On the Semi-Local Convergence of a Traub-Type Method for Solving Equations

Abstract

The celebrated Traub’s method involving Banach space-defined operators is extended. The main feature in this study involves the determination of a subset of the original domain that also contains the Traub iterates. In the smaller domain, the Lipschitz constants are smaller too. Hence, a finer analysis is developed without the usage of additional conditions. This methodology applies to other methods. The examples justify the theoretical results.

1. Introduction

The purpose of this article is to locate a solution $x^{*}$ of equation

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

(1)

provided that $F:\mathsf{\Omega }\subset E_1⟶E_2$ is derivable according to Fréchet. Moreover, $E_1,E_2$ stand for Banach spaces, whereas $\mathsf{\Omega }$ is nonempty and open.

The famous quadratically convergent Newton–Kantorovich method is defined for all $j=0,1,2,\dots$ as

$$x_0\in \mathsf{\Omega },x_{j+1}=x_j-F^{\prime }{\left(x_j\right)}^{-1}F\left(x_j\right)$$

(2)

has been used extensively to produce sequence $\left\{x_j\right\}$ such that ${lim}_{j⟶\infty }{x}_j=x^{*}$ [1,2,3,4,5,6,7,8]. Although there is a plethora on convergence results for (2) there exist some problems. In particular, the convergence ball is in general small [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]. Hence, it is important to extend this ball but with no additional conditions. Other defects relate to the accuracy of bounds on $\parallel x_{j+1}-x_j\parallel$ or $\parallel x_j-x^{*}\parallel ,$ as well as the results on the location uniqueness of $x^{*}.$ The same defects appear in the study of high convergence order methods [19,20,21,22]. We have developed a technique that helps determine some $D\subseteq \mathsf{\Omega },$ where iterates can also be found. This way, using D instead of $\mathsf{\Omega }$, a finer analysis is possible with no additional conditions.

We demonstrate our techniques for a certain high convergence order, although it can similarly be used on other methods [12,15,16,17].

We extend the two step Traub method [21] (see also [18,22]) to the following three step fifth order method

$$\begin{array}{ccc}\hfill y_j& =& x_j-F^{\prime }{\left(x_j\right)}^{-1}F\left(x_j\right)\hfill \\ \hfill z_j& =& y_j-F^{\prime }{\left(x_j\right)}^{-1}F\left(y_j\right)\hfill \\ \hfill x_{j+1}& =& z_j-F^{\prime }{\left(x_j\right)}^{-1}F\left(z_j\right).\hfill \end{array}$$

(3)

Traub’s two-step method requires less computational effort than any third-order method utilizing the second derivative [2,4,5,14].

Let us provide the earlier results.

(i)

Convergence has been shown in Potra and Pták in [14] using

$$\parallel F^{\prime }{\left(x_0\right)}^{-1}F\left(x_0\right)\parallel \le \mu$$

$$\parallel F^{\prime }{\left(x_0\right)}^{-1}(F^{\prime }\left(w\right)-F^{\prime }\left(v\right))\parallel \le L_1\parallel w-v\parallel \mathrm{for} \mathrm{all}w,v\in \mathsf{\Omega }$$

(4)

$$T_1=2L_1\mu \le 1$$

(5)

and

$$U[x_0,{\rho }_0]\subset \mathsf{\Omega },$$

and

$${\rho }_0=\frac{1-\sqrt{1-T_1}}{L_1},$$

(6)

where $U[x_0,{\rho }_0]$ is a closed ball with radius ${\rho }_0$ and center at $x_0\in \mathsf{\Omega }.$ The center-Lipschitz condition is introduced by us as

$$\parallel F^{\prime }{\left(x_0\right)}^{-1}(F^{\prime }\left(w\right)-F^{\prime }\left(x_0\right))\parallel \le L_0\parallel w-x_0\parallel \mathrm{for} \mathrm{all}w\in \mathsf{\Omega }.$$

(7)

Define the set

$$D=U(x_0,\frac{1}{L_0})\cap \mathsf{\Omega }.$$

(8)

Moreover, we introduced the restricted-Lipschitz condition

$$\parallel F^{\prime }{\left(x_0\right)}^{-1}(F^{\prime }\left(w\right)-F^{\prime }\left(v\right))\parallel \le L\parallel w-v\parallel \mathrm{for} \mathrm{all}w,v\in D.$$

(9)

However, then we notice that

$$L_0\le L_1$$

(10)

and

$$L\le L_1$$

(11)

hold, since

$$D\subseteq \mathsf{\Omega }.$$

(12)

Suppose

$$T=2L\mu \le 1.$$

(13)

It follows that (9), (13), and

$${\overline{\rho }}_0=\frac{1-\sqrt{1-T}}{L}$$

(14)

can used for (4), (5), and ${\rho }_0,$ respectively, given in [14] (Theorem 5.2, p. 79). Hence,

$$T_1\le 1\Rightarrow T\le 1$$

(15)

and

$${\overline{\rho }}_0\le {\rho }_0$$

(16)

hold. So, the applicability of Traub’s method is extended. The parameters $L_0$ and L are special cases of $L_1,$ so no additional effort is used. It is also worth to mention that $L_1=L_1\left(\mathsf{\Omega }\right),L_0=L_0\left(\mathsf{\Omega }\right)$ but $L=L(\mathsf{\Omega },L_0).$ The proof in [14] (Theorem 5.2) utilized

$$\parallel F^{\prime }{\left(x_0\right)}^{-1}(F^{\prime }\left(w\right)-F^{\prime }\left(x_0\right))\parallel \le L_1\parallel w-x_0\parallel <1$$

leading to (by the Banach lemma [12] on linear invertible operators)

$$\parallel F^{\prime }{\left(w\right)}^{-1}{F}^{\prime }\left(x_0\right)\parallel \le \frac{1}{1-L_1\parallel w-x_0\parallel }.$$

However, we get

$$\parallel F^{\prime }{\left(x_0\right)}^{-1}(F^{\prime }\left(w\right)-F^{\prime }\left(x_0\right))\parallel \le L_0\parallel w-x_0\parallel <1$$

leading to tighter

$$\parallel F^{\prime }{\left(w\right)}^{-1}{F}^{\prime }\left(x_0\right)\parallel \le \frac{1}{1-L_0\parallel w-x_0\parallel }.$$

This modification in the proof brings the aforementioned advantages. In the numerical section one can find cases when (10)–(12) are strict.

