On the Convergence Order of Jarratt-Type Methods for Nonlinear Equations

1. Introduction

Several real-world problems can be mathematically modelled as an equation of the form

$$\mathcal{A}\left(v\right)=0,$$

(1)

where $\mathcal{A}:E\subseteq X\to Y$ is a nonlinear operator mapping between the Banach spaces X and Y and E is open convex set in $X.$ One of the most challenging problems appearing in real-world is to determine the solution $v^{*}$ of (1). Iterative methods are an alternate attractive technique to approximate solutions of nonlinear equations as obtaining the exact solution to these nonlinear equations becomes difficult. One of the most extensively used quadratically convergent iterative method is Newton’s method as it converges rapidly from any sufficiently good initial guess. Even though this method provides a good convergence rate, the need to compute and invert the derivative of the given operator function in each of the iterative step, limits the applicability of these method. To overcome this several Newton-like methods are available in the literature [1,3,5,13,19]. One such successful attempt was made by Ren et al., in [18] providing iterative method (see(2)) of order six. Recall [10] that a sequence $\left\{a_n\right\}$ in X with ${lim}_{n\to \infty }{a}_n=\alpha$ is said to be convergent of order $q>1$, if there exist a nonzero constant C such that

$$\begin{array}{c}\hfill \underset{n⟶\infty }{lim}{\displaystyle \frac{\parallel a_{n+1}-\alpha \parallel }{\parallel a_n{-\alpha \parallel }^q}}=C.\end{array}$$

Previous studies primarily used Taylor expansion to determine the order of convergence(OC), which necessitates the existence of higher-order derivatives. An alternative method involves employing the computational order of convergence (COC) [23], defined as:

$$\overline{\rho }={\displaystyle \frac{ln\left(∥{\displaystyle \frac{a_{n+1}-\alpha }{a_n-\alpha }}∥\right)}{ln\left(∥{\displaystyle \frac{a_n-\alpha }{a_{n-1}-\alpha }}∥\right)}},$$

where $a_{n-1},a_n,a_{n+1}$ are three consecutive iterates near root $\alpha$ or the approximate computational order of convergence (ACOC) defined as:

$$\overline{\rho }={\displaystyle \frac{ln\left(∥{\displaystyle \frac{a_{n+1}-a_n}{a_n-a_{n-1}}}∥\right)}{ln\left(∥{\displaystyle \frac{a_n-a_{n-1}}{a_{n-1}-a_{n-2}}}∥\right)}},$$

where $a_{n-2},a_{n-1},a_n,a_{n+1}$ are four consecutive iterates near root $\alpha$, to obtain the OC.

The limitation of COC and ACOC for iterative methods lies in their susceptibility to the oscillating behavior of approximations and slow convergence during early iterations [17]. As a result, COC and ACOC do not accurately reflect the true OC.

In [18], Ren et al., considered the following iterative scheme defined by

$$\begin{array}{ccc}\hfill w_n& =& v_n-{\displaystyle \frac{2}{3}}{\mathcal{A}}^{\prime }{\left(v_n\right)}^{-1}\mathcal{A}\left(v_n\right),\hfill \\ \hfill z_n& =& v_n-{\displaystyle \frac{1}{2}}{[3{\mathcal{A}}^{\prime }\left(w_n\right)-{\mathcal{A}}^{\prime }\left(v_n\right)]}^{-1}[3{\mathcal{A}}^{\prime }\left(w_n\right)+{\mathcal{A}}^{\prime }\left(v_n\right)]{\mathcal{A}}^{\prime }{\left(v_n\right)}^{-1}\mathcal{A}\left(v_n\right),\hfill \\ \hfill v_{n+1}& =& z_n-{\mathcal{A}}^{\prime }{\left(z_n\right)}^{-1}\mathcal{A}\left(z_n\right),n=0,1,2,3...\hfill \end{array}$$

(2)

for solving (1), when $X=Y={\mathbb{R}}^k.$

In [18], Taylor’s expansion is used to achieve a sixth-order convergence, but the analysis requires conditions on the derivatives of $\mathcal{A}$ up to the seventh order. These assumptions restrict the applicability of the method (2) to problems involving operators that are differentiable at least seven times.

In this article, we initially determine the OC of the method (refer to [9,13]) defined for all $n=0,1,2,\dots$ as follows:

$$\begin{array}{ccc}\hfill w_n& =& v_n-{\displaystyle \frac{2}{3}}{\mathcal{A}}^{\prime }{\left(v_n\right)}^{-1}\mathcal{A}\left(v_n\right),\hfill \\ \hfill v_{n+1}& =& v_n-{\displaystyle \frac{1}{2}}{[3{\mathcal{A}}^{\prime }\left(w_n\right)-{\mathcal{A}}^{\prime }\left(v_n\right)]}^{-1}[3{\mathcal{A}}^{\prime }\left(w_n\right)+{\mathcal{A}}^{\prime }\left(v_n\right)]{\mathcal{A}}^{\prime }{\left(v_n\right)}^{-1}\mathcal{A}\left(v_n\right),\hfill \end{array}$$

(3)

where X and Y are Banach spaces. The local convergence of certain Jarratt-type methods was analyzed in [2] by relying solely on assumptions about the derivative of order one of $\mathcal{A}$. However, the OC was determined using COC and ACOC, which, as previously noted, are not ideal for calculating the convergence order. This raises the question: can we establish a third-order convergence for (3) and a sixth-order convergence for (2) without using assumptions on the higher-order derivatives of $\mathcal{A}$ or Taylor expansion?

Additionally, we enhance the method to a fifth-order approach, given as follows:

$$\begin{array}{ccc}\hfill w_n& =& v_n-{\displaystyle \frac{2}{3}}{\mathcal{A}}^{\prime }{\left(v_n\right)}^{-1}\mathcal{A}\left(v_n\right),\hfill \\ \hfill z_n& =& v_n-{\displaystyle \frac{1}{2}}{[3{\mathcal{A}}^{\prime }\left(w_n\right)-{\mathcal{A}}^{\prime }\left(v_n\right)]}^{-1}[3{\mathcal{A}}^{\prime }\left(w_n\right)+{\mathcal{A}}^{\prime }\left(v_n\right)]{\mathcal{A}}^{\prime }{\left(v_n\right)}^{-1}\mathcal{A}\left(v_n\right),\hfill \\ \hfill v_{n+1}& =& z_n-{\mathcal{A}}^{\prime }{\left({\displaystyle \frac{3w_n-v_n}{2}}\right)}^{-1}\mathcal{A}\left(z_n\right),n=0,1,2,3...\hfill \end{array}$$

(4)

in Section 4.

In Section 2, we establish a third-order convergence for method (3), and in Section 3, we demonstrate a sixth-order convergence for (2), relying on assumptions about the derivatives of $\mathcal{A}$ up to the second order. Consequently, our analysis broadens the applicability of methods (3), (2), and (4) to problems that could not be addressed using the approaches in [4,14,18,20,21].

In Section 5, we examine the constraints of our approach and propose novel strategies to overcome these limitations for local as well as semi-local convergence scenarios. The convergence conditions are solely tied to the operators involved in the method for both the semi-local and local cases.

The remaining part of the paper includes the efficiency index in Section 6, numerical demonstration in Section 7, and basins of attraction in Section 8, concluding with a summary in Section 9.

2. Order of Convergence(OC) of (3):

The analysis of local convergence relies on the following assumptions:

(A1)

${\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}$&∃$L_0>0$ such that $\forall v,w\in E,$

$$\parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}({\mathcal{A}}^{\prime }\left(v\right)-{\mathcal{A}}^{\prime }\left(w\right))\parallel \le L_0\parallel v-w\parallel .$$

(A2)

s $L_1>0$ such that $\forall v,w\in E,$

$$\parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}({\mathcal{A}}^{\prime \prime }\left(v\right)-{\mathcal{A}}^{\prime \prime }\left(w\right))\parallel \le L_1\parallel v-w\parallel .$$

(A3)

∃$L_2>0$ such that

$$\parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}{\mathcal{A}}^{\prime \prime }\left(v\right)\parallel \le L_2,\forall v\in E.$$

and

(A4)

∃$L_3>0$ such that

$$\parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}{\mathcal{A}}^{\prime }\left(v\right)\parallel \le L_3,\forall v\in E.$$

Using the constants $L_0,L_1,L_2$ and $L_3$, we define continuous nondecreasing functions (CNF)${\mathsf{\Theta }}_1,H_1:[0,{\displaystyle \frac{1}{L_0}})⟶\mathbb{R}$ as follows;

$$\begin{array}{ccc}\hfill {\mathsf{\Theta }}_1\left(t\right)& =& {\displaystyle \frac{L_0}{2}}(3(1+{\displaystyle \frac{2L_3}{3(1-L_{0}t)}})+1)t\hfill \end{array}$$

and

$$H_1\left(t\right)={\mathsf{\Theta }}_1\left(t\right)-1.$$

Given that $H_1\left(0\right)=-1$ and $H_1\left(t\right)approaches\infty$ as t tends to ${{\displaystyle \frac{1}{L_0}}}^{-}$, it follows that $H_1\left(t\right)=0$ has a smallest positive root in the interval $[0,{\displaystyle \frac{1}{L_0}})$, which we denote by $\lambda$. Define CNF $g_1,h_1:[0,\lambda )⟶\mathbb{R}$ by

$$\begin{array}{ccc}\hfill g_1\left(t\right)& =& {\displaystyle \frac{1}{48(1-{\mathsf{\Theta }}_1\left(t\right)){(1-L_{0}t)}^2}}\hfill \\ & & \times [(93(L_1-L_{0}t)+32L_1{L}_3+36L_0{L}_2)(1-L_{0}t)+12L_0{L}_2{L}_3]\hfill \end{array}$$

and

$$h_1\left(t\right)=g_1\left(t\right)t^2-1.$$

Since $h_1\left(0\right)=-1$ and $h_1\left(t\right)approaches\infty$ as $t\to {\lambda }^{-}$, it follows that $h_1\left(t\right)=0$ has a smallest positive root in the interval $[0,\lambda )$, which is denoted by ${\lambda }_1$.

Let

$$r=min\{{\lambda }_1,{\displaystyle \frac{1}{2L_0}},1\}.$$

(5)

Then, we have

$$0\le g_1\left(t\right)t^2<1,\forall t\in [0,r).$$

(6)

Throughout the paper, we consider $B(v_0,\rho )=\{v\in X:\parallel v-v_0\parallel <\rho \}$ and $B[v_0,\rho ]=\{v\in X:\parallel v-v_0\parallel \le \rho \}$, for $\rho >0$ and $v_0\in X.$

Theorem 1.

Assuming (A1)-(A4) are true, the sequence $\left\{v_n\right\}$ given by (3) with initial value $v_0\in B(v^{*},r)$ converges to $v^{*}$, and the following estimate is valid:

$$\parallel v_{n+1}-v^{*}\parallel \le g_1\left(r\right){\parallel v_n-v^{*}\parallel }^3.$$

(7)

Proof. An inductive argument will be employed for the proof. As a first step, we will demonstrate that the operator $3{\mathcal{A}}^{\prime }\left(w\right)-{\mathcal{A}}^{\prime }\left(v\right)$ is invertible for all $v,w$ belonging to the open ball $B(v^{*},r).$ Note that, by (A1), we have

$$\begin{array}{ccc}\hfill \parallel {\left(2{\mathcal{A}}^{\prime }\left(v^{*}\right)\right)}^{-1}(3{\mathcal{A}}^{\prime }\left(w\right)-{\mathcal{A}}^{\prime }\left(v\right)-2{\mathcal{A}}^{\prime }\left(v^{*}\right))\parallel & \le & \parallel {\left(2{\mathcal{A}}^{\prime }\left(v^{*}\right)\right)}^{-1}[3({\mathcal{A}}^{\prime }\left(w\right)-{\mathcal{A}}^{\prime }\left(v^{*}\right))\hfill \\ & & -({\mathcal{A}}^{\prime }\left(v\right)-{\mathcal{A}}^{\prime }\left(v^{*}\right))]\parallel \hfill \\ & \le & {\displaystyle \frac{3}{2}}\parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}({\mathcal{A}}^{\prime }\left(w\right)-{\mathcal{A}}^{\prime }\left(v^{*}\right))\parallel \hfill \\ & & +{\displaystyle \frac{1}{2}}\parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}({\mathcal{A}}^{\prime }\left(v\right)-{\mathcal{A}}^{\prime }\left(v^{*}\right))\parallel \hfill \\ & \le & {\displaystyle \frac{L_0}{2}}(3\parallel w-v^{*}\parallel +\parallel v-v^{*}\parallel )\hfill \\ & \le & {\displaystyle \frac{3}{2}}{L}_{0}r+{\displaystyle \frac{1}{2}}{L}_{0}r=2L_{0}r<1.\hfill \end{array}$$

