we obtain

$$\begin{array}{cc}\hfill ∥{\Gamma }_{x_0}\mathcal{P}\left(x_1\right)∥& \le \left[p_{\alpha }\left(a_0\right)+{\displaystyle \frac{2}{3}}{a}_0{p}_{\alpha }\left(a_0\right)+{\displaystyle \frac{2}{9}}{b}_0{p}_{\alpha }\left(a_0\right)+{\displaystyle \frac{2}{3}}{a}_0{p}_{\alpha }\left(a_0\right)+{\displaystyle \frac{1}{3}}+{\displaystyle \frac{2}{9}}{a}_0+{\displaystyle \frac{2}{27}}{b}_0\right.\hfill \\ \hfill & \left.+{\displaystyle \frac{2}{9}}{a}_0+{\displaystyle \frac{1}{2}}{a}_0{g}_{\alpha }^2\left(a_0\right)+{\displaystyle \frac{1}{6}}{b}_0{g}_{\alpha }^3\left(a_0\right)\right]{\eta }_0,\hfill \\ \hfill ∥{\Gamma }_{x_0}\mathcal{P}\left(x_1\right)∥& \le \left[p_{\alpha }\left(a_0\right)+{\displaystyle \frac{4}{3}}{a}_0{p}_{\alpha }\left(a_0\right)+{\displaystyle \frac{2}{9}}{b}_0{p}_{\alpha }\left(a_0\right)+{\displaystyle \frac{1}{3}}+{\displaystyle \frac{4}{9}}{a}_0+{\displaystyle \frac{2}{27}}{b}_0+{\displaystyle \frac{1}{2}}{a}_0{g}_{\alpha }^2\left(a_0\right)\right.\hfill \\ \hfill & \left.+{\displaystyle \frac{1}{6}}{b}_0{g}_{\alpha }^3\left(a_0\right)\right]{\eta }_0.\hfill \end{array}$$

Thus,

$$∥{\Gamma }_{x_0}\mathcal{P}\left(x_1\right)∥\le \left[\left({\displaystyle \frac{1}{3}}+p_{\alpha }\left(a_0\right)\right)\left(1+{\displaystyle \frac{4}{3}}{a}_0+{\displaystyle \frac{2}{9}}{b}_0\right)+{\displaystyle \frac{1}{2}}{a}_0{g}_{\alpha }^2\left(a_0\right)+{\displaystyle \frac{1}{6}}{b}_0{g}_{\alpha }^3\left(a_0\right)\right]{\eta }_0,$$

(16)

Then, taking (11) and (16), we obtain

$$\begin{array}{cc}\hfill ∥{\Gamma }_{x_1}\mathcal{P}\left(x_1\right)∥& =∥{\Gamma }_{x_1}{\mathcal{P}}^{\prime }\left(x_0\right){\Gamma }_{x_0}\mathcal{P}\left(x_1\right)∥\le ∥{\Gamma }_{x_1}{\mathcal{P}}^{\prime }\left(x_0\right)∥∥{\Gamma }_{x_0}\mathcal{P}\left(x_1\right)∥\hfill \\ \hfill & \le f_{\alpha }\left(a_0\right)\left[\left({\displaystyle \frac{1}{3}}+p_{\alpha }\left(a_0\right)\right)\left(1+{\displaystyle \frac{4}{3}}{a}_0+{\displaystyle \frac{2}{9}}{b}_0\right)+{\displaystyle \frac{1}{2}}{a}_0{g}_{\alpha }^2\left(a_0\right)+{\displaystyle \frac{1}{6}}{b}_0{g}_{\alpha }^3\left(a_0\right)\right]{\eta }_0\hfill \\ \hfill & =f_{\alpha }\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right){\eta }_0,\hfill \end{array}$$

where by definition

$${\phi }_{\alpha }\left(t,u\right)=\left({\displaystyle \frac{1}{3}}+p_{\alpha }\left(t\right)\right)\left(1+{\displaystyle \frac{4}{3}}t+{\displaystyle \frac{2}{9}}u\right)+{\displaystyle \frac{1}{2}}t{g}_{\alpha }^2\left(t\right)+{\displaystyle \frac{1}{6}}u{g}_{\alpha }^3\left(t\right).$$

(17)

In the following section, we will establish the recurrence relations and present some technical lemmas, which will be crucial for the demonstration of the convergence properties of the iterative scheme (1).

3. Recurrence Relations for the Scheme

Consider the following sequences:

$$a_{k+1}=a_k{f}_{\alpha }^2\left(a_k\right){\phi }_{\alpha }\left(a_k,b_k\right),$$

(18)

$$b_{k+1}=b_k{f}_{\alpha }^3\left(a_k\right){\phi }_{\alpha }^2\left(a_k,b_k\right),$$

(19)

$${\eta }_{k+1}={\eta }_k{f}_{\alpha }\left(a_k\right){\phi }_{\alpha }\left(a_k,b_k\right),$$

(20)

for $k=0,1,2,\dots$

Now, we present the following lemmas:

Lemma 3.

Let $p_{\alpha },g_{\alpha },f_{\alpha },and{\phi }_{\alpha }$ be the functions defined by (8), (10), (13) and (17), respectively. Suppose that $a_0\in \left]0,{\displaystyle \frac{3}{4}}\right[$ and $f_{\alpha }^2\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)<1$. Then, the following can be stated:

(i) 

$p_{\alpha }\left(t\right),g_{\alpha }\left(t\right)$, and $f_{\alpha }\left(t\right)$ are increasing functions and $p_{\alpha }\left(t\right)>0,g_{\alpha }\left(t\right)>0,f_{\alpha }\left(t\right)>1\forall t\in \left]0,{\displaystyle \frac{3}{4}}\right[$.

(ii) 

${\phi }_{\alpha }\left(t,u\right)$ is increasing as a function of t and increasing as a function of $u\forall t\in \left]0,{\displaystyle \frac{3}{4}}\right[$ and $u>0$.

(iii) 

The sequences $a_k,b_k,and{\eta }_k$ are decreasing; $a_k{g}_{\alpha }\left(a_k\right)<1$ and $f_{\alpha }^2\left(a_k\right){\phi }_{\alpha }\left(a_k,b_k\right)<1$.

Proof of Lemma 3.

Let $p_{\alpha }\left(t\right),g_{\alpha }\left(t\right),f_{\alpha }\left(t\right)$, and ${\phi }_{\alpha }\left(t,u\right)$ be the functions defined by (8), (10), (13) and (17), respectively.

(i)

Considering $t\in \left]0,{\displaystyle \frac{3}{4}}\right[$, we determine that the derivative of $p_{\alpha }\left(t\right)$ is

$$\begin{array}{cc}\hfill p_{\alpha }^{\prime }\left(t\right)& ={\displaystyle \frac{-3\left|\alpha \right|\left(-2\right)}{{\left(3-2t\right)}^2}}+{\displaystyle \frac{\left|\alpha \right|}{3}}{\displaystyle \frac{-2\left(3-4t\right)-\left(3-2t\right)\left(-4\right)}{3{\left(3-4t\right)}^2}}+\left|{\displaystyle \frac{3}{8}}-{\displaystyle \frac{\alpha }{3}}\right|{\displaystyle \frac{9\left(-2\right)\left(-2\right)}{{\left(3-2t\right)}^3}}\hfill \\ \hfill p_{\alpha }^{\prime }\left(t\right)& ={\displaystyle \frac{6\left|\alpha \right|}{{\left(3-2t\right)}^2}}+{\displaystyle \frac{2\left|\alpha \right|}{{\left(3-4t\right)}^2}}+\left|{\displaystyle \frac{3}{8}}-{\displaystyle \frac{\alpha }{3}}\right|{\displaystyle \frac{36}{{\left(3-2t\right)}^3}}.\hfill \end{array}$$

We have determined that $p_{\alpha }^{\prime }\left(t\right)>0$ for all t in the interval $\left]0,{\displaystyle \frac{3}{4}}\right[$. Therefore, $p_{\alpha }^{\prime }\left(t\right)$ is increasing for $t\in \left]0,{\displaystyle \frac{3}{4}}\right[$. Also, the following can be determined:

$$p_{\alpha }\left(0\right)=\left|{\displaystyle \frac{1}{24}}+\alpha \right|+\left|\alpha \right|+{\displaystyle \frac{\left|\alpha \right|}{3}}+\left|{\displaystyle \frac{3}{8}}-{\displaystyle \frac{\alpha }{3}}\right|>0$$

and $p_{\alpha }\left(t\right)>0$.

