An Improved Convergence Analysis of a Multi-Step Method with High-Efficiency Indices

Abstract

A multi-step method introduced by Raziyeh and Masoud for solving nonlinear systems with convergence order five has been considered in this paper. The convergence of the method was studied using Taylor series expansion, which requires the function to be six times differentiable. However, our convergence study does not depend on the Taylor series. We use the derivative of F up to two only in our convergence analysis, which is presented in a more general Banach space setting. Semi-local analysis is also discussed, which was not given in earlier studies. Unlike in earlier studies (where two sets of assumptions were used), we used the same set of assumptions for semi-local analysis and local convergence analysis. We discussed the dynamics of the method and also gave some numerical examples to illustrate theoretical findings.

1. Introduction

Numerous problems in science and engineering can be effectively modeled using linear and nonlinear mathematical equations [1,2,3]. However, unlike linear equations, finding exact solutions to nonlinear equations is not feasible. Therefore, we have to rely on iterative methods to find approximate solutions to such equations. One of the most prominent iterative methods is Newton’s method [4], which has a quadratic convergence.

In recent years, considerable efforts have been made to improve the convergence properties of iterative methods, aiming for higher convergence rate and better computational efficiency [1,5,6,7,8,9]. Many of these methods are extensions or modifications of Newton’s method [10,11,12] and have been applied to both single-variable and multi-variable nonlinear equations.

The order of convergence of an iterative method is a key measure of its efficiency, which indicates how rapidly the method converges to the solution. We say that a sequence $\left\{s_n\right\}$ converges to $s^{*}$ with the order of convergence at least $p>0$ [13,14,15] if there exists a constant $Q>0$ such that

$$\parallel s_{n+1}-s^{*}\parallel \le Q\parallel s_n-s^{*}{\parallel }^p.$$

(1)

The efficiency index (EI) and informational efficiency (IE) are two metrics to evaluate and compare the performance of iterative methods. Ostrowski [16] has introduced EI as

$$\begin{array}{c}\hfill EI=p^{{\displaystyle \frac{1}{\eta }}},\end{array}$$

and Traub [14] has introduced IE as

$$\begin{array}{c}\hfill IE={\displaystyle \frac{p}{\eta }},\end{array}$$

where p is the order of convergence and $\eta$ is the total number of functions and derivative evaluations per iteration.

We consider that the primary goal of the iterative method is to improve the convergence speed while enhancing the accuracy and overall efficiency of the method.

In this paper, we focus on the convergence analysis of iterative methods for solving equations of the form

$$\begin{array}{c}\hfill \AE (s)=0,\end{array}$$

(2)

where $\AE :\mathrm{\Omega }\subset X\to Y$ is a nonlinear operator from Banach space X into Banach space Y, and $\mathrm{\Omega }$ is a non-empty open convex set.

The multi-step method for solving nonlinear systems given by Raziyeh and Masoud [17] is defined for $s_0\in \mathrm{\Omega }$ and $n=0,1,2,\dots$ such that

$$\begin{array}{ccc}\hfill t_n& =& s_n-{\text{Æ}}^{\prime }{(s_n)}^{-1}\text{Æ}(s_n)\hfill \\ \hfill s_{n+1}& =& t_n-2{\text{Æ}}^{\prime }{(s_n)}^{-1}\left[\text{Æ}(v_n)+{\displaystyle \frac{3}{2}}\text{Æ}(z_n)-{\displaystyle \frac{3}{2}}\text{Æ}(t_n)\right],\hfill \end{array}$$

(3)

where $v_n=t_n+{\text{Æ}}^{\prime }{(s_n)}^{-1}\text{Æ}(t_n)\mathrm{and}{z}_n=t_n-{\text{Æ}}^{\prime }{(s_n)}^{-1}\text{Æ}(t_n).$

The iterative method (3) has a convergence order 5, which is studied in [17] using Taylor series expansion, and requires the assumption that the operator F is at least six times differentiable. These restrictions are the motivations for our study. It is shown in [17] that the method (3) is highly efficient and superior to the earlier methods. A comparison with other methods in terms of efficiency has been given in [17]. If we follow the analysis made in [17], then the method (3) cannot be used to approximate the solution of (2) if F cannot be differentiated six times. For example, consider $q:[-2,2]\to \mathbb{R}$ defined as

$$q(s)=\left\{\begin{array}{c}a{s}^{6}log{s}^2+b{s}^7+c{s}^8\mathrm{if} s\ne 0\hfill \\ 0 \mathrm{if} s=0,\hfill \end{array}\right.$$

where $a\ne 0$ and $a,b,c$ are real parameters. Notice that $s^{*}=1\in [-2,2]$ solves the equation $q(s)=0.$ Further, notice the unboundedness of $q^{(6)}$ on the interval $[-2,2]$ since $q^{(6)}(s)$ does not exists at $s=0.$ Thus, the method (3) cannot assure the convergence of $s_n$ to the solution $s^{*}=0$ if we use the analysis in [17]. However, method (3) does converge to $s^{*}$ if, e.g., $a=1$, $b=-2$, and $c=1$ and the initial guess $s_0=0.95.$ We have studied the convergence of the method (3) without using the Taylor series. Thus, our study relaxes the condition that F has to be six times differentiable and requires F to be just two times differentiable. The innovative aspects and advantages of our analysis are as follows:

• Our analysis is discussed in the Banach space setting.
• We have given the semi-local analysis in our studies, which was not given in earlier work [17].
• Earlier studies [17,18] rely on assumptions involving the solution for local convergence analysis. However, our assumptions for attaining the convergence order are independent of the solution.
• In the existing studies, the assumptions depend on the actual solution $x^{*}$, but our assumptions are independent of the solution $x^{*}$. Using the information about $x^{*}$ obtained from semi-local analysis, we study the convergence order using local convergence.

This paper is arranged as follows: Section 2 contains the semi-local convergence analysis of the method (3). Section 3 contains the local convergence of the method (3) without using the Taylor series expansion. Section 4 and Section 5 contain numerical examples and the basins of attraction of the method, respectively. The conclusion is given in Section 6.

2. Semi-Local Analysis

We will define scalar majorizing sequences for our semi-local analysis [4].

For ${\alpha }_0=0,{\beta }_0\ge 0$ and $L_0,L_1\ge 0,$ define the scalar sequences $\left\{{\alpha }_n\right\}$ and $\left\{{\beta }_n\right\}$ such that

$$\begin{array}{ccc}\hfill {\alpha }_{n+1}& =& {\beta }_n+{\displaystyle \frac{L_1}{1-L_0{\alpha }_n}}[{\displaystyle \frac{1}{1-L_0{\alpha }_n}}\left(2+L_0{\beta }_n+{\displaystyle \frac{L_0{L}_1{({\beta }_n-{\alpha }_n)}^2}{4(1-L_0{\alpha }_n)}}\right)\hfill \\ & & +{\displaystyle \frac{3}{2(1-L_0{\alpha }_n)}}\left(L_1({\beta }_n-{\alpha }_n)+{\displaystyle \frac{L_1^2{({\beta }_n-{\alpha }_n)}^2}{4(1-L_0{\alpha }_n)}}\right)+{\displaystyle \frac{3}{2}}]{({\beta }_n-{\alpha }_n)}^2,\hfill \\ \hfill {\delta }_{n+1}& =& {\displaystyle \frac{L_1}{2}}{({\alpha }_{n+1}-{\alpha }_n)}^2+(1+L_0{\alpha }_n)({\alpha }_{n+1}-{\beta }_n)\hfill \\ \hfill \mathrm{and}\\ \hfill {\beta }_{n+1}& =& {\alpha }_{n+1}+{\displaystyle \frac{{\delta }_{n+1}}{1-L_o{\alpha }_{n+1}}}.\hfill \end{array}$$

(4)

Lemma 1.

Assume there exists $\mu \ge 0$ such that

$$L_0{\alpha }_n<1 \mathrm{and} {\alpha }_n\le \mu \forall n\in \mathbb{N}\cup \left\{0\right\}.$$

(5)

Then, the sequences $\left\{{\alpha }_n\right\}$ and $\left\{{\beta }_n\right\}$ defined by (4) are convergent to some $\lambda \in [{\beta }_0,\mu ]$ and $0\le {\alpha }_n\le {\beta }_n\le {\alpha }_{n+1}\le \lambda .$

Proof. 

The scalar sequences $\left\{{\alpha }_n\right\}$ and $\left\{{\beta }_n\right\}$ are non-decreasing and bounded above $\mu$. Hence, $\exists \lambda \in [{\beta }_0,\mu ]$ such that $\left\{{\alpha }_n\right\}$ and $\left\{{\beta }_n\right\}$ converges to $\lambda .$ □

Let $U(s,r)$ be the ball centered at s with radius r and $\overline{U}(s,r)$ be its closure.

For the convergence analysis, we use the following assumptions.

$(a_1)$

∃ an initial point $s_0\in \mathrm{\Omega }$ such that $\parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(s_0)\parallel <{\beta }_0$.

$(a_2)$

There exist an operator $\mathcal{G}\in \mathcal{B}(X,Y)$ (the set of all bounded linear operators from X to Y) and constant $L_0>0$ with

$$\begin{array}{c}\hfill \parallel {\mathcal{G}}^{-1}({\text{Æ}}^{\prime }(s)-\mathcal{G})\parallel \le L_0\parallel s-s_0\parallel ,\forall s\in \mathrm{\Omega }.\end{array}$$

Set ${\mathrm{\Omega }}_1=\mathrm{\Omega }\cap U(s_0,{\displaystyle \frac{1}{L_0}})$.

$(a_3)$

There exists a constant $L_1>0$ with

$$\begin{array}{c}\hfill \parallel {\mathcal{G}}^{-1}({\text{Æ}}^{\prime }(s)-{\text{Æ}}^{\prime }(t))\parallel \le L_1\parallel s-t\parallel ,\forall s,t\in {\mathrm{\Omega }}_1.\end{array}$$

$(a_4)$

$\overline{U}(s_0,\lambda )\subset \mathrm{\Omega }$.

From assumption $(a_2)$, we have $\forall s\in {\mathrm{\Omega }}_1$

$$\begin{array}{ccc}\hfill \parallel {\mathcal{G}}^{-1}({\text{Æ}}^{\prime }(s)-\mathcal{G})\parallel & \le & L_0\parallel s-s_0\parallel <1\hfill \end{array}$$

(6)

and hence by Banach Lemma (BL) on invertible operators [4], we get ${\text{Æ}}^{\prime }{(s)}^{-1}\in \mathcal{B}(Y,X)$ and

$$\begin{array}{c}\hfill \parallel {\text{Æ}}^{\prime }{(s)}^{-1}\mathcal{G}\parallel \le {\displaystyle \frac{1}{1-L_0\parallel s-s_0\parallel }},\forall s\in {\mathrm{\Omega }}_1.\end{array}$$

(7)

Further, we will be using the following mean value theorem (MVT) [4]:

$$\begin{array}{c}\hfill \text{Æ}({\nu }_1)-\text{Æ}({\nu }_2)={\int }_0^1{\text{Æ}}^{\prime }({\nu }_2+\tau ({\nu }_1-{\nu }_2))d\tau ({\nu }_1-{\nu }_2),\forall {\nu }_1,{\nu }_2\in \mathrm{\Omega }.\end{array}$$

(8)

Next, the main semi-local result uses the conditions $(a_1)$ – $(a_4)$.

