On the Semi-Local Convergence of an Ostrowski-Type Method for Solving Equations

Abstract

Symmetries play a crucial role in the dynamics of physical systems. As an example, microworld and quantum physics problems are modeled on principles of symmetry. These problems are then formulated as equations defined on suitable abstract spaces. Then, these equations can be solved using iterative methods. In this article, an Ostrowski-type method for solving equations in Banach space is extended. This is achieved by finding a stricter set than before containing the iterates. The convergence analysis becomes finer. Due to the general nature of our technique, it can be utilized to enlarge the utilization of other methods. Examples finish the paper.

1. Introduction

We are concerned with finding $x_{*}$ solving

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

(1)

where $F:D\subset E⟶E_1$ is an operator acting between Banach spaces E and $E_1$ with $D\ne \varnothing .$

The famous Ostrowski-type method is defined for $x_0\in D$ and each $n=0,1,2,\dots$ by

$$\begin{array}{ccc}\hfill y_k& =& x_k-F^{\prime }{\left(x_k\right)}^{-1}F\left(x_k\right)\hfill \\ \hfill x_{k+1}& =& y_k-A_{k}F\left(x_k\right),\hfill \end{array}$$

(2)

where $A_k=2{[y_k,x_k;F]}^{-1}-F^{\prime }{\left(x_k\right)}^{-1}$, with $[.,.,F]:D\times D\to L(E,E_1)$. There are numerous results for the convergence of iterative methods utilizing the information $(D,x_0,F,F^{\prime })$ and higher order derivatives [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39]. However, higher order derivatives cannot be found on method (2). Moreover, these results do not give uniqueness ball or estimates on $\parallel x_k-x_{*}\parallel$ or $\parallel x_{k+1}-x_k\parallel$. That is why we are motivated to write this paper, where only hypotheses on the derivative and divided differences of order one are used. Notice that only these operators appear on method (2).

The method (2) is shown to be of order four using Taylor expansion and assumptions on the fifth order derivative of $F,$ which is not on these schemes [5]. So, the assumptions on the sixth derivative reduce the applicability of this method.

For example: Let $E=E_1=\mathbb{R},D=[-0.5,1.5].$ Define $\lambda$ on D by

$$\lambda \left(t\right)=\left\{\begin{array}{cc}{t}^{3}log{t}^2+t^5-t^4& ift\ne 0\\ 0& ift=0.\end{array}\right.$$

Then, we get $t^{*}=1,$ and

$${\lambda }^{‴}\left(t\right)=6log{t}^2+60t^2-24t+22.$$

Obviously, ${\lambda }^{‴}\left(t\right)$ is not bounded on $D.$ So, the convergence of method (2) is not guaranteed by the previous analyses in [5].

The rest of the study is organized as follows: Section 2 contains results on majorizing sequences. In Section 3, we develop the semi-local convergence analysis based on majorizing sequences. The local convergence analysis can be found in Section 4. Numerical examples can be found in Section 5. The paper ends with some concluding remarks in Section 6.

2. Majorizing Sequences

We recall the definition of a majorizing sequences.

Definition 1.

Let $\left\{v_k\right\}$ be a sequence in a complete normed space. Then, a non-decreasing scalar sequence $\left\{d_k\right\}$ is called majorizing for $\left\{v_k\right\}$ if

$$\parallel v_{k+1}-v_k\parallel \le d_{k+1}-d_k\mathrm{for} \mathrm{each}k=0,1,2,\dots .$$

Then, the convergence of sequence $\left\{v_k\right\}$ reduces to studying that of $\left\{d_k\right\}$ [40].

Let $\eta \ge 0$ and $L,L_i,i=0,1,2,3,4$ be positive parameters. Set $M_0=\frac{L_0{L}_2}{2},$$M=\frac{L}{2},M_1=\frac{L{L}_2}{2}$ and $M_2=\frac{L{L}_3}{2}.$ Define sequences $\left\{t_k\right\},\left\{s_k\right\},\left\{{\alpha }_k\right\}$ and $\left\{{\beta }_k\right\}$ for each $k=0,1,2,\dots$ by $t_0=0,s_0=\eta$

$$\begin{array}{ccc}\hfill t_1& =& s_0+\frac{M_0{s}_0^2}{1-L_1{s}_0}+\frac{M_2{s}_0^3}{(1-L_1{s}_0)(1-L_0{s}_0)}+\frac{M{s}_0^2}{1-L_0{s}_0},\hfill \\ \hfill s_1& =& t_1+\frac{M{(t_1-t_0)}^2+t_1-s_0}{1-L_0{t}_1}\hfill \\ \hfill t_{k+1}& =& s_k+{\alpha }_k(s_k-t_k),\hfill \\ \hfill s_{n+1}& =& t_{n+1}+\frac{M{(t_{n+1}-t_n)}^2+L_4{t}_n(t_{n+1}-s_n)}{1-L_0{t}_{n+1}},\hfill \\ \hfill {\alpha }_k& =& \frac{M_1(s_k-t_k)}{(1-L_1(s_k+t_k))(1-L_0{t}_k)}\hfill \\ \hfill & & +\frac{M_2(s_k-t_k)}{(1-L_1(s_k+t+k))(1-L_0{s}_k)}+\frac{M(s_k-t_k)}{1-L_0{s}_k},\hfill \\ \hfill {\beta }_k& =& \frac{M(t_{k+1}-t_k)+L_4{t}_k}{1-L_0{t}_{k+1}}.\hfill \end{array}$$

(3)

