On an iterative algorithm of Uľm-type for solving equations

Ioannis K. Argyros$^\ast $ Sanjay K. Khattri$^\S $

September 12, 2012.

$^\ast $Department of Mathematical Sciences, Cameron University, Lawton, Oklahoma 73505-6377, USA, email: iargyros@cameron.edu.

$^\S $Department of Engineering, Stord Haugesund University College, Norway, email: sanjay.khattri@hsh.no

We provide a semilocal convergence analysis of an iterative algorithm for solving nonlinear operator equations in a Banach space setting. This algorithm is of order $1.839\ldots $, and has already been studied in [ 3 , 8 , 18 , 20 ] . Using our new idea of recurrent functions we show that a finer analysis is possible with sufficient convergence conditions that can be weaker than before, and under the same computational cost. Numerical examples are also provided in this study.

MSC. 65G99; 65H10; 65B05; 47H17; 49M15

Keywords. Banach space, iterative algorithm, semilocal convergence, divided difference of operator, Fréchet-derivative.

1 Introduction

In this study, we are concerned with the problem of approximating a locally unique solution $x^\star $ of equation:

\begin{equation} \label{eq:11} F(x) = 0, \end{equation}

where $F$ is a nonlinear operator defined on an open subset $D$ of a Banach space $X$ with values in a Banach space $Y$. Many problems in computational mathematics can be written in the form 1 [ 8 , 14 , 16 ] . Potra in [ 18 ] used the Uľm-type method [ 20 ] (UTM):

\begin{alignat}{3} x_{n+1}& =x_n-A_{n}^{-1}F(x_n)\quad (n\ge {0})\quad (x_{-2},x_{-1},x_{0}\in {D}),\label{eq:12} \intertext {where,} A_n & =[x_n,x_{n-1};F]+[x_{n-2},x_n;F]-[x_{n-2},x_{n-1};F],\label{eq:utm2} \end{alignat}

to provide a local as well as a semilocal convergence analysis under hypotheses on the first $[\cdot ,\cdot ;F]$ and second $[\cdot ,\cdot ,\cdot ;F]$ order divided differences of operator $F$.

Here, an operator belonging to the space $L(X,Y)$ (the Banach space of linear and bounded operators from $X$ into $Y$) is called a divided difference of order one for the operator $F:X\rightarrow {Y}$ on the points $x,y\in {X}$ if the following properties hold:

\begin{alignat}{4} [x,y;F](y-x) & = F(y)-F(x)\quad \text{for}\quad x\neq {y};\label{eq:prop1} \intertext {if $F$ is Fr\'{e}chet-differentiable at $x\in X$, then} [x,x;F] & = F^\prime (x).\label{eq:prop2} \intertext {An operator belonging to the space $L(X,L(X,Y))$ denoted by $[x,y,z;F]$ is called a divided difference of order two for the operator $F:X\rightarrow {Y}$ on the points $x,y,z\in {X}$ if:} [x,y,z;F](z-x) & = [y,z;F]-[x,y;F],\label{eq:prop3} \intertext {for the distinct points $x,y,z$ if $F$ is twice Fr\'{e}chet-differentiable at $x\in {X}$, then} [x,x,x;F] & =\tfrac {1}{2}F^{\prime \prime }(x). \label{eq:prop4} \end{alignat}

Potra showed that the $R-$order of the method is given by the positive solution of the scalar equations:

\begin{equation} t^3-t^2-t-1=0, \end{equation}

which is approximately $1.839\ldots $. Other methods using divided differences of order can be found in [1-21], and references therein.

Here, we are motivated by optimization considerations, and we show that it is possible to provide under the same computational cost an analysis with the following advantages:

Semilocal case:

finer error bounds on he distances $\Vert {x_{n+1}-x_n}\Vert $, $\Vert {x_{n}-x^\star }\Vert $ ($n\ge {0}$),
weaker sufficient convergence conditions and,
an at least as precise information on the location of the solution $x^\star $.

Local case:

finer error bounds on the distances involved,
and at least as large radius of convergence.

The semilocal convergence is provided in §2 followed by local in §3. Numerical examples are also provided in §4.

2 Semilocal Convergence Analysis for (UTM)

We need the following result on majorizing sequence for (UTM).

