Local Convergence of an Efficient Multipoint Iterative Method in Banach Space

Abstract

We discuss the local convergence of a derivative-free eighth order method in a Banach space setting. The present study provides the radius of convergence and bounds on errors under the hypothesis based on the first Fréchet-derivative only. The approaches of using Taylor expansions, containing higher order derivatives, do not provide such estimates since the derivatives may be nonexistent or costly to compute. By using only first derivative, the method can be applied to a wider class of functions and hence its applications are expanded. Numerical experiments show that the present results are applicable to the cases wherein previous results cannot be applied.

1. Introduction

We study local criteria for obtaining a unique solution $u_{\ast }$ of the nonlinear model

$$F\left(u\right)=0,$$

(1)

for Banach space valued mappings with $F:\mathsf{\Omega }\subset B\to B$, where F is differentiable in the sense of Fréchet [1,2]. For a good survey of literature on local and semilocal convergence criteria of iterative methods see [3,4,5,6,7,8,9,10,11,12,13].

The most popular numerical method for approximating a solution $u_{\ast }$ of Equation (1) is the quadratically convergent Newton’s method, which is expressed as

$$u_{n+1}=u_n-F^{\prime }{\left(u_n\right)}^{-1}F\left(u_n\right),\mathrm{for}\mathrm{each}n=0,1,2,\dots$$

In quest of efficient higher order method, a number of improved, multipoint Newton’s or Newton-like iterative schemes have been proposed in literature; see, for example [3,5,8,9,10,12,13,14,15,16,17,18,19] and references cited therein.

In particular, Amiri et al. [16] have recently developed an eighth order method for solving $F\left(u\right)=0$ using a derivative-free composite scheme. The method is of order eight using only divided differences, derivatives up to the order nine and Taylor expansions in the special case when $B={\mathbb{R}}^i$ and $Q\left(u\right)=\left(f_1^m\left(u\right),f_2^m\left(u\right),\cdots ,f_i^m\left(u\right)\right)$, $m\ge 2$, where $Q:B\to B$, $Q\left(u_{\ast }\right)=F\left(u_{\ast }\right)=0$. We study this method in the more general setting of a Banach space setting:

$$\begin{array}{cc}\hfill y_n=& u_n-{[u_n+Q\left(u_n\right),u_n;F]}^{-1}F\left(u_n\right),\hfill \\ \hfill z_n=& y_n-\left[\frac{13}{4}I-A_n\left(\frac{7}{2}I-\frac{5}{4}{A}_n\right)\right]{[u_n+Q\left(u_n\right),u_n;F]}^{-1}F\left(y_n\right),\hfill \\ \hfill u_{n+1}=& z_n-\left[\frac{7}{2}I-A_n\left(4I-\frac{3}{2}{A}_n\right)\right]{[u_n+Q\left(u_n\right),u_n;F]}^{-1}F\left(z_n\right),\hfill \end{array}$$

(2)

where $u_0\in \mathsf{\Omega }$ is an initial point, $A_n={[u_n+Q\left(u_n\right),u_n;F]}^{-1}[y_n+Q\left(y_n\right),y_n;F]$, $[.,.;F]:\mathsf{\Omega }\times \mathsf{\Omega }\to L(B,B)$ is divided difference of order one with

$$[u,y;F](u-y)=F\left(u\right)-F\left(y\right)$$

for each $u,y\in \mathsf{\Omega }$ with $u\ne y$ and $[u,u;F]=F^{\prime }\left(u\right)$ for each $u\in \mathsf{\Omega }$, if F is differentiable at u. Here $L(B,B)$ is the set of bounded linear operators from B into B.

The benefits of using method (2) over others in the literature have been well explained in [16]. Then to avoid repetitions, we refer the reader to [16]. But there are drawbacks when it comes to using method (2) limiting its applicability. These are: The existence of the ninth derivative is needed to show the order of convergence; the upper bounds on $\parallel u_n-u_{\ast }\parallel$ or results on the uniqueness of the solution are not given; the initial point is a shot in the dark; the method is restricted only on the i-dimensional Euclidean space and higher order derivatives do not appear on the method. Notice that the method cannot even guarantee convergence, if we consider the scalar function $\phi$ on $\mathsf{\Omega }=[-\frac{1}{2},\frac{3}{2}]$ given as

$$\phi \left(x\right)=\left\{\begin{array}{c}{x}^{3}ln{x}^2+x^5-x^4,x\ne 0\hfill \\ 0,x=0.\hfill \end{array}\right.$$

Then, clearly ${\phi }^{‴}\left(x\right)$ is unbounded on $\mathsf{\Omega }$. Hence, there is no assurance that ${lim}_{n\to \infty }{u}_n=u_{\ast }$ under the conditions in [16]. The novelty of this work is that we deal with all these drawbacks using only conditions on the divided difference of order one which actually used in Equation (2). Hence, we extend its applicability and for operators valued on Banach space.