(8)

Therefore, by Banach lemma(BL)on invertible operators, $3{\mathcal{A}}^{\prime }\left(w\right)-{\mathcal{A}}^{\prime }\left(v\right)$ is invertible and by (8), we have

$$\parallel {(3{\mathcal{A}}^{\prime }\left(w\right)-{\mathcal{A}}^{\prime }\left(v\right))}^{-1}{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel \le {\displaystyle \frac{1}{2(1-\frac{L_0}{2}(3\parallel w-v^{*}\parallel +\parallel v-v^{*}\parallel \left)\right)}}.$$

(9)

Similarly, one can prove that

$$\parallel {\mathcal{A}}^{\prime }{\left(v\right)}^{-1}{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel \le {\displaystyle \frac{1}{1-L_0\parallel v-v^{*}\parallel }}.$$

(10)

Next, by the method (3), we have,

$$\begin{array}{ccc}\hfill v_1-v^{*}& =& v_0-v^{*}-{\displaystyle \frac{1}{2}}{(3{\mathcal{A}}^{\prime }\left(w_0\right)-{\mathcal{A}}^{\prime }\left(v_0\right))}^{-1}(3{\mathcal{A}}^{\prime }\left(w_0\right)+{\mathcal{A}}^{\prime }\left(v_0\right)){\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\mathcal{A}\left(v_0\right)\hfill \\ & =& {(3{\mathcal{A}}^{\prime }\left(w_0\right)-{\mathcal{A}}^{\prime }\left(v_0\right))}^{-1}[(3{\mathcal{A}}^{\prime }\left(w_0\right)-{\mathcal{A}}^{\prime }\left(v_0\right))(v_0-v^{*})\hfill \\ & & -{\displaystyle \frac{1}{2}}(3{\mathcal{A}}^{\prime }\left(w_0\right)+{\mathcal{A}}^{\prime }\left(v_0\right)){\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\mathcal{A}\left(v_0\right)].\hfill \end{array}$$

(11)

Note that,

$$\mathcal{A}\left(v_0\right)=\mathcal{A}\left(v_0\right)-\mathcal{A}\left(v^{*}\right)={\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+t(v_0-v^{*}))dt(v_0-v^{*}).$$

(12)

For convenience, let $P={(3{\mathcal{A}}^{\prime }\left(w_0\right)-{\mathcal{A}}^{\prime }\left(v_0\right))}^{-1}.$ In order to prove (7), we rearrange the equation (11) as follows:

$$\begin{array}{ccc}\hfill v_1-v^{*}& =& P\left\{[(3{\mathcal{A}}^{\prime }\left(w_0\right)-{\mathcal{A}}^{\prime }\left(v_0\right))-2\mathcal{A}\left(v_0\right)\right.\hfill \\ & & +\left.(2I-{\displaystyle \frac{1}{2}}(3{\mathcal{A}}^{\prime }\left(w_0\right)+{\mathcal{A}}^{\prime }\left(v_0\right)){\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1})\mathcal{A}\left(v_0\right)\right\}\hfill \\ & =& P\left\{3{\int }_0^1({\mathcal{A}}^{\prime }\left(w_0\right)-{\mathcal{A}}^{\prime }(v^{*}+t(v_0-v^{*})))dt(v_0-v^{*})\right.\hfill \\ & & +{\int }_0^1({\mathcal{A}}^{\prime }(v^{*}+t(v_0-v^{*}))-{\mathcal{A}}^{\prime }\left(v_0\right))dt(v_0-v^{*})\left(\mathrm{by}\left(12\right)\right)\hfill \\ & & \left.+(2{\mathcal{A}}^{\prime }\left(v_0\right)-{\displaystyle \frac{3}{2}}{\mathcal{A}}^{\prime }\left(w_0\right)-{\displaystyle \frac{1}{2}}{\mathcal{A}}^{\prime }\left(v_0\right)){\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\mathcal{A}\left(v_0\right)\right\}\hfill \\ & =& P\left\{3{\int }_0^1{\int }_0^1{\mathcal{A}}^{\prime \prime }(v^{*}+t(v_0-v^{*})+\theta (w_0-v^{*}-t(v_0-v^{*})))d\theta \right.\hfill \\ & & \times (w_0-v^{*}-t(v_0-v^{*}))dt(v_0-v^{*})\hfill \\ & & -{\int }_0^1{\int }_0^1{\mathcal{A}}^{\prime \prime }(v^{*}+t(v_0-v^{*})+\theta (1-t)(v_0-v^{*}))dt(1-t){(v_0-v^{*})}^2\hfill \\ & & +\left.{\displaystyle \frac{3}{2}}{\int }_0^1{\mathcal{A}}^{\prime \prime }(w_0+\theta (v_0-w_0))d\theta (v_0-w_0){\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\mathcal{A}\left(v_0\right)\right\}\hfill \\ & =& P\left\{3{\int }_0^1{\int }_0^1{\mathcal{A}}^{\prime \prime }(v^{*}+t(v_0-v^{*})+\theta (w_0-v^{*}-t(v_0-v^{*})))d\theta \right.\hfill \\ & & \times \left(({\displaystyle \frac{1}{3}}-t){(v_0-v^{*})}^2\right)dt\hfill \\ & & +3{\int }_0^1{\int }_0^1{\mathcal{A}}^{\prime \prime }(v^{*}+t(v_0-v^{*})+\theta (w_0-v^{*}-t(v_0-v^{*})))d\theta \hfill \\ & & \times {\displaystyle \frac{2}{3}}(v_0-v^{*}-{\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\mathcal{A}\left(v_0\right))dt(v_0-v^{*})\hfill \\ & & (\mathrm{because}{w}_0-v^{*}={\displaystyle \frac{1}{3}}(v_0-v^{*})+{\displaystyle \frac{2}{3}}(v_0-v^{*}-{\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\mathcal{A}\left(v_0\right)))\hfill \\ & & -{\int }_0^1{\int }_0^1{\mathcal{A}}^{\prime \prime }(v^{*}+t(v_0-v^{*})+\theta (1-t)(v_0-v^{*}))dt(1-t){(v_0-v^{*})}^2\hfill \\ & & +\left.{\int }_0^1{\mathcal{A}}^{\prime \prime }(w_0+\theta (v_0-w_0))d\theta {\left({\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\mathcal{A}\left(v_0\right)\right)}^2\right\}\hfill \end{array}$$

(13)

Let $S_1(\theta ,t)={\mathcal{A}}^{\prime \prime }(v^{*}+t(v_0-v^{*})+\theta (w_0-v^{*}-t(v_0-v^{*}))),$$S_2(\theta ,t)={\mathcal{A}}^{\prime \prime }(v^{*}+t(v_0-v^{*})+\theta (1-t)(v_0-v^{*}))$ and $S_3\left(\theta \right)={\mathcal{A}}^{\prime \prime }(w_0+\theta (v_0-w_0)).$ Then, by (13), we have

$$\begin{array}{ccc}\hfill v_1-v^{*}& =& P\left\{{\int }_0^1{\int }_0^1{S}_1(\theta ,t)d\theta \left((1-2t){(v_0-v^{*})}^2\right)dt\right.\hfill \\ & & +2{\int }_0^1{\int }_0^1{S}_1(\theta ,t)d\theta dt\hfill \\ & & \times {\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}({\mathcal{A}}^{\prime }\left(v_0\right)-{\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*}))d\tau ){(v_0-v^{*})}^2\hfill \\ & & -{\int }_0^1{\int }_0^1{S}_2(\theta ,t)d\theta dt{(v_0-v^{*})}^2\hfill \\ & & +{\int }_0^1{\int }_0^1(S_2(\theta ,t)-S_1(\theta ,t))tdt{(v_0-v^{*})}^2\hfill \\ & & +\left.{\int }_0^1{S}_3\left(\theta \right)d\theta {\left({\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\mathcal{A}\left(v_0\right)\right)}^2\right\}\hfill \\ & =& P\left\{{\int }_0^1{\int }_0^1{S}_1(\theta ,t)d\theta \left((1-2t){(v_0-v^{*})}^2\right)dt\right.\hfill \\ & & +2{\int }_0^1{\int }_0^1{S}_1(\theta ,t)d\theta dt\hfill \\ & & \times {\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}({\mathcal{A}}^{\prime }\left(v_0\right)-{\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*}))d\tau ){(v_0-v^{*})}^2\hfill \\ & & +{\int }_0^1{\int }_0^1(S_2(\theta ,t)-S_1(\theta ,t))tdt{(v_0-v^{*})}^2\hfill \\ & & +{\int }_0^1{\int }_0^1(S_3\left(\theta \right)-S_2(\theta ,t))d\theta dt{(v_0-v^{*})}^2\hfill \\ & & +\left.{\int }_0^1{S}_3\left(\theta \right)d\theta [{\left({\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}{\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*}))\right)}^2-I]{(v_0-v^{*})}^2\right\}.\hfill \\ & =:& I_1+I_2+I_3+I_4+I_5,\hfill \end{array}$$

(14)

where,

$$\begin{array}{ccc}\hfill I_1& =& P{\int }_0^1{\int }_0^1{S}_1(\theta ,t)d\theta \left((1-2t){(v_0-v^{*})}^2\right)dt,\hfill \\ \hfill I_2& =& 2P{\int }_0^1{\int }_0^1{S}_1(\theta ,t)d\theta dt{\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}({\mathcal{A}}^{\prime }\left(v_0\right)-{\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*}))d\tau ){(v_0-v^{*})}^2,\hfill \\ \hfill I_3& =& P{\int }_0^1{\int }_0^1(S_2(\theta ,t)-S_1(\theta ,t))tdt{(v_0-v^{*})}^2,\hfill \\ \hfill I_4& =& P{\int }_0^1{\int }_0^1(S_3\left(\theta \right)-S_2(\theta ,t))d\theta dt{(v_0-v^{*})}^2\hfill \end{array}$$

and

$$I_5=P{\int }_0^1{S}_3\left(\theta \right)d\theta [{\left({\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}{\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*}))\right)}^2-I]{(v_0-v^{*})}^2.$$

Next, we estimate the norms of $I_1,I_2,I_3,I_4$ and $I_5.$ Note that,

$$\begin{array}{ccc}\hfill \parallel I_1\parallel & =& ∥P{\int }_0^1{\int }_0^1{S}_1(\theta ,t)d\theta (1-2t){(v_0-v^{*})}^{2}dt∥\hfill \\ & \le & \parallel P{\int }_0^1{\int }_0^1[{\mathcal{A}}^{\prime \prime }(v^{*}+t(v_0-v^{*})+\theta (w_0-v^{*}-t(v_0-v^{*})))-{\mathcal{A}}^{\prime \prime }\left(v^{*}\right)]d\theta \hfill \\ & & \times (1-2t){(v_0-v^{*})}^{2}dt\hfill \\ & & +{(3{\mathcal{A}}^{\prime }\left(w_0\right)-{\mathcal{A}}^{\prime }\left(v_0\right))}^{-1}{\int }_0^1{\int }_0^1{\mathcal{A}}^{\prime \prime }\left(v^{*}\right)d\theta (1-2t){(v_0-v^{*})}^{2}dt\parallel \hfill \\ & \le & \parallel P{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel {\int }_0^1{\int }_0^1\parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}[{\mathcal{A}}^{\prime \prime }(v^{*}+t(v_0-v^{*})+\theta (w_0-v^{*}-t(v_0-v^{*})))\hfill \\ & & -{\mathcal{A}}^{\prime \prime }\left(v^{*}\right)]\parallel d\theta |1-2t|{\parallel v_0-v^{*}\parallel }^{2}dt\hfill \\ & & +\parallel {(3{\mathcal{A}}^{\prime }\left(w_0\right)-{\mathcal{A}}^{\prime }\left(v_0\right))}^{-1}{\mathcal{A}}^{\prime \prime }\left(v^{*}\right){\int }_0^1(1-2t){(v_0-v^{*})}^{2}dt\parallel \hfill \\ & \le & L_1\parallel P{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel {\int }_0^1{\int }_0^1\parallel t(v_0-v^{*})+\theta (w_0-v^{*}-t(v_0-v^{*}))\parallel \hfill \\ & & \times |1-2t|{\parallel v_0-v^{*}\parallel }^{2}d\theta dt\hfill \\ & \le & L_1\parallel P{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel \left[{\int }_0^1{\int }_0^1\left|t\right||1-2t|\parallel v_0-v^{*}\parallel \right.\hfill \\ & & +\left.\left|\theta \right||1-2t|\parallel w_0-v^{*}\parallel +\left|\theta t\right||1-2t|\parallel v_0-v^{*}\parallel \right]\hfill \\ & \times & {\parallel v_0-v^{*}\parallel }^{2}d\theta dt\hfill \\ & \le & L_1\parallel P{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel \left[{\int }_0^1\left|t\right||1-2t|\parallel v_0-v^{*}\parallel \right.\hfill \\ & & +\left.{\displaystyle \frac{|1-2t|}{2}}\parallel w_0-v^{*}\parallel +{\displaystyle \frac{\left|t\right||1-2t|}{2}}\parallel v_0-v^{*}\parallel \right]{\parallel v_0-v^{*}\parallel }^{2}d\theta dt\hfill \\ & \le & L_1\parallel P{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel \left[{\int }_0^1{\displaystyle \frac{3\left|t\right||1-2t|}{2}}\parallel v_0-v^{*}\parallel +{\displaystyle \frac{|1-2t|}{2}}\parallel w_0-v^{*}\parallel \right]{\parallel v_0-v^{*}\parallel }^{2}d\theta dt\hfill \\ & \le & L_1\parallel P{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel \left[{\displaystyle \frac{3}{8}}{\parallel v_0-v^{*}\parallel }^3+{\displaystyle \frac{1}{4}}\parallel w_0-v^{*}\parallel {\parallel v_0-v^{*}\parallel }^2\right],\hfill \end{array}$$