Lemma 1

Let $\alpha ,\phi ,\gamma ,a,b,c,p,$ and $q$ be given non-negative constants. Assume:

\begin{align} \phi +\gamma {\lt}p+q\, c,\label{eq:21} \intertext {then, the polynomial $g$ given by} g(s) = (qc+\alpha +\phi )s^2+(p+\alpha +\gamma )s+\phi +\gamma -p-qc,\label{eq:22} \intertext {has a unique positive root $\delta \in (0,1)$;} \intertext {moreover, suppose} \alpha (b+c)+\phi (a+c)+\gamma (a+b){\lt}1; \label{eq:23}\\ \delta _0\le {\delta };\\ f_2({\delta })\le {0}; \label{eq:25} \end{align}

where,

\begin{align} \delta _{0}& =\tfrac {pc+qb(a+b)}{1-\left[\alpha (b+c)+\phi (a+c)+\gamma (a+b)\right]}, \label{eq:26} \intertext {and} f_2(s) & = psc + q(s+1)s^2 + \alpha \left[(1+s+s^2)c+(1+s)c+b\right]\nonumber \\ & {}\qquad +\phi \left[(1+s+s^2)c+a+b+c\right]+\gamma \left[ (1+s)c+a+2b+c\right]-1.\nonumber \end{align}

Then, scalar sequence $\left\{ t_n\right\} $ $(n\ge {-2})$ given by

\begin{gather} t_{-2}=0,\quad t_{-1}=a,\quad t_{0}=a+b, \quad t_1=a+b+c,\nonumber \\ t_{n+2}=t_{n+1}+M_{n+1}\left(t_{n+1}-t_n\right), \nonumber \intertext {where} M_{n+1} = \tfrac {p\left(t_{n+1}-t_n\right)+q\left(t_n-t_{n-2}\right)(t_n-t_{n-1})/\mu _n}{\mu _{n+1}}, \end{gather}

is non-decreasing, bounded from above by

\begin{equation} \label{eq:28} t^{\star \star } = \tfrac {c}{1-\delta }+a+b, \end{equation}

and converges to its unique least upper bound $t^\star $ such that $t\in \left[0,t^{\star \star }\right]$. Moreover, the following estimates hold for all $n\ge {0}:$

\begin{gather} 0 {\le } {t_{n+2}-t_{n+1}} {\le }\delta \left(t_{n+1}-t_n\right)\le \delta ^{n+1}c \quad \left(n\ge {0}\right)\label{eq:29} \intertext {and} 0{\le } {t^\star -t_n} {\le } \tfrac {\delta ^n\, c}{1-\delta }\quad (n\ge {1}), \label{eq:210} \end{gather}

where,

\begin{align} \mu _{n+1} =& 1-\left[\alpha \left(t_{n+1}+t_n-2a-b\right)+\phi \left(t_{n+1}+t_{n-1}-a-b \right)\right.\\ & \left. +\gamma \left(t_n+t_{n-1}-a-b-c\right)\right].\nonumber \end{align}

Proof â–¼

We shall show using induction that

\begin{align} M_{n+1}& \le \delta \label{eq:212} \intertext {and} \mu _{n+1}& {\lt} 1,\label{eq:213} \end{align}

hold for all $n$. It will then follow that (2.9) also holds. Estimates 11 and 11 hold true for $n=0$, by (2.3) and (2.4), respectively. Let us assume 9, 11 and 11 hold for all $k\le {n}$. It then follows from the induction hypotheses:

\begin{align*} t_{k+2}& \le t_{k+1}+\delta \left(t_{k+1}-t_k\right)\le t_k+\delta \left(t_k-t_{k-1}\right)+ \delta \left(t_{k+1}-t_{k}\right) \\ & \le t_1 + \delta \left(t_1-t_0\right)+\cdots +\delta \left(t_{k+1}-t_k\right) \\ & \le a+ b + c + \delta c + \cdots + \delta ^{k+1}c \\ & = a+b+\tfrac {1-\delta ^{k+2}}{1-\delta }c {\lt} a+b+\tfrac {c}{1-\delta }= t^{\star \star }. \end{align*}

Estimates 11 and 11 will be true if

\begin{align*} p\delta ^k c & + q\delta ^{k-1}c\left(\delta ^{k-1}+\delta ^{k-2}\right)c \le \delta - \delta \left[\alpha \left(t_{k+1}+t_k-t_0-t_{-1}\right)\right.\nonumber \\ & \left.+\phi \left(t_{k+1}+t_{k-1}-t_{0}-t_{-2} \right)+\gamma \left(t_k+t_{k-1}-t_{-1}-t_{-2}\right)\right] \nonumber \end{align*}

\begin{multline} \label{eq:214} p\delta ^{k}c+q\delta ^{k-1}\left(\delta ^{k-1}+\delta ^{k-2}\right)c^2+\delta \left[ \alpha \left(\tfrac {1-\delta ^{k+1}}{1-\delta }c+\tfrac {1-\delta ^{k}}{1-\delta } c+b\right)+\right.\\ \left.+\phi \left(\tfrac {1-\delta ^{k+1}}{1-\delta }c+\tfrac {1-\delta ^{k-1}}{1-\delta } +a+b\right)+\gamma \left(\tfrac {1-\delta }{1-\delta }c+\tfrac {1-\delta ^{k-1}}{1-\delta } c+a+2b\right)\right]-\delta \le 0. \end{multline}