2. Local Convergence Analysis

Certain real functions and parameters appearing in the local convergence analysis of Equation (2) are introduced. Set $S=[0,\infty ]$, let $w_0:S\times S\to S$, $w_1:S\to S$ be continuous and increasing functions with $w_0(0,0)=0$. Suppose that equation

$$w_0(w_1\left(x\right)x,x)=1$$

(3)

has at least one positive solution. Let ${\rho }_0$ be the smallest such solution. Set $S_0=[0,{\rho }_0)$. Let also $w:S_0\times S_0\to S$, $v:S_0\to S$ and $w_2:S_0\to S$ be continuous and increasing functions with $w(0,0)=0$. Define functions $p_1$ and ${\overline{p}}_1$ in the interval $S_0$ by

$$p_1\left(x\right)=\frac{w(w_1\left(x\right)x,x)}{a\left(x\right)}$$

and

$${\overline{p}}_1\left(x\right)=p_1\left(x\right)-1,$$

where $a\left(x\right)=1-w_0(w_1\left(x\right)x,x)$.

We have ${\overline{p}}_1\left(0\right)=-1<0$ and ${\overline{p}}_1\left(x\right)\to \infty$ as $x\to {\rho }_0^{-}$. The intermediate value theorem implies that equation ${\overline{p}}_1\left(x\right)=0$ has at least one solution in $(0,{\rho }_0)$. Let $r_1$ be the smallest such solution. Further assume that equation

$$w_0\left(w_1\left(p_1\left(x\right)x\right)p_1\left(x\right)x,p_1\left(x\right)x\right)=1$$

(4)

possesses at least one positive solution. Denote by ${\rho }_1$ the smallest such solution. Set ${\rho }_2=\min \{{\rho }_0,{\rho }_1\}$ and $S_1=[0,{\rho }_2)$. Define functions $p_2$ and ${\overline{p}}_2$ on the interval $S_1$ by

$$p_2\left(x\right)=\left(p_1\left(p_1\left(x\right)x\right)+\frac{d\left(x\right)}{a\left(x\right)b\left(x\right)}v\left(p_1\left(x\right)x\right)+\frac{1}{4}\left[4h\left(x\right)+5h^2\left(x\right)\right]\frac{v\left(p_1\left(x\right)x\right)}{a\left(x\right)}\right)p_1\left(x\right)$$

and

$${\overline{p}}_2\left(x\right)=p_2\left(x\right)-1,$$

where

$$\begin{array}{cc}\hfill & b\left(x\right)=1-w_0(w_1\left(p_1\left(x\right)x\right)p_1\left(x\right)x,p_1\left(x\right)x),\hfill \\ \hfill & d\left(x\right)=w_0(w_1\left(x\right)x,x)+w_0(w_1\left(p_1\left(x\right)x\right),p_1\left(x\right)x)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & h\left(x\right)=\frac{d\left(x\right)}{a\left(x\right)}.\hfill \end{array}$$

Then, we also get ${\overline{p}}_2\left(0\right)=-1$ and ${\overline{p}}_2\left(x\right)\to \infty$ as $x\to {\rho }_2^{-}$. Denote by $r_2$ the smallest solution of equation ${\overline{p}}_2\left(x\right)=0$ in $(0,{\rho }_2)$.

Assume that equation

$$w_0\left(w_1\left(p_2\left(x\right)x\right)p_2\left(x\right)x,p_2\left(x\right)x\right)=1$$

(5)

possesses at least one positive solution. Denote by ${\rho }_3$ the smallest such solution. Set $\rho =\min \{{\rho }_2,{\rho }_3\}$ and $S_2=[0,\rho )$. Define functions $p_3$ and ${\overline{p}}_3$ on the interval $S_2$ by

$$p_3\left(x\right)=\left(p_1\left(p_2\left(x\right)x\right)+\frac{e\left(x\right)v\left(p_2\left(x\right)x\right)}{a\left(x\right)c\left(x\right)}+\frac{1}{2}\left[2h\left(x\right)+3h^2\left(x\right)\right]\frac{v\left(p_2\left(x\right)x\right)}{a\left(x\right)}\right)p_2\left(x\right)$$

and

$${\overline{p}}_3\left(x\right)=p_3\left(x\right)-1,$$

where

$$\begin{array}{cc}\hfill & c\left(x\right)=1-w_0(w_1\left(p_2\left(x\right)x\right)p_2\left(x\right)x,p_2\left(x\right)x)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & e\left(x\right)=w_0(w_1\left(x\right)x,x)+w_0(w_1\left(p_2\left(x\right)x\right),p_2\left(x\right)x).\hfill \end{array}$$