(15)

which is obtained using (A2) and (10)(with $v=v_0$). Note that $\parallel w_0-v^{*}\parallel \le \parallel w_0-v_0\parallel +\parallel v_0-v^{*}\parallel$ and using (10) we have,

$$\begin{array}{ccc}\hfill \parallel w_0-v_0\parallel & =& \parallel {\displaystyle \frac{2}{3}}{\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\mathcal{A}\left(v_0\right)\parallel \hfill \\ & \le & {\displaystyle \frac{2}{3}}\parallel {\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel \parallel {\int }_0^1{\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}{\mathcal{A}}^{\prime }(v^{*}+t(v_0-v^{*}))dt(v_0-v^{*})\parallel \hfill \\ & \le & {\displaystyle \frac{2L_3}{3(1-L_0\parallel v_0-v^{*}\parallel )}}\parallel v_0-v^{*}\parallel .\hfill \end{array}$$

(16)

Therefore,

$$\parallel w_0-v^{*}\parallel \le \left(1+{\displaystyle \frac{2L_3}{3(1-L_0\parallel v_0-v^{*}\parallel )}}\right)\parallel v_0-v^{*}\parallel$$

(17)

and hence by (9) (with $w=w_0$ and $v=v_0$), we have

$$\parallel P{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel \le {\displaystyle \frac{1}{2(1-\frac{L_0}{2}(3(1+\frac{2L_3}{3(1-L_0\parallel v_0-v^{*}\parallel )})+1)\parallel v-v^{*}\parallel )}}.$$

(18)

Therefore, on using (17) in (15) we get,

$$\begin{array}{ccc}\hfill \parallel I_1\parallel & \le & {\displaystyle \frac{L_1(3+2(1+\frac{2L_3}{3(1-L_0\parallel v_0-v^{*}\parallel )}))}{16(1-\frac{L_0}{2}(3(1+\frac{2L_3}{3(1-L_0\parallel v_0-v^{*}\parallel )})+1)\parallel v_0-v^{*}\parallel )}}{\parallel v_0-v^{*}\parallel }^3.\hfill \end{array}$$

(19)

Next,

$$\begin{array}{ccc}\hfill \parallel I_2\parallel & =& ∥2P{\int }_0^1{\int }_0^1{S}_1(\theta ,t)d\theta dt\hfill \\ & & \times {\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}({\mathcal{A}}^{\prime }\left(v_0\right)-{\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*})))d\tau {(v_0-v^{*})}^2∥\hfill \\ & =& \parallel 2P{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel ∥{\int }_0^1{\int }_0^1{\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}{S}_1(\theta ,t)d\theta dt{\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}{\mathcal{A}}^{\prime }\left(v^{*}\right)\hfill \\ & & \times {\int }_0^1{\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}({\mathcal{A}}^{\prime }\left(v_0\right)-{\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*})))d\tau {(v_0-v^{*})}^2∥.\hfill \end{array}$$

Therefore, by (9), (10), (A1) and (A3), we have

$$\parallel I_2\parallel \le {\displaystyle \frac{L_0{L}_2}{2(1-\frac{L_0}{2}(3(1+\frac{2L_3}{3(1-L_0\parallel v_0-v^{*}\parallel )})+1)\parallel v_0-v^{*}\parallel \left)\right(1-L_0\parallel v_0-v^{*}\parallel )}}{\parallel v_0-v^{*}\parallel }^3.$$

(20)

By using (9) and (A2), we have

$$\begin{array}{ccc}\hfill \parallel I_3\parallel & =& \parallel P{\mathcal{A}}^{\prime }\left(v^{*}\right){\int }_0^1{\int }_0^1{\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}(S_2(\theta ,t)-S_1(\theta ,t))td\theta dt{(v_0-v^{*})}^2\parallel \hfill \\ & \le & L_1\parallel P{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel \parallel {\int }_0^1{\int }_0^1(v^{*}+t(v_0-v^{*})+\theta (1-t)(v_0-v^{*})\hfill \\ & & -(v^{*}+t(v_0-v^{*})+\theta (w_0-v^{*}-t(v_0-v^{*}))))td\theta dt\parallel \parallel v_0-v^{*}{\parallel }^2\hfill \\ & \le & L_1\parallel P{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel {\int }_0^1{\int }_0^1\left|\theta t\right|(\parallel w_0-v^{*}\parallel +\parallel v_0-v^{*}\parallel )d\theta dt\parallel v_0-v^{*}{\parallel }^2\hfill \\ & \le & {\displaystyle \frac{L_1}{8(1-\frac{L_0}{2}(3(1+\frac{2L_3}{3(1-L_0\parallel v_0-v^{*}\parallel )})+1)\parallel v_0-v^{*}\parallel )}}\hfill \\ & & \times (\parallel v_0-v^{*}\parallel +\parallel w_0-v^{*}\parallel )\parallel v_0-v^{*}{\parallel }^2.\hfill \end{array}$$

(21)

Thus, by (16) and (17), we have

$$\begin{array}{ccc}\hfill \parallel I_3\parallel & \le & {\displaystyle \frac{L_1}{8(1-\frac{L_0}{2}(3(1+\frac{2L_3}{3(1-L_0\parallel v_0-v^{*}\parallel )})+1)\parallel v_0-v^{*}\parallel )}}\hfill \\ & & \times \left(2+{\displaystyle \frac{2L_3}{3(1-L_0\parallel v_0-v^{*}\parallel )}}\right){\parallel v_0-v^{*}\parallel }^3\hfill \\ & \le & {\displaystyle \frac{L_1\left(3\right(1-L_0\parallel v_0-v^{*}\parallel )+L_3)}{12(1-\frac{L_0}{2}(3(1+\frac{2L_3}{3(1-L_0\parallel v_0-v^{*}\parallel )})+1)\parallel v_0-v^{*}\parallel \left)\right(1-L_0\parallel v_0-v^{*}\parallel )}}{\parallel v_0-v^{*}\parallel }^3.\hfill \end{array}$$

(22)

Similarly, we have

$$\begin{array}{ccc}\hfill \parallel I_4\parallel & =& \parallel P{\mathcal{A}}^{\prime }\left(v^{*}\right){\int }_0^1{\int }_0^1{\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}(S_3\left(\theta \right)-S_2(\theta ,t))d\theta dt{(v_0-v^{*})}^2\parallel \hfill \\ & \le & {\displaystyle \frac{L_1}{2(1-\frac{L_0}{2}(3(1+\frac{2L_3}{3(1-L_0\parallel v_0-v^{*}\parallel )})+1)\parallel v_0-v^{*}\parallel )}}\hfill \\ & & \times ∥{\int }_0^1{\int }_0^1((w_0+\theta (v_0-w_0))\hfill \\ & & -(v^{*}+t(v_0-v^{*})+\theta (1-t)(v_0-v^{*})))d\theta dt∥{\parallel v_0-v^{*}\parallel }^2\hfill \\ & \le & {\displaystyle \frac{L_1}{2(1-\frac{L_0}{2}(3(1+\frac{2L_3}{3(1-L_0\parallel v_0-v^{*}\parallel )})+1)\parallel v_0-v^{*}\parallel )}}\hfill \\ & & {\int }_0^1{\int }_0^1(\parallel w_0-v^{*}\parallel +\theta \parallel v_0-w_0\parallel \hfill \\ & & +|t+\theta (1-t)|\parallel v_0-v^{*}\parallel )d\theta dt\parallel \parallel v_0-v^{*}{\parallel }^2\hfill \\ & \le & {\displaystyle \frac{L_1}{2(1-\frac{L_0}{2}(3(1+\frac{2L_3}{3(1-L_0\parallel v_0-v^{*}\parallel )})+1)\parallel v_0-v^{*}\parallel )}}\hfill \\ & & \times ({\displaystyle \frac{3}{4}}\parallel v_0-v^{*}\parallel +\parallel w_0-v^{*}\parallel +{\displaystyle \frac{1}{2}}\parallel v_0-w_0\parallel )\parallel v_0-v^{*}{\parallel }^2\hfill \\ & \le & {\displaystyle \frac{L_1}{2(1-\frac{L_0}{2}(3(1+\frac{2L_3}{3(1-L_0\parallel v_0-v^{*}\parallel )})+1)\parallel v_0-v^{*}\parallel )}}\hfill \\ & & \times ({\displaystyle \frac{5}{4}}\parallel v_0-v^{*}\parallel +{\displaystyle \frac{3}{2}}\parallel w_0-v^{*}\parallel )\parallel v_0-v^{*}{\parallel }^2.\hfill \end{array}$$

So, by using (16) and (17), we have

$$\begin{array}{ccc}\hfill \parallel I_4\parallel & \le & {\displaystyle \frac{L_1}{2(1-\frac{L_0}{2}(3(1+\frac{2L_3}{3(1-L_0\parallel v_0-v^{*}\parallel )})+1)\parallel v_0-v^{*}\parallel )}}\hfill \\ & & \times \left({\displaystyle \frac{5}{4}}+{\displaystyle \frac{3}{2}}(1+{\displaystyle \frac{2L_3}{3(1-L_0\parallel v_0-v^{*}\parallel )}})\right){\parallel v_0-v^{*}\parallel }^3\hfill \\ & =& {\displaystyle \frac{L_1(11-11L_0\parallel v_0-v^{*}\parallel +4L_3)}{8(1-\frac{L_0}{2}(3(1+\frac{2L_3}{3(1-L_0\parallel v_0-v^{*}\parallel )})+1)\parallel v_0-v^{*}\parallel \left)\right(1-L_0\parallel v_0-v^{*}\parallel )}}{\parallel v_0-v^{*}\parallel }^3.\hfill \end{array}$$

(23)

Next, we shall obtain an estimate for $\parallel I_5\parallel .$ Observe that

$$\begin{array}{ccc}\hfill \parallel I_5\parallel & =& \parallel P{\mathcal{A}}^{\prime }\left(v^{*}\right){\int }_0^1{\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}{S}_3\left(\theta \right)d\theta \hfill \\ & & \times \left[{\left({\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}{\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*}))d\tau \right)}^2-I\right]{(v_0-v^{*})}^2\parallel \hfill \\ & \le & \parallel P{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel \parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}{S}_3\left(\theta \right)\parallel \hfill \\ & & \times ∥\left[{\left({\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}{\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*}))d\tau \right)}^2-I\right]{(v_0-v^{*})}^2∥\hfill \\ & \le & \parallel P{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel \parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}{S}_3\left(\theta \right)\parallel \hfill \\ & & \times ∥{\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\left[{\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*}))d\tau \right.\hfill \\ & & \times \left.{\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}{\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*}))d\tau -{\mathcal{A}}^{\prime }\left(v_0\right)\right]{(v_0-v^{*})}^2∥\hfill \end{array}$$

$$\begin{array}{ccc}& \le & \parallel P{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel \parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}{S}_3\left(\theta \right)\parallel \hfill \\ & & \times ∥{\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\left[{\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*}))d\tau -{\mathcal{A}}^{\prime }\left(v_0\right)\right]\hfill \\ & & +{\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}{\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*}))d\tau \hfill \\ & & \times \left[{\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}{\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*}))d\tau -I\right]{(v_0-v^{*})}^2∥\hfill \\ & \le & \parallel P{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel \parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}{S}_3\left(\theta \right)\parallel \hfill \\ & & \times ∥{\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\left[{\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*}))d\tau -{\mathcal{A}}^{\prime }\left(v_0\right)\right]\hfill \\ & & +{\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}{\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*}))d\tau \hfill \\ & & \times {\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\left[{\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*}))d\tau -{\mathcal{A}}^{\prime }\left(v_0\right)\right]{(v_0-v^{*})}^2∥\hfill \\ & \le & \parallel P{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel \parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}{S}_3\left(\theta \right)\parallel \parallel {\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel \hfill \\ & & \times \left[{\int }_0^1\parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}({\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*}))-{\mathcal{A}}^{\prime }\left(v_0\right))\parallel d\tau \right.\hfill \\ & & +{\int }_0^1\parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}{\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*}))\parallel d\tau \parallel {\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel \hfill \\ & & \left.\times {\int }_0^1\parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}({\mathcal{A}}^{\prime }(v^{*}+\tau (v_0-v^{*}))-{\mathcal{A}}^{\prime }\left(v_0\right))\parallel d\tau \right]{\parallel v_0-v^{*}\parallel }^2.\hfill \end{array}$$

