Extended convergence of two-step iterative methods for solving equations with applications

Abstract.

The convergence of two-step iterative methods of third and fourth order of convergence are studied under weaker hypotheses than in earlier works using our new idea of the restricted convergence region. This way, we obtain a finer semilocal and local convergence analysis, and under the same or weaker hypotheses. Hence, we extend the applicability of these methods in cases not covered before. Numerical examples are used to compare our results favorably to earlier ones.

1. Introduction

Let $B_1$,$B_2$ stand for Banach spaces and $\mathrm{\Omega }\subseteq B_1$ be a nonempty, convex and open set. By $LB(B_1,B_2)$ we denote the space of bounded linear operators from $B_1$ to $B_2$.

There is a plethora of problems in various disciplines that can be written using mathematical modeling like

(1) $$F(x)=0$$

where $F:\mathrm{\Omega }\to B_2$ is differentiable in the sense of Fréchet. Therefore finding a solution $x_{\ast }$ of equation (1) is of great importance and challenge [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21]. One wishes $x_{\ast }$ to be found in closed form but this is done only in special cases. This is why, we resort to iterative methods approximating $x_{\ast }$. Numerous studies have been published on the local as well as the semilocal convergence of iterative methods. Among these methods is the single step Newton’s method defined by

(2) $$z_0\in \mathrm{\Omega },z_{n+1}=z_n-F^{\prime }(z_{n-1})F(z_n)$$

for each $n=0,1,2,\mathrm{\dots }$, which is considered the most popular.

Iterative methods converge under certain hypotheses. However, their convergence region is small in general. Finding a more precise than $\mathrm{\Omega }$ set D containing the iterates is very important, since the Lipschitz constants in D will be at least as tight as in $\mathrm{\Omega }$. This will in turn lead to a finer convergence analysis of these methods. We pursue this goal in the present study by studying the two-step fourth convergence order Newton’s method defined as

(3) $x_0\in \mathrm{\Omega },y_n$ $=$ $x_n-F^{\prime }{(x_n)}^{-1}F(x_n)$

$x_{n+1}$ $=$ $y_n-F^{\prime }{(y_n)}^{-1}F(y_n)$

as well as the two-step third order Traub method [21]

(4) $\overline{x_0}\in \mathrm{\Omega },\overline{y_n}$ $=$ $\overline{x_n}-F^{\prime }{(\overline{x_n})}^{-1}F(\overline{x_n})$

$\overline{x_{n+1}}$ $=$ $\overline{y_n}-F^{\prime }{(\overline{x_n})}^{-1}F(\overline{y_n})$

The local as well as the semilocal convergence of method (3) and (4) is carried out under the same set of hypotheses.

The layout of the rest of the article involves the semilocal and local convergence of these methods in Section 2 and Section 3, respectively. Numerical examples are given in Section 4.

2. Semilocal convergence analysis

We present first the semilocal convergence analysis for method (3). First, we need to show an auxiliary result on majorizing sequences for method (3). The proof is an extension of the corresponding one given by us in [9].

Let $l>0,l_0>0$ and $\eta >0$ be given parameters. Define parameters $\gamma ,s_0$ and $t_1$ by

(5) $$\gamma =\frac{2l}{l+\sqrt{l^2+8l_{0}l}},s_0=\eta ,\text{and}{t}_1=s_0[1+\frac{l_0{s}_0}{2(1-l_0{s}_0)}],\text{for}{l}_0{s}_0\ne 1,$$

and the scalar sequence $t_n$ for each n=1,2,…by

$t_0=0,t_1$ $=$ $s_0+\frac{l{s}_0^2}{2(1-l_0{s}_0)},$

$s_n$ $=$ $t_n+\frac{l{(t_n-s_{n-1})}^2}{2(1-l_0{t}_n)},$

(6) $t_{n+1}$ $=$ $s_n+\frac{l{(s_n-t_n)}^2}{2(1-l_0{s}_n)}.$

Lemma 1.

Let $>0,l_0>0$ and $\eta >0$ be given parameters. Suppose that

(7)

Then, the sequence $t_n$ is nondecreasing, bounded from above by $t_{\ast \ast }=\frac{\eta }{1-\gamma }$, and converges to its unique least upper bound $t_{\ast }$ satisfying $\eta \le t_{\ast }\le t_{\ast \ast }$. Moreover, the following items hold for each $n=1,2,\mathrm{\dots }$

(8)

(9)

and

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

Proof.