(ii)

In [12], they used

$$\parallel F^{\prime }{\left(x_0\right)}^{-1}F\left(x_0\right)\parallel \le \mu ,$$

$$\parallel F^{\prime }{\left(x_0\right)}^{-1}{F}^{\prime \prime }\left(x_0\right)\parallel \le \tau ,$$

$$\parallel F^{\prime }{\left(x_0\right)}^{-1}(F^{\prime \prime }\left(w\right)-F^{\prime \prime }\left(v\right))\parallel \le K_1\parallel w-v\parallel \mathrm{for} \mathrm{all}w,v\in \mathsf{\Omega },$$

$$\mu \le \frac{{({\tau }^2+2K_1)}^{\frac{3}{2}}-\tau ({\tau }^2+3K_1)}{3K_1}$$

and

$$U[x_0,{\rho }_1]\subset \mathsf{\Omega },$$

with ${\rho }_1$ denoting the minimal positive solution of equation

$$\frac{K_1}{6}{t}^3+\frac{\tau }{2}{t}^2-t+\mu =0.$$

In our case we use

$$\parallel F^{\prime }{\left(x_0\right)}^{-1}(F^{\prime \prime }\left(w\right)-F^{\prime \prime }\left(v\right))\parallel \le K\parallel w-v\parallel \mathrm{for} \mathrm{all}w,v\in D_0,$$

where

$$D_0=U(x_0,\frac{1}{L_0})\cap \mathsf{\Omega }$$

or

$$D_0=U(x_0,{\rho }_2)\cap \mathsf{\Omega },$$

if

$$\parallel F^{\prime }{\left(x_0\right)}^{-1}(F^{\prime \prime }\left(w\right)-F^{\prime \prime }\left(x_0\right))\parallel \le K_0\parallel w-x_0\parallel \mathrm{for} \mathrm{all}w\in \mathsf{\Omega },$$

is used, instead, where ${\rho }_2$ is the minimal positive solution of equation

$$\frac{K}{6}{t}^3+\frac{\tau }{2}{t}^2-t+\mu =0.$$

Then, condition

$$\mu \le \frac{{({\tau }^2+2K)}^{\frac{3}{2}}-\tau ({\tau }^2+3K)}{3K}$$

is the corresponding and weaker sufficient convergence criterion. Notice again that

$$K\le K_1,$$

$${\rho }_2\le {\rho }_1,$$

$K_1=K_1\left(\mathsf{\Omega }\right),K_0=K_0\left(\mathsf{\Omega }\right)$, and $K=K(\mathsf{\Omega },K_0).$ The old estimate in [12] involving the bounds on $\parallel F^{\prime }{\left(w\right)}^{-1}{F}^{\prime }\left(x_0\right)\parallel$ is

$$\parallel F^{\prime }{\left(w\right)}^{-1}{F}^{\prime }\left(x_0\right)\parallel \le \frac{1}{1-(\tau \parallel w-x_0\parallel +\frac{K_1}{2}\parallel w-x_0{\parallel }^2)},$$

whereas, we use

$$\parallel F^{\prime }{\left(w\right)}^{-1}{F}^{\prime }\left(x_0\right)\parallel \le \frac{1}{1-(\tau \parallel w-x_0\parallel +\frac{K}{2}\parallel w-x_0{\parallel }^2)},$$

which is more precise.

The rest of the paper is organized as follows: Next, Section 2 contains the convergence analysis, whereas the numerical examples and conclusions can be found in Section 3 and Section 4, respectively.

2. Majorizing Sequences

We introduce some auxiliary results on scalar majorizing sequences.

Definition 1.

Let $\left\{p_j\right\}$ be a Banach space valued sequence. Then, a nondecreasing scalar sequence $\left\{q_j\right\}$ is majorizing for $\left\{p_j\right\},$ if

$$\parallel p_{j+1}-p_j\parallel \le q_{j+1}-q_j\mathrm{for} \mathrm{each}j=0,1,2,\dots .$$

So, the convergence of sequence $\left\{p_j\right\}$ reduces to studying that of $\left\{q_j\right\}$ [14]. Set $H=[0,\infty )$ and $H_0=[0,b)$ for some $b>0.$

Let $\mu \ge 0$ be a parameter and ${\phi }_0:H⟶H,\phi :H_0⟶H$ be continuous and non-decreasing function. We shall use scalar sequences $\left\{t_j\right\},\left\{s_j\right\}$ and $\left\{u_j\right\}$ defined for each $j=0,1,2,\dots$ by $t_0=0,s_0=\mu$

$$\begin{array}{ccc}\hfill u_j& =& s_j+\frac{{\int }_0^1\overline{\phi }\left((1-\theta )(s_j-t_j)\right)d\theta (s_j-t_j)}{1-{\phi }_0\left(t_j\right)}\hfill \\ \hfill t_{j+1}& =& u_j+\frac{{\int }_0^1\overline{\phi }(s_j-t_j+\theta (u_j-s_j))d\theta (u_j-s_j)}{1-{\phi }_0\left(t_j\right)}\hfill \\ \hfill s_{j+1}& =& t_{j+1}+\frac{{\int }_0^1\overline{\phi }(u_j-t_j+\theta (t_{j+1}-u_j))d\theta (t_{j+1}-u_j)}{1-{\phi }_0\left(t_{j+1}\right)},\hfill \\ \hfill \end{array}$$

(17)

where $\overline{\phi }=\left\{\begin{array}{cc}{\phi }_0,& j=0\\ \phi ,& j=1,2,\dots \end{array}\right.$

Next, we present a convergence result for a sequence $\left\{t_j\right\}$ under very general conditions

Lemma 1.