Moreover, define quadratic polynomials and functions on the interval $[0,1]$ for some $b>1$

$$p_1\left(t\right)=t^2-(1-L_0{t}_1)t+L_4{t}_1,$$

$$p_2\left(t\right)=t^2+t-(1-\frac{2b{L}_1{t}_1}{b-1}),$$

$$p_3\left(t\right)=t^2+t-(1-L_0{t}_1),$$

$$g_1\left(t\right)=(M_{1}b+M_{2}b+M)t$$

and

$$g_2{\left(t\right)=Mt((1+t)t-1)(1+t))+(1+t)}^2{t}^2(L_4+L_0{t}^2(1+t)).$$

Denote by ${\gamma }_0,{\gamma }_1$ or ${\gamma }_2$ or ${\gamma }_3,{\gamma }_4,$ the non-negative zeros of $p_1,p_2,p_3$ if they exist. Furthermore, define sequences of functions on the interval $[0,1]$ for $\delta =\delta \left(t\right)=(1+t)t$ by

$$f_n^{\left(1\right)}\left(t\right)=M_{1}bt{\delta }^{n-1}{t}_1+M_{2}bt{\delta }^{n-1}{t}_1+Mt{\delta }^{n-1}{t}_1+L_{0}t(1+\delta +\dots +{\delta }^n)t_1-t$$

and

$$f_n^{\left(2\right)}\left(t\right)=M(t^2{\delta }^{n-1}+t{\delta }^{n-1})t_1+L_4(1+\delta +\dots +{\delta }^n)t_1+L_{0}t(1+\delta +\dots +{\delta }^n)t_1-t.$$

Next, we present two results on the majorizing sequence for method (2).

Lemma 1.

Suppose that for each $k=0,1,2,\dots$, items

$$s_k\le t_{k+1}<\frac{1}{L_0}$$

(4)

and

$$s_k+t_k<\frac{1}{L_1}$$

(5)

hold. Then, sequences $\left\{s_k\right\}$ and $\left\{t_k\right\}$ are increasing, bounded from above by $\frac{1}{L_0}$ and converge to their unique least upper bound $s_{*}\in [0,\frac{1}{L_0}].$

Proof. 

It follows from (3)–(5) that sequences $\left\{s_k\right\},\left\{t_k\right\}$ are increasing, bounded from above by $\frac{1}{L_0}$ and as such they converge to $s_{*}.$ □

Remark 1.

Conditions (4) and (5) hold only in some special cases. This is why we present stronger conditions that can be verified more easily.

We shall use the following set of conditions denoted by (A) in our second result on majorizing sequences for method (2).

Suppose: there exists $\gamma \in S:=(0,\frac{\sqrt{5}-1}{2}),b>0,\eta >0$ satisfying

$$\begin{array}{ccc}\hfill 0& \le & {\alpha }_0\le \gamma ,0\le {\beta }_0\le \gamma ,\hfill \\ \hfill L_0\eta & <& 1,L_1\eta <1,L_1{t}_1<\frac{b-1}{2b},\hfill \\ \hfill {\gamma }_0& \le & \gamma \le {\gamma }_1\mathrm{if}{g}_2\left(t\right)\ge 0\mathrm{for} \mathrm{each}t\in S,\hfill \end{array}$$

or

$$\begin{array}{ccc}\hfill f_1^{\left(2\right)}\left(\gamma \right)& \le & 0\mathrm{if}{g}_2\left(t\right)\le 0\mathrm{for} \mathrm{each}t\in S,\hfill \\ \hfill \gamma & \le & {\gamma }_2\hfill \\ \hfill \mathrm{and}& & \\ \hfill {\gamma }_3& \le & \gamma \le {\gamma }_4.\hfill \end{array}$$

Then, under the preceding notation and conditions (A), we can show.

Lemma 2.

Under conditions (A), the conclusions of Lemma 1 hold for sequences $\left\{s_k\right\},\left\{t_k\right\}.$ Moreover, the following assertions hold for each $k=0,1,2,\dots$

$$0\le s_k-t_k\le {\gamma }^k{(1+\gamma )}^{k-1}(t_1-t_0),$$

(6)

$$0\le t_{k+1}-s_k\le {\gamma }^{k+1}{(1+\gamma )}^{k-1}(t_1-t_0),$$

(7)

$$0\le s_k\le \frac{1-{\delta }^{k+1}}{1-\delta }{t}_1$$

(8)

and

$$0\le t_{k+1}\le \frac{1-{\delta }^{n+2}}{1-\delta }{t}_1.$$

(9)

Recall that $\delta =\gamma (\gamma +1)$ and $\delta \left(t\right)=\delta .$

Proof. 

We shall show using induction on n that the following hold.

$$0\le {\alpha }_n\le \gamma ,$$

(10)

$$0\le {\beta }_n\le \gamma ,$$

(11)

$$L_0{t}_n\le 1,L_0{s}_n<1,L_1(s_n+t_n)<1,$$

(12)

and

$$t_n\le s_n\le t_{n+1}.$$

(13)

Estimates (10)–(13) hold for $n=0,$ by the initial conditions and conditions (A). We also have

$$\begin{array}{ccc}\hfill 0& \le & s_0-t_0\le \eta ,0\le t_1-s_0\le \gamma \eta ,0\le s_1-t_1\le \gamma (t_1-t_0),\hfill \\ \hfill 0& \le & t_2-s_1\le {\gamma }^2(t_1-t_0),\hfill \\ \hfill 0& \le & s_2-t_2\le {\gamma }^2(1+\gamma )(t_1-t_0),\hfill \\ \hfill 0& \le & t_3-s_2\le {\gamma }^3(1+\gamma )(t_1-t_0),\hfill \\ \hfill & ⋮& \\ \hfill 0& \le & s_n-t_n\le {\gamma }^n{(1+\gamma )}^{n-1}(t_1-t_0),\hfill \end{array}$$

(14)

$$\begin{array}{ccc}\hfill 0& \le & t_{n+1}-s_n\le {\gamma }^{n+1}{(1+\gamma )}^{n-1}(t_1-t_0),\hfill \end{array}$$

