# A Note on the Fixed Point Method and the Linear Complementarity Problem$$^\ast$$

December 12, 2022; accepted: February 24, 2023; published online: July 30, 2023.

In this paper, we propose an extended form of a fixed point method for processing the large and sparse linear complementarity problem (LCP). We obtain an equivalent form of LCP by using two positive diagonal matrices and prove the equivalence. For the proposed method, we provide some convergence conditions when the system matrix is a $$P$$-matrix or an $$H_+$$-matrix or a symmetric positive definite matrix.

MSC. 65F10, 65F50.

Keywords. Linear complementarity problem, $$P$$-matrix, $$H_{+}$$-matrix, symmetric positive definite matrix, matrix splitting, convergence.

$$^\ast$$The first author is thankful to the University Grants Commission (UGC), Government of India under the SRF fellowship Program No. 1068/(CSIR-UGC NET DEC.2017).

$$^{\ast \ast }$$Mathematics Discipline, PDPM-Indian Institute of Information Technology, Design and Manufacturing, Jabalpur, M.P., India, e-mail: bharatnishad.kanpu@gmail.com, dmrai23@gmail.com.

$$^\S$$Indian Statistical Institute, 203 B.T. Road, Kolkata-700108, India, e-mail: akdas@isical.ac.in

# 1 Introduction

In the literature on mathematical programming, the linear complementarity problem has received considerable attention. It also appears in a number of applications in operations research applications, control theory, mathematical economics, optimization theory, stochastic optimal control, the American option pricing problem, economics, elasticity theory, the free boundary problem, and the Nash equilibrium point of the bimatrix game. For details, see [ 18 ] , [ 19 ] . For recent works on this problem, see [ 14 ] , [ 15 ] and references therein.

Consider $$A_1\in \mathbb {R}^{n\times n}$$ and a vector $$\, q\, \in \, \mathbb {R}^{n}.$$ The linear complementarity problem denoted as LCP$$(q, A_{1})$$ is to find the solution $$z\; \in \mathbb {R}^{n}\;$$ to the following system

\begin{eqnarray} \label{eq1} z\geq 0, ~ ~ ~ ~ A_{1}z +q \geq 0,~ ~ ~ ~ z^T(A_{1}z +q)=0. \end{eqnarray}

There are various methods of solving the LCP using an iterative process, namely the projected methods [ 4 ] , [ 9 ] , [ 12 ] , [ 16 ] , [ 20 ] , [ 21 ] , the modulus algorithm [ 2 ] , [ 3 ] , [ 5 ] , [ 10 ] and the modulus based matrix splitting iterative methods [ 8 ] , [ 11 ] , [ 22 ] .

A general fixed point method (GFP) is proposed by Fang [ 6 ] assuming the case where $$\phi = \omega A_{1D}^{-1}$$ with $$\omega$$ $$\textgreater$$ $$0$$ and $$A_{1D}$$ is a diagonal matrix of $$A_{1}$$. The GFP approach takes less iterations than the modulus-based successive over-relaxation (MSOR) iteration method [ 2 ] . In this paper we present an extended form of the fixed point method [ 6 ] that incorporates projected type iteration techniques by including two positive diagonal parameter matrices $$\phi _1$$ and $$\phi _2$$. We also show that the fixed point equation and the linear complementarity problem are equivalent and discuss the convergence conditions as well as provide some convergence domains for the proposed method.

The rest of this paper is organized as follows: In section 2, we present some notation, definitions and lemmas in order to establish our key findings. Section 3 discusses an extended form of a fixed point method for solving LCP$$(q, A_1)$$ with convergence analysis. In the last section, we give the conclusion.

# 2 Preliminaries

Some notations, introductory definitions and required lemmas are introduced. For details, see [ 2 ] , [ 6 ] .