Theorem 1.

Under the assumptions $(a_1)-(a_4)$, the sequence $\left\{s_n\right\}$ defined by method (3) with $s_0$ satisfies

$$\begin{array}{c}\hfill \parallel t_n-s_n\parallel \le {\beta }_n-{\alpha }_n\end{array}$$

(9)

and

$$\begin{array}{c}\hfill \parallel s_{n+1}-t_n\parallel \le {\alpha }_{n+1}-{\beta }_n.\end{array}$$

(10)

Moreover, the sequence $s_n$ converges to some $s^{*}\in \overline{U}(s_0,\lambda )$ with $\text{Æ}(s^{*})=0.$

Proof. 

We will be using mathematical induction to prove the result.

Using $(a_1)$ and first step of (3),

$$\begin{array}{c}\hfill \parallel t_0-s_0\parallel =\parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(s_0)\parallel <{\beta }_0={\beta }_0-{\alpha }_0\le \lambda .\end{array}$$

Thus, $t_0\in U(s_0,\lambda )$ and (9) is true for $n=0.$

Using MVT, (8) and the first step of (3), we have

$$\begin{array}{ccc}\hfill \parallel \text{Æ}{(s_0)}^{-1}\text{Æ}(t_0)\parallel & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\left[\text{Æ}(t_0)-\text{Æ}(s_0)+\text{Æ}(s_0)\right]\parallel \hfill \\ & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\left[{\int }_0^1{\text{Æ}}^{\prime }(s_0+\tau (t_0-s_0))d\tau (t_0-s_0)+\text{Æ}(s_0)\right]\parallel \hfill \\ & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\left[{\int }_0^1{\text{Æ}}^{\prime }(s_0+\tau (t_0-s_0))d\tau (t_0-s_0)-{\text{Æ}}^{\prime }(s_0)(t_0-s_0)\right]\parallel \hfill \\ & \le & {\int }_0^1\parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\left[{\text{Æ}}^{\prime }(s_0+\tau (t_0-s_0))-{\text{Æ}}^{\prime }(s_0)\right]\parallel d\tau \parallel t_0-s_0\parallel .\hfill \end{array}$$

Using the assumption $(a_3)$ and (7), we get

$$\begin{array}{ccc}\hfill \parallel \text{Æ}{(s_0)}^{-1}\text{Æ}(t_0)\parallel & \le & \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel {\int }_0^1\parallel {\mathcal{G}}^{-1}\left[{\text{Æ}}^{\prime }(s_0+\tau (t_0-s_0))-{\text{Æ}}^{\prime }(s_0)\right]\parallel d\tau \parallel t_0-s_0\parallel \hfill \\ & \le & {\displaystyle \frac{L_1}{2(1-L_0\parallel s_0-s_0\parallel )}}\parallel t_0-s_0{\parallel }^2={\displaystyle \frac{L_1}{2(1-L_0{\alpha }_0)}}{\parallel t_0-s_0\parallel }^2.\hfill \end{array}$$

(11)

Similarly,

$$\begin{array}{ccc}\hfill \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(v_0)\parallel & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}[\text{Æ}(v_0)-\text{Æ}(t_0)+\text{Æ}(t_0)]\parallel \hfill \\ & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\left[{\int }_0^1{\text{Æ}}^{\prime }(t_0+\tau (v_0-t_0))d\tau (v_0-t_0)+\text{Æ}(t_0)\right]\parallel \hfill \\ & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\left[{\int }_0^1{\text{Æ}}^{\prime }(t_0+\tau (v_0-t_0))d\tau (v_0-t_0)+{\text{Æ}}^{\prime }(s_0)(v_0-t_0)\right]\parallel \hfill \\ & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\left[{\int }_0^1{\text{Æ}}^{\prime }(t_0+\tau (v_0-t_0))d\tau +{\text{Æ}}^{\prime }(s_0)\right](v_0-t_0)\parallel \hfill \\ & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\left[{\int }_0^1{\text{Æ}}^{\prime }(t_0+\tau (v_0-t_0))d\tau +{\text{Æ}}^{\prime }(s_0)\right]{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)\parallel ,\hfill \end{array}$$

So, by using assumption ($a_2$), (7) and (11), we have

$$\begin{array}{ccc}\hfill \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(v_0)\parallel & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel [{\int }_0^1\parallel {\mathcal{G}}^{-1}({\text{Æ}}^{\prime }(t_0+\tau (v_0-t_0))-\mathcal{G}+\mathcal{G})\parallel d\tau \hfill \\ & & +\parallel {\mathcal{G}}^{-1}({\text{Æ}}^{\prime }(s_0)-\mathcal{G}+\mathcal{G})\parallel ]\parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)\parallel \hfill \\ & \le & {\displaystyle \frac{1}{(1-L_0{\alpha }_0)}}[1+L_0(\parallel t_0-s_0\parallel +{\displaystyle \frac{1}{2}}\parallel v_0-t_0\parallel )\hfill \\ & & +L_0\parallel s_0-s_0\parallel +1]{\displaystyle \frac{L_1}{2(1-L_0{\alpha }_0)}}{\parallel t_0-s_0\parallel }^2\hfill \\ & \le & {\displaystyle \frac{1}{(1-L_0{\alpha }_0)}}[2+L_0\parallel t_0-s_0\parallel +{\displaystyle \frac{L_0{L}_1}{4(1-L_0{\alpha }_0)}}{\parallel t_0-s_0\parallel }^2\hfill \\ & & +L_0{\alpha }_0]{\displaystyle \frac{L_1}{2(1-L_0{\alpha }_0)}}{\parallel t_0-s_0\parallel }^2,\hfill \end{array}$$

(12)

where we used $\parallel v_0-t_0\parallel ={\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0).$

Similarly using assumption $(a_3)$, (7) and (11) we get

$$\begin{array}{ccc}\hfill \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(z_0)\parallel & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}[\text{Æ}(z_0)-\text{Æ}(t_0)+\text{Æ}(t_0)]\parallel \hfill \\ & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\left[{\int }_0^1{\text{Æ}}^{\prime }(t_0+\tau (z_0-t_0))d\tau (z_0-t_0)+\text{Æ}(t_0)\right]\parallel \hfill \\ & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\left[{\int }_0^1{\text{Æ}}^{\prime }(t_0+\tau (z_0-t_0))d\tau -{\text{Æ}}^{\prime }(s_0)\right](z_0-t_0)\parallel \hfill \\ & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel \left[{\int }_0^1\parallel {\mathcal{G}}^{-1}({\text{Æ}}^{\prime }(t_0+\tau (z_0-t_0))-{\text{Æ}}^{\prime }(s_0))\right]\parallel d\tau \hfill \\ & & \times \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)\parallel \hfill \\ & \le & {\displaystyle \frac{1}{(1-L_0{\alpha }_0)}}\left[L_1\left(\parallel t_0-s_0\parallel +{\displaystyle \frac{L_1}{4(1-L_0{\alpha }_0)}}{\parallel t_0-s_0\parallel }^2\right)\right]\hfill \\ & & \times {\displaystyle \frac{L_1}{2(1-L_0{\alpha }_0)}}{\parallel t_0-s_0\parallel }^2.\hfill \end{array}$$

(13)

Next, using (11)–(13) in (3), we get

$$\begin{array}{ccc}\hfill \parallel s_1-t_0\parallel & =& 2\parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(v_0)\parallel +3\parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(z_0)+3\parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)\parallel \hfill \\ & \le & {\displaystyle \frac{2}{(1-L_0{\alpha }_0)}}[2+L_0\parallel t_0-s_0\parallel +{\displaystyle \frac{L_0{L}_1}{4(1-L_o{\alpha }_0)}}{\parallel t_0-s_0\parallel }^2\hfill \\ & & +L_0{\alpha }_0]{\displaystyle \frac{L_1}{2(1-L_0{\alpha }_0)}}\parallel t_0-s_0{\parallel }^2+{\displaystyle \frac{3}{(1-L_0{\alpha }_0)}}[L_1(\parallel t_0-s_0\parallel \hfill \\ & & +{\displaystyle \frac{L_1}{4(1-L_0{\alpha }_0)}}\parallel t_0-s_0{\parallel }^2)]{\displaystyle \frac{L_1}{2(1-L_0{\alpha }_0)}}{\parallel t_0-s_0\parallel }^2\hfill \\ & & +{\displaystyle \frac{3L_1}{2(1-L_0{\alpha }_0)}}{\parallel t_0-s_0\parallel }^2\hfill \\ & \le & {\displaystyle \frac{L_1}{(1-L_0{\alpha }_0)}}[{\displaystyle \frac{1}{1-L_0{\alpha }_0}}(2+L_0\parallel t_0-s_0\parallel \hfill \\ & & +{\displaystyle \frac{L_0{L}_1}{4(1-L_0{\alpha }_0)}}{\parallel t_0-s_0\parallel }^2+L_0{\alpha }_0)\hfill \\ & & +{\displaystyle \frac{3L_1}{2(1-L_0{\alpha }_0}}\left(\parallel t_0-s_0\parallel +{\displaystyle \frac{L_1}{4(1-L_0{\alpha }_0)}}{\parallel t_0-s_0\parallel }^2\right)+{\displaystyle \frac{3}{2}}]{\parallel t_0-s_0\parallel }^2\hfill \\ & \le & {\displaystyle \frac{L_1}{(1-L_0{\alpha }_0)}}[{\displaystyle \frac{1}{1-L_0{\alpha }_0}}(2+L_0({\beta }_0-{\alpha }_0)\hfill \\ & & +{\displaystyle \frac{L_0{L}_1}{4(1-L_0{\alpha }_0)}}{({\beta }_0-{\alpha }_0)}^2+L_0{\alpha }_0)+{\displaystyle \frac{3L_1}{2(1-L_0{\alpha }_0)}}(({\beta }_0-{\alpha }_0)\hfill \\ & & +{\displaystyle \frac{L_1}{4(1-L_0{\alpha }_0)}}{({\beta }_0-{\alpha }_0)}^2)+{\displaystyle \frac{3}{2}}]{({\beta }_0-{\alpha }_0)}^2\hfill \\ & =& {\alpha }_1-{\beta }_0.\hfill \end{array}$$

(14)

Note that $\parallel s_1-s_0\parallel \le \parallel s_1-t_0\parallel +\parallel t_0-s_0\parallel \le {\alpha }_1-{\beta }_0+{\beta }_0-{\alpha }_0\le \lambda .$ Therefore, $s_1\in U(s_0,\lambda )$ and (10) is true for $n=0.$

Assume that (9) and (10) are true for all $n=0,1,\dots ,j.$ This implies that $\parallel s_n-t_{n-1}\parallel \le {\alpha }_n-{\beta }_{n-1}$ and $\parallel t_n-s_n\parallel \le {\beta }_n-{\alpha }_n$ for all $n=1,2,\dots ,j.$

