A Unified Convergence Analysis for Some Two-Point Type Methods for Nonsmooth Operators

Abstract

The aim of this paper is the approximation of nonlinear equations using iterative methods. We present a unified convergence analysis for some two-point type methods. This way we compare specializations of our method using not necessarily the same convergence criteria. We consider both semilocal and local analysis. In the first one, the hypotheses are imposed on the initial guess and in the second on the solution. The results can be applied for smooth and nonsmooth operators.

1. Introduction

One of the most important techniques in order to approximate nonlinear equations are iterative methods [1,2,3,4,5,6]. In this paper, we present a unified approach for two-point Newton-type methods for smooth and nonsmooth operators [7,8,9,10]. We will consider two types of convergence. The semilocal convergence is where the hypotheses are imposed on the initial guess; and local convergence is where the hypotheses are imposed on the solution. Our family includes a great variety of methods. We are interested also in the application of these methods in practice (nonlinear systems, boundary problems and image processing).

For a greater generality, in this study, let X and Y be two Banach spaces and D a nonempty, open, and convex set; let $F_1:D\subset X\to Y$ and $F_2:D\subset X\to Y$ be continuous operators. Moreover, we assume that the operator $F_1$ has a continuous Fréchet derivative and $F_2$ is a continuous operator whose differentiability is not assumed. We consider the equation

$$F(x)=F_1(x)+F_2(x)=0.$$

(1)

To solve this equation, we use the two-point Newton-type methods defined by

$$x_{k+1}=x_k-L_{k-1,k}^{-1}(F_1(x_k)+F_2(x_k))$$

(2)

for each $k=0,1,2,\dots$, where $x_{-1},x_0\in D$ are the initial points, $L(.,.):D\times D\to \mathfrak{L}(X,Y)$ and $\mathfrak{L}(X,Y)$ is the space of bounded linear operators from X into Y. We have denoted by $L_{k-1,k}=L(x_{k-1},x_k)$.

If $F_2(x)\ne 0$, we have that the operator F is not Fréchet differentiable. In general, to approximate a solution of (1) in this situation, derivative-free iterative methods are used [11,12,13,14]. To obtain this type of iterative processes, it is common to approximate derivatives by difference divided. Remember that, given an operator $H:D\subset X\to Y,$ we call $\left[x,y;H\right]\in \mathfrak{L}(X,Y)$ a first order divided differences operator for H on the points x and y $(x\ne y)$ in D if

$$\left[x,y;H\right](x-y)=H(x)-H(y).$$

(3)

So, to solve (1) with iterative methods given from (2), we can consider at least two different procedures. Firstly, we have the Zincenko method [15], given by the following algorithm:

$$\left\{\begin{array}{c}\mathrm{Given}{x}_{-1},x_0\in D,\hfill \\ x_{k+1}=x_k-{\left[F_1^{\prime }(x_k)\right]}^{-1}F(x_k),n\ge 0,\hfill \end{array}\right.$$

(4)

where we directly eliminate the nondifferentiable part of F, i.e., $F_2$. So, in this case, $L_{k-1,k}=F_1^{\prime }(x_k)$ in (2). Secondly, we can consider an approximation of $F^{\prime }$ by divided differences, the secant-type methods [16,17]:

$$\left\{\begin{array}{c}\mathrm{Given}{x}_{-1},x_0\in D,\hfill \\ y_k=\lambda x_k+(1-\lambda )x_{k-1},\lambda \in [0,1),\hfill \\ x_{k+1}=x_k-{\left[y_k,x_k;F\right]}^{-1}F(x_k),n\ge 0,\hfill \end{array}\right.$$

(5)

where the secant method, for $\lambda =0$, is obtained. So, in this case, $L_{k-1,k}=\left[y_k,x_k;F\right]$ in (2). But, if we consider a better approximation of the derivative of F, an approximation of second order, we have the Kurchatov method [18]:

$$\left\{\begin{array}{c}\mathrm{Given}{x}_{-1},x_0\in D,\hfill \\ x_{k+1}=x_k-{\left[x_{k-1},2x_k-x_{k-1};F\right]}^{-1}F(x_n),n\ge 0,\hfill \end{array}\right.$$

(6)

in this case, $L_{k-1,k}=\left[x_{k-1},2x_k-x_{k-1};F\right]$ in (2).