Therefore, by (9), (A1)-(A4), we have

$$\begin{array}{ccc}\hfill \parallel I_5\parallel & \le & {\displaystyle \frac{L_2}{2(1-\frac{L_0}{2}(3(1+\frac{2L_3}{3(1-L_0\parallel v_0-v^{*}\parallel )})+1)\parallel v_0-v^{*}\parallel \left)\right(1-L_0\parallel v_0-v^{*}\parallel )}}\hfill \\ & & \times ({\displaystyle \frac{L_0}{2}}+{\displaystyle \frac{L_3{L}_0}{2(1-L_0\parallel v_0-v^{*}\parallel )}}){\parallel v_0-v^{*}\parallel }^3\hfill \\ & =& {\displaystyle \frac{1}{4(1-\frac{L_0}{2}(3(1+\frac{2L_3}{3(1-L_0\parallel v_0-v^{*}\parallel )})+1)\parallel v_0-v^{*}\parallel \left)\right(1-L_0\parallel v_0-v^{*}{\parallel )}^2}}\hfill \\ & & \times L_2{L}_0(1-L_0\parallel v_0-v^{*}\parallel +L_3)\parallel v_0-v^{*}{\parallel }^3.\hfill \end{array}$$

(24)

Thus, from (14)-(24), we have

$$\begin{array}{ccc}\hfill \parallel v_1-v^{*}\parallel & \le & g_1(\parallel v_0-v^{*}\parallel )\parallel v_0-v^{*}{\parallel }^3.\hfill \end{array}$$

Therefore, the iterate $v_1\in B(v^{*},r)$ because $g_1(\parallel v_0-v^{*}\parallel )\parallel v_0-v^{*}{\parallel }^3\le g_1\left(r\right)r^2\parallel v_0-v^{*}\parallel \le \parallel v_0-v^{*}\parallel \le r.$

Simply replace $v_0,w_0,v_1$ in the preceding arguments by $v_n,w_n,v_{n+1}$ to complete the induction for (7). □

Theorem 2.

The method defined by (3) exhibits a convergence order of 3.

Proof. The proof follows a similar argument to that of Theorem 3 in [6]. However, we include it here for completeness. Let $e_n=\parallel v_n-v^{*}\parallel .$ Let q be maximal such that for some $C>0,$

$$\underset{n⟶\infty }{lim}{\displaystyle \frac{e_{n+1}}{e_n^q}}=C.$$

(25)

Then, since $e_n<r<1,$ by (5) (for large enough n), we have

$$e_{n+1}\approx g_1\left(r\right)e_n^3.$$

(26)

So, by (25) and (26), we get

$${\displaystyle \frac{e_{n+1}}{e_n^3}}\approx g_1\left(r\right).$$

Thus, by (25), we get

$$q=3\mathrm{and}C=g_1\left(r\right).$$

Thus convergence order $q=3.$

□

3. Order of Convergence(OC) of (2):

This section examines the OC of method (2). For our analysis we require some more CNF:

Let ${\mathsf{\Theta }}_2,H_2:[0,\lambda )⟶\mathbb{R}$ defined by

$${\mathsf{\Theta }}_2\left(t\right)=L_0{g}_1\left(t\right)t^3$$

and

$$H_2\left(t\right)={\mathsf{\Theta }}_2\left(t\right)-1.$$

Given that $H_2\left(0\right)=-1$ and $H_2\left(t\right)⟶\infty$ and $t⟶{\lambda }^{-}$, we can conclude that the equation $H_2\left(t\right)=0$ possesses a smallest positive solution within $[0,\lambda )$. This solution is denoted as ${\lambda }_2.$

Let $g_2,h_2:[0,{\lambda }_2)⟶\mathbb{R}$ be CNF defined by

$$g_2\left(t\right)={\displaystyle \frac{L_0}{2(1-L_0{g}_1\left(t\right)t^3)}}{g}_1{\left(t\right)}^2$$

and

$$h_2\left(t\right)=g_2\left(t\right)t^5-1.$$

Then, $h_2\left(0\right)=-1$ and $h_2\left(t\right)⟶\infty$ as $t⟶{\lambda }_2^{-}.$ Therefore $h_2\left(t\right)=0$ has a smallest positive solution in $[0,{\lambda }_2)$ denoted by ${\lambda }_3.$

Let

$$R=min\{{\lambda }_3,{\displaystyle \frac{1}{2L_0}},1\}.$$

(27)

Then, for all $t\in [0,R],$

$$0\le g_2\left(t\right)t^5<1.$$

(28)

Theorem 3.

Assuming (A1)-(A4) are true, the sequence $\left\{v_n\right\}$ given by (2) with initial value $v_0\in B(v^{*},R)$ converges to $v^{*}$, and the following estimate is valid:

$$\parallel v_{n+1}-v^{*}\parallel \le g_2\left(R\right){\parallel v_n-v^{*}\parallel }^6.$$

(29)

Proof. Adopting the same proof strategy as in Theorem 1, we find that:

$$\parallel z_n-v^{*}\parallel \le g_1(\parallel v_n-v^{*}\parallel )\parallel v_n-v^{*}{\parallel }^3.$$

(30)

Note that by (10) and (A1)

$$\begin{array}{ccc}\hfill \parallel v_{n+1}-v^{*}\parallel & =& \parallel z_n-v^{*}-{\mathcal{A}}^{\prime }{\left(z_n\right)}^{-1}\mathcal{A}\left(z_n\right)\parallel \hfill \\ & =& \parallel {\mathcal{A}}^{\prime }{\left(z_n\right)}^{-1}({\mathcal{A}}^{\prime }\left(z_n\right)-{\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+t(z_n-v^{*}))dt)(z_n-v^{*})\parallel \hfill \\ & =& \parallel {\mathcal{A}}^{\prime }{\left(z_n\right)}^{-1}{\mathcal{A}}^{\prime }\left(v^{*}\right){\int }_0^1{\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}({\mathcal{A}}^{\prime }\left(z_n\right)-{\mathcal{A}}^{\prime }(v^{*}+t(z_n-v^{*})))dt(z_n-v^{*})\parallel \hfill \\ & \le & {\displaystyle \frac{L_0}{2(1-L_0\parallel z_n-v^{*}\parallel )}}{\parallel z_n-v^{*}\parallel }^2\hfill \\ & \le & {\displaystyle \frac{L_0}{2(1-L_0{g}_1(\parallel v_n-v^{*}\parallel )\parallel v_n-v^{*}{\parallel }^3)}}{g}_1(\parallel v_n-v^{*}{\parallel )}^2{\parallel v_n-v^{*}\parallel }^6\hfill \\ & \le & g_2\left(R\right){\parallel v_n-v^{*}\parallel }^6.\hfill \end{array}$$

Now since $g_2\left(R\right)\parallel v_n-v^{*}{\parallel }^6\le g_2\left(R\right)R^5\parallel v_n-v^{*}\parallel \le \parallel v_n-v^{*}\parallel \le R,$ the iterate $v_{n+1}\in B(v^{*},R).$

□

Theorem 4.

The method defined by (2) exhibits a convergence order of $6.$

Proof. Employing a proof strategy analogous to that of Theorem 2.

□

4. Order of Convergence(OC) of (4):

We analyze the OC of method (4) in this section. We require some more CNFs as in previous sections:

Let $g_3,h_3:[0,{\displaystyle \frac{\sqrt{3}-1}{L_0}})⟶\mathbb{R}$ be CNFs defined by

$$g_3\left(t\right)={\displaystyle \frac{L_0}{2(1-\frac{L_0^2{t}^2}{2(1-L_{0}t)})}}\left[{\displaystyle \frac{L_0}{1-L_{0}t}}+g_1\left(t\right)t\right]g_1\left(t\right)$$

and

$$h_3\left(t\right)=g_3\left(t\right)t^4-1.$$

Then, $h_3\left(0\right)=-1$ and $h_3\left(t\right)⟶\infty$ as $t⟶{{\displaystyle \frac{\sqrt{3}-1}{L_0}}}^{-}.$ Therefore $h_3\left(t\right)=0$ has a smallest positive solution in $[0,{\displaystyle \frac{\sqrt{3}-1}{L_0}})$ denoted by ${\lambda }_4.$

Let

$$R_1=min\{{\lambda }_4,{\displaystyle \frac{1}{2L_0}},1\}.$$

(31)

Then, for all $t\in [0,R_1],$

$$0\le g_3\left(t\right)t^4<1.$$

(32)

Theorem 5.

Assuming (A1)-(A4) are true, the sequence $\left\{v_n\right\}$ given by (4) with initial value $v_0\in B(v^{*},R_1)$ converges to $v^{*}.$, and the following estimate is valid:

$$\parallel v_{n+1}-v^{*}\parallel \le g_3\left(R_1\right){\parallel v_n-v^{*}\parallel }^5.$$

(33)

Proof. In imitation of the proof presented for Theorem 1, we obtain:

$$\parallel z_n-v^{*}\parallel \le g_1(\parallel v_n-v^{*}\parallel )\parallel v_n-v^{*}{\parallel }^3.$$

(34)

Note that by (10) and (A1)

$$\begin{array}{ccc}\hfill \parallel v_{n+1}-v^{*}\parallel & =& ∥z_n-v^{*}-{\mathcal{A}}^{\prime }{\left({\displaystyle \frac{3w_n-v_n}{2}}\right)}^{-1}\mathcal{A}\left(z_n\right)∥\hfill \\ & =& ∥{\mathcal{A}}^{\prime }{\left({\displaystyle \frac{3w_n-v_n}{2}}\right)}^{-1}({\mathcal{A}}^{\prime }\left({\displaystyle \frac{3w_n-v_n}{2}}\right)\hfill \\ & & -{\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+t(z_n-v^{*}))dt)(z_n-v^{*})∥\hfill \\ & =& ∥{\mathcal{A}}^{\prime }{\left({\displaystyle \frac{3w_n-v_n}{2}}\right)}^{-1}{\mathcal{A}}^{\prime }\left(v^{*}\right)\hfill \\ & & \times {\int }_0^1{\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}({\mathcal{A}}^{\prime }\left({\displaystyle \frac{3w_n-v_n}{2}}\right)-{\mathcal{A}}^{\prime }(v^{*}+t(z_n-v^{*})))dt(z_n-v^{*})∥\hfill \\ & \le & {\displaystyle \frac{L_0}{1-L_0\parallel \frac{3w_n-v_n}{2}-v^{*}\parallel }}\left[\parallel {\displaystyle \frac{3w_n-v_n}{2}}-v^{*}\parallel +{\displaystyle \frac{1}{2}}\parallel z_n-v^{*}\parallel \right]\parallel z_n-v^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{L_0}{2(1-L_0(\parallel v_n-v^{*}-{\mathcal{A}}^{\prime }{\left(v_n\right)}^{-1}\mathcal{A}\left(v_n\right)\parallel \left)\right)}}\hfill \\ & & \times \left[2\parallel v_n-v^{*}-{\mathcal{A}}^{\prime }{\left(v_n\right)}^{-1}\mathcal{A}\left(v_n\right)\parallel +\parallel z_n-v^{*}\parallel \right]\parallel z_n-v^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{L_0}{2(1-\frac{L_0^2}{2(1-L_0\parallel v_n-v^{*}\parallel )}\parallel v_n-v^{*}{\parallel }^2)}}\hfill \\ & & \times \left[{\displaystyle \frac{L_0}{1-L_0\parallel v_n-v^{*}\parallel }}\parallel v_n-v^{*}{\parallel }^2+\parallel z_n-v^{*}\parallel \right]\parallel z_n-v^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{L_0}{2(1-\frac{L_0^2}{2(1-L_0\parallel v_n-v^{*}\parallel )}\parallel v_n-v^{*}{\parallel }^2)}}\hfill \\ & & \times \left[{\displaystyle \frac{L_0}{1-L_0\parallel v_n-v^{*}\parallel }}+g_1(\parallel v_n-v^{*}\parallel )\parallel v_n-v^{*}\parallel \right]g_1(\parallel v_n-v^{*}\parallel )\parallel v_n-v^{*}{\parallel }^5\hfill \\ & \le & g_3\left(R_1\right){\parallel v_n-v^{*}\parallel }^5.\hfill \end{array}$$

Here, we used the inequality

$$\begin{array}{ccc}& & \parallel v_n-v^{*}-{\mathcal{A}}^{\prime }{\left(v_n\right)}^{-1}\mathcal{A}\left(v_n\right)\parallel \hfill \\ & =& \parallel {\mathcal{A}}^{\prime }{\left(v_n\right)}^{-1}({\mathcal{A}}^{\prime }\left(v_n\right)-{\int }_0^1\mathcal{A}(v^{*}+t(v_n-v^{*}))dt\parallel \parallel v_n-v^{*}\parallel \hfill \\ & \le & \parallel {\mathcal{A}}^{\prime }{\left(v_n\right)}^{-1}{\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel \parallel {\int }_0^1{\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}[\mathcal{A}(v^{*}+t(v_n-v^{*}))-{\mathcal{A}}^{\prime }\left(v_n\right)]dt\parallel \parallel v_n-v^{*}\parallel \hfill \\ & \le & {\displaystyle \frac{L_0}{2(1-L_0\parallel v_n-v^{*}\parallel )}}{\parallel v_n-v^{*}\parallel }^2.\hfill \end{array}$$

Now since $g_3\left(R_1\right)\parallel v_n-v^{*}{\parallel }^5\le g_3\left(R_1\right)R_1^4\parallel v_n-v^{*}\parallel \le \parallel v_n-v^{*}\parallel \le R_1,$ the iterate $v_{n+1}\in B(v^{*},R_1).$

□

Theorem 6.