To show that (9) is true for all $n=1,2,...,$ we consider $\text{Æ}(s_j),j\ne 0,$ and using the first step of (3) and MVT, we have

$$\begin{array}{ccc}\hfill \text{Æ}(s_j)& =& \text{Æ}(s_j)-\text{Æ}(s_{j-1})+\text{Æ}(s_{j-1})+{\text{Æ}}^{\prime }(s_{j-1})(s_j-s_{j-1})-{\text{Æ}}^{\prime }(s_{j-1})(s_j-s_{j-1})\hfill \\ & =& {\int }_0^1{\text{Æ}}^{\prime }\left(s_{j-1}+\theta (s_j-s_{j-1})\right)d\theta (s_j-s_{j-1})+{\text{Æ}}^{\prime }(s_{j-1})(s_{j-1}-t_{j-1})\hfill \\ & & +{\text{Æ}}^{\prime }(s_{j-1})(s_j-s_{j-1})-{\text{Æ}}^{\prime }(s_{j-1})(s_j-s_{j-1})\hfill \\ & =& {\int }_0^1{\text{Æ}}^{\prime }\left(s_{j-1}+\theta (s_j-s_{j-1})\right)d\theta (s_j-s_{j-1})-{\text{Æ}}^{\prime }(s_{j-1})(s_j-s_{j-1})\hfill \\ & & +{\text{Æ}}^{\prime }(s_{j-1})(s_j-t_{j-1})\hfill \\ & =& {\int }_0^1[{\text{Æ}}^{\prime }\left(s_{j-1}+\theta (s_j-s_{j-1})-{\text{Æ}}^{\prime }(s_{j-1})\right]d\theta (s_j-s_{j-1})\hfill \\ & & +{\text{Æ}}^{\prime }(s_{j-1})(s_j-t_{j-1}).\hfill \end{array}$$

(15)

Using assumptions $(a_2)-(a_3)$ in the above equation, we get

$$\begin{array}{ccc}\hfill \parallel {\mathcal{G}}^{-1}\text{Æ}(s_j)\parallel & \le & {\int }_0^1\parallel {\mathcal{G}}^{-1}\left[{\text{Æ}}^{\prime }{s}_{j-1}+\theta (s_j-s_{j-1}))-{\text{Æ}}^{\prime }(s_{j-1})\right]\parallel d\theta \parallel s_j-s_{j-1}\parallel \hfill \\ & & +\parallel {\mathcal{G}}^{-1}{\text{Æ}}^{\prime }(s_{j-1})\parallel \parallel s_j-t_{j-1}\parallel \hfill \\ & \le & {\displaystyle \frac{L_1}{2}}\parallel s_j-s_{j-1}{\parallel }^2+\parallel {\mathcal{G}}^{-1}({\text{Æ}}^{\prime }(s_{j-1})-\mathcal{G}+\mathcal{G})\parallel \parallel s_j-t_{j-1}\parallel \hfill \\ & \le & {\displaystyle \frac{L_1}{2}}\parallel s_j-s_{j-1}{\parallel }^2+(1+L_0\parallel s_{j-1}-s_0\parallel )\parallel s_j-t_{j-1}\parallel \hfill \\ & \le & {\displaystyle \frac{L_1}{2}}{({\alpha }_j-{\alpha }_{j-1})}^2+(1+L_0{\alpha }_{j-1})({\alpha }_j-{\beta }_{j-1})\hfill \\ & \le & {\delta }_j.\hfill \end{array}$$

(16)

Therefore, we have

$$\begin{array}{ccc}\hfill \parallel t_j-s_j\parallel & \le & \parallel {\text{Æ}}^{\prime }{(s_j)}^{-1}\text{Æ}(s_j)\parallel \hfill \\ & \le & \parallel {\text{Æ}}^{\prime }{(s_j)}^{-1}\mathcal{G}\parallel \parallel {\mathcal{G}}^{-1}\text{Æ}(s_j)\parallel \hfill \\ & \le & {\displaystyle \frac{{\delta }_j}{1-L_0\parallel s_j-s_0\parallel }}\hfill \\ & \le & {\beta }_j-{\alpha }_j\hfill \end{array}$$

(17)

and

$$\begin{array}{ccc}\hfill \parallel t_j-s_0\parallel & \le & \parallel t_j-s_j\parallel +\parallel s_j-s_0\parallel \hfill \\ & \le & {\beta }_j-{\alpha }_j+{\alpha }_j-{\alpha }_0\hfill \\ & =& {\beta }_j<\lambda .\hfill \end{array}$$

(18)

So, $t_j\in U(s_0,\lambda ),$ and the inequality (9) holds for all $j.$

The proof is completed by replacing $s_0$, $t_0$ and $s_1$ with $s_j$, $t_j$ and $s_{j+1}$, respectively, in the above argument.

Since $\left\{{\alpha }_n\right\}$ and $\left\{{\beta }_n\right\}$ are Cauchy sequences, $\left\{s_n\right\}$ and $\left\{t_n\right\}$ are also Cauchy by (9) and (10). Hence, we have $s_n\to s^{*}\in \overline{U}(s_0,\lambda )$ as $n\to \infty .$

Now, by (3) we get

$$\begin{array}{ccc}\hfill \parallel \text{Æ}(s_n)\parallel & =& \parallel {\text{Æ}}^{\prime }(s_n)(t_n-s_n)\parallel \le M({\beta }_n-{\alpha }_n),\hfill \end{array}$$

(19)

where $\parallel {\text{Æ}}^{\prime }(s_n)\parallel \le M$ for some $M>0$. By letting $n\to \infty$ in (19), we have $\text{Æ}(s^{*})=0.$

□

Next, we study the uniqueness of the solution.

Proposition 1.

Suppose $d\in U(s_0,a_0)$ is a simple solution of the Equation (2) for some $a_0>0$ and $a\ge a_0$ such that

$$L_0(a_0+a)<2.$$

Set ${\mathrm{\Omega }}_1=\mathrm{\Omega }\cap \overline{U}(s_0,a)$. Then, d is unique in the region ${\mathrm{\Omega }}_1.$

Proof. 

We can see the proof of the proposition from [19]. □

3. Local Convergence Analysis

We will be using the following extra assumptions in our local analysis.

$(a_5)$

The condition (5) of the lemma holds for $\mu ={\displaystyle \frac{1}{2L_0}}$.

($a_6$)

$\parallel {\mathcal{G}}^{-1}({\text{Æ}}^{″}(s)-{\text{Æ}}^{″}(t))\parallel \le L_2\parallel s-t\parallel$ for some $L_2>0$ and $s,t\in {\mathrm{\Omega }}_1.$

($a_7$)

$\parallel {\mathcal{G}}^{-1}{\text{Æ}}^{\prime }(s)\parallel \le K_1$ for some $K_1>0$ and $s\in {\mathrm{\Omega }}_1.$

($a_8$)

$\parallel {\mathcal{G}}^{-1}{\text{Æ}}^{″}(s)\parallel \le K_2$ for some $K_2>0$ and $s\in {\mathrm{\Omega }}_1.$

We obtained from our semi-local analysis that the solution $s^{*}\in \overline{U}(s_0,\lambda )\subset U(s_0,{\displaystyle \frac{1}{2L_0}})$. Then, by $(a_2)$, for all $s\in U(s_0,{\displaystyle \frac{1}{2L_0}}),$ we can get

$$\begin{array}{ccc}\hfill \parallel {\mathcal{G}}^{-1}({\text{Æ}}^{\prime }(s)-\mathcal{G})\parallel & \le & L_0\parallel s-s_0\parallel \le L_0(\parallel s-s^{*}\parallel +\parallel s^{*}-s_0\parallel )\hfill \\ & \le & L_0(\parallel s-s^{*}\parallel +{\displaystyle \frac{1}{2L_0}})\hfill \\ & =& {\displaystyle \frac{1}{2}}+L_0\parallel s-s^{*}\parallel .\hfill \end{array}$$

(20)

Now, using BL, ${\text{Æ}}^{\prime }(s)$ is invertible for all $s\in U(s_0,{\displaystyle \frac{1}{2L_0}}),$ and

$$\begin{array}{c}\hfill \parallel {\text{Æ}}^{\prime }{(s)}^{-1}\mathcal{G}\parallel \le {\displaystyle \frac{2}{1-2L_0\parallel s-s^{*}\parallel }}.\end{array}$$

(21)

We will use the following inequality in our study. For all $s,t\in {\mathrm{\Omega }}_1,$ by MVT, we get

$$\begin{array}{ccc}\hfill \parallel {\text{Æ}}^{\prime }{(s)}^{-1}\text{Æ}(t)\parallel & =& {\text{Æ}}^{\prime }{(s)}^{-1}(\text{Æ}(t)-\text{Æ}(s^{*}))\hfill \\ & \le & \parallel {\text{Æ}}^{\prime }{(s)}^{-1}{\int }_0^1{\text{Æ}}^{\prime }(s^{*}+\tau (t-s^{*}))d\tau (t-s^{*})\parallel \hfill \\ & \le & \parallel {\text{Æ}}^{\prime }{(s)}^{-1}\mathcal{G}\parallel {\int }_0^1\parallel {\mathcal{G}}^{-1}{\text{Æ}}^{\prime }(s^{*}+\tau (t-s^{*}))\parallel d\tau \parallel t-s^{*}\parallel .\hfill \end{array}$$

Moreover, using assumption $(a_7)$ and (21), for $s,t\in U(s^{*},{\displaystyle \frac{1}{2L_0}}),$ we obtain

$$\begin{array}{c}\hfill \parallel {\text{Æ}}^{\prime }{(s)}^{-1}\text{Æ}(t)\parallel \le {\displaystyle \frac{2K_1}{1-2L_0\parallel s-s^{*}\parallel }}\parallel t-s^{*}\parallel .\end{array}$$

(22)

Remark 1.

We study the local convergence in the ball $U(s^{*},r),$ which satisfies

$$\begin{array}{c}\hfill U(s^{*},r)\subset U(s^{*},{\displaystyle \frac{1}{2L_0}})\subset {\mathrm{\Omega }}_1.\end{array}$$

(23)

Hence, hereafter we select $s_0$ from $U(s^{*},r).$

Graphical representation of (23) is given in Figure 1.

As a consequence of (23), all the assumptions we have made remain valid in the local convergence ball. So, we can continue our local analysis independently under the same set of assumptions.

We need the following theorem for our local convergence analysis.

Theorem 2

([20]). Let $F:\mathrm{\Omega }\subset X\to Y$ be twice differentiable at the point $a,$ then

$$\begin{array}{c}\hfill ({\text{Æ}}^{″}(a)·h)·k=({\text{Æ}}^{″}(a)·k)·h,\forall h,k\in \mathrm{\Omega }.\end{array}$$

Proposition 2.

