A Study of at Least Sixth Convergence Order Methods Without or with Memory and Divided Differences for Equations Under Generalized Continuity

Abstract

Multistep methods typically use Taylor series to attain their convergence order, which necessitates the existence of derivatives not naturally present in the iterative functions. Other issues are the absence of a priori error estimates, information about the radius of convergence or the uniqueness of the solution. These restrictions impose constraints on the use of such methods, especially since these methods may converge. Consequently, local convergence analysis emerges as a more effective approach, as it relies on criteria involving only the operators of the methods. This expands the applicability of such methods, including in non-Euclidean space scenarios. Furthermore, this work uses majorizing sequences to address the more challenging semi-local convergence analysis, which was not explored in earlier research. We adopted generalized continuity constraints to control the derivatives and obtain sharper error estimates. The sufficient convergence criteria are demonstrated through examples.

1. Introduction

A scalar nonlinear equation (NE) or a system of nonlinear equations (SNE) can be derived by transforming models from science, engineering, and nature. Such equations are crucial in mathematics [1,2,3,4,5,6]. Solutions to these nonlinear problems enable us to forecast weather, analyze fluid dynamics, model populations, and predict financial markets. Therefore, to evaluate the following SNE by approximating its solution, we choose the initial approximation $x_0\in \mathbb{D}$

$$F\left(x\right)=0.$$

(1)

The above defined operator F maps an open subset $\mathbb{D}$ of a Banach space $\mathbb{X}$ into $\mathbb{X}$. It is usually impossible to find analytical solutions to equations such as (1). As a result, we are limited to using iterative methods. For instance, we have one of the most often used iterative techniques, the Newton–Raphson method, which help us to obtain an approximate solution of such problems.

Using an iterative technique, researchers apply the Newton–Raphson method or similar approaches to achieve the desired accuracy of the solution. Additionally, they study the convergence, extensions or modifications, stability, and basin of attraction. The following are popular single-step iterative methods for solving (1):

$$\begin{array}{cc}\hfill x_{n+1}& =x_n-F{\left(x_n\right)}^{-1}F\left(x_n\right),\hfill \\ \hfill x_{n+1}& =x_n-{[x_{n-1},x_n;F]}^{-1}F\left(x_n\right),\hfill \\ \mathrm{and}\hfill & \\ \hfill x_{n+1}& =x_n-{[x_n-F\left(x_n\right),x_n+F\left(x_n\right);F]}^{-1}F\left(x_n\right),\hfill \end{array}$$

These methods are called Newton’s secant and Steffensen’s method, respectively. But their convergence order (CO) does not exceed two [1,2,4,5,7,8]. The CO can be increased if we add more steps. That is why we have a sixth-order method of convergence without memory and a 6.60 CO with memory, as proposed by Cordero et al. [9]. These schemes are defined for $x_0\in \mathbb{X},$ and each $n=0,1,2,3,...$ by:

(1)

Sixth CO without memory:

$$\begin{array}{cc}\hfill y_n& =x_n-A_n^{-1}F\left(x_n\right),\hfill \\ \hfill z_n& =y_n-A_n^{-1}F\left(y_n\right),\hfill \\ \hfill x_{n+1}& =z_n-\left(3I+A_n^{-1}{B}_n\left(-3I+A_n^{-1}{B}_n\right)\right)A_n^{-1}F\left(z_n\right),\hfill \end{array}$$

(2)

where $A_n=[u_n,v_n;F]$, $B_n=[w_n,s_n;F]$, $u_n=x_n-aF\left(x_n\right)$, $v_n=x_n+bF\left(x_n\right)$, $w_n=z_n-cF\left(z_n\right)$, $s_n=z_n+dF\left(z_n\right)$ and $a,b,c,d\in \mathbb{R}$.

Method (2) specializes to the preceding ones provided that we stop only at the first substep and choose $a=b=0$ for Newton’s, $A_n=[x_{n-1},x_n;F]$ for secant and $a=b=1$ for Steffensen’s method. Thus, the convergence study of (2) includes the preceding methods.

(2)

6.60 CO with memory:

$$\begin{array}{cc}\hfill y_n& =x_n-K_n^{-1}F\left(x_n\right),\hfill \\ \hfill z_n& =y_n-K_n^{-1}F\left(y_n\right),\hfill \\ \hfill x_{n+1}& =z_n-\left(qI+K_n^{-1}{B}_n\left((3-2q)I+(q-2)K_n^{-1}{B}_n\right)\right)K_n^{-1}F\left(z_n\right),\hfill \end{array}$$

(3)

where $K_n=[u_n,v_n;F]$, $u_n=x_n+\delta L_{n}F\left(x_n\right)$, $v_n=x_n-\gamma L_{n}F\left(x_n\right)$, ($L:\mathbb{D}\times \mathbb{D}\to L\left(\mathbb{X},\mathbb{X}\right),[.,.;F]:\mathbb{D}\times \mathbb{D}\to L\left(\mathbb{X},\mathbb{X}\right)$. The parameters $q,\delta ,\gamma \in \mathbb{R}$ and are usually chosen to optimize the method. Authors usually choose $q=\delta =\gamma =1$. Notice that the notation $[.,.;F]$ stands for the divided difference of order one such that $[x,y;F](y-x)=F\left(y\right)-F\left(x\right)$, if $x\ne y$ and $[x,x;F]=F^{\prime }\left(x\right)$, if F is a differentiable x.

The CO is shown in [9] by adopting Taylor series and the fifth derivative of F. However, this approach has certain constraints.

• Motivation:

(P₁)

The inverses of derivatives, divided differences, or high-order derivatives are typically needed for the local convergence analysis. Local analysis of convergence (LAC) in [9] shows that the proof requires derivatives up to the fifth order, which is not present in the method. Their application in the case when $\mathbb{X}={\mathbb{R}}^m$ is restricted by these limitations. As an example, we choose the following fundamental and inspirational function F on $\mathbb{X}=\mathbb{R}$. If $\mathbb{X}=[-2,2]$, defined as

$$f\left(t\right)=\left\{\begin{array}{ccc}{s}_1{t}^{3}log{t}^6+s_2{t}^6+s_3{t}^5\hfill & & t\ne 0\hfill \\ 0,\hfill & & t=0,\hfill \end{array}\right.$$

where $s_1,s_2,s_3\in \mathbb{R}$ are three parameters with $s_1\ne 0,s_2+s_3=1$.

We can easily see that the function $f^{‴}\left(t\right)$ is not bounded on $\mathbb{X}$ at $t=0\in \mathbb{X}$ and $f\left(1\right)=0$. Thus, the convergence of procedures (2) and (3) is not guaranteed by the local convergence findings in [9]. However, if, for instance, $x_0=0.1\in \mathbb{X}$, $a=b=c=d=\gamma =\delta =1,q=3$ and $s_1=s_2=1,s_3=-1$, then the iterative schemes (2) and (3) converge to $x_{*}=1\in \mathbb{X}$. This observation implies that these criteria can be weakened.

(P₂)

The results are applicable only on ${\mathbb{R}}^m$.

(P₃)

There is no subset that exclusively contains $x_{*}$ as its solution of (1).

(P₄)

The more important to obtain and difficult semi-local analysis (SLAC) is not given previously [9].

(P₅)

There is no indication or information regarding the integer j that satisfies the condition such that $\parallel e_n\parallel <ϵ(ϵ>0)$ for each $n\ge j$, where $e_n=x_n-x_{*},(n=0,1,2,3\cdots )$.

We note that in the previous research [9], we had to deal with the above-mentioned issues $\left(P_1\right)$–$\left(P_5\right)$. This is the primary reason we are conducting this investigation. Our method addresses these issues. The generality of the new technique makes it useful on extending the applicability of other methods in the same way [3,6,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23]. Method (1.2) outperforms existing methods in terms of residual error and the absolute error difference between consecutive iterations. Additionally, it requires significantly less computational time and exhibits a more stable CO than the existing methods. That is the novelty of our paper.

The rest of the article includes LAC and SLAC of the method in Section 2 and Section 3, respectively. Section 4 provides numerical examples to illustrate the practical implementation and effectiveness of the methods. These examples demonstrate the accuracy, efficiency, radii of convergence, and convergence of the iterative schemes in solving SNE. Section 5 concludes this study with a summary of findings, key contributions, and potential directions for future research.

2. Local Convergence Analysis

The analysis uses some conditions.

• Suppose:

(1)

There exist functions $\lambda ,\mu ,{\psi }_0$, continuous and nondecreasing, defined on the interval $[0,+\infty )$, $[0,+\infty )$ and $[0,+\infty )\times [0,+\infty )$ so that the equation

$${\psi }_0\left(\lambda \left(t\right),\mu \left(t\right)\right)-1=0$$

has a positive smallest solution (SS), which is denoted by $r_0$.

(2)

There exist majorant functions $\mu ,{\mu }^1,\lambda ,{\lambda }^1,p,p^1$ (see conditions (Q)) defined on the interval $[0,r_0)$, continuous and nondecreasing, so that the equation

$$g_1\left(t\right)-1=0$$

has an SS $c_1\in (0,r_0)$, where

$$g_1\left(t\right)={\displaystyle \frac{\psi \left({\mu }^1\left(t\right),\mu \left(t\right)\right)}{1-{\psi }_0\left(\lambda \left(t\right),\mu \left(t\right)\right)}}.$$

(3)

The equation $g_2\left(t\right)-1=0$ has an SS $c_2\in (0,r_0)$, for function $g_2:[0,r_0)\to \mathbb{R}$ is given by

$$g_2\left(t\right)={\displaystyle \frac{\psi \left(\lambda \left(t\right)+g_1\left(t\right)t,\mu \left(t\right)\right)g_1\left(t\right)}{1-{\psi }_0\left(\lambda \left(t\right),\mu \left(t\right)\right)}}.$$

(4)

The equation $g_3\left(t\right)-1=0$ has an SS $c_3\in (0,r_0)$, for function $g_3:[0,r_0)\to \mathbb{R}$ is given as

$$\begin{array}{cc}\hfill g_3\left(t\right)& ={\displaystyle \frac{\psi \left({\mu }^1\left(t\right)+g_2\left(t\right)t,\mu \left(t\right)\right)}{1-{\psi }_0\left(\lambda \left(t\right),\mu \left(t\right)\right)}}+[{\left({\displaystyle \frac{\psi \left(p\left(t\right),p^1\left(t\right)\right)}{1-{\psi }_0\left(\lambda \left(t\right),\mu \left(t\right)\right)}}\right)}^2\hfill \\ \hfill & +\left({\displaystyle \frac{\psi \left(p\left(t\right),p^1\left(t\right)\right)}{1-{\psi }_0\left(\lambda \left(t\right),\mu \left(t\right)\right)}}\right)\left({\displaystyle \frac{{\lambda }^1\left(t\right)}{1-{\psi }_0\left(\lambda \left(t\right),\mu \left(t\right)\right)}}\right)]g_2\left(t\right).\hfill \end{array}$$

Let us introduce the parameter $r^{*}$ by

$$\begin{array}{c}\hfill r^{*}=min\left\{c_i\right\},i=1,2,3.\end{array}$$

(4)

The parameter $r^{*}$ is established as the radius of convergence for the method, as demonstrated in Theorem 1. This result confirms that $r^{*}$ represents the threshold within which the method always converges.

Clearly, expression (4) implies that for each $t\in [0,r^{*})$,

$$\begin{array}{cc}\hfill 0\le & {\psi }_0\left(\lambda \left(t\right),\mu \left(t\right)\right)<1,\hfill \end{array}$$

(5)

$$\mathrm{and}$$

$$\begin{array}{cc}\hfill 0\le & g_i\left(t\right)<1.\hfill \end{array}$$

(6)

Let us denote by $\mathbb{B}(\overline{x},\overline{r}),\mathbb{B}[\overline{x},\overline{r}]$ the open and closed balls in $\mathbb{X}$, respectively, with center $\overline{x}\in \mathbb{X}$ and of radius $\overline{r}>0$. We shall use the same symbol $\parallel ·\parallel$ for the norm of linear operators involved as well as that of the elements of the Banach space $\mathbb{X}$ to simplify the presentation.

The real functions and the parameter $r^{*}$ are connected to the divided differences $F^{\prime }$ on method (2).

• Suppose:

(H₁)

There exists a solution $x_{\in }\mathbb{D}$ to $F\left(x\right)=0$, along with a linear operator M that is invertible. This means that the equation has at least one solution within the domain $\mathbb{D}$, specifically the value $x_{*}$. Additionally, the operator M is guaranteed to be invertible, implying that it possesses a unique inverse.