Consequently, $g_{\alpha }^{\prime }\left(t\right)=p_{\alpha }^{\prime }\left(t\right)>0$ is increasing; $g_{\alpha }\left(0\right)=p_{\alpha }\left(0\right)+{\displaystyle \frac{2}{3}}>0$ and $g_{\alpha }\left(t\right)>0$. Thus,

$$f_{\alpha }^{\prime }\left(t\right)={\displaystyle \frac{t{g}_{\alpha }^{\prime }\left(t\right)+g_{\alpha }\left(t\right)}{{\left(1-t{g}_{\alpha }\left(t\right)\right)}^2}},$$

since $g_{\alpha }\left(t\right)>0,g_{\alpha }^{\prime }\left(t\right)>0$, then $f_{\alpha }^{\prime }\left(t\right)>0$. Therefore, $f_{\alpha }\left(t\right)$ is increasing; $f_{\alpha }\left(0\right)=1$ and $f_{\alpha }\left(t\right)>1$.

(ii)

From (17), we have determined that

$${\phi }_{\alpha }\left(t,u\right)=\left({\displaystyle \frac{1}{3}}+p_{\alpha }\left(t\right)\right)\left(1+{\displaystyle \frac{4}{3}}t+{\displaystyle \frac{2}{9}}u\right)+{\displaystyle \frac{1}{2}}t{g}_{\alpha }^2\left(t\right)+{\displaystyle \frac{1}{6}}u{g}_{\alpha }^3\left(t\right),$$

then

$$\begin{array}{cc}\hfill {\displaystyle \frac{\partial {\phi }_{\alpha }\left(t,u\right)}{\partial t}}& =\left({\displaystyle \frac{1}{3}}+p_{\alpha }\left(t\right)\right){\displaystyle \frac{4}{3}}+p_{\alpha }^{\prime }\left(t\right)\left(1+{\displaystyle \frac{4}{3}}t+{\displaystyle \frac{2}{9}}u\right)+t{g}_{\alpha }\left(t\right)g_{\alpha }^{\prime }\left(t\right)+{\displaystyle \frac{1}{2}}{g}_{\alpha }^2\left(t\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}u{g}_{\alpha }^2\left(t\right)g_{\alpha }^{\prime }\left(t\right).\hfill \end{array}$$

Consequently, ${\phi }_{\alpha }\left(t,u\right)$ is increasing as a function of t and ${\phi }_{\alpha }\left(0,u\right)>0$. In addition,

$${\displaystyle \frac{\partial {\phi }_{\alpha }\left(t,u\right)}{\partial u}}=\left({\displaystyle \frac{1}{3}}+p_{\alpha }\left(t\right)\right){\displaystyle \frac{2}{9}}++{\displaystyle \frac{1}{6}}{g}_{\alpha }^3\left(t\right),$$

Therefore, ${\phi }_{\alpha }\left(t,u\right)$ is increasing as a function of u.

It is concluded that ${\phi }_{\alpha }\left(t,u\right)$ is increasing as a function of t and also increasing as a function of u, $\forall t\in \left]0,{\displaystyle \frac{3}{4}}\right[$, and $u>0$.

(iii)

Suppose that

$$f_{\alpha }^2\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)<1.$$

Then,

$$a_1=a_0{f}_{\alpha }^2\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)<a_0$$

and

$$\begin{array}{cc}\hfill b_1& =b_0{f}_{\alpha }^3\left(a_0\right){\phi }_{\alpha }^2\left(a_0,b_0\right)\hfill \\ \hfill & =b_0{f}_{\alpha }^2\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)f_{\alpha }\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)\hfill \\ \hfill & =b_0{f}_{\alpha }\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)<b_0,\hfill \end{array}$$

as $f_{\alpha }^2\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)<1$ and $f_{\alpha }^2\left(a_0\right)>1$ implies that $f_{\alpha }\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)<1$.

Furthermore, we have determined that

$${\eta }_1={\eta }_0{f}_{\alpha }\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)<{\eta }_0.$$

(21)

By induction, suppose that $\forall k>0$, we have

$$\begin{array}{cc}\hfill a_k<a_{k-1}& \Rightarrow f_{\alpha }^2\left(a_{k-1}\right){\phi }_{\alpha }\left(a_{k-1},b_{k-1}\right)<1\hfill \\ \hfill b_k<b_{k-1}& \Rightarrow f_{\alpha }^3\left(a_{k-1}\right){\phi }_{\alpha }^2\left(a_{k-1},b_{k-1}\right)<1\hfill \\ \hfill {\eta }_k<{\eta }_{k-1}& \Rightarrow f_{\alpha }\left(a_{k-1}\right){\phi }_{\alpha }\left(a_{k-1},b_{k-1}\right)<1.\hfill \end{array}$$

For the above reasons and the fact that $f_{\alpha }$ and ${\phi }_{\alpha }$ are increasing, we have

$$\begin{array}{cc}\hfill f_{\alpha }^2\left(a_k\right){\phi }_{\alpha }\left(a_k,b_k\right)& <f_{\alpha }^2\left(a_{k-1}\right){\phi }_{\alpha }\left(a_{k-1},b_{k-1}\right)<1\hfill \\ \hfill f_{\alpha }^3\left(a_k\right){\phi }_{\alpha }^2\left(a_k,b_k\right)& <f_{\alpha }^3\left(a_{k-1}\right){\phi }_{\alpha }^2\left(a_{k-1},b_{k-1}\right)<1\hfill \\ \hfill f_{\alpha }\left(a_k\right){\phi }_{\alpha }\left(a_k,b_k\right)& <f_{\alpha }\left(a_{k-1}\right){\phi }_{\alpha }\left(a_{k-1},b_{k-1}\right)<1.\hfill \end{array}$$

Thus,

$$\begin{array}{cc}\hfill a_k& =a_{k-1}{f}_{\alpha }^2\left(a_{k-1}\right){\phi }_{\alpha }\left(a_{k-1},b_{k-1}\right)<a_{k-1}\hfill \\ \hfill b_k& =b_{k-1}{f}_{\alpha }^3\left(a_{k-1}\right){\phi }_{\alpha }^2\left(a_{k-1},b_{k-1}\right)<b_{k-1}\hfill \\ \hfill {\eta }_k& ={\eta }_{k-1}{f}_{\alpha }\left(a_{k-1}\right){\phi }_{\alpha }\left(a_{k-1},b_{k-1}\right)<{\eta }_{k-1}.\hfill \end{array}$$

Consequently, $a_k,b_k,$ and ${\eta }_k$ are decreasing sequences.

Since $g_{\alpha }\left(a_k\right)$ is increasing and $a_k<a_{k-1}<\dots <a_0,\forall k\le 0$, then by induction $g_{\alpha }\left(a_k\right)<g_{\alpha }\left(a_{k-1}\right)<\dots <g_{\alpha }\left(a_0\right)$ and $a_k{g}_{\alpha }\left(a_k\right)<a_0{g}_{\alpha }\left(a_0\right)<1$ are obtained. Hence, $a_k{g}_{\alpha }\left(a_k\right)<1$ and $f_{\alpha }^2\left(a_k\right){\phi }_{\alpha }\left(a_k,b_k\right)<1$.

□

Lemma 4.

