Expanding the applicability of inexact Newton methods using restricted convergence domains

Abstract

We provide a new semilocal convergence analysis for the inexact Newton method in order to approximate a solution of a nonlinear equation in a Banach space setting. Using a new idea of restricted convergence domains we present a convergence analysis with the following advantages over earlier studies: larger convergence domain, tighter error bounds on the distances involved and an at least as precise information on the location of the solution. This way we expand the applicability of the inexact Newton method. Special cases and numerical examples are also provided.

Inexact Newton’s method for generalized operator equations in Banach spaces

“…Since our assumptions on F and G in generalized operator equation 9are fairly general, our main result covers a wide variety of nonlinear operator equations. In fact, our main result provides an affirmative answer of Question 1 and also significantly improves the corresponding results of [7,8,10].…”

“…Lemma 1 [10] Let R, r, λ , ω, b and ρ be real numbers such that 0 < r < R, λ > 0, ω ≥ 1, ρ ≥ 0 and 0 < b ≤ 1. Let L and L 0 be two positive nondecreasing integrable functions defined on any involve intervals.…”

“…Clearly L 0 (t) ≤ L(t) for all 0 < t < 1 γ and functions f and g defined by (10) and (11) reduces to…”

“…In this paper we improved and extend the inexact Newton-like methods [7,8,10] in the context of differentiability of involved operator and introduced an inexact Newton-like algorithm for solving the generalized operator equations. We discussed the semilocal convergence analysis of our algorithm under the weak Lipschitz condition.…”

“…Assuming the residual controls (7) with k = 1, the γ-condition and the Smale's α-theory have been established by Shen and Li in [7] for the inexact Newton method (3). Recently, Xu et al [8], Argyros and Santosh George [10] have studied semilocal convergence analysis of the inexact Newton method (3) under the residual controls (7) with k = 1 and F −1…”