Estimations (8)–(10) hold true, if

(11)

(12)

and

(13)

Estimations (11)–(13) hold true for $m=0$, by (5)–(7). Suppose (11)–(13) hold true for $m=1,2,\mathrm{\dots },n$.

By (8) and (9), we get

(14) $s_m$ $\le$ $t_m+{\gamma }^{2m}\eta \le s_{m-1}+{\gamma }^{2m-1}\eta +{\gamma }^{2m}\eta$

$\le$

and

(15) $t_{m+1}$ $\le$ $t_m+{\gamma }^{2m+1}\eta \le t_m+{\gamma }^{2m}\eta +{\gamma }^{2m+1}\eta$

$\le$

Then, (11) shall hold, if

(16) $$\frac{l{\gamma }^{2m}\eta }{2(1-l_0(\frac{1-{\gamma }^{2m+1}}{1-\gamma })\eta )}\le \gamma$$

or

(17) $$\frac{l}{2}{\gamma }^{2m}\eta +l_0\gamma \frac{1-{\gamma }^{2m+1}}{1-\gamma }\eta \le \gamma$$

or

(18) $$\frac{l}{2}{\gamma }^{2m-1}\eta +l_0(1+\gamma +\mathrm{\dots }+{\gamma }^{2m})\eta -1\le 0.$$

We are motivated by (18) to define recurrent polynomials $f_m$ defined on the interval $[0,1)$ by

(19) $$f_m(t)=\frac{l}{2}{t}^{2m-1}\eta +l_0(1+t+\mathrm{\dots }+t^{2m})\eta -1,$$

which satisfies

(20) $$f_{m+1}(t)=f_m(t)+p(t)t^{2m-1}\eta ,$$

where, polynomial $p$ is given by

(21) $$p(t)=l_0{t}^3+(\frac{l}{2}+l_0)t^2-1=(t+1)(l_0{t}^2+\frac{l}{2}t-1),$$

so $p(\gamma )=0.$

Notice, in particular from (20) that

(22) $$f_{m+1}(\gamma )=f_m(\gamma )=\underset{m\to \mathrm{\infty }}{lim}{f}_m(\gamma )=f_{\mathrm{\infty }}(\gamma ).$$

Evidently by (21), (18) is true, if

(23) $$f_{\mathrm{\infty }}(\gamma )\le 0$$

But

(24) $$f_{\mathrm{\infty }}(\gamma )=\frac{l_0\eta }{1-\gamma }-1$$

Then, by (19) and (24), we see that (23) is satisfied.

Similarly, to show (12), we must have

(25) $$\frac{l{\gamma }^{2m+1}\eta }{2(1-l_0\frac{1-{\gamma }^{2m+2}}{1-\gamma }\eta )}\le \gamma$$

or

(26) $$\frac{l}{2}{\gamma }^{2m}\eta +l_0(1+\gamma +\mathrm{\dots }+{\gamma }^{2m+1})\eta -1\le 0$$

leading to the definition of recurrent functions $g_m$ defined on the interval $[0,1)$ by

$$g_m(t)=\frac{l}{2}{t}^{2m}\eta +l_0(1+t+\mathrm{\dots }+t^{2m+1})\eta -1,$$

which satisfies

$$g_{m+1}(t)=g_m(t)+p(t)t^{2m}\eta ,$$

so again

$$g_{m+1}(\gamma )=g_m(\gamma )==\underset{m\to \mathrm{\infty }}{lim}{g}_m(\gamma )=g_{\mathrm{\infty }}(\gamma )$$

Item (26) holds, if

(27) $$g_{\mathrm{\infty }}(\gamma )\le 0.$$

But

(28) $$g_{\mathrm{\infty }}(\gamma )=\frac{l_0\eta }{1-\gamma }-1,$$

so (27) holds true by (7) and (28). Item (13) also holds true by (6),(11) and (12). Hence, the induction for (11)–(13) is completed, and items (8)–(10) hold for each $n=1,2,\mathrm{\dots }$. It follows from (13)-(15) that sequence $t_n$ is nondecreasing, bounded from above by $t_{\ast \ast }$ and as such it converges to $t_{\ast }$. ∎

Remark 2.

