"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\left(x\right)=\theta

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

(2)

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

(3)

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

In relations ( 2 ) and ( 3 ) $Q_{1}\$ And $Q_{2}$ are two iterative operators attached to equation ( 1 ) and by $\left[x,y:P\right]$ we designated the divided difference of the operator $P$ on the knots $x,y\in X,$ [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 $\bar{x}=Q_{1}\left(\bar{x}\right)$ And $\bar{x}=Q_{2}\left(\bar{x}\right)$ 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 $\beta_{1}>0,\ \beta_{2}>0$ such that for each $x\in X$ we have the following inequalities:

$\left\|Q_{1}\left(x\right)-x\right\|\leq\beta_{1}\left\|P\left(x\right)\right% \|,\ \;\;\left\|Q_{2}\left(x\right)-x\right\|\leq\beta_{2}\left\|P\left(x% \right)\right\|;$
c)

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

$\left\|P\left(Q_{1}\left(x\right)\right)\right\|\leq\alpha_{1}\left\|P\left(x% \right)\right\|^{k_{1}},\ \ \left\|P\left(Q_{2}\left(x\right)\right)\right\|% \leq\alpha_{2}\left\|P\left(x\right)\right\|^{k_{2}}.$

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 $\left(x_{n}\right)_{n=0}^{\infty}$ obtained using method ( 2 ) or ( 3 ).

Theorem 1 .

Either $x_{0}$ $\in X,~{}\delta>0$ And $S=\{x\in X:\left\|x-x_{0}\right\|\leq\delta\}$

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

i)

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

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

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

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

there is the number $M>0$ , such that for each $x,y,z\in S$ on a $\left\|\left[x,y,z;P\right]\right\|\leq M$ ;
we)

$\varepsilon_{0}=\left(MB^{2}\alpha_{1}\alpha_{2}\right)^{1/\left(q-1\right)}% \cdot\left\|P\left(x_{0}\right)\right\|<1$ Or $q=k_{1}+k_{2};$
vii)

$\rho^{\frac{1}{q-1}}\displaystyle\sum_{i=1}^{\infty}\varepsilon_{0}^{q^{i-1}}% \Big{(}B\alpha_{1}\rho^{\frac{k_{1}-1}{1-q}}\varepsilon_{0}^{q^{i-1}\left(k_{1% }-1\right)}+\beta_{1}\Big{)}\leq\delta$ Or $\rho=MB^{2}\alpha_{1}\cdot\alpha_{2},$

then we have the following properties:

j)

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

if we designate by $\varepsilon_{n}$ -the expression $\rho^{\frac{1}{q-1}}\left\|P\left(x_{n}\right)\right\|,$ so we have $\varepsilon_{n}\leq\varepsilon_{0}^{q^{n}}$ for each $n=0,1,\ldots;$
jjj)

we have the following inequality:

$\left\|\bar{x}-x_{n}\right\|\leq B\cdot\varepsilon_{0}^{q^{n}}\cdot\rho^{\frac% {1}{1-q}},$

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

Demonstration.

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

Indeed, from ( 2 ) we deduce:

	$\displaystyle\left\\|x_{1}-x_{0}\right\\|$	$\displaystyle\leq\left\\|x_{1}-Q_{1}\left(x_{0}\right)\right\\|+\left\\|Q_{1}% \left(x_{0}\right)-x_{0}\right\\|$
		$\displaystyle\leq\beta_{1}\left\\|P\left(x_{0}\right)\right\\|+B\left\\|P\left(Q_% {1}\left(x_{0}\right)\right)\right\\|$
		$\displaystyle\leq\big{(}\beta_{1}+B\alpha_{1}\left\\|P\left(x_{0}\right)\right% \\|^{k_{1}-1}\big{)}\left\\|P\left(x_{0}\right)\right\\|$
		$\displaystyle=\varepsilon_{0}\rho^{\frac{1}{1-q}}\Big{(}\beta_{1}+B\alpha_{1}% \varepsilon_{0}^{k_{1}-1}\cdot\rho^{\frac{k_{1}-1}{1-q}}\Big{)}\leq\delta$