Estimate 10 motivates us to introduce functions $f_k \ (k\ge {2})$ on $[0,\infty )$ (for $\delta =s$) by:

\begin{multline} \label{eq:215} f_k(s) = pcs^{k-1}+qc^2 s^{k-1}(s+1) + \alpha \left[(1+s+\cdots +s^k)c \right. \\ \left. +(1+s+\cdots +s^{k-1})c +b\right] +\phi \left[(1+s+\cdots +s^k)c\right. \\ \left. +(1+s+\cdots +s^{k-2})c+a+b\right]\\ +\gamma \left[(1+s+\cdots +s^{k-1})c+(1+s+\cdots +s^{k-2})c+a+2b\right]-1. \end{multline}

We shall show instead of 11 and 11 that

\begin{equation} \label{eq:216} f_k(\delta )\le {0}. \end{equation}

We need a relationship between two consecutive functions $f_k:$

\begin{align} f_{k+1}(s)& = pcs^k+pcs^{k-1}-pcs^{k-1}+qc^2s^k\left(s+1\right)\nonumber \\ & \quad +qc^2s^{k-1}\left(s+1\right)-qc^2s^{k-1} \left(s+1\right)\nonumber \\ & \quad +\alpha \left[(1+s+\cdots +s^k+s^{k+1})c+(1+s+\cdots +s^{k-1 } +s^k)c+b\right] \nonumber \\ & \quad +\phi \left[(1+s+\cdots +s^k+s^{k+1})c \right. \nonumber \\ & \quad \left.+(1+s+\cdots +s^{k-2}+s^{k-1})c+a+b\right] \nonumber \\ & \quad +\gamma \left[(1+s+\cdots +s^k+s^{k+1})c\nonumber \right. \\ & \quad \left. +(1+s+\cdots +s^{k-2}+s^{k-1} )c+a+2b\right]-1\nonumber \\ & = f_k(s)+g(s)s^{k-1}c,\label{eq:217} \end{align}

where, function $g$ is given by 5. In view of 5 and 17 we get

\begin{equation} \label{eq:218} f_k(\delta ) = f_2(\delta )\quad (k\ge {2}). \end{equation}

Hence, 16 is true if $f_2(\delta )\le {0}$, which is true by 7.

Define function $f_\infty $ on $[0,1)$ by

\begin{equation} f_\infty (s)=\lim _{n\to \infty }f_n(s).\label{eq:219} \end{equation}

Then, we have by 16 and 19:

\begin{equation} f_{\infty }(\infty )=\lim _{n\to \infty }f_n(\delta )\le \lim _{n\to \infty }{0}=0.\label{eq:220} \end{equation}

The induction for 9, 11 and 11 is completed. Hence, we showed sequence $\{ t_n\} $ is non-decreasing, bounded above by $t^{\star \star }$ and as such it converges to $t^\star $. Finally, estimate 9 follows from

\begin{alignat}{3} 0& \le {t_{k+m}-t_n}=\left(t_{k+m}-t_{k+m-1}\right)\nonumber \\ & \quad +\left(t_{k+m-1}-t_{k+m-2}\right)+\cdots +\left(t_{k+1}-t_{k}\right)\nonumber \\ & \le \left(\delta ^{k+m-1}+\delta ^{k+m-2}+\cdots +\delta ^{k}\right)c = \tfrac {1-\delta ^{k+m}}{1-\delta }\delta ^kc,\label{eq:221} \end{alignat}

by letting $m\to \infty $. That completes the proof of the lemma.

Proof â–¼

We can show the main semilocal convergence result for (UTM).

Theorem 2

Let $F$ be a nonlinear operator defined on an open subset $D$ of a Banach space $X$ with values in a Banach space $Y$. Let $\left[\cdot ,\cdot ;F\right]$, $\left[\cdot ,\cdot ,\cdot ;F\right]$ be divided differences of first and second order of $F$ on $D$, respectively. Let $x_{-2}, x_{-1}, x_{0}\in {D}$ be three given points from $D$, and assume $A_0$ is invertible. Let $a, b, c, p, q, \alpha , \phi , \gamma $ be non-negative numbers such that for all $x, y, z, u, v\in {D}:$

