Improving Convergence Analysis of the Newton–Kurchatov Method under Weak Conditions

Abstract

The technique of using the restricted convergence region is applied to study a semilocal convergence of the Newton–Kurchatov method. The analysis is provided under weak conditions for the derivatives and the first order divided differences. Consequently, weaker sufficient convergence criteria and more accurate error estimates are retrieved. A special case of weak conditions is also considered.

1. Introduction

Nonlinear equations, in particular systems of nonlinear algebraic or transcendental equations, arise often when numerical methods are used for solving applied problems. A popular method for solving such equations is Newton’s [1,2,3]. However, it requires differentiability of the nonlinear function. This is not a requirement for difference methods [1,2,3,4]. They can be applied to equations with a nondifferentiable function [5]. For some problems, the nonlinear function can be represented as the sum of the differentiable and nondifferentiable parts. In this case, methods with operator decomposition are often used [1,2,6,7,8,9]. Numerical examples show that the convergence is faster than difference methods and the Newton-type method [10,11,12].

Let us consider an equation

$$H\left(x\right)\equiv F\left(x\right)+G\left(x\right)=0,$$

(1)

where F and $G:D\subset X\to Y$. Here F is a differentiable operator, G is a continuous operator, D is an open convex set, X and Y are Banach spaces.

We use Newton–Kurchatov method [8,13,14,15,16] for solving Equation (1) numerically

$$x_{n+1}=x_n-A_n^{-1}H\left(x_n\right),n\ge 0,$$

(2)

where $A_n=F^{\prime }\left(x_n\right)+G(2x_n-x_{n-1};x_{n-1})$. It is a combination of Newton and Kurchatov methods [3,4,17]. Their formulas for solving equation $F\left(x\right)=0$ are of the form

$$x_{n+1}=x_n-F^{\prime }{\left(x_n\right)}^{-1}F\left(x_n\right),n\ge 0$$

and

$$x_{n+1}=x_n-F{(2x_n-x_{n-1};x_{n-1})}^{-1}F\left(x_n\right),n\ge 0,$$

respectively; $G(·;·)$ and $F(·;·)$ denote a first-order divided difference. Let $\tilde{x}\in D$. Denote

$$B(\tilde{x},R)=\{x\in X:\parallel x-\tilde{x}\parallel <R\},$$

and

$$\overline{B(\tilde{x},R)}=\{x\in X:\parallel x-\tilde{x}\parallel \le R\}.$$

Our semilocal convergence is based on some generalized conditions. Suppose that for each $x,y\in D$:

$$\parallel A_0^{-1}(F^{\prime }\left(x\right)-F^{\prime }\left(x_0\right))\parallel \le {\omega }_1^0(\parallel x-x_0\parallel ),$$

(3)

$$\parallel A_0^{-1}(G(x;y)-G(2x_0-x_{-1};x_{-1}))\parallel \le {\omega }_2^0(\parallel x-(2x_0-x_{-1})\parallel ,\parallel y-x_{-1}\parallel ),$$

(4)

where ${\omega }_1^0:{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$ and ${\omega }_2^0:{\mathbb{R}}_{+}\times {\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$ are nondecreasing functions. Let $a>0$. Suppose that equation

$${\omega }_1^0\left(u\right)+{\omega }_2^0(3u+a,u+a)=1$$

(5)

has at least one positive solution. Denote by $R_0$ the smallest such solution. Set $D_0=D\cap B(x_0,3R_0)$ and suppose that for each $x,y,u,v\in D_0$ with $2y-x\in D_0$

$$\parallel A_0^{-1}(F^{\prime }\left(x\right)-F^{\prime }\left(y\right))\parallel \le {\omega }_1(\parallel x-y\parallel ),$$

(6)

$$\parallel A_0^{-1}(G(2y-x;x)-G(u;v))\parallel \le {\omega }_2(\parallel 2y-x-u\parallel ,\parallel x-v\parallel ),$$

(7)

where ${\omega }_1:{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$ and ${\omega }_2:{\mathbb{R}}_{+}\times {\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$ are nondecreasing functions. Moreover, ${\omega }_1\left(tr\right)\le h\left(t\right){\omega }_1\left(r\right),t\in [0,1],r\in [0,R]$, $h:[0,1]\to \mathbb{R}$ [11].

