The convergence theorem for fourth-order super-Halley method in weaker conditions

Abstract

In this paper, we establish the Newton-Kantorovich convergence theorem of a fourth-order super-Halley method under weaker conditions in Banach space, which is used to solve the nonlinear equations. Finally, some examples are provided to show the application of our theorem.

1 Introduction

For a number of problems arising in scientific and engineering areas one often needs to find the solution of nonlinear equations in Banach spaces

$$ F(x)=0, $$

(1)

where F is a third-order Fréchet-differentiable operator defined on a convex subset Ω of a Banach space X with values in a Banach space Y.

There are kinds of methods to find a solution of equation (1). Generally, iterative methods are often used to solve this problem [1]. The best-known iterative method is Newton’s method

$$ x_{n+1}=x_{n}-F'(x_{n})^{-1}F(x_{n}), $$

(2)

which has quadratic convergence. Recently a lot of research has been carried out to provide improvements. Third-order iterative methods such as Halley’s method, Chebyshev’s method, super-Halley’s method, Chebyshev-like’s method etc. [2–12] are used to solve equation (1). To improve the convergence order, fourth-order iterative methods are also discussed in [13–19].

Kou et al. [20] presented a variant of the super-Halley method which improves the order of the super-Halley method from three to four by using the values of the second derivative at $( x_{n}-\frac{1}{3}f(x_{n})/f'(x_{n}) )$ instead of $x_{n}$. Wang et al. [15] established the semilocal convergence of the fourth-order super-Halley method in Banach spaces by using recurrence relations. This method in Banach spaces can be given by

$$ x_{n + 1}= x_{n} - \biggl[I+\frac{1}{2}K_{F}(x_{n}) \bigl[I-K_{F}(x_{n}) \bigr]^{-1} \biggr] \Gamma_{n} F(x_{n}), $$

(3)

where $\Gamma_{n}=[F'(x_{n})]^{-1}$, $K_{F}(x_{n})=\Gamma_{n}F''(u_{n})\Gamma_{n}F(x_{n})$, and $u_{n}=x_{n}-\frac{1}{3}\Gamma_{n}F(x_{n})$.

Let $x_{0} \in\Omega$ and the nonlinear operator $F: \Omega\subset X\rightarrow Y$ be continuously third-order Fréchet differentiable where Ω is an open set and X and Y are Banach spaces. Assume that

• (C1) $\Vert \Gamma_{0}F(x_{0})\Vert \leq\eta$,

• (C2) $\Vert \Gamma_{0} \Vert \leq\beta$,

• (C3) $\Vert F^{\prime\prime}(x) \Vert \leq M, x \in\Omega$,

• (C4) $\Vert F'''(x)\Vert \leq N, x \in\Omega$,

• (C5) there exists a positive real number L such that

  $$\bigl\Vert F^{\prime\prime\prime}(x)- F^{\prime\prime\prime}(y) \bigr\Vert \leq L \Vert x-y \Vert ,\quad \forall x, y \in\Omega. $$

Under the above assumptions, we apply majorizing functions to prove the semilocal convergence of the method (3) to solve nonlinear equations in Banach spaces and establish its convergence theorems in [21]. The main results is as follows.

Theorem 1

[21]

Let X and Y be two Banach spaces and $F:\Omega\subseteq X \rightarrow Y$ be a third-order Fréchet differentiable on a non-empty open convex subset Ω. Assume that all conditions (C1)-(C5) hold and $x_{0} \in \Omega$, $h=K \beta\eta\leq1/2$, $\overline{B(x_{0}, t^{*})} \subset\Omega$, then the sequence $\{x_{n}\}$ generated by the method (3) is well defined, $x_{n} \in \overline{B(x_{0}, t^{*})}$ and converges to the unique solution $x^{*}\in B(x_{0}, t^{**})$ of $F(x)$, and $\Vert x_{n}-x^{*}\Vert \leq t^{*}-t_{n}$, where

$$\begin{aligned} &t^{*} =\frac{1-\sqrt{1-2h}}{h}\eta,\qquad t^{**} = \frac{1+\sqrt{1-2h}}{h} \eta, \\ & K \geq M \biggl[ 1 + \frac{N}{M^{2} \beta} +\frac{35 L}{36 M^{3} \beta^{2}} \biggr]. \end{aligned}$$

(4)

We know the conditions of Theorem 1 cannot be satisfied by some general nonlinear operator equations. For example,

$$ F(x)=\frac{1}{6}x^{3}+\frac{1}{6}x^{2}- \frac{5}{6}x+\frac{1}{3}=0. $$

(5)

Let the initial point $x_{0}=0$, $\Omega= [-1, 1]$. Then we know

$$\begin{aligned} &\beta= \bigl\Vert F'(x_{0})^{-1} \bigr\Vert = \frac{6}{5}, \qquad\eta= \bigl\Vert F'(x_{0})^{-1}F(x_{0}) \bigr\Vert =\frac{2}{5}, \\ &M=\frac{4}{3},\qquad N=1,\qquad L=0. \end{aligned}$$

From (4), we can get $K \geq M$, so

$$h=K \beta\eta\geq M \beta\eta=\frac{16}{25}> \frac{1}{2}. $$

The conditions of Theorem 1 cannot be satisfied. Hence, we cannot know whether the sequence $\{x_{n}\}$ generated by the method (3) converges to the solution $x^{*}$.