Let $p_{\alpha },g_{\alpha },f_{\alpha },and{\phi }_{\alpha }$ be the functions defined by (8), (10), (13) and (17), respectively. Suppose that $\delta \in \left]0,1\right[$. Then, $p_{\alpha }\left(\delta t\right)<p_{\alpha }\left(t\right),g_{\alpha }\left(\delta t\right)<g_{\alpha }\left(t\right),f_{\alpha }\left(\delta t\right)<f_{\alpha }\left(t\right)$, and ${\phi }_{\alpha }\left(\delta t,{\delta }^{2}u\right)<\delta {\phi }_{\alpha }\left(t,u\right)$.

Proof of Lemma 4.

For the proof of this lemma, we have considered the definitions of the functions.

Suppose that $\delta \in \left]0,1\right[$. Then,

$$\begin{array}{cc}\hfill {\phi }_{\alpha }\left(\delta t,{\delta }^{2}u\right)& =\left({\displaystyle \frac{1}{3}}+p_{\alpha }\left(\delta t\right)\right)\left(1+{\displaystyle \frac{4}{3}}\delta t+{\displaystyle \frac{2}{9}}{\delta }^{2}u\right)+{\displaystyle \frac{1}{2}}\delta t{g}_{\alpha }^2\left(\delta t\right)+{\displaystyle \frac{1}{6}}{\delta }^{2}u{g}_{\alpha }^3\left(\delta t\right)\hfill \\ \hfill & \le \left({\displaystyle \frac{1}{3}}+p_{\alpha }\left(t\right)\right)\left(1+{\displaystyle \frac{4}{3}}\delta t+{\displaystyle \frac{2}{9}}{\delta }^{2}u\right)+{\displaystyle \frac{1}{2}}\delta t{g}_{\alpha }^2\left(t\right)+{\displaystyle \frac{1}{6}}{\delta }^{2}u{g}_{\alpha }^3\left(t\right)\hfill \\ \hfill & \le \delta \left({\displaystyle \frac{1}{3}}+p_{\alpha }\left(t\right)\right)\left(1+{\displaystyle \frac{4}{3}}t+{\displaystyle \frac{2}{9}}u\right)+{\displaystyle \frac{1}{2}}\delta t{g}_{\alpha }^2\left(t\right)+{\displaystyle \frac{1}{6}}\delta u{g}_{\alpha }^3\left(t\right)=\delta {\phi }_{\alpha }\left(t,u\right).\hfill \end{array}$$

□

Lemma 5.

Under the hypothesis of Lemma 3, if we define $\sigma ={\displaystyle \frac{a_1}{a_0}}=f_{\alpha }^2\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)$ and $\mu ={\displaystyle \frac{1}{f_{\alpha }\left(a_0\right)}}$, the following are obtained:

(i) 

$\sigma \in \left]0,1\right[$.

(ii) 

$a_k\le {\sigma }^{2^{k-1}}{a}_{k-1}\le {\sigma }^{2^k-1}{a}_0$,

$b_k\le {\sigma }^{2\left(2^{k-1}\right)}{b}_{k-1}\le {\sigma }^{2\left(2^k-1\right)}{b}_0$.

(iii) 

$f_{\alpha }\left(a_k\right){\phi }_{\alpha }\left(a_k,b_k\right)\le {\sigma }^{2^k}\mu ,\forall k\in \mathbb{N}$.

(iv) 

${\eta }_k\le {\eta }_0{\sigma }^{2^k-1}{\mu }^k$.

(v) 

${\displaystyle \sum _{k=m}^{m+n-1}{\eta }_k\le {\mu }^m{\sigma }^{2^m-1}{\eta }_0\frac{1-{\left({\sigma }^{2^m}\mu \right)}^n}{1-{\sigma }^{2^m}\mu }}$.

Proof of Lemma 5.

Suppose that $\sigma ={\displaystyle \frac{a_1}{a_0}}=f_{\alpha }^2\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)$ and $\mu ={\displaystyle \frac{1}{f_{\alpha }\left(a_0\right)}}$. Then,

(i)

$\sigma \in \left]0,1\right[$, because $f_{\alpha }^2\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)<1$.

(ii)

$a_1=a_0{f}_{\alpha }^2\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)=\sigma a_0$, $\begin{array}{cc}\hfill a_2& =a_1{f}_{\alpha }^2\left(a_1\right){\phi }_{\alpha }\left(a_1,b_1\right)=\sigma a_0{f}_{\alpha }^2\left(\sigma a_0\right){\phi }_{\alpha }\left(\sigma a_0,{\sigma }^2{b}_0\right)\le \sigma a_0{f}_{\alpha }^2\left(a_0\right)\sigma {\phi }_{\alpha }\left(a_0,b_0\right)\hfill \\ & ={\sigma }^2{a}_0{f}_{\alpha }^2\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)={\sigma }^2{a}_1={\sigma }^3{a}_0.\hfill \end{array}$

Using mathematical induction, we can ascertain that

$$\begin{array}{cc}\hfill a_k& =a_{k-1}{f}_{\alpha }^2\left(a_{k-1}\right){\phi }_{\alpha }\left(a_{k-1},b_{k-1}\right)\hfill \\ & \le {\sigma }^{2^{k-2}}{a}_{k-2}{f}_{\alpha }^2\left({\sigma }^{2^{k-2}}{a}_{k-2}\right){\phi }_{\alpha }\left({\sigma }^{2^{k-2}}{a}_{k-2},{\sigma }^{{\left(2^{k-2}\right)}^2}{b}_{k-2}\right)\hfill \\ & \le {\sigma }^{2^{k-2}}{a}_{k-2}{f}_{\alpha }^2\left(a_{k-2}\right){\sigma }^{2^{k-2}}{\phi }_{\alpha }\left(a_{k-2},b_{k-2}\right)\hfill \\ & ={\sigma }^{\left(2^{k-2}\right)2}{a}_{k-2}{f}_{\alpha }^2\left(a_{k-2}\right){\phi }_{\alpha }\left(a_{k-2},b_{k-2}\right)={\sigma }^{\left(2^{k-2}\right)2}{a}_{k-1}={\sigma }^{2^{k-1}}{a}_{k-1}.\hfill \\ \hfill a_k& \le {\sigma }^{2^{k-1}}{a}_{k-1}\le \dots \le {\sigma }^{2^{k-1}}{\sigma }^{2^{k-2}}\dots \sigma a_0={\sigma }^{2^k-1}{a}_0.\hfill \end{array}$$

Similarly,

$$\begin{array}{cc}\hfill b_1& =b_0{f}_{\alpha }^3\left(a_0\right){\phi }_{\alpha }^2\left(a_0,b_0\right)=b_0{f}_{\alpha }^2\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)f_{\alpha }\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)\le {\sigma }^2{b}_0,\hfill \\ \hfill b_2& =b_1{f}_{\alpha }^3\left(a_1\right){\phi }_{\alpha }^2\left(a_1,b_1\right)\le {\sigma }^2{b}_0{f}_{\alpha }^3\left(\sigma a_0\right){\phi }_{\alpha }^2\left(\sigma a_0,{\sigma }^2{b}_0\right)\hfill \\ & \le {\sigma }^2{b}_0{f}_{\alpha }^3\left(a_0\right){\sigma }^2{\phi }_{\alpha }^2\left(a_0,b_0\right)={\sigma }^4{b}_0{f}_{\alpha }^3\left(a_0\right){\phi }_{\alpha }^2\left(a_0,b_0\right)={\sigma }^4{b}_1\le {\sigma }^6{b}_0,\hfill \end{array}$$

and using mathematical induction, we determine that

$$\begin{array}{cc}\hfill b_k& =b_{k-1}{f}_{\alpha }^3\left(a_{k-1}\right){\phi }_{\alpha }^2\left(a_{k-1},b_{k-1}\right)\hfill \\ & \le {\sigma }^{2\left(2^{k-2}\right)}{b}_{k-2}{f}_{\alpha }^3\left({\sigma }^{2^{k-2}}{a}_{k-2}\right){\phi }_{\alpha }^2\left({\sigma }^{2^{k-2}}{a}_{k-2},{\sigma }^{2\left(2^{k-2}\right)}{b}_{k-2}\right)\hfill \\ & \le {\sigma }^{2\left(2^{k-2}\right)}{b}_{k-2}{f}_{\alpha }^3\left(a_{k-2}\right){\sigma }^{{\left(2^{k-2}\right)}^2}{\phi }_{\alpha }^2\left(a_{k-2},b_{k-2}\right)\hfill \\ & ={\sigma }^{\left(2^{k-1}\right)}{b}_{k-2}{f}_{\alpha }^3\left(a_{k-2}\right){\sigma }^{\left(2^{k-1}\right)}{\phi }_{\alpha }^2\left(a_{k-2},b_{k-2}\right)={\sigma }^{2\left(2^{k-1}\right)}{b}_{k-1}.\hfill \\ \hfill b_k& ={\sigma }^{2\left(2^{k-1}\right)}{b}_{k-1}\le \dots \le {\sigma }^{2\left(2^{k-1}\right)}{\sigma }^{2\left(2^{k-2}\right)}\dots {\sigma }^2{b}_0\le {\sigma }^{2\left(2^k-1\right)}{b}_0.\hfill \end{array}$$