It is worth noticing that if

$$l_0\eta \le 1/2$$

then

$$\frac{l(t_1-s_0)}{2(1-l_0{t}_1)}\le \frac{l(s_0-t_0)}{2(1-l_0{s}_0)}$$

That is (7) holds, if

or equivalently, if

(29) $$h_A=l_1\eta \le \frac{1}{2},$$

where

(30) $$l_1=\frac{1}{8}(l+4l_0+\sqrt{l^2+8l_{0}l})$$

Hence, (29) and (30) can replace (7) in 1 and in what follows from now on. The sufficient convergence criterion (29) is similar to the corresponding one for Newton’s method given by us in [11], if $l$ replaces $l_1$. But $l\le l_1$, where $l_1$ is the Lipschitz constant on $\mathrm{\Omega }$. Hence, the sufficient convergence criteria for Newton’s method in [11] are also improved.

Let $S(x,a)$ stand for the open ball in $B_1$ with center $x\in B_1$ and of radius $a>0$. By $\overline{S}(x,a)$, we denote the closure of $S(x,a)$.

The semilocal convergence of method (3) uses the conditions (A):

• $F:\mathrm{\Omega }\subset B_1\to B_2$ is a continuously differentiable operator in the sense of, Fréchet, and there exists $x_0\in \mathrm{\Omega }$ such that $F^{\prime }{(x_0)}^{-1}\in LB(B_2,B_1)$ with $\Vert F^{\prime }{(x_0)}^{-1}F(x_0)\Vert =\eta$

• There exists $l_0>0$ such that for each $x\in \mathrm{\Omega }$

  $$\Vert F^{\prime }{(x_0)}^{-1}(F^{\prime }(x)-F^{\prime }(x_0))\Vert \le l_0\Vert x-x_0\Vert .$$

  Define $S_0=\mathrm{\Omega }\cap S(x_1,\frac{1}{l_0}-\eta ),$ where $x_1=x_0-F^{\prime }{(x_0)}^{-1}F(x_0)$, and by (29) and (30).

• There exists $l>0$ such that for each $x,y\in S_0$

  $$\Vert F^{\prime }{(x_0)}^{-1}(F^{\prime }(y)-F^{\prime }(x))\Vert \le l\Vert y-x\Vert .$$

• Hypotheses of 1 hold with (7) replaced by (29) and (30).

• $\overline{S}(x_0,t_{\ast })\subset \mathrm{\Omega }$, where $t_{\ast }$ is given in 1.

• There exists $t_{\ast }^1\ge t_{\ast }$ such that
  

Set $S_1=\mathrm{\Omega }\cap \overline{S}(x_0,t_{\ast }^1).$ Next, we can show the semilocal convergence result for methos (3)

Theorem 3.

Assume that the conditions (A) hold. Then, $x_n\subset S(y_0,\frac{1}{l_0}-\eta ),n=1,2\mathrm{\dots }$ and converges to some $x_{\ast }$ which is the only solution of equation $F(x)=0$ in the set $S_1$.

Proof.

We must prove using mathematical induction that

(31) $$\Vert x_{m+1}-y_m\Vert \le t_{m+1}-s_m$$

and

(32) $$\Vert y_m-x_m\Vert \le s_m-t_m$$

By (a1) and (29), we have

$$\Vert y_0-x_0\Vert =\Vert F^{\prime }{(x_0)}^{-1}F(x_0)\Vert \le \eta \le \frac{1}{l_0}-\eta ,$$

so $x_0\in S(y_0,\frac{1}{l_0}-\eta )$, and (32) holds for $\eta =0$. By method (3) for $\eta =0$, (a1),(6) and (a2) we have in turn that

$\Vert x_1-y_0\Vert$ $=$ $\Vert F^{\prime }{(x_0)}^{-1}[F(y_0)-F(x_0)-F^{\prime }(x_0)(y_0-x_0)]\Vert$

$=$ $\Vert {\displaystyle {\int }_0^1}{F}^{\prime }{(x_0)}^{-1}(F^{\prime }(x_0+\tau (y_0-x_0))-F^{\prime }(x_0))𝑑\tau (y_0-x_0)\Vert$

$\le$ $\frac{l_0}{2}{\Vert y_0-x_0\Vert }^2$

$\le$ $\frac{l_0}{2}{(s_0-t_0)}^2$

$=$

so $x_1\in \overline{S}(y_0,\frac{1}{l_0}-\eta )$.