By using this procedure of decomposition for operator F, we see that we can also consider the application of iterative methods that use derivatives when F is nondifferentiable. So, if we consider decomposition of F given in (1), we can use the Newton-secant-type algorithm:

$$\left\{\begin{array}{c}\mathrm{Given}{x}_{-1},x_0\in D,\hfill \\ y_k=\lambda x_k+(1-\lambda )x_{k-1},\lambda \in [0,1),\hfill \\ x_{k+1}=x_k-{\left(F_1^{\prime }(x_k)+\left[y_k,x_k;F\right]\right)}^{-1}F(x_k),n\ge 0,\hfill \end{array}\right.$$

(7)

where $L_{k-1,k}=F_1^{\prime }(x_k)+\left[y_k,x_k;F\right]$ in (2). The other possibility, from the decomposition method, is to consider the Newton–Kurchatov [19] algorithm:

$$\left\{\begin{array}{c}\mathrm{Given}{x}_{-1},x_0\in D,\hfill \\ x_{k+1}=x_k-{\left(F_1^{\prime }(x_k)+\left[x_{k-1},2x_k-x_{k-1};F\right]\right)}^{-1}F(x_k),n\ge 0,\hfill \end{array}\right.$$

(8)

where $L_{k-1,k}=F_1^{\prime }(x_k)+\left[x_{k-1},2x_k-x_{k-1};F\right]$ in (2). Another possibility is to consider Steffensen-type methods, that is, the methods associated to divided differences like $[x_k,x_k+F(x_k);F].$

As we can see, there are a lot of iterative methods that can be written as algorithms (2).

The main aim of this paper is to obtain a general study for the convergence, local and semilocal, for these Newton-type of iterative methods given in (2).

2. Convergence Analysis for Two-Point Newton-Type Methods

In this section, we present both semilocal and local convergence analysis. In the first one, the hypotheses are imposed on the initial guess; and in the second, on the solution. The results can be applied for smooth and nonsmooth operators.

2.1. Local Convergence Analysis

We start the local analysis of method (2). Let $v_0:[0,+\infty )\times [0,+\infty )\to [0,+\infty )$ be a nondecreasing continuous function. Assume that the equation

$$v_0(t,t)=1$$

(9)

has at least one positive root $r_0$. Let also $v:[0,r_0)\times [0,r_0)\to [0,+\infty )$ be a nondecreasing continuous function. Define function $\overline{v}$ on the interval $[0,r_0)$ by

$$\overline{v}(t)=\frac{v(t,t)}{1-v_0(t,t)}-1.$$

(10)

Assume that the equation

$$\overline{v}(t)=0$$

(11)

has a minimal positive solution r. It follows that for each $t\in [0,r)$

$$0\le v_0(t,t)<1$$

(12)

and

$$0\le \overline{v}(t)<1.$$

(13)

Our analysis of method (2) will use the conditions (A):

• $(a_1)$ There exist a solution $x^{*}\in D$ of Equation (1), and $B\in \mathfrak{L}(X,Y)$ such that $B^{-1}\in \mathfrak{L}(Y,X)$.
• $(a_2)$ Condition (9) holds and for each $x,u\in D$

  $$\parallel B^{-1}(L(x,u)-B)\parallel \le v_0(\parallel x-x^{*}\parallel ,\parallel u-y^{*}\parallel ),$$

  where $v_0$ is defined previously, and $r_0$ is given in (9).
  Set $D_0=D\cap \overline{U}(x^{*},r_0)$.
• $(a_3)$ For each $x,z\in D_0$, and any solution $y\in D$ of Equation (1)

  $$\parallel B^{-1}(F_1(x)+F_2(x)-L(z,x)y)\parallel \le v(\parallel z-y\parallel ,\parallel x-y\parallel )\parallel x-y\parallel ,$$

  where v is defined previously, and $L(·,·):D_0\times D_0\to \mathfrak{L}(X,Y)$.
• $(a_4)$$\overline{U}(x^{*},r)\subset D$, where r is given in (10).
• $(a_5)$

  $$\xi :=\frac{v(r,r)}{1-v_0(r,r)}\in [0,1).$$

We are able to perform our local analysis of method (2) based on the aformentioned conditions (A).

Theorem 1.

Assume that the conditions (A) hold. Then, sequence $x_k$, defined by method (2) for $x_{-1},x_0\in U(x^{*},r)-x^{*}$, is well defined in $U(x^{*},r)$; remains in $U(x^{*},r)$; and converges to $x^{*}$. Finally, the following estimates hold.

$$\parallel x_{k+1}-x^{*}\parallel \le \frac{v(\parallel x_{k-1}-x^{*}\parallel ,\parallel x_k-x^{*}\parallel )}{1-v_0(\parallel x_{k-1}-x^{*}\parallel ,\parallel x_k-x^{*}\parallel )}\parallel x_k-x^{*}\parallel \le \parallel x_k-x^{*}\parallel <r.$$

(14)

The vector $x^{*}$ is the only solution of Equation (1) in $\overline{U}(x^{*},r)$.

Proof. 

We will use mathematical induction on k.

Let $x,u\in U(x^{*},r)$.