In this article, we also consider $\epsilon$-type conditions for $x,y\in D$

$$\parallel A_0^{-1}(F^{\prime }\left(x\right)-F^{\prime }\left(x_0\right))\parallel \le {\epsilon }_1^0,$$

(8)

$$\parallel A_0^{-1}(G(x;y)-G(2x_0-x_{-1};x_{-1}))\parallel \le {\epsilon }_2^0,$$

(9)

for $\overline{x},\overline{y},u,v\in D_0$ with $2\overline{y}-\overline{x}\in D_0$

$$\parallel A_0^{-1}(F^{\prime }\left(x\right)-F^{\prime }\left(y\right))\parallel \le {\epsilon }_1,$$

(10)

$$\parallel A_0^{-1}(G(2\overline{y}-\overline{x};\overline{x})-G(u;v))\parallel \le {\epsilon }_2,$$

(11)

where ${\epsilon }_1^0$, ${\epsilon }_1$, ${\epsilon }_2^0$ and ${\epsilon }_2$ are positive constants, for some $D_0\subseteq D$.

2. Semilocal Analysis

Set $\mathsf{\Phi }={\int }_0^{1}h\left(t\right)dt$.

Theorem 1.

Let F and G be nonlinear operators with the specified properties. Assume that:

• linear operator $A_0$, where $x_{-1}$ and $x_0\in D$, is invertible;
• $$\parallel A_0^{-1}(F\left(x_0\right)+G\left(x_0\right))\parallel \le \eta ,\parallel x_0-x_{-1}\parallel \le \alpha ;$$

  (12)
• Equations (3) and (4) hold on D, Equations (6) and (7) hold on $D_0$;
• equation

  $$u\left(1-\frac{m}{1-{\omega }_1^0\left(u\right)-{\omega }_2^0(3u+\alpha ,u+\alpha )}\right)-\eta =0,$$

  (13)

  where $m=\mathsf{\Phi }{\omega }_1\left(\eta \right)+max\{{\omega }_2(\eta +\alpha ,\alpha ),{\omega }_2(2\eta ,\eta )\}$, has at least one positive solution greater than η and α. Denote by R the smallest such solution;
• ${\omega }_1^0\left(R\right)+{\omega }_2^0(3R+\alpha ,R+\alpha )<1$, $M={\displaystyle \frac{m}{1-{\omega }_1^0\left(R\right)-{\omega }_2^0(3R+\alpha ,R+\alpha )}}<1$, $B(x_0,3R)\subset D$.

Then, the sequence ${\left\{x_n\right\}}_{n\ge 0}$, generated by Equation (2), is well-defined, remains in $B(x_0,R)$ and converges to a unique solution $x^{*}\in \overline{B(x_0,R)}$ of Equation (1), and $R<R_0$.

Proof. 

The proof of Theorem 1 is carried out by mathematical induction and is similar to the one in [8] but there are some differences. By Equations (2) and (12), for $n=0$ we have

$$\parallel x_1-x_0\parallel \le \parallel A_0^{-1}(F\left(x_0\right)+G\left(x_0\right))\parallel \le \eta <R$$

and $x_1\in B(x_0,R)$. Using the conditions in Equations (3) and (4), we get

$$\parallel I-A_0^{-1}{A}_1\parallel =\parallel A_0^{-1}(A_0-A_1)\parallel$$

$$\le {\omega }_1^0(\parallel x_1-x_0\parallel )+{\omega }_2^0(\parallel 2x_0-x_{-1}-2x_1+x_0\parallel ,\parallel x_{-1}-x_0\parallel )$$

$$\le {\omega }_1^0\left(\eta \right)+{\omega }_2^0(2\eta +\alpha ,\alpha )\le {\omega }_1^0\left(R\right)+{\omega }_2^0(2R+\alpha ,\alpha )$$

$$\le {\omega }_1^0\left(R\right)+{\omega }_2^0(3R+\alpha ,R+\alpha )<1.$$

According to the Banach lemma on inverse operators [1] $A_1^{-1}{A}_0$ exists and

$$\parallel A_1^{-1}{A}_0\parallel \le \frac{1}{1-{\omega }_1^0\left(R\right)-{\omega }_2^0(3R+\alpha ,R+\alpha )}.$$

Then, we have

$$A_0^{-1}(F\left(x_1\right)+G\left(x_1\right))$$

$$=A_0^{-1}(F\left(x_1\right)-F\left(x_0\right)-F^{\prime }\left(x_0\right)(x_1-x_0))$$

$$+A_0^{-1}(G\left(x_1\right)-G\left(x_0\right)-G(2x_0-x_{-1};x_{-1})(x_1-x_0))$$

$$={\int }_0^1{A}_0^{-1}(F^{\prime }(x_0+t(x_1-x_0))-F^{\prime }\left(x_0\right))dt(x_1-x_0)$$

$$+A_0^{-1}(G(x_1;x_0)-G(2x_0-x_{-1};x_{-1}))(x_1-x_0).$$

Hence, by the conditions in Equations (6) and (7), we obtain

$$\parallel x_2-x_1\parallel =\parallel A_1^{-1}(F\left(x_1\right)+G\left(x_1\right))\parallel \le \parallel A_1^{-1}{A}_0\parallel \parallel A_0^{-1}(F\left(x_1\right)+G\left(x_1\right))\parallel$$

