Bul. Ştiinţ. Univ, Baia Mare, Ser. B,

Matematică-Informatică, Vol. XVI (2000) Nr. 2, 219–224

Dedicated to Maria S. Pop on her 60^th anniversary

On the Chebyshev Method for Approximating the Solutions of Polynomial Operator Equations of Degree 2

Ion Păvăloiu

Abstract.

In this paper, the Chebyshev method for approximating the solutions of polynomial operator equations of degree 2 is presented. The convergence of the Chebyshev method is studied.

MSC 2000: 47H60.

Keywords: polynomial operator, Chebyshev method.

1. Introduction

The polynomial operator equations represent an important class of operator equations [1]. Among them, the polynomial operator equations of degree 2 have a special importance because the convergence hypotheses for the usual methods (Newton method, chord method, Steffensen method, Chebyshev method, etc.) are much simplified compared to the general case [6]. In this note we shall study the convergence of the Chebyshev method for the mentioned equations.

Let $X$ be a Banach space and consider a mapping $f:X\rightarrow X$ . We remind that the mapping $f$ is a polynomial operator of degree two if

a)

$f$ is three times differentiable;
b)

$f^{\prime\prime\prime}\left(x\right)=\theta_{3},\forall x\in X,$ where $\theta_{3}$ is the trilinear null operator.

2. The convergence of the Chebyshev method

Consider the equation

(2.1)

f\left(x\right)=\theta,

where $\theta\in X$ is the zero element. Given an initial approximation $u_{0}\in X$ of a solution $\bar{u}$ of the above equation, the Chebyshev method generates the sequence $\left(u_{n}\right)_{n\geq 0}$ by

u_{n+1}=u_{n}-\Gamma_{n}f\left(u_{n}\right)-\tfrac{1}{2}\Gamma_{n}f^{\prime% \prime}\left(u_{n}\right)\left(\Gamma_{n}f\left(u_{n}\right)\right)^{2},\ n=0,% 1,2,\ldots,\ u_{0}\in X

where $\Gamma_{n}=f^{\prime}\left(u_{n}\right)^{-1}$ .

Consider $r>0$ and denote $S=\{u\in X:\left\|u-u_{0}\right\|\leq r\}$ . Since $f^{\prime\prime\prime}\left(x\right)=\theta_{3},\forall x\in X$ , it is clear that $f^{\prime\prime}\left(x\right)$ does not depend on $x$ , so we may take $m_{2}=\left\|f^{\prime\prime}\left(x\right)\right\|.$

From the Taylor formula we obtain

(2.2)		$\displaystyle f\left(u\right)=$	$\displaystyle f\left(u_{0}\right)+f^{\prime}\left(u_{0}\right)\left(u-u_{0}% \right)+\tfrac{1}{2}f^{\prime\prime}\left(u_{0}\right)\left(u-u_{0}\right)^{2}$
(2.3)		$\displaystyle f^{\prime}\left(u\right)=$	$\displaystyle f^{\prime}\left(u_{0}\right)+f^{\prime\prime}\left(u_{0}\right)% \left(u-u_{0}\right).$

By (2.3) it follows

\left\|f^{\prime}\left(u\right)\right\|\leq\left\|f^{\prime}\left(u_{0}\right)% \right\|+m_{2}r,\ \forall u\in S,

which implies

(2.4)

\sup_{u\in S}\left\|f^{\prime}\left(u\right)\right\|\leq\left\|f^{\prime}\left% (u_{0}\right)\right\|+m_{2}r.

Similarly (2.2), leads to

(2.5)

\sup_{u\in S}\left\|f\left(u\right)\right\|\leq\left\|f\left(u_{0}\right)% \right\|+r\left\|f^{\prime}\left(u_{0}\right)\right\|+\tfrac{1}{2}m_{2}r^{2}.

We shall make the following notations:

(2.6)

m_{0}=\|f(u_{0})\|+r\|f^{\prime}(u_{0})\|+\tfrac{1}{2}m_{2}r^{2}

(2.7)

\mu=\tfrac{1}{2}m_{2}^{2}b^{4}\left(1+\tfrac{1}{4}m_{2}m_{0}b^{2}\right),

(2.8)

v=b\left(1+\tfrac{1}{2}m_{2}m_{0}b^{2}\right),

where

(2.9)

b=\tfrac{b_{0}}{1-m_{2}b_{0}r}\ \text{and }b_{0}=\left\|f^{\prime}\left(u_{0}% \right)^{-1}\right\|.

With the above notations, the following result holds:

Theorem 1.

If for some $u_{0}\in X$ and $r>0,$ the mapping $f$ satisfies

i.

$\exists f^{\prime}\left(u_{0}\right)^{-1};$
ii.