Using (2), $(a_1)$ and $(a_2)$, we obtain

$$\parallel B^{-1}(L(x,u)-B)\parallel \le v_0(\parallel x-x^{*}\parallel ,\parallel u-x^{*}\parallel )\le v_0(r,r)<1.$$

(15)

Using the Banach lemma on invertible operators [20] and (15), we deduce that $L{(x,u)}^{-1}\in \mathfrak{L}(Y,X)$, and

$${\parallel L(x,u)}^{-1}B\parallel \le \frac{1}{1-v_0(\parallel x-x^{*}\parallel ,\parallel u-x^{*}\parallel )}.$$

(16)

In particular, estimate (16) holds for $x=x_0$, so $x_1$ is well defined by method (2) for $k=0$.

Using the definition of the method (2) (for $k=0$); $(a_1)$, $(a_3)$, (13), and (16) (for $k=0$) that

$$\begin{array}{ccc}\hfill \parallel x_1-x^{*}\parallel & =& \parallel x_0-x^{*}-L{(x_{-1},x_0)}^{-1}(x_{-1},x_0)(F_1(x_0)+F_2(x_0))\parallel \hfill \\ & =& \parallel [L{(x_{-1},x_0)}^{-1}B][B^{-1}(F_1(x_0)+F_2(x_0)-L(x_{-1},x_0)(x_0-x^{*}))]\parallel \hfill \\ & \le & \parallel L{(x_{-1},x_0)}^{-1}B\parallel \parallel B^{-1}(F_1(x_0)+F_2(x_0)-L(x_{-1},x_0)(x_0-x^{*}))\parallel \hfill \\ & \le & \frac{v(\parallel x_{-1}-x^{*}\parallel ,\parallel x_0-x^{*}\parallel )}{1-v_0(\parallel x_{-1}-x^{*}\parallel ,\parallel x_0-x^{*}\parallel )}\parallel x_0-x^{*}\parallel \le \parallel x_0-x^{*}\parallel <r,\hfill \end{array}$$

(17)

which shows estimate (14) for $k=0$ and $x_1\in U(x^{*},r)$.

Replace $x_0$, $x_1$ by $x_i$, $x_{i+1}$ in the preceding estimates to complete the induction for estimate (14). Then, from the estimate

$$\parallel x_{i+1}-x^{*}\parallel \le \mu \parallel x_i-x^{*}\parallel <r,$$

(18)

where

$$\mu =\frac{v(\parallel x_{-1}-x\parallel ,\parallel x_0-x^{*}\parallel )}{1-v_0(\parallel x_{-1}-x^{*}\parallel ,\parallel x_0-x^{*}\parallel )}\in [0,1),$$

thus, ${lim}_{i\to +\infty }{x}_i=x^{*}$ and $x_{i+1}\in U(x^{*},r)$. Moreover, for the uniqueness part, let $y^{*}\in \overline{U}(x^{*},r)$ with $F_1(y^{*})+F_2(y^{*})=0$. Using $(a3)$, $(a5)$, and estimate (17), we obtain in turn that

$$\begin{array}{ccc}\hfill \parallel x_{i+1}-y^{*}\parallel & \le & \parallel L{(x_{i-1},x_i)}^{-1}B\parallel \parallel B^{-1}(F_1(x_i)+F_2(x_i)-L(x_{i-1},x_i)(x_i-y^{*}))\parallel \hfill \\ & \le & \frac{v(\parallel x_{i-1}-y^{*}\parallel ,\parallel x_i-y^{*}\parallel )}{1-v_0(\parallel x_{i-1}-y^{*}\parallel ,\parallel x_i-y^{*}\parallel )}\parallel x_i-y^{*}\parallel \hfill \\ & \le & \mu \parallel x_i-y^{*}\parallel <{\mu }^{i+1}\parallel x_0-y^{*}\parallel ,\hfill \end{array}$$

(19)

which shows ${lim}_{i\to +\infty }{x}_i=y^{*}$—but, we showed ${lim}_{i\to +\infty }{x}_i=x^{*}$. Hence, we conclude that $x^{*}=y^{*}$.  □

Remark 1.

• Condition $(a_3)$ can be replaced by the stronger: for each $x,y\in D_0$

  $$\parallel B^{-1}(F_1(x)+F_2(x)-L(x)(x-y))\parallel \le v_1(\parallel x-y\parallel )\parallel x-y\parallel ,$$
  where function $v_1$ is as v. However, for each $t,t^{{}^{\prime }}\ge 0$

  $$v(t,t^{{}^{\prime }})\le v_1(t,t^{{}^{\prime }}).$$
• Linear operator B does not necessarily depend on q, where $q=x^{*}$ or $q=x_0$. It is used to determine the invertibility of linear operator $L(·,·)$ appearing in the method. The invertibility of B can be assured by an additional condition of the form $\left|\right|I-B\left|\right|<1$ or in some other way. A possible choice for B is $B=B(q)$ or $B=F_1^{{}^{\prime }}(q)$.
• It follows from the definition of $r_0$ and r that $r_0\ge r$.

2.2. Semilocal Convergence Analysis

For the semilocal case, we also define some functions and parameters. Let $w_0:[0,+\infty )\times [0,+\infty )\to [0,+\infty )$ be a continuous and nondecreasing function.