In this paper, we consider weaker conditions and establish a new Newton-Kantorovich convergence theorem. The paper is organized as follows: in Section 2 the convergence analysis based on weaker conditions is given and in Section 3, a new Newton-Kantorovich convergence theorem is established. In Section 4, some numerical examples are worked out. We finish the work with some conclusions and references.

2 Analysis of convergence

Let $x_{0} \in\Omega$ and nonlinear operator $F: \Omega\subset X\rightarrow Y$ be continuously third-order Fréchet differentiable where Ω is an open set and X and Y are Banach spaces. We assume that:

• (C6) $\Vert \Gamma_{0}F(x_{0})\Vert \leq\eta$,

• (C7) $\Vert F'(x_{0})^{-1}F^{\prime\prime}(x_{0}) \Vert \leq\gamma$,

• (C8) $\Vert F'(x_{0})^{-1}F'''(x)\Vert \leq N, x \in\Omega$,

• (C9) there exists a positive real number L such that

  $$ \bigl\Vert F'(x_{0})^{-1} \bigl[F^{\prime\prime\prime}(x)- F^{\prime\prime\prime}(y) \bigr] \bigr\Vert \leq L \Vert x-y \Vert ,\quad \forall x, y \in\Omega. $$

  (6)

Denote

$$ g(t)=\frac{1}{6}Kt^{3}+\frac{1}{2}\gamma t^{2}-t+\eta, $$

(7)

where $K, \gamma, \eta$ are positive real numbers and

$$ \frac{5L}{12N \eta+ 36\gamma}+N \leq K. $$

(8)

Lemma 1

[19]

Let $\alpha=\frac{2}{\gamma+\sqrt{\gamma^{2}+2K}}, \beta=\alpha-\frac{1}{6}K \alpha^{3}-\frac{1}{2}\gamma \alpha^{2}=\frac{2 (\gamma+2\sqrt{\gamma^{2}+2K} )}{3 (\gamma+\sqrt{\gamma^{2}+2K} )}$. If $\eta\leq \beta$, then the polynomial equation $g(t)$ has two positive real roots $r_{1}, r_{2}$ ($r_{1}\leq r_{2}$) and a negative root $-r_{0}$ ($r_{0}>0$).

Lemma 2

Let $r_{1}, r_{2}, -r_{0}$ be three roots of $g(t)$ and $r_{1} \leq r_{2}, r_{0}>0$. Write $u=r_{0}+t$, $a=r_{1}-t$, $b=r_{2}-t$, and

$$ q(t)=\frac{{b - u}}{{a - u}} \cdot\frac{{(a - u)b^{3} + u(u - a)b^{2} + u^{2} (a - u)b - au^{3} }}{{(b - u)a^{3} + u(u - b)a^{2} + u^{2} (b - u)a - bu^{3} }} . $$

(9)

Then as $0\leq t\leq r_{1}$, we have

$$ q(0)\leq q(t) \leq q(r_{1})\leq1. $$

(10)

Proof

Since $g(t)=\frac{K}{6}abu$ and $g''(t)\geq 0$ ($t\geq0$), we have

$$ u-a-b \geq0. $$

(11)

Differentiating q and noticing $q'(t)\geq0$ ($0\leq t\leq r_{1}$), we obtain

$$ q(0)\leq q(t)\leq q(r_{1}). $$

(12)

On the other hand, since

$$q(t)-1 \leq0, $$

the lemma is proved.

Now we consider the majorizing sequences $\{t_{n}\}$, $\{ s_{n} \}$ $(n\geq0)$, $t_{0}=0$,

$$ \textstyle\begin{cases} s_{n} = t_{n} - \frac{{g(t_{n} )}}{{g'(t_{n} )}}, \\ h_{n} = - g'(t_{n} )^{ - 1} g''(r_{n} ) (s_{n} - t_{n} ), \\ t_{n + 1} = t_{n} - [ {1 + \frac{1}{2} \frac{{h_{n} }}{{1 - h_{n} }}} ]\frac{{g(t_{n} )}}{{g'(t_{n} )}} , \end{cases} $$

(13)

where $r_{n}=t_{n}+1/3(s_{n}-t_{n})$. □

Lemma 3

Let $g(t)$ be defined by (7) and satisfy the condition $\eta\leq\beta$, then we have

$$ \frac{{(\sqrt[3]{\lambda_{2}} \theta)^{4^{n} } }}{{\sqrt[3]{\lambda_{2}} - (\sqrt[3]{\lambda_{2}} \theta)^{4^{n} } }}(r_{2} - r_{1} ) \le r_{1} - t_{n} \le\frac{{(\sqrt[3]{\lambda_{1}} \theta)^{4^{n} } }}{{\sqrt[3]{\lambda_{1}} - (\sqrt[3]{\lambda_{1}} \theta)^{4^{n} } }}(r_{2} - r_{1} ),\quad n = 0,1, \ldots, $$

(14)

where $\theta = \frac{{r_{1} }}{{r_{2} }}, \lambda_{1} = q(r_{1} ), \lambda_{2} = q(0)$.