(iii)

$f_{\alpha }\left(a_k\right){\phi }_{\alpha }\left(a_k,b_k\right)\le f_{\alpha }\left({\sigma }^{2^k-1}{a}_0\right){\phi }_{\alpha }\left({\sigma }^{2^k-1}{a}_0,{\sigma }^{2\left(2^k-1\right)}{b}_0\right)$

$$\begin{array}{cc}& \le {\sigma }^{2^k-1}{f}_{\alpha }\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)={\displaystyle \frac{{\sigma }^{2^k-1}{f}_{\alpha }^2\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)}{f_{\alpha }\left(a_0\right)}}\hfill \\ & ={\displaystyle \frac{{\sigma }^{2^k-1}\sigma }{f_{\alpha }\left(a_0\right)}}={\displaystyle \frac{{\sigma }^{2^k}}{f_{\alpha }\left(a_0\right)}}={\sigma }^{2^k}\mu .\hfill \end{array}$$

(iv)

As a direct consequence of the previous statement, we have

$$\prod _{j=0}^{k-1}{f}_{\alpha }\left(a_j\right){\phi }_{\alpha }\left(a_j,b_j\right)\le \prod _{j=0}^{k-1}{\displaystyle \frac{{\sigma }^{2^j}}{f_{\alpha }\left(a_0\right)}},$$

and the product expansion is

$$P=\prod _{j=0}^{k-1}{\displaystyle \frac{{\sigma }^{2^j}}{f_{\alpha }\left(a_0\right)}}={\displaystyle \frac{{\displaystyle \prod _{j=0}^{k-1}{\sigma }^{2^j}}}{{\displaystyle \prod _{j=0}^{k-1}{f}_{\alpha }\left(a_0\right)}}}={\displaystyle \frac{{\sigma }^{{\displaystyle \sum _{j=0}^{k-1}{2}^j}}}{{\displaystyle \prod _{j=0}^{k-1}{f}_{\alpha }\left(a_0\right)}}}={\displaystyle \frac{{\sigma }^{\frac{2^k-1}{2-1}}}{f_{\alpha }^k\left(a_0\right)}}={\displaystyle \frac{{\sigma }^{2^k-1}}{f_{\alpha }^k\left(a_0\right)}}={\sigma }^{2^k-1}{\mu }^k.$$

By definition of ${\eta }_k$,

$${\eta }_k={\eta }_{k-1}{f}_{\alpha }\left(a_{k-1}\right){\phi }_{\alpha }\left(a_{k-1},b_{k-1}\right)\le {\eta }_0\prod _{j=0}^{k-1}{f}_{\alpha }\left(a_j\right){\phi }_{\alpha }\left(a_j,b_j\right)\le {\eta }_0{\sigma }^{2^k-1}{\mu }^k.$$

(v)

${\displaystyle \begin{array}{cc}& \sum _{k=m}^{m+n-1}{\eta }_k\le \sum _{k=m}^{m+n-1}{\eta }_0{\sigma }^{2^k-1}{\mu }^k\le \sum _{k=m}^{n-1}{\eta }_0{\sigma }^{2^{k+m}-1}{\mu }^{k+m}\\ & {\displaystyle =\sum _{k=m}^{n-1}{\eta }_0{\left({\sigma }^{2^m}\right)}^{\left(\frac{2^m\left(2^k-1\right)}{2^m}\right)}{\left({\sigma }^{2^m}\right)}^{\left(\frac{2^m-1}{2^m}\right)}{\mu }^k{\mu }^m}\hfill \\ & {\displaystyle ={\mu }^m{\sigma }^{2^m-1}\sum _{k=m}^{n-1}{\eta }_0{\left({\sigma }^{2^m}\right)}^{\left(2^k-1\right)}{\mu }^k.}\hfill \end{array}}$

Since $k\le 2^k-1$ and $\sigma \in \left]0,1\right[$, we have determined that ${\left({\sigma }^{2m}\right)}^{2^k-1}\le {\left({\sigma }^{2m}\right)}^k$; therefore,

$$\sum _{k=m}^{m+n-1}{\eta }_k\le {\mu }^m{\sigma }^{2^m-1}\sum _{k=m}^{n-1}{\eta }_0{\left({\sigma }^{2^m}\right)}^k{\mu }^k={\mu }^m{\sigma }^{2^m-1}\sum _{k=m}^{n-1}{\eta }_0{\left({\sigma }^{2^m}\mu \right)}^k.$$

We then normalize the terms of the geometric progression to obtain

$$\sum _{k=m}^{m+n-1}{\eta }_k\le {\mu }^m{\sigma }^{2^m-1}{\eta }_0{\displaystyle \frac{1-{\left({\sigma }^{2^m}\mu \right)}^n}{1-{\sigma }^{2^m}\mu }}.$$

□

Furthermore, as a consequence of the last statement from the aforementioned Lemma 5, since $\mu <1$ and $\sigma <1$, we obtain a convergent series which establishes that

$$\sum _{k=0}^{\infty }{\eta }_k={\displaystyle \frac{1}{1-\sigma \mu }}{\eta }_0.$$

Lemma 6.

Under the hypothesis of Lemma 3 and conditions (C₁)–(C₄), it is verified, for all $k\ge 0$, that

(i) 

${\Gamma }_{x_k}$ exists and $∥{\Gamma }_{x_k}∥\le ∥{\Gamma }_{x_{k-1}}∥f_{\alpha }\left(a_{k-1}\right)$.

(ii) 

$∥{\Gamma }_{x_k}\mathcal{P}\left(x_k\right)∥\le {\eta }_k$.

(iii) 

$\rho ∥{\Gamma }_{x_k}∥∥{\Gamma }_{x_k}\mathcal{P}\left(x_k\right)∥\le a_k$.

(iv) 

$\vartheta ∥{\Gamma }_{x_k}∥{∥{\Gamma }_{x_k}\mathcal{P}\left(x_k\right)∥}^2\le b_k$.

(v) 

$∥x_{k+1}-x_k∥\le g_{\alpha }\left(a_k\right){\eta }_k,y_k\in \overline{B\left(x_0,\left({\displaystyle \frac{2}{3}}+R\right){\eta }_0\right)}$, and $x_{k+1}\in \overline{B\left(x_0,R{\eta }_0\right)}.$

(vi) 

$R={\displaystyle \frac{1}{a_0}},\forall \alpha \in \mathbb{R}.$

Proof of Lemma 6.

By following the previously outlined development and employing an inductive method, the proof of this lemma can be obtained.