Suppose:

(a)

for all $j=0,1,2,\dots$

$${\phi }_0\left(t_j\right)<1$$

(18)

and

$$t_j\le a\mathrm{for} \mathrm{some}a>0.$$

(19)

(b)

Function ${\phi }_0:H⟶H$ is continuous, increasing and

$$t_j\le {\phi }_0^{-1}\left(1\right)$$

(20)

for each $j=0,1,2,\dots .$ Then, sequences $\left\{t_j\right\},\left\{s_j\right\},\left\{u_j\right\}$ converge monotonically to $s_{**},$ which is their unique upper bound (least).

Proof. 

(a) By (17)–(19): $0\le t_j\le s_j\le u_j\le t_{j+1},$ are bounded by a so, they converge to $s_{*}\in [\mu ,a].$

(b) By (17) and (20) one has again

$$\underset{j⟶\infty }{lim}{t}_j=\underset{j⟶\infty }{lim}{s}_j=\underset{j⟶\infty }{lim}{u}_j=s_{*}.$$

□

Remark 1.

Conditions (18)–(20) can be replaced by stronger, which however are easier to satisfy. This is why we give alternative criteria (but stronger) that can easier be verified.

We introduce, sequences functions and sequences of functions as follows

$$\begin{array}{ccc}\hfill a_j& =& \frac{{\int }_0^1\overline{\phi }\left((1-\theta )(s_j-t_j)\right)d\theta }{1-{\phi }_0\left(t_j\right)},\hfill \\ \hfill b_j& =& \frac{{\int }_0^1\overline{\phi }(u_j-t_j+\theta (u_j-s_j))d\theta }{1-{\phi }_0\left(t_j\right)},\hfill \\ \hfill c_j& =& \frac{{\int }_0^1\overline{\phi }(u_j-t_j+\theta (t_{j+1}-u_j))d\theta }{1-{\phi }_0\left(t_{j+1}\right)},\hfill \end{array}$$

$$\begin{array}{ccc}\hfill {\overline{h}}_j^{\left(1\right)}\left(t\right)& =& {\int }_0^1\phi \left((1-\theta )(s_j-t_j)\right)d\theta +t{\phi }_0\left(t_j\right)-t,\hfill \\ \hfill h_j^{\left(1\right)}\left(t\right)& =& {\int }_0^1\phi \left((1-\theta )t^j\mu \right)d\theta \hfill \\ \hfill & & +t{\phi }_0\left(\frac{1-t^{j+1}}{1-t}\mu \right)-t,\hfill \\ \hfill h^{\left(1\right)}\left(t\right)& =& {\int }_0^1\phi \left((1-\theta )\mu \right)d\theta +t{\phi }_0\left(\frac{\mu }{1-t}\right)-t,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill {\overline{h}}_j^{\left(2\right)}\left(t\right)& =& {\int }_0^1\phi (s_j-t_j+\theta (u_j-s_j))d\theta +t{\phi }_0\left(t_j\right)-t,\hfill \\ \hfill h_j^{\left(2\right)}\left(t\right)& =& {\int }_0^1\phi \left((1+\theta t)t^{2j}\mu \right)d\theta \hfill \\ \hfill & & +t{\phi }_0\left(\frac{1-t^{2j+1}}{1-t}\mu \right)-t,\hfill \\ \hfill h^{\left(2\right)}\left(t\right)& =& {\int }_0^1\phi \left((1+\theta t)\mu \right)d\theta +t{\phi }_0\left(\frac{\mu }{1-t}\right)-t,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill {\overline{h}}_j^{\left(3\right)}\left(t\right)& =& {\int }_0^1\phi (u_j-t_j+\theta (t_{j+1}-u_j))d\theta +t{\phi }_0\left(t_{j+1}\right)-t,\hfill \\ \hfill h_j^{\left(3\right)}\left(t\right)& =& {\int }_0^1\phi ((1+t+\theta t^2)t^{2j}\mu d\theta +t{\phi }_0\left(\frac{1-t^{2j+3}}{1-t}\mu \right)-t\hfill \\ \hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill h^{\left(3\right)}\left(t\right)& =& {\int }_0^1\phi \left((1+t+\theta t^2)\mu \right)d\theta +t{\phi }_0\left(\frac{\mu }{1-t}\right)-t.\hfill \end{array}$$

Next, we present a second convergence result for $\left\{t_j\right\}.$

Lemma 2.