Proof

Let $a_{n}=r_{1}-t_{n}, b_{n}=r_{2}-t_{n}, u_{n}=r_{0}+t_{n}$, then

$$\begin{aligned} & g(t_{n} ) = \frac{K}{6}a_{n} b_{n} u_{n}, \end{aligned}$$

(15)

$$\begin{aligned} &g'(t_{n} ) = -{\frac{K}{6}} [ {a_{n} u_{n} + b_{n} u_{n} - a_{n} b_{n} } ], \end{aligned}$$

(16)

$$\begin{aligned} &g''(t_{n} ) = { \frac{K}{3}} [ {u_{n} - b_{n} - a_{n} } ]. \end{aligned}$$

(17)

Write $\varphi(t_{n})= a_{n} u_{n} + b_{n} u_{n} - a_{n} b_{n}$, then we have

$$\begin{aligned} & a_{n + 1} = a_{n} + \biggl[ {1 + \frac{1}{2}\frac{{h_{n} }}{{1 - h_{n} }}} \biggr]\frac{{g(t_{n} )}}{{g'(t_{n} )}} \\ &\phantom{a_{n + 1}} = \frac{{a_{n}^{4} (b_{n} - u_{n} ) [ {(a_{n} - u_{n} )b_{n}^{3} + u_{n} (u_{n} - a_{n} )b_{n}^{2} + u_{n}^{2} (a_{n} - u_{n} )b_{n} - a_{n} u_{n}^{3} } ]}}{\varphi^{2} (t_{n} )(a_{n}^{2} u_{n}^{2} + b_{n}^{2} u_{n}^{2} + a_{n}^{2} b_{n}^{2} ) - 2a_{n}^{2} b_{n}^{2} \varphi(t_{n} )}, \end{aligned}$$

(18)

$$\begin{aligned} & b_{n + 1}= b_{n} + \biggl[ {1 + \frac{1}{2}\frac{{h_{n} }}{{1 - h_{n} }}} \biggr]\frac{{g(t_{n} )}}{{g'(t_{n} )}} \\ &\phantom{b_{n + 1}} = \frac{{b_{n}^{4} (a_{n} - u_{n} ) [ {(b_{n} - u_{n} )a_{n}^{3} + u_{n} (u_{n} - b_{n} )b_{n}^{2} + u_{n}^{2} (b_{n} - u_{n} )a_{n} - b_{n} u_{n}^{3} } ]}}{\varphi^{2} (t_{n} )(a_{n}^{2} u_{n}^{2} + b_{n}^{2} u_{n}^{2} + a_{n}^{2} b_{n}^{2} ) - 2a_{n}^{2} b_{n}^{2} \varphi(t_{n} )}. \end{aligned}$$

(19)

We can obtain

$$ \frac{{a_{n + 1} }}{{b_{n + 1} }} = \frac{{b_{n} - u_{n} }}{{a_{n} - u_{n} }} \cdot\frac{{(a_{n} - u_{n} )b_{n}^{3} + u_{n} (u_{n} - a_{n} )b_{n}^{2} + u_{n}^{2} (a_{n} - u_{n} )b_{n} - a_{n} u_{n}^{3} }}{{(b_{n} - u_{n} )a_{n}^{3} + u_{n} (u_{n} - b_{n} )a_{n}^{2} + u_{n}^{2} (b_{n} - u_{n} )a_{n} - b_{n} u_{n}^{3} }} \cdot \biggl( {\frac{{a_{n} }}{{b_{n} }}} \biggr)^{4}. $$

(20)

From Lemma 2, we have $\lambda_{2}\leq q(t)\leq\lambda_{1}$. Thus

$$ \frac{{a_{n} }}{{b_{n} }} \le\lambda_{1} \biggl( { \frac{{a_{n - 1} }}{{b_{n - 1} }}} \biggr)^{4} \le \cdots \le ( {\lambda_{1}} )^{1 + 4 + \cdots + 4^{n - 1} } \biggl( {\frac{{a_{0} }}{{b_{0} }}} \biggr)^{4^{n} } = \frac{1}{{\sqrt[3]{\lambda_{1}} }} \bigl( {\sqrt[3]{\lambda_{1}} \theta} \bigr)^{4^{n} }. $$

