1991
DOI: 10.1016/0022-247x(91)90408-r
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Leighton's bounds for Sturm-Liouville eigenvalues with eigenvalue parameter in the boundary conditions

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Cited by 9 publications
(5 citation statements)
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“…For all stable positions, all the eigenvalues are negative (Figs 10b and 11b); for all unstable positions, the largest eigenvalue is positive, and all others are negative (Figs 10d and 11d). The eigenfunctions corresponding to the largest eigenvalues for both stable and unstable steady-state configurations do not have zeros, consistent with the theorem of Linden (1991).…”
Section: Appendix a Schoof's Boundary Layersupporting
confidence: 77%
See 1 more Smart Citation
“…For all stable positions, all the eigenvalues are negative (Figs 10b and 11b); for all unstable positions, the largest eigenvalue is positive, and all others are negative (Figs 10d and 11d). The eigenfunctions corresponding to the largest eigenvalues for both stable and unstable steady-state configurations do not have zeros, consistent with the theorem of Linden (1991).…”
Section: Appendix a Schoof's Boundary Layersupporting
confidence: 77%
“…Amrein and others, 2005). In particular, one can show using theorem 1 of Linden (1991) that for the Sturm-Lioville problem (B8) with boundary conditions (B6b) and (B6c), the kth eigenfunction has k − 1 zeros, and consequently, the eigenfunction corresponding to the largest eigenvalue does not have zero crossings, provided that in addition to the continuity, positivity and homogeneity conditions of the regular problem, the inequality…”
Section: Appendix a Schoof's Boundary Layermentioning
confidence: 99%
“…The evolution of the grounding line in the corresponding linearized perturbation problem (Supplementary Methods 2 ) is described by , where Λ is an eigenvalue. A stability condition for the case of relies on properties of the eigenfunctions of the perturbation problem 27 and is determined by the sign of the largest eigenvalue—either the fastest growing, in the case of an unstable grounding line position, or the slowest decaying in the case of a stable grounding line position 20 . In the case of no inferences about either the sign or magnitude of eigenvalues can be made; and the perturbation eigenvalue problem has to be solved explicitly to determine the largest eigenvalue.…”
Section: Resultsmentioning
confidence: 99%
“…The sign of the first term in square brackets on the right-hand side is positive when A 2 is positive. This is due to the fact that, the eigenfunction h corresponding to the largest value of Λ has no zeros in accordance with Theorem 1 of Linden (1991), provided A 2 > 0; and consequently the term xc 0 hdx ′ / h is positive. The sign of the denominator ĥx − h cx is known for a given calving rule (17-19) and a particular steady-state configuration (12).…”
Section: Appendix a A Linear Stability Analysismentioning
confidence: 85%