(H₂)

$∥M^{-1}\left([x,y;F]-M\right)∥\le {\psi }_0(\parallel x-x_{*}\parallel ,\parallel y-x_{*}\parallel )$ for each $x,y\in \mathbb{D}$.

Set ${\mathbb{D}}_0=\mathbb{D}\cap \mathbb{B}[x_{*},r_0)$.

(H₃)

$∥M^{-1}\left([x,y;F]-[z,x_{*};F]\right)∥\le \psi (\parallel x-z\parallel ,\parallel y-x_{*}\parallel )$ for each $x,y,z\in {\mathbb{D}}_0$.

and

(H₄)

$\mathbb{B}[x_{*},R]\subset \mathbb{D}$ for some $R>0$ to be given later.

• Moreover, we consider:

(Q)

All the iterates on the method (2) exist and

$$\begin{array}{cc}\hfill & \parallel u_n-x_{*}\parallel \le \lambda (\parallel e_n\parallel )\le {\lambda }_n<r^{*},\hfill \\ \hfill & ∥M^{-1}F\left(z_n\right)∥\le {\lambda }^1(\parallel e_n\parallel )\le {\lambda }_n^1,\hfill \\ \hfill & \parallel v_n-x_{*}\parallel \le \mu (\parallel e_n\parallel )\le {\mu }_n<r^{*},\hfill \\ \hfill & \parallel u_n-x_n\parallel \le {\mu }^1(\parallel e_n\parallel )\le {\mu }_n^1,\hfill \\ \hfill & \parallel w_n-x_{*}\parallel \le \xi (\parallel e_n\parallel )\le {\xi }_n<r^{*},\hfill \\ \hfill & \parallel s_n-x_{*}\parallel \le \rho (\parallel e_n\parallel )\le {\rho }_n<r^{*},\hfill \\ \hfill & \parallel u_n-w_n\parallel \le p(\parallel e_n\parallel )\le p_n<r^{*},\hfill \\ \hfill & \parallel v_n-s_n\parallel \le p^1(\parallel e_n\parallel )\le p_n^1<r^{*},\hfill \end{array}$$

where the functions $\lambda ,\mu ,{\mu }^1,\xi ,\rho ,p,p^1,{\lambda }^1$ are continuous and nondecreasing and the sequences

$\left\{{\lambda }_n\right\},\left\{{\mu }_n\right\},\left\{{\mu }_n^1\right\},\left\{{\xi }_n\right\},\left\{{\rho }_n\right\},\left\{p_n\right\},\left\{p_n^1\right\},\left\{{\lambda }_n^1\right\}$ are nonnegative sequences.

• The functions and sequences appearing in the conditions $\left(Q\right)$ are specialized later in terms of the conditions $\left(H_1\right)-\left(H_4\right)$ (see Remarks 1 and 2).

The local convergence result for (2) follows from the following conditions.

Theorem 1. 

Under the conditions $\left(H_1\right)-\left(H_4\right)$ and $\left(Q\right)$, sequence $x_n$ converges to $x_{*}$, as long as the initial value $x_0$ lies within the ball $\mathbb{B}(x_{*},r^{*})-\left\{x_{*}\right\}$ but is not equal to $x_{*}$ itself.

Proof. 

Mathematical induction is employed to demonstrate the validity of the following items:

$$\begin{array}{cc}\hfill & \left\{x_n\right\}\subset \mathbb{B}(x_{*},r^{*}),\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill & \parallel y_n-x_{*}\parallel \le g_1(\parallel e_n\parallel )\parallel e_n\parallel \le \parallel e_n\parallel <r^{*},\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill & \parallel z_n-x_{*}\parallel \le g_2(\parallel e_n\parallel )\parallel e_n\parallel \le \parallel e_n\parallel ,\hfill \end{array}$$

(9)

$$\mathrm{and}$$

$$\begin{array}{cc}\hfill & \parallel x_{n+1}-x_{*}\parallel \le g_3(\parallel e_n\parallel )\parallel e_n\parallel \le \parallel e_n\parallel ,\hfill \end{array}$$

(10)

where the parameter $r^{*}$ is defined by (4).

Mathematical induction shall establish the validity of the assertions (7)–(10).

• By hypothesis, $x_0\in \mathbb{B}(x_{*},r^{*})-\left\{x_{*}\right\}$. Then, the conditions $\left(H_1\right),\left(H_2\right)$ and (4) give in turn

$$\begin{array}{c}\hfill ∥M^{-1}\left([u_0,v_0;F]-M\right)∥\le {\psi }_0\left(\parallel u_0-x_{*}\parallel ,\parallel v_0-x_{*}\parallel \right)\le {\psi }_0(r^{*},r^{*})<1.\end{array}$$

(11)

The estimate (11) and the standard lemma on invertible operators due to Banach [2,4,5,6,8,14,19] give that the linear operator $A_0$ is invertible and

$$\begin{array}{c}\hfill ∥A_0^{-1}M∥\le {\displaystyle \frac{1}{1-{\psi }_0(\parallel u_0-x_{*}\parallel ,\parallel v_0-x_{*}\parallel )}}.\end{array}$$

(12)

The first substep of the method (2) implies

$$\begin{array}{cc}\hfill y_0-x_{*}=& e_0-A_0^{-1}F\left(x_0\right)\hfill \\ \hfill =& A_0^{-1}([u_0,v_0;F]-[x_0,x_{*};F])\left(e_0\right).\hfill \end{array}$$

(13)

It follows in turn by (4), (6), $\left(H_3\right)$, (12) and (13),

$$\begin{array}{cc}\hfill \parallel y_0-x_{*}\parallel \le & {\displaystyle \frac{\psi (\parallel u_0-x_0\parallel ,\parallel v_0-x_{*}\parallel )\parallel e_0\parallel }{1-{\psi }_0(\parallel u_0-x_0\parallel ,\parallel v_0-x_{*}\parallel )}}\hfill \\ \hfill \le & g_1(\parallel e_0\parallel )\parallel e_0\parallel \hfill \\ \hfill \le & \parallel e_0\parallel <r^{*}.\hfill \end{array}$$

So, the iterate $y_0\in \mathbb{B}(x_{*},r^{*})$, and the expression (8) holds if $n=0$.

The second substep of the method (2) gives, as previously,

$$\begin{array}{cc}\hfill z_0-x_{*}=& y_0-x_{*}-A_0^{-1}F\left(y_0\right)=A_0^{-1}(A_0-[y_0,x_{*};F])(y_0-x_{*}).\hfill \\ \hfill \parallel z_0-x_{*}\parallel \le & {\displaystyle \frac{\psi (\parallel u_0-y_0\parallel ,\parallel v_0-x_{*}\parallel )\parallel y_0-x_{*}\parallel }{1-{\psi }_0(\parallel u_0-x_{*}\parallel ,\parallel v_0-x_{*}\parallel )}}\hfill \\ \hfill \le & {\displaystyle \frac{\psi (\parallel u_0-x_{*}\parallel +\parallel y_0-x_{*}\parallel ,\parallel v_0-x_{*}\parallel )\parallel y_0-x_{*}\parallel }{1-{\psi }_0\left(\lambda \right(\parallel e_0\parallel ),\mu (\parallel v_0-x_{*}\parallel \left)\right)}}\hfill \\ \hfill \le & {\displaystyle \frac{\psi ({\lambda }_0+g_1(\parallel e_0\parallel )\parallel e_0\parallel ,{\mu }_0)\parallel y_0-x_{*}\parallel }{1-{\psi }_0\left(\lambda \right(\parallel e_0\parallel ),\mu (\parallel v_0-x_{*}\parallel \left)\right)}}\hfill \\ \hfill \le & g_2(\parallel e_0\parallel )\parallel e_0\parallel \le \parallel e_0\parallel .\hfill \end{array}$$

Hence, the iterate $z_0\in \mathbb{B}(x_{*},r^{*})$, and the assertion (9) holds if $n=0$.

Next, the third substep of the method (2) gives

$$\begin{array}{cc}\hfill x_1-x_{*}=& z_0-x_{*}-\left[3I+A_0^{-1}{B}_0\left(-3I+A_0^{-1}{B}_0\right)\right]A_0^{-1}F\left(z_0\right)\hfill \\ \hfill =& z_0-x_{*}-A_0^{-1}F\left(z_0\right)-\left(2I-3A_0^{-1}{B}_0+A_0^{-1}{B}_0{A}_0^{-1}{B}_0\right)A_0^{-1}F\left(z_0\right)\hfill \\ \hfill =& z_0-x_{*}-A_0^{-1}F\left(z_0\right)-\left[{\left(A_0^{-1}{B}_0-I\right)}^2-\left(A_0^{-1}{B}_0-I\right)\right]A_0^{-1}F\left(z_0\right).\hfill \end{array}$$

(14)

The following estimate is needed:

$$\begin{array}{cc}\hfill ∥A_0^{-1}{B}_0-I∥=& ∥A_0^{-1}(B_0-A_0)∥\le {\displaystyle \frac{\psi (\parallel u_0-z_0\parallel ,\parallel v_0-x_{*}\parallel )}{1-{\psi }_0(\parallel u_0-x_{*}\parallel ,\parallel v_0-x_{*}\parallel )}},\hfill \end{array}$$

(15)

so by (14) and (15),

$$\begin{array}{cc}\hfill \parallel x_1-x_{*}\parallel \le & {\displaystyle \frac{\psi (\parallel u_0-z_0\parallel ,\parallel v_0-x_{*}\parallel )}{1-{\psi }_0(\parallel u_0-x_{*}\parallel ,\parallel v_0-x_{*}\parallel )}}+[{\left({\displaystyle \frac{\psi (\parallel u_0-w_0\parallel ,\parallel v_0-r_0\parallel )}{1-{\psi }_0(\parallel u_0-x_{*}\parallel ,\parallel v_0-x_{*}\parallel )}}\right)}^2\hfill \\ \hfill & +\left({\displaystyle \frac{\psi (\parallel u_0-w_0\parallel ,\parallel v_0-r_0\parallel )}{1-{\psi }_0(\parallel u_0-x_{*}\parallel ,\parallel v_0-x_{*}\parallel )}}\right)\left({\displaystyle \frac{{\lambda }_0^1}{1-{\psi }_0(\parallel u_0-x_{*}\parallel ,\parallel v_0-x_{*}\parallel )}}\right)]\parallel z_0-x_{*}\parallel \hfill \\ \hfill \le & g_3(\parallel e_0\parallel )\parallel e_0\parallel \le \parallel e_0\parallel ,\hfill \end{array}$$

where we also used the conditions $\left(Q\right)$,

$$\parallel u_0-z_0\parallel \le \parallel u_0-x_{*}\parallel +\parallel z_0-x_{*}\parallel$$

and

$$∥M^{-1}F\left(z_0\right)∥=∥M^{-1}[z_0,x_{*};F](z_0-x_{*})∥\le {\lambda }_0^1\parallel z_0-x_{*}\parallel .$$

Hence, the iterate $x_1\in \mathbb{B}(x_{*},r^{*})$ and the assertion (10) holds if $n=0$. The proof for the statements in Equations (7)–(10) can be quickly finished by replacing $x_0,y_0,z_0,x_1$ with $x_m,y_m,z_m,x_{m+1}$ in the previous steps of the calculation. This substitution allows us to apply the same reasoning to the next case in the sequence. Additionally, based on the calculation for $c=g_3(\parallel e_0\parallel )\in [0,1)$, we have

$$\parallel x_{m+1}-x_{*}\parallel \le c\parallel x_m-x_{*}\parallel <r^{*},$$

and it follows that the iterate $x_{m+1}\in \mathbb{B}(x_{*},r^{*})$ and ${\displaystyle \underset{m\to \infty }{lim}{x}_m=x_{*}}$. □

Proposition 1. 