(21)

In a similar way,

$$ \frac{{a_{n} }}{{b_{n} }} \ge\frac{1}{{\sqrt[3]{\lambda_{2}} }} \bigl( {\sqrt[3]{ \lambda_{2}} \theta} \bigr)^{4^{n} }. $$

(22)

That completes the proof of the lemma. □

Lemma 4

Suppose $t_{n}, s_{n}$ are generated by (13). If $\eta< \beta$, then the sequences $\{t_{n}\}, \{s_{n}\}$ increase and converge to $r_{1}$, and

$$ 0\leq t_{n} \leq s_{n} \leq t_{n+1} < r_{1}. $$

(23)

Proof

Let

$$\begin{aligned} &U(t) = t - g'(t)^{ - 1} g(t), \\ &H(t) = \bigl( {g'(t)^{ - 1} } \bigr)^{2} g''(T)g(t), \\ &V(t) = t + \biggl[ {1 + \frac{1}{2}\frac{{H(t)}}{{1 - H(t)}}} \biggr] \bigl( {U(t) - t} \bigr), \end{aligned}$$

(24)

where $T= (2t+U(t) ) /3$.

When $0 \leq t \leq r_{1}$, we can obtain $g(t)\geq0$, $g'(t)< 0$, $g''(t)> 0$. Hence

$$ U(t)=t-\frac{g(t)}{g'(t)} \geq t \geq0. $$

(25)

So $\forall t \in[0,r_{1}]$, we always have $U(t)\geq t$.

Since $T=\frac{2t+U(t)}{3}\geq t \geq0$, we have

$$ H(t)= \frac{g''(T)g(t)}{g'(t)^{2}} \geq0. $$

(26)

On the other hand $g''(T)g(t)-g'(t)^{2}>0$, then

$$ 0\leq H(t)= \frac{g''(T)g(t)}{g'(t)^{2}}< 1. $$

(27)

Thus

$$V(t)=U(t)+\frac{1}{2}\frac{{H(t)}}{{1 - H(t)}} \bigl( {U(t) - t} \bigr) \geq0, $$

and $\forall t \in[0, r_{1}]$, we always have $V(t)\geq U(t)$.

Since

$$\begin{aligned} V'(t)={}& \frac{g(t) [ 3g'(t)^{2} ( {g''(T) - g''(t)} ) ( {g''(T)g(t) - 2g'(t)^{2} } )- Kg(t)^{2} g'(t)g''(t) }{{6g'(t)^{2} [ {g''(T)g(t) - g'(t)^{2} } ]^{2} }} \\ &{} +\frac{ - Kg(t)^{2} g''(t)g'(t) + 3g(t)^{2} g''(t)g''(T) ]}{{6g'(t)^{2} [ {g''(T)g(t) - g'(t)^{2} } ]^{2} }} \\ ={}& \frac{{g(t) [ { - Kg(t)^{2} g'(t)g''(t) - Kg(t)^{2} g''(t)g'(t) + 3g(t)^{2} g''(t)g''(T)} ]}}{{6g'(t)^{2} [ {g''(T)g(t) - g'(t)^{2} } ]^{2} }}, \end{aligned}$$

(28)

we know $V'(t) > 0$ for $0\leq t \leq r_{1}$. That is to say that $V(t)$ is monotonically increasing. By this we will inductively prove that

$$ 0\leq t_{n}\leq s_{n} \leq t_{n+1} < V(r_{1})=r_{1}. $$

(29)

In fact, (29) is obviously true for $n=0$. Assume (29) holds until some n. Since $t_{n+1}< r_{1}$, $s_{n+1}, t_{n+2}$ are well defined and $t_{n+1}\leq s_{n+1}\leq t_{n+2}$. On the other hand, by the monotonicity of $V(t)$, we also have

$$ t_{n+2}=V(t_{n+1})< V(r_{1})=r_{1}. $$

Thus, (29) also holds for $n+1$.

From Lemma 3, we can see that $\{t_{n}\}$ converges to $r_{1}$. That completes the proof of the lemma. □

Lemma 5

Assume F satisfies the conditions (C6)-(C9), then $\forall x \in\overline{B(x_{0}, r_{1})}$, $F'(x)^{-1}$ exists and satisfies the inequality

• (I) $\Vert F'(x_{0})^{-1}F''(x)\Vert \leq g''(\Vert x-x_{0}\Vert )$,

• (II) $\Vert F'(x)^{-1}F'(x_{0})\Vert \leq -g'(\Vert x-x_{0}\Vert )^{-1}$.