We obtain again ${\overline{p}}_3\left(0\right)=-1$ and ${\overline{p}}_3\left(x\right)\to \infty$ as $x\to {\rho }^{-}$. Denote by $r_3$ the smallest solution of equation ${\overline{p}}_3\left(x\right)=0$ in $(0,\rho )$. Define parameter r by

$$r=\min \left\{r_j\right\},j=1,2,3.$$

(6)

This parameter shall be shown to be a radius of convergence for Equation (2) in Theorem 1. Then, we have that for each $x\in [0,r)$

$$\begin{array}{cc}\hfill & a\left(x\right)>0,\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill & b\left(x\right)>0,\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill & c\left(x\right)>0,\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill & d\left(x\right)\ge 0,\hfill \end{array}$$

(10)

$$\begin{array}{cc}\hfill & e\left(x\right)\ge 0,\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill & h\left(x\right)\ge 0,\hfill \end{array}$$

(12)

and

$$\begin{array}{cc}\hfill & 0\le p_i\left(x\right)\le 1.\hfill \end{array}$$

(13)

By $U(\mu ,\lambda )$, $\overline{U}(\mu ,\lambda )$ we denote the open and closed balls in B, respectively with center $\mu \in B$ and of radius $\lambda >0$. In order to study the local convergence of Equation (2), we need to rewrite the three steps.

Lemma 1.

Suppose that Equation (2) is well defined for each $n=0,1,2,....$ and ${[u+Q\left(u\right),u;F]}^{-1}\in L(B,B)$ for each $u\in \mathsf{\Omega }$. Then, the following assertions hold

$$\begin{array}{cc}\hfill y_n-u_{\ast }=& {[u_n+Q\left(u_n\right),u_n;F]}^{-1}\left([u_n+Q\left(u_n\right),u_n;F]-[u_n,u_{\ast };F]\right)(u_n-u_{\ast }),\hfill \end{array}$$

(14)

$$\begin{array}{cc}\hfill z_n-u_{\ast }=& y_n-u_{\ast }+{[y_n+Q\left(y_n\right),y_n;F]}^{-1}F\left(y_n\right)+\left({[y_n+Q\left(y_n\right),y_n;F]}^{-1}-{[u_n+Q\left(u_n\right),u_n;F]}^{-1}\right)F\left(y_n\right)\hfill \\ & -\frac{1}{4}[4(I-A_n)+5{(I-A_n)}^2]{[u_n+Q\left(u_n\right),u_n;F]}^{-1}F\left(y_n\right)\hfill \end{array}$$

(15)

and

$$\begin{array}{cc}\hfill u_{n+1}-u_{\ast }=& z_n-u_{\ast }-{[z_n+Q\left(z_n\right),z_n;F]}^{-1}F\left(z_n\right)+\left({[z_n+Q\left(z_n\right),z_n;F]}^{-1}-{[u_n+Q\left(u_n\right),u_n;F]}^{-1}\right)F\left(z_n\right)\hfill \\ & -\frac{1}{2}[2(I-A_n)+3{(I-A_n)}^2]{[u_n+Q\left(u_n\right),u_n;F]}^{-1}F\left(z_n\right).\hfill \end{array}$$

(16)

Proof. 

We have in turn by the first substep of Equation (2) and the definition of the divided difference

$$\begin{array}{cc}\hfill y_n-u_{\ast }& =u_n-u_{\ast }-{[u_n+Q\left(u_n\right),u_n;F]}^{-1}F\left(u_n\right)\hfill \\ & ={[u_n+Q\left(u_n\right),u_n;F]}^{-1}\left([u_n+Q\left(u_n\right),u_n;F]-[u_n,u_{\ast };F]\right)(u_n-u_{\ast })\hfill \end{array}$$

which shows Equation (14).

Then, similarly from the second substep of Equation (2)

$$\begin{array}{cc}\hfill z_n-u_{\ast }=& y_n-u_{\ast }-{[y_n+Q\left(y_n\right),y_n;F]}^{-1}F\left(y_n\right)+\left({[y_n+Q\left(y_n\right),y_n;F]}^{-1}-{[u_n+Q\left(u_n\right),u_n;F]}^{-1}\right)F\left(y_n\right)\hfill \\ \hfill & -\frac{1}{4}(9I-14A_n+5A_n^2){[u_n+Q\left(u_n\right),u_n;F]}^{-1}F\left(y_n\right)\hfill \\ \hfill =& y_n-u_{\ast }-{[y_n+Q\left(y_n\right),y_n;F]}^{-1}F\left(y_n\right)+\left({[y_n+Q\left(y_n\right),y_n;F]}^{-1}-{[u_n+Q\left(u_n\right),u_n;F]}^{-1}\right)F\left(y_n\right)\hfill \\ \hfill & -\frac{1}{4}[4(I-A_n)+5{(I-A_n)}^2]{[u_n+Q\left(u_n\right),u_n;F]}^{-1}F\left(y_n\right),\hfill \end{array}$$

which shows Equation (15). Finally, from the third substep of Equation (2), we obtain in turn that