Assume that the equation

$$w_0(t,t)=1,$$

(20)

has one smallest positive root that we denote by ${\rho }_0$. Let $w:[0,{\rho }_0)\times [0,{\rho }_0)\times [0,{\rho }_0)\to [0,+\infty )$ be a nondecreasing continuous function. Moreover, for $\eta ,\overline{\eta }\ge 0$, define parameters $C_1$ and $C_2$ by

$$\begin{array}{ccc}\hfill C_1& =& \frac{w(\overline{\eta },\eta ,0)}{1-w_0(0,\eta )},\hfill \\ \hfill C_2& =& \frac{w(0,\frac{\eta }{1-C_1},\eta )}{1-w_0(\eta ,\frac{\eta }{1-C_1})}\hfill \end{array}$$

and function $C:[0,{\rho }_0)\to [0,+\infty )$ by $C(t)=\frac{w(t,t,t)}{1-w_0(t,t)}$. Assume that the equation

$$(\frac{C_1{C}_2}{1-C(t)}+C_1+1)\eta -t=0$$

(21)

has one smallest positive root that we denote by $\rho$.

The semilocal convergence analysis of method (2) will be based on conditions (H):

• $(h1)$ There exists $x_{-1},x_0\in D$, and $B\in \mathfrak{L}(X,Y)$ such that $B^{-1}\in \mathfrak{L}(Y,X)$.
• $(h2)$ Condition (20) holds, and for each $x\in D$

  $$\parallel B^{-1}(L(z,x)-B)\parallel \le w_0(\parallel z-x_0\parallel ,\parallel x-x_0\parallel ),$$

  where $w_0$ is defined previously and ${\rho }_0$ is given in (20).
  Set $D_1=D\bigcap \overline{U}(x_0,{\rho }_0)$.
• $(h3)$ For $L(·,·):D_1\times D_1\to \mathfrak{L}(X,Y)$, and each $x,y,z\in D_1$

  $$\begin{array}{c}\hfill \parallel B^{-1}(F_1(y)-F_1(x)+F_2(y)-F_2(x)-L(z,x)(y-x))\parallel \\ \hfill \le w(\parallel z-x_0\parallel ,\parallel y-x_0\parallel ,\parallel x-x_0\parallel )\parallel y-x\parallel ,\end{array}$$

  where w is defined previously.
• $(h4)$$\overline{U}(x_0,\rho )\subseteq D$ and condition (21) holds for $\rho$, where $\parallel x_1-x_0\parallel \le \eta$ and $\parallel x_{-1}-x_0\parallel \le \overline{\eta }$.

Then, using the hypotheses (H), we obtain the estimates:

$$\begin{array}{ccc}\hfill \parallel x_2-x_1\parallel & \le & \frac{w(\parallel x_{-1}-x_0\parallel ,\parallel x_1-x_0\parallel ,\parallel x_0-x_0\parallel )\parallel x_1-x_0\parallel }{1-w_0(\parallel x_0-x_0\parallel ,\parallel x_1-x_0\parallel )}=C_1\parallel x_1-x_0\parallel ,\hfill \\ \hfill \parallel x_2-x_0\parallel & \le & \parallel x_2-x_1\parallel +\parallel x_1-x_0\parallel \le (1+C_1)\parallel x_1-x_0\parallel \hfill \\ & =& \frac{1-C_1^2}{1-C_1}\parallel x_1-x_0\parallel \hfill \\ & <& \frac{\eta }{1-C_1}<\rho ,\hfill \\ \hfill \parallel x_3-x_2\parallel & \le & \frac{w(\parallel x_0-x_0\parallel ,\parallel x_2-x_0\parallel ,\parallel x_1-x_0\parallel )}{1-w_0(\parallel x_1-x_0\parallel ,\parallel x_2-x_0\parallel )}\parallel x_2-x_1\parallel \hfill \\ & \le & \frac{w(0,\frac{\eta }{1-C_1},\eta )}{1-w_0(\eta ,\frac{\eta }{1-C_1})}\parallel x_2-x_1\parallel =C_2\parallel x_2-x_1\parallel ,\hfill \\ \hfill \parallel x_3-x_0\parallel & \le & \parallel x_3-x_2\parallel +\parallel x_2-x_1\parallel +\parallel x_1-x_0\parallel \hfill \\ & \le & C_2\parallel x_2-x_1\parallel +C_1\parallel x_1-x_0\parallel +\parallel x_1-x_0\parallel \hfill \\ & \le & (C_2{C}_1+C_1+1)\parallel x_1-x_0\parallel ,\hfill \\ \hfill \parallel x_4-x_3\parallel & \le & \frac{w(\parallel x_1-x_0\parallel ,\parallel x_3-x_0\parallel ,\parallel x_2-x_0\parallel )}{1-w_0(\parallel x_2-x_0\parallel ,\parallel x_3-x_0\parallel )}\parallel x_3-x_2\parallel \hfill \\ & \le & C(\rho )\parallel x_3-x_2\parallel \le C(\rho )C_2\parallel x_2-x_1\parallel \hfill \\ & \le & C(\rho )C_2{C}_1\parallel x_1-x_0\parallel ,\hfill \end{array}$$