(i)

Let $x_0\in D$; then, ${\Gamma }_{x_0}={\left[{\mathcal{P}}^{\prime }\left(x_0\right)\right]}^{-1}$ exists by the Banach Lemma and ${\Gamma }_{x_0}\le \beta$ with ${\beta }_0={\displaystyle \frac{a_0}{\rho {\eta }_0}}$. Moreover, there exists ${\Gamma }_{x_1}$, and from (16), we have

$$∥{\Gamma }_{x_1}∥\le ∥{\Gamma }_{x_0}∥f_{\alpha }\left(a_0\right).$$

By induction, ${\Gamma }_{x_k}$ exists and

$$∥{\Gamma }_{x_k}∥\le ∥{\Gamma }_{x_{k-1}}∥f_{\alpha }\left(a_{k-1}\right).$$

(ii)

From (C₂), we have determined that $∥{\Gamma }_{x_0}\mathcal{P}\left(x_0\right)∥\le \eta$, with $∥{\Gamma }_{x_0}\mathcal{P}\left(x_0\right)∥={\eta }_0$. Thus,

$$∥{\Gamma }_{x_1}\mathcal{P}\left(x_1\right)∥\le ∥{\Gamma }_{x_1}∥∥\mathcal{P}\left(x_1\right)∥\le ∥{\Gamma }_{x_0}∥f_{\alpha }\left(a_0\right)∥\mathcal{P}\left(x_1\right)∥={\eta }_1={\eta }_0{f}_{\alpha }\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right).$$

Using induction, the following is obtained:

$$\begin{array}{cc}\hfill ∥{\Gamma }_{x_k}\mathcal{P}\left(x_k\right)∥& \le ∥{\Gamma }_{x_k}∥∥\mathcal{P}\left(x_k\right)∥\le ∥{\Gamma }_{x_{k-1}}∥f_{\alpha }\left(a_{k-1}\right)∥\mathcal{P}\left(x_k\right)∥={\eta }_k\hfill \\ \hfill & ={\eta }_{k-1}{f}_{\alpha }\left(a_{k-1}\right){\phi }_{\alpha }\left(a_{k-1},b_{k-1}\right)\hfill \end{array}$$

$$∥{\Gamma }_{x_k}\mathcal{P}\left(x_k\right)∥\le {\eta }_{k-1}{f}_{\alpha }\left(a_{k-1}\right){\phi }_{\alpha }\left(a_{k-1},b_{k-1}\right).$$

(iii)

We have determined that $\rho ={\displaystyle \frac{a_0}{{\beta }_0{\eta }_0}}$, with $∥{\Gamma }_{x_0}∥={\beta }_0$ and $∥{\Gamma }_{x_0}\mathcal{P}\left(x_0\right)∥={\eta }_0$. Therefore,

$$\rho ∥{\Gamma }_{x_0}∥∥{\Gamma }_{x_0}\mathcal{P}\left(x_0\right)∥=a_0.$$

Then,

$$\begin{array}{cc}\hfill \rho ∥{\Gamma }_{x_1}∥∥{\Gamma }_{x_1}\mathcal{P}\left(x_1\right)∥\le \rho ∥{\Gamma }_{x_0}∥f_{\alpha }\left(a_0\right){\eta }_1& ={\displaystyle \frac{a_0}{{\beta }_0{\eta }_0}}{\beta }_0{f}_{\alpha }\left(a_0\right){\eta }_0{f}_{\alpha }\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)\hfill \\ \hfill & =a_0{f}_{\alpha }^2\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)=a_1.\hfill \end{array}$$

Using this method, $\rho ∥{\Gamma }_{x_1}∥∥{\Gamma }_{x_1}\mathcal{P}\left(x_1\right)∥=a_1$, and by induction,

$$\rho ∥{\Gamma }_{x_k}∥∥{\Gamma }_{x_k}\mathcal{P}\left(x_k\right)∥=a_k.$$

(iv)

Similarly to the previous item,

$$\vartheta ∥{\Gamma }_{x_0}∥{∥{\Gamma }_{x_0}\mathcal{P}\left(x_0\right)∥}^2=\vartheta {\beta }_0{\eta }_0^2={\displaystyle \frac{a_0}{{\beta }_0{\eta }_0}}{\beta }_0{\eta }_0^2=a_0{\eta }_0=b_0.$$

Then,

$$\begin{array}{cc}\hfill \vartheta ∥{\Gamma }_{x_1}∥{∥{\Gamma }_{x_1}\mathcal{P}\left(x_1\right)∥}^2& \le \vartheta ∥{\Gamma }_{x_0}∥f_{\alpha }\left(a_0\right){\eta }_1^2={\displaystyle \frac{a_0}{{\beta }_0{\eta }_0}}{\beta }_0{f}_{\alpha }\left(a_0\right){\eta }_0^2{f}_{\alpha }^2\left(a_0\right){\phi }_{\alpha }^2\left(a_0,b_0\right)\hfill \\ \hfill & =a_0{\eta }_0{f}_{\alpha }^3\left(a_0\right){\phi }_{\alpha }^2\left(a_0,b_0\right)=b_0{f}_{\alpha }^3\left(a_0\right){\phi }_{\alpha }^2\left(a_0,b_0\right)=b_1.\hfill \end{array}$$

Using this method, $\vartheta ∥{\Gamma }_{x_1}∥{∥{\Gamma }_{x_1}\mathcal{P}\left(x_1\right)∥}^2=b_1$, and by induction,

$$\vartheta ∥{\Gamma }_{x_k}∥{∥{\Gamma }_{x_k}\mathcal{P}\left(x_k\right)∥}^2=b_k.$$

(v)

For this proof, we proceed as follows:

$$∥x_{k+1}-x_k∥\le \sum _{i=0}^k∥x_{i+1}-x_i∥\le \sum _{i=0}^k{g}_{\alpha }\left(a_i\right){\eta }_i\le g_{\alpha }\left(a_0\right)\sum _{i=0}^k{\eta }_i\le R{\eta }_0,$$

with $R=g_{\alpha }\left(a_0\right){\displaystyle \frac{1}{1-\mu \sigma }}$.

For the sequence $y_k$, we have

$$\begin{array}{cc}\hfill ∥y_k-x_0∥& \le ∥y_k-x_k∥+∥x_k-x_0∥\le {\displaystyle \frac{2}{3}}{\eta }_k+\sum _{i=0}^k{g}_{\alpha }\left(a_i\right){\eta }_i\hfill \\ \hfill & \le {\displaystyle \frac{2}{3}}{\eta }_k+g_{\alpha }\left(a_0\right)\sum _{i=0}^k{\eta }_i\le {\displaystyle \frac{2}{3}}{\eta }_0+R{\eta }_0\le \left({\displaystyle \frac{2}{3}}+R\right){\eta }_0.\hfill \end{array}$$

Therefore, $y_k\in \overline{B\left(x_0,\left({\displaystyle \frac{2}{3}}+R\right){\eta }_0\right)}$.

(vi)

From the definition of $R,\mu$, and $\sigma$, the following can be obtained:

$$R=g_{\alpha }\left(a_0\right){\displaystyle \frac{1}{1-\mu \sigma }}={\displaystyle \frac{g_{\alpha }\left(a_0\right)}{1-f_{\alpha }\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)}}\le {\displaystyle \frac{g_{\alpha }\left(a_0\right)}{1-f_{\alpha }\left(a_0\right)}}={\displaystyle \frac{g_{\alpha }\left(a_0\right)}{1-{\displaystyle \frac{1}{1-a_0{g}_{\alpha }\left(a_0\right)}}}}\le {\displaystyle \frac{1}{a_0}},$$

where ${\phi }_{\alpha }\left(a_0,b_0\right)<1$ and $a_0{g}_{\alpha }\left(a_0\right)<1$ by the hypothesis of Lemma 3 and the definitions of $g_{\alpha },f_{\alpha }$, and ${\phi }_{\alpha }$.

□

4. Semilocal Convergence Analysis

With the results previously obtained, we will establish the domains of existence and uniqueness of the solution.

Theorem 1.