$$\le \frac{\mathsf{\Phi }{\omega }_1(\parallel x_1-x_0\parallel )+{\omega }_2(\parallel 2x_0-x_{-1}-x_1\parallel ,\parallel x_{-1}-x_0\parallel )}{1-{\omega }_1\left(R\right)-{\omega }_2(3R+\alpha ,R+\alpha )}\parallel x_1-x_0\parallel$$

$$\le \frac{\mathsf{\Phi }{\omega }_1\left(\eta \right)+{\omega }_2(\eta +\alpha ,\alpha )}{1-{\omega }_1^0\left(R\right)-{\omega }_2^0(3R+\alpha ,R+\alpha )}\parallel x_1-x_0\parallel \le M\parallel x_1-x_0\parallel <\eta .$$

On the other hand,

$$\parallel x_2-x_0\parallel \le \parallel x_2-x_1\parallel +\parallel x_1-x_0\parallel$$

$$\le (M+1)\parallel x_1-x_0\parallel \le (M+1)\eta =\frac{1-M^2}{1-M}\eta <\frac{1}{1-M}\eta =R.$$

Therefore, $x_2\in B(x_0,R)$. Suppose that for $k=1,\dots ,n-1$ holds:

• $A_k^{-1}{A}_0$ exists and $\parallel A_k^{-1}{A}_0\parallel \le {\displaystyle \frac{1}{1-{\omega }_1^0\left(R\right)-{\omega }_2^0(3R+\alpha ,R+\alpha )}}$;
• $\parallel x_{k+1}-x_k\parallel \le M\parallel x_k-x_{k-1}\parallel \le M^k\parallel x_1-x_0\parallel <\eta$;
• $x_{k+1}\in B(x_0,R)$.

Then, using the conditions in Equations (3) and (4), for $k=n$ we have

$$\parallel I-A_0^{-1}{A}_n\parallel =\parallel A_0^{-1}(A_0-A_n)\parallel$$

$$\le {\omega }_1^0(\parallel x_0-x_n\parallel )+{\omega }_2^0(\parallel 2x_0-x_{-1}-2x_n+x_{n-1}\parallel ,\parallel x_{-1}-x_{n-1}\parallel )$$

$$\le {\omega }_1^0\left(R\right)+{\omega }_2^0(3R+\alpha ,R+\alpha )<1.$$

According to the Banach lemma on inverse operators [1] $A_n^{-1}{A}_0$ exists and

$$\parallel A_n^{-1}{A}_0\parallel \le \frac{1}{1-{\omega }_1^0\left(R\right)-{\omega }_2^0(3R+\alpha ,R+\alpha )}.$$

By equality

$$A_0^{-1}(F\left(x_n\right)+G\left(x_n\right))=A_0^{-1}(F\left(x_n\right)-F\left(x_{n-1}\right)-F^{\prime }\left(x_{n-1}\right)(x_n-x_{n-1}))$$

$$+A_0^{-1}(G\left(x_n\right)-G\left(x_{n-1}\right)-G(2x_{n-1}-x_{n-2};x_{n-2})(x_n-x_{n-1}))$$

$$={\int }_0^1{A}_0^{-1}(F^{\prime }(x_{n-1}+t(x_n-x_{n-1}))-F^{\prime }\left(x_{n-1}\right))dt(x_n-x_{n-1})$$

$$+A_0^{-1}(G(x_n;x_{n-1})-G(2x_{n-1}-x_{n-2};x_{n-2}))(x_n-x_{n-1})$$

and the conditions in Equations (6) and (7), we have

$$\parallel x_{n+1}-x_n\parallel =\parallel A_n^{-1}(F\left(x_n\right)+G\left(x_n\right))\parallel \le \parallel A_n^{-1}{A}_0\parallel \parallel A_0^{-1}(F\left(x_n\right)+G\left(x_n\right))\parallel$$

$$\le \frac{1}{1-{\omega }_1^0\left(R\right)-{\omega }_2^0(3R+\alpha ,R+\alpha )}(\mathsf{\Phi }{\omega }_1(\parallel x_n-x_{n-1}\parallel )$$

$$+{\omega }_2(\parallel 2x_{n-1}-x_{n-2}-x_n\parallel ,\parallel x_{n-1}-x_{n-2}\parallel \left)\right)\parallel x_n-x_{n-1}\parallel$$

$$\le \frac{\mathsf{\Phi }{\omega }_1\left(\eta \right)+{\omega }_2(2\eta ,\eta )}{1-{\omega }_1^0\left(R\right)-{\omega }_2^0(3R+\alpha ,R+\alpha )}\parallel x_n-x_{n-1}\parallel$$

$$\le M\parallel x_n-x_{n-1}\parallel \le M^n\parallel x_1-x_0\parallel <\eta .$$