Proof

(I) From the above assumptions, we have

$$\begin{aligned} \bigl\Vert F'(x_{0})^{-1}F''(x) \bigr\Vert &= \bigl\Vert F'(x_{0})^{-1}F''(x_{0})+F'(x_{0})^{-1} \bigl[F''(x)-F''(x_{0}) \bigr] \bigr\Vert \\ &\leq\gamma+ N\Vert x-x_{0}\Vert \leq\gamma+ K\Vert x-x_{0}\Vert \\ &= g'' \bigl(\Vert x-x_{0}\Vert \bigr). \end{aligned}$$

(II) When $t\in[0,r_{1})$, we know $g'(t)<0$. Hence when $x \in \overline{B(x_{0}, r_{1})}$,

$$\begin{aligned} &\bigl\Vert F'(x_{0})^{-1}F'(x)-I \bigr\Vert \\ &\quad= \bigl\Vert F'(x_{0})^{-1} \bigl[F'(x)-F'(x_{0})-F''(x_{0}) (x-x_{0})+F''(x_{0}) (x-x_{0}) \bigr] \bigr\Vert \\ &\quad\le \biggl\Vert { \int_{0}^{1} {F'(x_{0} )^{ - 1} \bigl[ {F'' \bigl( {x_{0} + t(x - x_{0} )} \bigr) - F''(x_{0} )} \bigr]\,dt(x - x_{0} )} } \biggr\Vert + \gamma \Vert {x - x_{0} } \Vert \\ &\quad\le \biggl\Vert { \int_{0}^{1} {Nt\,dt(x - x_{0} )^{2} } } \biggr\Vert + \gamma \Vert {x - x_{0} } \Vert \le\frac{1}{2}K\Vert {x - x_{0} } \Vert ^{2} + \gamma \Vert {x - x_{0} } \Vert \\ &\quad=1+g' \bigl(\Vert x-x_{0}\Vert \bigr)< 1. \end{aligned}$$

By the Banach lemma, we know $(F'(x_{0})^{-1}F'(x) )^{-1}=F'(x)^{-1}F'(x_{0})$ exists and

$$\bigl\Vert {F'(x)^{ - 1} F'(x_{0} )} \bigr\Vert \le\frac{1}{{1 - \Vert {I - F'(x_{0} )^{ - 1} F'(x)} \Vert }} \le - g' \bigl( {\Vert {x - x_{0} } \Vert } \bigr)^{ - 1}. $$

That completes the proof of the lemma. □

Lemma 6

[21]

Assume that the nonlinear operator $F:\Omega\subset X \rightarrow Y$ is continuously third-order Fréchet differentiable where Ω is an open set and X and Y are Banach spaces. The sequences $\{x_{n}\}$, $\{y_{n}\}$ are generated by (3). Then we have

$$\begin{aligned} F(x_{n+1})={}& \frac{1}{2}F^{\prime\prime}(u_{n}) (x_{n+1}-y_{n})^{2}+\frac {1}{6}F'''(x_{n}) (x_{n+1}-y_{n}) (x_{n+1}-x_{n})^{2} \\ &{}- \frac{1}{6} \int^{1}_{0} \biggl[F''' \biggl(x_{n}+\frac{1}{3}t(y_{n}-x_{n}) \biggr)-F'''(x_{n}) \biggr]\,dt(y_{n}-x_{n}) (x_{n+1}-x_{n})^{2} \\ &{}+\frac{1}{2} \int^{1}_{0} \bigl[F''' \bigl(x_{n}+t(x_{n+1}-x_{n}) \bigr)-F'''(x_{n}) \bigr](1-t)^{2}\,dt(x_{n+1}-x_{n})^{3} , \end{aligned}$$

where $y_{n}=x_{n}-\Gamma_{n}F(x_{n})$ and $u_{n}=x_{n}+\frac{1}{3}(y_{n}-x_{n})$.

3 Newton-Kantorovich convergence theorem

Now we give a theorem to establish the semilocal convergence of the method (3) in weaker conditions, the existence and uniqueness of the solution and the domain in which it is located, along with a priori error bounds, which lead to the R-order of convergence of at least four of the iterations (3).

Theorem 2

Let X and Y be two Banach spaces, and $F:\Omega\subseteq X \rightarrow Y$ be a third-order Fréchet differentiable on a non-empty open convex subset Ω. Assume that all conditions (C6)-(C9) hold true and $x_{0} \in\Omega$. If $\eta< \beta$, $\overline{B(x_{0}, r_{1})}\subset\Omega$, then the sequence $\{ x_{n}\}$ generated by (3) is well defined, $\{x_{n}\} \in\overline{B(x_{0}, r_{1})}$ and converges to the unique solution $x^{*} \in B(x_{0}, \alpha)$, and $\Vert x_{n}-x^{*}\Vert \leq r_{1}-t_{n}$. Further, we have

$$ \bigl\Vert x_{n}-x^{*} \bigr\Vert \leq \frac{{(\sqrt[3]{\lambda_{1}} \theta)^{4^{n} } }}{{\sqrt[3]{\lambda_{1}} - (\sqrt[3]{\lambda_{1}} \theta)^{4^{n} } }}(r_{2} - r_{1} ),\quad n = 0,1, \ldots, $$

(30)

where $\theta=\frac{{r_{1} }}{{r_{2} }}, \lambda_{1} = q(r_{1} )$, $\alpha=\frac{2}{\gamma+\sqrt{\gamma^{2}+2K}}$.