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

Taking into account identity

	$\displaystyle P\left(x_{1}\right)=$	$\displaystyle P\left(Q_{1}\left(x_{0}\right)\right)+\left[Q_{1}\left(x_{0}% \right),Q_{2}\left(x_{0}\right);P\right]\left(x_{1}-Q_{1}\left(x_{0}\right)\right)$
		$\displaystyle+\left[Q_{1}\left(x_{0}\right),Q_{2}\left(x_{0}\right),x;P\right]% \left(x_{1}-Q_{1}\left(x_{0}\right)\right)\left(x_{1}-Q_{2}\left(x_{0}\right)\right)$

and from ( 2 ), it results:

	$\displaystyle\left\\|P\left(x_{1}\right)\right\\|$	$\displaystyle\leq M\left\\|x_{1}-Q_{1}\left(x_{0}\right)\right\\|\cdot\left\\|x_{% 1}-Q_{2}\left(x_{0}\right)\right\\|$
		$\displaystyle\leq MB^{2}\alpha_{1}\alpha_{2}\cdot\left\\|P\left(x_{0}\right)% \right\\|^{q}\leq\rho\cdot\rho^{\frac{q}{1-q}}\cdot\varepsilon_{0}^{q}$
		$\displaystyle=\rho^{\frac{1}{1-q}}\cdot\varepsilon_{0}^{q}$

from which it results:

\rho^{\frac{1}{1-q}}\left\|P\left(x_{1}\right)\right\|\leq\varepsilon_{0}^{q}

that's to say

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

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

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

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

In fact we have:

\left\|x_{n+1}-x_{0}\right\|\leq\sum_{i=1}^{n+1}\left\|x_{i}-x_{i-1}\right\|.

	$\displaystyle\left\\|x_{i}-x_{i-1}\right\\|$	$\displaystyle\leq\left\\|x_{i}-Q_{1}\left(x_{i-1}\right)\right\\|+\left\\|Q_{1}% \left(x_{i-1}\right)-x_{i-1}\right\\|$
		$\displaystyle\leq B\alpha_{1}\left\\|P\left(x_{i-1}\right)\right\\|^{k_{1}}+% \beta_{1}\left\\|P\left(x_{i-1}\right)\right\\|$
		$\displaystyle\leq\varepsilon_{i-1}\cdot\rho^{\frac{1}{1-q}}\left(\beta_{1}+B% \alpha_{1}\varepsilon_{i-1}^{k_{1}-1}\cdot\rho^{\frac{k_{1}-1}{1-q}}\right)$
		$\displaystyle\leq\varepsilon_{0}^{q^{i-1}}\cdot\rho^{\frac{1}{1-q}}\left(\beta% _{1}+B\alpha_{1}\cdot\varepsilon_{0}^{q^{i-1_{\left(k_{1}-1\right)}}}\rho^{% \frac{k_{1}-1}{1-q}}\right)$

from which we deduce that

\left\|x_{n+1}-x_{0}\right\|\leq\delta

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

As a result we have:

\left\|P\left(x_{n+1}\right)\right\|\leq MB^{2}\alpha_{1}\alpha_{2}\left\|P% \left(x_{n}\right)\right\|^{q}\leq\rho\cdot\rho^{\frac{q}{1-q}}\cdot% \varepsilon_{n}^{q}

from which results the inequality

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

that's to say

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

what needed to be demonstrated.

