2013
DOI: 10.1051/cocv/2012014
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Convex shape optimization for the least biharmonic Steklov eigenvalue

Abstract: Abstract. The least Steklov eigenvalue d1 for the biharmonic operator in bounded domains gives a bound for the positivity preserving property for the hinged plate problem, appears as a norm of a suitable trace operator, and gives the optimal constant to estimate the L 2 -norm of harmonic functions. These applications suggest to address the problem of minimizing d1 in suitable classes of domains. We survey the existing results and conjectures about this topic; in particular, the existence of a convex domain of … Show more

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Cited by 10 publications
(24 citation statements)
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“…Nevertheless, in [7] the authors can prove that, among all convex domain of fixed measure there exists a minimizer, but nothing is known about the optimal shape, or if the convexity assumption can be relaxed. We also refer to [2,5,6] for other results on problem (5). Another Steklov problem for the biharmonic operator which has appeared very recently in [15] (see also [16]) is the following…”
Section: Introductionmentioning
confidence: 99%
“…Nevertheless, in [7] the authors can prove that, among all convex domain of fixed measure there exists a minimizer, but nothing is known about the optimal shape, or if the convexity assumption can be relaxed. We also refer to [2,5,6] for other results on problem (5). Another Steklov problem for the biharmonic operator which has appeared very recently in [15] (see also [16]) is the following…”
Section: Introductionmentioning
confidence: 99%
“…Now we observe that all the eigenfunctions show symmetry properties, i.e., they are either even or odd in the y variable, and symmetric or skew-symmetric with respect to x, see Theorem 2.1. In particular this implies that |D 2 v| 2 , (∆v) 2 and v 2 are equal for y = ± and symmetric with respect to x = π/2. This proves formula (13) and completes the proof of Theorem 3.3.…”
Section: Proof Of Theorem 33mentioning
confidence: 96%
“…The boundary conditions on the large edges model the fact that the deck is free to move vertically and they may also be written in the equivalent form (cf. [25]) (2) (1 − σ) ∂ 2 u ∂ν 2 + σ∆u = ∂∆u ∂ν + (1 − σ)div ∂Ω (ν · D 2 u) ∂Ω = 0 on (0, π) × {− , }.…”
Section: Introductionmentioning
confidence: 99%
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“…Let us summarize some properties which are well known by now. We refer to [1,3,4,5] and [6] for more details.…”
Section: Introductionmentioning
confidence: 99%