Then, by (a2), we have

(33)

so by (33) and the Banach lemma on invertible operators [18] $F^{\prime }{(x_1)}^{-1}\in LB(B_2,B_1)$, and

(34) $$\Vert F^{\prime }{(x_1)}^{-1}{F}^{\prime }(x_0)\Vert \le \frac{1}{1-l_0\Vert x_1-x_0\Vert }.$$

In view of method (3) and (a3), we get

$\Vert F^{\prime }{(x_0)}^{-1}F(x_1)\Vert$ $=\Vert F^{\prime }{(x_0)}^{-1}[F(x_1)-F(y_0)-F^{\prime }(y_0)(x_1-y_0)]\Vert$

$=\Vert {\displaystyle {\int }_0^1}{F}^{\prime }{(x_0)}^{-1}(F^{\prime }(y_0+\tau (x_1-y_0)-F^{\prime }(y_0)))\Vert d\tau \Vert x_1-y_0\Vert$

$\le \frac{l_0}{2}{\Vert x_1-y_0\Vert }^2\le \frac{l}{2}{(t_1-s_0)}^2$

so

$\Vert y_1-x_1\Vert$ $=\Vert [F^{\prime }{(x_1)}^{-1}{F}^{\prime }(x_0)][F^{\prime }{(x_0)}^{-1}F(x_1)]\Vert$

$\le \Vert [F^{\prime }{(x_1)}^{-1}{F}^{\prime }(x_0)]\Vert \Vert [F^{\prime }{(x_0)}^{-1}F(x_1)]\Vert$

$\le \frac{l{(t_1-s_0)}^2}{2(1-l_0{t}_1)}=s_1-t_1,$

and

so (32) holds for $m=1$ and $y_1\in \overline{S}(y_0,\frac{1}{l_0}-\eta )$.

Using method (3) as above, we have

$\Vert x_2-y_1\Vert$ $\le \Vert F^{\prime }{(y_0)}^{-1}{F}^{\prime }(x_0)\Vert \Vert F^{\prime }{(x_0)}^{-1}{F}^{\prime }(y_1)\Vert$

$\le \frac{l{\Vert y_1-x_1\Vert }^2}{2(1-l_0\Vert y_1-x_0\Vert )}$

$\le \frac{l{(s_1-t_1)}^2}{2(1-l_0{s}_1)}=t_2-s_1,$

and

so (31) holds for $m=1$ and $x_2\in \overline{S}(y_0,\frac{1}{l_0}-\eta )$.

Then, replace $x_1,y_1,x_2$ by $x_m,y_m,x_{m+1}$ to complete the induction for (31) and (32). Moreover, we have

(35) $$\Vert x_{m+1}-x_m\Vert \le \Vert x_{m+1}-y_m\Vert +\Vert y_m-x_m\Vert \le t_{m+1}-s_m+s_m-t_m=t_{m+1}-t_m,$$

and

(36)

It then follows from (35), (36) and 1 that sequence $t_m$ is complete in $B_1$, so there exists $x_{\ast }\in \overline{S}(y_0,\frac{1}{l_0}-\eta )$ with ${lim}_{m\to \mathrm{\infty }}{x}_m=x_{\ast }$.

Furthermore, by the second substep of method (3), we have

$$\Vert F^{\prime }{(x_0)}^{-1}F(x_{m+1})\Vert \le \frac{l}{2}\Vert x_{m+1}-y_m\Vert \le \frac{l}{2}{(t_{m+1}+s_m)}^2,$$

so $F(x_{\ast })=0$, by the continuity of $F$ and 1. The uniqueness of the solution part is given in [11]. ∎

Next, we study the semilocal convergence of method (4) in an analogous way.

Remark 4.

Let $p_1$ be a cubic polynomial defined by

$$p_1(t)=l_0{t}_3+\frac{l}{2}{t}_2+\frac{l}{2}t-l$$

We have and $p_1(1)=l_0>0$.

It follows by the intermediate value theorem that $P_1$ has at least one root in $(0,1)$.

But $p_1^{\prime }(t)=3l_0{t}_2+lt+\frac{l}{2}>0$, so $p_1$ increasing, so $p_1(t)=0$ has a unique root in $(0,1)$. Denote by $\delta$ this root.