(15)

$$\begin{array}{ccc}\hfill t_{n+1}& \le & s_n+{\gamma }^{n+1}{(1+\gamma )}^{n-1}(t_1-t_0)\le t_n+{\gamma }^n{(1+\gamma )}^{n-1}(t_1-t_0)\hfill \\ \hfill & & +{\gamma }^{n+1}{(1+\gamma )}^{n-1}(t_1-t_0)\hfill \\ \hfill & ⋮& \\ \hfill & \le & t_1+\gamma {(1+\gamma )}^{n-1}(t_1-t_0)+{\gamma }^2(1+\gamma )(t_1-t_0)+\hfill \\ \hfill & ⋮& \\ \hfill & & +{\gamma }^n{(1+\gamma )}^{n-1}(t_1-t_0)+{\gamma }^{n+1}{(1+\gamma )}^{n-1}(t_1-t_0)\hfill \\ \hfill & \le & (1+\delta +\dots +{\delta }^{n+1})t_1=\frac{1-{\delta }^{n+2}}{1-\delta }{t}_1\hfill \end{array}$$

(16)

and

$$s_n\le t_n+{\gamma }^n{(1+\gamma )}^{n-1}{t}_1\le \dots \le \frac{1-{\delta }^{n+1}}{1-\delta }{t}_1.$$

(17)

Suppose these estimates hold for all integers smaller or equal to $n.$ Then, evidently, (10) holds (since $\frac{1}{1-L_1(s_n+t_n)}\le b$), if we show instead using (14)–(18) that

$$\frac{M_{1}b(s_n-t_n)}{1-L_0{t}_n}+\frac{M_{2}b(s_n-t_n)}{1-L_0{s}_n}+\frac{M(s_n-t_n)}{1-L_0{s}_n}\le \gamma$$

(18)

or

$$\begin{array}{ccc}\hfill & & M_{1}b\gamma {\delta }^{n-1}{t}_1+M_{2}b\gamma {\delta }^{n-1}{t}_1+M\gamma {\delta }^{n-1}{t}_1\hfill \\ \hfill & & +\gamma L_0(1+\delta +\dots +{\delta }^n)t_1-\gamma \le 0.\hfill \end{array}$$

(19)

Notice that expression (19) is obtained if we replace $s_n-t_n,t_n,s_n$ by the right hand sides of (14), (15) and (17), respectively, in (18), remove denominators and move all terms at the right hand side of the inequality.

Estimate (19) motivates us to define functions $f_n^{\left(1\right)}$ on the interval $[0,1]$ and show instead of (19)

$$f_n^{\left(1\right)}\left(t\right)\le 0att=\gamma .$$

(20)

We shall find a relationship between two consecutive functions $f_n^{\left(1\right)}.$ We can write in turn that

$$\begin{array}{ccc}\hfill f_{n+1}^{\left(1\right)}\left(t\right)& =& M_{1}bt{\delta }^n{t}_1+M_{2}bt{\delta }^n{t}_1+Mt{\delta }^n{t}_1+L_{0}t(1+\delta +\dots +{\delta }^{n+1})t_1-t\hfill \\ & & -M_{1}bt{\delta }^{n-1}{t}_1-M_{2}bt{\delta }^{n-1}{t}_1-Mt{\delta }^{n-1}{t}_1\hfill \\ & & -L_{0}t(1+\delta +\dots +{\delta }^n)t_1+t+f_n^{\left(1\right)}\left(t\right)\hfill \\ & =& f_n^{\left(1\right)}\left(t\right)+(M_{1}bt{\delta }^n{t}_1-M_{1}bt{\delta }^{n-1}{t}_1)\hfill \\ & & +(M_{2}bt{\delta }^n{t}_1-M_{2}bt{\delta }^{n-1}{t}_1)+(Mt{\delta }^n{t}_1-Mt{\delta }^{n-1}{t}_1)+t{L}_0{\delta }^{n+1}{t}_1\hfill \\ & =& f_n^{\left(1\right)}\left(t\right)+(\delta -1)t(M_{1}b+M_{2}b+M){\delta }^{n-1}{t}_1\hfill \\ & =& f_n^{\left(1\right)}\left(t\right)+(\delta -1)g_1\left(t\right){\delta }^{n-1}{t}_1\hfill \\ & \le & f_n^{\left(1\right)}\left(t\right),\hfill \end{array}$$

since $t\in [0,\frac{\sqrt{5}-1}{2}],$ so

$$f_{n+1}^{\left(1\right)}\left(t\right)\le f_n^{\left(1\right)}\left(t\right).$$

(21)

Define function

$$f_{\infty }^{\left(1\right)}\left(t\right)=\underset{k⟶+\infty }{lim}{f}_k^{\left(1\right)}\left(t\right).$$

(22)

By the definition of functions $f_n^{\left(1\right)}$ and $f_{\infty }^{\left(1\right)},$ we get

$$f_{\infty }^{\left(1\right)}\left(t\right)=\frac{t{L}_0{t}_1}{1-\delta }-t.$$

(23)

Then, we can show instead of (20) that

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

(24)

which is true by the definition of $p_3$ and ${\gamma }_3\le \gamma \le {\gamma }_4.$ Similarly, (11) holds if

$$\begin{array}{ccc}\hfill & & M({\gamma }^2{\delta }^{n-1}+\gamma {\delta }^{n-1})t_1+L_4(1+\delta +\dots +{\delta }^n)t_1\hfill \\ \hfill & & +L_0\gamma (1+\delta +\dots +{\delta }^{n+1})t_1-\gamma \le 0\hfill \end{array}$$

(25)

or

$$f_n^{\left(2\right)}\left(t\right)\le 0att=\gamma .$$

(26)