\begin{align} \left\Vert {x_{-1}-x_0}\right\Vert \le {a}, \quad \Vert {x_{-2}-x_{-1}}\Vert & \le {b},\quad \left\Vert {A_0^{-1}F(x_0)}\right\Vert \le {c},\label{eq:222}\\ \left\Vert {A_{0}^{-1}\left(\left[x,y;F\right]-\left[u,v;F\right]\right) }\right\Vert & \le {p}\left(\left\Vert {x-u}\right\Vert +\left\Vert {y-v}\right\Vert \right),\label{eq:223} \\ \left\Vert {A_{0}^{-1}\left(\left[x,y,z;F\right]-\left[u,v,z;F\right]\right) }\right\Vert & \le {q}\left\Vert {x-u}\right\Vert ,\label{eq:224}\\ \left\Vert {A_{0}^{-1}\left(\left[x,y;F\right]-\left[x_{0},x_{-1},;F\right]\right) }\right\Vert & \le {\alpha } \left(\left\Vert {x-x_{0}}\right\Vert +\left\Vert {y-x_{-1}}\right\Vert \right), \label{eq:225} \\ \left\Vert {A_{0}^{-1}\left(\left[x,y;F\right]-\left[x_{-2},x_{0};F\right]\right) }\right\Vert & \le {\phi }\left(\left\Vert {x-x_{-2}}\right\Vert +\left\Vert {y-x_{0}}\right\Vert \right), \label{eq:226} \\ \left\Vert {A_{0}^{-1}\left(\left[x,y;F\right]-\left[x_{-2},x_{-1};F\right]\right) }\right\Vert & \le {\gamma }\left(\left\Vert {x-x_{-2}}\right\Vert +\left\Vert {y-x_{-1}}\right\Vert \right), \label{eq:227} \end{align}

hypotheses of Lemma 1 hold and

\begin{equation} \label{eq:228} \overline{U}(x_0,t^\star ) {:=} \left\{ x\in {X}:\left\Vert {x-x^\star }\right\Vert \subseteq {t^\star }\right\} \subseteq {D}. \end{equation}

Then, sequence $\{ x_n\} $ $(n\ge {-2})$ generated by (UTM) is well defined, remains in $\overline{U}(x_0,t^\star )$ for all $n\ge {0}$ and converges to a solution $x^\star \in {\overline{U}(x_0,t^\star )}$ of equation $F(x)=0$. Moreover, the following estimates hold for all $n\ge {0}:$

\begin{alignat}{3} \left\Vert {x_{n+2}-x_{n+1}}\right\Vert & \le L_{n+1}{\left\Vert {x_{n+1}-x_{n}}\right\Vert }\le t_{n+2}-t_{n+1},\label{eq:229}\\ \left\Vert {x_{n}-x^\star }\right\Vert & \le t^\star -t_n,\label{eq:230} \end{alignat}

where

\begin{alignat}{3} L_{n+1} & = \frac{p{\left\Vert {x_{n+1}-x_{n}}\right\Vert }+q{\left\Vert {x_{n}-x_{n-2}}\right\Vert }\, \left\Vert { x_{n}-x_{n-1}}\right\Vert }{d_{n+1}},\nonumber \\ d_{n+1} & = 1\! -\! \left[\alpha \left(\left\Vert {x_{n+1}-x_{0}}\right\Vert + \left\Vert {x_{n}-x_{-1}}\right\Vert \right) \! +\! \phi \left(\left\Vert {x_{n-1}-x_{-2}} \right\Vert \! +\! \left\Vert {x_{n+1}-x_{0}}\right\Vert \right)\right.\nonumber \\ & \left. \quad + \gamma \left(\left\Vert {x_{n-1}-x_{-2}}\right\Vert + \left\Vert {x_{n}-x_{-1}}\right\Vert \right)\right]. \end{alignat}

Furthermore, if there exists $r\ge {t^\star }$ such that

\begin{alignat}{3} U(x_0,r) & \subseteq {D}\label{eq:232} \intertext {and} \phi (t^\star +r)+(\phi +\gamma )(a+b) & \le 1,\label{eq:233} \end{alignat}

then, the solution $x^\star $ is unique in $U(x_0,r)$.