The following estimate is needed:

(37)

Evidently, (37) holds, if

$$0\le 1+(\frac{l}{2}-l_0)\eta +\frac{l_0}{2}{\eta }^2(l_0+l)$$

Estimations (37) is true, if $l\ge 2l_0$. Moreover, if , then $\frac{l_0}{2}(l_0+l)t^2+(\frac{l}{2}-l_0)t+1=0$, since it has two negative roots by the Descarte’s rule of signs, so (33) holds in this case too.

We need an auxiliary result on majorizing sequences for method (4) similar to 1.

Define sequence ${\overline{t}}_n$ for each $n=1,2,\mathrm{\dots }$ by ${\overline{t}}_0=0,{\overline{t}}_1=\eta (1+l_0\eta )$

${\overline{s}}_n={\overline{t}}_n+\frac{l({\overline{t}}_n+{\overline{s}}_{n-1}-2{\overline{t}}_{n-1})({\overline{t}}_n-{\overline{s}}_{n-1})}{1-l_0{t}_n}$

${\overline{t}}_{n+1}={\overline{s}}_n+\frac{l({\overline{s}}_n-{\overline{t}}_n)}{2(1-l_0{\overline{t}}_n)}({\overline{s}}_n-{\overline{t}}_n)$

Lemma 5.

Let $l_0>0,l>0$ and $\eta >0$ be positive parameters. Assume that

(38)

Then, the conclusions of 1 hold with sequence ${\overline{t}}_n$ defined by (38) replacing sequence $t_n$ given by (6).

Proof.

As in 1, we must show

(39)

and

$$0\le {\overline{t}}_m\le {\overline{s}}_m\le {\overline{s}}_{m+1}.$$

But

and

Notice that

and

so it suffices to show only (39), or

$$\frac{l(\frac{1-{\delta }^{2m+2}}{1-\delta }-\frac{1-{\delta }^{2m}}{1-\delta })\eta }{1-l_0(1+\mathrm{\dots }+{\delta }^{2m+1})\eta }\le \delta$$

or

$$f_m(t)=\frac{l}{2}{t}^{2m-1}(t+2)\eta +l_0(1+\mathrm{\dots }+t^{2m+1})\eta -1\le 0$$

But

$$f_{m+1}(t)=f_m(t)+p_1(t)t^{2m-1}(t+1)\eta ,$$

so

$$f_{m+1}(\delta )=f_m(\delta )=\underset{m\to \mathrm{\infty }}{lim}{f}_m(\delta )=f_{\mathrm{\infty }}(\delta )=\frac{l\eta }{1-\delta }-1$$

The rest follows as in 1 with ${\overline{t}}_{\ast }={lim}_{n\to \mathrm{\infty }}{\overline{t}}_n$ and ${\overline{t}}_{\ast \ast }=\frac{\eta }{1-\delta }$.

Replace $x_m,y_m,t_{\ast },t_{\ast \ast },t_{\ast }^1,t_n,s_n$ by ${\overline{x}}_m,{\overline{y}}_m,{\overline{t}}_{\ast },{\overline{t}}_{\ast \ast },{\overline{t}}_{\ast }^1,{\overline{t}}_n,{\overline{s}}_n$, hypotheses of 1, method (3) by 5, method (4) respectively. Call the resulting hypotheses (A)’. Then in an analogous to 3 way, we arrive at:

Theorem 6.

Suppose that the conditions (A)’ hold. Then, the conclusions of 3 hold but with method (4) replacing method (3).

Proof.

Notice that the only difference in the proof is that we use

$$\Vert {\overline{y}}_m-{\overline{x}}_m\Vert \le \frac{l({\overline{t}}_m+{\overline{s}}_{m-1}-2{\overline{s}}_{m-1})}{1-l_0{\overline{t}}_m}({\overline{t}}_m-{\overline{s}}_{m-1}),$$

$\Vert {\overline{x}}_{m+1}-{\overline{y}}_m\Vert$ $=\Vert (F^{\prime }{({\overline{x}}_m)}^{-1}{F}^{\prime }(x_0))(F^{\prime }{(x_0)}^{-1}F({\overline{y}}_m))\Vert \le$

$\le \Vert (F^{\prime }{({\overline{x}}_m)}^{-1}{F}^{\prime }(x_0))\Vert \Vert (F^{\prime }{(x_0)}^{-1}F({\overline{y}}_m))\Vert$