If $s_0\in U(s^{*},\tilde{r})$, where $\tilde{r}=min\{{\displaystyle \frac{1}{L_1+2L_0}},{\rho }_1\}$, ${\rho }_1$ is the smallest zero of $h_1$ on $[0,{\displaystyle \frac{1}{2L_0}})$,

$$\begin{array}{ccc}\hfill {\varphi }_1(u)& =& {\displaystyle \frac{L_{1}u}{(1-2L_{0}u)}}\left(1+{\displaystyle \frac{2K_1}{1-2L_{0}u}}\right)\hfill \\ \hfill \mathit{and}\\ \hfill h_1(u)& =& {\varphi }_1(u)u-1.\hfill \end{array}$$

Then, under assumptions $(a_1)-(a_3)$, we have $t_0,v_0\in U(s,\tilde{r})$ with

$$\begin{array}{ccc}\hfill \parallel t_0-s^{*}\parallel & \le & {\displaystyle \frac{L_1}{(1-2L_0\parallel s_0-s^{*}\parallel )}}{\parallel s_0-s^{*}\parallel }^2\hfill \\ \hfill \mathit{and}\\ \hfill \parallel v_0-s^{*}\parallel & \le & {\varphi }_1(\parallel s_0-s^{*}\parallel )\parallel s_0-s^{*}{\parallel }^2.\hfill \end{array}$$

Proof. 

Note that ${\varphi }_1,h_1$ are non-decreasing continuous functions (NDCF) on $[0,{\displaystyle \frac{1}{2L_0}})$, with

$$h_1(0)=-1 \mathrm{and}\underset{u\to {{\displaystyle \frac{1}{2L_0}}}^{-}}{lim}{h}_1(u)=+\infty .$$

So, by Intermediate Value Theorem (IVT), there exists a smallest ${\rho }_1\in (0,{\displaystyle \frac{1}{2L_0}})$ such that $h_1({\rho }_1)=0$.

Note that

$$\begin{array}{ccc}\hfill t_0-s^{*}& =& s_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(s_0)\hfill \\ & =& {\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1({\text{Æ}}^{\prime }(s_0)-{\text{Æ}}^{\prime }(s^{*}+\tau (s_0-s^{*})))d\tau (s_0-s^{*}).\hfill \end{array}$$

So, by the assumption $(a_3)$ and (21), we obtain

$$\begin{array}{ccc}\hfill \parallel t_0-s^{*}\parallel & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel {\int }_0^1\parallel {\mathcal{G}}^{-1}({\text{Æ}}^{\prime }(s_0)-{\text{Æ}}^{\prime }(s^{*}+\tau (s_0-s^{*})))\parallel d\tau \parallel s_0-s^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{L_1{\parallel s_0-s^{*}\parallel }^2}{(1-2L_0\parallel s_0-s^{*}\parallel )}}<\parallel s_0-s^{*}\parallel .\hfill \end{array}$$

Thus, the iterate $t_0\in U(s^{*},\tilde{r})$.

Also, by (22) and the fact that $\varphi (t)t<1,\forall t\in (0,{\rho }_1),$ we have

$$\begin{array}{ccc}\hfill v_0-s^{*}& =& t_0-s^{*}+{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)\hfill \\ \hfill \parallel v_0-s^{*}\parallel & \le & {\displaystyle \frac{L_1}{(1-2L_0\parallel s_0-s^{*}\parallel )}}{\parallel s_0-s^{*}\parallel }^2\hfill \\ & & +{\displaystyle \frac{2K_1}{1-2L_0\parallel s_0-s^{*}\parallel }}{\displaystyle \frac{L_1}{(1-2L_0\parallel s_0-s^{*}\parallel )}}{\parallel s_0-s^{*}\parallel }^2\hfill \\ & \le & {\displaystyle \frac{L_1}{(1-2L_0\parallel s_0-s^{*}\parallel )}}\left(1+{\displaystyle \frac{2K_1}{1-2L_0\parallel s_0-s^{*}\parallel }}\right){\parallel s_0-s^{*}\parallel }^2\hfill \\ & =& {\varphi }_1(\parallel s_0-s^{*}\parallel )\parallel s_0-s^{*}{\parallel }^2<\parallel s_0-s^{*}\parallel .\hfill \end{array}$$

Hence, the iterate $v_0\in U(s^{*},\tilde{r})$. □

Proposition 3.

If $s_0\in U(s^{*},\tilde{\tilde{r}})$, where $\tilde{\tilde{r}}=min\{\tilde{r},{\rho }_2\}$, ${\rho }_2$ is the smallest zero of $h_2$ on $[0,{\displaystyle \frac{1}{2L_0}})$,

$$\begin{array}{ccc}\hfill {\varphi }_2(u)& =& {\displaystyle \frac{2L_1^2}{{(1-2L_{0}u)}^2}}\left(1+{\displaystyle \frac{L_{1}u}{2(1-2L_{1}u)}}\right)\hfill \\ \hfill \mathrm{and}\\ \hfill h_2(u)& =& {\varphi }_2(u)u^2-1.\hfill \end{array}$$

Then, under assumptions $(a_1)-(a_3),$ we have $z_0\in U(s,\tilde{\tilde{r}})$ with

$$\begin{array}{ccc}\hfill \parallel z_0-s^{*}\parallel & =& {\varphi }_2(\parallel s_0-s^{*}\parallel )\parallel s_0-s^{*}{\parallel }^3.\hfill \end{array}$$

Proof. 

Note that ${\varphi }_2,h_2$ are NDCF on $[0,{\displaystyle \frac{1}{2L_0}})$, with

$$h_2(0)=-1\mathrm{and}\underset{u\to {{\displaystyle \frac{1}{2L_0}}}^{-}}{lim}{h}_2(u)=+\infty .$$

So, by IVT, there exists a smallest ${\rho }_2\in (0,{\displaystyle \frac{1}{2L_0}})$ such that $h_2({\rho }_2)=0$.

Using the assumption $(a_3)$ and (21), we obtain

$$\begin{array}{ccc}\hfill z_0-s^{*}& =& t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)\hfill \\ & =& {\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\text{Æ}}^{\prime }(s_0)-{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))d\tau (t_0-s^{*})\hfill \\ \hfill \parallel z_0-s^{*}\parallel & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel {\int }_0^1\parallel {\mathcal{G}}^{-1}({\text{Æ}}^{\prime }(s_0)-{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*})))\parallel d\tau \parallel t_0-s^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{2L_1}{(1-2L_0\parallel s_0-s^{*}\parallel )}}\left(\parallel s_0-s^{*}\parallel +{\displaystyle \frac{1}{2}}\parallel t_0-s^{*}\parallel \right)\parallel t_0-s^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{2L_1^2}{(1-2L_0\parallel s_0-s^{*}{\parallel )}^2}}\left(1+{\displaystyle \frac{L_1}{2(1-2L_1\parallel s_0-s^{*}\parallel )}}\parallel s_0-s^{*}\parallel \right){\parallel s_0-s^{*}\parallel }^3\hfill \\ & =& {\varphi }_2(\parallel s_0-s^{*}\parallel )\parallel s_0-s^{*}{\parallel }^3<\parallel s_0-s^{*}\parallel ,\hfill \end{array}$$

and hence $z_0\in U(s^{*},\tilde{\tilde{r}}).$ □

For the next lemma, we introduced NDCF $\varphi ,h:[0,{\displaystyle \frac{1}{2L_0}})\to \mathbb{R}$ defined as

$$\begin{array}{ccc}\hfill \varphi (u)& =& {\displaystyle \frac{2L_1{L}_{2}u}{3{(1-2L_{0}u)}^2}}\left[{\varphi }_1^2(u)+2{\left({\displaystyle \frac{L_1}{1-2L_{0}u)}}\right)}^2+{\varphi }_2^2(u)u^2\right]\hfill \\ & & +{\displaystyle \frac{2K_1{L}_{1}u}{3{(1-2L_{0}u)}^3}}\left[{\varphi }_1^2(u)+{\varphi }_2^2(u)u^2\right]\hfill \\ & & +{\displaystyle \frac{L_1^2{L}_2}{{(1-2L_{0}u)}^3}}\left[{\varphi }_2(u)u^2+{\displaystyle \frac{1}{3}}{\varphi }_2^2(u)u^4+1\right]\left(2+{\displaystyle \frac{L_{1}u}{1-2L_{0}u}}\right)\hfill \\ & & +{\displaystyle \frac{K_2{L}_1^{2}u}{{(1-2L_{0}u)}^3}}{\varphi }_2(u)\left(2+{\displaystyle \frac{L_{1}u}{1-2L_{0}u}}\right)+{\displaystyle \frac{32K_1^2{K}_2{L}_1{L}_2}{{(1-2L_{0}u)}^5}}\hfill \\ & & +{\displaystyle \frac{2K_2{L}_1{L}_2}{{(1-2L_{0}u)}^3}}\left[{\displaystyle \frac{2L_{1}u}{1-2L_{0}u}}+{\displaystyle \frac{L_1}{1-2L_{0}u}}+1\right]\hfill \\ & & +{\displaystyle \frac{2K_2^2{L}_1^2}{{(1-2L_{0}u)}^4}}+4\left({\displaystyle \frac{K_1{K}_2{L}_1^2+K_2^3{L}_1}{{(1-2L_{0}u)}^4}}\right)\left(2+{\displaystyle \frac{L_{1}u}{(1-2L_{0}u)}}\right)\hfill \end{array}$$

and

$$h(u)=\varphi (u)u^4-1.$$

Notice that

$$h(0)=-1 \mathrm{and}\underset{u\to {\displaystyle \frac{1}{2L_0}}}{lim}h(u)=+\infty .$$

So, by IVT, there exists a smallest $\rho \in (0,{\displaystyle \frac{1}{2L_0}})$ such that $h(\rho )=0$.

• Let $r=min\{\tilde{\tilde{r}},\rho \}$.

Lemma 2.

If the assumptions $(a_2)-(a_8)$ hold and $s_0\in U(s^{*},r)-\left\{s^{*}\right\}$. Then, we have $s_1\in U(s^{*},r)$ and

$$\begin{array}{c}\hfill \parallel s_1-s^{*}\parallel \le \varphi (r)\parallel s_0-s^{*}{\parallel }^5.\end{array}$$

(24)

Proof. 

