2020
DOI: 10.1002/mma.7047
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Estimation of the smallest eigenvalue of an nth‐order linear boundary value problem

Abstract: This paper describes a recursive procedure to estimate the smallest eigenvalue of an nth‐order boundary value problem under a wide set of boundary conditions. The procedure yields lower and upper bounds for that eigenvalue as well as an estimation of the associated eigenfunction, both of which are shown to converge to their exact values as the recursion index grows. A simpler version of the procedure is also displayed for the self‐adjoint case.

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Cited by 4 publications
(7 citation statements)
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“…From here and [25, Lemma 1.16], one gets (33). Likewise, following the same reasoning as in [32,Theorems 6 and 7], one can get to (35) and, noting that for some j 0 > 0, M j p u ∈ int(K p ) for all j ≥ j 0 , also to (34).…”
Section: The Operator M P and Its Properties In The Conesmentioning
confidence: 56%
See 2 more Smart Citations
“…From here and [25, Lemma 1.16], one gets (33). Likewise, following the same reasoning as in [32,Theorems 6 and 7], one can get to (35) and, noting that for some j 0 > 0, M j p u ∈ int(K p ) for all j ≥ j 0 , also to (34).…”
Section: The Operator M P and Its Properties In The Conesmentioning
confidence: 56%
“…. From the sign-regularity of G(x, t), (7) and (32), one has that the signs of ∆ p (M p u) and ∆ p (u) coincide in Ω * . This completes the proof.…”
Section: The Operator M P and Its Properties In The Conesmentioning
confidence: 99%
See 1 more Smart Citation
“…If E j [a, b] > 1 for some j > K(α , β ), from Lemma 1 and ( 7) one would get z n-1 [a, b] > 1, and by Rolle's theorem there would be a change of signs of y (n) in (a, b), contradicting ( 28)-( 29). Therefore, E j [a, b] = 1 for all j > K(α , β ), and from (7), it follows z j [a, b] = Z j {α , β }j + 1. This completes the proof.…”
Section: Theorem 3 Let Us Assume That Formentioning
confidence: 99%
“…Therefore, the knowledge of the sign of G(x, t) and its derivatives can provide information about the sign of the solution y(x) and these same derivatives, at least when f does not change sign on (a, b). Likewise, there is a large amount of literature ( [4][5][6][7][8], and [9]) on the use of the sign of G(x, t) to define cones that, by means of Krein and Rutman's [10] works, allow finding information about the eigenvalues and eigenfunctions of the general problem Ly = λ μ l=0 c l (x)y (l) (x), x ∈ (a, b), y (α i ) (a) = 0, α i ∈ α; y (β i ) (b) = 0, β i ∈ β; (4) with μ ≤ n -1, c l (x) ∈ C(J) for 0 ≤ l ≤ μ, and even calculate them.…”
Section: Introductionmentioning
confidence: 99%