$\le \frac{l{({\overline{s}}_m-{\overline{t}}_m)}^2}{2(1-l_0{\overline{t}}_m)}$

instead of

$$\Vert x_{m+1}-y_m\Vert \le \frac{l{(s_m-t_m)}^2}{2(1-l_0{s}_m)}$$

and

$$\Vert y_m-x_m\Vert \le \frac{l{(t_m-s_{m-1})}^2}{2(1-l_0{t}_m)},$$

respectively. ∎

3. Local convergence analysis

The local convergence analysis for both methods uses the hypotheses (H):

• $F:\mathrm{\Omega }\subset B_1\to B_2$ is differentiable in the sense of Fr’echet and there exist $x_{\ast }\in \mathrm{\Omega }$ such that $F^{\prime }{(x_{\ast })}^{-1}\in LB(B_2,B_1)$ and $F(x_{\ast })=0$.

• There exists $L_0>0$ such that for each $x\in \mathrm{\Omega }$

  $$\Vert F^{\prime }{(x_{\ast })}^{-1}(F^{\prime }(x)-F^{\prime }(x_{\ast }))\Vert \le L_0\Vert x-x_{\ast }\Vert .$$

  Set $S_2=\mathrm{\Omega }\cap S(x,\frac{1}{L_0})$.

• There exists $L>0$ such that for each $x,y\in S_2$

  $$\Vert F^{\prime }{(x_{\ast })}^{-1}(F^{\prime }(y)-F^{\prime }(x))\Vert \le L\Vert y-x\Vert .$$

• $S(x_{\ast },\rho )\subseteq \mathrm{\Omega }$, where

• There exists $\overline{r}\ge \rho$ such that .

  Set $S_3=\mathrm{\Omega }\cap \overline{S}(x_{\ast },\overline{r})$.

The proofs of the following two results are omitted, since they follow as the corresponding ones for single step Newton’s method (2) given in [9, 11].

Theorem 7.

Under the hypotheses (H) starting from $x_0\in S(x_{\ast },{\mu }_A)-x_{\ast }$ sequence $x_n$ produced by method (3) converges to $x_{\ast }$ which is the only solution of equation $F(x)=0$ in the set $S_3$. Moreover, the following items hold:

(40) $$\Vert y_n-x_{\ast }\Vert \le \frac{L{\Vert x_n-x_{\ast }\Vert }^2}{2(1-L_0\Vert x_n-x_{\ast }\Vert )}$$

and

(41) $$\Vert x_{n+1}-x_{\ast }\Vert \le \frac{L{\Vert y_n-x_{\ast }\Vert }^2}{2(1-L_0\parallel y_n-x_{\ast }\parallel }$$

Theorem 8.

Under the hypotheses (H) starting from $x_0\in S(x_{\ast },R)-x_{\ast }$ sequence ${\overline{x}}_n$ produced by method (4) converges to $x_{\ast }$ which is the only solution of equation $F(x)=0$ in the set $S_3$. Moreover the following items hold:

(42) $$\Vert y_n-x_{\ast }\Vert \le \frac{L{\Vert x_n-x_{\ast }\Vert }^2}{2(1-L_0\Vert x_n-x_{\ast }\Vert )}$$

and

(43) $$\Vert y_{n+1}-x_{\ast }\Vert \le \frac{L(2\parallel x_n-x_{\ast }\parallel +\parallel y_n-x_{\ast })}{2(1-L_0\Vert x_n-x_{\ast }\Vert )}\Vert y_n-x_{\ast }\Vert$$

Lemma 9.

The radius of convergence in [9, 11] for the single step Newton’s method was given by

(44) $${\overline{\mu }}_A=\frac{2}{2L_0+L_1},$$

where $L_1$ is the Lipschitz constant on $\mathrm{\Omega }$. Then, since $s_0\subseteq \mathrm{\Omega }$, we get

(45) $$L\le L_1$$

so

(46) $${\overline{\mu }}_A\le {\mu }_A$$

The error bounds are tighter too, since $L_1$ is used in (40) and (41).

4. Numerical Examples

Example 10.