The method defined by (4) exhibits a convergence order of $5.$

Proof. Resembling the proof of Theorem 2.

□

The subsequent result addresses the uniqueness property of the solutions derived from the methods (3), (2), and (4).

Theorem 7.

Suppose Assumption (A1) holds and the equation $\mathcal{A}\left(v\right)=0$, has a simple solution $v^{*}$. Then, for the equation $\mathcal{A}\left(v\right)=0$ the only solution in the set $E_1=E\cap B[v^{*},\rho ]$ is $v^{*}$ provided that

$$L_0<{\displaystyle \frac{2}{\rho }}.$$

(35)

Proof. Suppose $c\in E_1$ is such that $\mathcal{A}\left(c\right)=0.$ Define the operator $N:={\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+\gamma (c-v^{*}))d\gamma .$ Then by Assumption (A1) and (35), we have

$$\begin{array}{ccc}\hfill \parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}(N-{\mathcal{A}}^{\prime }\left(v^{*}\right))\parallel & \le & L_0{\int }_0^1\parallel v^{*}+\gamma (c-v^{*})-v^{*}\parallel d\gamma \hfill \\ & \le & L_0{\int }_0^1\gamma \parallel v^{*}-c\parallel d\gamma \hfill \\ & \le & {\displaystyle \frac{L_0}{2}}\rho <1.\hfill \end{array}$$

So by BL, N is invertible and hence we get $c=v^{*}$ from the identity $0=\mathcal{A}\left(c\right)-\mathcal{A}\left(v^{*}\right)=N(c-v^{*}).$

5. Convergence Under Generalized Conditions

The applicability of method (3) and the method (4) can be extended. Notice that the second condition (A2) can can be violated easily even for simple scalar functions. Define the function

$$\mathcal{A}\left(t\right)=\left\{\begin{array}{cc}{t}^{2}logt+5t^5-5t^4,& t\ne 0\\ 0,& t=0.\end{array}\right.$$

Since $v^{*}=1$ and ${\mathcal{A}}^{\prime \prime }\left(t\right)$ is discontinuous at $t=0.$, condition (A2) is violated in any neighborhood containing 0 and 1. This necessitates a convergence analysis based on generalized conditions and the operators inherent to the methods.

First the local convergence is considered under some conditions. Set $T=[0,+\infty ).$

Presume:

(H1)

Consider a CNF ${\psi }_0:T⟶T$ for which the smallest positive solution to ${\psi }_0\left(t\right)-1=0$ is ${\rho }_0.$. Let $T_0$ be the interval $[0,{\rho }_0).$.

(H2)

Let ${\rho }_1\in T_0-\left\{0\right\},$ be the SPS of ${\sigma }_1\left(t\right)-1=0$, where the function ${\sigma }_1:T_0⟶T$ is given by

$${\sigma }_1\left(t\right)={\displaystyle \frac{{\int }_0^1\psi \left((1-\theta )t\right)d\theta +\frac{1}{3}(1+{\int }_0^1{\psi }_0\left(\theta t\right)d\theta )}{1-{\psi }_0\left(t\right)}}$$

for some CNF $\psi :T_0⟶T.$

(H3)

The equation $p\left(t\right)-1=0$ has a SPS denoted by ${\rho }_p\in T_0-\left\{0\right\},$ where $p:T_0⟶T$ is given by

$$p\left(t\right)={\displaystyle \frac{3}{2}}\psi \left(({\sigma }_1\left(t\right)+1)t\right)+{\psi }_0\left(t\right).$$

Let $T_1=[0,{\rho }_p).$

(H4)

The equation ${\sigma }_2\left(t\right)-1=0$ has a SPS denoted by ${\rho }_2\in T_1-\left\{0\right\},$ where ${\sigma }_2:T_1⟶T$ is given by

$$\begin{array}{ccc}\hfill {\sigma }_2\left(t\right)& =& {\displaystyle \frac{{\int }_0^1\psi \left((1-\theta )t\right)d\theta }{1-{\psi }_0\left(t\right)}}\hfill \\ & & +{\displaystyle \frac{3\overline{\psi }\left(t\right)(1+{\int }_0^1{\psi }_0\left(\theta t\right)d\theta )}{4(1-p\left(t\right))(1-{\psi }_0\left(t\right))}},\hfill \end{array}$$

where

$$\overline{\psi }\left(t\right)=\left\{\begin{array}{c}\psi \left((1+{\sigma }_1\left(t\right))t\right)\\ {\psi }_0\left({\sigma }_1\left(t\right)t\right)+{\psi }_0\left(t\right).\end{array}\right.$$

(H5)

The equation $q\left(t\right)-1=0$ has a SPS denoted by ${\rho }_q\in T_1-\left\{0\right\},$ here $q:T_1⟶T$ is given by

$$q\left(t\right)={\psi }_0\left({\displaystyle \frac{(3{\sigma }_1\left(t\right)+1)t}{2}}\right).$$

Let $T_2=[0,{\rho }_q).$

(H6)

The equation ${\sigma }_3\left(t\right)-1=0$ has a SPS denoted by ${\rho }_3,$ where ${\sigma }_3:T_2⟶T$ is given as

$$\begin{array}{ccc}\hfill {\sigma }_3\left(t\right)& =& \left[{\displaystyle \frac{{\int }_0^1\psi \left((1-\theta ){\sigma }_2\left(t\right)t\right)d\theta }{1-{\psi }_0\left({\sigma }_2\left(t\right)t\right)}}\right.\hfill \\ & & +\left.{\displaystyle \frac{3\overline{\overline{\psi }}\left(t\right)(1+{\int }_0^1{\psi }_0\left(\theta {\sigma }_2\left(t\right)t\right)d\theta )}{2(1-q\left(t\right))(1-{\psi }_0\left({\sigma }_2\left(t\right)t\right))}}\right],\hfill \end{array}$$

where

$$\overline{\overline{\psi }}\left(t\right)={\displaystyle \frac{q\left(t\right)+{\psi }_0\left({\sigma }_2\left(t\right)t\right)}{1-q\left(t\right)}}$$

Let

$$\rho *=min\left\{{\rho }_i\right\},i=1,2,3.$$

(36)

The developed functions ${\psi }_0$ and $\psi$ relate to the operators on the method (4).

(H7)

There exist an invertible linear operator L and $v^{*}\in E$ solving the equation $\mathcal{A}\left(x\right)=0$ such that for each $x\in E$

$$\parallel L^{-1}({\mathcal{A}}^{\prime }\left(x\right)-L)\parallel \le {\psi }_0(\parallel x-v^{*}\parallel ).$$

Notice that under condition (H1) and (36)

$$\parallel L^{-1}({\mathcal{A}}^{\prime }\left(v^{*}\right)-L)\parallel \le {\psi }_0\left(0\right)<1.$$

Thus ${\mathcal{A}}^{\prime }\left(v^{*}\right)$ is invertible. Let $E_1=E\cap B(v^{*},{\rho }_0).$

(H8)

$\parallel L^{-1}({\mathcal{A}}^{\prime }\left(y\right)-{\mathcal{A}}^{\prime }\left(x\right))\parallel \le \psi (\parallel y-x\parallel )$ for each $x,y\in E_1.$

and

(H9)

$B[v^{*},\rho *]\subset E.$

The main local analysis for the method (4) follows in the next result.

Theorem 8.

Let the conditions (H1)-(H9) hold. Then, the following assertions are satisfied provided that $v_0\in B(v^{*},\rho *)-\left\{v^{*}\right\}$

$$\left\{v_n\right\}\subset B(v^{*},\rho *),$$

(37)

$$\parallel w_n-v^{*}\parallel \le {\sigma }_1(\parallel v_n-v^{*}\parallel )\parallel v_n-v^{*}\parallel \le \parallel v_n-v^{*}\parallel <\rho *,$$

(38)

$$\parallel z_n-v^{*}\parallel \le {\sigma }_2(\parallel v_n-v^{*}\parallel )\parallel v_n-v^{*}\parallel \le \parallel v_n-v^{*}\parallel ,$$

(39)

$$\parallel v_{n+1}-v^{*}\parallel \le {\sigma }_3(\parallel v_n-v^{*}\parallel )\parallel v_n-v^{*}\parallel \le \parallel v_n-v^{*}\parallel$$

(40)

and ${lim}_{n⟶\infty }{v}_n=v^{*},$ where the functions ${\sigma }_i$ are provided previously and the radius $\rho *$ is defined by the formula (36).

Proof. Let $T^{*}=[0,\rho *).$ It follows that for each $t\in T^{*}$

$$0\le {\psi }_0\left(t\right)<1,$$

(41)

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

(42)

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

(43)

and

$$0\le {\sigma }_i\left(t\right)<1.$$

(44)

The assertions (37)-(40) are shown by induction. Let $u\in E^{*}=B(v^{*},\rho *)$ but be arbitrary. The condition (H1) and the formula (36) give

$$\parallel L^{-1}({\mathcal{A}}^{\prime }\left(u\right)-L)\parallel \le {\psi }_0(\parallel u-v^{*}\parallel )\le {\psi }_0(\rho *)<1.$$

Thus, ${\mathcal{A}}^{\prime }\left(u\right)$ is invertible,

$$\parallel {\mathcal{A}}^{\prime }{\left(u\right)}^{-1}L\parallel \le {\displaystyle \frac{1}{1-{\psi }_0(\parallel u-v^{*}\parallel )}}$$

(45)

and the iterate $w_0$ exists by the method (4) if $n=0.$ Moreover, the first substep gives

$$\begin{array}{ccc}\hfill w_0-v^{*}& =& v_0-v^{*}-{\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\mathcal{A}\left(v_0\right)+{\displaystyle \frac{1}{3}}{\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\mathcal{A}\left(v_0\right)\hfill \\ & =& {\int }_0^1{\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}[{\mathcal{A}}^{\prime }\left(v_0\right)-{\mathcal{A}}^{\prime }(v^{*}+\theta (v_0-v^{*}))]d\theta (v_0-v^{*})\hfill \\ & & +{\displaystyle \frac{1}{3}}{\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\mathcal{A}\left(v_0\right).\hfill \end{array}$$

(46)

Using (36), (44) (for $i=1$), (H8), (45) and (46)

$$\begin{array}{ccc}\hfill \parallel w_0-v^{*}\parallel & \le & {\displaystyle \frac{1}{1-{\psi }_0(\parallel v_0-v^{*}\parallel )}}\left[{\int }_0^1\psi ((1-\theta )\parallel v_0-v^{*}\parallel )d\theta \parallel v_0-v^{*}\parallel \right.\hfill \\ & & +\left.{\displaystyle \frac{1}{3}}(1+{\int }_0^1{\psi }_0(\theta \parallel v_0-v^{*}\parallel \left)d\theta \right)\parallel v_0-v^{*}\parallel \right]\hfill \\ & \le & {\sigma }_1(\parallel v_0-v^{*}\parallel )\parallel v_0-v^{*}\parallel \le \parallel v_0-v^{*}\parallel <\rho *.\hfill \end{array}$$