$m_{2}b_{0}r<1;$
iii.

the numbers $\mu$ and $v$ given by (2.7) and (2.8) verify

$\displaystyle\rho_{0}$ $\displaystyle=\sqrt{\mu}\cdot\left\|f\left(u_{0}\right)\right\|<1\ \text{and}$

$\displaystyle\tfrac{v\cdot\rho_{0}}{\sqrt{\mu\left(1-\rho_{0}\right)}}$ $\displaystyle\leq r,$

then the following relations hold:

j.

the sequence $\left(u_{n}\right)_{n\geq 0}$ generated by the Chebyshev method converges;
jj.

denoting $\bar{u}=\lim\limits_{n\rightarrow\infty}u_{n},$ then $\bar{u}\in S$ and $f\left(\bar{u}\right)=\theta;$
jjj.

$\left\|u_{n+1}-u_{n}\right\|\leq\frac{v\rho_{0}^{3^{n}}}{\sqrt{\mu}},\quad n=0% ,1,\ldots;$
jv.

$\left\|\bar{u}-u_{n}\right\|\leq\frac{v\rho_{0}^{3^{n}}}{\sqrt{\mu}\left(1-% \rho_{0}^{3^{n}}\right)},\quad n=0,1,\ldots$

Proof.

First we shall show that hypothesis $ii)$ implies the existence of the application $f^{\prime}\left(u\right)^{-1}$ for all $u\in S$ and, moreover, $\left\|f^{\prime}\left(u\right)^{-1}\right\|\leq b$ . Indeed, one has

\left\|f^{\prime}\left(u_{0}\right)^{-1}\left(f^{\prime}\left(u_{0}\right)-f^{% \prime}\left(u\right)\right)\right\|\leq m_{2}b_{0}r,\ \forall u\in S.

Applying the Banach Lemma and taking into account relation $ii)$ it follows the existence of $f^{\prime}\left(u\right)^{-1}$ for all $u\in S$ and, moreover,

(2.10)

\left\|f^{\prime}\left(u\right)^{-1}\right\|\leq\tfrac{b_{0}}{1-m_{2}b_{0}r}=b.

Denote by $g$ the mapping $g:S\rightarrow X$ given by

(2.11)

g\left(u\right)=-\Gamma\left(u\right)f\left(u\right)-\tfrac{1}{2}\Gamma\left(u% \right)f^{\prime\prime}\left(u\right)\left(\Gamma\left(u\right)f\left(u\right)% \right)^{2}

where $\Gamma\left(u\right)=f^{\prime}\left(u\right)^{-1}$ .

It can be easily seen that for all $u\in S$ , the following identity holds:

	$\displaystyle f\left(u\right)+f^{\prime}\left(u\right)g\left(u\right)+\tfrac{1% }{2}f^{\prime\prime}\left(u\right)g^{2}\left(u\right)=$
	$\displaystyle=\tfrac{1}{2}f^{\prime\prime}\left(u\right)\left(f^{\prime}\left(% u\right)^{-1}f\left(u\right),f^{\prime}\left(u\right)^{-1}f^{\prime\prime}% \left(u\right)\left(f^{\prime}\left(u\right)^{-1}f\left(u\right)\right)^{2}\right)$
	$\displaystyle\;\;\;+\tfrac{1}{8}f^{\prime\prime}\left(u\right)\left(f^{\prime}% \left(u\right)^{-1}f^{\prime\prime}\left(u\right)\left(f^{\prime}\left(u\right% )^{-1}f\left(u\right)\right)^{2}\right)^{2},$

whence

(2.12)

\left\|f\left(u\right)+f^{\prime}\left(u\right)g\left(u\right)+\tfrac{1}{2}f^{% \prime\prime}\left(u\right)g^{2}\left(u\right)\right\|\leq\mu\left\|f\left(u% \right)\right\|^{3},\qquad\forall u\in S.

Since $u_{n+1}=u_{n}+g\left(u_{n}\right),$ from the Taylor formula we get

	$\displaystyle f\left(u_{n+1}\right)$	$\displaystyle=f\left(u_{n}\right)+f^{\prime}\left(u_{n}\right)\left(u_{n+1}-u_% {n}\right)+\tfrac{1}{2}f^{\prime\prime}\left(u_{n}\right)\left(u_{n+1}-u_{n}% \right)^{2}$
		$\displaystyle=f\left(u_{n}\right)+f^{\prime}\left(u_{n}\right)g\left(u_{n}% \right)+\tfrac{1}{2}f^{\prime\prime}\left(u_{n}\right)g^{2}\left(u_{n}\right),$

and by (2.2),

(2.13)

\left\|f\left(u_{n+1}\right)\right\|\leq\mu\left\|f\left(u_{n}\right)\right\|^% {3},

provided that $u_{n}\in S$ . Since $u_{0}\in S$ one obtains

\left\|u_{1}-u_{0}\right\|=\left\|g\left(u_{0}\right)\right\|\leq v\left\|f% \left(u_{0}\right)\right\|\leq\tfrac{v\sqrt{\mu}\left\|f\left(u_{0}\right)% \right\|}{\sqrt{\mu}\left(1-\rho_{0}\right)}=\tfrac{v\rho_{0}}{\sqrt{\mu}\left% (1-\rho_{0}\right)}\leq r,

i.e. $u_{1}\in S.$

Suppose now that the following relations hold:

$\alpha)$

$u_{i}\in S,\ i=\overline{0,k};$
$\beta)$

$\left\|f\left(u_{i}\right)\right\|\leq\mu\left\|\left(u_{i-1}\right)\right\|^{% 3},\ i=\overline{1,k}.$

From $u_{k}\in S$ and (2.2) it results

(2.14)

\left\|f\left(u_{k+1}\right)\right\|\leq\mu\left\|f\left(u_{k}\right)\right\|^% {3}\ \

and

(2.15)

\|u_{k+1}-u_{k}\|\leq v\left\|f\left(u_{k}\right)\right\|.

Inequality (2.14) leads to

\left\|f\left(u_{i}\right)\right\|\leq\tfrac{1}{\sqrt{\mu}}\left(\sqrt{\mu}% \left\|f\left(u_{0}\right)\right\|\right)^{3^{i}},\qquad i=\overline{1,k+1.}

By (2.13) one gets

	$\displaystyle\left\\|u_{k+1}-u_{0}\right\\|$	$\displaystyle\leq\sum_{i=1}^{k+1}\left\\|u_{1}-u_{i-1}\right\\|\leq\sum_{i=1}^{k% +1}v\left\\|f\left(u_{i-1}\right)\right\\|$
		$\displaystyle\leq\tfrac{v}{\sqrt{\mu}}\sum_{i=1}^{k+1}\rho_{0}^{3^{i-1}}\leq% \tfrac{v\rho_{0}}{\sqrt{\mu}\left(1-\rho_{0}\right)},$

i.e., $u_{k+1}\in S$ .

It is easy to show that

(2.16)

\left\|u_{n+m}-u_{n}\right\|\leq\tfrac{v\rho_{0}^{3^{n}}}{\sqrt{\mu}\left(1-% \rho_{0}^{3^{n}}\right)},\qquad n=0,1,\ldots,m\in\mathbb{N}

and, since $\rho_{0}<1$ , it follows that the sequence $\left(u_{n}\right)_{n\geq 0}$ is Cauchy, so it converges. Denoting $\bar{u}=\lim u_{n}$ , it is clear that $f\left(\bar{u}\right)=\theta$ . Letting $m\rightarrow\infty$ in (2.16) leads us to $j v$ . ∎

The Chebyshev method may be applied with the aid of the following algorithm:

Let $u_{n}$ be an arbitrary approximation of the solution of (2.1), and which satisfies the hypotheses of Theorem 1. The next approximation $u_{n+1}$ may be obtained by

1. Solve the linear operator equation

f^{\prime}\left(u_{n}\right)p_{n}=f\left(u_{n}\right),

2. Solve the linear operator equation

f^{\prime}\left(u_{n}\right)q_{n}=f^{\prime\prime}\left(u_{n}\right)p_{n}^{2}

3. Compute

u_{n+1}=u_{n}-p_{n}-\tfrac{1}{2}q_{n}.

References

[1] Argyros, I.K., Polynomial Operator Equations in Abstract Spaces and Applications, CRC Press, Boca Raton, Boston (1998).
[2] Ciarlet, P.G., Introduction à l’Analyse Numérique Matricielle et à l’Optimisation, Mason, Paris (1990).
[3] Chatelin, F., Valeur Propres de Matrices, Mason, Paris (1998).
[4] Collatz, L., Functionalanalysis und Numerische Mathematick, Springer-Verlag, Berlin (1964).
[5] Kartîşov, V.S., Iuhno, F.L., O nekotoryh k Modifikatsiah Method Niutona dlea Resenia Nelineinoi Spektralnoi Zadaci, J. Vîcisl. matem. i mamem. fiz. (33) 9, (1973), 1403-1409.
[6] ^†^†margin: clickable $\rightarrow$ Păvăloiu, I. Sur les procédées itératifs à un ordre élevé de convergence, Mathematica, 12 (35) 2 (1970), 309–324.

Received 18.05.2000

”T. Popoviciu” Institute of Numerical Analysis

Str. Gh. Bilaşcu nr.37

C.P. 68, O.P. 1

3400 Cluj-Napoca

On the Chebyshev method for approximating the solutions of polynomial operator equations of degree 2

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On the Chebyshev Method for Approximating the Solutions of Polynomial Operator Equations of Degree 2

Abstract.

1. Introduction

2. The convergence of the Chebyshev method

Theorem 1.

Proof.

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	$\displaystyle\tfrac{v\cdot\rho_{0}}{\sqrt{\mu\left(1-\rho_{0}\right)}}$	$\displaystyle\leq r,$