Suppose:

There exists parameter $\gamma \in [0,1)$ such that

$$0\le a_0\le \gamma ,0\le b_0\le \gamma ,0\le c_0\le \gamma ,$$

(21)

$$0\le {\phi }_0\left(t_1\right)<1,$$

(22)

$$h^{\left(1\right)}\left(\gamma \right)\le 0,h^{\left(2\right)}\left(\gamma \right)\le 0\mathrm{and}{h}^{\left(3\right)}\left(\gamma \right)\le 0.$$

(23)

Then, sequences $\left\{t_j\right\},\left\{s_j\right\},\left\{u_j\right\}$ converge to $s_{*}\in [\mu ,s_{**}],$ where $s_{**}=\frac{1}{1-\gamma }\mu .$ Moreover, the following estimates hold

$$0\le s_j-t_j\le {\gamma }^j(t_j-s_{j-1})\le {\gamma }^{2j}\mu ,$$

(24)

$$0\le u_j-s_j\le \gamma (s_j-t_j)\le {\gamma }^{2j+1}\mu ,$$

(25)

$$0\le t_{j+1}-u_j\le \gamma (u_j-s_j)\le {\gamma }^{2j+2}\mu ,$$

(26)

$$t_{j+1}\le \frac{1-{\gamma }^{2j+3}}{1-\gamma }\mu ,$$

(27)

and

$$t_j\le s_j\le u_j\le t_{j+1}\le s_{**}.$$

(28)

Furthermore, we have ${\phi }_0\left(t_j\right)<1$ for each $j=0,1,2,\dots .$

Proof. 

Estimates (24)–(28) hold if

$$0\le a_m\le \gamma ,$$

(29)

$$0\le b_m\le \gamma ,$$

(30)

$$0\le c_m\le \gamma ,$$

(31)

$${\phi }_0\left(t_{m+1}\right)<1$$

(32)

and

$$t_m\le s_m\le u_m\le t_{m+1}$$

(33)

hold for each $m=0,1,2,\dots .$ However, they are true for $m=0,$ by (21)–(23). Notice that we have by the definition (17) and these conditions that

$$\begin{array}{ccc}\hfill t_{m+1}& \le & u_m+{\gamma }^{2m+2}\mu \le s_m+{\gamma }^{2m+1}\mu +{\gamma }^{2m+2}\mu \hfill \\ & \le & t_m+{\gamma }^{2m}\mu +{\gamma }^{2m+1}\mu +{\gamma }^{2m+2}\mu \hfill \\ & \dots & \mu +\gamma \mu +\dots +{\gamma }^{2m+2}\mu \hfill \\ & =& \frac{1-{\gamma }^{2m+3}}{1-\gamma }\mu \le \frac{\mu }{1-\gamma }=s_{**}.\hfill \end{array}$$

Suppose, estimates (24)–(28) hold for all integers smaller or equal to $k.$ Hence, by replacing $t_0,s_0,u_0,t_1$ by $t_m,s_m,u_m,t_{m+1}$ and using the induction hypotheses we see that (29)–(33) shall be true if

$${\overline{h}}_j{\left(\gamma \right)}^{\left(i\right)}\le 0$$

or

$$h_j{\left(\gamma \right)}^{\left(i\right)}\le 0$$

or

$$h{\left(\gamma \right)}^{\left(i\right)}\le 0,$$

for $i=1,2,3$ and $j=0,1,2,\dots ,m,$ which holds true by (23). The induction for (24)–(28) is terminated. The remaining of the proof can be found in Lemma 1. □

Remark 2.

(a) The conditions of Lemma 2 imply those of Lemma 1 but not necessarily vice versa.

(b) Consider functions “φ“ to be given by ${\phi }_0$ and φ in the interesting case ${\phi }_0\left(t\right)=L_{0}t$ and $\phi \left(t\right)=Lt$ for $L_0>0,$ and $L>0.$

Then, consider functions $f_j^{\left(i\right)}$ on $[0,1)$ given by

$$\begin{array}{ccc}\hfill f_j^{\left(1\right)}\left(t\right)& =& \frac{1}{2}{t}^{2j-1}\mu \hfill \\ & & +L_0(1+t+\dots +t^{2j})\mu -1,\hfill \\ \hfill f_j^{\left(2\right)}\left(t\right)& =& L(1+\frac{1}{2}t)t^{2j-1}\mu \hfill \\ & & +L_0(1+t+\dots +t^{2j})\mu -1,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill f_j^{\left(3\right)}\left(t\right)& =& L(1+t+\frac{t^2}{2})t^{2j-1}\mu \hfill \\ & & +L_0(1+t+\dots +t^{2j+2})\mu -1,\hfill \end{array}$$

$$g_1\left(t\right)=L_0{t}^3+(L_0+\frac{L}{2})t^2-\frac{L}{2},$$

$$g_2\left(t\right)=(L_0+\frac{L}{2})t^3+(L_0+L)t^2-\frac{L}{2}t-L$$

and

$$g_3\left(t\right)=L_0{t}^5+(L_0+\frac{L}{2})t^4+L{t}^3+\frac{L}{2}{t}^2-Lt-L.$$

By these definitions, we have

$$\begin{array}{ccc}\hfill g_1\left(0\right)& =& -\frac{L}{2}<0,g_1\left(1\right)=2L_0>0,\hfill \\ \hfill g_2\left(0\right)& =& -L,g_2\left(1\right)=2L_0\hfill \\ \hfill g_3\left(0\right)& =& -L\mathrm{and}{g}_3\left(1\right)=2L_0.\hfill \end{array}$$

It follows from the intermediate value theorem (IVT) that functions $g_i$ have zeros in $(0,1).$ Denote the minimal such zeros by ${\gamma }_i,$ respectively.

Define parameters

$${\lambda }_0=max\{a_0,b_0,c_0\},{\lambda }_1=min\{{\gamma }_1,{\gamma }_2,{\gamma }_3\}$$

(34)

and

$${\lambda }_2=max\{{\gamma }_1,{\gamma }_2,{\gamma }_3\}.$$

Then, we can show a third result on the convergence of sequence $\left\{t_j\right\}.$

Lemma 3.

Suppose:

There exists $\gamma \in [0,1)$ satisfying

$${\lambda }_0\le {\lambda }_1\le \gamma \le {\lambda }_2<1-L_0\mu .$$

(35)

Then, the conclusions of Lemma 2 for a sequence $\left\{t_j\right\}$ follow.

