Remarks on the secant method for the solution of nonlinear operator equations

Abstract

Let \(X_{1},X_{2}\) be two Banach spaces and \(f:X_{1}\rightarrow X_{2}\) a nonlinear equation. We study the chord method for solving the equation \(f\left( x\right) =0\). Assuming the first order divided differences of \(f\) satisfy a Holder type condition, we obtain sufficient convergence conditions and error estimations at each step.

Authors

Ion Păvăloiu
(Tiberiu Popoviciu Institute of Numerical Analysis)

Keywords

chord method; divided differences; Holder condition

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Cite this paper as:

I. Păvăloiu, Remarks on the secant method for the solution of nonlinear operatorial equations, Research Seminars, Seminar on Mathematical Analysis, Preprint no. 7 (1991), pp. 127-132.

About this paper

Journal

Seminar on mathematical analysis,
Preprint

Publisher Name

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

DOI

Not available yet.

References

[1] Argyros, I.K., The secant method and fixed points of nonlinear operators, Mh. Math. 106, 85 94 (1988).

[2] Dennis, J.E., Toward a unified convergence theory for Newton like methods, Nonlinear Functional analysis and Applications (Ed. by L.B. Rall), pp. 425–472, New York, John Wiley (1986).

[3] Pavaloiu, I., Introduction to the Theory of Approximation of Equations Solutions, Dacia Ed., Cluj-Napoca, 1976 (in Romanian)

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"Babeş-Bolyai" University

Faculty of Mathematics and Physics

Research Seminars

Seminar on Mathematical Analysis

Preprint Nr.7, 1991, pp.127-132

Remarks on the secant method for the solution of nonlinear operatorial equations

Ion Păvăloiu

This note has for purpose some refinements of the convergence conditions and error delimitations obtained by I.K. Argyros in [2] with respect to the secant method for the solution of the equation:

(1) f⁢(x)=0,

where f:X1→X2 is a nonlinear operator, while X1 and X2 are Banach spaces.

If we denote by [x,y;f] the divided difference of the mapping f on the point x and y, then for fixed x,y we have [x,y;f]∈ℒ⁢(X1,X2). It is known that in certain conditions the sequence (xn)n≥0 generated by the secant method:

(2) xn+1=xn−[xn−1,xn;f]−1⁢f⁢(xn),x0,x1∈X1,n=1,2,…

converges to the solution x∗ of equation (1).

1. Generalizing a result on J.E. Dennis [3], I.K. Argyros [2] studies the convergence of the method (2) with the assumptions that the operator f is Fréchet differentiable, while the derivative f′⁢(x) fulfils a Hölder-like condition on a set D⊂X1,namely there exist a constant C>0 and number p∈(0,1] such that the inequality:

(3) ‖f′⁢(x)−f′⁢(y)‖≤C⁢‖x−y‖p

holds for every x,y∈D. In this case we shall say that f′⁢(⋅)∈HD⁢(C,p).

In the quoted paper I.K. Argyros defines the divided difference operator [x,y;f] as a linear operator which fulfils the conditions:

(4) [x,y;f]⁢(y−x)=f⁢(y)−f⁢(x),∀x,y∈D,

and, in addition, for every x,y,u∈D the following inequality holds:

(5) ‖[x,y;f]−[y,u;f]‖≤l1⁢‖x−u‖p+l2⁢‖x−y‖p+l2⁢‖y−u‖p,

where l1≥0,l2≥0 are constants which do not depend on x,y and u, while p∈(0,1].

Let x∗ be a simple solution of (1). We mean by that the mapping f′⁢(x∗) admits a bounded inverse mapping, and if [x∗,x∗;f]=f′⁢(x∗) then [x∗,x∗;f] admits a bounded inverse mapping. Thus the continuity of the mapping [x,y;f] with respect to the variable x and y ensures the existence of a number ε>0 such that the mapping [x,y;f] admits a bounded inverse mapping for every x,y∈U⁢(x∗,ε), where U⁢(x∗,ε)={x∈X1:‖x−x∗‖<ε} that is, the set B⁢(x,y)=‖[x,y;f]−1‖ is uniformly bounded in U⁢(x∗,ε)={x∈X1:‖x−x∗‖≤ε}.

Theorem 1.

[2] Let f:X1→X2 and let D⊂X1 an open set. The following conditions are fulfilled:

  • (a)

    x∗∈D is a simple solution of the equation (1);

  • (b)

    there exist ε∈0,b>0 such that ‖[x,y;f]−1‖≤b for every x,y∈U⁢(x∗,ε);

  • (c)

    there exists a convex set D0⊂D such that x∗∈D0, and there exists ε1>0, with 0<ε1<ε such that f′⁢(⋅)∈HD0⁢(C,p) for every x,y∈D0 and U⁢(x∗,ε1)⊂D0.

Let r>0 such that:

(6) 0<r<min⁡{ε1,(q⁢(p))−1/p}

where:

(7) q⁢(p)=bp+1⁢[2p⁢(l1+l2)⁢(1+p)+C].

Then, if x0⁢x1∈U¯⁢(x∗,r), the iterates xn,n=2,3,…, generated by (2) are well defined and belong to the set U¯⁢(x∗,r), while the sequence (xn)n≥0 converges to the unique solution x∗ of equation (1).

Moreover, the following estimation:

(8) ‖xn+1−x∗‖≤γ1⁢‖xn−1−x∗‖p⋅‖xn−x∗‖+γ2⁢‖xn−x∗‖p+1

holds for sufficiently great n, where:

(9) γ1=b⁢(l1+l2)⁢2p,
(10) γ2=b⁢C1+p

while l1,l2 and p were precised by the relation (5).