Let $$A_{1}= (a_{ij}) \in \mathbb {R}^{n\times n}$$ and $$B_{1}=(b_{ij}) \in \mathbb {R}^{n\times n}$$. We use $$A_{1} \geq$$ $$(\textgreater )$$ $$B_{1}$$ to denote $$a_{ij}\geq (\textgreater )$$ $$b_{ij}$$ for all $$i, j$$. The comparison matrix $$\langle A_{1} \rangle = (\langle a_{ij} \rangle )$$ of $$A_{1}$$ is defined by $$\langle a_{ij} \rangle =|a_{ij}|$$ with $$i=j$$ and $$\langle a_{ij} \rangle =-|a_{ij}|$$ with $$i\neq j$$ for $$i,j=1,2,\ldots ,n$$. The matrix $$A_{1}$$ is called a $$Z$$-matrix if all of its non-diagonal elements are less than or equal to zero; an $$Z$$-matrix is called an $$M$$-matrix if $$A_{1}^{-1}\geq 0$$; an $$H$$-matrix if $$\langle A_{1} \rangle$$ is an $$M$$-matrix. The splitting $$A_{1}=M_{1}-N_{1}$$ is called an $$M$$-splitting if det$$(M_{1})\neq 0$$ and $$N_{1} \geq 0$$, an $$H$$-splitting if $$\langle M_{1} \rangle -|N_{1}|$$ is an $$M$$-matrix [ 7 ] . An $$H$$-matrix is called an $$H_{+}$$-matrix [ 1 ] if $${a}_{ii} ~ \textgreater ~ 0 ~$$ for$$~ i = 1,2,\ldots ,n$$. Let $$A_{1}\in \mathbb {R}^{n\times n}$$, then $$A_{1}$$ is said to be a $$P$$-matrix if all its principle minors are positive that is det$$({A_1}_{\alpha \alpha }) ~ \textgreater ~ 0$$ for all $$\alpha \subseteq \{ 1,2,\ldots , n\}$$. Suppose $$A_{1}=(a_{ij})\in \mathbb {R}^{n\times n}$$ be a square matrix, then $$| A_{1} |=(c_{ij})$$ is defined by $$c_{ij} = | {a}_{ij} |$$ $$\forall ~ i,j$$ and $$A_{1}^{T}$$ denotes the transpose of $$A_{1}.$$

Lemma 1 [ 13 ]

The LCP($$q, A_1$$) has a unique solution for any $$q \in \mathbb {R}^n$$ if $$A_1 \in \mathbb {R}^{n \times n}$$ is a $$P$$-matrix.

Lemma 2 [ 6 ]

Let $$A_{1}\in \mathbb {R}^{n\times n}$$ and $$A_{1}=M_{1}-N_{1}$$ be an $$M$$-splitting with $$M_{1}$$ an $$M$$-matrix and $$N_{1}\geq 0$$. Then  $$\rho (M_{1}^{-1}N_{1})$$ $$\textless$$ $$1$$.

# 3 Main results

For a given vector $$\xi \in \mathbb {R}^{n}$$, we indicate the vectors $$\xi _{+}$$=max$$\{ 0,\xi \}$$, $$\xi _{-}$$=
max$$\{ 0,-\xi \}$$. Since $$\xi _{+}\geq 0$$, $$\xi _{-}\geq 0$$, $$\xi =\xi _{+}-\xi _{-}$$, $$\xi ^T_{+}\xi _{-}=0.$$ Let $$z=\phi _{1}\xi _{+}$$ and $$w=\phi _{2}\xi _{-}$$, where $$\phi _{1}$$ and $$\phi _{2}$$ are positive diagonal matrices of order $$n$$. Now we convert the LCP into a fixed point formulation that is

\begin{eqnarray} \label{eq2} \xi =(I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1})\xi _{+}-\phi ^{-1}_{2}q, \end{eqnarray}

where $$I_{1}$$ is an identity matrix of order $$n$$.

In the following result, we provide an equivalent formulation of the LCP$$(q, A_{1})$$ for solving LCP$$(q, A_{1})$$.

Theorem 1

Suppose $$A_{1} \in \mathbb {R}^{n \times n}$$ and $$q\in \mathbb {R}^n$$. Then $$\xi ^{*}$$ is a solution of 2 if and only if $$z^{*}=\phi _{1}\xi ^{*}_{+}$$ is a solution of LCP$$(q, A_{1})$$.