$$\begin{array}{cc}\hfill u_{n+1}-u_{\ast }=& z_n-u_{\ast }-{[z_n+Q\left(z_n\right),z_n;F]}^{-1}F\left(z_n\right)+\left({[z_n+Q\left(z_n\right),z_n;F]}^{-1}-{[u_n+Q\left(u_n\right),u_n;F]}^{-1}\right)F\left(z_n\right)\hfill \\ \hfill & -\left(\frac{5}{2}I-A_n(4I-\frac{3}{2}{A}_n)\right){[u_n+Q\left(u_n\right),u_n;F]}^{-1}F\left(z_n\right)\hfill \\ \hfill =& z_n-u_{\ast }-{[z_n+Q\left(z_n\right),z_n;F]}^{-1}F\left(z_n\right)\left({[z_n+Q\left(z_n\right),z_n;F]}^{-1}-{[u_n+Q\left(u_n\right),u_n;F]}^{-1}\right)F\left(z_n\right)\hfill \\ \hfill & -\frac{1}{2}[2(I-A_n)+3{(I-A_n)}^2]{[u_n+Q\left(u_n\right),u_n;F]}^{-1}F\left(z_n\right),\hfill \end{array}$$

which completes the proof. □

The local convergence analysis is based on the following conditions (say, A):

$\left(a_1\right)$

$F:\mathsf{\Omega }\to B$ is continuously differentiable in the sense of Frèchet, $[.,.;F]:\mathsf{\Omega }\times \mathsf{\Omega }\to L(B,B)$, $[.,.;Q]:\mathsf{\Omega }\times \mathsf{\Omega }\to L(B,B)$ are a divided difference of order one and there exists $u_{\ast }\in \mathsf{\Omega }$ such that $F\left(u_{\ast }\right)=Q\left(u_{\ast }\right)=0$ and $F^{\prime }{\left(u_{\ast }\right)}^{-1}\in L(B,B).$

$\left(a_2\right)$

There exist continuous and increasing functions $w_0:S\times S\to S$ and $w_1:S\to S$ with $w_0(0,0)=0$ such that for each $u\in \mathsf{\Omega }$

$$\parallel F^{\prime }{\left(u_{\ast }\right)}^{-1}([u+Q\left(u\right),u;F]-F^{\prime }\left(u_{\ast }\right))\parallel \le w_0(\parallel u+Q\left(u\right)-u_{\ast }\parallel ,\parallel u-u_{\ast }\parallel ).$$

and

$$\parallel I+[u,u_{\ast };Q]\parallel \le w_1(\parallel u-u_{\ast }\parallel ).$$

Set ${\mathsf{\Omega }}_0=\mathsf{\Omega }\bigcap U(u_{\ast },{\rho }_0),$ where ${\rho }_0$ is given in Equation (3).

$\left(a_3\right)$

There exist continuous and increasing functions $w:S_0\times S_0\to S$, $w_2:S_0\to S$ and $v:S_0\to S$ such that for each $u\in {\mathsf{\Omega }}_0$

$$\begin{array}{cc}\hfill & \parallel F^{\prime }{\left(u_{\ast }\right)}^{-1}([u+Q\left(u\right),u;F]-[u,u_{\ast };F])\parallel \le w(\parallel Q\left(u\right)\parallel ,\parallel u-u_{\ast }\parallel ),\hfill \\ \hfill & \parallel Q\left(u\right)\parallel \le w_2(\parallel u-u_{\ast }\parallel )\parallel u-u_{\ast }\parallel \hfill \end{array}$$

and

$$\parallel F^{\prime }{\left(u_{\ast }\right)}^{-1}[u,u_{\ast };F]\parallel \le v(\parallel u-u_{\ast }\parallel ).$$

$\left(a_4\right)$

$\overline{U}(u_{\ast },\overline{r})\subseteq \mathsf{\Omega },$$\overline{r}=w_1\left(r\right)r,$${\rho }_0$, ${\rho }_1$ and ${\rho }_3$ given by Equations (3)–(5), respectively exist and r is defined in Equation (6).

$\left(a_5\right)$

There exists $r_{\ast }\ge r$ such that

$$v_0\left(r_{\ast }\right)<1,$$

where function $v_0:S_0\to S$ is continuous and increasing with $v_0\left(0\right)=0.$ Set ${\mathsf{\Omega }}_1=\mathsf{\Omega }\bigcap U(u_{\ast },r_{\ast }).$