Let $X,Y$ be Banach spaces and $\mathcal{P}:D\subseteq X\to Y$ a nonlinear twice continuously Fréchet differentiable operator in an open convex domain of D. Assume $x_0\in D$ and that conditions (C₁)–(C₄) hold. Consider $a_0=\rho {\beta }_0{\eta }_0$ and $b_0=\vartheta {\beta }_0{\eta }_0^2$ with $a_0=\left]0,{\displaystyle \frac{3}{4}}\right[$ and functions $p_{\alpha },g_{\alpha },f_{\alpha }$, and ${\phi }_{\alpha }$ defined by (8), (10), (13) and (17), respectively. If $\overline{B\left(x_0,\left({\displaystyle \frac{2}{3}}+R\right){\eta }_0\right)}\subseteq D$, with $R=g_{\alpha }\left(a_0\right){\displaystyle \frac{1}{1-\mu \sigma }}$, $\mu ={\displaystyle \frac{1}{f_{\alpha }\left(a_0\right)}}$, and $\sigma =f_{\alpha }^2\left(a_0\right){\phi }_{\alpha }\left(a_0,b_0\right)<1$, then the sequence $\left\{x_k\right\}$ defined by (1) converges to a solution $x^{*}$ of $\mathcal{P}\left(x\right)=0$. The R-order of convergence is at least two for any $\alpha \in \mathbb{R}$. The iterates $y_k$ belong to $\overline{B\left(x_0,\left({\displaystyle \frac{2}{3}}+R\right){\eta }_0\right)}$, the iterates $x_{k+1}$ and $x^{*}$ belong to $\overline{B\left(x_0,R{\eta }_0\right)}$, and $x^{*}$ is the only solution in $B\left(x_0,{\displaystyle \frac{2}{3\left(\rho \beta \right)}}-R{\eta }_0\right)\cap D$. Moreover, a priori error estimate is given by

$$∥x_k-x^{*}∥\le g_{\alpha }\left(a_0\right){\mu }^k{\sigma }^{2^k-1}{\eta }_0{\displaystyle \frac{1}{1-{\sigma }^{2^k}\mu }}.$$

Proof of Theorem 1.

From earlier results, it was established that the sequence $\left\{x_k\right\}$ is well defined. Now, we establish that $\left\{x_k\right\}$ is a Cauchy sequence. Using the fact that $g_{\alpha }$ is increasing and that $a_j\le a_k\forall j\le k$, we have $g_{\alpha }\left(a_j\right)\le g_{\alpha }\left(a_k\right)$. Applying the last statement of Lemma 5, we obtain

$$\begin{array}{cc}\hfill ∥x_{k+m}-x_k∥& \le \sum _{i=k}^{k+m-1}∥x_{i+1}-x_i∥\le \sum _{i=k}^{k+m-1}{g}_{\alpha }\left(a_i\right){\eta }_i\le g_{\alpha }\left(a_0\right)\sum _{i=k}^{k+m-1}{\eta }_i\hfill \\ \hfill & \le g_{\alpha }\left(a_0\right){\mu }^k{\sigma }^{2^k-1}{\eta }_0{\displaystyle \frac{1-{\left({\sigma }^{2^k}\mu \right)}^m}{1-{\sigma }^{2^k}\mu }},\hfill \end{array}$$

(22)

such that $x_k$ is a Cauchy sequence and therefore has a limit $x^{*}$. Taking $m\to \infty$, the a priori error estimate is

$$∥x_k-x^{*}∥\le g_{\alpha }\left(a_0\right){\mu }^k{\sigma }^{2^k-1}{\eta }_0{\displaystyle \frac{1}{1-{\sigma }^{2^k}\mu }}.$$

(23)

In (23), we take $k=0$ and the following is obtained:

$$\begin{array}{cc}\hfill ∥x_0-x^{*}∥& \le g_{\alpha }\left(a_0\right){\mu }^0{\sigma }^{2^0-1}{\eta }_0{\displaystyle \frac{1}{1-{\sigma }^{2^0}\mu }}=g_{\alpha }\left(a_0\right){\eta }_0{\displaystyle \frac{1}{1-\sigma \mu }}=R{\eta }_0.\hfill \end{array}$$

Thus, $x^{*}\in \overline{B\left(x_0,R{\eta }_0\right)}$. Moreover, $x^{*}$ is a solution of $\mathcal{P}\left(x\right)=0$ because

$$\begin{array}{cc}\hfill ∥{\mathcal{P}}^{\prime }\left(x_k\right)∥& \le ∥{\mathcal{P}}^{\prime }\left(x_0\right)∥+∥{\mathcal{P}}^{\prime }\left(x_k\right)-{\mathcal{P}}^{\prime }\left(x_0\right)∥\le ∥{\mathcal{P}}^{\prime }\left(x_0\right)∥+{\int }_{x_0}^{x_k}∥{\mathcal{P}}^{″}\left(x_k\right)∥dx\hfill \\ \hfill & \le ∥{\mathcal{P}}^{\prime }\left(x_0\right)∥+\rho {\int }_{x_0}^{x_k}∥dx∥\le ∥{\mathcal{P}}^{\prime }\left(x_0\right)∥+\rho ∥x_k-x_0∥.\hfill \end{array}$$

${\mathcal{P}}^{\prime }\left(x_k\right)$ is bounded (because ${\Gamma }_{x_k}$ exists in all steps of the scheme) and using item (iv) of Lemma 5, we obtain $∥{\Gamma }_{x_k}{\mathcal{P}}^{\prime }\left(x_k\right)∥\to 0$; then, it can be stated that

$$∥\mathcal{P}\left(x_k\right)∥=∥{\mathcal{P}}^{\prime }\left(x_k\right){\Gamma }_{x_k}\mathcal{P}\left(x_k\right)∥\le ∥{\mathcal{P}}^{\prime }\left(x_k\right)∥∥{\Gamma }_{x_k}\mathcal{P}\left(x_k\right)∥\le ∥{\mathcal{P}}^{\prime }\left(x_k\right)∥{\eta }_k\to 0.$$

Since operator $\mathcal{P}$ is continuous in D, we obtain

$$\underset{k\to \infty }{lim}\mathcal{P}\left(x_k\right)=\mathcal{P}\underset{k\to \infty }{lim}\left(x_k\right)=\mathcal{P}\left(x^{*}\right)=0.$$

Hence, $x^{*}$ is a solution of $\mathcal{P}\left(x\right)=0$.

Let $B\left(x_0,{\displaystyle \frac{2}{3\rho \beta }}-R{\eta }_0\right)$ be the ball centered in $x_0$ with radius ${\displaystyle \frac{2}{3\rho \beta }}-R{\eta }_0$. For the uniqueness proof, let us suppose that $y^{*}\in B\left(x_0,{\displaystyle \frac{2}{3\rho \beta }}-R{\eta }_0\right)\cap D$ is another solution of $\mathcal{P}\left(x\right)=0$. Then,

$$0=\mathcal{P}\left(y^{*}\right)-\mathcal{P}\left(x^{*}\right)={\int }_{x^{*}}^{y^{*}}{\mathcal{P}}^{\prime }\left(v\right)dv={\int }_0^1{\mathcal{P}}^{\prime }\left(x^{*}+t\left(y^{*}-x^{*}\right)\right)\left(y^{*}-x^{*}\right)dt,$$

where $v=x^{*}+t\left(y^{*}-x^{*}\right)$ and $dv=\left(y^{*}-x^{*}\right)dt$.