There is a solution $\overline{y}\in \mathbb{B}(x_{*},{\tau }_1)$ to $F\left(x\right)=0$ within a neighborhood of $x_{*}$, specifically in the ball $\mathbb{B}(x_{*},{\tau }_1)$ for some ${\tau }_1>0$ representing a certain radius. Then, we have

$$\begin{array}{c}\hfill ∥M^{-1}\left([x_{*},\overline{y};F]-M\right)∥\le {\overline{\psi }}_0(\parallel \overline{y}-x_{*}\parallel ),\end{array}$$

(16)

where ${\overline{\psi }}_0:[0,+\infty )\to \mathbb{R}$ is a continuous and nondecreasing function and there exists ${\tau }_2\ge {\tau }_1$ such that

$$\begin{array}{c}\hfill {\overline{\psi }}_0\left(s_4\right)<1.\end{array}$$

(17)

Let ${\mathbb{D}}_1=\mathbb{D}\cap \mathbb{B}[x_{*},{\tau }_2]$. Then, $x_{*}$ is the only solution of $F\left(x\right)=0$ in ${\mathbb{D}}_1$.

Proof. 

We consider the linear operator $T=[\overline{y},x_{*};F]$, if $\overline{y}\ne x_{*}$. By applying (16) and (17), we have

$$∥M^{-1}\left(T-M\right)∥\le {\overline{w}}_0(\parallel \overline{y}-x_{*}\parallel )\le {\overline{w}}_0\left(s_4\right)<1.$$

Thus, the inverse of T exists and

$$\overline{y}-x_{*}=T^{-1}\left(F\left(\overline{y}\right)-F\left(x_{*}\right)\right)=T^{-1}\left(0\right)=0.$$

Hence, we obtain $\overline{y}=x_{*}$. □

Remark 1. 

The LAC of the method (3) is analyzed in an analogous way in two interesting cases. Let L be any invertible linear operator.

(I) 

Let $L_n=L$. Suppose that the hypotheses $\left(H_1\right)-\left(H_4\right)$ and $\left(Q\right)$ hold but with $a=-\delta L^{-1}$ and $b=-\gamma L^{-1}$ and the functions ${\overline{g}}_1$ and ${\overline{g}}_2$ are ${\overline{g}}_1\left(t\right)=g_1\left(t\right)$ and ${\overline{g}}_2\left(t\right)=g_2\left(t\right)$. In order to define the corresponding function ${\overline{g}}_3$, notice in turn the calculation

$$\begin{array}{cc}\hfill x_{n+1}-x_{*}=& z_n-x_{*}-K_n^{-1}F\left(z_n\right)\hfill \\ \hfill & -\left((3-2q)\left(K_n^{-1}{B}_n-I\right)+(q-2){\left(K_n^{-1}{B}_n-I\right)}^2\right)K_n^{-1}F\left(z_n\right)\hfill \end{array}$$

leading to

$$\begin{array}{cc}\hfill \parallel x_{n+1}-x_{*}\parallel \le & [{\displaystyle \frac{\psi (\parallel u_n-z_n\parallel ,\parallel v_n-x_{*}\parallel )}{1-{\psi }_0(\parallel u_n-x_{*}\parallel ,\parallel v_n-x_{*}\parallel )}}\hfill \\ \hfill & +\left(\right|q-2|{\left({\displaystyle \frac{\psi (\parallel u_n-w_n\parallel ,\parallel v_n-s_n\parallel )}{1-{\psi }_0(\parallel u_n-x_{*}\parallel ,\parallel v_n-x_{*}\parallel )}}\right)}^2\hfill \\ \hfill & +|3-2q|\left({\displaystyle \frac{\psi (\parallel u_n-w_n\parallel ,\parallel v_n-s_n\parallel )}{1-{\psi }_0(\parallel u_n-x_{*}\parallel ,\parallel v_n-x_{*}\parallel )}}\right))\hfill \\ \hfill & \times {\displaystyle \frac{{\lambda }_n^1}{1-{\psi }_0(\parallel u_n-x_{*}\parallel ,\parallel v_n-x_{*}\parallel )}}]\parallel z_n-x_{*}\parallel .\hfill \end{array}$$

Therefore, we can choose

$$\begin{array}{cc}\hfill {\overline{g}}_3\left(t\right)=& \left[{\displaystyle \frac{\psi \left({\mu }^1\left(t\right)+g_2\left(t\right),\mu \left(t\right)\right)}{1-{\psi }_0\left(\lambda \left(t\right),\mu \left(t\right)\right)}}+\left(|q-2|{\left({\displaystyle \frac{\psi \left(p\left(t\right),p^1\left(t\right)\right)}{1-{\psi }_0\left(\lambda \left(t\right),\mu \left(t\right)\right)}}\right)}^2\right.\right.\hfill \\ \hfill & \left.\left.+|3-2q|\left({\displaystyle \frac{\psi \left(p\left(t\right),p^1\left(t\right)\right)}{1-{\psi }_0\left(\lambda \left(t\right),\mu \left(t\right)\right)}}\right)\right){\displaystyle \frac{{\lambda }_n^1\left(t\right)}{1-{\psi }_0\left(\lambda \left(t\right),\mu \left(t\right)\right)}}\right]{\overline{g}}_2\left(t\right).\hfill \end{array}$$

Let us assume that the equation $g_3\left(t\right)-1=0$ has an SS ${\overline{c}}_3\in (0,r_0)$.

(II) 

Let $L_n={[x_{n-1},x_n;F]}^{-1}$ for each $n=0,1,2,\dots$ Suppose that there exists $\tilde{\gamma }>0$ such that

$$∥M^{-1}∥\le \tilde{\gamma }$$

and

$${\psi }_0(r^{*},r^{*})<1.$$

Then, notice that

$$\begin{array}{cc}\hfill ∥-\delta {[x_{n-1},x_n;F]}^{-1}M{M}^{-1}∥& \le {\displaystyle \frac{\left|\delta \right|\tilde{\gamma }}{1-{\psi }_0(\parallel x_{n-1}-x_{*}\parallel ,\parallel x_n-x_{*}\parallel )}}\hfill \\ \hfill & \le {\displaystyle \frac{\left|\delta \right|\tilde{\gamma }}{1-{\psi }_0(r^{*},r^{*})}}=a,\hfill \end{array}$$

and similarly,

$$\parallel -\gamma {[x_{n-1},x_n;F]}^{-1}M{M}^{-1}\parallel \le {\displaystyle \frac{\left|\delta \right|\tilde{\gamma }}{1-{\psi }_0(r^{*},r^{*})}}=b.$$

Under these choices of $a,b$ and $L_n$, the conclusions of Theorem 1 hold for the method (2).

Remark 2. 

The conditions $\left(Q\right)$ shall be dropped by expressing the scalar sequences in terms of the conditions $\left(H_1\right)-\left(H_4\right)$ and

$$\begin{array}{cc}\hfill & \parallel I-aM\parallel \le a_0,\parallel I+bM\parallel \le b_0,\parallel M\parallel \le a_1,\parallel I-cM\parallel \le c_0,\parallel I+dM\parallel \le d_0,\hfill \\ \hfill & ∥M^{-1}([x,x_{*};F]-M)∥\le {\psi }_1(\parallel x-x_{*}\parallel ),\hfill \end{array}$$

for each $x\in {\mathbb{D}}_0$, $a_0,b_0,a_1,c_0,d_0$ are nonnegative parameters and ${\psi }_1:[0,s_0)\to \mathbb{R}$ is a continuous and nondecreasing function.

We first determine ${\lambda }_n$ and ${\lambda }_0$:

$$\begin{array}{cc}\hfill u_n-x_{*}=& e_n-aF\left(x_n\right)=e_n-a[x_n,x_{*};F]\left(e_n\right)\hfill \\ \hfill =& (I-a[x_n,x_{*};F])\left(e_n\right),\hfill \end{array}$$

but

$$\begin{array}{cc}\hfill I-aM{M}^{-1}([x_n,x_{*};F]-M+M)=& I-aM(M^{-1}([x_n,x_{*};F]-M)+I)\hfill \\ \hfill =& I-aM-aM{M}^{-1}([x_n,x_{*};F]-M),\hfill \end{array}$$

So, we have

$$\begin{array}{cc}\hfill ∥\left(I-a[x_n,x_{*};F]\right)\left(e_n\right)∥\le & \left(\parallel I-aM\parallel +\left|a\right|a_1{\psi }_1(\parallel e_n\parallel )\right)\parallel e_n\parallel \hfill \\ \hfill =& {\lambda }_n.\hfill \end{array}$$

Consequently, we obtain

$$\begin{array}{cc}\hfill \lambda \left(t\right)=& \left(a_0+\left|a\right|a_1{\psi }_1\left(t\right)\right)t,\hfill \\ \hfill \mu \left(t\right)=& \left(b_0+\left|b\right|a_1{\psi }_1\left(t\right)\right)t,\hfill \\ \hfill u_n-x_n=& -aF\left(x_n\right)=-a[x_n,x_{*};F]\left(e_n\right)\hfill \\ \hfill =& \left(-aM{M}^{-1}([x_n,x_{*};F]-M+M)\right)\left(e_n\right)\hfill \\ \hfill =& \left(-aM(M^{-1}([x_n,x_{*};F]-M)+I)\right)\left(e_n\right)\hfill \\ \hfill =& \left(-aM-aM{M}^{-1}([x_n,x_{*};F]-M)\right)\left(e_n\right).\hfill \end{array}$$

Thus,

$$\parallel u_n-x_n\parallel \le \left(a_1\left|a\right|+a_1\left|a\right|{\psi }_1(\parallel e_n\parallel )\right)\parallel e_n\parallel ={\mu }_n^1,$$

and

$${\mu }_1\left(t\right)=a_1\left|a\right|\left(1+{\psi }_1\left(t\right)\right)t,$$

$$\begin{array}{cc}\hfill F\left(z_n\right)=& F\left(z_n\right)-F\left(x_{*}\right)=[z_n,x_{*};F](z_n-x_{*})\hfill \\ \hfill =& M{M}^{-1}\left([z_n,x_{*};F]-M+M\right)(z_n-x_{*})\hfill \end{array}$$

so

$$\begin{array}{cc}\hfill ∥M^{-1}F\left(z_n\right)∥\le & (1+{\psi }_1(\parallel z_n-x_{*}\parallel \left)\right)\parallel z_n-x_{*}\parallel \hfill \\ \hfill \le & \left(1+{\psi }_1\left(g_2(\parallel e_n\parallel )\parallel e_n\parallel \right)\right)g_2(\parallel e_n\parallel )\parallel e_n\parallel \hfill \\ \hfill =& {\lambda }_n^1,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\lambda }^1\left(t\right)=& \left(1+{\psi }_1\left(g_2\left(t\right)t\right)\right)g_2\left(t\right)t,\hfill \\ \hfill w_n-x_{*}=& z_n-x_{*}+dF\left(z_n\right)=\left(I+d[z_n,x_{*};F]\right)(z_n-x_{*}),\hfill \end{array}$$

but

$$I+d[z_n,x_{*};F]=I+dM(M^{-1}\left([z_n,x_{*};F]-M)+I\right).$$

Thus, we obtain

$$\begin{array}{cc}\hfill \parallel w_n-x_{*}\parallel \le & \left(\parallel I+dM\parallel +a_1\left|d\right|{\psi }_1(\parallel z_n-x_{*}\parallel )\right)\parallel z_n-x_{*}\parallel \hfill \\ \hfill \le & \left(d_0+a_1\left|d\right|{\psi }_1\left(g_2(\parallel e_n\parallel )\parallel e_n\parallel \right)\right)g_2(\parallel e_n\parallel )\parallel e_n\parallel \hfill \\ \hfill =& {\xi }_n.\hfill \end{array}$$

In addition, we have

$$\xi \left(t\right)=\left(d_0+a_1\left|d\right|{\psi }_1\left(g_2\left(t\right)t\right)\right)g_2\left(t\right)t,$$

similarly,

$$\parallel e_n\parallel \le \left(\parallel I+cM\parallel +a_1\left|c\right|{\psi }_1(\parallel z_n-x_{*}\parallel )\right)\parallel z_n-x_{*}\parallel ={\rho }_n$$

and

$$\begin{array}{cc}\hfill \rho \left(t\right)=& \left(c_0+a_1\left|c\right|{\psi }_1\left(g_2\left(t\right)t\right)\right)g_2\left(t\right)t,\hfill \\ \hfill u_n-w_n=& (u_n-x_{*})+(x_{*}-w_n),\hfill \\ \hfill \parallel u_n-w_n\parallel \le & \parallel u_n-x_{*}\parallel +\parallel w_n-x_{*}\parallel \le {\mu }_n^1+r_n=p_n,\hfill \end{array}$$

so

$$p\left(t\right)={\mu }^1\left(t\right)+\xi \left(t\right).$$

In a similar way, we obtain

$$\parallel v_n-s_n\parallel \le \parallel v_n-x_{*}\parallel + \parallel s_n-x_{*}\parallel \le {\mu }_n+{\rho }_n=p_n^1.$$

Consequently, we set

$$p^1\left(t\right)=\mu \left(t\right)+\rho \left(t\right).$$

In view of these calculations, the parameter R appearing in condition $\left(H_4\right)$ is defined by

$$R=max\left\{r^{*},\lambda \left(r^{*}\right),\mu \left(r^{*}\right),r\left(r^{*}\right),\rho \left(r^{*}\right)\right\}.$$

(III) 

Possible choices for the linear operator $L_n$ can be

$L_n=-{[x_n,x_{n-1};F]}^{-1}$ (secant [2,4,5,7]);

$L_n=-[2x_n-x_{n-1},x_{n-1};F]$ (Kurchatov [8,10,19]);

$L_n=-[x_n,y_{n-1};F]$;

$L_n=-{[2x_n-y_{n-1},y_{n-1};F]}^{-1}$ (Kurchatov [8,10,19]).