Proof

We will prove the following formula by induction:

$(I_{n})$ $x_{n} \in\overline{B(x_{0}, t_{n})}$,

$(II_{n})$ $\Vert F'(x_{n})^{-1}F'(x_{0})\Vert \leq-g'(t_{n})^{-1}$,

$(III_{n})$ $\Vert F'(x_{0})^{-1}F''(x_{n}) \Vert \leq g''(\Vert x_{n}-x_{0}\Vert )\leq g''(t_{n})$,

$(IV_{n})$ $\Vert y_{n}-x_{n}\Vert \leq s_{n}-t_{n}$,

$(V_{n})$ $y_{n}\in\overline{B(x_{0}, s_{n})}$,

$(VI_{n})$ $\Vert x_{n+1}-y_{n}\Vert \leq t_{n+1}-s_{n}$.

Estimate that ($I_{n}$)-($VI_{n}$) are true for $n=0$ by the initial conditions. Now, assume that $(I_{n})$-$(VI_{n})$ are true for all integers $k \leq n$.

$(I_{n+1})$ From the above assumptions, we have

$$\begin{aligned} \Vert x_{n+1}-x_{0} \Vert & \leq \Vert x_{n+1}-y_{n}\Vert +\Vert y_{n}-x_{n}\Vert +\Vert x_{n}-x_{0} \Vert \\ & \leq(t_{n+1}-s_{n})+(s_{n}-t_{n})+(t_{n}-t_{0})=t_{n+1}. \end{aligned}$$

(31)

$(II_{n+1})$ From (II) of Lemma 5, we can obtain

$$ \bigl\Vert {F'(x_{n + 1} )^{ - 1} F'(x_{0} )} \bigr\Vert \le - g' \bigl( { \Vert {x_{n + 1} - x_{0} } \Vert } \bigr)^{ - 1} \le - g'(t_{n + 1} )^{ - 1}. $$

(32)

$(III_{n+1})$ From (I) of Lemma 5, we can obtain

$$ \bigl\Vert {F'(x_{0} )^{ - 1} F''(x_{n + 1} )} \bigr\Vert \le g'' \bigl( {\Vert {x_{n + 1} - x_{0} } \Vert } \bigr) \le g''(t_{n + 1} ). $$

(33)

$(IV_{n+1})$ From Lemma 5, we have

$$\begin{aligned} \bigl\Vert {F'(x_{0} )^{ - 1} F(x_{n + 1} )} \bigr\Vert \le{}& \frac{1}{2}g''(r_{n} ) (t_{n + 1} - s_{n} )^{2} + \frac{1}{6}N(t_{n + 1} - s_{n} ) (t_{n + 1} - t_{n} )^{2} \\ &{} + \frac{L}{{36}}(s_{n} - t_{n} )^{2} (t_{n + 1} - t_{n} )^{2} + \frac{L}{{24}}(t_{n + 1} - t_{n} )^{4} \\ \le{}&\frac{1}{2}g''(r_{n}) (t_{n + 1} - s_{n} )^{2} \\ &{}+ \frac{1}{6} \biggl( N + \frac{L}{{36}}\cdot\frac{(s_{n} - t_{n})^{2}}{t_{n + 1} - s_{n}} +\frac{L}{24} \cdot \frac{(t_{n + 1} - t_{n} )^{2} }{t_{n + 1} - s_{n} } \biggr) ( {t_{n + 1} - s_{n} } ) (t_{n + 1} - t_{n} )^{2} \\ \le{}& \frac{1}{2}g''(r_{n}) (t_{n + 1} - s_{n} )^{2} + \frac{1}{6}K ( {t_{n + 1} - s_{n} } ) (t_{n + 1} - t_{n} )^{2} \\ \le{}& g(t_{n+1}). \end{aligned}$$

(34)

Thus, we have

$$\begin{aligned} \Vert y_{n+1}-x_{n+1} \Vert & \leq \bigl\Vert -F'(x_{n+1})F'(x_{0}) \bigr\Vert \bigl\Vert F'(x_{0})^{-1}F(x_{n+1}) \bigr\Vert \\ &\leq- \frac{g(t_{n+1})}{g'(t_{n+1})}=s_{n+1}-t_{n+1}. \end{aligned}$$

(35)

$(V_{n+1})$ From the above assumptions and (35), we obtain

$$\begin{aligned} \Vert y_{n+1}-x_{0} \Vert & \leq \Vert y_{n+1}-x_{n+1}\Vert +\Vert x_{n+1}-y_{n}\Vert +\Vert y_{n}-x_{n} \Vert +\Vert x_{n}-x_{0}\Vert \\ & \leq (s_{n+1}-t_{n+1})+(t_{n+1}-s_{n})+(s_{n}-t_{n})+(t_{n}-t_{0})=s_{n+1}, \end{aligned}$$

(36)

so $y_{n+1}\in\overline{B(x_{0}, s_{n+1})}$.