This time, we have

$$\begin{array}{ccc}\hfill f_{n+1}^{\left(2\right)}\left(t\right)& =& M(t^2{\delta }^n+t{\delta }^n)t_1+L_4(1+\delta +\dots +{\delta }^{n+1})t_1\hfill \\ & & +L_{0}t((1+\delta +\dots +{\delta }^{n+2})t_1-t\hfill \\ & & -M(t^2{\delta }^{n-1}+t{\delta }^{n-1})t_1-L_4(1+\delta +\dots +{\delta }^n)t_1\hfill \\ & & -L_{0}t(1+\delta +\dots +{\delta }^{n+1})t_1+t+f_n^{\left(2\right)}\left(t\right)\hfill \\ & =& f_n^{\left(2\right)}+M(t^2{\delta }^n+t{\delta }^n-t^2{\delta }^{n-1}-t{\delta }^{n-1})t_1\hfill \\ & & +L_4{\delta }^{n+1}{t}_1+L_{0}t{\delta }^{n+2}{t}_1\hfill \\ & =& f_n^{\left(2\right)}\left(t\right)+g_2\left(t\right){\delta }^{n-1}{t}_1,\hfill \end{array}$$

so

$$f_{n+1}^{\left(2\right)}\left(t\right)=f_n^{\left(2\right)}\left(t\right)+g_2\left(t\right){\delta }^{n-1}{t}_1.$$

(27)

Define function

$$f_{\infty }^{\left(2\right)}\left(t\right)=\underset{k⟶+\infty }{lim}{f}_n^{\left(2\right)}\left(t\right).$$

(28)

Then, we get

$$f_{\infty }^{\left(2\right)}\left(t\right)=\frac{L_4{t}_1}{1-t}+\frac{L_{0}t{t}_1}{1-t}-t.$$

(29)

If ${\gamma }_0\le \gamma \le {\gamma }_1,$ then $g_2\left(t\right)\ge 0$ for each $t\in S$ and $f_{\infty }^{\left(2\right)}\left(t\right)\le 0$ holds at $t=\gamma .$ However, if $g_2\left(t\right)\le 0$ for each $t\in S,$ then

$$f_{n+1}^{\left(2\right)}\left(t\right)\le f_n^{\left(2\right)}\left(t\right).$$

(30)

In this case, (26) holds if $f_1^{\left(2\right)}\left(t\right)\le 0$ at $t=\gamma$, which is true. Therefore, the induction for (10)–(13) is completed. Hence, sequences $\left\{s_k\right\},\left\{t_k\right\}$ are non-decreasing, bounded from above by $\frac{t_1}{1-\delta }{t}_1$ and as such they converge to $s_{*}$ satisfying $s_{*}\in [\eta ,\frac{t_1}{1-\delta }].$ □

3. Semi-Local Convergence

We shall use conditions (H):

Suppose

(H1)

There exist $x_0\in D,\eta \ge 0$ such that $F^{\prime }\left(x_0\right)$ is invertible and

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

(H2)

For each $u\in D$

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

Set $D_0=U(x_{*},\frac{1}{L_0})\cap D.$

(H3)

For each $v,w\in D_0$

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

$$\parallel F^{\prime }{(x_0*)}^{-1}([v,w;F]-F^{\prime }\left(x_0\right))\parallel \le L_1(\parallel v-x_0\parallel +\parallel w-x_0\parallel ),$$

$$\parallel F^{\prime }{\left(x_0\right)}^{-1}([v,w;F]-F^{\prime }\left(w\right))\parallel \le L_2\parallel v-w\parallel$$

$$\parallel F^{\prime }{\left(x_0\right)}^{-1}([v,w;F]-F^{\prime }\left(v\right))\parallel \le L_3\parallel v-w\parallel .$$

and

$$\parallel F^{\prime }{\left(x_0\right)}^{-1}{F}^{\prime }\left(w\right)\parallel \le L_4\parallel w-x_0\parallel .$$

(H4)

$U[x_0,s_{*}]\subset D$

and

(H5)

Conditions of Lemma 1 or Lemma 2 hold.

Then, based on conditions (H), we present the semi-local convergence analysis of method (2).

Theorem 1.

Suppose hypotheses (H) hold. Then, sequences $\left\{x_k\right\},\left\{y_k\right\}$ generated by method (2) with starter $x_0$ are well defined in $U[x_0,s_{*}],$ remain in $U[x_0,s_{*}]$ for each $n=0,1,2,\dots$ and converge to a solution $x_{*}\in U[x_0,s_{*}]$ of equation $F\left(x\right)=0.$ Moreover, the following error estimates hold

$$\parallel x_k-x_{*}\parallel \le t_n-s_{*}.$$

(31)

Proof. 

Mathematical induction is employed to show

$$\parallel x_{k+1}-y_k\parallel \le t_{k+1}-s_k$$

(32)

and

$$\parallel y_k-x_k\parallel \le s_k-t_k.$$

(33)

Iterate $y_0$ is well defined by the first substep of method (2) and (H1). We can write

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

so $y_0\in U[x_0,s_{*}].$ Using (H3), we get in turn for $v,w\in U(x_0,s_{*})$

$$\begin{array}{ccc}\hfill \parallel F^{\prime }{\left(x_0\right)}^{-1}([v,w;F]-F^{\prime }\left(x_0\right))\parallel & \le & L_1(\parallel v-x_0\parallel +\parallel w-x_0\parallel )\hfill \\ \hfill & \le & L_1(s_{*}+s_{*})=2L_1{s}_{*}<1\hfill \end{array}$$

(34)

by the Lemma on invertible opertors due to Banach [41,42], leading to

$${\parallel [v,w;F]}^{-1}{F}^{\prime }\left(x_0\right)\parallel \le \frac{1}{1-L_1(\parallel v-x_0\parallel +\parallel w-x_0\parallel )}.$$

(35)

Similarly, iterate $x_1$ is well defined by the second substep of method (2). We also have by (H2) for $w\in U(x_0,s^{*})$