Moreover, for M, we can choose $M=I$ or $M=F^{\prime }\left(x_{*}\right)$ or $M=F^{\prime }\left(x_0\right)$. Other choices exist [4,5,14].

3. Semi-Local Convergence

Sequences are developed that are shown to be majorizing for the sequence $\left\{x_n\right\}$ given by the Formula (2). Let $\lambda ,\mu ,{\varphi }_0$ be functions such that ${\varphi }_0:[0,+\infty )\times [0,+\infty )\to \mathbb{R}$ is continuous and nondecreasing and $\lambda ,\mu :[0,+\infty )\to \mathbb{R}$ are continuous.

Suppose:

• $\left(C_0\right)$ the equation ${\varphi }_0\left(\lambda \left(t\right),\mu \left(t\right)\right)-1=0$ has a positive SS, say $R_0$. Define the scalar sequence $\left\{{\alpha }_n\right\}$, where $\left\{{\rho }_n\right\},\left\{{\sigma }_n\right\},\left\{{\lambda }_n\right\},\left\{{\mu }_n\right\},\left\{p_n\right\},\left\{q_n\right\}$ are given nonnegative sequences for ${\alpha }_0=0$, ${\beta }_0\ge 0$ and each $n=0,1,2,\dots$ by

$$\begin{array}{cc}\hfill {\gamma }_n=& {\beta }_n+{\displaystyle \frac{\varphi ({\rho }_n,{\sigma }_n)({\beta }_n-{\alpha }_n)}{1-{\varphi }_0({\lambda }_n,{\mu }_n)}},\hfill \\ \hfill e_n=& {\displaystyle \frac{\varphi (p_n,q_n)}{1-{\varphi }_0({\lambda }_n,{\mu }_n)}},\hfill \\ \hfill t_n=& \left(1+{\varphi }_0({\gamma }_n,{\beta }_n)\right)({\gamma }_n-{\beta }_n)+\varphi ({\rho }_n,{\sigma }_n)({\beta }_n-{\alpha }_n),\hfill \\ \hfill {\alpha }_{n+1}=& {\gamma }_n+{\displaystyle \frac{(1+e_n+e_n^2)t_n}{1-{\varphi }_0({\lambda }_n,{\mu }_n)}},\hfill \\ \hfill l_{n+1}=& \varphi ({\alpha }_{n+1}-{\beta }_n+{\delta }_n,{\sigma }_n)({\alpha }_{n+1}-{\alpha }_n)+\left(1+{\varphi }_0({\lambda }_n,{\mu }_n)\right)({\alpha }_{n+1}-{\beta }_n),\hfill \\ \hfill {\beta }_{n+1}=& {\alpha }_{n+1}+{\displaystyle \frac{l_{n+1}}{1-{\varphi }_0({\lambda }_{n+1},{\mu }_{n+1})}}.\hfill \end{array}$$

(18)

A convergence proof of $\left\{{\alpha }_n\right\}$ sequence is essential in order to establish the behavior and stability of the sequence as it progresses.

Lemma 1. 

Suppose that there exists $\overline{R}\in [0,R_0]$ such that for $\lambda \left({\alpha }_n\right)={\lambda }_n,\mu \left({\alpha }_n\right)={\mu }_n$ and each $n=0,1,2,\dots ,$

$$\begin{array}{c}\hfill {\varphi }_0({\lambda }_0,{\mu }_n)<1\mathrm{and}{\alpha }_n\le \overline{R}.\end{array}$$

(19)

Then, we have

$$\begin{array}{c}\hfill 0\le {\alpha }_n\le {\beta }_n\le {\gamma }_n\le {\alpha }_{n+1}\le \overline{R}.\end{array}$$

(20)

and there exists $R^{*}\in [0,\overline{R}]$ so that ${\displaystyle \underset{n\to \infty }{lim}{\alpha }_n=R^{*}}$.

Proof. 

The conditions (19) and the Formula (18) imply using induction on the assertion (20), from which the rest of the proof is completed. □

The functions ${\varphi }_0,\varphi$ and the parameter $R_0$ relate to the method (2) as follows:

• Suppose:

(C₁)

There exist a point $x_0\in \mathbb{D}$, a linear operator M which is invertible and

$∥A_0^{-1}F\left(x_0\right)∥\le {\beta }_0$.

(C₂)

$∥M^{-1}([x,y;F]-M)∥\le {\varphi }_0(\parallel x-x_0\parallel ,\parallel y-x_0\parallel )$ for each $x,y\in \mathbb{D}$.

It follows that the linear operator $A_0$ is invertible, since

$\parallel M^{-1}(A_0-M)\parallel \le {\varphi }_0\left(0,\mu \left(t\right)\right)<1$. Hence, we can choose $\parallel A_0^{-1}F\left(x_0\right)\parallel \le {\beta }_0$.

Set ${\mathbb{D}}_0=\mathbb{D}\cap \mathbb{B}(x_0,R_0)$.

(C₃)

$∥M^{-1}([x,y;F]-[z,w;F])∥\le \varphi (\parallel x-z\parallel ,\parallel y-w\parallel )$ for each $x,y,z,w\in {\mathbb{D}}_0$.

(C₄)

$\mathbb{B}[x_0,\overline{R}]\subset \mathbb{D}$, where the parameter $\overline{R}\ge R^{*}$ is specified later.

The following conditions are imposed on the iterates $\left\{x_n\right\}$.

(E)

The iterates $\left\{x_n\right\}$ exist and satisfy for each $n=0,1,2,\dots$, $\left\{x_n\right\}\subset \mathbb{D}$,

$$\begin{array}{cc}\hfill \parallel u_n-x_0\parallel \le & {\overline{\lambda }}_n\le {\lambda }_n<R^{*},\hfill \\ \hfill \parallel v_n-x_0\parallel \le & {\overline{\mu }}_n\le {\mu }_n<R^{*},\hfill \\ \hfill \parallel w_n-x_0\parallel \le & {\overline{\xi }}_n\le {\xi }_n<R^{*},\hfill \\ \hfill \parallel s_n-x_0\parallel \le & {\overline{\rho }}_n<R^{*},\hfill \\ \hfill \parallel y_n-u_n\parallel \le & {\overline{\delta }}_n\le {\delta }_n,\hfill \\ \hfill \parallel v_n-x_n\parallel \le & {\overline{\sigma }}_n\le {\sigma }_n,\hfill \\ \hfill \parallel w_n-u_n\parallel \le & {\overline{p}}_n\le p_n,\hfill \\ \hfill \parallel s_n-v_n\parallel \le & {\overline{q}}_n\le q_n,\hfill \end{array}$$

where $\left\{{\overline{p}}_n\right\},\left\{{\overline{\sigma }}_n\right\},\left\{{\overline{\delta }}_n\right\},\left\{{\overline{q}}_n\right\},\left\{{\overline{\lambda }}_n\right\},\left\{{\overline{\mu }}_n\right\},\left\{{\overline{\xi }}_n\right\},\left\{{\overline{\rho }}_n\right\}$ are nonnegative sequences. In addition, ${\psi }_0,\psi$ and $R_{0j}$ depending on the iterates of the method and $\left\{{\lambda }_n\right\},\left\{{\mu }_n\right\},\left\{{\xi }_n\right\},$$\left\{{\rho }_n\right\},\left\{{\delta }_n\right\},$$\left\{{\sigma }_n\right\},\left\{p_n\right\},\left\{q_n\right\}$ are nonnegative sequences that will later be expressed in terms of the conditions $\left(C_0\right)$–$\left(C_4\right)$.

• The main semi-local result for the method (2) follows.

Theorem 2. 

Given the conditions $\left(C_0\right)$ through $\left(C_4\right)$ and $\left(E\right)$, there exists a solution $x_{*}\in \mathbb{B}[x_0,R^{*}]$ to the expression $F\left(x\right)=0$. Further, the following properties hold:

$$\begin{array}{c}\hfill \parallel y_n-x_n\parallel \le {\beta }_n-{\alpha }_n,\end{array}$$

(21)

$$\begin{array}{c}\hfill \parallel z_n-y_n\parallel \le {\gamma }_n-{\beta }_n,\end{array}$$

(22)

$$\begin{array}{c}\hfill \parallel x_{n+1}-z_n\parallel \le {\alpha }_{n+1}-{\gamma }_n.\end{array}$$

(23)

Proof. 

The assertions (21)–(23) are established by induction. The assertions (21) hold if $n=0$ by (18), (2) and $\left(C_1\right)$, since

$$\parallel y_0-x_0\parallel =∥-A_0^{-1}F\left(x_0\right)∥=∥A_0^{-1}F\left(x_0\right)∥\le {\beta }_0={\beta }_0-{\alpha }_0<R^{*}.$$

Thus, the iterate $y_0\in \mathbb{B}(x_0,R^{*})$. Notice that

$$∥F^{\prime }{\left(x_0\right)}^{-1}(A_n-F^{\prime }\left(x_0\right))∥\le {\varphi }_0(\parallel u_n-x_0\parallel ,\parallel v_n-x_0\parallel )\le {\varphi }_0({\lambda }_n,{\mu }_n)<1,$$

so

$$\begin{array}{c}\hfill ∥A_n^{-1}{F}^{\prime }\left(x_0\right)∥\le {\displaystyle \frac{1}{1-{\varphi }_0({\lambda }_n,{\mu }_n)}},\end{array}$$

(24)

and iterates $y_n,z_n,x_{n+1}$ are well defined.

Then, from the second substep of the iterative scheme (2), we obtain

$$\begin{array}{cc}\hfill z_n-y_n& =-A_n^{-1}F\left(y_n\right)\hfill \\ \hfill & =-A_n^{-1}\left(F\left(y_n\right)-F\left(x_n\right)-A_n(y_n-x_n)\right)\hfill \\ \hfill & =-A_n^{-1}\left([y_n,x_n;F]-A_n\right)(y_n-x_n).\hfill \end{array}$$

(25)

So, by the conditions $\left(C_3\right)$, $\left(E\right)$, (24) and (25), we have

$$\begin{array}{cc}\hfill \parallel z_n-y_n\parallel \le & {\displaystyle \frac{\varphi (\parallel y_n-u_n\parallel ,\parallel x_n-v_n\parallel )\parallel y_n-x_n\parallel }{1-{\varphi }_0(\parallel u_n-x_0\parallel ,\parallel v_n-x_0\parallel )}}\le {\displaystyle \frac{\varphi ({\overline{\delta }}_n,{\overline{\sigma }}_n)\parallel y_n-x_n\parallel }{1-{\varphi }_0({\overline{\lambda }}_n,{\overline{\mu }}_n)}}\hfill \\ \hfill \le & {\displaystyle \frac{\varphi ({\delta }_n,{\sigma }_n)({\beta }_n-{\alpha }_n)}{1-{\varphi }_0({\lambda }_n,{\mu }_n)}}={\gamma }_n-{\beta }_n\hfill \end{array}$$

(26)

and

$$\parallel z_n-x_0\parallel \le \parallel z_n-y_n\parallel +\parallel y_n-x_0\parallel \le {\gamma }_n-{\beta }_n+{\beta }_n-{\alpha }_0={\gamma }_n<R^{*}.$$

Thus, the iterate $z_n\in \mathbb{B}(x_0,R^{*})$ and the assertion (22) holds.