Proof. 

We must show by Lemma 2

$$h_m^{\left(i\right)}\left(\gamma \right)\le 0.$$

(36)

But by the preceding definitions, we can show instead

$$f_m^{\left(i\right)}\left(\gamma \right)\le 0.$$

(37)

We must relate $f_{m+1}^{\left(i\right)}\left(t\right)$ to $f_m^{\left(i\right)}.$ We can write

$$\begin{array}{ccc}\hfill f_{m+1}^{\left(i\right)}\left(t\right)& =& \frac{L}{2}{t}^{2m+1}+L_0(1+t+\dots +t^{2m+2})\mu -1\hfill \\ & & -\frac{L}{2}{t}^{2m-1}\mu -L_0(1+t+\dots +t^{2m})\mu +1+f_m^{\left(1\right)}\left(t\right)\hfill \\ & =& f_m^{\left(1\right)}\left(t\right)+(\frac{L}{2}{t}^2+L_0(t^2+t^3)-\frac{L}{2})t^{2m-1}\mu \hfill \\ & =& f_m^{\left(1\right)}\left(t\right)+g_1\left(t\right)t^{2m-1}\mu ,\hfill \end{array}$$

so

$$f_{m+1}^{\left(1\right)}\left(t\right)=f_m^{\left(1\right)}\left(t\right)+g_1\left(t\right)t^{2m-1}\mu .$$

(38)

In particular, by (38) and the definition of $\gamma$

$$f_{m+1}^{\left(1\right)}\left(\gamma \right)\le f_m^{\left(1\right)}\left(\gamma \right).$$

(39)

Define function $f_{\infty }^{\left(1\right)}$ by

$$f_{\infty }^{\left(1\right)}\left(t\right)=\underset{m⟶\infty }{lim}{f}_m^{\left(1\right)}\left(t\right).$$

(40)

Then, we have by (40)

$$f_{\infty }^{\left(1\right)}\left(t\right)=\frac{L_0\mu }{1-t}-1.$$

(41)

So, we can show instead of (37) (for $i=1$) that

$$f_{\infty }^{\left(1\right)}\left(\gamma \right)\le 0,$$

(42)

which is true by (35). Similarly, we get

$$\begin{array}{ccc}\hfill f_{m+1}^{\left(2\right)}\left(t\right)& =& L(1+\frac{t}{2})t^{2m+1}\mu +L_0(1+t+\dots +t^{2m+2})\mu -1\hfill \\ & & -L(1+\frac{t}{2})(1+\frac{t}{2})t^{2m-1}\mu -L_0(1+t+\dots +t^{2m})+1+f_m^{\left(2\right)}\left(t\right)\hfill \\ & =& f_m^{\left(2\right)}\left(t\right)+[(1+\frac{t}{2})L{t}^2-(1+\frac{t}{2})L+L_0(t^2+t^3)]t^{2m-1}\mu \hfill \\ & =& f_m^{\left(2\right)}\left(t\right)+g_2\left(t\right)t^{2m-1}\mu ,\hfill \end{array}$$

so

$$f_{m+1}^{\left(2\right)}\left(t\right)=f_m^{\left(2\right)}\left(t\right)+=g_2\left(t\right)t^{2m-1}\mu .$$

In particular, we have

$$f_{m+1}^{\left(2\right)}\left(\gamma \right)\le f_m^{\left(2\right)}\left(\gamma \right),$$

and again

$$f_{\infty }^{\left(2\right)}\left(t\right)=\frac{L_0\mu }{1-t}-1.$$

Therefore, (37) (for $i=2$) reduces to showing

$$f_{\infty }^{\left(2\right)}\left(\gamma \right)\le 0,$$

which is true by (35). Moreover, we have analogously

$$\begin{array}{ccc}\hfill f_{m+1}^{\left(3\right)}\left(t\right)& =& L(1+t+\frac{t^2}{2})t^{2m+1}\mu +L_0(1+t+\dots +t^{2m+1})\mu -1\hfill \\ & & -L(1+t+\frac{t^2}{2})t^{2m-1}\mu -L_0(1+t+\dots +t^{2m+2})\mu +1+f_m^{\left(3\right)}\left(t\right)\hfill \\ & =& f_m^{\left(3\right)}\left(t\right)+[L(1+t+\frac{t^2}{2})t^2\hfill \\ & & +L_0(t^4+t^5)-L(1+t+\frac{t^2}{2})]t^{2m-1}\mu \hfill \\ & =& f_m^{\left(3\right)}\left(t\right)+g_3\left(t\right)t^{2m-1}\mu ,\hfill \end{array}$$

so

$$f_{m+1}^{\left(3\right)}\left(\gamma \right)\le f_m^{\left(3\right)}\left(\gamma \right),$$

Hence, we can show instead of (37) (for $i=3$) that

$$f_{\infty }^{\left(3\right)}\left(\gamma \right)\le 0,$$

which is true by (34), where

$$f_{\infty }^{\left(3\right)}\left(t\right)=\underset{m⟶\infty }{lim}{f}_m^{\left(3\right)}\left(t\right)=\frac{L_0\mu }{1-t}-1.$$

Therefore, sequence $\left\{t_j\right\}$ is nondecreasing and bounded from above by $s_{**}=\frac{1}{1-\gamma }\mu ,$ so it converges to $s_{*}.$ □

Next, we connect Lemmas 1 and 2 to method (3). We first consider conditions (A):

Suppose

(A1)

There exists $x_0\in \mathsf{\Omega },\mu \ge 0$ such that $F^{\prime }{\left(x_0\right)}^{-1}\in L(E_2,E_1)$ and

$$\parallel F^{\prime }{\left(x_0\right)}^{-1}F\left(x_0\right)\parallel \le \mu .$$