Proof â–¼

We shall show using induction on $k\ge {0}:$

\begin{equation} \label{eq:234} \left\Vert {t_{k+1}-t_k}\right\Vert \le {t_{k+1} - t_k}. \end{equation}

Estimate 33 holds for $k=-2,-1,0$ by 8, and 22. We also have have $x_{-2},x_{-1},x_{0}\in \overline{U}(x_0,t^\star )$. Let us assume 33, and $x_k\in \overline{U}(x_0,t^\star )$ hold for all $n\le {k+1}$. We have using 25-27:

\begin{align} & \left\Vert {A_0}^{-1}\left(A_{k+1}-A_0\right)\right\Vert =\nonumber \\ & = \left\Vert {A_0^{-1}}\left(\left[x_{k+1},x_k;F\right]-\left[x_{0},x_{-1};F\right]\right)\right. \nonumber \\ & {}\quad \left.+{A_0^{-1}}\left(\left[x_{k-1},x_{k+1};F\right]-\left[x_{-2},x_{0};F\right]\right) +{A_0^{-1}}\left(\left[x_{k-1},x_k;F\right]-\left[x_{-2},x_{-1};F\right] \right)\right\Vert \nonumber \\ & \le \alpha \left(\left\Vert {x_{k+1}-x_0}\right\Vert +\left\Vert {x_{k}-x_{-1}}\right\Vert \right) + \phi \left(\left\Vert {x_{k-1}-x_{-2}}\right\Vert +\left\Vert {x_{k+1}-x_{0}}\right\Vert \right) \nonumber \\ & {} \quad + \gamma \left(\left\Vert {x_{k-1}-x_{-2}}\right\Vert +\left\Vert {x_{k}-x_{-1}}\right\Vert \right) \nonumber \\ & \le \alpha \left(t_{k+1}+t_k-2a-b\right) +\phi \left(t_{k+1}+t_{k-1}-a-b\right)+\nonumber \\ & {} \qquad +\gamma \left(t_{k+1}+t_{k-1}-a-b-c\right) {\lt} 1 \qquad (\textrm{by}\eqref{eq:214}).\label{eq:235} \end{align}

It follows from 34 and the Banach lemma on invertible operators [ 8 , 14 ] that $A_{k+1}^{-1}$ exists and

\begin{align} \left\Vert {A_{k+1}^{-1}A_{0}}\right\Vert & \le d_{k+1}^{-1} \nonumber \\ & \le \big\{ 1-\left[\alpha \left(t_{k+1}+t_k-2a-b\right)+\phi \left(t_{k+1}+t_{k-1}-a-b\right)\right.\nonumber \\ & \left.\quad +\gamma \left(t_{k+1}+t_{k-1} -a-b-c\right)\right]\big\} ^{-1} \label{eq:236} \end{align}

In view of 2, 35, 36, we have:

\begin{align} \left\Vert {x_{k+2}-x_{k+1}}\right\Vert & = \left\Vert \left(A_{k+1}^{-1}A_{0}\right)\left(A_{0}^{-1}F(x_{k+1})\right) \right\Vert \nonumber \\ & \le \left\Vert {A_{k+1}^{-1}A_{0}} \right\Vert \left\Vert {A_{0}^{-1}F(x_{k+1})}\right\Vert \nonumber \\ & \le L_{k+1}\left\Vert {x_{k+1}-x_k}\right\Vert \le t_{k+2}-t_{k+1},\nonumber \end{align}

which shows 29 and 33 for all $n$. By Lemma 2.1, sequence $\{ x_n\} $ is Cauchy in a Banach space $X$ and as such it converges to some $x^\star \in \overline{U}(x_0,t^\star )$ (since $\overline{U}(x_0,t^\star )$ is a closed set). Estimate 30 follows from 29 by using standard majorization techniques [ 8 , 14 , 16 ] .

Using 23, 24 and the induction hypotheses, we obtain in turn

\begin{align} & \left\Vert {A_0^{-1}\left(\left[x_k,x_{k+1};F\right]-A_k\right)}\right\Vert = \nonumber \\ & = \left\Vert {A_0^{-1}} \left( \left[x_k,x_{k+1};F\right] - \left[x_k,x_{k};F\right] + \left[x_k,x_{k};F\right] -A_k \right) \right\Vert \nonumber \\ & = \Big\| {A_0^{-1}} \left\{ \left[x_k,x_{k+1};F\right] - \left[x_k,x_{k};F\right] - \left(\left[x_k,x_k,x_{k-1};F\right] \right. \right. \nonumber \\ & {}\quad \left. \left. - \left[x_{k-2},x_k,x_{k-1};F\right]\right) \left(x_k-x_{k-1}\right) \right\} \Big\| \nonumber \\ & \le p \left\Vert {x_{k+1}-x_k} \right\Vert + q \left\Vert {x_{k}-x_{k-2}} \right\Vert \left\Vert {x_{k}-x_{k-1}} \right\Vert \nonumber \\ & \le p \left(t_{k+1}-t_k\right) + q \left(t_{k}-t_{k-2}\right) \left(t_{k}-t_{k-1}\right). \label{eq:237} \end{align}