Similarly, from the third substep of the method (2), we obtain in turn, as in the local case,

$$\begin{array}{cc}\hfill x_{n+1}-z_n=& -(3I+A_n^{-1}{B}_n(-3I+A_n^{-1}{B}_n))A_n^{-1}F\left(z_n\right),\hfill \\ \hfill \parallel x_{n+1}-z_n\parallel =& ∥({(A_n^{-1}{B}_n-I)}^2-(A_n^{-1}{B}_n-I)+I)A_n^{-1}F\left(z_n\right)∥\hfill \\ \hfill \le & {\displaystyle \frac{(1+{\overline{e}}_n+{\overline{e}}_n^2){\overline{t}}_n}{1-{\varphi }_0({\lambda }_n,{\mu }_n)}}\le {\displaystyle \frac{(1+e_n+e_n^2)t_n}{1-{\varphi }_0({\lambda }_n,{\mu }_n)}}={\alpha }_{n+1}-{\gamma }_n\hfill \end{array}$$

(27)

and

$$\parallel x_{n+1}-x_0\parallel \le \parallel x_{n+1}-z_n\parallel +\parallel z_n-x_0\parallel \le {\alpha }_{n+1}-{\gamma }_n+{\gamma }_n-{\alpha }_0={\alpha }_{n+1}<R^{*},$$

where we used

$$\begin{array}{cc}\hfill F\left(z_n\right)=& F\left(z_n\right)-F\left(y_n\right)+F\left(y_n\right)\hfill \\ \hfill & =[z_n,y_n;F](z_n-y_n)+F\left(y_n\right)-F\left(x_n\right)-A_n(y_n-x_n)\hfill \\ \hfill =& [z_n,y_n;F](z_n-y_n)+([y_n,x_n;F]-A_n)(y_n-x_n),\hfill \\ \hfill ∥M^{-1}F\left(z_n\right)∥\le & (1+{\varphi }_0(\parallel z_n-x_0\parallel ,\parallel y_n-x_0\parallel \left)\right)\parallel z_n-y_n\parallel +\varphi (\parallel y_n-u_n\parallel ,\parallel x_n-v_n\parallel )\parallel y_n-x_n\parallel \hfill \\ \hfill =& {\overline{t}}_n\le (1+{\varphi }_0({\gamma }_n,{\beta }_n))({\gamma }_n-{\beta }_n)+\varphi ({\delta }_n,{\sigma }_n)({\beta }_n-{\alpha }_n)=t_n\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill ∥A_n^{-1}{B}_n-I∥=& ∥A_n^{-1}(B_n-A_n)∥=∥A_n^{-1}([w_n,s_n;F]-[u_n,v_n;F])∥\hfill \\ \hfill \le & {\displaystyle \frac{\varphi (\parallel w_n-u_n\parallel ,\parallel s_n-v_n\parallel )}{1-{\varphi }_0({\lambda }_n,{\mu }_n)}}={\overline{e}}_n\le {\displaystyle \frac{\varphi (p_n,q_n)}{1-{\varphi }_0({\lambda }_n,{\mu }_n)}}=e_n.\hfill \end{array}$$

Therefore, the iterate $x_{n+1}\in \mathbb{B}(x_0,R^{*})$ and the assertion (23) holds.

Using the first substep of method (2), we can express

$$\begin{array}{cc}\hfill F\left(x_{n+1}\right)=& F\left(x_{n+1}\right)-F\left(x_n\right)-A_n(y_n-x_n)\hfill \\ \hfill =& ([x_{n+1},x_n;F]-A_n)(x_{n+1}-x_n)+A_n(x_{n+1}-y_n)\hfill \end{array}$$

leading to

$$\begin{array}{cc}\hfill ∥M^{-1}F\left(x_{n+1}\right)∥\le & \varphi (\parallel x_{n+1}-u_n\parallel ,\parallel x_n-v_n\parallel )\parallel x_{n+1}-x_n\parallel \hfill \\ \hfill & +(1+{\varphi }_0(\parallel u_n-x_0\parallel ,\parallel v_n-x_0\parallel \left)\right)\parallel x_{n+1}-y_n\parallel ={\overline{\ell }}_{n+1}\hfill \\ \hfill \le & \varphi (\parallel x_{n+1}-y_n\parallel +\parallel y_n-u_n\parallel ,\parallel x_n-v_n\parallel )\parallel x_{n+1}-x_n\parallel \hfill \\ \hfill & +(1+{\varphi }_0({\lambda }_n,{\mu }_n))({\alpha }_{n+1}-{\beta }_n)\hfill \\ \hfill =& \varphi ({\alpha }_{n+1}-{\beta }_n+{\delta }_n,{\sigma }_n)({\alpha }_{n+1}-{\alpha }_n)+(1+{\varphi }_0({\lambda }_n,{\mu }_n))({\alpha }_{n+1}-{\beta }_n)\hfill \\ \hfill =& {\ell }_{n+1}.\hfill \end{array}$$

(28)

Hence, we have in turn

$$\begin{array}{cc}\hfill \parallel y_{n+1}-x_{n+1}\parallel \le & ∥A{\left(x_{n+1}\right)}^{-1}{F}^{\prime }\left(x_0\right)∥∥F^{\prime }{\left(x_0\right)}^{-1}F\left(x_{n+1}\right)∥\hfill \\ \hfill \le & {\displaystyle \frac{{\overline{e}}_{n+1}}{1-{\varphi }_0({\lambda }_n,{\mu }_n)}}\le {\displaystyle \frac{e_{n+1}}{1-{\varphi }_0({\lambda }_n,{\mu }_n)}}={\beta }_{n+1}-{\alpha }_{n+1}\hfill \end{array}$$

(29)

and

$$\begin{array}{cc}\hfill \parallel y_{n+1}-x_0\parallel \le & \parallel y_{n+1}-x_{n+1}\parallel +\parallel x_{n+1}-x_0\parallel \hfill \\ \hfill \le & {\beta }_{n+1}-{\alpha }_{n+1}+{\alpha }_{n+1}-{\alpha }_0={\beta }_{n+1}<R^{*}.\hfill \end{array}$$

The proof by induction for the assertions in Equations (21)–(23) has been completed, and it is shown that the sequence $\left\{x_n\right\}$ is contained within the ball $\mathbb{B}(x_0,R^{*})$. As a result, the sequence $\left\{x_n\right\}$ is a Cauchy sequence in the Banach space $\mathbb{X}$. Since every Cauchy sequence in a Banach space converges, it follows that $\left\{x_n\right\}$ converges to some limit $x_{*}$ lying in the ball $\mathbb{B}[x_0,R^{*}]$. Notice that ${\displaystyle \underset{n\to \infty }{lim}{\mathit{\ell }}_{n+1}=0}$. Hence, we obtain $F\left(x_{*}\right)=0$ by (28) and the continuity of F.

Finally, from the estimate

$$\parallel x_{n+m}-x_n\parallel \le {\alpha }_{n+m}-{\alpha }_n,$$

we obtain

$$\begin{array}{c}\hfill \parallel x_{*}-x_n\parallel \le {\alpha }^{*}-{\alpha }_n\end{array}$$

(30)

provided that $m\to +\infty$. □

The region of uniqueness for the solution is then identified and defined. This means that we will now establish the specific area or set within which the solution is guaranteed to be unique.

Proposition 2. 

Assume the following:

• There exists a solution $\overline{z}\in \mathbb{B}(x_0,k_3)$ to $F\left(x\right)=0$ for some $k_3>0$;
• The condition $\left(C_2\right)$ is satisfied within the ball $\mathbb{B}(x_0,k_3)$;
• There exists a constant $k_4\ge k_3$ such that

$${\varphi }_0(k_3,k_4)<1.$$

(31)

Set ${\mathbb{D}}_2=\mathbb{D}\cap \mathbb{B}[x_0,k_4]$.

Then, there are no other points within ${\mathbb{D}}_2$ satisfying the equation $F\left(x\right)=0$, ensuring that $\overline{z}$ is the unique root in this domain.

Proof. 

Let $z^{*}\in {\mathbb{D}}_2$ with $F\left(z^{*}\right)=0$ with $z^{*}=\overline{z}$. Define the linear operator $M_1=[\overline{z},z^{*};F]$. In view of condition $\left(C_2\right)$ and (31), one obtains

$$\parallel M^{-1}\left(M_1-F^{\prime }\left(x_0\right)\right)\parallel \le {\varphi }_0(\parallel \overline{z}-x_0\parallel ,\parallel z^{*}-x_0\parallel )\le {\varphi }_0(k_3,k_4)<1.$$

Thus, we deduce that $\overline{z}=z^{*}.$ □

Remark 3. 

Under the full set of conditions outlined in Theorem 2, which include all the necessary assumptions and requirements, set $z^{*}=x_{*}$ and $k_3=R^{*}$.

Remark 4. 

Replace the conditions $\left(E\right)$ by $\left(E^{\prime }\right)$ given by

(E′)

$\parallel I-aM\parallel \le \overline{\alpha }$,$\parallel F\left(x_0\right)\parallel \le \beta$, $\parallel M\parallel \le \overline{\gamma }$,$\parallel I+bM\parallel \le h$, $\parallel I-cM\parallel \le \mu$,

$\parallel I+dM\parallel \le h$,$\parallel I-(a+c)M\parallel \le l$ and $\parallel I+(b+d)M\parallel \le s$.

Next, the scalar sequences appearing in the condition $\left(E\right)$ are specialized. The following calculations are required:

$$\begin{array}{cc}\hfill u_n-x_0=& x_n-x_0-a\left(F\left(x_n\right)-F\left(x_0\right)\right)-aF\left(x_0\right)\hfill \\ \hfill =& (I-a[x_n,x_0;F])(x_n-x_0)-aF\left(x_0\right),\hfill \\ \hfill \parallel u_n-x_0\parallel \le & \parallel I-a[x_n,x_0;F]\parallel \parallel x_n-x_0\parallel +|a\parallel |F\left(x_0\right)\parallel \hfill \\ \hfill =& \parallel I-aM{M}^{-1}([x_n,x_0;F]-M+M)\parallel +|a\parallel |F\left(x_0\right)\parallel \hfill \\ \hfill \le & \left(\parallel I-aM\parallel +\left|a\right|\parallel M\parallel {\varphi }_0(0,\parallel x_n-x_0\parallel )\right)\parallel x_n-x_0\parallel +\left|a\right|\parallel F\left(x_0\right)\parallel \hfill \\ \hfill \le & (\overline{\alpha }+\left|a\right|\overline{\gamma }{\varphi }_0(0,\parallel x_n-x_0\parallel \left)\right)\parallel x_n-x_0\parallel +\left|a\right|\beta :={\overline{\lambda }}_n,\hfill \end{array}$$

and thus,

$${\lambda }_n:=(\overline{\alpha }+\left|a\right|\gamma {\varphi }_0(0,t_n))t_n+\left|a\right|\beta .$$

In a similar way, we have

$$\begin{array}{cc}\hfill v_n-x_0=& \left(I+b[x_n,x_0;F]\right)(x_n-x_0)+bF\left(x_0\right),\hfill \\ \hfill \parallel v_n-x_0\parallel \le & \left(\parallel I+bM\parallel +\left|b\right|\parallel M\parallel {\varphi }_0(0,\parallel x_n-x_0\parallel )\right)\parallel x_n-x_0\parallel +\left|b\right|\parallel F\left(x_0\right)\parallel \hfill \\ \hfill \le & \left(h+\left|b\right|\overline{\gamma }{\varphi }_0(0,\parallel x_n-x_0\parallel )\right)\parallel x_n-x_0\parallel +\left|b\right|\beta ={\overline{\mu }}_n,\hfill \end{array}$$

so, we obtain

$$\begin{array}{cc}\hfill {\mu }_n=& (h+|b|\overline{\gamma }{\varphi }_0(0,t_n))t_n+\left|b\right|\beta ;\hfill \\ \hfill v_n-x_n=& bF\left(x_n\right)=b\left(F\left(x_n\right)-F\left(x_0\right)+F\left(x_0\right)\right)\hfill \\ \hfill =& b[x_n,x_0;F](x_n-x_0)+bF\left(x_0\right)\hfill \\ \hfill =& bM{M}^{-1}([x_n,x_0;F]-M+M)(x_n-x_0)+bF\left(x_0\right)\hfill \\ \hfill =& bM\left(M^{-1}([x_n,x_0;F]-M)+I\right)(x_n-x_0)+bF\left(x_0\right)\hfill \\ \hfill =& bF\left(x_0\right)+\left(bM+bM{M}^{-1}([x_n,x_0;F]-M)\right)(x_n-x_0),\hfill \end{array}$$

$$\begin{array}{cc}\hfill \parallel v_n-x_n\parallel \le & \left|b\right|\beta +\left(\right|b|\gamma +\left|b\right|\overline{\gamma }{\varphi }_0(0,\parallel x_n-x_0\parallel \left)\right)\parallel x_n-x_0\parallel ={\overline{\sigma }}_n.\hfill \end{array}$$