(47)

Thus, the iterate $w_0\in E^{*}$ and the item (38) holds if $n=0.$

The following estimate establishes the invertability of the linear operator $(3{\mathcal{A}}^{\prime }\left(w_0\right)-\mathcal{A}\left(v_0\right))$ and iterate $z_0$ by the second substep of the method (4):

$$\begin{array}{ccc}& & {\parallel \left(2L\right)}^{-1}(3{\mathcal{A}}^{\prime }\left(w_0\right)-{\mathcal{A}}^{\prime }\left(v_0\right)-2L)\parallel \hfill \\ & \le & {\displaystyle \frac{1}{2}}[3\parallel L^{-1}({\mathcal{A}}^{\prime }\left(w_0\right)-{\mathcal{A}}^{\prime }\left(v_0\right))\parallel +\parallel L^{-1}({\mathcal{A}}^{\prime }\left(v_0\right)-L)\parallel \hfill \\ & \le & {\displaystyle \frac{1}{2}}\left(3\psi \right(\parallel w_0-v_0\parallel )+2{\psi }_0(\parallel v_0-v^{*}\parallel \left)\right)\hfill \\ & \le & p(\parallel v_0-v^{*}\parallel )<1,\hfill \end{array}$$

(48)

where we used the conditions (H3), (H7), formulas (36), (42) and (37). Hence, by (48)

$$\parallel {(3{\mathcal{A}}^{\prime }\left(w_0\right)-{\mathcal{A}}^{\prime }\left(v_0\right))}^{-1}L\parallel \le {\displaystyle \frac{1}{2(1-p(\parallel v_0-v^{*}\parallel \left)\right)}}.$$

(49)

Moreover, the second substep gives

$$\begin{array}{ccc}\hfill z_0-v^{*}& =& v_0-v^{*}-{\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\mathcal{A}\left(v_0\right)\hfill \\ & & +[I-{\displaystyle \frac{1}{2}}{(3{\mathcal{A}}^{\prime }\left(w_0\right)-{\mathcal{A}}^{\prime }\left(v_0\right))}^{-1}(3{\mathcal{A}}^{\prime }\left(w_0\right)+{\mathcal{A}}^{\prime }\left(v_0\right))]{\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\mathcal{A}\left(v_0\right).\hfill \end{array}$$

(50)

It follows by (36), (44) (for $i=2$), (45), (47), (49) and (50)

$$\begin{array}{ccc}\hfill \parallel z_0-v^{*}\parallel & \le & \left[{\displaystyle \frac{{\int }_0^1\psi ((1-\theta )\parallel v_0-v^{*}\parallel )d\theta }{1-{\psi }_0(\parallel v_0-v^{*}\parallel )}}\right.\hfill \\ & & +\left.{\displaystyle \frac{3\psi \left(\right(1+{\sigma }_1(\parallel v_0-v^{*}\parallel \left)\right)\parallel v_0-v^{*}\parallel \left)\right(1+{\int }_0^1{\psi }_0(\theta \parallel v_0-v^{*}\parallel \left)d\theta \right)}{4(1-p(\parallel v_0-v^{*}\parallel \left)\right)(1-{\psi }_0(\parallel v_0-v^{*}\parallel \left)\right)}}\right]\parallel v_0-v^{*}\parallel \hfill \\ & \le & {\sigma }_2(\parallel v_0-v^{*}\parallel )\parallel v_0-v^{*}\parallel \le \parallel v_0-v^{*}\parallel .\hfill \end{array}$$

(51)

Thus, the iterate $z_0\in E*$ and for $n=0$ the assertion (39) holds. Next the invertability of the linear operator ${\mathcal{A}}^{\prime }\left({\displaystyle \frac{3w_0-v_0}{2}}\right)$ establishes the existence of the iterate $v_1$ as follows:

$$\begin{array}{ccc}\hfill \parallel L^{-1}({\mathcal{A}}^{\prime }\left({\displaystyle \frac{3w_0-v_0}{2}}\right)-L)\parallel & \le & {\psi }_0\left(\parallel {\displaystyle \frac{3w_0-v_0}{2}}-v^{*}\parallel \right)\hfill \\ & \le & {\psi }_0\left({\displaystyle \frac{2\parallel w_0-v^{*}\parallel +\parallel w_0-v^{*}\parallel +\parallel v_0-v^{*}\parallel }{2}}\right)=q(\parallel v_0-v^{*}\parallel )<1,\hfill \\ & & \left(\mathrm{by}\right(36\left)\mathrm{and}\right(43\left)\right),\hfill \end{array}$$

so

$$\parallel {\mathcal{A}}^{\prime }{\left({\displaystyle \frac{3w_0-v_0}{2}}\right)}^{-1}L\parallel \le {\displaystyle \frac{1}{1-q(\parallel v_0-v^{*}\parallel )}}.$$

(52)

Then, the last substep of the method (4) gives in turn

$$\begin{array}{ccc}\hfill v_1-v^{*}& =& z_0-v^{*}-{\mathcal{A}}^{\prime }{\left(z_0\right)}^{-1}\mathcal{A}\left(z_0\right)\hfill \\ & & +({\mathcal{A}}^{\prime }{\left(z_0\right)}^{-1}-{\mathcal{A}}^{\prime }{\left({\displaystyle \frac{3w_0-v_0}{2}}\right)}^{-1})\mathcal{A}\left(z_0\right)\hfill \\ & =& z_0-v^{*}-{\mathcal{A}}^{\prime }{\left(z_0\right)}^{-1}\mathcal{A}\left(z_0\right)+{\mathcal{A}}^{\prime }{\left(z_0\right)}^{-1}\left({\mathcal{A}}^{\prime }\left({\displaystyle \frac{3w_0-v_0}{2}}\right)\right.\hfill \\ & & -\left.{\mathcal{A}}^{\prime }\left(z_0\right)\right){\mathcal{A}}^{\prime }{\left({\displaystyle \frac{3w_0-v_0}{2}}\right)}^{-1}\mathcal{A}\left(z_0\right).\hfill \end{array}$$

(53)

Using (36), (H8), (44) (for $i=3$), (51), (52) and (53)

$$\begin{array}{ccc}\hfill \parallel v_1-v^{*}\parallel & \le & \left[{\displaystyle \frac{{\int }_0^1\psi ((1-\theta )\parallel z_0-v^{*}\parallel )d\theta }{1-{\psi }_0(\parallel z_0-v^{*}\parallel )}}\right.\hfill \\ & & +\left.{\displaystyle \frac{\overline{\overline{\psi }}(\parallel z_0-v^{*}\parallel \left)\right(1+{\int }_0^1{\psi }_0(\theta \parallel z_0-v^{*}\parallel \left)d\theta \right)}{(1-{\psi }_0(\parallel z_0-v^{*}\parallel \left)\right)}}\right]\parallel z_0-v^{*}\parallel \hfill \\ & \le & {\sigma }_3(\parallel v_0-v^{*}\parallel )\parallel v_0-v^{*}\parallel \le \parallel v_0-v^{*}\parallel ,\hfill \end{array}$$

(54)

Hence, the iterate $v_1\in E^{*}$ and the assertion (40) holds for $n=0.$ The induction is terminated if $v_k,w_k,z_k,v_{k+1}$ replaces $v_0,w_0,z_0,v_1$ in the preceding calculations. Finally, from the estimate

$$\parallel v_{k+1}-v^{*}\parallel \le c\parallel v_k-v^{*}\parallel <\rho *,$$

(55)

where $c={\sigma }_3(\parallel v_0-v^{*}\parallel )\in [0,1).$ It follows ${lim}_{k⟶\infty }{v}_k=v^{*}$ and the iterate $v_{k+1}\in E^{*}.$

□

The isolation of the solution $v^{*}$ is discussed in the next result.

Proposition 1.

Suppose: there exists a solution $\overline{v}\in B(v^{*},{\rho }_4)$ for some ${\rho }_4>0;$ the condition (H7) holds in the ball $B(v^{*},{\rho }_4)$ and there exists ${\rho }_5\ge {\rho }_4$ such that

$${\int }_0^1{\psi }_0\left(\theta {\rho }_5\right)d\theta <1.$$

(56)

Let $E_2=E\cap B[v^{*},{\rho }_5].$

Then, the equation $\mathcal{A}\left(v\right)=0$ is uniquelly solvable by $v^{*}$ in the region $E_2.$

Proof. Define the linear operator $L_1={\int }_0^1{\mathcal{A}}^{\prime }(v^{*}+\theta (\overline{v}-v^{*}))d\theta .$ Then, by the condition (H7) in the ball $B(v^{*},{\rho }_4)$ and (56)

$$\parallel L(L_1-L)\parallel \le {\int }_0^1{\psi }_0(\theta \parallel \overline{v}-v^{*}\parallel )d\theta \le {\psi }_0\left({\rho }_5\right)<1.$$

Hence, $\overline{v}=v^{*}$ follows from the identity $\overline{v}-v^{*}=L_1^{-1}(\mathcal{A}\left(\overline{v}\right)-\mathcal{A}\left(v^{*}\right))=L_1^{-1}\left(0\right)=0.$ □

Remark 1.

(1) 

A possible choice for $L={\mathcal{A}}^{\prime }\left(v^{*}\right).$ In practice L shall be chosen to tighten the function ${\psi }_0.$ Notice also that it does not necessarily follow from (H7) that $v^{*}$ is a simple solution or that $\mathcal{A}$ is differentiable at $v^{*}.$

(2) 

The results for the method (3) are obtained by restriction to the first two substep of the method (4).

A analogous approach is followed in the semi-local analysis but the role of $v^{*}$ is exchanged by $v_0$ and that of function ${\psi }_0$ and $\psi$ by ${\phi }_0$ and $\phi ,$ respectively which are developed below.

Suppose:

(e1)

There exists CNF ${\phi }_0:T⟶T$ such that the equation ${\phi }_0\left(t\right)-1=0$ has a SPS denoted by ${\rho }_6\in T-\left\{0\right\}.$

Set $T_2=[0,{\rho }_6).$ Let $\phi :T_2⟶T$ be a CNF. Define the sequence $\left\{{\alpha }_n\right\}$ for ${\alpha }_0=0,{\beta }_0\ge 0$ and each $n=0,1,2,\dots$ by

$$\begin{array}{ccc}\hfill {\tilde{p}}_n& =& {\displaystyle \frac{3}{2}}\phi ({\beta }_n-{\alpha }_n)+{\phi }_0\left({\alpha }_n\right),\hfill \\ \hfill {\gamma }_n& =& {\beta }_n+{\displaystyle \frac{1}{8}}{\displaystyle \frac{(3\phi ({\beta }_n-{\alpha }_n)+4(1+{\phi }_0\left({\alpha }_n\right)))({\beta }_n-{\alpha }_n)}{1-{\tilde{p}}_n}},\hfill \\ \hfill {\tilde{\mu }}_n& =& (1+{\int }_0^1{\phi }_0({\alpha }_n+\theta ({\gamma }_n-{\alpha }_n))d\theta )({\gamma }_n-{\alpha }_n)\hfill \\ & & +{\displaystyle \frac{3}{2}}(1+{\phi }_0\left({\alpha }_n\right))({\beta }_n-{\alpha }_n),\hfill \\ \hfill {\tilde{q}}_n& =& {\phi }_0\left({\displaystyle \frac{3{\beta }_n-{\alpha }_n}{2}}\right),\hfill \\ \hfill {\alpha }_{n+1}& =& {\gamma }_n+{\displaystyle \frac{{\mu }_n}{1-{\tilde{q}}_n}},\hfill \\ \hfill {\xi }_{n+1}& =& (1+{\int }_0^1{\phi }_0({\alpha }_n+\theta ({\alpha }_{n+1}-{\alpha }_n))d\theta )({\alpha }_{n+1}-{\alpha }_n)\hfill \\ & & +{\displaystyle \frac{3}{2}}(1+{\phi }_0\left({\alpha }_n\right))({\beta }_n-{\alpha }_n),\hfill \end{array}$$

and

$${\beta }_{n+1}={\alpha }_{n+1}+{\displaystyle \frac{2}{3}}{\displaystyle \frac{{\xi }_{n+1}}{1-{\phi }_0\left({\alpha }_{n+1}\right)}}.$$

(e2)

There exists ${\rho }_7\in [0,{\rho }_6)$ such that for each $n=0,1,2,\dots$

$${\tilde{p}}_n<1,{\tilde{q}}_n<1,{\phi }_0\left({\alpha }_n\right)<1\mathrm{and}{\alpha }_n\le {\rho }_7.$$

It follows that $0\le {\alpha }_n\le {\beta }_n\le {\gamma }_n\le {\alpha }_{n+1}\le {\rho }_7$ and there exists ${\rho }_8\in [0,{\rho }_7)$ such that ${lim}_{n⟶\infty }{\alpha }_n={\rho }_8.$

The functions ${\phi }_0$ and $\phi$ are connected to the operators on the method (4).