(22)

similarly for $i=3,4,\dots$

$$\begin{array}{ccc}\hfill \parallel x_{i+1}-x_i\parallel & \le & C(\rho )\parallel x_i-x_{i-1}{\parallel \le C(\rho )}^{i-2}\parallel x_3-x_2\parallel ,\hfill \\ \hfill \parallel x_{i+1}-x_0\parallel & \le & \parallel x_{i+1}-x_i\parallel +\dots +\parallel x_4-x_3\parallel +\parallel x_3-x_0\parallel \hfill \\ & \le & C(\rho )\parallel x_i-x_{i-1}\parallel +\dots +C(\rho )\parallel x_3-x_2\parallel \hfill \\ & & +(C_2{C}_1+C_1+1)\parallel x_1-x_0\parallel \hfill \\ & \le & C{(\rho )}^{i-2}\parallel x_3-x_2\parallel +\dots +C(\rho )\parallel x_3-x_2\parallel \hfill \\ & & +(C_2{C}_1+C_1+1)\parallel x_1-x_0\parallel \hfill \\ & \le & (\frac{1-C{(\rho )}^{i-1}}{1-C(\rho )}{C}_2{C}_1+C_1+1)\parallel x_1-x_0\parallel \hfill \\ & <& (\frac{C_1{C}_2}{1-C(\rho )}+C_1+1)\eta \le \rho ,\hfill \end{array}$$

(23)

$$\begin{array}{ccc}\hfill \parallel x_{i+j}-x_i\parallel & \le & \parallel x_{i+j}-x_{i+j-1}\parallel +\parallel x_{i+j-1}-x_{i+j-2}\parallel +\dots +\parallel x_{i+1}-x_i\parallel \hfill \\ & \le & (C{(\rho )}^{i+j-3}+\dots +C{(\rho )}^{i-2})\parallel x_3-x_2\parallel \hfill \\ & \le & C{(\rho )}^{i-2}\frac{1-C{(\rho )}^{j-1}}{1-C(\rho )}\parallel x_3-x_2\parallel \hfill \\ & \le & C{(\rho )}^{i-2}\frac{1-C{(\rho )}^{j-1}}{1-C(\rho )}{C}_2{C}_1\parallel x_1-x_0\parallel \hfill \\ & \le & C{(\rho )}^{i-2}\frac{1-C{(\rho )}^{j-1}}{1-C(\rho )}{C}_2{C}_1\eta .\hfill \end{array}$$

(24)

It follows from (23) that $x_i\in U(x_0,\rho )$; and from (24) that sequence $x_i$ is complete in a Banach space X. In particular, it converges to some $x^{*}\in \overline{U}(x_0,\rho )$. By letting $i\to +\infty$ in the estimate

$$\parallel B^{-1}(F_1(x_i)+F_2(x_i))\parallel =\parallel B^{-1}(F_1(x_i)+F_2(x_i)-F_1(x_{i-1})-F_2(x_{i-1})-L_{i-2,i-1}(x_i-x_{i-1}))\parallel$$

$$\le \frac{w(\parallel x_{i-2}-x_0\parallel ,\parallel x_i-x_0\parallel ,\parallel x_{i-1}-x_0\parallel )\parallel x_i-x_{i-1}\parallel }{1-w_0(\parallel x_{i-1}-x_0\parallel ,\parallel x_i-x_0\parallel )}\le \frac{w(\rho ,\rho ,\rho )}{1-w_0(\rho ,\rho )}\parallel x_i-x_{i-1}\parallel ,$$

we obtain $F_1(x^{*})+F_2(x^{*})=0$. The uniqueness part is omitted as analogous to the one in the local convergence case.

Hence, we can present our semilocal convergence result associated to the method (2).

Theorem 2.

Assume that the conditions (H) hold. Then, sequence $x_k$, defined by the method (2) for $x_{-1},x_0\in D$, is well defined in $U(x_0,\rho )$; remains in $U(x_0,\rho )$; and converges to a solution $x^{*}\in \overline{U}(x_0,\rho )$ of Equation (1). On the other hand, the vector $x^{*}$ is the only solution of Equation (1) in $\overline{U}(x_0,\rho )$.

The same comments given in the previous remark hold.