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

so $F^{\prime }{\left(w\right)}^{-1}\in L(E_1,E)$ and

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

(36)

Hence, by (35) for $v=y_0$ and $w=x_0$ and (36) for $w=x_0,$ we have

$$\begin{array}{ccc}\hfill x_1-y_0& =& -{[y_0,x_0;F]}^{-1}(F^{\prime }\left(x_0\right)-[y_0,x_0;F])F^{\prime }{\left(x_0\right)}^{-1}F\left(y_0\right)\hfill \\ \hfill & & -{[y_0,x_0;F]}^{-1}(F^{\prime }\left(y_0\right)-[y_0,x_0;F])F^{\prime }{\left(y_0\right)}^{-1}F\left(y_0\right)\hfill \\ \hfill & & -F^{\prime }{\left(y_0\right)}^{-1}F\left(y_0\right).\hfill \end{array}$$

(37)

In view of (H3), (35), (36) (for $v=y_0,w=x_0$), (37) and triangle inequality, we get in turn

$$\begin{array}{ccc}\hfill \parallel x_1-y_0\parallel & \le & \frac{L{L}_2\parallel y_0-x_0\parallel \parallel y_0-x_0{\parallel }^2}{2(1-L_1(\parallel y_0-x_0\parallel +\parallel x_0-x_0\parallel \left)\right)(1-L_0\parallel x_0-x_0\parallel )}\hfill \\ \hfill & & +\frac{L{L}_3\parallel y_0-x_0\parallel \parallel y_0-x_0{\parallel }^2}{2(1-L_1(\parallel y_0-x_0\parallel +\parallel x_0-x_0\parallel \left)\right)(1-L_0\parallel y_0-x_0\parallel )}\hfill \\ \hfill & & +\frac{L\parallel y_0-x_0{\parallel }^2}{2(1-L_0\parallel y_0-x_0\parallel )}\le {\alpha }_0(s_0-t_0)=t_1-s_0,\hfill \end{array}$$

(38)

and

$$\parallel x_1-x_0\parallel \le \parallel x_1-y_0\parallel +\parallel y_0-x_0\parallel \le t_1-s_0+s_0-t_0=t_1\le s_{*},$$

so $x_1\in U[x_0,s_{*}].$ Thus, estimates (32) and (33) hold for $n=0,$ where we also used

$$\begin{array}{ccc}\hfill \parallel F^{\prime }{\left(x_0\right)}^{-1}F\left(y_0\right)\parallel & =& \parallel {\int }_0^1{F}^{\prime }{\left(x_0\right)}^{-1}(F^{\prime }(x_0+\theta (y_0-x_0))-F^{\prime }\left(x_0\right))(y_0-x_0)d\theta \parallel \hfill \\ \hfill & \le & \frac{L_0}{2}{\parallel y_0-x_0\parallel }^2\le \frac{L}{2}{(s_0-t_0)}^2.\hfill \end{array}$$

(39)

We know that (36) holds for $w=x_1,$ so iterate $y_1$ is well defined by the first substep of method (2) for $n=1,$ and we can write

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

(40)

Then, we obtain by method (2), (36) (for $w=x_1$), (40) and the triangle inequality

$$\begin{array}{ccc}\hfill \parallel y_1-x_1\parallel & \le & \frac{\frac{L}{2}\parallel x_1-x_0{\parallel }^2+\parallel x_1-y_0\parallel }{1-L_0\parallel x_1-x_0\parallel }\hfill \\ \hfill & \le & \frac{M{(t_1-t_0)}^2+t_1-s_0}{1-L_0{t}_1}=s_1-t_1.\hfill \end{array}$$

(41)

Then, we have

$$\parallel y_1-x_0\parallel \le \parallel y_1-x_1\parallel +\parallel x_1-x_0\parallel \le s_1-t_1+t_1=s_1\le s_{*},$$

so $y_1\in U[x_0,s_{*}].$ Suppose estimates (32) and (33) hold for all integers smaller or equal to $n-1.$ Then, simply repeat the preceding calculations with $x_0,y_0,x_1$ replaced by $x_m,y_m,x_{m+1},$ respectively, and use the induction hypotheses to terminate the proof for (32) and (33). By the Lemma sequence $\left\{t_k\right\}$ is Cauchy in a Banach space E and as such it converges to some $x_{*}\in U[x_0,s_{*}]$ since it is a closed set. Finally, using (40), we get

$$\begin{array}{ccc}\hfill \parallel F^{\prime }{\left(x_0\right)}^{-1}F\left(x_{k+1}\right)\parallel & \le & \frac{L}{2}{\parallel x_{k+1}-x_k\parallel }^2\hfill \\ & & +L_4\parallel x_k-x_0\parallel \parallel x_{k+1}-x_k\parallel \hfill \\ & \le & \frac{L}{2}{(t_{k+1}-t_k)}^2+L_4{t}_k(t_{k+1}-s_k)⟶0\hfill \end{array}$$

as $n⟶+\infty$ implying $F\left(x_{*}\right)=0$ (by the continuity of F). □

The point $s_{*}$ can be replaced by $\frac{1}{L_0}$ or $\frac{t_1}{1-\delta },$ respectively, given in closed form.

Next, a uniqueness of the solution $x_{*}$ of equation $F\left(x\right)=0$ is presented.

Proposition 1.

Suppose:

(a) 

There exists a solution $x_{*}\in D$ of equation $F\left(x\right)=0$;

(b) 

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

$$\frac{L_0}{2}(s+s_{*})<1.$$

(42)

Set $D_1=U[x_0,s_{*}]\cap D.$ Then, the only solution of equation $F\left(x\right)=0$ in the region $D_1$ is $x_{*}.$

Proof. 

