On a Steffensen type method for solving nonlinear operator equations

9 years ago

Abstract

Let $X$ be a Banach space, $Y$ a normed space and the nonlinear operator equation $P\left( x\right) =0$, where $P:X\rightarrow Y$. We consider two operators $Q_{1},Q_{2}:X\rightarrow X$ attached to $P$ and we study the convergence of the Steffensen type method \[x_{n+1}=Q_1(x_n)-[Q_1( x_n), Q_2( x_n);P]^{-1}P(Q_1(x_n)). \] We give some conditions ensuring the convergence of this sequence to the solution and we obtain the convergence order of the sequence in terms of the convergence orders of $Q_{1}$ and $Q_{2}$.

Authors

Ion Păvăloiu
(Tiberiu Popoviciu Institute of Numerical Analysis)

Title

Original title (in French)

Sur une méthode de type Steffensen utilisée pour la résolution des equations operationnelles non-linéaires

English translation of the title

On a Steffensen type method for solving nonlinear operator equations

Keywords

Steffensen type method; Banach space; iterative method; convergence order

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Cite this paper as:

I. Păvăloiu, Sur une méthode de type Steffensen utilisée pour la résolution des equations operationnelles non-linéaires, Seminar on functional analysis and numerical methods, Preprint no. 1 (1989), pp. 105-110 (in French).

About this paper

Journal

Seminar on functional analysis and numerical methods,
Preprint

Publisher Name

“Babes-Bolyai” University,
Faculty of Mathematics,
Research Seminars

DOI

Not available yet.

References

[1] Pavaloiu, I., Asupra operatorilor iterativi, Studii si Cercetari Matematice, 23 (1971), 10, 1537–1544.

[2] Pavaloiu, I., Introducere in teoria aproximarii solutiilor ecuatiilor, Ed. Dacia, 1976.

[3] Ul’m, S., Ob oboboscennyh rezdelennih reznostiak I, Izv. Akad. Nauk Estonskoi SSR 16 (1867), 1, 13–36.

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"Babeş-Bolyai" University

Faculty of Mathematics and Physics

Research Seminars

Seminar on Functional Analysis and Numerical Methods

Preprint Nr.1, 1989, pp.105-110

On a Steffensen type method for solving nonlinear operator equations

by
Ion Pavaloiu

Either $X$ a Banach space and $AND$ a normed linear space. To solve the equation

(1)

P (x) = i

Or $P : X \to AND$ is an operator, and $i$ is the zero element of the space $AND$ , we consider the following iterative methods:

(2)

x_{n + 1} = Q_{1} (x_{n}) - {[Q_{1} (x_{n}), Q_{2} (x_{n}); P]}^{- 1} P (Q_{1} (x_{n}))

(3)

x_{n + 1} = Q_{2} (x_{n}) - {[Q_{1} (x_{n}), Q_{2} (x_{n}); P]}^{- 1} P (Q_{2} (x_{n}))

In relations ( 2 ) and ( 3 ) $Q_{1}$ And $Q_{2}$ are two iterative operators attached to equation ( 1 ) and by $[x, and : P]$ we designated the divided difference of the operator $P$ on the knots $x, and \in X,$ [2], [3].

To clarify, we will impose on operators $Q_{1}$ And $Q_{2}$ the following conditions:

a)

And $\bar{x}$ is a solution of equation ( 1 ) then we have $\bar{x} = Q_{1} (\bar{x})$ And $\bar{x} = Q_{2} (\bar{x})$ and vice versa, if $\bar{x}$ is a fixed point for operators $Q_{1}$ And $Q_{2}$ SO $\bar{x}$ is a solution to equation ( 1 );
b)

there are numbers $b_{1} > 0, b_{2} > 0$ such that for each $x \in X$ we have the following inequalities:

$‖ Q_{1} (x) - x ‖ \leq b_{1} ‖ P (x) ‖, ‖ Q_{2} (x) - x ‖ \leq b_{2} ‖ P (x) ‖;$
c)

there are real and positive numbers $a_{1} < a_{2}$ and also natural numbers $k_{1}, k_{2}$ such that for each $x \in X$ we have the following inequalities:

$‖ P (Q_{1} (x)) ‖ \leq a_{1} {‖ P (x) ‖}^{k_{1}}, ‖ P (Q_{2} (x)) ‖ \leq a_{2} {‖ P (x) ‖}^{k_{2}} .$

We can easily see that in the case where we start from the same initial element $x_{0} \in X$ , the iterative methods ( 2 ) and ( 3 ) provide the same sequence of approximations to the solution of equation ( 1 ).