We also need the estimate:

\begin{align} \left\Vert {x_{k+2}-x_{k+1}}\right\Vert & = \left\Vert {\left(A_{k+1}^{-1}A_0\right)\left(A_{0}^{-1}F\left(x_{k+1}\right)\right)}\right\Vert \nonumber \\ & = \left\Vert {\left(A_{k+1}^{-1}A_0\right)A_0^{-1}\left(F\left(x_{k+1}\right)- F\left(x_{k}\right) - A_k\left(x_{k+1}-x_k\right)\right)}\right\Vert \nonumber \\ & \le \left\Vert {A_{k+1}^{-1}A_0}\right\Vert \left\Vert {A_0^{-1} \left(\left[x_k,x_{k+1};F\right]-A_k\right)}\right\Vert \left\Vert {x_{k+1}-x_k}\right\Vert .\label{eq:238} \end{align}

The fact that $x^\star $ is a solution of equation $F(x)=0$ follows by letting $k\to \infty $ in the estimate:

\begin{align} \left\Vert {A_{0}^{-1}F\left(x_{k+1}\right)}\right\Vert & = \left\Vert {A_{0}^{-1}\left(\left[x_k,_{k+1};F\right]-A_k\right)\left(x_{k+1}-x_k\right) } \right\Vert \nonumber \\ & \le p \left\Vert {x_{k+1}-x_k}\right\Vert ^2 + q \left\Vert {x_{k}-x_{k-2}}\right\Vert \, \left\Vert {x_{k}-x_{k-1}}\right\Vert \, \left\Vert {x_{k+1}-x_k}\right\Vert .\label{eq:239} \end{align}

Finally, to show the uniqueness part, let $y^\star \in {U(x_0,r)}$ be a solution of equation $F(x)=0$. We can write for $L=\left[y^\star ,x^\star ;F\right]:$

\begin{equation} \label{eq:240} F(y^\star )-F(x^\star ) = L\left(y^\star -x^\star \right). \end{equation}

We shall show linear operator $L$ is invertible. Using 25-27, 32 and 32, we have:

\begin{align} & \left\Vert {A_{0}^{-1}}\left(\left[x_{0},x_{-2};F\right] + \left[x_{-1},x_{0};F\right] - \left[y^\star ,x^\star ;F\right] - \left[x_{-2},x_{-1};F\right]\right)\right\Vert \le \nonumber \\ & {}\le \left\Vert {A_{0}^{-1}} \left( \left[x_{0},x_{-2};F\right] - \left[y^\star ,x^\star ;F\right] \right)\right\Vert \nonumber \\ & \quad + \left\Vert {A_{0}^{-1}} \left( \left[x_{-1},x_{0};F\right] - \left[x_{-2},x_{-1};F\right] \right)\right\Vert \nonumber \\ & {} \le \phi \left(\left\Vert {x_{0}-y^\star }\right\Vert + \left\Vert {x_{-2}-x^\star }\right\Vert \right) + \gamma \left(\left\Vert {x_{-1}-x_{-2}}\right\Vert + \left\Vert {x_{0}-x_{-1}}\right\Vert \right) \nonumber \\ & {} {\lt} \phi \left(r+t^\star +b+a\right) + \gamma \left(b+a\right)\le {1}.\label{eq:241} \end{align}

In view of 40 and the Banach lemma on invertible operators, $L^{-1}$ exists. We deduce from 39 that $x^\star =y^\star .$ That completes the proof of the Theorem.

Proof â–¼

Remark 3

$(a)$ A similar existence Theorem (without a uniqueness result) was provided in [ 18 , p.91 ] using conditions 22-23, a decreasing majorizing sequence, and some different sufficient convergence conditions. Therefore a direct comparison is not possible. However, in §4, we show that the results obtained in Theorem 2.2 can be weaker than the corresponding ones of Theorem 5.1 in [ 18 , p.91 ] .

$(b)$ Note that $t^{\star \star }$ given by 8 can replace $t^\star $ in hypotheses 28 and 32 of Theorem 2.2.â–¡

3 Local Convergence of (UTM)

We can show the local convergence result for (UTM).