Next, we show that $x_{n+1}\in B(x_0,R)$. Indeed,

$$\parallel x_{n+1}-x_0\parallel \le \parallel x_{n+1}-x_n\parallel +\parallel x_n-x_{n-1}\parallel +\dots +\parallel x_1-x_0\parallel$$

$$\le (M^n+M^{n-1}+\dots +M+1)\parallel x_1-x_0\parallel \le \frac{1-M^{n+1}}{1-M}\eta <\frac{1}{1-M}\eta =R,$$

and $x_{n+1}\in B(x_0,R)$. Moreover, we show that sequence ${\left\{x_n\right\}}_{n\ge 0}$ is fundamental. Indeed, for $p\ge 1$

$$\parallel x_{n+p}-x_n\parallel \le \parallel x_{n+p}-x_{n+p-1}\parallel +\parallel x_{n+p-1}-x_{n+p-2}\parallel +\dots +\parallel x_{n+1}-x_n\parallel$$

$$\le (M^{p-1}+M^{p-2}+\dots +1)\parallel x_{n+1}-x_n\parallel$$

$$=\frac{1-M^p}{1-M}\parallel x_{n+1}-x_n\parallel \le \frac{1-M^p}{1-M}{M}^n\eta <\frac{M^n}{1-M}\eta .$$

Therefore, ${\left\{x_n\right\}}_{n\ge 0}$ is a fundamental sequence and converges to $x^{*}\in \overline{B(x_0,R)}$. Furthermore, we prove that $x^{*}$ is a unique solution of Equation (1). Since

$$\parallel A_0^{-1}H\left(x_n\right)\parallel \le (\mathsf{\Phi }{\omega }_1\left(\eta \right)+{\omega }_2(2\eta ,\eta ))\parallel x_n-x_{n-1}\parallel$$

and $\parallel x_n-x_{n-1}\parallel \to 0$ for $n\to \infty$, so $H\left(x^{*}\right)=0$. Finally, suppose there exists $y^{*}\in \overline{B(x_0,R)}$, $y^{*}\ne x^{*}$ such that $H\left(y^{*}\right)=0$. Denote

$$A={\int }_0^1{F}^{\prime }(x^{*}+t(y^{*}-x^{*}))dt+G(y^{*};x^{*}).$$

Using the conditions in Equations (3) and (4), we get

$$\parallel A_0^{-1}(A_0-A)\parallel \le {\int }_0^1{\omega }_1^0(\parallel x_0-x^{*}-t(y^{*}-x^{*})\parallel )dt$$

$$+{\omega }_2^0(\parallel 2x_0-x_{-1}-y^{*}\parallel ,\parallel x_{-1}-x^{*}\parallel )$$

$$\le {\int }_0^1{\omega }_1^0((1-t)\parallel x_0-x^{*}\parallel +t\parallel x_0-y^{*}\parallel )dt$$

$$+{\omega }_2^0(\parallel x_0-x_{-1}\parallel +\parallel x_0-y^{*}\parallel ,\parallel x_{-1}-x_0\parallel +\parallel x_0-x^{*}\parallel )$$

$$\le {\omega }_1^0\left(R\right)+{\omega }_2^0(R+\alpha ,R+\alpha )<1.$$

According to the Banach lemma on inverse operators A is invertible and in view of

$$A(y^{*}-x^{*})=H\left(y^{*}\right)-H\left(x^{*}\right)$$

it follows $y^{*}=x^{*}$. □

Theorem 2.

Let F and G be nonlinear operators with the specified properties. Assume that

• linear operator $A_0$, where $x_{-1}$ and $x_0\in D$, is invertible;
• Equations (8)–(11) hold;
• numbers $\eta >0$, γ and $R>0$ such that

  $$\parallel A_0^{-1}(F\left(x_0\right)+G\left(x_0\right))\parallel \le \eta ,\parallel x_0-x_{-1}\parallel <R,0<{\epsilon }_1^0+{\epsilon }_2^0<1,$$

  $$\gamma =\frac{{\epsilon }_1+{\epsilon }_2}{1-({\epsilon }_1^0+{\epsilon }_2^0)}<1,\frac{\eta }{1-\gamma }<R,B(x_0,3R)\subset D.$$

Then, the sequence ${\left\{x_n\right\}}_{n\ge 0}$, generated by Equation (2), is well-defined, remains in $B(x_0,R)$ and converges to a unique solution $x^{*}\in \overline{B(x_0,R)}$ of Equation (1). Moreover, the following inequality holds for each $n\ge 0$

$$\parallel x_n-x^{*}\parallel \le \frac{{\gamma }^n}{1-\gamma }\eta .$$

(14)

Notice that one possibility is ${\epsilon }_1^0$ and ${\epsilon }_2^0$ in practice may be functions of $\parallel x-x_0\parallel$. That is

$${\epsilon }_1^0:[0,\infty )\to [0,\infty )\mathrm{and}{\epsilon }_2^0:[0,\infty )\to [0,\infty )$$

to be nondecreasing and continuous functions.