Let $s_0\in U(s^{*},r)$. Note that, by adding and subtracting ${\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0),$ we have by (3)

$$\begin{array}{ccc}\hfill s_1-s^{*}& =& t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)-{\text{Æ}}^{\prime }{(s_0)}^{-1}\left[2\text{Æ}(v_0)+3\text{Æ}(z_0)-4\text{Æ}(t_0)\right]\hfill \\ & =& t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)\hfill \\ & & -{\text{Æ}}^{\prime }(s_0)\left[2(\text{Æ}(v_0)-\text{Æ}(s^{*}))+3(\text{Æ}(z_0)-\text{Æ}(s^{*}))-4(\text{Æ}(t_0)-\text{Æ}(s^{*}))\right].\hfill \end{array}$$

So, by MVT, and the definition of $v_0$ and $z_0$, we have

$$\begin{array}{ccc}\hfill s_1-s^{*}& =& t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)\hfill \\ & & -{\text{Æ}}^{\prime }{(s_0)}^{-1}[2{\int }_0^1{\text{Æ}}^{\prime }(s^{*}+\tau (v_0-s^{*}))d\tau (t_0-s^{*}+{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))\hfill \\ & & +3{\int }_0^1{\text{Æ}}^{\prime }(s^{*}+\tau (z_0-s^{*}))d\tau (t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))\hfill \\ & & -4{\int }_0^1{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))d\tau (t_0-s^{*})].\hfill \end{array}$$

Then, by rearranging, we get

$$\begin{array}{ccc}\hfill s_1-s^{*}& =& t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)\hfill \\ & & -2{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1\left[{\text{Æ}}^{\prime }(s^{*}+\tau (v_0-s^{*}))-{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))\right]d\tau (t_0-s^{*})\hfill \\ & & -2{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1\left[{\text{Æ}}^{\prime }(s^{*}+\tau (z_0-s^{*}))-{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))\right]d\tau (t_0-s^{*})\hfill \\ & & -2{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1\left[{\text{Æ}}^{\prime }(s^{*}+\tau (v_0-s^{*}))-{\text{Æ}}^{\prime }(s^{*}+\tau (z_0-s^{*}))\right]d\tau ({\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))\hfill \\ & & -{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\text{Æ}}^{\prime }(s^{*}+\tau (z_0-s^{*}))d\tau (t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)).\hfill \end{array}$$

Combining the first and last terms, and adding and subtracting ${\text{Æ}}^{\prime }(s^{*})$ appropriately, we have

$$\begin{array}{ccc}\hfill s_1-s^{*}& =& \left[I-{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\text{Æ}}^{\prime }(s^{*}+\tau (z_0-s^{*}))d\tau \right](t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))\hfill \\ & & -2{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1[({\text{Æ}}^{\prime }(s^{*}+\tau (v_0-s^{*}))-{\text{Æ}}^{\prime }(s^{*}))\hfill \\ & & -({\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))-{\text{Æ}}^{\prime }(s^{*}))]d\tau (t_0-s^{*})\hfill \\ & & -2{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1[({\text{Æ}}^{\prime }(s^{*}+\tau (z_0-s^{*}))-{\text{Æ}}^{\prime }(s^{*}))\hfill \\ & & -({\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))-{\text{Æ}}^{\prime }(s^{*}))]d\tau (t_0-s^{*})\hfill \\ & & -2{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1[({\text{Æ}}^{\prime }(s^{*}+\tau (v_0-s^{*}))-{\text{Æ}}^{\prime }(s^{*}))\hfill \\ & & -({\text{Æ}}^{\prime }(s^{*}+\tau (z_0-s^{*}))-{\text{Æ}}^{\prime }(s^{*}))]d\tau ({\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)).\hfill \end{array}$$

Next, by applying MVT for first derivatives, we have

$$\begin{array}{ccc}\hfill s_1-s^{*}& =& \left[I-{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\text{Æ}}^{\prime }(s^{*}+\tau (z_0-s^{*}))d\tau \right](t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))\hfill \\ & & -2{\text{Æ}}^{\prime }{(s_0)}^{-1}[{\int }_0^1{\int }_0^1{\text{Æ}}^{″}(s^{*}+\theta \tau (v_0-s^{*}))\tau d\theta d\tau (v_0-s^{*})\hfill \\ & & -{\int }_0^1{\int }_0^1{\text{Æ}}^{″}(s^{*}+\theta \tau (t_0-s^{*}))\tau d\theta d\tau (t_0-s^{*})](t_0-s^{*})\hfill \\ & & -2{\text{Æ}}^{\prime }{(s_0)}^{-1}[{\int }_0^1{\int }_0^1{\text{Æ}}^{″}(s^{*}+\theta \tau (z_0-s^{*}))\tau d\theta d\tau (z_0-s^{*})\hfill \\ & & -{\int }_0^1{\int }_0^1{\text{Æ}}^{″}(s^{*}+\theta \tau (t_0-s^{*}))\tau d\theta d\tau (t_0-s^{*})](t_0-s^{*})\hfill \\ & & -2{\text{Æ}}^{\prime }{(s_0)}^{-1}[{\int }_0^1{\int }_0^1{\text{Æ}}^{″}(s^{*}+\theta \tau (v_0-s^{*}))\tau d\theta d\tau (v_0-s^{*})\hfill \\ & & -{\int }_0^1{\int }_0^1{\text{Æ}}^{″}(s^{*}+\theta \tau (z_0-s^{*}))\tau d\theta d\tau (z_0-s^{*})]{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0).\hfill \end{array}$$

(25)

Applying MVT on the first term, adding and subtracting ${\text{Æ}}^{″}(s^{*})$ in other terms, we get

$$\begin{array}{ccc}\hfill s_1-s^{*}& =& {\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\int }_0^1{\text{Æ}}^{″}(s^{*}+\tau (z_0-s^{*})+\theta (s_0-s^{*}-\tau (z_0-s^{*})))\hfill \\ & & \times (s_0-s^{*}-\tau (z_0-s^{*}))d\theta d\tau (t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))\hfill \\ & & -2{\text{Æ}}^{\prime }{(s_0)}^{-1}[{\int }_0^1{\int }_0^1\left({\text{Æ}}^{″}(s^{*}+\theta \tau (v_0-s^{*}))-{\text{Æ}}^{″}(s^{*})\right)\tau d\theta d\tau (v_0-s^{*})\hfill \\ & & -{\int }_0^1{\int }_0^1\left({\text{Æ}}^{″}(s^{*}+\theta \tau (t_0-s^{*}))-{\text{Æ}}^{″}(s^{*})\right)\tau d\theta d\tau (t_0-s^{*})](t_0-s^{*})\hfill \\ & & -{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*})(v_0-t_0)(t_0-s^{*})\hfill \\ & & -2{\text{Æ}}^{\prime }{(s_0)}^{-1}[{\int }_0^1{\int }_0^1\left({\text{Æ}}^{″}(s^{*}+\theta \tau (z_0-s^{*}))-{\text{Æ}}^{″}(s^{*})\right)\tau d\theta d\tau (z_0-s^{*})\hfill \\ & & -{\int }_0^1{\int }_0^1\left({\text{Æ}}^{″}(s^{*}+\theta \tau (t_0-s^{*}))-{\text{Æ}}^{″}(s^{*})\right)\tau d\theta d\tau (t_0-s^{*})](t_0-s^{*})\hfill \\ & & -{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*})(z_0-t_0)(t_0-s^{*})\hfill \\ & & -2{\text{Æ}}^{\prime }{(s_0)}^{-1}[{\int }_0^1{\int }_0^1\left({\text{Æ}}^{″}(s^{*}+\theta \tau (v_0-s^{*}))-{\text{Æ}}^{″}(s^{*})\right)\tau d\theta d\tau (v_0-s^{*})\hfill \\ & & -{\int }_0^1{\int }_0^1\left({\text{Æ}}^{″}(s^{*}+\theta \tau (z_0-s^{*}))-{\text{Æ}}^{″}(s^{*})\right)\tau d\theta d\tau (z_0-s^{*})]{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)\hfill \\ & & -{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*})(v_0-z_0){\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0).\hfill \end{array}$$

Note that ${\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*})(v_0-t_0)(t_0-s^{*})=-{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*})(z_0-t_0)(t_0-s^{*}).$ This can be seen by substituting for $(v_0-t_0)$ and $(z_0-t_0)$.

For convenience, let

$$\begin{array}{ccc}\hfill A_1& =& -2{\text{Æ}}^{\prime }{(s_0)}^{-1}[{\int }_0^1{\int }_0^1\left({\text{Æ}}^{″}(s^{*}+\theta \tau (v_0-s^{*}))-{\text{Æ}}^{″}(s^{*})\right)\tau d\theta d\tau (v_0-s^{*})\hfill \\ & & -{\int }_0^1{\int }_0^1\left({\text{Æ}}^{″}(s^{*}+\theta \tau (t_0-s^{*}))-{\text{Æ}}^{″}(s^{*})\right)\tau d\theta d\tau (t_0-s^{*})](t_0-s^{*}),\hfill \\ \hfill A_2& =& -2{\text{Æ}}^{\prime }{(s_0)}^{-1}[{\int }_0^1{\int }_0^1\left({\text{Æ}}^{″}(s^{*}+\theta \tau (z_0-s^{*}))-{\text{Æ}}^{″}(s^{*})\right)\tau d\theta d\tau (z_0-s^{*})\hfill \\ & & -{\int }_0^1{\int }_0^1\left({\text{Æ}}^{″}(s^{*}+\theta \tau (t_0-s^{*}))-{\text{Æ}}^{″}(s^{*})\right)\tau d\theta d\tau (t_0-s^{*})](t_0-s^{*}),\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill A_3& =& -2{\text{Æ}}^{\prime }{(s_0)}^{-1}[{\int }_0^1{\int }_0^1\left({\text{Æ}}^{″}(s^{*}+\theta \tau (v_0-s^{*}))-{\text{Æ}}^{″}(s^{*})\right)\tau d\theta d\tau (v_0-s^{*})\hfill \\ & & -{\int }_0^1{\int }_0^1\left({\text{Æ}}^{″}(s^{*}+\theta \tau (z_0-s^{*}))-{\text{Æ}}^{″}(s^{*})\right)\tau d\theta d\tau (z_0-s^{*})]{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0).\hfill \end{array}$$

Then, we obtain

$$\begin{array}{ccc}\hfill s_1-s^{*}& =& \sum _{i=1}^3{A}_i\hfill \\ & & +{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\int }_0^1\left({\text{Æ}}^{″}(s^{*}+\tau (z_0-s^{*})+\theta (s_0-s^{*}-\tau (z_0-s^{*})))-{\text{Æ}}^{″}(s^{*})\right)\hfill \\ & & \times (s_0-s^{*}-\tau (z_0-s^{*}))d\theta d\tau (t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))\hfill \\ & & +{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*})(s_0-s^{*}-{\displaystyle \frac{1}{2}}(z_0-s^{*}))(t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))\hfill \\ & & -2{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){({\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))}^2,\hfill \end{array}$$

where we have used the relation $v_0-z_0=2{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)$.