Let $B_1=B_2=ℝ$, $\mathrm{\Omega }=S(x_0,1-\alpha )$, $x_0=1$ and $\alpha \in I=[0,\frac{1}{2})$. Define function $f$ on $\mathrm{\Omega }$ by $f(x)=x^3-\alpha$.

Then, using hypotheses (a1)-(a3), we get $l_0=3-\alpha$,$l=\frac{2(6+5\alpha -2{\alpha }^2)}{3(3-\alpha )}$ and $l_1=2(2-\alpha ).$ Method (2) For $\alpha \in I_0=[0.371269,0.5]$ has solutions under our approach but no solutions according to Kantorovich, since $h_K=l_1\eta >\frac{1}{2}$ for each $\alpha \in [0,0.5].$ Method (3) has no solution in $[0,0.5].$

Example 11.

Let $B_1=B_2=C[0,1]$, where $C[0,1]$ stands for the space of continuous function on $[0,1]$. We shall use the mimum norm. Let ${\mathrm{\Omega }}_0=\{x\in C[0,1]:\Vert x\Vert \le d\}$.

Define operator $G$ on ${\mathrm{\Omega }}_0$ by

$G(x)(s)=x(s)-g(s)-b{\displaystyle {\int }_0^1}K(s,t)x{(t)}^3𝑑t,x\in C[0,1],s\in [0,1]$

where $g\in C[0,1]$ is a given function, $\xi$ is a real constant and the kernel $K$ is the Green’s function. In this case, for each $x\in D^{\ast }$, $F^{\prime }(x)$ is a linear operator defined on $D^{\ast }$ by the following expression: $[F^{\prime }(x)(v)](s)=v(s)-3\xi {\int }_0^{1}K(s,t)x{(t)}^{2}v(t)𝑑t,v\in C[0,1],s\in [0,1]$ If we choose $x_0(s)=f(s)=1$, it follows $\Vert I-F^{\prime }(x_0)\Vert \le \frac{3|\xi |}{8}$. Thus, if , $f^{\prime }{(x_0)}^{-1}$ is defined and

$$\Vert F^{\prime }{(x_0)}^{-1}\Vert \le \frac{8}{8-3|\xi |},\Vert F(x_0)\Vert \le \frac{|\xi |}{8},\eta =\Vert F^{\prime }{(x_0)}^{-1}F(x_0)\Vert \le \frac{|\xi |}{8-3|\xi |}.$$

Choosing $\xi =1.00$ and $x=3$, we have $\eta =0.2,T=3.8,b=2.6,L_1=2.28$, and $l=1.38154\mathrm{\dots }$

Using this values we obtain that conditions (28)-(30) are not satisfied, since the Kantorovich condition $h_K=l_1\eta \le \frac{1}{2}$, gives $h_K=0.76>\frac{1}{2}$. but condition (29) is satisfied since The convergence of the Newton’s method follows by 3.

Example 12.

Let $B_1=B_2={ℝ}^3$, $\mathrm{\Omega }=S(0,1),x^{\ast }={(0,0,0)}^T$ and define $G$ on $\mathrm{\Omega }$ by

(47) $$G(x)=F(x_1,x_2,x_3)={(e^{x_1}-1,\frac{e-1}{2}x_2^{}{}_{}{}^2+x_2,x_3)}^T.$$

For the points $u={(u_1,u_2,u_3)}^T$, the Fréchet derivative is given by

Then, $G^{\prime }(x^{\ast })=diag(1,1,1),$ we have $L_0=e-1,L=e^{\frac{1}{e-1}},L_1=e.$

Then, we obtain that

Example 13.

Let $B_1=B_2=C[0,1],$ the space of continuous functions defined on $[0,1]$ and be equipped with the max norm. Let $\mathrm{\Omega }=\overline{S}(0,1).$ Define function $G$ on $\mathrm{\Omega }$ by

(48) $$G(\phi )(x)=\phi (x)-5{\int }_0^{1}x\theta \phi {(\theta )}^3𝑑\theta .$$

We have that

$$G^{\prime }(\phi (\xi ))(x)=\xi (x)-15{\int }_0^{1}x\theta \phi {(\theta )}^2\xi (\theta )𝑑\theta ,\text{for each}\xi \in \mathrm{\Omega }.$$

Then, we get that $x^{\ast }=0,$ $L_0=7.5,L=15=L_1.$ This way, we have that

Acknowledgement.

We thank the referee for his remarks in improving this paper.