Suppose that equation

$${\epsilon }_1^0\left(u\right)+{\epsilon }_2^0\left(u\right)=1$$

has a minimal positive solution $\rho$. Then, the set $D_0$ can be defined as $D_0=D\cap B(x_0,\rho )$. Moreover, in this case we have

$$\parallel A_0^{-1}(A_0-A_n)\parallel \le {\epsilon }_1^0+{\epsilon }_2^0<1,$$

so $\parallel A_n^{-1}{A}_0\parallel \le {\displaystyle \frac{1}{1-({\epsilon }_1^0+{\epsilon }_2^0)}}$.

However, $D_0$ can also be defined in other ways too depending on the construction of F and G.

Let

$${\omega }_1^0(\parallel x-y\parallel )=2{\ell }_0\parallel x-y\parallel ,{\omega }_2^0(\parallel x-u\parallel ,\parallel y-v\parallel )=p_0(\parallel x-u\parallel +\parallel y-v\parallel ),$$

$${\omega }_1(\parallel x-y\parallel )=2\ell \parallel x-y\parallel ,{\omega }_2(\parallel x-u\parallel ,\parallel y-v\parallel )=p(\parallel x-u\parallel +\parallel y-v\parallel ).$$

Next, we obtain from Theorem 1 the convergence analysis of the method in Equation (2) under the Lipschitz conditions.

Corollary 1.

Let F and G be nonlinear operators with the specified properties. Assume that:

• linear operator $A_0$, where $x_{-1}$ and $x_0\in D$, is invertible;
• numbers $\eta >0$ and $\alpha >0$ such that Equation (12) is satisfied;
• the Lipschitz conditions are fulfilled for each $x,y\in D$

  $$\parallel A_0^{-1}(F^{\prime }\left(x_0\right)-F^{\prime }\left(y\right))\parallel \le 2{\ell }_0\parallel x-y\parallel ,$$

  $$\parallel A_0^{-1}(G(x;y)-G(2x_0-x_{-1};x_{-1}))\parallel \le p_0(\parallel x-(2x_0-x_{-1})\parallel +\parallel y-x_{-1}\parallel ),$$
  and for each $x,y,u,v\in D_0$

  $$\parallel A_0^{-1}(F^{\prime }\left(x\right)-F^{\prime }\left(y\right))\parallel \le 2\ell \parallel x-y\parallel ,$$

  $$\parallel A_0^{-1}(G(x;y)-G(u;v))\parallel \le p(\parallel x-u\parallel +\parallel y-v\parallel );$$
• equation

  $$s\left(1-\frac{m}{1-2{\ell }_{0}s-p_0(4s+2\alpha )}\right)-\eta =0,$$

  where $m=\ell \eta +max\left\{p\right(\eta +2\alpha ),3p\eta \}$ has at least one positive solution greater than η and α. Denote by R the smallest such solution;
• $2{\ell }_{0}R+p_0(4R+2\alpha )<1$, $M={\displaystyle \frac{m}{1-2{\ell }_{0}R-p_0(4R+2\alpha )}}<1$, $B(x_0,3R)\subset D$.

Then, the sequence ${\left\{x_n\right\}}_{n\ge 0}$, generated by Equation (2), is well-defined, remains in $B(x_0,R)$ and converges to a unique solution $x^{*}\in \overline{B(x_0,R)}$ of Equation (1).

Remark 1.

The corresponding Equations (6) and (7) to conditions in [8] are given for each $x,y,u,v\in D$ by

$$\parallel A_0^{-1}(F^{\prime }\left(x\right)-F^{\prime }\left(y\right))\parallel \le {\omega }_1^1(\parallel x-y\parallel ),$$

(15)

$$\parallel A_0^{-1}(G(x;y)-G(u;v))\parallel \le {\omega }_2^1(\parallel x-u\parallel ,\parallel y-v\parallel ).$$

(16)

Notice that ${\omega }_1={\omega }_1\left({\omega }_1^0\right)$ and ${\omega }_2={\omega }_2\left({\omega }_2^0\right)$, i.e., they are functions of ${\omega }_1^0$ and ${\omega }_2^0$, and since $D_0\subseteq D$

$$\begin{array}{ccc}\hfill {\omega }_1^0\left(t\right)& \le & {\omega }_1^1\left(t\right),\hfill \end{array}$$