Let

$$\begin{array}{ccc}\hfill A_4& =& {\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\int }_0^1\left({\text{Æ}}^{″}(s^{*}+\tau (z_0-s^{*})+\theta (s_0-s^{*}-\tau (z_0-s^{*})))-{\text{Æ}}^{″}(s^{*})\right)\hfill \\ & & \times (s_0-s^{*}-\tau (z_0-s^{*}))d\theta d\tau (t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))\hfill \\ \hfill \mathrm{and}\\ \hfill A_5& =& -{\displaystyle \frac{1}{2}}{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*})(z_0-s^{*})(t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)).\hfill \end{array}$$

Then, we have

$$\begin{array}{ccc}\hfill s_1-s^{*}& =& \sum _{i=1}^5{A}_i\hfill \\ & & +{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*})(s_0-s^{*})(t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))\hfill \\ & & -2{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){({\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))}^2,\hfill \end{array}$$

(26)

Since $\text{Æ}(t_0)={\int }_0^1{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))d\tau (t_0-s^{*}),$ we have

$$\begin{array}{ccc}\hfill s_1-s^{*}& =& \sum _{i=1}^5{A}_i\hfill \\ & & +{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*})(s_0-s^{*})(t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))\hfill \\ & & -2{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))d\tau (t_0-s^{*})({\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)).\hfill \end{array}$$

(27)

Note that

$$\begin{array}{ccc}\hfill t_0-s^{*}& =& s_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(s_0)\hfill \\ & =& {\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1({\text{Æ}}^{\prime }(s_0)-{\text{Æ}}^{\prime }(s^{*}+\tau (s_0-s^{*}))d\tau (s_0-s^{*})\hfill \\ & =& {\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\int }_0^1{\text{Æ}}^{″}(s_{(\theta ,\tau )})(1-\tau )d\tau {(s_0-s^{*})}^2,\hfill \end{array}$$

where $s_{(\theta ,\tau )}=(s^{*}+(\tau +\theta (1-\tau ))(s_0-s^{*}))$, and hence by (27) we have

$$\begin{array}{ccc}\hfill s_1-s^{*}& =& \sum _{i=1}^5{A}_i\hfill \\ & & +{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*})[(s_0-s^{*})(t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))\hfill \\ & & -2{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))d\tau {\text{Æ}}^{\prime }{(s_0)}^{-1}\hfill \\ & & \times {\int }_0^1{\int }_0^1{\text{Æ}}^{″}(s_{\theta ,\tau })(1-\tau )d\theta d\tau {(s_0-s^{*})}^2({\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))].\hfill \end{array}$$

Add and subtract ${\text{Æ}}^{″}(s^{*})$ in the last term appropriately to obtain

$$\begin{array}{ccc}\hfill s_1-s^{*}& =& \sum _{i=1}^6{A}_i\hfill \\ & & +{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*})[(s_0-s^{*})(t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))\hfill \\ & & -{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))d\tau {\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){(s_0-s^{*})}^2{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)],\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill A_6& =& -2{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1({\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))d\tau {\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\int }_0^1({\text{Æ}}^{″}(s_{(\theta ,\tau )})\hfill \\ & & -{\text{Æ}}^{″}(s^{*}))(1-\tau )d\theta d\tau {(s_0-s^{*})}^2({\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)).\hfill \end{array}$$

Using Theorem 2, with $h=s_0-s^{*}$ and $k=t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)$, we get ${\text{Æ}}^{″}(s^{*})(s_0-s^{*})(t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))={\text{Æ}}^{″}(s^{*})(t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))(s_0-s^{*}).$

Therefore, we can write

$$\begin{array}{ccc}\hfill s_1-s^{*}& =& \sum _{i=1}^6{A}_i\hfill \\ & & +{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*})[(t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))(s_0-s^{*})\hfill \\ & & -{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))d\tau {\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){(s_0-s^{*})}^2{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)].\hfill \\ \hfill \end{array}$$

(28)

Note that since

$$\begin{array}{ccc}\hfill t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)& =& {\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1({\text{Æ}}^{\prime }(s_0)-{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*})))d\tau (t_0-s^{*})\hfill \\ & =& {\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\int }_0^1{\text{Æ}}^{″}(s^{*}+\tau (t_0-s^{*})+\theta (s_0-s^{*}-\tau (t_0-s^{*}))\hfill \\ & & \times (s_0-s^{*}-\tau (t_0-s^{*}))d\tau d\theta (t_0-s^{*}),\hfill \end{array}$$

(29)

we have by (28)

$$\begin{array}{ccc}\hfill s_1-s^{*}& =& \sum _{i=1}^6{A}_i\hfill \\ & & +{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*})[{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\int }_0^1{\text{Æ}}^{″}(s^{*}+\tau (t_0-s^{*})+\theta (s_0-s^{*}-\tau (t_0-s^{*})))\hfill \\ & & \times (s_0-s^{*}-\tau (t_0-s^{*}))d\theta d\tau (t_0-s^{*})(s_0-s^{*})\hfill \\ & & -{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))d\tau {\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){(s_0-s^{*})}^2{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)].\hfill \end{array}$$

Add and subtract ${\text{Æ}}^{″}(s^{*})$ in the seventh term appropriately again to get

$$\begin{array}{ccc}\hfill s_1-s^{*}& =& \sum _{i=1}^6{A}_i+{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*})\hfill \\ & & \times [{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\int }_0^1\left({\text{Æ}}^{″}(s^{*}+\tau (t_0-s^{*})+\theta (s_0-s^{*}-\tau (t_0-s^{*})))-{\text{Æ}}^{″}(s^{*})\right)\hfill \\ & & \times (s_0-s^{*}-\tau (t_0-s^{*}))d\theta d\tau (t_0-s^{*})(s_0-s^{*})\hfill \\ & & +{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*})(s_0-s^{*}-{\displaystyle \frac{1}{2}}(t_0-s^{*}))(t_0-s^{*})(s_0-s^{*})\hfill \\ & & -{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))d\tau {\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){(s_0-s^{*})}^2{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)].\hfill \end{array}$$

Using Theorem 2, with $h=(s_0-s^{*})(t_0-s^{*})$ and $k=s_0-s^{*}$, we get ${\text{Æ}}^{″}(s^{*})(s_0-s^{*})(t_0-s^{*})(s_0-s^{*})={\text{Æ}}^{″}(s^{*}){(s_0-s^{*})}^2(t_0-s^{*}).$

Then

$$\begin{array}{ccc}\hfill s_1-s^{*}& =& \sum _{i=1}^8{A}_i+{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*})[{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){(s_0-s^{*})}^2(t_0-s^{*})\hfill \\ & & -{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))d\tau {\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){(s_0-s^{*})}^2{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)],\hfill \\ \hfill \end{array}$$

(30)

where

$$\begin{array}{ccc}\hfill A_7& =& {\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*})[{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\int }_0^1({\text{Æ}}^{″}(s^{*}+\tau (t_0-s^{*})+\theta (s_0-s^{*}-\tau (t_0-s^{*})))\hfill \\ & & -{\text{Æ}}^{″}(s^{*}))(s_0-s^{*}-\tau (t_0-s^{*}))d\theta d\tau (s_0-s^{*})(t_0-s^{*})]\hfill \\ \hfill \mathrm{and}\\ \hfill A_8& =& -{\displaystyle \frac{1}{2}}{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){(t_0-s^{*})}^2(s_0-s^{*}).\hfill \end{array}$$

Now, adding and subtracting $(t_0-s^{*})$ in the last term of (30), we get

$$\begin{array}{ccc}\hfill s_1-s^{*}& =& \sum _{i=1}^8{A}_i\hfill \\ & & +{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*})[{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){(s_0-s^{*})}^2(t_0-s^{*})\hfill \\ & & +{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))d\tau {\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){(s_0-s^{*})}^2\hfill \\ & & \times ((t_0-s^{*})-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0))\hfill \\ & & -{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*})){\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){(s_0-s^{*})}^2(t_0-s^{*})].\hfill \end{array}$$

Combine the terms to get

$$\begin{array}{ccc}\hfill s_1-s^{*}& =& \sum _{i=1}^9{A}_i+{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*})\left[I-{\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))d\tau \right]\hfill \\ & & \times {\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){(s_0-s^{*})}^2(t_0-s^{*}),\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill A_9& =& {\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))d\tau {\text{Æ}}^{\prime }{(s_0)}^{-1}\hfill \\ & & \times {\text{Æ}}^{″}(s^{*}){(s_0-s^{*})}^2((t_0-s^{*})-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)).\hfill \end{array}$$

Apply MVT again to get

$$\begin{array}{ccc}\hfill s_1-s^{*}& =& \sum _{i=1}^9{A}_i+{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){\text{Æ}}^{\prime }{(s_0)}^{-1}\hfill \\ & & \times [{\int }_0^1{\int }_0^1{\text{Æ}}^{″}(s^{*}+\tau (t_0-s^{*})+\theta (s_0-s^{*}-\tau (t_0-s^{*})))\hfill \\ & & \times (s_0-s^{*}-\tau (t_0-s^{*}))d\theta d\tau ]{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){(s_0-s^{*})}^2(t_0-s^{*})\hfill \\ & =& \sum _{i=1}^{10}{A}_i,\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill A_{10}& =& {\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){\text{Æ}}^{\prime }{(s_0)}^{-1}[{\int }_0^1{\int }_0^1{\text{Æ}}^{″}(s^{*}+\tau (t_0-s^{*})+\theta (s_0-s^{*}-\tau (t_0-s^{*})))\hfill \\ & & \times (s_0-s^{*}-\tau (t_0-s^{*}))d\theta d\tau ]{\text{Æ}}^{\prime }{(s_0)}^{-1}{\text{Æ}}^{″}(s^{*}){(s_0-s^{*})}^2(t_0-s^{*}).\hfill \end{array}$$

Using the assumptions and inequalities we already have, we calculate $\parallel A_i\parallel ,i=1,2,\dots ,10$ as follows.

Using (21) and assumption $(a_6)$, we get

$$\begin{array}{ccc}\hfill \parallel A_1\parallel & =& 2\parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel [{\int }_0^1{\int }_0^1\parallel {\mathcal{G}}^{-1}\left({\text{Æ}}^{″}(s^{*}+\theta \tau (v_0-s^{*}))-{\text{Æ}}^{″}(s^{*})\right)\parallel \tau d\theta d\tau \parallel v_0-s^{*}\parallel \hfill \\ & & +{\int }_0^1{\int }_0^1\parallel {\mathcal{G}}^{-1}\left({\text{Æ}}^{″}(s^{*}+\theta \tau (t_0-s^{*}))-{\text{Æ}}^{″}(s^{*})\right)\parallel \tau d\theta d\tau \parallel t_0-s^{*}\parallel ]\parallel t_0-s^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{2\times 2}{1-2L_0\parallel s_0-s^{*}\parallel }}[L_2{\int }_0^1{\int }_0^1\theta {\tau }^{2}d\tau d\theta {\parallel v_0-s^{*}\parallel }^2\hfill \\ & & +L_2{\int }_0^1{\int }_0^1\theta {\tau }^{2}d\tau d\theta \parallel t_0-s^{*}{\parallel }^2]\parallel t_0-s^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{4L_2}{1-2L_0\parallel s_0-s^{*}\parallel }}{\displaystyle \frac{1}{6}}\left(\parallel v_0-s^{*}{\parallel }^2+{\parallel t_0-s^{*}\parallel }^2\right)\parallel t_0-s^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{2L_1{L}_2}{3(1-2L_0\parallel s_0-s^{*}{\parallel )}^2}}\left[{\varphi }_1^2(\parallel s_0-s^{*}\parallel )+{\left({\displaystyle \frac{L_1}{1-2L_0\parallel s_0-s^{*}\parallel }}\right)}^2\right]{\parallel s_0-s^{*}\parallel }^6.\hfill \end{array}$$