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

Indeed for each $k\in{\mathbb{N}}$ on a:

	$\displaystyle\left\\|x_{n+k}-x_{n}\right\\|$	$\displaystyle\leq\sum_{i=n+1}^{n+k}\left\\|x_{i}-x_{i-1}\right\\|$
		$\displaystyle\leq\varepsilon_{0}^{q^{n}}\cdot\rho^{\frac{1}{1-q}}\sum_{i=n+1}^% {n+k}\varepsilon_{0}^{q^{i-1_{-q^{n}}}}\Big{(}B\alpha_{1}\rho^{\frac{k_{1}-1}{% 1-q}}\cdot\varepsilon_{0}^{q^{i-1}\left(k_{1}-1\right)}+\beta_{1}\Big{)}$

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

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

\left\|\bar{x}-x_{0}\right\|\leq\delta

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

Of inequality

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

it results

\lim_{n\rightarrow\infty}\left\|P\left(x_{n}\right)\right\|=0

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

Of identity

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

it follows that

\left\|\bar{x}-x_{n}\right\|\leq B\left\|P\left(x_{n}\right)\right\|\leq B% \cdot\rho^{\frac{1}{1-q}}\varepsilon_{0}^{q^{n}}.

The theorem is therefore proven. ∎∎

Bibliography

[1] ^†^†margin: clickable $\rightarrow$ Păvăloiu, I., On iterative operators , Mathematical Studies and Research, 23 (1971), 10, 1537–1544.
[2] Pavaloiu, I., ^†^†margin: clickable $\rightarrow$ 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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	$\displaystyle\left\\|x_{1}-x_{0}\right\\|$	$\displaystyle\leq\left\\|x_{1}-Q_{1}\left(x_{0}\right)\right\\|+\left\\|Q_{1}% \left(x_{0}\right)-x_{0}\right\\|$
		$\displaystyle\leq\beta_{1}\left\\|P\left(x_{0}\right)\right\\|+B\left\\|P\left(Q_% {1}\left(x_{0}\right)\right)\right\\|$
		$\displaystyle\leq\big{(}\beta_{1}+B\alpha_{1}\left\\|P\left(x_{0}\right)\right% \\|^{k_{1}-1}\big{)}\left\\|P\left(x_{0}\right)\right\\|$
		$\displaystyle=\varepsilon_{0}\rho^{\frac{1}{1-q}}\Big{(}\beta_{1}+B\alpha_{1}% \varepsilon_{0}^{k_{1}-1}\cdot\rho^{\frac{k_{1}-1}{1-q}}\Big{)}\leq\delta$

	$\displaystyle\left\\|P\left(x_{1}\right)\right\\|$	$\displaystyle\leq M\left\\|x_{1}-Q_{1}\left(x_{0}\right)\right\\|\cdot\left\\|x_{% 1}-Q_{2}\left(x_{0}\right)\right\\|$
		$\displaystyle\leq MB^{2}\alpha_{1}\alpha_{2}\cdot\left\\|P\left(x_{0}\right)% \right\\|^{q}\leq\rho\cdot\rho^{\frac{q}{1-q}}\cdot\varepsilon_{0}^{q}$
		$\displaystyle=\rho^{\frac{1}{1-q}}\cdot\varepsilon_{0}^{q}$

	$\displaystyle\left\\|x_{i}-x_{i-1}\right\\|$	$\displaystyle\leq\left\\|x_{i}-Q_{1}\left(x_{i-1}\right)\right\\|+\left\\|Q_{1}% \left(x_{i-1}\right)-x_{i-1}\right\\|$
		$\displaystyle\leq B\alpha_{1}\left\\|P\left(x_{i-1}\right)\right\\|^{k_{1}}+% \beta_{1}\left\\|P\left(x_{i-1}\right)\right\\|$
		$\displaystyle\leq\varepsilon_{i-1}\cdot\rho^{\frac{1}{1-q}}\left(\beta_{1}+B% \alpha_{1}\varepsilon_{i-1}^{k_{1}-1}\cdot\rho^{\frac{k_{1}-1}{1-q}}\right)$
		$\displaystyle\leq\varepsilon_{0}^{q^{i-1}}\cdot\rho^{\frac{1}{1-q}}\left(\beta% _{1}+B\alpha_{1}\cdot\varepsilon_{0}^{q^{i-1_{\left(k_{1}-1\right)}}}\rho^{% \frac{k_{1}-1}{1-q}}\right)$

On a Steffensen type method for solving nonlinear operator equations

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On a Steffensen type method for solving nonlinear operator equations

Theorem 1 .

Demonstration.

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