Thus, we have

$$\begin{array}{cc}\hfill {\sigma }_n=& \left|b\right|\beta +\left(\right|b|\overline{\gamma }+\left|b\right|\overline{\gamma }{\varphi }_0(0,t_n))t_n;\hfill \\ \hfill w_n-u_n=& \left(I-(a+c)[z_n,x_0;F]\right)(z_n-x_n)-c[x_n,x_0;F](x_n-x_0)-cF\left(x_0\right)\hfill \\ \hfill & +a\left(F\left(z_n\right)-F\left(x_0\right)\right)+aF\left(x_0\right)\hfill \\ \hfill =& \left((I-(a+c)M)-(a+c)M{M}^{-1}\left([z_n,x_0;F]-M\right)\right)(z_n-x_n)\hfill \\ \hfill & -cM{M}^{-1}\left([x_n,x_0;F]-M+M\right)(x_n-x_0)\hfill \\ \hfill & +aM{M}^{-1}\left([z_n,x_0;F]-M+M\right)(z_n-x_0)+(a-c)F\left(x_0\right),\hfill \end{array}$$

which further yields

$$\begin{array}{cc}\hfill \parallel w_n-u_n\parallel \le & \left(l+|a+c|\overline{\gamma }{\varphi }_0(0,\parallel z_n-x_0\parallel )\right)\parallel z_n-x_n\parallel \hfill \\ \hfill & +\left(\left|c\right|\gamma +\left|c\right|\overline{\gamma }{\varphi }_0(0,\parallel x_n-x_0\parallel )\right)\parallel x_n-x_0\parallel \hfill \\ \hfill & +\left(\left|a\right|\overline{\gamma }+\left|a\right|\overline{\gamma }{\varphi }_0(0,\parallel z_n-x_0\parallel )\right)\parallel z_n-x_0\parallel +|a-c|\parallel F\left(x_0\right)\parallel ={\overline{p}}_n.\hfill \end{array}$$

Therefore, we obtain

$$\begin{array}{cc}\hfill p_n=& \left(l+|a+c|\overline{\gamma }{\varphi }_0(0,{\gamma }_n)\right)({\gamma }_n-{\alpha }_n)+(\left|c\right|\overline{\gamma }\hfill \\ \hfill & +\left|c\right|{\varphi }_0(0,t_n))\left(t_n+\left(\left|a\right|\overline{\gamma }+\left|a\right|\overline{\gamma }{\varphi }_0(0,{\gamma }_n)\right)\right){\gamma }_n\hfill \\ \hfill & +|a-c|\beta ,\hfill \\ \hfill s_n-v_n=& \left(I+(b+d)M+(b+d)M{M}^{-1}\left([z_n,x_0;F]-M\right)\right)(z_n-x_n)\hfill \\ \hfill & +d\left(F\left(x_n\right)-F\left(x_0\right)\right)-b\left(F\left(z_n\right)-F\left(x_0\right)\right)+(d-b)F\left(x_0\right)\hfill \\ \hfill =& \left(\left(I+(b+d)M\right)+(b+d)M{M}^{-1}\left([z_n,x_0;F]-M\right)\right)(z_n-x_n)\hfill \\ \hfill & +dM{M}^{-1}\left([x_n,x_0;F]-M+M\right)(x_n-x_0)\hfill \\ \hfill & -bM{M}^{-1}\left([z_n,x_0;F]-M+M\right)(z_n-x_0)+(d-b)F\left(x_0\right).\hfill \end{array}$$

In the similar fashion, we have

$$\begin{array}{cc}\hfill \parallel w_n-x_0\parallel =& \parallel \left(I-cM-cM\left(M^{-1}\left([x_n,x_0;F]-M\right)\right)(x_n-x_0)-cF\left(x_0\right)\right)\parallel \hfill \\ \hfill \le & \left(\mu +\left|c\right|\overline{\gamma }{\varphi }_0(0,\parallel x_n-x_0\parallel )\right)\parallel x_n-x_0\parallel +\left|c\right|\beta ={\overline{\xi }}_n,\hfill \end{array}$$

so

$${\xi }_n=\left(\mu +\left|c\right|\overline{\gamma }{\varphi }_0(0,t_n)\right)t_n+\left|c\right|\beta .$$

Moreover, we obtain

$$\parallel s_n-x_0\parallel \le \left(\parallel I+dM\parallel +\left|d\right|\overline{\gamma }{\varphi }_0(0,\parallel x_n-x_0\parallel )\right)\parallel x_n-x_0\parallel +\left|d\right|\beta ={\overline{\xi }}_n$$

and

$${\rho }_n=\left(h+\left|d\right|\overline{\gamma }{\varphi }_0(0,t_n)\right)t_n+\left|d\right|\beta ,$$

$$\begin{array}{cc}\hfill y_n-u_n=& y_n-x_n+aF\left(x_n\right)=y_n-x_n-a{A}_n(y_n-x_n)=(I-a{A}_n)(y_n-x_n)\hfill \\ \hfill =& \left((I-aM)+(I-aM)M^{-1}(A_n-M)+M^{-1}(A_n-M)\right)(y_n-x_n),\hfill \\ \hfill \parallel y_n-u_n\parallel \le & \left(\alpha +\alpha {\varphi }_0(\parallel u_n-x_0\parallel ,\parallel v_n-x_0\parallel )+{\varphi }_0(\parallel u_n-x_0\parallel ,\parallel v_n-x_0\parallel )\right)\parallel y_n-x_n\parallel \hfill \\ \hfill \le & \left(\alpha +\alpha {\varphi }_0({\overline{\lambda }}_n,{\overline{\mu }}_n)+{\varphi }_0({\overline{\lambda }}_n,{\overline{\mu }}_n)\right)\parallel y_n-x_n\parallel ={\overline{\delta }}_n,\hfill \end{array}$$

so

$$\begin{array}{cc}\hfill s_n=& \left(\alpha +\alpha {\varphi }_0({\lambda }_n,{\mu }_n)+{\varphi }_0({\lambda }_n,{\mu }_n)\right)({\beta }_n-{\alpha }_n),\hfill \\ \hfill \parallel s_n-v_n\parallel \le & \left(s+|b+d|\overline{\gamma }{\varphi }_0(0,\parallel z_n-x_n\parallel )\right)\parallel z_n-x_n\parallel +\left|d\right|\overline{\gamma }{\varphi }_0(0,\parallel x_n-x_0\parallel )\parallel x_n-x_0\parallel \hfill \\ \hfill & +\left|b\right|{\varphi }_0(0,\parallel z_n-x_0\parallel )\parallel z_n-x_0\parallel +\left|d\right|\overline{\gamma }\parallel x_n-x_0\parallel +\left|b\right|\overline{\gamma }\parallel z_n-x_0\parallel \hfill \\ \hfill & +|d-b|\beta ={\overline{q}}_n.\hfill \end{array}$$

Hence, we obtain

$$\begin{array}{cc}\hfill q_n=& \left(s+|b+d|\overline{\gamma }{\varphi }_0(0,{\gamma }_n)\right){\gamma }_n+\left|d\right|\overline{\gamma }{\varphi }_0(0,t_n)t_n\hfill \\ \hfill & +\left|b\right|{\varphi }_0(0,{\gamma }_n){\gamma }_n+\left|d\right|\overline{\gamma }{t}_n+\left|b\right|\overline{\gamma }{\gamma }_n+|d-b|\beta .\hfill \end{array}$$

Under these choices,

$$\overline{R}=max\left\{R^{*},\lambda \left(R^{*}\right),\mu \left(R^{*}\right),r\left(R^{*}\right),\rho \left(R^{*}\right)\right\}.$$

Remark 5. 

Regarding the SLAC of the method (3), let us examine two cases as in the local case. Suppose $\parallel M^{-1}\parallel \le \tilde{\tilde{\gamma }}$.

(I) 

$L_n=L^{-1}$. Then, set $a=-\delta L^{-1}$ and $b=-\gamma L^{-1}$. This way, we have the following expressions from the third substep of (2):

$$\begin{array}{cc}\hfill x_{n+1}-z_n=& -\left[qI+{\overline{K}}_n^{-1}{B}_n\left((3-2q)I+(q-2){\overline{K}}_n^{-1}{B}_n\right)\right]{\overline{A}}_n^{-1}F\left(z_n\right)\hfill \\ \hfill =& -\left[(q-2){({\overline{K}}_n^{-1}{B}_n-I)}^2-\left({\overline{K}}_n^{-1}{B}_n-I\right)+I\right]{\overline{K}}_n^{-1}F\left(z_n\right),\hfill \\ \hfill \parallel x_{n+1}-z_n\parallel =& \left(|q-z|e_n^2+e_n+1\right){\displaystyle \frac{t_n}{1-{\varphi }_0({\lambda }_n,{\mu }_n)}}={\alpha }_{n+1}-{\gamma }_n.\hfill \end{array}$$

The iterates ${\gamma }_n,{\beta }_{n+1}$ are the same as in the method (2).

(II) 

$L_n={[x_{n-1},x_n;F]}^{-1}$. Notice that

$$\begin{array}{cc}\hfill ∥-\delta {[x_{n-1},x_n;F]}^{-1}{F}^{\prime }\left(x_0\right)F^{\prime }{\left(x_0\right)}^{-1}∥\le & {\displaystyle \frac{\left|\delta \right|\tilde{\tilde{\gamma }}}{1-{\varphi }_0(\parallel x_{n-1}-x_0\parallel ,\parallel x_n-x_0\parallel )}}={\overline{a}}_n\hfill \\ \hfill \le & {\displaystyle \frac{\left|\delta \right|\tilde{\tilde{\gamma }}}{1-{\varphi }_0(R,R)}}=a,\hfill \end{array}$$

similarly,

$$∥-\gamma {[x_{n-1},x_n;F]}^{-1}{F}^{\prime }\left(x_0\right)F^{\prime }{\left(x_0\right)}^{-1}∥\le b,$$

where $b={\displaystyle \frac{\left|\gamma \right|\tilde{\tilde{\gamma }}}{1-{\varphi }_0(R,R)}}$.

Then, under cases (I) and (II), the conclusions of Theorem 2 hold for the method (3).

4. Numerical Experiments

4.1. Local Area Convergence

In Example 1, we examined the LAC and presented the computational results in Table 1. These findings were derived from a system of differential equations, with a focus on analyzing LAC.

4.2. Semi-Local Area of Convergence

On the other hand, the other examples illustrate the SLAC. Table 2 provides the numerical results of SLAC for the boundary value problem in Example 2. We have chosen a larger SNE of order $60\times 60$ with 60 variables. In Example 3, we investigate another applied science problem, namely the Hammerstein integral equation, to demonstrate the applicability and efficacy of method (2). The values of abscissas $t_j$ and weights $w_j$ are depicted in Table 3 and numerical outcomes in Table 4. In the last Example 4, we consider an SNE, and the numerical results are shown in Table 5. We compared method (2) with existing sixth-order methods, selecting the following approaches for evaluation: method (8) from Abbasbandy et al. [24], method (14) from Hueso et al. [25], and method (6) from Wang and Li [26]. Finally, we included method (5) proposed by Lotfi et al. [27] in our comparison.

Moreover, we study the computational order of convergence $\left(COC\right)$ that has been determined by the following formulas:

$$\eta ={\displaystyle \frac{ln\frac{\parallel x_{\beta +1}-x_{*}\parallel }{|x_{\beta }-x_{*}\parallel }}{ln\frac{\parallel x_{\beta }-x_{*}\parallel }{\parallel x_{\beta -1}-x_{*}\parallel }}},for\beta =1,2,\dots$$

or approximated COC $\left(ACOC\right)$ [15,16] by:

$${\eta }^{*}={\displaystyle \frac{ln\frac{\parallel x_{\beta +1}-x_{\beta }\parallel }{\parallel x_{\beta }-x_{\beta -1}\parallel }}{ln\frac{\parallel x_{\beta }-x_{\beta -1}\parallel }{\parallel x_{\beta -1}-x_{\beta -2}\parallel }}},for\beta =2,3,\dots$$

The conditions for terminating the program are specified as follows:

(i)

The difference between successive iterations satisfies $\parallel x_{\beta +1}-x_{\beta }\parallel <ϵ;$

(ii)

The norm of the operator at the current point meets the condition $\parallel J\left(x_{\beta }\right)\parallel <ϵ$.