(31)

and

$$\begin{array}{ccc}\hfill \parallel A_2\parallel & =& 2\parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel [{\int }_0^1{\int }_0^1\parallel {\mathcal{G}}^{-1}\left({\text{Æ}}^{″}(s^{*}+\theta \tau (z_0-s^{*}))-{\text{Æ}}^{″}(s^{*})\right)\parallel \tau d\theta d\tau \parallel (z_0-s^{*})\parallel \hfill \\ & & +{\int }_0^1{\int }_0^1\parallel {\mathcal{G}}^{-1}\left({\text{Æ}}^{″}(s^{*}+\theta \tau (t_0-s^{*}))-{\text{Æ}}^{″}(s^{*})\right)\parallel \tau d\theta d\tau \parallel (t_0-s^{*})\parallel ]\parallel t_0-s^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{2\times 2}{1-2L_0\parallel s_0-s^{*}\parallel }}\left[{\displaystyle \frac{1}{6}}{L}_2\parallel z_0-s^{*}{\parallel }^2+{\displaystyle \frac{1}{6}}{L}_2{\parallel t_0-s^{*}\parallel }^2\right]\parallel t_0-s^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{2L_1{L}_2}{3(1-2L_0\parallel s_0-s^{*}{\parallel )}^2}}\left[{\varphi }_2^2(\parallel s_0-s^{*}\parallel )\parallel s_0-s^{*}{\parallel }^2+{\left({\displaystyle \frac{L_1}{1-2L_0\parallel s_0-s^{*}\parallel }}\right)}^2\right]{\parallel s_0-s^{*}\parallel }^6.\hfill \end{array}$$

(32)

Moreover, using (2), (22) and assumption $(a_6),$ we get

$$\begin{array}{ccc}\hfill \parallel A_3\parallel & =& 2\parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel [{\int }_0^1{\int }_0^1\parallel {\mathcal{G}}^{-1}\left({\text{Æ}}^{″}(s^{*}+\theta \tau (v_0-s^{*}))-{\text{Æ}}^{″}(s^{*})\right)\parallel \tau d\theta d\tau \parallel v_0-s^{*}\parallel \hfill \\ & & +{\int }_0^1{\int }_0^1\parallel {\mathcal{G}}^{-1}\left({\text{Æ}}^{″}(s^{*}+\theta \tau (z_0-s^{*}))-{\text{Æ}}^{″}(s^{*})\right)\parallel \tau d\theta d\tau \parallel z_0-s^{*}\parallel ]\hfill \\ & & \times \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)\parallel \hfill \\ & \le & {\displaystyle \frac{2\times 2}{1-2L_0\parallel s_0-s^{*}\parallel }}\left[{\displaystyle \frac{L_2}{6}}\parallel v_0-s^{*}{\parallel }^2+{\displaystyle \frac{L_2}{6}}{\parallel z_0-s^{*}\parallel }^2\right]{\displaystyle \frac{2K_1}{1-2L_0\parallel s_0-s^{*}\parallel }}\parallel t_0-s^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{2K_1{L}_1}{3(1-2L_0\parallel s_0-s^{*}{\parallel )}^3}}\left[{\varphi }_1^2(\parallel s_0-s^{*}\parallel )+{\varphi }_2^2(\parallel s_0-s^{*}\parallel )\parallel s_0-s^{*}{\parallel }^2\right]{\parallel s_0-s^{*}\parallel }^6.\hfill \end{array}$$

(33)

By (29), we have

$$\begin{array}{ccc}\hfill \parallel t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)\parallel & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}{\int }_0^1\left[{\text{Æ}}^{\prime }(s_0)-{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))\right]d\tau (t_0-s^{*})\parallel \hfill \\ & \le & \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel \hfill \\ & & \times {\int }_0^1\parallel {\mathcal{G}}^{-1}\left[{\text{Æ}}^{\prime }(s_0)-{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))\right]\parallel d\tau \parallel t_0-s^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{2L_1}{1-2L_0\parallel s_0-s^{*}\parallel }}(\parallel s_0-s^{*}\parallel +{\displaystyle \frac{1}{2}}\parallel t_0-s^{*}\parallel )\parallel t_0-s^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{L_1^2}{(1-2L_0\parallel s_0-s^{*}{\parallel )}^2}}\left(2+{\displaystyle \frac{L_1\parallel s_0-s^{*}\parallel }{1-2L_0\parallel s_0-s^{*}\parallel }}\right){\parallel s_0-s^{*}\parallel }^3.\hfill \end{array}$$

(34)

Furthermore, by (21), (34) and assumption $(a_6)$,

$$\begin{array}{ccc}\hfill \parallel A_4\parallel & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel {\int }_0^1{\int }_0^1\parallel \left({\text{Æ}}^{″}(s^{*}+\tau (z_0-s^{*})+\theta (s_0-s^{*}-\tau (z_0-s^{*})))-{\text{Æ}}^{″}(s^{*})\right)\parallel \hfill \\ & & \times \parallel (s_0-s^{*}-\tau (z_0-s^{*}))\parallel d\theta d\tau \parallel t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)\parallel \hfill \\ & \le & {\displaystyle \frac{2L_2}{1-2L_0\parallel s_0-s^{*}\parallel }}{\int }_0^1{\int }_0^1\left((1-\theta )\tau \parallel z_0-s^{*}\parallel +\theta \parallel s_0-s^{*}\parallel \right)\hfill \\ & & \left(\parallel s_0-s^{*}\parallel +\tau \parallel z_0-s^{*}\parallel \right)d\tau d\theta \parallel t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)\parallel \hfill \\ & \le & {\displaystyle \frac{2L_2}{1-2L_0\parallel s_0-s^{*}\parallel }}{\int }_0^1{\int }_0^1[\tau \parallel z_0-s^{*}\parallel \parallel s_0-s^{*}\parallel +(1-\theta ){\tau }^2{\parallel z_0-s^{*}\parallel }^2\hfill \\ & & +\theta \parallel s_0-s^{*}{\parallel }^2]d\tau d\theta \hfill \\ & & \times \parallel t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)\parallel \hfill \\ & \le & {\displaystyle \frac{2L_2}{1-2L_0\parallel s_0-s^{*}\parallel }}\left[{\displaystyle \frac{1}{2}}\parallel z_0-s^{*}\parallel \parallel s_0-s^{*}\parallel +{\displaystyle \frac{1}{6}}\parallel z_0-s^{*}{\parallel }^2+{\displaystyle \frac{1}{2}}{\parallel s_0-s^{*}\parallel }^2\right]\hfill \\ & & \times {\displaystyle \frac{L_1^2}{(1-2L_0\parallel s_0-s^{*}{\parallel )}^2}}\left(2+{\displaystyle \frac{L_1\parallel s_0-s^{*}\parallel }{1-2\parallel s_0-s^{*}\parallel }}\right){\parallel s_0-s^{*}\parallel }^3\hfill \\ & \le & {\displaystyle \frac{L_1^2{L}_2}{(1-2L_0\parallel s_0-s^{*}{\parallel )}^3}}\left[{\varphi }_2(\parallel s_0-s^{*}\parallel )\parallel s_0-s^{*}{\parallel }^2+{\displaystyle \frac{1}{3}}{\varphi }_2^2(\parallel s_0-s^{*}\parallel )\parallel s_0-s^{*}{\parallel }^4+1\right]\hfill \\ & & \times \left(2+{\displaystyle \frac{L_1\parallel s_0-s^{*}\parallel }{1-2L_0\parallel s_0-s^{*}\parallel }}\right){\parallel s_0-s^{*}\parallel }^5.\hfill \end{array}$$

(35)

Similarly, using (21) and assumption $(a_8),$

$$\begin{array}{ccc}\hfill \parallel A_5\parallel & =& {\displaystyle \frac{1}{2}}\parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel \parallel {\mathcal{G}}^{-1}{\text{Æ}}^{″}(s^{*})\parallel \parallel (z_0-s^{*})\parallel \parallel t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)\parallel \hfill \\ & \le & {\displaystyle \frac{1}{2}}{\displaystyle \frac{2K_2}{1-2L_0\parallel s_0-s^{*}\parallel }}{\varphi }_2(\parallel s_0-s^{*}\parallel )\parallel s_0-s^{*}{\parallel }^3\hfill \\ & & \times {\displaystyle \frac{L_1^2}{(1-2L_0\parallel s_0-s^{*}{\parallel )}^2}}\left(2+{\displaystyle \frac{L_1\parallel s_0-s^{*}\parallel }{1-2L_0\parallel s_0-s^{*}\parallel }}\right){\parallel s_0-s^{*}\parallel }^3\hfill \\ & \le & {\displaystyle \frac{K_2{L}_1^2}{(1-2L_0\parallel s_0-s^{*}{\parallel )}^3}}{\varphi }_2(\parallel s_0-s^{*}\parallel )\left(2+{\displaystyle \frac{L_1\parallel s_0-s^{*}\parallel }{1-2L_0\parallel s_0-s^{*}\parallel }}\right){\parallel s_0-s^{*}\parallel }^6.\hfill \end{array}$$

(36)

Then, use (21), (22) and assumptions $(a_6)$ and $(a_7)$, we get

$$\begin{array}{ccc}\hfill \parallel A_6\parallel & =& 2\parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel \parallel {\mathcal{G}}^{-1}{\text{Æ}}^{″}(s^{*})\parallel \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel {\int }_0^1\parallel {\mathcal{G}}^{-1}{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))\parallel d\tau \hfill \\ & & \times \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel {\int }_0^1{\int }_0^1\parallel {\mathcal{G}}^{-1}({\text{Æ}}^{″}(s^{*}+\tau (s_0-s^{*})+\theta (1-\tau )(s_0-s^{*}))\hfill \\ & & -{\text{Æ}}^{″}(s^{*}))\parallel (1-\tau )d\theta d\tau \parallel s_0-s^{*}{\parallel }^2\parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)\parallel \hfill \\ & \le & {\displaystyle \frac{16K_1{K}_2}{(1-2L_0\parallel s_0-s^{*}{\parallel )}^3}}{\int }_0^1{\int }_0^1{L}_2\parallel \tau (s_0-s^{*})+\theta (1-\tau )(s_0-s^{*})\parallel (1-\tau )d\tau d\theta \hfill \\ & & \times \parallel s_0-s^{*}{\parallel }^2\parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)\parallel \hfill \\ & \le & {\displaystyle \frac{32K_1^2{K}_2{L}_1{L}_2}{(1-2L_0\parallel s_0-s^{*}{\parallel )}^5}}{\parallel s_0-s^{*}\parallel }^5.\hfill \end{array}$$

(37)

Then, using (21), assumptions $(a_6)$ and $(a_8),$ we get

$$\begin{array}{ccc}\hfill \parallel A_7\parallel & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel \parallel {\mathcal{G}}^{-1}{\text{Æ}}^{″}(s^{*})\parallel \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel \hfill \\ & & \times {\int }_0^1{\int }_0^1{\mathcal{G}}^{-1}\left({\text{Æ}}^{″}(s^{*}+\tau (t_0-s^{*})+\theta (s_0-s^{*}-\tau (t_0-s^{*})))-{\text{Æ}}^{″}(s^{*})\right)\parallel \hfill \\ & & \times \parallel s_0-s^{*}-\tau (t_0-s^{*})\parallel d\theta d\tau \parallel s_0-s^{*}\parallel \parallel t_0-s^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{4K_2}{(1-2L_0\parallel s_0-s^{*}{\parallel )}^2}}[{\int }_0^1{\int }_0^1{L}_2\parallel \tau (t_0-s^{*})+\theta (s_0-s^{*})+\theta \tau (t_0-s^{*})\parallel \hfill \\ & & (\parallel s_0-s^{*}\parallel +\tau \parallel t_0-s^{*}\parallel )d\tau d\theta ]\parallel s_0-s^{*}\parallel \parallel t_0-s^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{4K_2{L}_2}{(1-2L_0\parallel s_0-s^{*}{\parallel )}^2}}\left[\parallel t_0-s^{*}\parallel \parallel s_0-s^{*}\parallel +{\displaystyle \frac{1}{2}}\parallel t_0-s^{*}{\parallel }^2+{\displaystyle \frac{1}{2}}{\parallel s_0-s^{*}\parallel }^2\right]\hfill \\ & & \times \parallel s_0-s^{*}\parallel \parallel t_0-s^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{2K_2{L}_1{L}_2}{(1-2L_0\parallel s_0-s^{*}{\parallel )}^3}}\left[{\displaystyle \frac{2L_1\parallel s_0-s^{*}\parallel }{1-2L_0\parallel s_0-s^{*}\parallel }}+{\displaystyle \frac{L_1}{1-2L_0\parallel s_0-s^{*}\parallel }}+1\right]{\parallel s_0-s^{*}\parallel }^5.\hfill \end{array}$$