3. Numerical Experiment

Consider the nondifferentiable system of equations

$$\left\{\begin{array}{c}3x_1^2{x}_2+x_2^2-1+{|x_1-1|}^{3/2}=0,\hfill \\ x_1^4+x_1{x}_2^3-1+{|x_2|}^{3/2}=0.\hfill \end{array}\right.$$

(25)

We therefore have an operator $F:R^2\to R^2$ such that $F=F_1+F_2$, as in (1), with $F_1,F_2:R^2\to R^2$, $F_1=({F_1}^1,{F_1}^2)$, $F_2=({F_2}^1,{F_2}^2)$, being

$$F_1^1(x_1,x_2)=3x_1^2{x}_2+x_2^2-1\mathrm{and}{F}_1^2(x_1,x_2)=x_1^4+x_1{x}_2^3-1,$$

$$F_2^1(x_1,x_2)={|x_1-1|}^{3/2}\mathrm{and}{F}_2^2(x_1,x_2)={|x_2|}^{3/2},$$

where the operator $F_1$ is continuously Fréchet-differentiable and $F_2$ is continuous but is a Fréchet nondifferentiable operator.

For $\mathbf{u}=(u_1,u_2^T),\mathbf{v}={(v_1,v_2)}^T\in R^2$, we consider the divided difference of first order defined by $[\mathbf{u},\mathbf{v};F]={({[\mathbf{u},\mathbf{v};F]}_{ij})}_{i,j=1}^2\in \mathcal{L}(R^2,R^2)$, where

$$\begin{array}{ccc}\hfill {[\mathbf{u},\mathbf{v};F]}_{i1}& =& \left\{\begin{array}{c}{\displaystyle \frac{F_i(u_1,u_2,\dots ,v_m)-F_i(u_1,v_2,\dots ,v_m)}{u_1-v_1},\mathrm{if}{u}_2\ne v_2,}\hfill \\ 0,if{u}_1=v_1,\hfill \end{array}\right.\hfill \\ \hfill \\ \hfill \\ \hfill {[\mathbf{u},\mathbf{v};F]}_{i2}& =& \left\{\begin{array}{c}{\displaystyle \frac{F_i(u_1,u_2,\dots ,v_m)-F_i(u_1,v_2,\dots ,v_m)}{u_2-v_2},\mathrm{if}{u}_2\ne v_2,}\hfill \\ 0,if{u}_2=v_2,\hfill \end{array}\right.\hfill \\ \hfill \\ \hfill \end{array}$$

for $i=1,2.$

The iterative processes given by (1) allow us to consider direct iterative processes, such as (5) and (6); as well as iterative processes that use the decomposition method, such as (7) and (8). In this experiment, for the nondifferentiable system (25), we check that the application of the iterative processes that use the decomposition method have better behavior than the direct methods.

To carry out this study, we will consider as an approximate solution of system (25):

$${\mathbf{x}}^{*}=\left(0.9383410452297656,0.3312445136375143\right),$$

the starting points ${\mathbf{x}}_{-\mathbf{1}}=(5,5)$ and ${\mathbf{x}}_{\mathbf{0}}=(1,0)$, and use a tolerance $\parallel {\mathbf{x}}_{\mathbf{n}+\mathbf{1}}-{\mathbf{x}}_{\mathbf{n}}\parallel \le {10}^{-16}.$ In these conditions, in

Table 1. $\parallel {\mathbf{x}}^{*}-{\mathbf{x}}_{\mathbf{n}}\parallel$ for secant-type methods (5) and different values of the parameter $\lambda$.

n	$\mathit{\lambda }=0$	$\mathit{\lambda }=0.5$	$\mathit{\lambda }=0.99$
1	$3.18484\times {10}^{-1}$	$2.965\times {10}^{-1}$	$4.54388\times {10}^{-2}$
2	$5.21264\times {10}^{-2}$	$4.13083\times {10}^{-2}$	$3.74494\times {10}^{-3}$
3	$3.66108\times {10}^{-3}$	$2.35344\times {10}^{-3}$	$2.94716\times {10}^{-5}$
4	$2.59348\times {10}^{-4}$	$7.2935\times {10}^{-5}$	$3.69966\times {10}^{-9}$
5	$1.30031\times {10}^{-6}$	$1.24012\times {10}^{-7}$	$2.10942\times {10}^{-15}$
6	$4.42187\times {10}^{-10}$	$6.07747\times {10}^{-12}$
7	$1.11022\times {10}^{-15}$	$1.11022\times {10}^{-16}$

and

Table 2. Kurchatov method (6).

n	$\parallel {\mathit{x}}^{*}-{\mathit{x}}_{\mathit{n}}\parallel$
1	$2.99754\times {10}^{-1}$
2	$1.07269\times {10}^{-1}$
3	$4.20963\times {10}^{-2}$
4	$8.37098\times {10}^{-3}$
5	$2.78931\times {10}^{-4}$
6	$9.0784\times {10}^{-8}$
7	$2.9826\times {10}^{-11}$

we can see the results of the application of the direct iterative processes, the secant-type, and Kurchatov methods. Whereas in

Table 3. $\parallel {\mathbf{x}}^{*}-{\mathbf{x}}_{\mathbf{n}}\parallel$ for Newton-secant-type methods (7) and different values of the parameter $\lambda$.