Let $x^{**}\in D_1$ with $F\left(x^{**}\right)=0.$ Set $M={\int }_0^1{F}^{\prime }(x^{**}+\theta (x_{*}-x^{**}))d\theta .$ Using (H2) and (42), we obtain in turn that

$$\begin{array}{ccc}\hfill \parallel F^{\prime }{\left(x_0\right)}^{-1}(M-F^{\prime }\left(x_0\right))\parallel & \le & L_0{\int }_0^1((1-\theta )\parallel x_{*}-x_0\parallel +\theta \parallel x^{**}-x_0\parallel )d\theta \hfill \\ \hfill & \le & L_0{\int }_0^1(1-\theta )sd\theta +L_0{\int }_0^1\theta s_{*}d\theta \hfill \\ \hfill & \le & \frac{L_0}{2}(s+s_{*})<1,\hfill \end{array}$$

(43)

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

4. Local Convergence

Let $\ell ,{\ell }_j,j=0,1,2,3,4$ be positive parameters. Define function ${\psi }_1:[0,\frac{1}{{\ell }_0})⟶[0,+\infty )$ by

$${\psi }_1\left(t\right)=\frac{\ell t}{2(1-{\ell }_{0}t)}$$

and set

$${\rho }_A=\frac{2}{2{\ell }_0+\ell }.$$

(44)

Define functions $q:[0,\frac{1}{{\ell }_0})⟶[0,+\infty )$ by

$$q\left(t\right)={\ell }_4(1+{\psi }_1\left(t\right))t-1.$$

By this definition, we have $q\left(0\right)=-1$ and $q\left(t\right)⟶+\infty$ as $t⟶{\frac{1}{{\ell }_0}}^{-}.$ It then follows from the intermediate value theorem that function q has zeros in $(0,\frac{1}{{\ell }_0}).$ Denote by ${\rho }_q$ the smallest such zero. Similarly, denote by ${\rho }_p$ the smallest zero of function $p:[0,\frac{1}{{\ell }_0})⟶[0,+\infty )$ defined by $p\left(t\right)={\ell }_0{\psi }_1\left(t\right)t-1.$ Set $\overline{\rho }=min\{{\rho }_q,{\rho }_p\}.$ Moreover, define function ${\psi }_2:[0,\overline{\rho })⟶[0,+\infty )$ by

$$\begin{array}{ccc}\hfill {\psi }_2\left(t\right)& =& \left[\frac{\ell {\psi }_1\left(t\right)t}{2(1-{\ell }_0{\psi }_1\left(t\right)t)}\right.\hfill \\ & & +\frac{{\ell }_3{\ell }_4(1+{\psi }_1\left(t\right)t){\psi }_1\left(t\right)t}{(1-{\ell }_0{\psi }_1\left(t\right)t)(1-{\ell }_1(1+{\psi }_1\left(t\right))t)}\hfill \\ & & +\left.\frac{{\ell }_2{\ell }_4(1+{\psi }_1\left(t\right))}{(1-{\ell }_1(1+{\psi }_1\left(t\right))t){)}^2}\right]{\psi }_1\left(t\right)t.\hfill \end{array}$$

Set

$$\mu \left(t\right)={\psi }_2\left(t\right)-1.$$

We have again $\mu \left(0\right)=-1$ and $\mu \left(t\right)⟶+\infty$ as $t⟶{\overline{\rho }}^{-}.$ Denote by ${\rho }_{\mu }$ the smallest zero of function $\mu$ in $(0,\overline{\rho }).$ We shall show that

$${\rho }_{*}=min\{{\rho }_A,{\rho }_{\mu }\}$$

(45)

is a convergence radius for method (2). Set $T=[0,{\rho }_{*}).$ Then, it follows from these definitions that for each $t\in T$

$$0\le {\ell }_{0}t<1,$$

(46)

$$0\le q\left(t\right)<1,$$

(47)

$$0\le p\left(t\right)<1,$$

(48)

and

$$0\le {\psi }_i\left(t\right)<1,i=1,2.$$

(49)

The conditions (C) shall be used together with the preceding notation provided that $x_{*}$ is a simple solution of equation $F\left(x\right)=0.$

Suppose:

(C1)

For each $u\in D$

$$\parallel F^{\prime }{\left(x_{*}\right)}^{-1}(F^{\prime }\left(u\right)-F^{\prime }\left(x_{*}\right))\parallel \le {\ell }_0\parallel u-x_0\parallel .$$

Set $D_2=U(x_{*},\frac{1}{{\ell }_0})\cap D.$

(C2)

For each $v,w\in D_2$

$$\parallel F^{\prime }{\left(x_{*}\right)}^{-1}(F^{\prime }\left(w\right)-F^{\prime }\left(v\right))\parallel \le \ell \parallel w-v\parallel ,$$

$$\parallel F^{\prime }{\left(x_{*}\right)}^{-1}{F}^{\prime }\left(v\right)\parallel \le {\ell }_1\parallel v-x_{*}\parallel ,$$

$$\parallel F^{\prime }{\left(x_{*}\right)}^{-1}([w,v;F]-F^{\prime }\left(v\right))\parallel \le {\ell }_2(\parallel w-v\parallel ),$$

$$\parallel F^{\prime }{\left(x_{*}\right)}^{-1}([w,v;F]-F^{\prime }\left(w\right))\parallel \le {\ell }_3\parallel w-v\parallel$$

and

$$\parallel F^{\prime }{\left(x_{*}\right)}^{-1}([w,v;F]-F^{\prime }\left(x_{*}\right))\parallel \le {\ell }_4(\parallel w-x_{*}\parallel +\parallel v-x_{*}\parallel ).$$

(C3)

$U[x_{*},{\rho }_{*}]\subset D.$

Next, we present the local convergence analysis of method (2).

Theorem 2.