Here $ϵ={10}^{-300}$, represents an extremely small tolerance level, ensuring high precision and stability in the computational results. All computations used Mathematica 11 with multi-precision arithmetic, and the computer’s configuration details are listed below. Device Name: HP; Windows 10 Enterprise; OS Build: 19045.2006; Processor: Intel(R) Core(TM) i7-4790 CPU @ 3.60GHz 3.60 GHz; Installed RAM: 8.00 GB (7.89 GB usable); System type: 64-bit operating system, x64-based processor; Version: 22H2.

Example 1. 

Let us examine the following system of differential equations:

$$\begin{array}{cc}\hfill & l_1^{\prime }\left(\lambda \right)-l_1\left(\lambda \right)-1=0\hfill \\ \hfill & l_2^{\prime }\left(\eta \right)-(e-1)\eta -1=0\hfill \\ \hfill & l_3^{\prime }\left(\theta \right)-1=0,\hfill \end{array}$$

(32)

which are defined as the movement of a particle in three dimensions with $\lambda ,\eta ,\theta \in \Gamma$ for $l_1\left(0\right)=l_2\left(0\right)=l_3\left(0\right)=0$. The solution $v={(\lambda ,\eta ,\theta )}^{tr}$ is then associated with $\overline{l}=(l_1,l_2,l_3):\Gamma \to {\mathbb{R}}^3$ given as

$$L\left(v\right)={\left(e^{\lambda }-1,{\displaystyle \frac{e-1}{2}}{\eta }^2+\eta ,\theta \right)}^{tr}.$$

(33)

By (33), we obtain

$$L^{\prime }\left(v\right)=\left[\begin{array}{ccc}{e}^{\lambda }& 0& 0\\ 0& (e-1)\eta +1& 0\\ 0& 0& 1\end{array}\right].$$

For $A=L^{\prime }\left(\overline{l}\right)=L^{\prime }\left({(0,0,0)}^{tr}\right)=I$ and $[y,x;L]={\int }_0^1{L}^{\prime }\left(x+\theta (y-x)\right)d\theta$, we have

$$\begin{array}{cc}\hfill {\varphi }_0(s,t)& ={\displaystyle \frac{1}{2}}(e-1)(s+t),\varphi (s,t)={\displaystyle \frac{1}{2}}\left(e^{{\displaystyle \frac{1}{e-1}}}s+(e-1)t\right)\hfill \\ \hfill {\varphi }_1\left(t\right)& ={\displaystyle \frac{1}{2}}(e-1)t,a_0=|1-a|,b_0=|1+b|,c_0=|1-c|\mathrm{and}{d}_0=|1+d|.\hfill \end{array}$$

Then, compute $r^{*}$ for both methods (see Remarks 1 and 2). For the selection of the function in the condition $\left(Q\right)$, other parameters are given below:

$$a_1=a=b=c=d=\gamma =\delta =1,\mathit{and}q=3.$$

(34)

The radii of convergence methods (2) and (3) for Example 1 are depicted in Table 1.

Table 1. Distinct convergence radii for Example 1.

Methods	${\mathit{c}}_0$	${\mathit{c}}_1$	${\mathit{c}}_2$	${\mathit{c}}_3$	${\overline{\mathit{c}}}_3$	$min\{{\mathit{c}}_0,{\mathit{c}}_1,{\mathit{c}}_2,{\mathit{c}}_3,{\overline{\mathit{c}}}_3\}$
Method (2)	0.2616	0.1536	0.1393	0.1164	-	0.1164
Method (3)	0.2616	0.1536	0.1393	-	0.1125	0.1125

Table 1. Distinct convergence radii for Example 1.

Methods	${\mathit{c}}_0$	${\mathit{c}}_1$	${\mathit{c}}_2$	${\mathit{c}}_3$	${\overline{\mathit{c}}}_3$	$min\{{\mathit{c}}_0,{\mathit{c}}_1,{\mathit{c}}_2,{\mathit{c}}_3,{\overline{\mathit{c}}}_3\}$
Method (2)	0.2616	0.1536	0.1393	0.1164	-	0.1164
Method (3)	0.2616	0.1536	0.1393	-	0.1125	0.1125

Example 2. 

Boundary value problems (BVPs) [17] are fundamental in applied science, involving differential equations with conditions at distinct points, often boundaries. These issues represent real-world processes such as fluid dynamics, nuclear reactors, heat transfer, optimization, quantum mechanics, and applied science, leading us to choose the following BVP (for details, see [18]):

$$v^{″}+a^2{v}^{\prime 2}+1=0$$

(35)

with $v\left(0\right)=0,v\left(1\right)=1$. Divide the interval $[0,1]$ into ℓ parts, which further provides

$${\gamma }_0=0<{\gamma }_1<{\gamma }_2<\cdots <{\gamma }_{\ell -1}<{\gamma }_{\ell },{\gamma }_{\ell +1}={\gamma }_{\ell }+j,j={\displaystyle \frac{1}{\ell }}.$$

Set $v_0=v\left({\gamma }_0\right)=0,v_1=v\left({\gamma }_1\right),\dots ,v_{\ell -1}=v\left({\gamma }_{\ell -1}\right),v_{\ell }=v\left({\gamma }_{\ell }\right)=1$. Then, we have

$$v_{\tau }^{\prime }={\displaystyle \frac{v_{\tau +1}-v_{\tau -1}}{2j}},v_{\tau }^{″}={\displaystyle \frac{v_{\tau -1}-2v_{\tau }+v_{\tau +1}}{j^2}},\tau =1,2,3,\dots ,p-1,$$

by discretization. Thus, we obtain a system of $(\tau -1)\times (\tau -1)$

$$v_{\tau -1}-2v_{\tau }+v_{\tau +1}+{\displaystyle \frac{{\mu }^2}{4}}{(v_{\tau +1}-v_{\tau -1})}^2+j^2=0.$$

(36)

For example, we have an SNE of order $60\times 60$ with $p=61$ and $\mu ={\displaystyle \frac{1}{2}}$. The following is the required solution for the system (36) $x_{*}$ mentioned above:

$$\begin{array}{cc}\hfill x_{*}=(& 0.02757\cdots ,0.05468\cdots ,0.08134\cdots ,0.1076\cdots ,0.1333\cdots ,0.1587\cdots ,0.1836\cdots ,\hfill \\ \hfill & 0.2081\cdots ,0.2322\cdots ,0.2559\cdots ,0.2791\cdots ,0.3020\cdots ,0.3245\cdots ,\hfill \\ \hfill & 0.3465\cdots ,0.3682\cdots ,0.3895\cdots ,0.4104\cdots ,0.4310\cdots ,0.4511\cdots ,\hfill \\ \hfill & 0.4710\cdots ,0.4904\cdots ,0.5095\cdots ,0.5282\cdots ,0.5466\cdots ,0.5646\cdots ,\hfill \\ \hfill & 0.5822\cdots ,0.5996\cdots ,0.6165\cdots ,0.6332\cdots ,0.6495\cdots ,0.6654\cdots ,,\hfill \\ \hfill & 0.6811\cdots ,0.6964\cdots ,0.7114\cdots ,0.7260\cdots ,0.7404\cdots ,0.7544\cdots ,,\hfill \\ \hfill & 0.7681\cdots ,0.7815\cdots ,0.7945\cdots ,0.8073\cdots ,0.8198\cdots ,0.8319\cdots ,,\hfill \\ \hfill & 0.8437\cdots ,0.8553\cdots ,0.8665\cdots ,0.8775\cdots ,0.8881\cdots ,0.8985\cdots ,,\hfill \\ \hfill & 0.9085\cdots ,0.9183\cdots ,0.9277\cdots ,0.9369\cdots ,0.9458\cdots ,0.9544\cdots ,,\hfill \\ \hfill & 0.9627\cdots ,0.9707\cdots ,0.9785\cdots ,0.9859\cdots ,0.9931\cdots {)}^{tr}\hfill \end{array}$$

Table 2 provides a detailed overview of the performance for Example 2, including the Coefficient of Convergence (COC), CPU time, number of iterations, residual errors, and the error differences between consecutive iterations. These computational results offer us valuable insight into the efficiency and accuracy of the required solution. This table provide us both computational time and the convergence behavior as the iterative methods with other existing methods.

Table 2. Computational results of Example 2.

Methods	${\mathit{x}}_0$	$\parallel \mathit{F}\left({\mathit{x}}_3\right)\parallel$	$\parallel {\mathit{x}}_4-{\mathit{x}}_3\parallel$	n	${\mathit{\eta }}^{*}$	CPU Timing
Lotfi et al. [27]	${(1.1,1.1,\stackrel{60}{\cdots },1.1)}^{tr}$	$1.2\times {10}^{-134}$	$4.7\times {10}^{-133}$	4	5.0541	135.752
Wang and Li [26]	${(1.1,1.1,\stackrel{60}{\cdots },1.1)}^{tr}$	$2.1\times {10}^{-177}$	$1.2\times {10}^{-176}$	4	6.0281	127.473
Abbasbandy [24]	${(1.1,1.1,\stackrel{60}{\cdots },1.1)}^{tr}$	$7.6\times {10}^{-168}$	$4.3\times {10}^{-167}$	4	6.0293	347.9
Hueso et al. [25]	${(1.1,1.1,\stackrel{60}{\cdots },1.1)}^{tr}$	$6.6\times {10}^{-125}$	$3.9\times {10}^{-124}$	4	5.0358	222.306
Method (2)	${(1.1,1.1,\stackrel{60}{\cdots },1.1)}^{tr}$	$3.2\times {10}^{-182}$	$1.7\times {10}^{-181}$	4	6.0616	104.265

CPU timing and ${\eta }^{*}$ are calculated based on the number of iterations required to reach the desired accuracy.

Table 2. Computational results of Example 2.

Methods	${\mathit{x}}_0$	$\parallel \mathit{F}\left({\mathit{x}}_3\right)\parallel$	$\parallel {\mathit{x}}_4-{\mathit{x}}_3\parallel$	n	${\mathit{\eta }}^{*}$	CPU Timing
Lotfi et al. [27]	${(1.1,1.1,\stackrel{60}{\cdots },1.1)}^{tr}$	$1.2\times {10}^{-134}$	$4.7\times {10}^{-133}$	4	5.0541	135.752
Wang and Li [26]	${(1.1,1.1,\stackrel{60}{\cdots },1.1)}^{tr}$	$2.1\times {10}^{-177}$	$1.2\times {10}^{-176}$	4	6.0281	127.473
Abbasbandy [24]	${(1.1,1.1,\stackrel{60}{\cdots },1.1)}^{tr}$	$7.6\times {10}^{-168}$	$4.3\times {10}^{-167}$	4	6.0293	347.9
Hueso et al. [25]	${(1.1,1.1,\stackrel{60}{\cdots },1.1)}^{tr}$	$6.6\times {10}^{-125}$	$3.9\times {10}^{-124}$	4	5.0358	222.306
Method (2)	${(1.1,1.1,\stackrel{60}{\cdots },1.1)}^{tr}$	$3.2\times {10}^{-182}$	$1.7\times {10}^{-181}$	4	6.0616	104.265

CPU timing and ${\eta }^{*}$ are calculated based on the number of iterations required to reach the desired accuracy.

Example 3. 