(A2)

For all $w\in \mathsf{\Omega }$

$$\parallel F^{\prime }{\left(x_0\right)}^{-1}(F^{\prime }\left(w\right)-F^{\prime }\left(x_0\right))\parallel \le {\phi }_0(\parallel w-x_0\parallel ).$$

(A3)

Function ${\phi }_0\left(t\right)-1$ has a smallest positive solution $\rho .$ Set

$${\mathsf{\Omega }}_0=U(x_0,\rho )\cap \mathsf{\Omega }.$$

(A4)

For each $w,v\in {\mathsf{\Omega }}_0$

$$\parallel F^{\prime }{\left(x_0\right)}^{-1}(F^{\prime }\left(w\right)-F^{\prime }\left(v\right))\parallel \le \phi (\parallel w-v\parallel )$$

(A5)

Hypotheses of Lemma 1 or Lemma 2 or Lemma 3 hold and

(A6)

$U[w_0,s_{*}]\subset \mathsf{\Omega }$ (or $U[w_0,s_{**}]\subset \mathsf{\Omega }$).

Next, we prove the first semi-local convergence theorem for sequence $\left\{w_j\right\}.$

Theorem 1.

Suppose hypotheses (A) hold. Then, sequences $\left\{w_j\right\}$ produced by method (3) is well defined in $U[w_0,s_{*}],$ remains in $U[w_0,s_{*}]$ for each $j=0,1,2,\dots$ and converges to a solution $w_{*}\in U[w_0,s_{*}]$ (or $w_{*}\in U[w_0,s_{**}]$) of equation $F\left(w\right)=0.$

Proof. 

Using condition (A1) and the first substep of method (3) for $j=0,$ we see that $y_0$ is well defined and

$$\parallel y_0-x_0\parallel =\parallel F^{\prime }{\left(x_0\right)}^{-1}F\left(x_0\right)\parallel \le \mu =s_0-t_0=s_0\le s_{*},$$

so $y_0\in U(x_0,s_{*}).$ Iterate $z_0$ is exists by (A1) and (3) for $j=0.$ So, by (3) and (A3) one has

$$\begin{array}{ccc}\hfill \parallel z_0-y_0\parallel & =& \parallel F^{\prime }{\left(x_0\right)}^{-1}F\left(y_0\right)\parallel \hfill \\ & =& \parallel F^{\prime }{\left(x_0\right)}^{-1}(F\left(y_0\right)-F\left(x_0\right)-F^{\prime }\left(x_0\right)(y_0-x_0))\parallel \hfill \\ & =& \parallel {\int }_0^1{F}^{\prime }{\left(x_0\right)}^{-1}(F^{\prime }\left(x\right)-F^{\prime }(x_0+\theta (y_0-x_0)))d\theta (y_0-x_0)\parallel \hfill \\ & \le & {\int }_0^1\overline{\phi }((1-\theta )\parallel y_0-x_0\parallel )d\theta \parallel y_0-x_0\parallel \hfill \\ & \le & {\int }_0^1\overline{\phi }\left((1-\theta )(s_0-t_0)\right)d\theta (s_0-t_0)\le u_0-s_0.\hfill \end{array}$$

We also have $\parallel z_0-x_0\parallel \le \parallel z_0-y_0\parallel +\parallel y_0-x_0\parallel \le u_0-s_0+s_0-t_0=u_0-t_0<s_{*},$ so $z_0\in U(x_0,s_{*}).$ By condition (A1) and (3) for $j=0,$$x_1$ exists, and we can write

$$\begin{array}{ccc}\hfill x_1-y_0& =& z_0-x_0-F^{\prime }{\left(x_0\right)}^{-1}(F\left(z_0\right)-F\left(x_0\right))\hfill \\ \hfill & =& F^{\prime }{\left(x_0\right)}^{-1}{\int }_0^1[F^{\prime }\left(x_0\right)-F^{\prime }(x_0+\theta (z_0-x_0))]d\theta (z_0-x_0).\hfill \end{array}$$

The condition (A2) and (17) give in turn that

$$\begin{array}{ccc}\hfill \parallel x_1-y_0\parallel & \le & {\int }_0^1{\phi }_0((1-\theta )\parallel z_0-x_0\parallel )d\theta \parallel z_0-x_0\parallel \hfill \\ \hfill & \le & \frac{{\int }_0^1{\phi }_0((1-\theta )\parallel z_0-x_0\parallel )d\theta \parallel z_0-x_0\parallel }{1-{\phi }_0(\parallel x_0-x_0\parallel )}\hfill \\ \hfill & \le & \frac{{\int }_0^1{\phi }_0\left((1-\theta )(u_0-t_0)\right)d\theta (u_0-t_0)}{1-{\phi }_0\left(t_0\right)}\hfill \\ \hfill & =& t_1-s_0.\hfill \end{array}$$

(43)

We also have

$$\begin{array}{ccc}\hfill \parallel x_1-x_0\parallel & \le & \parallel x_1-y_0\parallel +\parallel y_0-x_0\parallel \hfill \\ & \le & t_1-s_0+s_0-t_0=t_1\le s_{*},\hfill \end{array}$$

so $w_1\in U(w_0,s_{*}).$ Let $w\in U(w_0,s_{*}).$ Using (A2) one obtains

$$\parallel F^{\prime }{\left(w_0\right)}^{-1}(F^{\prime }\left(w\right)-F^{\prime }\left(w_0\right))\parallel \le {\phi }_0(\parallel w-w_0\parallel )\le {\phi }_0\left(s_{*}\right)<1,$$

so the Banach lemma for linear invertible operators [5] assures the existence of $F^{\prime }{\left(w\right)}^{-1}$ and

$$\parallel F^{\prime }{\left(w\right)}^{-1}{F}^{\prime }\left(w_0\right)\parallel \le \frac{1}{1-{\phi }_0(\parallel w-w_0\parallel )}.$$