In order to prove this theorem the author uses the following two lemmas:

Lemma 1.

[2]. Let f:X1→X2 and D⊂X1. Suppose that D is an open set and f′⁢(⋅) does exist in every point of D. If, for a convex set D0⊆D,f′⁢(⋅)∈HD0⁢(C,p), then for every x,y∈D0 the following inequality holds:

‖f⁢(x)−f⁢(y)−f′⁢(x)⁢(y−x)‖≤C1+p⁢‖x−y‖1+p.
Lemma 2.

[2]. If [x,y;f] fulfils the conditions (4) and (5), the following relations hold:

  • (a)

    [x,x;f]=f′⁢(x) for every x∈D0;

  • (b)

    f′⁢(⋅)∈HD0⁢(2⁢(l1+l2),p).

From the proof of Theorem 1 follows, for the error estimation and for the convergence speeds of the sequence (xn)n≥0, the inequality:

(11) ‖xn+1−x∗‖≤(M⁢(r))n+1⁢‖x0−x∗‖

where one shows that M⁢(r)∈(0,1).

2. We shall make further down some remarks upon the above exposed results, showing that the hypotheses imposed in [2] can lead to more rich conclusions with respect to both the convergency order of the secant method and the error estimation.

Suppose that x0 and x1 fulfil the conditions:

  • (a’)

    ‖x∗−x0‖≤α⁢d0;

  • (b’)

    ‖x∗−x1‖≤min⁡{α⁢d0t1,‖x∗−x0‖}

where 0<d0<1,α=(q⁢(p))−1, while t1 is the positive root of the equation:

(12) t2−t−p =0
namely ⁢t1 =1+(1+4⁢p)1/22.

Using the condition (4) and (5), Lemmas 1 and 2, and the hypotheses of 1, it results easily from (2),for n=1, the inequality [2]:

(13) ‖x2−x∗‖≤γ1⁢‖x0−x∗‖p⁢‖x1−x∗‖+γ2⁢‖x1−x∗‖p+1

from which, using (a’) and (b’) and the fact that t1 is a root of equation (12), we obtain:

‖x2−x∗‖ ≤γ1⁢αp⁢d0p⁢α⁢d0t1+γ2⁢α1+p⁢d0t1⁢(1+p)
=α1+p⁢(γ1⁢d0t1+p+γ2⁢d0t1⁢(1+p))
=α1+p⁢d0t1+p⁢(γ1+γ2⁢d0p⁢(t1−1))
=α⁢d0t12⁢(γ1+γ2⁢d0p⁢(t1−1))⁢αp.

But

(γ1+γ2⁢d0p⁢(t1−1))⁢αp=γ1+γ2⁢d0p⁢(t1−1)γ1+γ2<1,

then the following inequality holds

‖x2−x∗‖≤α⁢d0t12.

We prove now that ‖x2−x∗‖≤‖x1−x∗‖. From the inequality (13) we obtain:

‖x2−x∗‖⁢(γ1⁢αp⁢d0p+γ2⁢αp⁢d0t1⁢p)⁢‖x1−x∗‖≤
≤αp⁢d0p⁢(γ1+γ2⁢d0p⁢(t1−1))⁢‖x1−x∗‖<‖x1−x∗‖

since d0p<1 and, as we saw above, αp⁢(γ1+γ2⁢d0p⁢(t1−1))<1.

Assume now that for n∈ℕ, n≥2, the following relations hold:

  • (a”)

    ‖xn−1−x∗‖≤α⁢d0t1n−1;

  • (b”)

    ‖xn−x∗‖≤min⁡{α⁢d0t1n,‖xn−1−x∗‖}

Proceeding as in the case of x2, and taking into account (a”), (b”) and (8), we obtain:

‖xn+1−x∗‖ ≤α1+p⁢d0t1n+1⋅(γ1+γ2⁢d0p⁢t1n−1⁢(t1+1))=
=α⁢d0t1n+1⋅αp⁢(γ1+γ2⁢d0p⁢t1n−1⁢(t1−1))≤α⁢d0t1n+1,

since, as previously, it is easy to show that:

αp⁢(γ1+γ2⁢d0p⁢t1n−1⁢(t1−1))<1

In order to complete the proof, we shall show that:

‖xn+1−x∗‖≤‖xn−x∗‖

Indeed, form (8) we deduce:

‖xn+1−x∗‖≤(γ1⁢αp⁢d0p⁢t1n−1+γ2⁢αp⁢d0p⁢t1n)⁢‖xn−x∗‖.

But d0<1 and αp⁢(γ1+γ2⁢d0p⁢(t1−1))<1, therefore:

‖xn+1−x∗‖≤‖xn−x∗‖.

We proved in this way the following theorem:

Theorem 2.

If the conditions of Theorem 1 are fulfilled, with the difference that x0 and x1 are chosen in such a manner to verify the relations (a’) and (b’), where α=(q⁢(p))−1/p and d0∈(0,1), then, for every n∈ℕ,xn∈U={x∈X1|‖x−x∗‖<α} and the following inequality holds:

(14) ‖xn+1−x∗‖≤α⁢d0t1n+1,n=0,1,…
Remark.

The inequality (14) contains in its right-hand side a number substantially smaller than that yielded by relation (11).

References


Institutul de Calcul

Oficiul Postal 1

C.P. 68

3400 Cluj-Napoca

Romania

This paper is in final form and no version of it is or will be submitted for publication elsewhere.

1991

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