(38)

Using (21) and assumption $(a_8),$

$$\begin{array}{ccc}\hfill \parallel A_8\parallel & =& {\displaystyle \frac{1}{2}}\parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel \parallel {\mathcal{G}}^{-1}{\text{Æ}}^{″}(s^{*})\parallel \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel \parallel {\mathcal{G}}^{-1}{\text{Æ}}^{″}(s^{*})\parallel \parallel t_0-s^{*}{\parallel }^2\parallel s_0-s^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{2K_2^2{L}_1^2}{(1-2L_0\parallel s_0-s^{*}{\parallel )}^4}}{\parallel s_0-s^{*}\parallel }^5.\hfill \end{array}$$

(39)

Using (21), (34) and the assumptions $(a_7)$ and $(a_8)$, we get

$$\begin{array}{ccc}\hfill \parallel A_9\parallel & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel {\int }_0^1\parallel {\mathcal{G}}^{-1}{\text{Æ}}^{\prime }(s^{*}+\tau (t_0-s^{*}))\parallel d\tau \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel \parallel {\mathcal{G}}^{-1}{\text{Æ}}^{″}(s^{*})\parallel \hfill \\ & & \times \parallel s_0-s^{*}{\parallel }^2\parallel t_0-s^{*}-{\text{Æ}}^{\prime }{(s_0)}^{-1}\text{Æ}(t_0)\parallel \hfill \\ & \le & {\displaystyle \frac{4K_1{K}_2{L}_1^2}{(1-2L_0\parallel s_0-s^{*}{\parallel )}^4}}\left(2+{\displaystyle \frac{L_1\parallel s_0-s^{*}\parallel }{1-2L_0\parallel s_0-s^{*}\parallel }}\right){\parallel s_0-s^{*}\parallel }^5.\hfill \end{array}$$

(40)

Finally, using (21) and assumption $(a_8)$, we get

$$\begin{array}{ccc}\hfill \parallel A_{10}\parallel & =& \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel \parallel {\mathcal{G}}^{-1}{\text{Æ}}^{″}(s^{*})\parallel \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel \hfill \\ & & \times {\int }_0^1{\int }_0^1\parallel {\mathcal{G}}^{-1}{\text{Æ}}^{″}(s^{*}+\tau (t_0-s^{*})+\theta (s_0-s^{*}-\tau (t_0-s^{*})))\hfill \\ & & \times \parallel s_0-s^{*}-\tau (t_0-s^{*})\parallel d\theta d\tau \parallel {\text{Æ}}^{\prime }{(s_0)}^{-1}\mathcal{G}\parallel \parallel {\mathcal{G}}^{-1}{\text{Æ}}^{″}(s^{*})\parallel \parallel s_0-s^{*}{\parallel }^2\parallel t_0-s^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{4K_2^3{L}_1}{(1-2L_0\parallel s_0-s^{*}{\parallel )}^4}}\left(2+{\displaystyle \frac{L_1\parallel s_0-s^{*}\parallel }{(1-2L_0\parallel s_0-s^{*}\parallel )}}\right){\parallel s_0-s^{*}\parallel }^5.\hfill \end{array}$$

(41)

Combining the inequalities $(3.16)-(3.25),$ we get

$$\begin{array}{ccc}\hfill \parallel s_1-s^{*}\parallel & \le & \sum _{i=1}^{10}\parallel A_i\parallel \hfill \\ & \le & \varphi (\parallel s_0-s^{*}\parallel )\parallel s_0-s^{*}{\parallel }^5.\hfill \end{array}$$

(42)

Now, since $\varphi (\parallel s_0-s^{*}\parallel )\parallel s_0-s^{*}{\parallel }^5<\parallel s_0-s^{*}\parallel ,$ we have $s_1\in U(s^{*},r).$ □

Theorem 3.

If the assumptions $(a_2)-(a_8)$ hold, then the sequence $(s_n)$ defined by (3) with $s_0\in U(s^{*},r)-\left\{s^{*}\right\}$ is well defined and

$$\begin{array}{c}\hfill \parallel s_{n+1}-s^{*}\parallel \le \varphi (r)\parallel s_n-s^{*}{\parallel }^5.\end{array}$$

(43)

In particular, $s_n\in U(s^{*},r)$ for all $n\in \mathbb{N}\cup \left\{0\right\}$ and $(s_n)$ converges to $s^{*}$ with order of convergence five.

Proof. 

Proof of the theorem follows inductively from the previous lemma by replacing $s_0,t_0$ and $s_1$ with $s_n,t_n$ and $s_{n+1}$, respectively. □

Next, we study the uniqueness of $s^{*}.$

Proposition 4.

Suppose there exists

(i)

a simple solution $s^{*}\in U(s^{*},\nu )$ of (2) and assumption $(a_3)$ holds.

(ii)

a $\delta >\nu$ such that

$$\begin{array}{c}\hfill \delta <{\displaystyle \frac{2}{L}}.\end{array}$$

(44)

Set ${\mathrm{\Omega }}_2=\overline{U}(s^{*},\delta )\cap \mathrm{\Omega }.$ Then (2) has a unique solution $s^{*}$ in ${\mathrm{\Omega }}_2.$

Proof. 

We can see the proof of the proposition from [19].

□

4. Numerical Examples

In this section, we examine two examples to calculate the parameters we have discussed in our theoretical part.

Example 1.

Let $X={\mathbb{R}}^3$ with $\mathrm{\Omega }=\overline{U}(0,1)$. $F:\mathrm{\Omega }\subset X\to X$ is defined for $\omega =({\omega }_1,{\omega }_2,{\omega }_3)$ by

$$\text{Æ}(\omega )=\left({\displaystyle \frac{1}{3}}sin{\omega }_1,{\displaystyle \frac{{\omega }_2^2}{15}}+{\displaystyle \frac{{\omega }_2}{3}},{\displaystyle \frac{{\omega }_3}{3}}\right)$$

The first derivative will be

$${\text{Æ}}^{\prime }(\omega )=\left[\begin{array}{ccc}{\displaystyle \frac{cos{\omega }_1}{3}}& 0& 0\\ 0& {\displaystyle \frac{2{\omega }_2}{15}}+{\displaystyle \frac{1}{3}}& 0\\ 0& 0& {\displaystyle \frac{1}{3}}\end{array}\right],$$

and the second derivative will be

$${\text{Æ}}^{″}(w)=\left[\begin{array}{ccccccccc}{\displaystyle \frac{-sin{\omega }_1}{3}}& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& {\displaystyle \frac{2}{15}}& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right].$$

Consider the solution $s^{*}=(0,0,0)$. Start with the initial point $(0,0,{\displaystyle \frac{1}{3}})$. Choosing $G=I$, we have our solution $s^{*}\in U((0,0,{\displaystyle \frac{1}{3}}),{\displaystyle \frac{1}{2}})$. By comparing with the assumptions $(a_2)-(a_3)$ and $(a_6)-(a_8)$, the constants can be found to be $L_0=L_1=L_2=K_1=K_1={\displaystyle \frac{1}{3}}.$ Then the parameters are ${\rho }_1=0.7780542$, ${\rho }_2=0.8549823$, and $\rho =0.588058733$. Thus, $r=0.588058733.$

Example 2.

Let us consider the trajectory of an electron in the air gap between two parallel plates, described by the expression

$$\begin{array}{c}\hfill \text{Æ}(s)={\displaystyle \frac{\pi }{4}}-{\displaystyle \frac{1}{2}}cos(s)+s.\end{array}$$

Let the domain be $\mathrm{\Omega }=[-1,1]$ and take the initial point $s_0=0$. Choosing $\mathcal{G}=I,$ the iterated solution is found to be $s^{*}=0.30909327$[21]. By comparing with the assumptions $(a_2)-(a_3)$ and $(a_6)-(a_8)$, the constants can be found to be $L_0=L_1=L_2=K_1=K_1={\displaystyle \frac{1}{2}}.$ Then the parameters are ${\rho }_1=0.53918887$, ${\rho }_2=0.60668011$, and $\rho =0.38631554$. Thus $r=0.38631554$.

5. Basins of Attraction

To verify the numerical stability of the method, we analyze the dynamics of the method (3). The set of all initial points that converge to a specific root is called the Basin of Attraction (BA) [22].

Example 3.

$\left\{\begin{array}{c}{a}^5-b=0\hfill \\ b^5-a=0\hfill \end{array}\right.$

with roots $(-1,-1),(0,0)$, $(1,1)$.

Example 4.

$\left\{\begin{array}{c}3a^{2}b-b^3=0\hfill \\ a^3-3a{b}^2-1=0\hfill \end{array}\right.$

with roots $(-{\displaystyle \frac{1}{2}},-{\displaystyle \frac{\sqrt{3}}{2}}),(-{\displaystyle \frac{1}{2}},{\displaystyle \frac{\sqrt{3}}{2}})$, $(1,0)$.

Example 5.

$\left\{\begin{array}{c}{a}^2+b^2=4\hfill \\ 3a^2+7b^2=16\hfill \end{array}\right.$

with roots $(\sqrt{3},-1)$, $(-\sqrt{3},-1),$$(\sqrt{3},1),(-\sqrt{3},1).$

The BA for the roots of the given nonlinear equations is given (Figure 2) in a $401\times 401$ equidistant grid points within a rectangular domain $D=\{(a,b)\in {\mathbb{R}}^2:-2\le a\le 2,-2\le b\le 2\}.$ Each initial point is given a color corresponding to the root, which the iterative method converges. If the method fails to converge or diverges, the point is marked as black. The BA is shown with a tolerance of ${10}^{-8}$, and a maximum of 45 iterations is considered.

6. Conclusions

We have analyzed both semi-local and local convergence analyses of the fifth-order method (3) in a more general Banach space setting. Apart from the previous studies, we have used the same set of assumptions for both semi-local and local analysis, which is independent of the solution $x^{*}$. Even though our analysis is limited to the Lipschitz-type assumptions, it improves the applicability of the considered method. We have given numerical examples to calculate the parameters we have discussed in our theoretical part. To visualize the convergence behavior of the method, the dynamics of the method is presented through examples.