(44)

In particular for $w=w_1,F^{\prime }{\left(w_1\right)}^{-1}$ exists, so does iterate $y_1.$ Then, we can write by the first substep of method (3) for $k=1$

$$\begin{array}{ccc}\hfill \parallel y_1-x_1\parallel & =& \parallel F^{\prime }{\left(x_1\right)}^{-1}F\left(x_1\right)\parallel \hfill \\ & \le & \parallel F^{\prime }{\left(x_1\right)}^{-1}{F}^{\prime }\left(x_0\right)\parallel \parallel F^{\prime }{\left(x_0\right)}^{-1}F\left(x_1\right)\parallel \hfill \\ \hfill & \le & \frac{\parallel {\int }_0^1{F}^{\prime }{\left(x_0\right)}^{-1}(F^{\prime }(z_0+\theta (x_1-z_0))-F^{\prime }\left(x_0\right))d\theta (x_1-z_0)\parallel }{1-{\phi }_0(\parallel x_1-x_0\parallel )}\hfill \\ & \le & \frac{{\int }_0^1{\phi }_0(\parallel z_0-x_0\parallel +\theta \parallel x_1-z_0\parallel )d\theta \parallel x_1-z_0\parallel }{1-{\phi }_0(\parallel x_1-x_0\parallel )}\hfill \\ \hfill & \le & \frac{{\int }_0^1\overline{\phi }(u_0-t_0+\theta (t_1-u_0))d\theta (t_1-u_0)}{1-{\phi }_0\left(t_1\right)},\hfill \\ \hfill \end{array}$$

where we also used by the definition of the method

$$\begin{array}{ccc}\hfill \parallel F^{\prime }{\left(x_0\right)}^{-1}F\left(x_1\right)\parallel & =& \parallel F^{\prime }{\left(x_0\right)}^{-1}(F\left(x_1\right)-F\left(z_0\right)+F\left(z_0\right))\hfill \\ \hfill & \le & {\int }_0^1{F}^{\prime }{\left(x_0\right)}^{-1}(F^{\prime }(z_0+\theta (x_1-z_0))-F^{\prime }\left(x_0\right))d\theta (x_1-z_0)\parallel \hfill \\ & \le & {\int }_0^1{\phi }_0(\parallel z_0-x_0\parallel +\theta \parallel x_1-z_0\parallel )d\theta \parallel x_1-z_0\parallel \hfill \\ \hfill & \le & {\int }_0^1\overline{\phi }(u_0-t_0+\theta (t_1-u_0))d\theta (t_1-u_0)\hfill \\ \hfill & =& s_1-t_1,\hfill \end{array}$$

(45)

and

$$\parallel x_1-x_0\parallel \le \parallel x_1-y_0\parallel +\parallel y_0-x_0\parallel \le t_1-s_0+s_0-t_0=t_1-t_0.$$

Hence, we showed so far

$$\parallel y_m-x_m\parallel \le s_m-t_m,m=0,1,$$

(46)

$$\parallel z_m-y_m\parallel \le u_m-s_m,m=0,$$

(47)

$$\parallel x_{m+1}-z_m\parallel \le t_{m+1}-u_m,m=0,$$

(48)

and

$$x_m,y_m,z_m,x_{m+1}\in U(x_0,s_{*})$$

(49)

for $m=0.$ Consider these estimates are true for all $m\le j-1.$ Then, simply replace $x_0,y_0,z_0,x_1$ by $x_m,y_m,z_m,x_{m+1},$ to terminate the induction for items (46)–(49). So, $\left\{x_j\right\}$ is fundamental in a Banach space $E_1.$ Hence, ${lim}_{j⟶\infty }{x}_j=x^{*}\in U[x_0,s_{*}].$ Then, by letting $j⟶\infty$ in the estimate (see also (45))

$$\begin{array}{ccc}\hfill \parallel F^{\prime }{\left(x_0\right)}^{-1}F\left(x_{j+1}\right)\parallel & \le & {\int }_0^1\phi (u_j-t_j+\theta (t_{j+1}-u_j))d\theta (t_{j+1}-u_j),\hfill \end{array}$$

and the continuity of $F,$ we conclude $F\left(x_{*}\right)=0.$ □

A uniqueness of the solution result follows.

Proposition 1.

Suppose:

(i)

There exists a simple solution $x^{*}\in \mathsf{\Omega }$ of equation $F\left(x\right)=0.$

(ii)

There exists $\alpha \ge s_{*}$ such that

$${\int }_0^1{\phi }_0((1-\theta )\alpha +\theta s_{*})d\theta <1.$$

Set ${\mathsf{\Omega }}_1=U[x_0,\alpha ]\cap \mathsf{\Omega }.$ Then, $x^{*}$ is unique in ${\mathsf{\Omega }}_1.$

Proof. 

Let $w^{*}\in {\mathsf{\Omega }}_1$ with $F\left(w^{*}\right)=0.$ Set $M={\int }_0^1{F}^{\prime }(w^{*}+\theta (x^{*}-w^{*}))d\theta .$ Then, in view of (A2), we obtain in turn that

$$\begin{array}{ccc}\hfill \parallel F^{\prime }{\left(x_0\right)}^{-1}(M-F^{\prime }\left(x_0\right))\parallel & \le & {\int }_0^1{\phi }_0((1-\theta )\parallel w^{*}-x_0\parallel +\theta \parallel x_{*}-x_0\parallel )d\theta \hfill \\ & \le & {\int }_0^1{\phi }_0((1-\theta )\alpha +\theta s_{*})d\theta <1,\hfill \end{array}$$

so $w^{*}=x_{*}$ follows from the invertability of M and the estimate $M(x_{*}-w^{*})=F\left(x_{*}\right)-F\left(w^{*}\right)=0-0=0.$ □

3. Examples

We present examples to further justify the theoretical results.