Under the conditions (C) further suppose that $x_0\in U(x_{*},{\rho }_{*})-\left\{x_{*}\right\}.$ Then, we have ${lim}_{k⟶+\infty }{x}_k=x_{*}.$

Proof. 

We shall use mathematical induction to show

$$\parallel y_n-x_{*}\parallel \le {\psi }_1(\parallel x_n-x_{*}\parallel )\parallel x_n-x_{*}\parallel \le \parallel x_n-x_{*}\parallel <{\rho }_{*}$$

(50)

and

$$\parallel x_{n+1}-x_{*}\parallel \le {\psi }_2(\parallel x_n-x_{*}\parallel )\parallel x_n-x_{*}\parallel \le \parallel x_n-x_{*}\parallel .$$

(51)

where functions ${\psi }_i$ are given previously and radius ${\rho }_{*}$ is defined by (45). Let $z\in U(x_{*},{\rho }_{*})-\left\{x_{*}\right\}.$ Then, using (C1), (45) and (46), we obtain

$$\parallel F^{\prime }{\left(x_{*}\right)}^{-1}(F^{\prime }\left(z\right)-F^{\prime }\left(x_{*}\right))\parallel \le {\ell }_0\parallel z-x_{*}\parallel \le {\ell }_0{\rho }_{*}<1,$$

so $F^{\prime }\left(z\right)$ is invertible with

$$\parallel F^{\prime }{\left(z\right)}^{-1}{F}^{\prime }\left(x_{*}\right)\parallel \le \frac{1}{1-{\ell }_0\parallel z-x_{*}\parallel },$$

(52)

and iterate $y_0$ exists by (52) for $z=x_0.$ Then, we can write

$$y_0-x_{*}={\int }_0^1{F}^{\prime }{\left(x_0\right)}^{-1}(F^{\prime }(x_{*}+\theta (x_0-x_{*}))-F^{\prime }\left(x_0\right))d\theta (x_0-x_{*}),$$

so by (C1), (C2) and (52) (for $z=x_0$), we get

$$\begin{array}{ccc}\hfill \parallel y_0-x_{*}\parallel & \le & \frac{\ell \parallel x_0-x_{*}{\parallel }^2}{2(1-{\ell }_0||x_0-x_{*}\parallel )}\hfill \\ \hfill & \le & {\psi }_1(\parallel x_0-x_{*}\parallel )\parallel x_0-x_{*}\parallel \hfill \\ \hfill & \le & \parallel x_0-x_{*}\parallel <{\rho }_{*},\hfill \end{array}$$

(53)

so $y_0\in U(x_{*},r)$ and (50) hold for $n=0.$ As in (52), we also show

$$\parallel F^{\prime }{\left(y_0\right)}^{-1}{F}^{\prime }\left(x_{*}\right)\parallel \le \frac{1}{1-{\ell }_0\parallel y_0-x_{*}\parallel }$$

(54)

and

$$\parallel {[y_0,x_0;F]}^{-1}{F}^{\prime }\left(x_{*}\right)\parallel \le \frac{1}{1-{\ell }_4(\parallel y_0-x_{*}\parallel +\parallel x_0-x_{*}\parallel )},$$

(55)

so iterate $x_1$ exists. Then, we can write in turn by the second substep of method (2) that

$$\begin{array}{ccc}\hfill x_1-x_{*}& =& y_0-x_{*}-F^{\prime }{\left(y_0\right)}^{-1}F\left(y_0\right)\hfill \\ \hfill & & +(F^{\prime }{\left(y_0\right)}^{-1}-{[y_0,x_0;F]}^{-1})F\left(y_0\right)\hfill \\ \hfill & & +({[y_0,x_0;F]}^{-1}-F^{\prime }{\left(x_0\right)}^{-1})F\left(y_0\right)\hfill \\ \hfill & =& y_0-x_{*}-F^{\prime }{\left(y_0\right)}^{-1}F\left(y_0\right)\hfill \\ \hfill & & +F^{\prime }{\left(y_0\right)}^{-1}([y_0,x_0;F]-F^{\prime }\left(y_0\right)){[y_0,x_0;F]}^{-1}F\left(y_0\right)\hfill \\ \hfill & & +{[y_0,x_0;F]}^{-1}(F^{\prime }\left(x_0\right)-[y_0,x_0;F])F^{\prime }{\left(x_0\right)}^{-1}F\left(y_0\right).\hfill \end{array}$$

(56)

Then, in view of (45), (49) (C2), (52) (for $z=y_0$) and (54)–(56), we get in turn that

$$\begin{array}{ccc}\hfill \parallel x_1-x_{*}\parallel & \le & \frac{\ell \parallel y_0-x_{*}{\parallel }^2}{2(1-{\ell }_0\parallel y_0-x_{*}\parallel )}\hfill \\ \hfill & & +\frac{{\ell }_3\parallel y_0-x_0\parallel {\ell }_4\parallel y_0-x_{*}\parallel }{(1-{\ell }_0\parallel y_0-x_{*}\parallel \left)\right(1-{\ell }_1(\parallel y_0-x_{*}\parallel +\parallel x_0-x_{*}\parallel \left)\right)}\hfill \\ \hfill & & +\frac{{\ell }_2\left|\right|y_0-x_0\parallel {\ell }_4\parallel y_0-x_{*}\parallel }{(1-{\ell }_1(\parallel y_0-x_{*}\parallel +\parallel x_0-x_{*}{\parallel \left)\right)}^2}\hfill \\ \hfill & \le & {\psi }_2(\parallel x_0-x_{*}\parallel )\parallel x_0-x_{*}\parallel \le \parallel x_0-x_{*}\parallel ,\hfill \end{array}$$