(17)

$$\begin{array}{ccc}\hfill {\omega }_2^0\left(t\right)& \le & {\omega }_2^1\left(t\right),\hfill \end{array}$$

(18)

$$\begin{array}{ccc}\hfill {\omega }_1\left(t\right)& \le & {\omega }_1^1\left(t\right),\hfill \end{array}$$

(19)

$$\begin{array}{ccc}\hfill {\omega }_2\left(t\right)& \le & {\omega }_2^1\left(t\right),\hfill \end{array}$$

(20)

$$\begin{array}{ccc}\hfill m& \le & m^1,\hfill \end{array}$$

(21)

$$\begin{array}{ccc}\hfill {\epsilon }_1^0& \le & {\epsilon }_1^1,{\epsilon }_2^0\le {\epsilon }_2^1,{\epsilon }_1\le {\epsilon }_1^1,{\epsilon }_2\le {\epsilon }_2^1,\hfill \end{array}$$

(22)

$$\begin{array}{ccc}\hfill l_0& \le & l^1,\hfill \end{array}$$

(23)

$$\begin{array}{ccc}\hfill and{p}_0& \le & p^1,\hfill \end{array}$$

(24)

where $m^1=\mathsf{\Phi }{\omega }_1^1\left(\eta \right)+max\{{\omega }_2^1(\eta +\alpha ,\alpha ),{\omega }_2^1(2\eta ,\eta )\}$, $M^1={\displaystyle \frac{m^1}{1-{\omega }_1^1\left(R^1\right)-{\omega }_2^1(3R^1+\alpha ,R^1+\alpha )}}$.

It’s easy to see that if $R^1\le R$ then $M^1\le M$, and if $R^1\ge R$ then $M^1\ge M$.

If ${\omega }_1^1$, ${\omega }_2^1$ are constant functions

$${\omega }_1^1(\parallel x-y\parallel )=2{\ell }^1\parallel x-y\parallel ,{\omega }_2^1(\parallel x-u\parallel ,\parallel y-v\parallel )=p^1(\parallel x-u\parallel +\parallel y-v\parallel ).$$

It follows from the above that we obtain the following improvements:

1. Weaker sufficient convergence criteria, since

$$\begin{array}{c}{\omega }_1^0(\parallel x_0-x_n\parallel )+{\omega }_2^0(\parallel 2x_0-x_{-1}-2x_n+x_{n-1}\parallel ,\parallel x_{-1}-x_{n-1}\parallel )\\ \le {\omega }_1^1(\parallel x_0-x_n\parallel )+{\omega }_2^1(\parallel 2x_0-x_{-1}-2x_n+x_{n-1}\parallel ,\parallel x_{-1}-x_{n-1}\parallel )\end{array}$$

and

$$\begin{array}{c}\mathsf{\Phi }{\omega }_1(\parallel x_n-x_{n-1}\parallel )+{\omega }_2(\parallel 2x_{n-1}-x_{n-2}-x_n\parallel ,\parallel x_{n-1}-x_{n-2}\parallel )\\ \le \mathsf{\Phi }{\omega }_1^1(\parallel x_n-x_{n-1}\parallel )+{\omega }_2^1(\parallel 2x_{n-1}-x_{n-2}-x_n\parallel ,\parallel x_{n-1}-x_{n-2}\parallel )\\ \Rightarrow M_n\le M_n^1\Rightarrow M_n\parallel x_n-x_{n-1}\parallel \le M_n^1\parallel x_n-x_{n-1}\parallel ,\end{array}$$

where

$$M_n={\displaystyle \frac{\mathsf{\Phi }{\omega }_1(\parallel x_n-x_{n-1}\parallel )+{\omega }_2(\parallel 2x_{n-1}-x_{n-2}-x_n\parallel ,\parallel x_{n-1}-x_{n-2}\parallel )}{1-{\omega }_1^0(\parallel x_0-x_n\parallel )-{\omega }_2^0(\parallel 2x_0-x_{-1}-2x_n+x_{n-1}\parallel ,\parallel x_{-1}-x_{n-1}\parallel )}}\le M,$$

$$M_n^1={\displaystyle \frac{\mathsf{\Phi }{\omega }_1^1(\parallel x_n-x_{n-1}\parallel )+{\omega }_2^1(\parallel 2x_{n-1}-x_{n-2}-x_n\parallel ,\parallel x_{n-1}-x_{n-2}\parallel )}{1-{\omega }_1^1(\parallel x_0-x_n\parallel )-{\omega }_2^1(\parallel 2x_0-x_{-1}-2x_n+x_{n-1}\parallel ,\parallel x_{-1}-x_{n-1}\parallel )}}\le M^1,$$