n	$\mathit{\lambda }=0$	$\mathit{\lambda }=0.5$	$\mathit{\lambda }=0.99$
1	$2.3538\times {10}^{-1}$	$1.00278\times {10}^{-1}$	$4.29554\times {10}^{-2}$
2	$3.48717\times {10}^{-1}$	$2.88094\times {10}^{-2}$	$2.53626\times {10}^{-3}$
3	$1.47537\times {10}^{-1}$	$1.90518\times {10}^{-3}$	$9.06208\times {10}^{-6}$
4	$3.4371\times {10}^{-2}$	$8.39107\times {10}^{-6}$	$1.9925\times {10}^{-10}$
5	$3.08399\times {10}^{-3}$	$4.78016\times {10}^{-9}$
6	$4.63665\times {10}^{-5}$	$7.38298\times {10}^{-15}$
7	$4.05776\times {10}^{-8}$
8	$2.96929\times {10}^{-13}$

and

Table 4. Newton–Kurchatov method (8).

n	$\parallel \overline{{\mathit{x}}^{*}}-\overline{{\mathit{x}}_{\mathit{n}}}\parallel$
1	$6.1659\times {10}^{-1}$
2	$4.75269\times {10}^{-3}$
3	$6.29174\times {10}^{-5}$
4	$3.37027\times {10}^{-9}$

we can see the results of the application of the iterative processes that use the decomposition method, Newton-secant-type and Newton–Kuchatov methods. Observing the results obtained, it is evident that the best behavior of the iterative processes is given by (2) using the decomposition method.

Remark 2.

In the above example, we have selected the initial guess in a region where the operator is not smooth. The methods can be applied to systems where the operator is not smooth at the solution.

For instance, for the system:

$$\left\{\begin{array}{c}|x_1^2-1|+x_2-1=0,\hfill \\ x_1+x_2^2-1=0,\hfill \end{array}\right.$$

(26)

the solution is $(1,1)$. If we take as initial guess $(0.5,0.5)$, the Steffensen method gives as errors $3.12\times {10}^{-2}$, $5.48\times {10}^{-4}$, $1.92\times {10}^{-7}$, $2.15\times {10}^{-14}$, we observe its second order.

4. Boundary Value Problem: Discretization via the Multiple Shooting Method

We will use the multiple shooting method for the discretization of boundary problems of the type

$$y^{″}(t)=f(t,y(t),y^{\prime }(t)),y(a)=\alpha ,y(b)=\beta .$$

(27)

Thus, we should find the solution of the following nonlinear system of equations $F(s)=0$, where $F:{\mathbb{R}}^N⟶{\mathbb{R}}^N$ and

$$\left\{\begin{array}{ccc}\hfill F_1(s_0,s_1,\dots ,s_{N-1})& =& s_1-y^{\prime }(t_1;s_0)\hfill \\ \hfill F_2(s_0,s_1,\dots ,s_{N-1})& =& s_2-y^{\prime }(t_2;s_0,s_1)\hfill \\ & ⋮& \\ \hfill F_{N-1}(s_0,s_1,\dots ,s_{N-1})& =& s_{N-1}-y^{\prime }(t_{N-1};s_0,s_1,\dots ,s_{N-2})\hfill \\ \hfill F_N(s_0,s_1,\dots ,s_{N-1})& =& \beta -y(t_N;s_0,s_1,s_{N-2},s_{N-1}).\hfill \end{array}\right.$$

for a discretization of $[a,b]$ with N subintervals,

$$TNj,T=b-a,j=0,1,\dots ,N.$$

We consider the secant-type method

$$\left\{\begin{array}{c}\mathrm{Given}{y}_{-1},y_0\in D,\hfill \\ z_n={\lambda }_n{y}_n+(1-{\lambda }_n)y_{n-1},{\lambda }_n\in [0,1),\hfill \\ y_{n+1}=y_n-{\left[z_n,y_n;F\right]}^{-1}F(y_n),n\ge 0,\hfill \end{array}\right.$$

(28)

where ${\lambda }_n$ is such that $\left|\right|z_n-x_n\left|\right|\le Tol$ for a given tolerance, and Newton’s method

$$\left\{\begin{array}{c}\mathrm{Given}{y}_0\in D,\hfill \\ y_{n+1}=y_n-{F(y_n)}^{-1}F(y_n),n\ge 0.\hfill \end{array}\right.$$

(29)

We perform a numerical comparison between both methods. As we can see, in the multiple shooting method, the iterative schemes are used as black boxes.