By Lemma 1, we can use the notations

$$\begin{array}{cc}\hfill a_n=& 1-w_0\left(w_1(\parallel u_n-u_{\ast }\parallel )\parallel u_n-u_{\ast }\parallel ,\parallel u_n-u_{\ast }\parallel \right),\hfill \\ \hfill b_n=& 1-w_0\left(w_1(p_1(\parallel u_n-u_{\ast }\parallel )\parallel u_n-u_{\ast }\parallel ),p_1(\parallel u_n-u_{\ast }\parallel )\parallel u_n-u_{\ast }\parallel \right),\hfill \\ \hfill c_n=& 1-w_0\left(w_1(p_2(\parallel u_n-u_{\ast }\parallel )\parallel u_n-u_{\ast }\parallel )p_2(\parallel u_n-u_{\ast }\parallel )\parallel u_n-u_{\ast }\parallel ,p_2(\parallel u_n-u_{\ast }\parallel )\parallel u_n-u_{\ast }\parallel \right),\hfill \\ \hfill d_n=& w_0\left(w_1(\parallel u_n-u_{\ast }\parallel )\parallel u_n-u_{\ast }\parallel ,\parallel u_n-u_{\ast }\parallel \right)+w_0\left(w_1(p_1(\parallel u_n-u_{\ast }\parallel )\parallel u_n-u_{\ast }\parallel ),p_1(\parallel u_n-u_{\ast }\parallel )\parallel u_n-u_{\ast }\parallel \right),\hfill \\ \hfill e_n=& w_0\left(w_1(\parallel u_n-u_{\ast }\parallel )\parallel u_n-u_{\ast }\parallel ,\parallel u_n-u_{\ast }\parallel \right)+w_0\left(w_1(p_2(\parallel u_n-u_{\ast }\parallel )\parallel u_n-u_{\ast }\parallel ),p_2(\parallel u_n-u_{\ast }\parallel )\parallel u_n-u_{\ast }\parallel \right)\hfill \end{array}$$

and

$$h_n=\frac{d_n}{a_n}.$$

Next, we present the local convergence of Equation (2) using the conditions (A) and the notations mentioned above.

Theorem 1.

Suppose that the conditions (A) hold. Then, sequence $\left\{u_n\right\}$ generated for $u_0\in U(u_{\ast },r)-\left\{u_{\ast }\right\}$ is well defined, remains in $U(u_{\ast },r)$ for each $n=0,1,2......$ and converges to $u_{\ast }$, so that

$$\begin{array}{cc}\hfill \parallel y_n-u_{\ast }\parallel & \le p_1(\parallel u_n-u_{\ast }\parallel )\parallel u_n-u_{\ast }\parallel \le \parallel u_n-u_{\ast }\parallel <r,\hfill \end{array}$$

(17)

$$\begin{array}{cc}\hfill \parallel z_n-u_{\ast }\parallel & \le p_2(\parallel u_n-u_{\ast }\parallel )\parallel u_n-u_{\ast }\parallel \le \parallel u_n-u_{\ast }\parallel \hfill \end{array}$$

(18)

and

$$\begin{array}{cc}\hfill \parallel u_{n+1}-u_{\ast }\parallel & \le p_3(\parallel u_n-u_{\ast }\parallel )\parallel u_n-u_{\ast }\parallel \le \parallel u_n-u_{\ast }\parallel ,\hfill \end{array}$$

(19)

where the functions $p_i$ are given previously and r is defined in Equation (6).

Proof. 

We shall show estimates for Equations (17)–(19) using mathematical induction. Let $u\in U(u_{\ast },r)-\left\{u_{\ast }\right\}.$ By $\left(a_1\right)$, $\left(a_2\right)$, $\left(a_4\right)$ and Equation (6), we obtain in turn

$$\begin{array}{cc}\hfill \parallel F^{\prime }{\left(u_{\ast }\right)}^{-1}\left([u+Q\left(u\right),u;F]-F^{\prime }\left(u_{\ast }\right)\right)\parallel & \le w_0(\parallel u+Q\left(u\right)-u_{\ast }\parallel ,\parallel u-u_{\ast }\parallel )\hfill \\ & \le w_0\left(\parallel (I+[u,u_{\ast }])(u-u_{\ast })\parallel ,\parallel u-u_{\ast }\parallel \right)\hfill \\ & \le w_0\left(w_1(\parallel u-u_{\ast }\parallel )\parallel u-u_{\ast }\parallel ,\parallel u-u_{\ast }\parallel \right)\hfill \\ & <w_0(w_1\left(r\right)r,r)r<1,\hfill \end{array}$$

(20)

where we also used that

$$\parallel u+Q\left(u\right)-u_{\ast }\parallel =\parallel (I+[u,u_{\ast },Q])(u-u_{\ast })\parallel \le w_1(\parallel u-u_{\ast }\parallel )\parallel u-u_{\ast }\parallel \le w_1\left(r\right)r=\overline{r},$$

so $u+Q\left(u\right)-u_{\ast }\in U(u_{\ast },\overline{r}).$

It follows from Equation (20) and the Banach perturbation lemma on invertible operators [8] that ${[u+Q\left(u\right),u;F]}^{-1}\in L(B,B)$ and

$${\parallel [u+Q\left(u\right),u;F]}^{-1}{F}^{\prime }\left(u_{\ast }\right)\parallel \le \frac{1}{a(\parallel u-u_{\ast }\parallel )}$$