$(VI_{n+1})$ Since

$$\begin{aligned} \bigl\Vert I-K_{F}(x_{n+1}) \bigr\Vert &\geq1- \bigl\Vert K_{F}(x_{n+1}) \bigr\Vert \geq 1- \bigl(-g'(t_{n+1}) \bigr)^{-1}g''(r_{n+1}) (s_{n+1}-t_{n+1}) \\ &=1+g'(t_{n+1})^{-1}g''(r_{n+1}) (s_{n+1}-t_{n+1})=1-h_{n+1}, \end{aligned}$$

we have

$$\begin{aligned} \Vert x_{n+2}-y_{n+1}\Vert & = \frac{1}{2} \bigl\Vert K_{F}(x_{n+1}) \bigl[I-K_{F}(x_{n+1}) \bigr]^{-1} \bigr\Vert \bigl\Vert F'(x_{n+1})^{-1}F(x_{n+1}) \bigr\Vert \\ & \leq \frac{1}{2}\frac{h_{n+1}}{1-h_{n+1}}\cdot\frac {g(t_{n+1})}{-g'(t_{n+1})}=t_{n+2}-s_{n+1}. \end{aligned}$$

(37)

Further, we have

$$ \Vert x_{n+2}-x_{n+1}\Vert \leq \Vert x_{n+2}-y_{n+1}\Vert +\Vert y_{n+1}-x_{n+1} \Vert \leq t_{n+2}-t_{n+1}, $$

(38)

and when $m > n$

$$ \Vert x_{m}-x_{n}\Vert \leq \Vert x_{m}-x_{m-1}\Vert +\cdots+\Vert x_{n+1}-x_{n} \Vert \leq t_{m}-t_{n}. $$

(39)

It then follows that the sequence $\{x_{n}\}$ is convergent to a limit $x^{*}$. Take $n \rightarrow\infty$ in (34), we deduce $F(x^{*})=0$. From (39), we also get

$$ \bigl\Vert x^{*}-x_{n} \bigr\Vert \leq r_{1}-t_{n}. $$

(40)

Now, we prove the uniqueness. Suppose $x^{**}$ is also the solution of $F(x)$ on $B(x_{0}, \alpha)$. By Taylor expansion, we have

$$ 0=F \bigl(x^{**} \bigr)-F \bigl(x^{*} \bigr)= \int^{1}_{0}F^{\prime} \bigl((1-t)x^{*}+tx^{**} \bigr)\,dt \bigl(x^{**}-x^{*} \bigr). $$

(41)

Since

$$\begin{aligned} & \biggl\Vert F'(x_{0})^{-1} \int^{1}_{0} \bigl[F' \bigl((1-t)x^{*}+tx^{**} \bigr)-F'(x_{0}) \bigr]\,dt \biggr\Vert \\ &\quad \leq \biggl\Vert F'(x_{0})^{-1} \int^{1}_{0}{ \int^{1}_{0}F'' \bigl[x_{0}+t \bigl(x^{*}-x_{0} \bigr)+t \bigl(x^{**}-t^{*} \bigr) \bigr]}\,ds\,dt \bigl[x^{*}-x_{0}+t \bigl(x^{**}-x^{*} \bigr) \bigr] \biggr\Vert \\ &\quad \leq \int^{1}_{0}{ \int^{1}_{0}g'' \bigl[s \bigl\Vert x^{*}-x_{0}+t \bigl(x^{**}-x^{*} \bigr) \bigr\Vert \bigr]}\,ds\,dt \bigl\Vert x^{*}-x_{0}+t \bigl(x^{**}-x^{*} \bigr) \bigr\Vert \\ &\quad = \int^{1}_{0}g' \bigl( \bigl\Vert \bigl(x^{*}-x_{0} \bigr)+t \bigl(x^{**}-x^{*} \bigr) \bigr\Vert \bigr)\,dt-g'(0) \\ &\quad = \int^{1}_{0}g' \bigl( \bigl\Vert (1-t) \bigl(x^{*}-x_{0} \bigr)+t \bigl(x^{**}-x_{0} \bigr) \bigr\Vert \bigr)+1 \\ &\quad < \frac{g'(r_{1})+g'(\alpha)}{2}+1 \leq1, \end{aligned}$$

(42)

we can find that the inverse of $\int^{1}_{0}F' ((1-t)x^{*}+tx^{**} )\,dt$ exists, so $x^{**}=x^{*}$.