Proof â–¼
Let $$\xi ^{*}$$ be a solution of 2. Then
\begin{align*} \xi ^{*}& =(I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1})\xi ^{*}_{+}-\phi ^{-1}_{2}q,\\ \phi _{2}\xi ^{*}_{-} & =A_{1}\phi _{1}\xi ^{*}_{+} + q. \end{align*}

Since $$\phi _{2}\xi ^{*}_{-} \geq 0$$,

$A_{1}\phi _{1}\xi ^{*}_{+} + q \geq 0.$

Moreover,

$(\phi _{1}\xi ^{*}_{+})^T(A_{1}\phi _{1}\xi ^{*}_{+} + q) = (\phi _{1}\xi ^{*}_{+})^T(\phi _{2}\xi ^{*}_{-}) = 0,$

and $$\phi _{1}\xi ^{*}_{+}\geq 0$$. Therefore $$z^{*}=\phi _{1}\xi ^{*}_{+}$$ is a solution of LCP$$(q,A_{1})$$.

Let $$z^{*}=\phi _{1}\xi ^{*}_{+}$$ and $$w^{*}=\phi _{2}\xi ^{*}_{-}$$, and $$\xi ^{*}=\xi ^{*}_{+}-\xi ^{*}_{-}$$. Now from LCP$$(q,A_{1})$$

\begin{align*} \phi _{2}\xi ^{*}_{-} & =A_{1}\phi _{1}\xi ^{*}_{+} + q,\\ \xi ^{*}& =\xi ^{*}_{+}-\phi ^{-1}_{2}(A_{1}\phi _{1}\xi ^{*}_{+}+ q),\\ \xi ^{*}& =(I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1})\xi ^{*}_{+}-\phi ^{-1}_{2}q. \end{align*}

Thus, $$\xi ^{*}$$ is a solution of 2.

Proof â–¼

Now we show that the solution of $$\eqref{eq2}$$ is unique when the matrix $$A_1$$ is a $$P$$-matrix.

Theorem 2

For any positive diagonal matrices $$\phi _1$$ and $$\phi _2$$, 2 has a unique solution if $$A_{1} \in \mathbb {R}^{n \times n}$$ is a $$P$$-matrix.

Proof â–¼
Since $$A_{1}$$ is a $$P$$-matrix, for any $$q \in \mathbb {R}^{n}$$ LCP$$(q, A_{1})$$ has a unique solution. Let $$y^{*}$$ and $$u^{*}$$ be the solutions of 2. Then
\begin{eqnarray*} & y^{*}=y^{*}_{+}-\phi ^{-1}_{2}(A_{1}\phi _{1}y^{*}_{+}+ q ). \end{eqnarray*}

and

\begin{eqnarray*} & u^{*}=u^{*}_{+}-\phi ^{-1}_{2}(A_{1}\phi _{1}u^{*}_{+}+ q ). \end{eqnarray*}

As $$y^{*}_{+} =u^{*}_{+}$$,

$$y^{*}+\phi ^{-1}_{2}(A_{1}\phi _{1}y^{*}_{+}+ q )=u^{*}+\phi ^{-1}_{2}(A_{1}\phi _{1}u^{*}_{+}+ q ).$$

Hence

$y^{*}=u^{*}.$

Proof â–¼

Based on 2, we obtain an extended form of the fixed point method which is referred to as method 1.

Method 1

Let $$A_{1}\in \mathbb {R}^{n \times n}$$ and $$q \in \mathbb {R}^n$$. Suppose $$\xi ^{(0)}\in \mathbb {R}^{n}$$ an initial vector and the sequence $$\{ z^{(k)}\} ^{+\infty }_{k=1} \subset \mathbb {R}^{n}$$. Let Residue be an Euclidean norm of the error vector and define the Residue as

$\operatorname {Res}(z^{(k)})=\| \min (z^{(k)}, A_{1}z^{(k)}+q) \| _{2},$

where $$z^{(k)}$$ is the $$k^{th}$$ approximate solution of the LCP$$(q,A_{1})$$. The iteration process stop if $$(z^{(k)})$$ $$\textless$$ $$10^{-5}$$ or the number of iteration reached 900. For computing $$\xi ^{(k+1)}\in \mathbb {R}^{n}$$ is as follows:

1. Given an initial vector $$\xi ^{(0)} \in \mathbb {R}^{n}$$, error $$\epsilon$$ $$\textgreater$$ $$0$$ and set $$k=0$$.