For the initial slope ${\overrightarrow{s}}_0=\left(s_0^0,s_1^0,\dots ,s_{N-1}^0\right)$, we propose

$$\left\{\begin{array}{ccc}\hfill s_0^0& =& {\displaystyle \frac{\beta -\alpha }{b-a}=\frac{y(t_N)-y(t_0)}{t_N-t_0},}\hfill \\ \hfill s_1^0& =& {\displaystyle \frac{y(t_N)-y(t_1;s_0)}{t_N-t_1},}\hfill \\ \hfill s_2^0& =& {\displaystyle \frac{y(t_N)-y(t_2;s_0,s_1)}{t_N-t_2},}\hfill \\ & ⋮& \\ \hfill s_{N-1}^0& =& {\displaystyle \frac{y(t_N)-y(t_{N-1};s_0,s_1,\dots ,s_{N-2})}{t_N-t_{N-1}}.}\hfill \end{array}\right.$$

We analyze this particular example ([21], p. 554):

$$\begin{array}{ccc}\hfill y^{{}^{″}}(t)& =& \tau ·sinh(\tau ·y(t)),\hfill \\ \hfill y(0)& =& 0,y(1)=1.\hfill \end{array}$$

We take $\tau =2.5$ and $N=4$ subintervals.

This problem admits the solution:

$${\displaystyle y(t)=\frac{2}{\tau }argsinh\left(\frac{s}{2}·\frac{\mathrm{sn}(\tau t,1-s^2/4)}{\mathrm{cn}(\tau t,1-s^2/4)}\right),}$$

where

$$s=y^{\prime }(0)=0.3713363932677645$$

and $\mathrm{sn}(·,·)$ and $\mathrm{cn}(·,·)$ are the Jacobi elliptic functions.

$$\begin{array}{c}\mathrm{Newton}’\mathrm{s} \mathrm{method} (29),\hfill \\ \begin{array}{}\\ n& \parallel F({\overrightarrow{s}}_n){\parallel }_{\infty }& \parallel y(t)-y_n{\parallel }_{\infty }& \parallel y^{\prime }(t)-y_n^{\prime }{\parallel }_{\infty }\\ 0& {10}^0& {10}^{-1}& {10}^0\\ 1& {10}^{-1}& {10}^{-1}& {10}^{-1}\\ 2& {10}^{-2}& {10}^{-2}& {10}^{-2}\\ 3& {10}^{-4}& {10}^{-4}& {10}^{-4}\\ 4& {10}^{-7}& {10}^{-7}& {10}^{-7}\\ 5& {10}^{-15}& {10}^{-15}& {10}^{-14}\\ \end{array}\hfill \end{array}$$

$$\begin{array}{c}\mathrm{Secant}-\mathrm{type} \mathrm{method} (28),\hfill \\ \begin{array}{}\\ n& \parallel F({\overrightarrow{s}}_n){\parallel }_{\infty }& \parallel y(t)-y_n{\parallel }_{\infty }& \parallel y^{\prime }(t)-y_n^{\prime }{\parallel }_{\infty }& \parallel F^{{}^{\prime }}(y_n)-[y_n,x_n;F]{\parallel }_{\infty }\\ 0& {10}^0& {10}^{-1}& {10}^0& {10}^{-6}\\ 1& {10}^{-1}& {10}^{-1}& {10}^{-1}& {10}^{-6}\\ 2& {10}^{-2}& {10}^{-2}& {10}^{-2}& {10}^{-7}\\ 3& {10}^{-4}& {10}^{-4}& {10}^{-4}& {10}^{-6}\\ 4& {10}^{-7}& {10}^{-7}& {10}^{-7}& {10}^{-6}\\ 5& {10}^{-15}& {10}^{-15}& {10}^{-14}& {10}^{-6}\\ \end{array}\hfill \end{array}$$

The methods using Jacobians obtain their order of convergence. However, in this example, the computation of the derivatives involves the approximation of a more complicated problem. For this reason, the methods free of derivatives are preferred, see [21]. Of course, we need to compute a good approximation to the Jacobian, this is the motivation of our parameters ${\lambda }_n$. For more similar examples and conclusions, we refer [22].

Remark 3.

In many cases, when we manipulate an image, some random noise appears. This noise makes the later steps of processing the image difficult and inaccurate.

Let $f:\mathsf{\Omega }\to \mathbb{R}$ be a noise signal or image.

Introducing the variable w:

$$w=\frac{\nabla u}{\sqrt{{|\nabla u|}^2}},$$

the Total-Variation model is equivalent to the nonlinear and nondifferentiable system:

$$\begin{array}{ccc}\hfill -\nabla ·w+\lambda (u-f)& =& 0,\hfill \\ \hfill w\sqrt{{|\nabla u|}^2}-\nabla u& =& 0.\hfill \end{array}$$

This system should be discretized using finite differences and the associated nonlinear system of equations can be approximated by our family (see [23] for more details).

5. Conclusions

This paper was devoted to the analysis of a general family of two-point Newton-type methods for smooth and nonsmooth operators. We have considered two types of convergence—semilocal and local. The family includes a great number of methods. We have applied the schemes to several interesting problems, in particular to nonsmooth nonlinear systems, boundary problems, and image denoising models.