Let us denote the operator ${\int }_0^1{\mathcal{P}}^{\prime }\left(x^{*}+t\left(y^{*}-x^{*}\right)\right)dt$ as $\mathcal{Q}$. It is easy to prove that ${\mathcal{Q}}^{-1}$ exists, given that

$$\begin{array}{cc}\hfill ∥{\Gamma }_{x_0}\mathcal{Q}-I∥& =∥{\Gamma }_{x_0}\left(\mathcal{Q}-{\mathcal{P}}^{\prime }\left(x_0\right)\right)∥\le ∥{\Gamma }_{x_0}∥∥\left(\mathcal{Q}-{\mathcal{P}}^{\prime }\left(x_0\right)\right)∥\hfill \\ \hfill & \le ∥{\Gamma }_{x_0}∥{\int }_0^1∥{\mathcal{P}}^{\prime }\left(x^{*}+t\left(y^{*}-x^{*}\right)\right)-{\mathcal{P}}^{\prime }\left(x_0\right)∥dt\hfill \\ \hfill & \le \beta \rho {\int }_0^1∥x^{*}+t\left(y^{*}-x^{*}\right)-x_0∥dt\le \beta \rho {\int }_0^1∥x^{*}+t{y}^{*}-t{x}^{*}+t{x}_0-t{x}_0-x_0∥dt\hfill \\ \hfill & \le \beta \rho {\int }_0^1∥\left(1-t\right)\left(x^{*}-x_0\right)+t\left(y^{*}-x_0\right)∥dt\hfill \\ \hfill & \le \beta \rho \left(∥x^{*}-x_0∥{\left[t-{\displaystyle \frac{t^2}{2}}\right]}_0^1+∥y^{*}-x_0∥{\left[{\displaystyle \frac{t^2}{2}}\right]}_0^1\right)\le \beta \rho \left[{\displaystyle \frac{1}{2}}R\eta +{\displaystyle \frac{1}{2}}\left({\displaystyle \frac{2}{3\beta \rho }}-R\eta \right)\right]\hfill \\ \hfill & ={\displaystyle \frac{\beta \rho }{2}}\left(R\eta +{\displaystyle \frac{2}{3\beta \rho }}-R\eta \right)={\displaystyle \frac{\beta \rho }{2}}\left({\displaystyle \frac{2}{3\beta \rho }}\right)={\displaystyle \frac{1}{3}}<1.\hfill \end{array}$$

Since $∥{\Gamma }_{x_0}\mathcal{Q}-I∥<1$, the Banach Lemma can be applied and the operator $\mathcal{Q}$ has an inverse. Taking $\mathcal{Q}\left(y^{*}-x^{*}\right)=0$ and applying ${\mathcal{Q}}^{-1}$, we determine that $y^{*}-x^{*}=0$ and, consequently, $y^{*}=x^{*}$. Hence, the solution is unique. □

5. Numerical Examples

In this section, we will apply the theoretical results to applied problems, some of which have already been cited in earlier studies.

Example 1

(see [17,19]). Consider the nonlinear integral equation $\mathcal{P}\left(x\right)=0$, where

$$\mathcal{P}\left(x\right)\left(\theta \right)=x\left(\theta \right)-1+{\displaystyle \frac{1}{2}}{\int }_0^1\theta cos\left(x\left(t\right)\right)dt,$$

with $\theta \in \left[0,1\right]$ and $x\in D=B\left(0,2\right)\subset X$. This is a Chandrasekhar integral equation. These equations represent a category of integral equations used to tackle problems related to radiative transfer theory in a plane-parallel atmosphere. Additionally, these equations are applicable in other areas of research, such as traffic modeling, queuing theory, neutron transport, and the kinetic theory of gases. Let $X=C\left[0,1\right]$, the space of continuous functions defined on $\left[0,1\right]$ with the norm

$$∥x∥=\underset{\theta \in \left[0,1\right]}{max}\left|x\left(\theta \right)\right|.$$

The first and second derivatives of $\mathcal{P}$ are, respectively,

$${\mathcal{P}}^{\prime }\left(x\right)y\left(\theta \right)=y\left(\theta \right)-{\displaystyle \frac{1}{2}}\theta {\int }_0^{1}sin\left[x\left(t\right)\right]y\left(t\right)dt$$

and

$${\mathcal{P}}^{″}\left(x\right)yz\left(\theta \right)=-{\displaystyle \frac{1}{2}}\theta {\int }_0^{1}cos\left[x\left(t\right)\right]y\left(t\right)z\left(t\right)dt,$$

for $y,z\in D$.

The second derivative verifies

$$∥{\mathcal{P}}^{″}\left(x\right)∥\le {\displaystyle \frac{1}{2}}=\rho ,x\in D$$

and using the mean value theorem,

$${\mathcal{P}}^{″}\left(x\right)-{\mathcal{P}}^{″}\left(y\right)={\mathcal{P}}^{‴}\left(x\right)\left(x-y\right)$$

(24)

with

$${\mathcal{P}}^{‴}\left(x\right)yz\varphi \left(\theta \right)={\displaystyle \frac{1}{2}}\theta {\int }_0^{1}sin\left[x\left(t\right)\right]y\left(t\right)z\left(t\right)\varphi \left(t\right)dt,y,z,\varphi \in D.$$

Taking the norm in (24),

$$∥{\mathcal{P}}^{″}\left(x\right)-{\mathcal{P}}^{″}\left(y\right)∥=∥{\mathcal{P}}^{‴}\left(x\right)\left(x-y\right)∥\le ∥{\mathcal{P}}^{‴}\left(x\right)∥∥\left(x-y\right)∥$$

we consider the Lipschitz condition

$$∥{\mathcal{P}}^{″}\left(x\right)-{\mathcal{P}}^{″}\left(y\right)∥={\displaystyle \frac{1}{2}}∥x-y∥,\mathrm{for}x,y\in D$$

Thus, $\vartheta ={\displaystyle \frac{1}{2}}$.

Choosing an initial estimate of the solution, $x_0\left(t\right)=1.55$, we obtain

$$∥\mathcal{P}\left(x_0\right)∥=∥{\displaystyle \frac{1}{2}}{\int }_0^1\theta cos\left[x_0\left(t\right)\right]dt∥\le {\displaystyle \frac{1}{2}}\theta {\int }_0^1∥cos\left[x_0\left(t\right)\right]dt∥\le {\displaystyle \frac{1}{2}}cos\left[1.55\right]$$

and

$$∥I-{\mathcal{P}}^{\prime }\left(x_0\right)∥=∥1-y\left(\theta \right)+{\displaystyle \frac{1}{2}}{\int }_0^1\theta sin\left[x_0\left(t\right)\right]dt∥\le {\displaystyle \frac{1}{2}}\theta {\int }_0^1∥sin\left[x_0\left(t\right)\right]dt∥\le {\displaystyle \frac{1}{2}}sin\left[1.55\right]<1.$$

Therefore, by the Banach Lemma, ${\Gamma }_{x_0}$ exists and

$$∥{\Gamma }_{x_0}∥\le {\displaystyle \frac{1}{1-∥I-{\mathcal{P}}^{\prime }\left(x_0\right)∥}}={\displaystyle \frac{2}{2-sin\left[1.55\right]}}=1.99956={\beta }_0.$$

Thus,

$$\begin{array}{cc}\hfill & ∥{\Gamma }_{x_0}\mathcal{P}\left(x_0\right)∥\le ∥{\Gamma }_{x_0}∥∥\mathcal{P}\left(x_0\right)∥\le \left({\displaystyle \frac{cos\left[1.55\right]}{2}}\right)\left({\displaystyle \frac{2}{2-sin\left[1.55\right]}}\right)={\displaystyle \frac{cos\left[1.55\right]}{2-sin\left[1.55\right]}}=0.02079={\eta }_0,\hfill \\ \hfill & a_0=\rho {\beta }_0{\eta }_0=0.02078<{\displaystyle \frac{3}{4}},\hfill \\ \hfill & b_0=\vartheta {\beta }_0{\eta }_0^2=4.32144\times {10}^{-4}.\hfill \end{array}$$

The auxiliary functions and recurrence relations of the scheme for some chosen values of parameter $\alpha$ are shown in Table 1 and Table 2.

Table 1. Numerical results for Example 1 with starting condition $x_0=1.55$.

	$\mathit{\alpha }=-0.05$	$\mathit{\alpha }=-0.03$	$\mathit{\alpha }=0$	$\mathit{\alpha }=0.01$	$\mathit{\alpha }=0.03$	$\mathit{\alpha }=0.05$
Convergence radius	$0.19569$	$0.15481$	$0.13534$	$0.15402$	$0.21179$	$0.33590$
Uniqueness radius	$0.48498$	$0.52586$	$0.54533$	$0.52666$	$0.46888$	$0.34477$

Table 2. Recurrence relations for $\alpha =-0.05,0,0.05$ in Example 1.