2. Using the following scheme, create the sequence $$\xi ^{(k)}$$:

\begin{eqnarray} \label{eq3} & \xi ^{(k+1)}=(I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1})\xi ^{(k)}_{+}-\phi ^{-1}_{2}q \end{eqnarray}

and set $$z^{(k+1)}=\phi _{1}\xi ^{(k+1)}_{+}$$.

3. If $$(z^{(k)})$$ $$\textless$$ $$\epsilon$$ then stop; otherwise, set $$k=k+1$$ and return to step 2.

Remark 3

Fang [ 6 ] introduced a fixed point method, which is a special case of 3 with $$\phi _{2}=\phi ^{-1}$$ and $$\phi _{1}=I_{1}$$, where $$\phi$$ is a positive diagonal matrix. â–¡

In the following result, we discuss the convergence condition when the system matrix $$A_{1}$$ is a $$P$$-matrix.

Theorem 4

Let $$A_{1} \in \mathbb {R}^{n \times n}$$ be a $$P$$-matrix. Let $$\rho (|I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1}|)$$ $$\textless$$ $$1$$ and $$\xi ^{*}$$ be the solution of 2. Then the sequence $$\{ z^{(k)}\} ^{+\infty }_{k=1}$$ generated by $$\cref{mthd1}$$ converges to $$z^{*}$$ for any initial vector $$\xi ^{(0)}\in \mathbb {R}^{n}$$.

Proof â–¼
Suppose $$A_{1}$$ is a $$P$$-matrix. Then $$\xi ^{*}$$ is a unique solution of 2. Thus
$\xi ^{*}=(I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1})\xi ^{*}_{+}-\phi ^{-1}_{2}q.$

From 3, it results

$\xi ^{(k+1)}-\xi ^{*}=(I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1})(\xi _{+}^{(k)}-\xi ^{*}_{+}).$

It follows that

\begin{equation*} \begin{split} |\xi ^{(k+1)}-\xi ^{*}|& =|(I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1})|\cdot |\xi _{+}^{(k)}-\xi ^{*}_{+}|\\ & \leq |(I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1})|\cdot |\xi ^{(k)}-\xi ^{*}|. \end{split}\end{equation*}

Since $$\rho (|I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1}|)$$ $$\textless$$ $$1$$. Hence, for any initial vector $$\xi ^{(0)}\in \mathbb {R}^{n}$$ the sequence $$\{ z^{(k)}\} ^{+\infty }_{k=1}$$ converges to the $$z^{*}$$.

Proof â–¼

In the following result, we provide the convergence conditions for $$\cref{mthd1}$$ when the system matrix is an $$H_{+}$$-matrix.

Theorem 5

Assume $$A_{1} \in \mathbb {R}^{n \times n}$$ is an $$H_{+}$$-matrix, $$A_{1D}=diag(A_{1})$$ and $$B=A_{1D}-A_{1} \in \mathbb {R}^{n \times n}$$. Let $$\phi _{1}=\alpha _{1}D_{1}$$, $$\phi _{2}=\omega ^{-1}A_{1D}$$ and

$\rho \Big({A_{1D}^{-1}|B|D_{1}}\Big) \leq \rho \Big({A_{1D}^{-1}|B|}\Big)\rho (D_{1}),$

where $$\alpha _{1}$$, $$\omega$$ are positive constants and $$D_{1}$$ is a positive diagonal matrix. Let $$\omega \alpha _{1}=\beta$$ and $$\xi ^{*}$$ be the solution of 2. Then the sequence $$\{ z^{(k)}\} ^{+\infty }_{k=1}$$ generated by $$\cref{mthd1}$$ converges to $$z^{*}$$ for any initial vector $$\xi ^{(0)}\in \mathbb {R}^{n}$$ if