Example 1.

Consider

$$\psi \left(t\right)=b_{0}t+b_1+b_{2}sin{b}_{3}t,t_0=0,$$

where $b_j,j=0,1,2,3$ are parameters. Then, clearly for $b_3$ large and $b_2$ small, $\frac{L_0}{L_1}$ can be small (arbitrarily). Notice that as $\frac{L_0}{L_1}⟶0,$$\frac{T}{T_1}⟶0.$

Example 2.

If $E_1=E_2=\mathbb{R},x_0=1$ and $\mathsf{\Omega }=U[1,1-p]$ for $p\in (0,\frac{2}{3}),$ define polynomial ψ on $\mathsf{\Omega }$ as

$$\psi \left(t\right)=t^3-p.$$

If we consider case 1 of Newton’s method, then, we obtain $L_0=3-p,L=L_1=2(2-p)$ and $\mu =\frac{1}{3}(1-p).$ But then, $T_1>\frac{1}{2}$ for all $p\in (0,\frac{1}{2}).$ So, Theorem 5.2 in [14] cannot assure convergence. However, we have $T\le \frac{1}{2}$ for all $p\in I=[0.4271907643,\frac{1}{2}).$ Hence, our result guarantees convergence to $x^{*}=\sqrt[3]{p}$ as long as $p\in I.$

Example 3.

Let $E_1=E_2=H\left([0,1]\right)$ the domain of functions given on $[0,1]$ which are continuous. We consider the max-norm. Choose $\mathsf{\Omega }=U(0,d),$$d>1.$ Define G on Ω be

$$G\left(x\right)\left(s\right)=x\left(s\right)-w\left(s\right)-\delta {\int }_0^{1}N(s,t)x^3\left(t\right)dt,$$

(50)

$x\in E_1,s\in [0,1],w\in E_1$ is given, δ is a parameter and N is the Green’s kernel given by

$$N(b_2,b_1)=\left\{\begin{array}{cc}(1-b_2)b_1,\hfill & b_1\le b_2\hfill \\ b_2(1-b_1),\hfill & b_2\le b_1.\hfill \end{array}\right.$$

By (50), we have

$$\left(G^{\prime }\left(x\right)\left(z\right)\right)\left(s\right)=z\left(s\right)-3\delta {\int }_0^{1}N(s,t)x^2\left(t\right)z\left(t\right)dt,$$

$t\in E_1,s\in [0,1].$ Consider $x_0\left(s\right)=w\left(s\right)=1$ and $\left|\delta \right|<\frac{8}{3}.$ We get

$$\parallel I-G^{\prime }\left(x_0\right)\parallel <\frac{3}{8}\left|\delta \right|,G^{\prime }{\left(x_0\right)}^{-1}\in L(E_2,E_1),$$

$$\parallel F^{\prime }{\left(x_0\right)}^{-1}\parallel \le \frac{8}{8-3\left|\delta \right|},\mu =\frac{\left|\delta \right|}{8-3\left|\delta \right|},L_0=\frac{12\left|\delta \right|}{8-3\left|\delta \right|},$$

and $L_1=L=\frac{6\mu \left|\delta \right|}{8-3\left|\delta \right|}.$

In Table 1 that follows we have listed the results on the convergence criteria for various values of the parameter involved.

Table 1. Comparison table of criteria.

$\mathit{\mu }$	${\mathit{\delta }}^{*}$	${\mathit{T}}_1$	T
2.09899	0.9976613778	1.007515200	0.9639223786
2.19897	0.9831766058	1.055505600	0.9678118280
2.29597	0.9698185659	1.102065600	0.9715205068
3.095467	0.87963113211	1.485824160	1.000082409

Example 4.

Let $E_1,E_2$, and $\mathsf{\Omega }$ be as in the Example 3. It is well known that the boundary value problem [16]

$$\psi \left(0\right)=0,\left(1\right)=1,$$

$${\psi }^{\prime \prime }=-\psi -\ell {\psi }^2$$

can be presented as a Hammerstein-like nonlinear integral equation [12]

$$\psi \left(s\right)=s+{\int }_0^{1}K(s,t)({\psi }^3\left(t\right)+\ell {\psi }^2\left(t\right))dt$$

for ℓ being a parameter, consider $F:\mathsf{\Omega }⟶E_2$ given by

$$\left[F\left(x\right)\right]\left(s\right)=x\left(s\right)-s-{\int }_0^{1}K(s,t)(x^3\left(t\right)+\ell x^2\left(t\right))dt.$$

Choose ${\psi }_0\left(s\right)=s$ and $\mathsf{\Omega }=U({\psi }_0,{\rho }_0).$ Then, clearly $U({\psi }_0,{\rho }_0)\subset U(0,{\rho }_0+1),$ since $\parallel {\psi }_0\parallel =1.$ Suppose $2\ell <5.$ Then, conditions (A) are satisfied for

$$L_0=\frac{2\ell +3{\rho }_0+6}{8},L_1=L=\frac{\ell +6{\rho }_0+3}{4}$$

and $\mu =\frac{1+\ell }{5-2\ell }.$ Notice that $L_0<L_1.$

4. Conclusions

Two different techniques and a new domain D included in the original one are introduced. This change in the analysis gives a finer convergence with no additional conditions.