(e3)

There exists $v_0\in E$ such that

$$\parallel L^{-1}({\mathcal{A}}^{\prime }\left(v\right)-L)\parallel \le {\phi }_0(\parallel v-v_0\parallel ).$$

Let $E_3=E\cap B(v^{*},{\rho }_6).$ Notice that (e1) and (e3) imply that the linear operator ${\mathcal{A}}^{\prime }\left(v_0\right)$ is invertible. Let $\parallel {\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\mathcal{A}\left(v_0\right)\parallel \le {\displaystyle \frac{3}{2}}{\beta }_0.$

(e4)

$\parallel L^{-1}({\mathcal{A}}^{\prime }\left(w\right)-{\mathcal{A}}^{\prime }\left(v\right))\parallel \le \phi (\parallel w-v\parallel )$ for each $v,w\in E_3$ and

(e5)

$B[v_0,{\rho }_8]\subset E.$

As in the local case we obtain in turn and induction the estimates

$$\parallel w_0-v_0\parallel ={\displaystyle \frac{2}{3}}\parallel {\mathcal{A}}^{\prime }{\left(v_0\right)}^{-1}\mathcal{A}\left(v_0\right)\parallel \le {\beta }_0={\beta }_0-{\alpha }_0<{\rho }_8,$$

(57)

$$\begin{array}{ccc}\hfill z_n-w_n& =& [{\displaystyle \frac{2}{3}}I-{\displaystyle \frac{1}{2}}{(3{\mathcal{A}}^{\prime }\left(w_n\right)-{\mathcal{A}}^{\prime }\left(v_n\right))}^{-1}\hfill \\ & & (3{\mathcal{A}}^{\prime }\left(w_n\right)+{\mathcal{A}}^{\prime }\left(v_n\right))]{\mathcal{A}}^{\prime }{\left(v_n\right)}^{-1}\mathcal{A}\left(v_n\right),\hfill \\ & =& {\displaystyle \frac{1}{6}}{(3{\mathcal{A}}^{\prime }\left(w_n\right)-{\mathcal{A}}^{\prime }\left(v_n\right))}^{-1}[4(3{\mathcal{A}}^{\prime }\left(w_n\right)-{\mathcal{A}}^{\prime }\left(v_n\right))\hfill \\ & & -3(3{\mathcal{A}}^{\prime }\left(w_n\right)+{\mathcal{A}}^{\prime }\left(v_n\right))](-{\displaystyle \frac{3}{2}}(w_n-v_n)),\hfill \\ \hfill \parallel z_n-w_n\parallel & \le & {\displaystyle \frac{1}{8}}{\displaystyle \frac{[3\parallel L^{-1}({\mathcal{A}}^{\prime }\left(w_n\right)-{\mathcal{A}}^{\prime }\left(v_n\right))\parallel +4\parallel L^{-1}{\mathcal{A}}^{\prime }\left(v_n\right)\parallel ]}{1-{\tilde{p}}_n}}\parallel w_n-v_n\parallel \hfill \end{array}$$

(58)

$$\begin{array}{ccc}& \le & {\gamma }_n-{\beta }_n,\hfill \\ \hfill \parallel z_n-v_0\parallel & \le & \parallel z_n-w_n\parallel +\parallel w_n-v_0\parallel \hfill \end{array}$$

(59)

$$\begin{array}{ccc}& \le & {\gamma }_n-{\beta }_n+{\beta }_n-{\alpha }_0={\gamma }_n<{\rho }_8,\hfill \end{array}$$

(60)

where ${\tilde{p}}_n={\displaystyle \frac{3}{2}}\phi ({\beta }_n-{\alpha }_n)+{\phi }_0\left({\alpha }_n\right),$

$$\begin{array}{ccc}\hfill \mathcal{A}\left(z_n\right)& =& \mathcal{A}\left(z_n\right)-\mathcal{A}\left(v_n\right)+\mathcal{A}\left(v_n\right)\hfill \\ & =& {\int }_0^1{\mathcal{A}}^{\prime }(v_n+\theta (z_n-v_n))d\theta (z_n-v_n)-{\displaystyle \frac{3}{2}}{\mathcal{A}}^{\prime }\left(v_n\right)(w_n-v_n),\hfill \\ \hfill \parallel L^{-1}\mathcal{A}\left(z_n\right)\parallel & \le & \left(1+{\int }_0^1{\phi }_0(\parallel v_n-v_0\parallel )+\theta \parallel z_n-v_n\left)\right)d\theta \right)\parallel z_n-v_n\parallel \hfill \\ & & +{\displaystyle \frac{3}{2}}(1+{\phi }_0(\parallel v_n-v_0\parallel \left)\right)\parallel w_n-v_n\parallel \hfill \\ & \le & \left(1+{\int }_0^1{\phi }_0({\alpha }_n+\theta ({\gamma }_n-{\alpha }_n))d\theta \right)({\gamma }_n-{\alpha }_n)\hfill \\ & & +{\displaystyle \frac{3}{2}}(1+{\phi }_0\left({\alpha }_n\right))({\beta }_n-{\alpha }_n)={\mu }_n,\hfill \end{array}$$

$$\parallel v_{n+1}-z_n\parallel \le {\displaystyle \frac{{\mu }_n}{1-{\tilde{q}}_n}}={\alpha }_{n+1}-{\gamma }_n,$$

(61)

where

$$\begin{array}{ccc}\hfill {\tilde{q}}_n& =& {\phi }_0\left({\displaystyle \frac{3{\beta }_n-{\alpha }_n}{2}}\right)\hfill \\ \hfill {\phi }_0\left(\parallel {\displaystyle \frac{3w_n-v_n}{2}}-v_0\parallel \right)& \le & {\phi }_0\left({\displaystyle \frac{2\parallel w_n-v_0\parallel +\parallel w_n-v_n\parallel }{2}}\right)\hfill \\ & \le & {\phi }_0\left({\displaystyle \frac{2{\beta }_n+{\beta }_n-{\alpha }_n}{2}}\right)={\tilde{q}}_n<1,\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \parallel v_{n+1}-v_0\parallel & \le & \parallel v_{n+1}-z_n\parallel +\parallel z_n-v_0\parallel \hfill \end{array}$$

$$\begin{array}{ccc}& \le & {\alpha }_{n+1}-{\beta }_n+{\beta }_n={\alpha }_{n+1}<{\rho }_8,\hfill \\ \hfill \mathcal{A}\left(v_{n+1}\right)& =& \mathcal{A}\left(v_{n+1}\right)-\mathcal{A}\left(v_n\right)-{\displaystyle \frac{3}{2}}{\mathcal{A}}^{\prime }\left(v_n\right)(w_n-v_n)\hfill \end{array}$$

(62)

$$\begin{array}{ccc}& =& {\int }_0^1{\mathcal{A}}^{\prime }(v_n+\theta (v_{n+1}-v_n))d\theta (v_{n+1}-v_n)\hfill \\ & & +{\displaystyle \frac{3}{2}}{\mathcal{A}}^{\prime }\left(v_n\right)(w_n-v_n),\hfill \end{array}$$

(63)

$$\begin{array}{ccc}\hfill \parallel L^{-1}\mathcal{A}\left(v_{n+1}\right)\parallel & \le & \left(1+{\int }_0^1{\phi }_0(\parallel v_n-v_0\parallel +\theta \parallel v_{n+1}-v_n\parallel )d\theta \right)\parallel v_{n+1}-v_n\parallel \hfill \\ & & +{\displaystyle \frac{3}{2}}(1+{\phi }_0(\parallel v_n-v_0\parallel \left)\right)\parallel w_n-v_n\parallel \hfill \\ & \le & \left(1+{\int }_0^1{\phi }_0({\alpha }_n+\theta ({\alpha }_{n+1}-{\alpha }_n))d\theta \right)({\alpha }_{n+1}-{\alpha }_n)\hfill \\ & & +{\displaystyle \frac{3}{2}}(1+{\phi }_0\left({\alpha }_n\right))({\beta }_n-{\alpha }_n)={\xi }_{n+1},\hfill \end{array}$$

$$\begin{array}{ccc}\hfill \parallel w_{n+1}-v_{n+1}\parallel & \le & {\displaystyle \frac{2}{3}}\parallel {\mathcal{A}}^{\prime }{\left(v_{n+1}\right)}^{-1}L\parallel \parallel L^{-1}\mathcal{A}\left(v_{n+1}\right)\parallel \hfill \\ & \le & {\displaystyle \frac{2}{3}}{\displaystyle \frac{{\xi }_{n+1}}{1-{\phi }_0(\parallel v_{n+1}-v_0\parallel )}}\hfill \\ & \le & {\displaystyle \frac{2}{3}}{\displaystyle \frac{{\xi }_{n+1}}{1-{\phi }_0\left({\alpha }_{n+1}\right)}}={\beta }_{n+1}-{\alpha }_{n+1},\hfill \end{array}$$

(64)

and

$$\begin{array}{ccc}\hfill \parallel w_{n+1}-v_0\parallel & \le & \parallel w_{n+1}-v_{n+1}\parallel +\parallel v_{n+1}-v_0\parallel \hfill \\ & \le & {\beta }_{n+1}-{\alpha }_{n+1}+{\alpha }_{n+1}-{\alpha }_0={\beta }_{n+1}<{\rho }_8.\hfill \end{array}$$

(65)

It follows by (57)-(65) that the sequence $\left\{x_n\right\}$ is complete, since $\left\{{\alpha }_n\right\}$ is convergent by the condition (e2). But X is a Banach space. Hence, there exists $v^{*}\in B[x_0,{\rho }_8]$ such that ${lim}_{n⟶\infty }{v}_n=v^{*}.$ Then, by letting $n⟶\infty$ in

$$\parallel L^{-1}\mathcal{A}\left(v_{n+1}\right)\parallel \le {\xi }_{n+1},$$

we deduce that $\mathcal{A}\left(v^{*}\right)=0.$ Finally, notice that for $j=0,1,2,\dots$

$$\parallel v_{n+j}-v_n\parallel \le {\alpha }_{n+j}-{\alpha }_n$$

thus for $j⟶\infty$

$$\parallel v^{*}-v_n\parallel \le {\rho }_8-{\alpha }_n.$$

Hence, the semi-local result for the method (4) is achieved.

Theorem 9.

Let that the conditions (e1)-(e5) hold. Then, there exists $v^{*}\in B[v_0,{\rho }_8]$ solving the equation $\mathcal{A}\left(v\right)=0.$ Moreover, the following assertions hold

$$\left\{v_n\right\}\subset B(v_0,{\rho }_8),$$

$$\parallel w_n-v_n\parallel \le {\beta }_n-{\alpha }_n,$$

$$\parallel z_n-w_n\parallel \le {\gamma }_n-{\beta }_n,$$

$$\parallel v_{n+1}-z_n\parallel \le {\alpha }_{n+1}-{\gamma }_n$$

and

$$\parallel v^{*}-v_n\parallel \le {\rho }_8-{\alpha }_n.$$

The uniqueness property of the solution is specified in the next result.

Proposition 2.

Suppose: There exists a solution $\overline{v}\in B(v_0,{\rho }_9)$ of the equation $\mathcal{A}\left(v\right)=0$ for some ${\rho }_9>0,$ the condition (e3) holds in the ball $B(v_0,{\rho }_9)$ and there exists ${\rho }_{10}\ge {\rho }_9$ such that

$${\int }_0^1{\phi }_0((1-\theta ){\rho }_9+\theta {\rho }_{10})d\theta <1.$$

(66)

Let $E_4=E\cap B[v_0,{\rho }_{10}].$ Then, the only possible solution of the equation $\mathcal{A}\left(v\right)=0$ in the region $E_4$ is $\overline{v}.$

Proof. Let $\overline{z}\in E_4$ with $\mathcal{A}\left(\overline{z}\right)=0$ and the linear operator $L_2={\int }_0^1{\mathcal{A}}^{\prime }(\overline{v}+\theta (\overline{z}-\overline{v}))d\theta .$ It follows

$$\begin{array}{ccc}\hfill \parallel L^{-1}(L_2-L)\parallel & \le & {\int }_0^1{\phi }_0((1-\theta )\parallel \overline{v}-v_0\parallel +\theta \parallel \overline{z}-v_0\parallel )d\theta \hfill \\ & \le & {\int }_0^1{\phi }_0((1-\theta ){\rho }_9+\theta {\rho }_{10})d\theta <1.\hfill \end{array}$$

Thus, we deduce $\overline{z}=\overline{v}.$ □

Remark 2.