$0{\lt} \beta {\lt} \tfrac {2}{\big(1+\rho (A_{1D}^{-1}|B|)\big)\rho (D_1)}.$

Proof â–¼
We have $$A_{1}$$ is an $$H_{+}$$-matrix, $$A_{1D}=diag(A_{1})$$, $$B=A_{1D}-A_{1}$$ and $$\rho (A_{1D}^{-1}|B|)$$ $$\textless$$ $$1$$. For $$\phi _{1}=\alpha _{1}D_{1}$$ and $$\phi _{2}=\omega ^{-1}A_{1D}$$, we obtain
\begin{align*} |I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1}| & = |I_{1}-(\omega ^{-1}A_{1D})^{-1}A_{1}\alpha _{1}D_{1}|\\ & = |I_{1}-(\omega ^{-1}A_{1D})^{-1}(A_{1D}-B)\alpha _{1}D_{1}|\\ & = |I_{1}-\omega \alpha _{1}D_{1}+\omega A_{1D}^{-1}B\alpha _{1}D_{1}|\\ & \leq |I_{1}-\omega \alpha _{1}D_{1}|+|\omega A_{1D}^{-1}B\alpha _{1}D_{1}|\\ & \leq |I_{1}-\beta D_{1}|+\beta A_{1D}^{-1}|B|D_{1}. \end{align*}

It follows that

\begin{equation*} |I_{1}-\beta D_{1}|+\beta A_{1D}^{-1}|B|D_{1} = \begin{cases} (I_{1}-\beta D_{1})+\beta A_{1D}^{-1}|B|D_{1}, & \text{if $0\textless \beta D_{1}\leq I_{1}$},\\ (\beta D_{1}-I_{1})+\beta A_{1D}^{-1}|B|D_{1}, & \text{if $\beta D_{1}\textgreater I_{1}$}. \end{cases}\end{equation*}

Now we write

$$\label{eq4} \rho (|I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1}|) \leq \begin{cases} 1- (1- \rho (A_{1D}^{-1}|B|)) \beta \rho (D_{1}), & \text{if 0\textless \beta \rho (D_{1})\leq 1 },\\ (1+ \rho (A_{1D}^{-1}|B|))\beta \rho (D_{1})-1, & \text{if \beta \rho (D_{1})\textgreater 1 }. \end{cases}$$
8

From 8 we can seen that $$\rho (|I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1}|)$$ $$\textless$$ $$1$$ for $$\beta \rho (D_{1})\in (0, 1]$$ and for $$\beta \rho (D_1)$$ $$\textgreater$$ $$1$$, $$\rho (|I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1}|)$$ $$\textless$$ $$1$$ if and only if

$\big(1+ \rho (A_{1D}^{-1}|B|)\big)\beta \rho (D_{1})-1 {\lt} 1$

such that $$\beta {\lt} \tfrac {2}{\big(1+\rho (A_{1D}^{-1}|B|)\big)\rho (D_1)}$$. Therefore, if

$0 {\lt} \beta {\lt} \tfrac {2}{\big(1+\rho (A_{1D}^{-1}|B|)\big)\rho (D_1)},$

for any initial vector $$\xi ^{(0)}\in \mathbb {R}^{n}$$, the sequence $$\{ z^{(k)}\} ^{+\infty }_{k=1}$$ converges to $$z^{*}$$.

Proof â–¼

In the following result, we provide the convergence conditions for $$\cref{mthd1}$$ when the system matrix is a symmetric positive definite (SPD) matrix.

Theorem 6

Let $$A_{1} \in \mathbb {R}^{n \times n}$$ be the SPD matrix. Let $$\phi _{2}=\omega ^{-1}I_{1}$$ and $$\phi _{1}=\alpha _{1}D_{1}$$, where $$D_{1}$$ is a scalar matrix and denote the minimum and the maximum eigenvalues of $$A_{1}D_{1}$$ by $$\nu _{min}$$ and $$\nu _{max}$$ respectively. Let $$\xi ^{*}$$ be the solution of 2. Then the sequence $$\{ z^{(k)}\} ^{+\infty }_{k=1}$$ generated by $$\cref{mthd1}$$ converges to $$z^{*}$$ for any initial vector $$\xi ^{(0)}\in \mathbb {R}^{n}$$ if $$0$$ $$\textless$$ $$\beta$$ $$\textless$$ $$\frac{2}{\nu _{max}}$$.