but not necessarily vice versa unless, if

$${\omega }_1^1={\omega }_1^0\mathrm{and}{\omega }_2^1={\omega }_2^0.$$

2. Fewer iterates to obtain a desired error accuracy on $\parallel x_{n+1}-x_n\parallel$.

3. Better information on the location of the solution $x^{*}$.

Notice that (${\omega }_1^0$, ${\omega }_1$), (${\omega }_2^0$, ${\omega }_2$) are special cases of the old functions ${\omega }_1^1$, ${\omega }_2^1$, respectively. So no additional information or computational effort are required to obtain these improvements. This technique of using the restricted convergence region can be used to extend the applicability of other iterative methods along the same lines. Finally, Lipschitz functions and parameters can become at least as small, if $D_0$ is replaced by $D_1:=D\cap B(x_1,R_0-b)$, $b=max\{\eta ,\alpha \}$. Notice that $D_1\subseteq D_0$. The results then can be adjusted in this setting. Numerical examples where Equations (17)–(24) hold as strict inequalities can be found in [1,6].

3. Numerical Results

In this Section, we test the old and the new convergence criteria.

Let $X=Y=\mathbb{R}$. In this case $\parallel x\parallel =\left|x\right|$ for $x\in X$ or $x\in Y$, $D=(a,b)$, $D_0=(a_0,b_0)$. Let us define function $F+G:\mathbb{R}\to \mathbb{R}$, where

$$F\left(x\right)=e^{x-0.5}+x^3-1.3,G\left(x\right)=0.2x|x^2-2|.$$

The exact solution of $F\left(x\right)+G\left(x\right)=0$ is $x_{*}=0.5$. Let $D=(0,1.4)$. Then, we have

$$F^{\prime }\left(x\right)=e^{x-0.5}+3x^2,$$

$${\displaystyle G(x,y)=\frac{0.2x(2-x^2)-0.2y(2-y^2)}{x-y}=0.2(2-x^2-xy-y^2).}$$

$$A_0=e^{x_0-0.5}+3x_0^2+0.2(2-x_{-1}^2-x_{-1}{x}_0-x_0^2),$$

and

$${\displaystyle |A_0^{-1}(F^{\prime }\left(x\right)-F^{\prime }\left(y\right))|\le \frac{e^{b-0.5}+3|x+y|}{|A_0|}|x-y|,}$$

$${\displaystyle |A_0^{-1}(G(x,y)-G(u,v))|\le \frac{0.2}{|A_0|}\left(|u+x+y||u-x|+|v+y+u||v-y|\right).}$$

In view of this, we can write

$${\displaystyle {\omega }_1^0\left(\right|x-x_0\left|\right)=\frac{e^{b-0.5}+3|1.4+x_0|}{|A_0|}|x-x_0|,}$$

$${\displaystyle {\omega }_2^0\left(\right|x-(2x_0-x_{-1})|,|y-x_{-1}\left|\right)=\frac{0.2}{|A_0|}\left(|2x_0-x_{-1}+2b\left|\right|x-(2x_0-x_{-1})|+|2x_0+b\left|\right|y-x_{-1}|\right),}$$

$${\displaystyle {\omega }_1\left(\right|x-y\left|\right)=\frac{e^{b_0-0.5}+6b_0}{|A_0|}|x-y|,{\omega }_2\left(\right|2y-x-u\parallel ,|x-v\left|\right)=\frac{0.6b_0}{|A_0|}\left(\right|2y-x-u|+|x-v\left|\right),}$$

$${\displaystyle {\omega }_1^1\left(\right|x-y\left|\right)=\frac{e^{b-0.5}+6b}{|A_0|}|x-y|,{\omega }_2^1\left(\right|x-u|,|y-v\left|\right)=\frac{0.6b}{|A_0|}\left(|x-u|+|y-v|\right).}$$

Let $x_0=0.55$, $x_{-1}=0.551$. Then, we get $\alpha =0.001$, $\eta \approx 0.0479$, $R_0\approx 0.2011$, $m\approx 0.1431$, $m^1\approx 0.1750$, $D_0=D\cap B(x_0,3R_0)=(0,1.153)$. To get the radius of convergence we solve Equation (13) and similar one from [8]. Every such equation has two positive solutions. The smallest solutions satisfy conditions of appropriate theorems. So, we get $R\approx 0.0602$, $R^1\approx 0.0714$, $M\approx 0.2043$, $M^1\approx 0.3283$, $B(x_0,3R)=(0.3693,0.7307)\subset D$ and $B(x_0,3R^1)=(0.3359,0.7641)\subset D$. The error estimates are given in