(1) 

A possible choice for $L={\mathcal{A}}^{\prime }\left(v_0\right).$

(2) 

Suppose that the conditions (e1)-(e5) hold. Then, set ${\rho }_9={\rho }_8$ and $\overline{v}=v^{*}$ in Proposition 2.

(3) 

Replace the limit point ${\rho }_8$ by ${\rho }_6$ in the condition (e5).

(4) 

Clearly the results for the method (3) are obtained by simply restricting in the first two substeps of the method (4).

6. Efficiency Indices

There are several measures for comparing iterative methods other than OC, one of them is efficiency of the method. Recall the informational efficiency, introduced by Traub [22] is given by $E.I=o^{1/s},$ where o is the order of the methods and s is the number of function evaluations. Ostowski [16], introduced a term before Traub called efficiency index or computational efficiency defined as $C.E={\varpi }^{{\displaystyle \frac{1}{{\theta }_f}}},$ where $\varpi$ is the OC of the method and ${\theta }_f$ is the number of function evaluations. Thus, the E.I and the C.E of the method (2) are ${\displaystyle \frac{6}{5}}=1.2$ and $6^{{\displaystyle \frac{1}{5}}}=1.4310,$ the E. I and C. E of the method (3) are ${\displaystyle \frac{4}{3}}=1.33$ and $4^{{\displaystyle \frac{1}{3}}}=1.587$ and E. I and C. E of the method (4) are ${\displaystyle \frac{5}{5}}=1$ and $5^{{\displaystyle \frac{1}{5}}}=1.3797.$

7. Numerical Example

Example 1.

Consider $X=Y={\mathbb{R}}^3$, $v_0={(0,0,0)}^T$, $E=B[0,1].$ Define function $\mathcal{A}$ on E for $u={(v,w,z)}^T$ by

$$\mathcal{A}\left(u\right)={\left(e^v-1,{\displaystyle \frac{e-1}{6}}{w}^3+w,{\displaystyle \frac{z^3}{6}}+z\right)}^T.$$

Then, the first and second Fréchet derivatives are as follows:

$${\mathcal{A}}^{\prime }\left(u\right)=\left[\begin{array}{ccc}{e}^v& 0& 0\\ 0& {\displaystyle \frac{e-1}{2}}{w}^2+1& 0\\ 0& 0& {\displaystyle \frac{z^2}{2}}+1\end{array}\right]$$

and

$${\mathcal{A}}^{\prime \prime }\left(u\right)=\left[\begin{array}{ccccccccccc}{e}^v& 0& 0& |& 0& 0& 0& |& 0& 0& 0\\ 0& 0& 0& |& 0& (e-1)w& 0& |& 0& 0& 0\\ 0& 0& 0& |& 0& 0& 0& |& 0& 0& z\end{array}\right].$$

Now, we can observe that ${\mathcal{A}}^{\prime }\left(v^{*}\right)={\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}=\mathrm{diag}(1,1,1).$ Thus we get $\parallel {\mathcal{A}}^{\prime }\left(v^{*}\right)\parallel =1.$ Thus,

$$\parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}({\mathcal{A}}^{\prime }\left(v\right)-{\mathcal{A}}^{\prime }\left(v^{*}\right))\parallel \le (e-1)\parallel v-v^{*}\parallel ,$$

$$\parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}({\mathcal{A}}^{\prime \prime }\left(v\right)-{\mathcal{A}}^{\prime \prime }\left(v^{*}\right))\parallel \le (e-1)\parallel v-v^{*}\parallel ,$$

$$\parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}{\mathcal{A}}^{\prime \prime }\left(v\right)\parallel \le e.$$

$$\parallel {\mathcal{A}}^{\prime }{\left(v^{*}\right)}^{-1}{\mathcal{A}}^{\prime }\left(v\right)\parallel \le e^{{\displaystyle \frac{1}{e-1}}},$$

Hence $L_0=L_1=(e-1)$, $L_2=e$ and $L_3=e^{{\displaystyle \frac{1}{e-1}}}$. With respect to ${\lambda }_1=0.1146,{\lambda }_2=0.7115,{\lambda }_3=0.6959,$ and ${\lambda }_4=0.3001$, we get $r=R=R_1=0.0667$.

Example 2.

Consider the non-linear integral equation of the Hammerstein-type given by

$$\mathcal{A}\left(v\right)\left(\theta \right)=v\left(\theta \right)-5H\left(v\right)\left(\theta \right),$$

where H is any function such that

$$H^{\prime }\left(v\right)\left(\theta \right)=\theta {\int }_0^1\beta v^3\left(\beta \right)d\left(\beta \right),$$

defined on $X=Y=C[0,1]$, the space of all continuous functions on the on the interval $[0,1]$ let $E=B[0,1].$ Then, we obtain first Fréchet derivatives as

$${\mathcal{A}}^{\prime }\left(v\left(\psi \right)\right)\left(\theta \right)=\psi \left(\theta \right)-15\theta {\int }_0^1\gamma v^2\left(\gamma \right)\psi \left(\gamma \right)d\gamma ,\mathrm{for}\mathrm{all}v\in E.$$

we can observe that $v^{*}=v^{*}\left(\theta \right)=0$ is a solution of $\mathcal{A}\left(v\right).$ Then, by applying the conditions $\left(A1\right)-\left(A4\right)$ we have $L_0=L_1=L_2=7.5$ and $L_3=2$. With respect to ${\lambda }_1=.5028,{\lambda }_2=0.0129,{\lambda }_3=0.0523,$ and ${\lambda }_4=0.1333$, we get $r=0.0667,R=0.0523,$ and $R_1=0.0667$.

In the next example, we compare the iteration and the convergence order of methods (3), (2) and (4) with that of following methods:

Noor Waseem-type methods [10]: given for $n=0,1,2,\cdots$ as

$$\begin{array}{ccc}\hfill w_n& =& v_n-{\mathcal{A}}^{\prime }{\left(v_n\right)}^{-1}\mathcal{A}\left(v_n\right)\hfill \\ \hfill v_{n+1}& =& v_n-4G_n^{-1}\mathcal{A}\left(v_n\right),\hfill \end{array}$$

(67)

where $G_n=3{\mathcal{A}}^{\prime }\left({\displaystyle \frac{2v_n+w_n}{3}}\right)+A^{\prime }\left(w_n\right),$

$$\begin{array}{ccc}\hfill w_n& =& v_n-{\mathcal{A}}^{\prime }{\left(v_n\right)}^{-1}\mathcal{A}\left(v_n\right)\hfill \\ \hfill z_n& =& v_n-4G_n^{-1}\mathcal{A}\left(v_n\right)\hfill \\ \hfill v_{n+1}& =& z_n-{\mathcal{A}}^{\prime }{\left(w_n\right)}^{-1}\mathcal{A}\left(z_n\right)\hfill \end{array}$$

(68)

and

$$\begin{array}{ccc}\hfill w_n& =& v_n-{\mathcal{A}}^{\prime }{\left(v_n\right)}^{-1}\mathcal{A}\left(v_n\right)\hfill \\ \hfill z_n& =& v_n-4G_n^{-1}\mathcal{A}\left(v_n\right)\hfill \\ \hfill v_{n+1}& =& z_n-{\mathcal{A}}^{\prime }{\left(z_n\right)}^{-1}\mathcal{A}\left(z_n\right).\hfill \end{array}$$

(69)

Newton Simpson-type methods [11]: given for $n=0,1,2,\cdots$ as

$$\begin{array}{cc}\hfill w_n& =v_n-{\mathcal{A}}^{\prime }{\left(v_n\right)}^{-1}\mathcal{A}\left(v_n\right)\hfill \\ \hfill v_{n+1}& =v_n-6G_n^{-1}\mathcal{A}\left(v_n\right),\hfill \end{array}$$

(70)

where $G_n={\mathcal{A}}^{\prime }\left(v_n\right)+4{\mathcal{A}}^{\prime }\left({\displaystyle \frac{v_n+w_n}{2}}\right)+{\mathcal{A}}^{\prime }\left(w_n\right),$

$$\begin{array}{cc}\hfill w_n& =v_n-{\mathcal{A}}^{\prime }{\left(v_n\right)}^{-1}\mathcal{A}\left(v_n\right)\hfill \\ \hfill z_n& =v_n-6G_n^{-1}\mathcal{A}\left(v_n\right)\hfill \\ \hfill v_{n+1}& =z_n-{\mathcal{A}}^{\prime }{\left(w_n\right)}^{-1}\mathcal{A}\left(z_n\right)\hfill \end{array}$$

(71)

and

$$\begin{array}{cc}\hfill w_n& =v_n-{\mathcal{A}}^{\prime }{\left(v_n\right)}^{-1}\mathcal{A}\left(v_n\right)\hfill \\ \hfill z_n& =v_n-6G_n^{-1}\mathcal{A}\left(v_n\right)\hfill \\ \hfill v_{n+1}& =z_n-{\mathcal{A}}^{\prime }{\left(z_n\right)}^{-1}\mathcal{A}\left(z_n\right).\hfill \end{array}$$

(72)

Example 3.

Let $\mathbb{T}={\mathbb{T}}_1={\mathbb{R}}^2.$ Consider the system of equations [12]

$$\begin{array}{ccc}\hfill 3t_1^2{t}_2+t_2^2& =& 1\hfill \\ \hfill t_1^4+t_1{t}_2^3& =& 1.\hfill \end{array}$$

Observe, $a_1=(-1,0.2),a_2=(-0.4,-1.3)$ and $a_3=(0.9,0.3)$ are the solutions of the above system of equations. The approximation to the solution $a_3$ using the methods (67)-(72), (3), (2) and (4) starting with $v_0=(2,-1)$ is given. The results are displayed in Table 1- Table 3.

8. Basins of Attraction

For an iterative method, the set of all initial points which converges to a solution of an equation is known as Basins of attraction [7,8]. Using the approach of the basins of attractions we obtain the convergence area of the methods (2), (3) and (4) when applied to the following examples;

Example 4.

$\left\{\begin{array}{c}{\alpha }^3-\beta =0\hfill \\ {\beta }^3-\alpha =0\hfill \end{array}\right.$ with solutions $\left\{\right(-1,-1),(0,0),(1,1\left)\right\}.$

Example 5.

$\left\{\begin{array}{c}3{\alpha }^2\beta -{\beta }^3=0\hfill \\ {\alpha }^3-3\alpha {\beta }^2-1=0\hfill \end{array}\right.$ with solutions

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

Example 6.

$\left\{\begin{array}{c}{\alpha }^2+{\beta }^2-4=0\hfill \\ 3{\alpha }^2+7{\beta }^2-16=0\hfill \end{array}\right.$ with solutions

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

Corresponding to the roots of system of nonlinear equations, the basins of attraction are generated in a rectangle domain $w=\{(\alpha ,\beta )\in {\mathbb{R}}^2:-2\le \alpha \le 2,-2\le \beta \le 2\}$ with equidistant grid points of $401\times 401$. According to root, each initial point $({\alpha }_0,{\beta }_0)\in {\mathbb{R}}^2$ is assigned a color, to which the corresponding iterative method converges, starting from $({\alpha }_0,{\beta }_0)$. If either the method converges to infinity or it does not converge, then the point is marked black. In a maximum of 100 iterations a tolerance of ${10}^{-8}$ is used.

Figure 1. Dynamical plane of the method (2) with Basins of attraction for Example 4(left), Example 5(middle) and Example 6(right).

Figure 1. Dynamical plane of the method (2) with Basins of attraction for Example 4(left), Example 5(middle) and Example 6(right).

Figure 2. Dynamical plane of the method (3) with Basins of attraction for Example 4(left), Example 5(middle) and Example 6(right).

Figure 2. Dynamical plane of the method (3) with Basins of attraction for Example 4(left), Example 5(middle) and Example 6(right).

Figure 3. Dynamical plane of the method (4) with Basins of attraction for Example 4(left), Example 5(middle) and Example 6(right).

Figure 3. Dynamical plane of the method (4) with Basins of attraction for Example 4(left), Example 5(middle) and Example 6(right).

9. Conclusion

We studied Jarrat-type method of convergence order three and its two extensions with convergence order six and five, respectively. As mentioned in the introduction, we used assumptions on $A^{\prime }$ and $A^{\prime \prime }$ only, so these methods (2), (3) and (4) can be used to solve problems which were not possible if we use the earlier convergence analysis using Taylor expansion. We discussed the limitations of our approach and developed new ways to overcome these limitations in Section 5. Finally, we compare the methods with other similar methods using an example. Also using Basins of attraction approach the convergence areas of the methods (2), (3) and (4) are given. In future research our ideas shall be applied on other methods to obtain similar benefits analogously [1,2,3,4,5,6,7,8,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24]