We investigate a widely recognized problem, the Hammerstein integral equation, as detailed in [2] (pp. 19–20). The primary objective is to evaluate and contrast the effectiveness and practicality of iterative scheme (2) against the earlier ones of the same CO. The Hammerstein integral equation is given below:

$$x\left(s\right)=1+{\displaystyle \frac{1}{5}}{\int }_0^{1}G(s,t)x^3\left(t\right)dt,\mathrm{for}x\in C[0,1]s,t\in [0,1].$$

The kernel G is given by

$$G(s,t)=\left\{\begin{array}{cc}\hfill & s(1-t),s\le t\hfill \\ \hfill & (1-s)t,t\le s.\hfill \end{array}\right.$$

To convert the given equation into a finite-dimensional problem, we utilize the Gauss–Legendre quadrature formula (GLQF), which allows us to approximate integrals with greater accuracy. Then, we obtain

$${\int }_0^{1}g\left(t\right)dt\simeq \sum _{j=1}^{10}{w}_{j}g\left(t_j\right).$$

The values $t_j$ (abscissas) and $w_j$ (weights) are calculated using the GLQF for $j=10$ and shown in the Table 3. Let $x_i$ be defined as the approximations of $x\left(t_i\right)$ by $x_i(i=1,2,\dots ,10)$. This leads to an SNE, which is defined as follows:

$$5x_i-5-\sum _{j=1}^{10}{a}_{ij}{x}_j^3=0,i=1,2,\dots ,10$$

where

$$a_{ij}=\left\{\begin{array}{c}{w}_j{t}_j(1-t_i),j\le i,\hfill \\ w_j{t}_i(1-t_j),i<j.\hfill \end{array}\right.$$

Table 3. The values $t_j$ (abscissas) and $w_j$ (weights).

j	${\mathit{t}}_{\mathit{j}}$	${\mathit{w}}_{\mathit{j}}$
1	$0.01304673574141413996101799\dots$	$0.03333567215434406879678440\dots$
2	$0.06746831665550774463395165\dots$	$0.07472567457529029657288816\dots$
3	$0.16029521585048779688283632\dots$	$0.10954318125799102199776746\dots$
4	$0.28330230293537640460036703\dots$	$0.13463335965499817754561346\dots$
5	$0.42556283050918439455758700\dots$	$0.14776211235737643508694649\dots$
6	$0.57443716949081560544241300\dots$	$0.14776211235737643508694649\dots$
7	$0.71669769706462359539963297\dots$	$0.13463335965499817754561346\dots$
8	$0.83970478414951220311716368\dots$	$0.10954318125799102199776746\dots$
9	$0.93253168334449225536604834\dots$	$0.07472567457529029657288816\dots$
10	$0.98695326425858586003898201\dots$	$0.03333567215434406879678440\dots$

Table 3. The values $t_j$ (abscissas) and $w_j$ (weights).

j	${\mathit{t}}_{\mathit{j}}$	${\mathit{w}}_{\mathit{j}}$
1	$0.01304673574141413996101799\dots$	$0.03333567215434406879678440\dots$
2	$0.06746831665550774463395165\dots$	$0.07472567457529029657288816\dots$
3	$0.16029521585048779688283632\dots$	$0.10954318125799102199776746\dots$
4	$0.28330230293537640460036703\dots$	$0.13463335965499817754561346\dots$
5	$0.42556283050918439455758700\dots$	$0.14776211235737643508694649\dots$
6	$0.57443716949081560544241300\dots$	$0.14776211235737643508694649\dots$
7	$0.71669769706462359539963297\dots$	$0.13463335965499817754561346\dots$
8	$0.83970478414951220311716368\dots$	$0.10954318125799102199776746\dots$
9	$0.93253168334449225536604834\dots$	$0.07472567457529029657288816\dots$
10	$0.98695326425858586003898201\dots$	$0.03333567215434406879678440\dots$

Table 4 provides a detailed overview of the performance for Example 3, including the Coefficient of Convergence (COC), CPU time, number of iterations, residual errors, and the error differences between consecutive iterations. These computational results offer us a valuable insight into the efficiency and accuracy of the required solution. This table provide us both computational time and the convergence behavior as the iterative methods with other existing methods. Our required solution is given below:

$$\begin{array}{cc}\hfill x_{*}=& (1.001\dots ,1.006\dots ,1.014\dots ,1.021\dots ,1.026\dots ,1.026\dots ,1.021\dots ,1.014\dots ,\hfill \\ \hfill & 1.006\dots ,1.0013\dots {)}^{tr}.\hfill \end{array}$$

Table 4. Computational results of Example 3.

Methods	${\mathit{x}}_0$	$\parallel \mathit{F}\left({\mathit{x}}_3\right)\parallel$	$\parallel {\mathit{x}}_4-{\mathit{x}}_3\parallel$	n	$\mathit{\eta }$	CPU Timing
Lotfi et al. [27]	${(1.1,1.1,\stackrel{10}{\cdots },1.1)}^{tr}$	$3.3\times {10}^{-316}$	$6.7\times {10}^{-317}$	3	5.1023	2.86834
Wang and Li [26]	${(1.1,1.1,\stackrel{10}{\cdots },1.1)}^{tr}$	$4.2\times {10}^{-422}$	$9.0\times {10}^{-423}$	3	6.0410	2.82447
Abbasbandy [24]	${(1.1,1.1,\stackrel{10}{\cdots },1.1)}^{tr}$	$5.1\times {10}^{-394}$	$1.1\times {10}^{-394}$	3	6.0428	3.67107
Hueso et al. [25]	${(1.1,1.1,\stackrel{10}{\cdots },1.1)}^{tr}$	$9.8\times {10}^{-251}$	$2.1\times {10}^{-251}$	4	5.0099	10.559
Method (2)	${(1.1,1.1,\stackrel{10}{\cdots },1.1)}^{tr}$	$4.7\times {10}^{-409}$	$1.0\times {10}^{-409}$	3	5.9995	3.30025

CPU timing and $\eta$ are calculated based on the number of iterations required to reach the desired accuracy.

Table 4. Computational results of Example 3.

Methods	${\mathit{x}}_0$	$\parallel \mathit{F}\left({\mathit{x}}_3\right)\parallel$	$\parallel {\mathit{x}}_4-{\mathit{x}}_3\parallel$	n	$\mathit{\eta }$	CPU Timing
Lotfi et al. [27]	${(1.1,1.1,\stackrel{10}{\cdots },1.1)}^{tr}$	$3.3\times {10}^{-316}$	$6.7\times {10}^{-317}$	3	5.1023	2.86834
Wang and Li [26]	${(1.1,1.1,\stackrel{10}{\cdots },1.1)}^{tr}$	$4.2\times {10}^{-422}$	$9.0\times {10}^{-423}$	3	6.0410	2.82447
Abbasbandy [24]	${(1.1,1.1,\stackrel{10}{\cdots },1.1)}^{tr}$	$5.1\times {10}^{-394}$	$1.1\times {10}^{-394}$	3	6.0428	3.67107
Hueso et al. [25]	${(1.1,1.1,\stackrel{10}{\cdots },1.1)}^{tr}$	$9.8\times {10}^{-251}$	$2.1\times {10}^{-251}$	4	5.0099	10.559
Method (2)	${(1.1,1.1,\stackrel{10}{\cdots },1.1)}^{tr}$	$4.7\times {10}^{-409}$	$1.0\times {10}^{-409}$	3	5.9995	3.30025

CPU timing and $\eta$ are calculated based on the number of iterations required to reach the desired accuracy.

Example 4. 

Further, we analyze another larger SNE of order 100 by 100 variables, to demonstrate the method’s effectiveness and scalability when applied to more complex and large-scale systems. The analysis highlights the method’s capacity to tackle significant computational challenges and its applicability to a wide range of practical problems involving large-scale nonlinear systems. Thus, we consider the following system:

$$P\left(X\right)=\left\{\begin{array}{cc}\hfill & x_j^2{x}_{j+1}-1=0,1\le j\le 299,\hfill \\ \hfill & x_{300}^2{x}_1-1=0,otherwise.\hfill \end{array}\right.$$

(37)

The required zero for this problem is $x_{*}={(1,1,1,\stackrel{100}{\cdots },1)}^{tr}$. In Table 5, we provide the Coefficient of Convergence (COC), CPU time, number of iterations, residual errors, and the error differences between two iterations for Example 4.

Table 5. Computational results of Example 4.

Methods	${\mathit{x}}_0$	$\parallel \mathit{F}\left({\mathit{x}}_4\right)\parallel$	$\parallel {\mathit{x}}_5-{\mathit{x}}_4\parallel$	n	${\mathit{\eta }}^{*}$	CPU Timing
Lotfi et al. [27]	${\left({\displaystyle \frac{96}{100}},{\displaystyle \frac{96}{100}},\stackrel{100}{\cdots },{\displaystyle \frac{96}{100}}\right)}^{tr}$	$1.1\times {10}^{-238}$	$3.7\times {10}^{-239}$	4	6.0252	48.3817
Wang and Li [26]	${\left({\displaystyle \frac{96}{100}},{\displaystyle \frac{96}{100}},\stackrel{100}{\cdots },{\displaystyle \frac{96}{100}}\right)}^{tr}$	$1.2\times {10}^{-231}$	$3.9\times {10}^{-232}$	4	6.0260	51.2171
Abbasbandy [24]	${\left({\displaystyle \frac{96}{100}},{\displaystyle \frac{96}{100}},\stackrel{100}{\cdots },{\displaystyle \frac{96}{100}}\right)}^{tr}$	$1.8\times {10}^{-224}$	$6.0\times {10}^{-225}$	4	6.0268	78.1555
Hueso et al. [25]	${\left({\displaystyle \frac{96}{100}},{\displaystyle \frac{96}{100}},\stackrel{100}{\cdots },{\displaystyle \frac{96}{100}}\right)}^{tr}$	$7.3\times {10}^{-158}$	$2.4\times {10}^{-158}$	4	5.0318	100.235
Method (2)	${\left({\displaystyle \frac{96}{100}},{\displaystyle \frac{96}{100}},\stackrel{100}{\cdots },{\displaystyle \frac{96}{100}}\right)}^{tr}$	$2.0\times {10}^{-246}$	$6.7\times {10}^{-247}$	4	6.0244	50.3581

CPU timing and ${\eta }^{*}$ are calculated based on the number of iterations required to reach the desired accuracy.

Table 5. Computational results of Example 4.

Methods	${\mathit{x}}_0$	$\parallel \mathit{F}\left({\mathit{x}}_4\right)\parallel$	$\parallel {\mathit{x}}_5-{\mathit{x}}_4\parallel$	n	${\mathit{\eta }}^{*}$	CPU Timing
Lotfi et al. [27]	${\left({\displaystyle \frac{96}{100}},{\displaystyle \frac{96}{100}},\stackrel{100}{\cdots },{\displaystyle \frac{96}{100}}\right)}^{tr}$	$1.1\times {10}^{-238}$	$3.7\times {10}^{-239}$	4	6.0252	48.3817
Wang and Li [26]	${\left({\displaystyle \frac{96}{100}},{\displaystyle \frac{96}{100}},\stackrel{100}{\cdots },{\displaystyle \frac{96}{100}}\right)}^{tr}$	$1.2\times {10}^{-231}$	$3.9\times {10}^{-232}$	4	6.0260	51.2171
Abbasbandy [24]	${\left({\displaystyle \frac{96}{100}},{\displaystyle \frac{96}{100}},\stackrel{100}{\cdots },{\displaystyle \frac{96}{100}}\right)}^{tr}$	$1.8\times {10}^{-224}$	$6.0\times {10}^{-225}$	4	6.0268	78.1555
Hueso et al. [25]	${\left({\displaystyle \frac{96}{100}},{\displaystyle \frac{96}{100}},\stackrel{100}{\cdots },{\displaystyle \frac{96}{100}}\right)}^{tr}$	$7.3\times {10}^{-158}$	$2.4\times {10}^{-158}$	4	5.0318	100.235
Method (2)	${\left({\displaystyle \frac{96}{100}},{\displaystyle \frac{96}{100}},\stackrel{100}{\cdots },{\displaystyle \frac{96}{100}}\right)}^{tr}$	$2.0\times {10}^{-246}$	$6.7\times {10}^{-247}$	4	6.0244	50.3581

CPU timing and ${\eta }^{*}$ are calculated based on the number of iterations required to reach the desired accuracy.

5. Concluding Remarks

A new methodology has been proposed in this study, which expands the use of multistep approaches without relying on Taylor series expansions that impose derivative-based conditions which are not present in the method. Other drawbacks involve the absence of a priori and computable error distances and uniqueness of the solution results. These concerns are all positively addressed in this paper. Indeed, the sufficient convergence requirements of our approach depend solely on the operators involved in the technique. Additionally, our approach establishes the uniqueness of the solution, determines the radii of convergence, and provides upper a priori estimates for the error distances involved in the methods (2) and (3). Furthermore, the SLAC, which is not discussed in earlier papers, is also presented in this work. Method (2) outperforms existing methods in terms of residual error and the absolute difference in error between two consecutive iterations. Furthermore, method (1.2) consumes significantly less time compared to the existing methods. Moreover, it demonstrates a stable CO compared to the existing ones. In future research, we will explore how this methodology can be utilized to extend other iterative methods in a similar manner due to its generality [7,8,10,11,12,13,14,15,16,17,18,19,28,29,30,31,32,33,34,35,36,37,38,39].