(21)

and $y_0$, $z_0$, $u_1$ are well defined by Equation (2) for $n=0.$

Then, by Equations (6) and (13) (for $j=1$), (14) and (21) (for $u=u_0$), and $\left(a_3\right)$, we get, in turn, that

$$\begin{array}{cc}\hfill \parallel y_0-u_{\ast }\parallel \le & \parallel {[u_0+Q\left(u_0\right),u_0;F]}^{-1}{F}^{\prime }\left(u_{\ast }\right)\parallel \parallel F^{\prime }{\left(u_{\ast }\right)}^{-1}\left([u_0+Q\left(u_0\right),u_0;F]-[u_0,u_{\ast };F]\right)\parallel \parallel u-u_{\ast }\parallel \hfill \\ \hfill \le & \frac{w\left(w_1(\parallel u_0-u_{\ast }\parallel )\parallel u_0-u_{\ast }\parallel ,\parallel u_0-u_{\ast }\parallel \right)\parallel u_0-u_{\ast }\parallel }{a(\parallel u_0-u_{\ast }\parallel )}\hfill \\ \hfill =& p_1(\parallel u_0-u_{\ast }\parallel )\parallel u_0-u_{\ast }\parallel \le \parallel u_0-u_{\ast }\parallel <r,\hfill \end{array}$$

(22)

so Equation (17) holds for $n=0$ and $y_0\in U(u_{\ast },r).$ We need the estimate obtained using $\left(a_2\right)$ and (20)

$$\begin{array}{cc}\hfill \parallel I-A_0\parallel =& \parallel \left({[u_0+Q\left(u_0\right),u_0;F]}^{-1}{F}^{\prime }\left(u_{\ast }\right)\right)F^{\prime }{\left(u_{\ast }\right)}^{-1}\hfill \\ \hfill & \times \left[\left([u_0+Q\left(u_0\right),u_0;F]-F^{\prime }\left(u_{\ast }\right)\right)+\left(F^{\prime }\left(u_{\ast }\right)-[y_0+Q\left(y_0\right),y_0;F]\right)\right]\parallel \hfill \\ \hfill \le & \frac{d(\parallel u-u_{\ast }\parallel )}{a(\parallel u-u_{\ast }\parallel )}\hfill \end{array}$$

(23)

and the estimate using $\left(a_1\right)$ and $\left(a_3\right)$ that

$$F\left(u\right)-F\left(u_{\ast }\right)=[u,u_{\ast };F](u-u_{\ast }),$$

so

$$\parallel F^{\prime }{\left(u_{\ast }\right)}^{-1}[u,u_{\ast };F]\parallel \le v(\parallel u-u_{\ast }\parallel ).$$

(24)

Then, by Equations (6), (13) (for $j=2$), (15), and Equations (21)–(24) (for $u=y_0$), we have, in turn, that

$$\begin{array}{cc}\hfill \parallel z_0-u_{\ast }\parallel \le & (p_1\left(p_1(\parallel u_0-u_{\ast }\parallel )\parallel u_0-u_{\ast }\parallel \right)+\frac{d(\parallel u_0-u_{\ast }\parallel )v\left(p_1(\parallel u_0-u_{\ast }\parallel )\parallel u_0-u_{\ast }\parallel \right)}{a(\parallel u_0-u_{\ast }\parallel \left)b\right(\parallel u_0-u_{\ast }\parallel )}\hfill \\ \hfill & +\frac{1}{4}\left[4\frac{d(\parallel u_0-u_{\ast }\parallel )}{a(\parallel u_0-u_{\ast }\parallel )}+5{\left(\frac{d(\parallel u_0-u_{\ast }\parallel )}{a(\parallel u_0-u_{\ast }\parallel )}\right)}^2\right]\frac{v(p_1(\parallel u_0-u_{\ast }\parallel )\parallel u_0-u_{\ast }\parallel )}{a\parallel u_0-u_{\ast }\parallel })p_1(\parallel u_0-u_{\ast }\parallel )\hfill \\ \hfill =& p_2(\parallel u_0-u_{\ast }\parallel )\parallel u_0-u_{\ast }\parallel \le \parallel u_0-u_{\ast }\parallel ,\hfill \end{array}$$