Table 1. Results for $\epsilon ={10}^{-15}$.

n	$|{\mathit{x}}_{\mathit{n}+1}-{\mathit{x}}_{\mathit{n}}|$	${\mathit{M}}_{\mathit{n}}|{\mathit{x}}_{\mathit{n}}-{\mathit{x}}_{\mathit{n}-1}|$	$\mathit{M}|{\mathit{x}}_{\mathit{n}}-{\mathit{x}}_{\mathit{n}-1}|$	${\mathit{M}}_{\mathit{n}}^1|{\mathit{x}}_{\mathit{n}}-{\mathit{x}}_{\mathit{n}-1}|$	${\mathit{M}}^1|{\mathit{x}}_{\mathit{n}}-{\mathit{x}}_{\mathit{n}-1}|$
1	2.0602 × 10${}^{-3}$	6.8518 × 10${}^{-3}$	9.7951 × 10${}^{-3}$	9.1409 × 10${}^{-3}$	1.5738× 10${}^{-2}$
2	3.1428 × 10${}^{-6}$	8.9551 × 10${}^{-5}$	4.2097 × 10${}^{-4}$	1.1958 × 10${}^{-4}$	6.7636 × 10${}^{-4}$
3	7.0617 × 10${}^{-12}$	5.2824 × 10${}^{-9}$	6.4218 × 10${}^{-7}$	7.0457 × 10${}^{-9}$	1.0318 × 10${}^{-6}$
4	0	1.8032 × 10${}^{-17}$	1.4429 × 10${}^{-12}$	2.4050 × 10${}^{-17}$	2.3184 × 10${}^{-12}$

. For error $|x_{n+1}-x_n|$, $n\ge 1$, holds

$$|x_{n+1}-x_n|\le M_n|x_n-x_{n-1}|\le M|x_n-x_{n-1}|$$

and

$$|x_{n+1}-x_n|\le M_n^1|x_n-x_{n-1}|\le M^1|x_n-x_{n-1}|.$$

Let $x_0=0.53$, $x_{-1}=0.6$. Then, we get $\alpha =0.07$, $\eta \approx 0.0293$, $R_0\approx 0.1892$, $m\approx 0.1114$, $m^1\approx 0.1431$, $D_0=D\cap B(x_0,3R_0)=(0,1.0975)$. In this case only the largest solutions satisfy conditions of appropriate theorems. So, we get $R\approx 0.1624$, $R^1\approx 0.1109$, and $M\approx 0.8199$, $M^1\approx 0.7362$. Moreover, $B(x_0,3R)=(0.3676,0.6924)\subset D$ and $B(x_0,3R^1)=(0.4191,0.6409)\subset D$. The error estimates are given in

Table 2. Results for $\epsilon ={10}^{-15}$.

n	$|{\mathit{x}}_{\mathit{n}+1}-{\mathit{x}}_{\mathit{n}}|$	${\mathit{M}}_{\mathit{n}}|{\mathit{x}}_{\mathit{n}}-{\mathit{x}}_{\mathit{n}-1}|$	$\mathit{M}|{\mathit{x}}_{\mathit{n}}-{\mathit{x}}_{\mathit{n}-1}|$	${\mathit{M}}_{\mathit{n}}^1|{\mathit{x}}_{\mathit{n}}-{\mathit{x}}_{\mathit{n}-1}|$	${\mathit{M}}^1|{\mathit{x}}_{\mathit{n}}-{\mathit{x}}_{\mathit{n}-1}|$
1	7.3779 × 10${}^{-4}$	3.1227 × 10${}^{-3}$	2.3990 × 10${}^{-2}$	4.2444 × 10${}^{-3}$	2.1541 × 10${}^{-2}$
2	3.9991 × 10${}^{-7}$	1.6564 × 10${}^{-5}$	6.0489 × 10${}^{-4}$	2.2443 × 10${}^{-5}$	5.4314 × 10${}^{-4}$
3	1.1419 × 10${}^{-13}$	2.1230 × 10${}^{-10}$	3.2787 × 10${}^{-7}$	2.8737 × 10${}^{-10}$	2.9440 × 10${}^{-7}$
4	0	3.2808 × 10${}^{-20}$	9.3616 × 10${}^{-14}$	4.4410 × 10${}^{-20}$	8.4060 × 10${}^{-14}$

.

So, more accurate error estimates are retrieved because $M_n|x_n-x_{n-1}|\le M_n^1|x_n-x_{n-1}|$. Although $M|x_n-x_{n-1}|\le M^1|x_n-x_{n-1}|$ in the first case and $M|x_n-x_{n-1}|\ge M^1|x_n-x_{n-1}|$ in the second case.