Theorem 4

Let $F\colon {D\subseteq {X\rightarrow {Y}}}$ and let $x^\star \in {D}$ be such that ${{F^\prime (x^\star )^{-1}}}$ exists. Assume that for all $x,y,u,v,z\in {D}:$

\begin{align} \left\Vert {{F^\prime (x^\star )^{-1}}} \left(\left[x^\star ,x^\star ;F\right]-\left[x,x^\star ; F\right]\right)\right\Vert & \le p_0\left\Vert {x^\star -x}\right\Vert , \label{eq:31} \\ \left\Vert {{F^\prime (x^\star )^{-1}}} \left(\left[z,x^\star ;F\right]-\left[z,x; F\right]\right)\right\Vert & \le p_1\left\Vert {x^\star -x}\right\Vert ,\label{eq:32} \end{align}

\begin{align} \left\Vert {{F^\prime (x^\star )^{-1}}} \left(\left[x,x^\star ,y;F\right]-\left[z,x^\star ,y; F\right]\right)\right\Vert & \le q_0\left\Vert {x-z}\right\Vert ,\label{eq:33} \\ \left\Vert {{F^\prime (x^\star )^{-1}}} \left(\left[u,x,y;F\right]-\left[v,x,y; F\right]\right)\right\Vert & \le q_\star \left\Vert {u-v}\right\Vert \label{eq:34} \end{align}

and

\begin{align} {{U\left(x^\star ,r^\star \right)}} & \subseteq {D},\label{eq:35} \end{align}

where

\begin{align} & r^\star = \tfrac {2}{p_0+2p_1+\sqrt{\left(p_0+2p_1\right)^2+16\left(q_0+q_\star \right)}}.\label{eq:36} \end{align}

Then, sequence $\left\{ x_n\right\} $ generated by (UTM) is well defined, remains in $U\left(x^\star ,r^\star \right)$ for all $n\ge {0}$ and converges to $x^\star $, provided that $x_0\in U(x^\star ,r^\star )$. Moreover, the following estimates hold:

\begin{equation} \left\Vert {x_{n+1}-x^\star }\right\Vert \le \tfrac {e_n}{h_n}\left\Vert {x_n-x^\star }\right\Vert ,\label{eq:47} \end{equation}

where

\begin{equation} e_n = p_1\left\Vert {x_n-x^\star }\right\Vert + q_\star \left\Vert {x_n-x_{n-2}}\right\Vert \left\Vert {x_n-x_{n-1}}\right\Vert \label{eq:48}\qquad \end{equation}

and

\begin{equation} h_n = 1- \left(p_0+p_1\right) \left\Vert {x_n-x^\star }\right\Vert - q_0 \left\Vert {x_n-x_{n-2}}\right\Vert \left\Vert {x_{n-1}-x^\star }\right\Vert .\label{eq:49} \end{equation}

Proof â–¼

It follows as the proof of Theorem 4.1 in [ 8 , p.87 ] but uses the needed conditions 41-44 instead of:

\begin{align} \left\Vert {F^\prime (x^\star )^{-1}}\left(\left[x,y;F\right]-\left[u,v;F\right]\right)\right\Vert & \le p_\star \left(\left\Vert {x-u}\right\Vert +\left\Vert {y-v}\right\Vert \right)\label{eq:310} \intertext {and} \left\Vert {F^\prime (x^\star )^{-1}}\left(\left[u,x,y;F\right]-\left[v,x,y;F\right]\right)\right\Vert & \le q_\star \left\Vert {u-v}\right\Vert .\label{eq:411} \end{align}

Remark 5

$(a)$ Clearly

\begin{align} p_0 & \le p_\star , \label{eq:312}\\ p_1 & \le p_\star , \label{eq:313}\\ q_0 & \le q_\star , \label{eq:314} \end{align}

hold in general, and $p_\star /p_0$, $p_\star /p_1$ and $q_\star /q_0$ can be arbitrarily large [ 7 , 8 ] . If equality holds in 51-53, then our results reduce to the ones in [ 18 ] . Otherwise they constitute an improvement with advantages as noted in the Introduction of this study. $(b)$ The radius of convergence $r^\star $ obtained in Theorem 3.1 is smaller in general than the corresponding one of Newton’s method. Indeed from the hypotheses 50 it follows that $F$ is Fréchet-differentiable on $D$ and its Fréchet derivative satisfies

\begin{align} \left\Vert {F^\prime (x^\star )^{-1}} \left(F^\prime (x)-F^\prime (y)\right) \right\Vert & \le 2p_\star \left\Vert {x-y}\right\Vert \label{eq:315} \intertext {and} \left\Vert {F^\prime (x^\star )^{-1}} \left(F^\prime (x)-F^\prime (x^\star )\right) \right\Vert & \le 2p_2 \left\Vert {x-x^\star }\right\Vert . \label{eq:316} \end{align}