(57)

showing (51) for $n=0$ and $x_1\in U(x_{*},{\rho }_{*}),$ where we also used (53) and

$$\begin{array}{ccc}\hfill \parallel y_0-x_0\parallel & \le & \parallel y_0-x_{*}\parallel +\parallel x_{*}-x_0\parallel \hfill \\ & \le & {\psi }_1(\parallel x_0-x_{*}\parallel )\parallel x_0-x_{*}\parallel +\parallel x_0-x_{*}\parallel \hfill \\ & =& (1+{\psi }_1(\parallel x_0-x_{*}\parallel \left)\right)\parallel x_0-x_{*}\parallel .\hfill \end{array}$$

If we exchange $x_0,y_0,x_1$ by $x_m,y_m,x_{m+1},$ respectively, in the previous calculations we complete the induction for (50) and (51). Then, from the estimate

$$\parallel x_{m+1}-x_{*}\parallel \le {\lambda }_1\parallel x_m-x_{*}\parallel ,$$

(58)

where ${\lambda }_1={\psi }_2(\parallel x_0-x_{*}\parallel )\in [0,1),$ we conclude ${lim}_{m⟶+\infty }{x}_m=x_{*}.$ We also have

$$\parallel y_m-x_{*}\parallel \le {\lambda }_2\parallel x_m-x_{*}\parallel <{\rho }_{*},$$

(59)

where ${\lambda }_2={\psi }_1(\parallel x_0-x_{*}\parallel )\in [0,1),$ so ${lim}_{m⟶+\infty }{y}_m=x_{*}.$ □

Next, we present a uniqueness of the solution result.

Proposition 2.

Suppose:

(a) 

$x_{*}\in D$ is a simple solution of equation $F\left(x\right)=0.$

(b) 

There exists $\tilde{s}\ge 0$ such that

$$\frac{{\ell }_0}{2}\tilde{s}<1.$$

(60)

Set $D_4=D\cap U[x_{*},\tilde{s}].$ Then, the only solution of equation $F\left(x\right)=0$ in the region $D_4$ is $x_{*}.$

Proof. 

Let $x^{**}\in D_4$ with $F\left(x^{**}\right)=0.$ Set $Q={\int }_0^1{F}^{\prime }(x_{*}+\theta (x^{**}-x_{*}))d\theta .$ Then, using (C1) and (60), we obtain

$$\begin{array}{ccc}\hfill \parallel F^{\prime }{\left(x_{*}\right)}^{-1}(Q-F^{\prime }\left(x_{*}\right))\parallel & \le & \frac{{\ell }_0}{2}\parallel x^{**}-x_{*}\parallel \hfill \\ & \le & \frac{{\ell }_0}{2}\tilde{s}<1,\hfill \end{array}$$

so $x^{**}=x_{*},$ since $Q^{-1}\in L(E_1,E)$ and $Q(x^{**}-x_{*})=F\left(x^{**}\right)-F\left(x_{*}\right)=0-0=0.$ □

5. Numerical Experiments

We provide some examples, showing that the old convergence criteria are not verified, but ours are. The divided difference is chosen by

$$[u,v;F]={\int }_0^1{F}^{\prime }(v+\theta (u-v))d\theta .$$

Example 1.

Define function

$$h\left(t\right)=c_{0}t+c_1+c_{2}sin{c}_{3}t,t_0=0,$$

where $c_j,j=0,1,2,3$ are parameters. Then, clearly for $c_3$ large and $c_2$ small, $\frac{L_0}{L}$ can be small (arbitrarily).

Example 2.

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

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

(61)

$x\in E,s\in [0,1],w\in E$ is given, ϵ is a parameter and P is the Green’s kernel given by

$$P({ϵ}_2,{ϵ}_1)=\left\{\begin{array}{cc}(1-{ϵ}_2){ϵ}_1,\hfill & {ϵ}_1\le {ϵ}_2\hfill \\ {ϵ}_2(1-{ϵ}_1),\hfill & {ϵ}_2\le {ϵ}_1.\hfill \end{array}\right.$$

By (31), we have

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

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

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

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

and $L=\frac{6\eta \left|ϵ\right|}{8-3\left|ϵ\right|}.$

Example 3.

Let $E,E_1$ and D be as in the Example 5.3. It is well known that the boundary value problem [4]

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

$${\xi }^{″}=-\xi -\lambda {\xi }^2$$

can be given as a Hammerstein-like nonlinear integral equation

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

where λ is a parameter. Then, define $F:D⟶E_1$ by

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

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

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

and $\eta =\frac{1+\lambda }{5-2\lambda }.$ Notice that $L_0<L.$

Example 4.

Consider the motion system

$$T_1^{\prime }\left(x\right)=e^x,T_2^{\prime }\left(y\right)=(e-1)y+1,T_3^{\prime }\left(z\right)=1$$

with $T_1\left(0\right)=T_2\left(0\right)=T_3\left(0\right)=0.$ Let $T=(T_1,T_2,T_3).$ Let $E=E_1={\mathbb{R}}^3,D=B[0,1],x_{*}={(0,0,0)}^T.$ Define function T on D for $w={(x,y,z)}^T$ by

$$T\left(w\right)={(e^x-1,\frac{e-1}{2}{y}^2+y,z)}^T.$$

Then, we get

$$T^{\prime }\left(v\right)=\left[\begin{array}{ccc}{e}^x& 0& 0\\ 0& (e-1)y+1& 0\\ 0& 0& 1\end{array}\right],$$

so ${\ell }_0=e-1,\ell =e^{\frac{1}{e-1}}={\ell }_1,{\ell }_2={\ell }_3=\frac{\ell }{2},{\ell }_4=\frac{{\ell }_0}{2}.$ Then, the radii are:

$${\rho }_A=0.3827={\rho }^{*},{\rho }_{\mu }=1.7156.$$

6. Conclusions

A finer convergence analysis is presented for method (2) utilizing generalized conditions. This analysis includes weaker criteria of convergence and computable error bounds not given in earlier papers.