Subsequently we will study the convergence of the sequence ${(x_{n})}_{n = 0}^{\infty}$ obtained using method ( 2 ) or ( 3 ).

Theorem 1 .

Either $x_{0}$ $\in X, d > 0$ And $S = {x \in X : ‖ x - x_{0} ‖ \leq d}$

If we can choose the initial element $x_{0}$ , the real number $d$ and applications $Q_{1}$ And $Q_{2}$ such as:

i)

applications $Q_{1}$ And $Q_{2}$ meet condition a);
ii)

$Q_{1} (S) \subseteq S, Q_{2} (S) \subseteq S;$
iii)

applications $Q_{1}, Q_{2}$ And $P$ meet conditions b) and c) for each $x \in S;$
iv)

for each $x, and \in S$ it exists ${[x, and; P]}^{- 1}$ and there is the number $B > 0$ , such that for each $x, and \in S$ we have $‖ {[x, and; P]}^{- 1} ‖ \leq B;$
in)

there is the number $M > 0$ , such that for each $x, and, With \in S$ on a $‖ [x, and, With; P] ‖ \leq M$ ;
we)

$e_{0} = {(M B^{2} a_{1} a_{2})}^{1 / (q - 1)} \cdot ‖ P (x_{0}) ‖ < 1$ Or $q = k_{1} + k_{2};$
vii)

$r^{\frac{1}{q - 1}} \sum_{i = 1}^{\infty} e_{0}^{q^{i - 1}} (B a_{1} r^{\frac{k_{1} - 1}{1 - q}} e_{0}^{q^{i - 1} (k_{1} - 1)} + b_{1}) \leq d$ Or $r = M B^{2} a_{1} \cdot a_{2},$

then we have the following properties:

j)

the sequel ${(x_{n})}_{n \in N}$ obtained using method ( 2 ) or using method ( 3 ) is convergent and if we denote by $\bar{x}$ the limit of the sequence ${(x_{n})}_{n = 0}^{\infty},$ so we have $P (\bar{x}) = i;$
jj)

if we designate by $e_{n}$ -the expression $r^{\frac{1}{q - 1}} ‖ P (x_{n}) ‖,$ so we have $e_{n} \leq e_{0}^{q^{n}}$ for each $n = 0, 1, \dots;$
jjj)

we have the following inequality:

$‖ \bar{x} - x_{n} ‖ \leq B \cdot e_{0}^{q^{n}} \cdot r^{\frac{1}{1 - q}},$

for each $n = 0, 1, 2, \dots$

Demonstration.

Let us first prove that in the hypotheses of the theorem the elements of the sequence ${(x_{n})}_{n \in N}$ belong to the whole $S$ .

Indeed, from ( 2 ) we deduce:

	$‖ x_{1} - x_{0} ‖$	$\leq ‖ x_{1} - Q_{1} (x_{0}) ‖ + ‖ Q_{1} (x_{0}) - x_{0} ‖$
		$\leq b_{1} ‖ P (x_{0}) ‖ + B ‖ P (Q_{1} (x_{0})) ‖$
		$\leq (b_{1} + B a_{1} {‖ P (x_{0}) ‖}^{k_{1} - 1}) ‖ P (x_{0}) ‖$
		$= e_{0} r^{\frac{1}{1 - q}} (b_{1} + B a_{1} e_{0}^{k_{1} - 1} \cdot r^{\frac{k_{1} - 1}{1 - q}}) \leq d$

from which it follows that $x_{1} \in S$ .

Taking into account identity

	$P (x_{1}) =$	$P (Q_{1} (x_{0})) + [Q_{1} (x_{0}), Q_{2} (x_{0}); P] (x_{1} - Q_{1} (x_{0}))$
		$+ [Q_{1} (x_{0}), Q_{2} (x_{0}), x; P] (x_{1} - Q_{1} (x_{0})) (x_{1} - Q_{2} (x_{0}))$

and from ( 2 ), it results:

	$‖ P (x_{1}) ‖$	$\leq M ‖ x_{1} - Q_{1} (x_{0}) ‖ \cdot ‖ x_{1} - Q_{2} (x_{0}) ‖$
		$\leq M B^{2} a_{1} a_{2} \cdot {‖ P (x_{0}) ‖}^{q} \leq r \cdot r^{\frac{q}{1 - q}} \cdot e_{0}^{q}$
		$= r^{\frac{1}{1 - q}} \cdot e_{0}^{q}$

from which it results:

r^{\frac{1}{1 - q}} ‖ P (x_{1}) ‖ \leq e_{0}^{q}

that's to say

e_{1} \leq e_{0}^{q} .