Proof â–¼
From theorem 4, we have
\begin{eqnarray*} & \xi ^{(k+1)}-\xi ^{*}=(I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1})(\xi _{+}^{(k)}-\xi ^{*}_{+}). \end{eqnarray*}

This implies that

\begin{eqnarray*} & \| \xi ^{(k+1)}-\xi ^{*}\| _{2}=\| (I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1})(\xi _{+}^{(k)}-\xi ^{*}_{+})\| _{2}. \end{eqnarray*}

Since $$\| (\xi _{+}^{(k)}-\xi ^{*}_{+})\| _{2} \leq \| (\xi ^{(k)}-\xi ^{*})\| _{2}$$,

\begin{eqnarray*} \begin{split} \| \xi ^{(k+1)}-\xi ^{*}\| _{2}& \leq \| (I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1})\| _{2} \| (\xi ^{(k)}-\xi ^{*})\| _{2} \\ & \leq \| (I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1})\| _{2}\| (\xi ^{(k)}-\xi ^{*})\| _{2}. \end{split}\end{eqnarray*}

If $$\| I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1}\| _{2}$$ $$\textless$$ $$1$$, then $$\cref{mthd1}$$ is convergent. Therefore

\begin{align*} \| I_{1}-\phi ^{-1}_{2}A_{1}\phi _{1}\| _{2}& =\| I_{1}-\omega I_{1} A_{1}\alpha _{1}D_{1}\| _{2} \\ & =\| I_{1}-\beta A_{1}D_{1}\| _{2}.\\ \end{align*}

We have

$\| I-\beta A_{1}D_{1}\| _{2}=\max \{ |1-\beta \nu _{\min }|,|1-\beta \nu _{\max }|\} .$

It follows that

\begin{equation*} \| I_{1}-\beta A_{1}D_{1}\| _{2} = \begin{cases} |1-\beta \nu _{\min }|, & \text{if $|1-\beta \nu _{\min }| \geq |1-\beta \nu _{\max }|$},\\ |1-\beta \nu _{\max }|, & \text{if $|1-\beta \nu _{\max }| \geq |1-\beta \nu _{\min }|$}. \end{cases}\end{equation*}

Thus $$\| I_{1}-\beta A_{1}D_{1}\| _{2}\textless 1$$ if and only if

\begin{equation*} (a) \begin{cases} |1-\beta \nu _{\min }|\textless 1, & \text{}\\ |1-\beta \nu _{\min }| \geq |1-\beta \nu _{\max }|,& \text{} \end{cases}\end{equation*}

and

\begin{equation*} (b) \begin{cases} |1-\beta \nu _{\max }|\textless 1, & \text{}\\ |1-\beta \nu _{\max }| \geq |1-\beta \nu _{\min }|. & \text{} \end{cases}\end{equation*}

From $$(a)$$ and $$(b)$$ we obtain the convergence condition of $$\cref{mthd1}$$ that is $$0$$ $$\textless$$ $$\beta$$ $$\leq$$ $$\tfrac {2}{\nu _{min}+\nu _{max}}$$ and $$\tfrac {2}{\nu _{min}+\nu _{max}}$$ $$\leq$$ $$\beta$$ $$\textless$$ $$\tfrac {2}{\nu _{max}}$$, from these two inequalities we obtain

$$0$$ $$\textless$$ $$\beta$$ $$\textless$$ $$\tfrac {2}{\nu _{max}}.$$

Proof â–¼

# 4 Conclusion

We introduced an extended form of a fixed point method for solving the linear complementarity problem LCP$$(q, A _1)$$ with parameter matrices $$\phi _1$$ and $$\phi _2$$. Also, we have shown how the iterative form relates to the parameter matrices $$\phi _1$$ and $$\phi _2$$. We have presented some convergence conditions and some sufficient convergence domains for the proposed method.

Acknowledgements

The authors are grateful to the editor associated with this paper for their excellent suggestions. Also, the authors thank the anonymous referees who spent their precious time for reviewing this paper. The authors would like to acknowledge their contribution, due to which there is a significant improvement in the paper.