$\mathit{\alpha }$	k	${\mathit{a}}_{\mathit{k}}$	${\mathit{b}}_{\mathit{k}}$	${\mathit{n}}_{\mathit{k}}$
	0	$0.02079$	$4.32144\times {10}^{-4}$	$0.02079$
	1	$0.01850$	$3.42481\times {10}^{-4}$	$0.01807$
$-0.05$	2	$0.01466$	$2.15106\times {10}^{-4}$	$0.01398$
	3	$0.00921$	$8.48558\times {10}^{-5}$	$0.00857$
	4	$0.00363$	$1.32051\times {10}^{-5}$	$0.00330$
	0	$0.02079$	$4.32144\times {10}^{-4}$	$0.02079$
	1	$0.01729$	$2.98918\times {10}^{-4}$	$0.01690$
0	2	$0.01196$	$1.43021\times {10}^{-4}$	$0.01142$
	3	$0.00572$	$3.27412\times {10}^{-5}$	$0.00534$
	4	$0.00131$	$1.71587\times {10}^{-6}$	$0.00119$
	0	$0.02079$	$4.32144\times {10}^{-4}$	$0.02079$
	1	$0.01967$	$3.87049\times {10}^{-4}$	$0.01919$
$0.05$	2	$0.01762$	$3.10484\times {10}^{-4}$	$0.01676$
	3	$0.01413$	$1.99797\times {10}^{-4}$	$0.01311$
	4	$0.00910$	$8.27341\times {10}^{-5}$	$0.00823$

Next, let us discretize the integral equation using Simpson’s quadrature, with the aim of transforming it into a large finite-dimensional problem. Let us denote the nodes as $t_i=ih$ for $i=0,1,\dots ,n$ and the weights as $p_j={\displaystyle \frac{h}{3}}\left(1,4,2,\dots ,2,4,1\right)\in {\mathbb{R}}^{n+1}$, where $h={\displaystyle \frac{1}{n}}$ is the step size and n is an even number of subintervals. Expressing $x_i=x\left(t_i\right)$, we obtain the (nonlinear) system of equations as follows:

$$x_i-1+{\displaystyle \frac{t_i}{2}}\sum _{j=0}^n{p}_{j}cos\left(x_j\right)=0;i=0,1,\dots ,n.$$

Table 3 displays the results of a comparative study between a selected member of the family of methods (1) and three other iterative methods, including the Newton [1], Chun [28], and Jarratt [29] methods. The solution obtained by the schemes is $x^{*}={\left[1,0.9966,\dots ,0.6669,0.6634\right]}^T$ for $n=100$, where the number of iterations (iter) required to converge to the solution is presented, as well as the execution time (ex-time) in seconds required for convergence to the solution, which is determined as the average over 10 consecutive runs for each scheme. The stopping criterion is $∥x_{k+1}-x_k∥+∥f\left(x_{k+1}\right)∥<ϵ$, where $∥x_{k+1}-x_k∥$ represents the error estimate between two consecutive iterations, $∥f\left(x_{k+1}\right)∥$ denotes the residual error, and the error tolerance is set to $ϵ={10}^{-10}$. The approximated computational order of convergence (ACOC) (see [30]) quantifies the rate at which successive iterates approach the solution, providing an empirical estimate of the method’s order of convergence. It is employed to verify the validity of the theoretical convergence order p.

Table 3. Numerical performance of iterative methods for nonlinear equations for Example 1.

Method	$∥{\mathit{x}}_{\mathit{k}+1}-{\mathit{x}}_{\mathit{k}}∥$	$∥\mathit{f}\left({\mathit{x}}_{\mathit{k}+1}\right)∥$	Iter	ACOC	Ex-Time
Newton	$1.6009\times {10}^{-18}$	$2.7006\times {10}^{-38}$	5	$2.0000$	$186.6108$
Method (1)α=0	$3.7590\times {10}^{-14}$	$3.0992\times {10}^{-14}$	3	$3.9180$	$200.6128$
Chun	$1.2825\times {10}^{-36}$	$1.0292\times {10}^{-36}$	4	$4.0050$	$252.7789$
Jarratt	$5.5763\times {10}^{-13}$	$4.5976\times {10}^{-13}$	3	$3.9893$	$193.0262$

Example 2.

Consider the system of equations of size $n=100$ with the following expression:

$$x_i-cos\left(2x_i-x_1-x_2-x_3-x_4\right)=0;i=1,2,\dots ,n$$

with $x^{*}={(0.5149,0.5149,\dots ,0.5149)}^T$ and $x_i\in D=\left[-1,1\right]\subset X$ with the norm

$$∥F\left(x\right)∥=\underset{i}{max}\left|F_i\left(x\right)\right|.$$

The first derivative of $\mathcal{P}$ is

$${\mathcal{P}}^{\prime }\left(x\right)=\left\{\begin{array}{lll}1+2sin\left(2x_i-x_1-x_2-x_3-x_4\right)& ,& i=j\\ -sin\left(2x_i-x_1-x_2-x_3-x_4\right)& ,& i\ne j,j\in \left\{1,2,3,4\right\}\\ 0& ,& i\ne j,j\ge 5.\end{array}\right.$$

Choosing an initial estimate of the solution, $x_0={\left[0.25,0.25\dots ,0.25,0.25\right]}^T$ and denoting $S=x_1+x_2+x_3+x_4$, we obtain

$$∥I-{\mathcal{P}}^{\prime }\left(x_0\right)∥=2.397>1.$$

Therefore, the system fails to meet the conditions established in Banach’s Lemma. For this example, the results obtained for $n=100$ are presented in Table 4.

Table 4. Numerical performance of iterative methods for nonlinear equations for Example 2.

Method	$∥{\mathit{x}}_{\mathit{k}+1}-{\mathit{x}}_{\mathit{k}}∥$	$∥\mathit{f}\left({\mathit{x}}_{\mathit{k}+1}\right)∥$	Iter	ACOC	Ex-Time
Newton	$5.0513\times {10}^{-13}$	$2.6278\times {10}^{-26}$	5	$2.0000$	$2.2428$
Method (1)α=0	$2.8382\times {10}^{-42}$	$7.7043\times {10}^{-42}$	4	$3.9992$	$3.2802$
Chun	$1.0218\times {10}^{-30}$	$2.7738\times {10}^{-30}$	4	$3.9922$	$3.1338$
Jarratt	$2.5793\times {10}^{-44}$	$7.0013\times {10}^{-44}$	4	$3.9994$	$3.4073$

This iterative scheme is not limited to systems with standard algebraic or differential structures; rather, it extends naturally to a broader class of nonlinear operator equations, including those that include coefficients involving domain integrals, such as terms of the form $f_{\Omega }{x}^2$, provided that certain conditions are met. These include that the operator is well defined in a suitable Banach space, the existence and continuity of its Fréchet derivative, and the satisfaction of regularity conditions such as the Lipschitz continuity. Integral terms are common in nonlinear and functional analysis, and their presence does not rule out the use of iterative methods, especially when the operator remains differentiable and bounded. Therefore, with appropriate analytical justification, such operators are compatible with standard semilocal and local convergence theories.

6. Conclusions

In this paper, we have studied the semilocal convergence of a family of iterative schemes in Banach spaces. The analysis is conducted by applying the Lipschitz condition to the second derivative. Using recurrence relations, the semilocal convergence theory for the iterative scheme is established. The key results are summarized in Theorem 1. Moreover, the existence and uniqueness domain of the solution is obtained, and it has been proven that the sequence $x_k$ converges to the solution. The R-order of convergence is at least two. Finally, numerical experiments were conducted to further validate our theoretical findings. Example 1, a Chandrasekhar integral equation, satisfies the conditions established in Theorem 1.

Future research based on these results will aim to extend the Banach Lemma to situations where the system of equations does not meet the conditions required by the lemma.