Let us assume that the elements $x_{1}, x_{2}, \dots, x_{n} \in S,$

e_{i} \leq e_{0}^{q^{i}}

and demonstrate that $x_{n + 1} \in S$ And $e_{n + 1} \leq e_{0}^{q^{n + 1}} .$

In fact we have:

‖ x_{n + 1} - x_{0} ‖ \leq \sum_{i = 1}^{n + 1} ‖ x_{i} - x_{i - 1} ‖ .

	$‖ x_{i} - x_{i - 1} ‖$	$\leq ‖ x_{i} - Q_{1} (x_{i - 1}) ‖ + ‖ Q_{1} (x_{i - 1}) - x_{i - 1} ‖$
		$\leq B a_{1} {‖ P (x_{i - 1}) ‖}^{k_{1}} + b_{1} ‖ P (x_{i - 1}) ‖$
		$\leq e_{i - 1} \cdot r^{\frac{1}{1 - q}} (b_{1} + B a_{1} e_{i - 1}^{k_{1} - 1} \cdot r^{\frac{k_{1} - 1}{1 - q}})$
		$\leq e_{0}^{q^{i - 1}} \cdot r^{\frac{1}{1 - q}} (b_{1} + B a_{1} \cdot e_{0}^{q^{i - 1_{(k_{1} - 1)}}} r^{\frac{k_{1} - 1}{1 - q}})$

from which we deduce that

‖ x_{n + 1} - x_{0} ‖ \leq d

that is to say that $x_{n + 1} \in S$ .

As a result we have:

‖ P (x_{n + 1}) ‖ \leq M B^{2} a_{1} a_{2} {‖ P (x_{n}) ‖}^{q} \leq r \cdot r^{\frac{q}{1 - q}} \cdot e_{n}^{q}

from which results the inequality

e_{n + 1} \leq e_{n}^{q}

that's to say

e_{n + 1} \leq e_{0}^{q^{n + 1}}

what needed to be demonstrated.

We will now demonstrate that the following ${(x_{n})}_{n = 0}^{\infty}$ provided by relation ( 2 ) is fundamental.

Indeed for each $k \in ℕ$ on a:

	$‖ x_{n + k} - x_{n} ‖$	$\leq \sum_{i = n + 1}^{n + k} ‖ x_{i} - x_{i - 1} ‖$
		$\leq e_{0}^{q^{n}} \cdot r^{\frac{1}{1 - q}} \sum_{i = n + 1}^{n + k} e_{0}^{q^{i - 1_{- q^{n}}}} (B a_{1} r^{\frac{k_{1} - 1}{1 - q}} \cdot e_{0}^{q^{i - 1} (k_{1} - 1)} + b_{1})$

which expresses that the following ${(x_{n})}_{n = 0}^{\infty}$ is fundamental.

Either $\bar{x} = lim_{n \to \infty} x_{n};$ then from the inequality above where we pose $n = 0$ and let's do $k \to \infty$ , it follows that

‖ \bar{x} - x_{0} ‖ \leq d

that's to say $\bar{x} \in S .$

Of inequality

e_{n} \leq e_{0}^{q^{n}}

it results

lim_{n \to \infty} ‖ P (x_{n}) ‖ = 0

that's to say $P (\bar{x}) = i,$ equality which expresses the fact that $\bar{x}$ is a solution to equation ( 1 ).

Of identity

P (\bar{x}) - P (x_{n}) = [\bar{x}, x_{n}; P] (\bar{x} - x_{n})

it follows that

‖ \bar{x} - x_{n} ‖ \leq B ‖ P (x_{n}) ‖ \leq B \cdot r^{\frac{1}{1 - q}} e_{0}^{q^{n}} .

The theorem is therefore proven. ∎∎

Bibliography

[1] ^†^†margin: clickable $\to$ Păvăloiu, I., On iterative operators , Mathematical Studies and Research, 23 (1971), 10, 1537–1544.
[2] Pavaloiu, I., ^†^†margin: clickable $\to$ Introduction to the theory of approximation of solutions of equations , Ed. Dacia, 1976.
[3] Ul'm, S., Ob oboboscennyh rezdelennîh reznostiak I , Izv. Akad. Nauk Estonskoi SSR 16 (1867), 1, 13–36.

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On a Steffensen type method for solving nonlinear operator equations

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On a Steffensen type method for solving nonlinear operator equations

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