(25)

so Equation (18) holds for $n=0$ and $z_0\in U(u_{\ast },r).$ Next, from Equations (6) and (13) (for $j=3$), (21) (for $u=z_0$), and Equations (22)–(25), we obtain in turn that

$$\begin{array}{cc}\hfill \parallel u_1-u_{\ast }\parallel \le & (p_1\left(p_2(\parallel u_0-u_{\ast }\parallel )\parallel u_0-u_{\ast }\parallel \right)+\frac{e(\parallel u_0-u_{\ast }\parallel )v\left(p_2(\parallel u_0-u_{\ast }\parallel )\parallel u_0-u_{\ast }\parallel \right)}{a(\parallel u_0-u_{\ast }\parallel \left)c\right(\parallel u_0-u_{\ast }\parallel )}\hfill \\ \hfill & +\frac{1}{2}\left[2\frac{d(\parallel u_0-u_{\ast }\parallel )}{a(\parallel u_0-u_{\ast }\parallel )}+3{\left(\frac{d(\parallel u_0-u_{\ast }\parallel )}{a(\parallel u_0-u_{\ast }\parallel )}\right)}^2\right]\frac{v(p_2(\parallel u_0-u_{\ast }\parallel )\parallel u_0-u_{\ast }\parallel )}{a\parallel u_0-u_{\ast }\parallel })p_2(\parallel u_0-u_{\ast }\parallel )\hfill \\ \hfill =& p_3(\parallel u_0-u_{\ast }\parallel )\parallel u_0-u_{\ast }\parallel ,\hfill \end{array}$$

(26)

so Equation (19) holds for $n=0$ and $u_1\in U(u_{\ast },r).$ The induction for estimates of Equations (17)–(19) is terminated, if we replace $u_0$, $y_0$, $z_0$, $u_1$ by $u_m$, $y_m$, $z_m$, $u_{m+1}$ in the preceding computations.

Then, from the estimate

$$\parallel u_{m+1}-u_{\ast }\parallel \le q\parallel u_m-u_{\ast }\parallel <r,q=p_3(\parallel u_0-u_{\ast }\parallel )\in [0,1),$$

(27)

we get that ${\lim }_{m\to \infty }{u}_m=u_{\ast }$ and $u_{m+1}\in U(u_{\ast },r).$ Finally, set $G=[y_{\ast },u_{\ast };F]$ for some $y_{\ast }\in {\mathsf{\Omega }}_1$ for $F\left(y_{\ast }\right)=0.$ Using $\left(a_5\right),$ we have

$$\parallel F^{\prime }{\left(u_{\ast }\right)}^{-1}(G-F^{\prime }\left(u_{\ast }\right))\parallel \le v_0(\parallel y_{\ast }-u_{\ast }\parallel )\le v\left(r_{\ast }\right)<1,$$

so $G^{-1}$ is invertible. Then, we obtain $u_{\ast }=y_{\ast }$ via identity

$$0=F\left(u_{\ast }\right)-F\left(y_{\ast }\right)=G(u_{\ast }-y_{\ast }).$$

□

3. Numerical Results

It is noted that in all examples $r_i$ are found by solving scalar equations ${\overline{p}}_i\left(x\right)=0$, $i=1,2,3$. Then, r is obtained using Equation (6). The parameters $r_i$ have been shown to exist above Lemma 1. The divided difference in all examples is chosen as

$$[x,y,F]={\int }_0^1{F}^{\prime }(y+\theta (x-y))\mathrm{d}\theta .$$

(28)

All computations are performed in Mathematica software using multi-precision arithmetic.

Example 1.

Let $B=C[0,1]$, $\mathsf{\Omega }=\overline{U}(u_{\ast },1)$. Consider the Hammerstein-type problem as

$$u\left(s\right)={\int }_0^{1}K(s,x)\frac{u{\left(x\right)}^2}{2}dx,$$

where

$$K(s,x)=\left\{\begin{array}{c}(1-s)x,x\le s,\hfill \\ s(1-x),s\le x.\hfill \end{array}\right.$$

Let $F:\mathcal{C}[0,1]$ be defined as

$$F\left(u\right)\left(s\right)=u\left(s\right)-{\int }_0^{1}K(s,x)\frac{u{\left(x\right)}^2}{2}dx.$$

But, we get

$$\parallel {\int }_0^{1}K(s,x)dx\parallel \le \frac{1}{8},$$

leading to

$$F^{\prime }\left(u\right)y\left(s\right)=y\left(s\right)-{\int }_0^{1}K(s,x)u\left(x\right)dx,$$

since $F^{\prime }\left(u_{\ast }\left(s\right)\right)=I$,

$$\parallel F{\left(u_{\ast }\right)}^{-1}(F\left(u\right)-F^{\prime }\left(y\right))\parallel <\frac{1}{8}\parallel u-y\parallel .$$

By Equation (28), we select $w_0(s,x)=w(s,x)=\frac{s+x}{16}$, $v\left(x\right)=w_1\left(x\right)=8$. The parameters are

$$r_1=0.888889,r_2=0.118175,r_3=1.42264\times {10}^{-2}\mathrm{and}r=1.42264\times {10}^{-2}.$$

Example 2.

