2024
DOI: 10.3390/sym16010120
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Memory-Accelerating Methods for One-Step Iterative Schemes with Lie Symmetry Method Solving Nonlinear Boundary-Value Problem

Chein-Shan Liu,
Chih-Wen Chang,
Chung-Lun Kuo

Abstract: In this paper, some one-step iterative schemes with memory-accelerating methods are proposed to update three critical values f′(r), f″(r), and f‴(r) of a nonlinear equation f(x)=0 with r being its simple root. We can achieve high values of the efficiency index (E.I.) over the bound 22/3=1.587 with three function evaluations and over the bound 21/2=1.414 with two function evaluations. The third-degree Newton interpolatory polynomial is derived to update these critical values per iteration. We introduce relaxati… Show more

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Cited by 3 publications
(6 citation statements)
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“…The many decimal numbers after this point guarantees that x c is a solution of Equation ( 8), thus satisfying the error tolerance with ε = 10 −15 . Next, we solve Equations ( 1) and ( 2) with F = 3u 2 /2, a = 4, and b = 1, and compare the results with the solution u(y) = 4/(y + 1) 2 . In Equation (18), we take H(t) = q 0 (1 − e −t ), and with a 0 = 2.27615, b 0 = −0.149978, and q 0 = −0.15 through eight iterations, we obtain the slope u ′ (0) = −8.000000000000028, which is very close to the exact one r = −8; 5.33 × 10 −15 is the maximum error obtained for u(y).…”
Section: Numerical Resultsmentioning
confidence: 99%
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“…The many decimal numbers after this point guarantees that x c is a solution of Equation ( 8), thus satisfying the error tolerance with ε = 10 −15 . Next, we solve Equations ( 1) and ( 2) with F = 3u 2 /2, a = 4, and b = 1, and compare the results with the solution u(y) = 4/(y + 1) 2 . In Equation (18), we take H(t) = q 0 (1 − e −t ), and with a 0 = 2.27615, b 0 = −0.149978, and q 0 = −0.15 through eight iterations, we obtain the slope u ′ (0) = −8.000000000000028, which is very close to the exact one r = −8; 5.33 × 10 −15 is the maximum error obtained for u(y).…”
Section: Numerical Resultsmentioning
confidence: 99%
“…where F(y, u(y), u ′ (y)) : [0, 1] × R 2 → R is a given nonlinear continuous function, and a and b are given constants. We integrate Equation (1), starting from the initial values u(0) = a and u ′ (0) = x, where x is an unknown value determined by u(1) = b in Equation (2), which results in a nonlinear equation:…”
Section: Introductionmentioning
confidence: 99%
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“…In this nonlinear problem, when we apply the NM to solve f (x) = u(1, x) − 1 = 0, we encounter a difficulty to calculate f ′ (x). Recently, Liu et al [9] proposed a single-step memory-dependent method to solve f (x) = 0 by a Lie-symmetry formulation of Equation (18). Consider the following one [10]:…”
Section: Introductionmentioning
confidence: 99%