The radius of convergence $r^\star $ is then given for $q_0=q_\star =0:$

\begin{align} r_A^\star & = \tfrac {1}{2p_2+p_\star } \label{eq:317} \intertext {for} p_2 & \le {p_\star },\label{eq:318} \intertext {whereas the one obtained by Theorem 4.1 in \cite{eighteen} is given by} r_R^\star & = \tfrac {1}{3p_\star },\label{eq:319} \end{align}

found by Rheinboldt in [ 16 ] . Note that

\begin{equation} r_R^\star \le r_A^\star . \label{eq:320} \end{equation}

If strict inequality holds in 55, then so does in 56.â–¡

4 A Numerical Example

We provide a numerical example to show that Theorem 2.2 can be used to solve equation 1 but not corresponding Theorem 5.1 in [ 18 ] . Let $X=Y=R^2$ be equipped with the max-norm, $x_0=(1,1)^T,$ $D=U(x_0,1-\lambda )$, $\lambda \in [0,1/2)$ and define function $F$ on $D$ for $x=\left(\mu _1,\mu _2\right)$ by

\begin{equation} \label{eq:41} F(x) = \left(\mu _1^3-\lambda ,\mu _2^3-\lambda \right)^T. \end{equation}

Using 57 we obtain the Fréchet-derivative

\begin{equation} \label{eq:42} F^\prime (x)=\left[\begin{matrix} 3\mu _1^2 & 0 \\ 0 & 3\mu _2^2 \end{matrix}\right]. \end{equation}

Define

\begin{equation} \label{eq:43} [x,y;F] = \int _{0}^1F^\prime (y+t(x-y)){\rm dt}. \end{equation}

Let $x_{-2} = \left(1.02,1.02\right)^T$, $x_{-1}=\left(1.01,1.01\right)$, $\lambda = 0.49$. Using 22-27 and 59, we get

\begin{gather*} a=b=0.1,\quad c = 0.170011334, \quad \left\Vert {A_0}^{-1}\right\Vert = 0.33335557, \nonumber \\ p = 3 \left\Vert {A_{0}^{-1}}\right\Vert (2-\lambda ), \quad q = 3\left\Vert {A_{0}^{-1}}\right\Vert , \nonumber \\ \alpha = \tfrac {1}{2} \left(1-\lambda +a + \left\Vert {x_0+x_{-1} }\right\Vert \right) \left\Vert {A_{0}^{-1}}\right\Vert , \\ \phi = \tfrac {1}{2} \left(1-\lambda + 2\left\Vert {x_0}\right\Vert + \left\Vert {x_{-2}-x_{0} }\right\Vert \right) \left\Vert {A_{0}^{-1}}\right\Vert , \quad \nonumber \\ \gamma = \tfrac {1}{2} \left(1-\lambda +b + \left\Vert {x_0+x_{-1} }\right\Vert \right) \left\Vert {A_{0}^{-1}}\right\Vert ,\nonumber \intertext {so} p = 1.510100673, \quad q = 1.000066671, \quad \alpha = \phi = \gamma = 0.42169478. \end{gather*}

Moreover, using 5-7 and 8, we obtain 5, 5 become

\begin{align} 0.84338956 & {\lt} 1.680123342, \nonumber \\ 0.160253576 & {\lt} 1,\nonumber \end{align}

respectively,

\begin{gather*} \delta _0 = 0.305966463, \quad \delta = 0.310470973{\gt} \delta _0, \quad f_2(\delta ) = -0.310669379 {\lt} 0 \end{gather*}

and

\begin{gather*} t^{\star \star } = 0.267566675.\nonumber \end{gather*}

That is, the hypotheses of Theorem 2.2 are satisfied. Moreover, using Remark 2.3(b) and 32, we see that we can set $r = 1-\lambda = 0.51.$ Hence, there exists a unique solution:

\begin{equation} \nonumber x^\star = \left(\sqrt[3]{0.49},\sqrt[3]{0.49}\right)^T = \left(0.788373516,0.78373516\right)^T \end{equation}

of equation $F(x)=0$ in $\overline{U}(x_0,r)$, which can be obtained as the limit of (UTM). However, hypotheses of Theorem 5.1 in [ 18 ] do not hold. The sufficient convergence condition corresponding to 5 is given by

\begin{gather} c \le \lambda = \tfrac {1}{3}\tfrac {p+qa+2\lambda _0}{\left(p+qa+\lambda _0\right)^2}\left[1-qa(a+b)\right]^2,\label{eq:44} \intertext {where} \lambda _0 = \left\{ \left(p+qa\right)^2+3q\left(1-qa(a+b)\right)\right\} ^{1/2}.\nonumber \end{gather}

We have

\begin{gather} \lambda _0 = 1.520496025,\quad \lambda = 0.08533979\nonumber \\ \intertext {{{and, so \eqref{eq:44} is violated, since}}} c = 0.170011334{\gt}\lambda = 0.08533979.\nonumber \end{gather}

Hence, there is no guarantee that (UTM) starting at $x_0$ converges to $x^\star $.

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