From Lemma 3, we get

$$ \bigl\Vert x_{n}-x^{*} \bigr\Vert \leq \frac{{(\sqrt[3]{\lambda_{1}} \theta)^{4^{n} } }}{{\sqrt[3]{\lambda_{1}} - (\sqrt[3]{\lambda_{1}} \theta)^{4^{n} } }}(r_{2} - r_{1} ), \quad n = 0,1, \ldots. $$

(43)

This completes the proof of the theorem. □

4 Numerical examples

In this section, we illustrate the previous study with an application to the following nonlinear equations.

Example 1

Let $X=Y=R$, and

$$ F(x)=\frac{1}{6}x^{3}+\frac{1}{6}x^{2}- \frac{5}{6}x+\frac{1}{3}=0. $$

(44)

We consider the initial point $x_{0}=0$, $\Omega= [-1, 1]$, we can get

$$\eta=\gamma=\frac{2}{5},\qquad N=\frac{6}{5},\qquad L=0. $$

Hence, from (8), we have $K=N=\frac{6}{5}$ and

$$\beta= \frac{{2 ( {\gamma+ 2\sqrt{\gamma^{2} + 2K} } )}}{{3 ( {\gamma + \sqrt{\gamma^{2} + 2K} } )^{2} }} = \frac{3}{5},\quad \eta< \beta. $$

This means that the hypotheses of Theorem 2 are satisfied, we can get the sequence $\{x_{n}\}_{(n \geq 0)}$ generated by the method (3) is well defined and converges.

Example 2

Consider an interesting case as follows:

$$ x(s)=1+\frac{1}{4}x(s) \int^{1}_{0}\frac{s}{s+t}x(t)\,dt, $$

(45)

where we have the space $X=C[0,1]$ with norm

$$\Vert x \Vert = \mathop{\max } _{0 \le s \le1} \bigl\vert {x(s)} \bigr\vert . $$

This equation arises in the theory of the radiative transfer, neutron transport and the kinetic theory of gases.

Let us define the operator F on X by

$$ F(x)=\frac{1}{4}x(s) \int^{1}_{0}{\frac{s}{s+t}x(t)\,dt}-x(s)+1. $$

(46)

Then for $x_{0}=1$ we can obtain

$$\begin{aligned} &N=0, \qquad L=0,\qquad K=0,\qquad \bigl\Vert F'(x_{0})^{-1} \bigr\Vert =1.5304,\qquad \eta=0.2652, \\ &\gamma = \bigl\Vert {F'(x_{0} )^{ - 1} F''(x_{0} )} \bigr\Vert = 1.5304 \times 2 \cdot\frac{1}{4}\mathop{\max} _{0 \le s \le1} \biggl\vert { \int _{0}^{1} {\frac{s}{{s + t}}\,dt} } \biggr\vert = 1.5304 \times\frac{{\ln2}}{2} = 0.5303, \\ &\frac{{2 ( {\gamma + 2\sqrt{\gamma^{2} + 2K} } )}}{{3 ( {\gamma + \sqrt{\gamma^{2} + 2K} } )^{2} }} = 0.9429 > \eta. \end{aligned}$$

That means that the hypotheses of Theorem 2 are satisfied.

Example 3

Consider the problem of finding the minimizer of the chained Rosenbrock function [22]:

$$ g(\mathbf{x})=\sum_{i=1}^{m} \bigl[4 \bigl(x_{i}-x^{2}_{i+1} \bigr)^{2}+ \bigl(1-x^{2}_{i+1} \bigr) \bigr],\quad \mathbf{x} \in \mathbf{R}^{m} . $$

(47)

For finding the minimum of g one needs to solve the nonlinear system $F(\mathbf{x})=0$, where $F(\mathbf{x})=\nabla g(\mathbf{x})$. Here, we apply the method (3), and compare it with Chebyshev method (CM), the Halley method (HM), and the super-Halley method (SHM).

In a numerical tests, the stopping criterion of each method is $\Vert \mathbf{x}_{k}-\mathbf{x}^{*}\Vert _{2} \leq1e-15$, where $\mathbf{x}^{*}=(1,1,\ldots,1)^{T}$ is the exact solution. We choose $m = 30$ and $x_{0} = 1.2\mathbf{ x}^{*}$. Listed in Table 1 are the iterative errors ($\Vert \mathbf{x}_{k}-\mathbf{x}^{*}\Vert _{2}$) of various methods. From Table 1, we know that, as tested here, the performance of the method (3) is better.

Table 1 The iterative errors ( $\pmb{\Vert \mathbf{x}_{n}-\mathbf{x}^{*}\Vert _{2}}$ ) of various methods

5 Conclusions

In this paper, a new Newton-Kantorovich convergence theorem of a fourth-order super-Halley method is established. As compared with the method in [21], the differentiability conditions of the method in the paper are mild. Finally, some examples are provided to show the application of the convergence theorem.