Consider the three-dimensional system

$$\begin{array}{cc}\hfill & f_1^{\prime }\left(u_1\right)-f_1\left(u_1\right)-1=0,\hfill \\ \hfill & f_2^{\prime }\left(u_2\right)-(e-1)u_2-1=0,\hfill \\ \hfill & f_3^{\prime }\left(u_3\right)-1=0,\hfill \end{array}$$

with $u_1,u_2,u_3\in \mathsf{\Omega }$ for $f_1\left(0\right)=f_2\left(0\right)=f_3\left(0\right)=0$. Then, the system for $u={(u_1,u_2,u_3)}^T$ with $F:=(f_1,f_2,f_3):\mathsf{\Omega }\to {\mathbb{R}}^3$ is

$$F\left(u\right)={\left(e^{u_1}-1,\frac{e-1}{2}{u}_2{}^2+u_2,u_3\right)}^T.$$

Hence, we obtain

$$F^{\prime }\left(u\right)=\left[\begin{array}{ccc}{e}^{u_1}& 0& 0\\ 0& (e-1)u_2+1& 0\\ 0& 0& 1\end{array}\right].$$

But $u_{\ast }={(0,0,0)}^T$ and $F^{\prime }\left(u_{\ast }\right)=I$, by the definition in Equation (28), we select $w_0(s,x)=\frac{L_0}{2}(s+x)$, $w(s,x)=\frac{L}{2}(s+x)$, and $w_1\left(x\right)=v\left(x\right)=\frac{1}{2}(1+e^{\frac{1}{L_0}})$, where $L_0=e-1$, $L=e$. Then, we have

$$r_1=1.88242\times {10}^{-1},r_2=3.70077\times {10}^{-2},r_3=7.34314\times {10}^{-3}\mathrm{and}r=7.34314\times {10}^{-3}.$$

Example 3.

Consider $F:=(f_1,f_2,f_3):\mathsf{\Omega }\to {\mathbb{R}}^3$ be defined by

$$F\left(u\right)={(10u_1+sin(u_1+u_2)-1,8u_2-{cos}^2(u_3-u_2)-1,12u_3+sin\left(u_3\right)-1)}^T,$$

where $u={(u_1,u_2,u_3)}^T$.

Then, we obtain

$$F^{\prime }\left(u\right)=\left[\begin{array}{ccc}10+cos(u_1+u_2)& cos(u_1+u_2)& 0\\ 0& 8+sin2(u_2-u_3)& -sin2(u_2-u_3)\\ 0& 0& 12+cos\left(u_3\right)\end{array}\right].$$

Hence, by Equation (28) $w_1\left(x\right)=v\left(x\right)=0.269812x$, and $w(s,x)=w_0(s,x)=\frac{1.08139}{2}(s+x)$. Then we obtain the parameters as

$$r_1=0.766299,r_2=0.658762,r_3=0.637403,r=0.637403.$$

Example 4.

Define function F on $\mathsf{\Omega }=\overline{U}(0,1)$, given as

$$F\left(\varphi \right)\left(s\right)=\varphi \left(s\right)-10{\int }_0^{1}sx\varphi \left(x\right)dx.$$

Then, we get

$$F^{\prime }\left(\varphi \left(\xi \right)\right)\left(s\right)=\xi \left(s\right)-30{\int }_0^{1}sx\varphi {\left(x\right)}^2\xi \left(x\right)dx,forall\xi \in \mathsf{\Omega }.$$

By Equation (28), we can choose $w_1\left(x\right)=v\left(x\right)=30$ and $w(s,x)=w_0(s,x)=\frac{15}{2}(s+x)$. The parameter values are given as

$$r_1=2.15054\times {10}^{-3},r_2=2.24959\times {10}^{-9},r_3=9.81657\times {10}^{-16},r=9.81657\times {10}^{-16}.$$

Example 5.

The Van der Waals equation of state for a vapor is (see [16])

$$\left(P+\frac{a}{V^2}\right)(V-b)=RT,$$

This equation leads to,

$${PV}^3-(Pb+RT)V^2+aV-ab=0$$

in V, where all constants have a physical meaning whose values can be found in [16]. Choose P = 10,000 kPa and T = 800 K. Then, $u_{\ast }=36.9167\dots$. By Equation (28), we can set $w_1\left(x\right)=v\left(x\right)=10$ and $w(s,x)=w_0(s,x)=0.386121(s+x)$. The parameter values are given as

$$r_1=1.17721\times {10}^{-1},r_2=8.79936\times {10}^{-4},r_3=2.96866\times {10}^{-6},r=2.96866\times {10}^{-6}.$$

4. Conclusions

In the forgoing study, the local convergence of an eighth order derivative-free method is discussed comprehensively in Banach space. Far from other methods that depend on higher derivatives and Taylor series, we have considered only the first derivative in our procedure. In this sense, the method can be applied to a wider class of functions and hence its applications are expanded. Another advantage of analyzing the convergence is the computation of a convergence ball (wherein the iterates lie) and error estimates. Theoretical results of analysis so derived are confirmed through numerical testing on some practical problems.