2019
DOI: 10.1007/s40316-019-00120-7
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Courant-sharp Robin eigenvalues for the square: the case with small Robin parameter

Abstract: We consider the cases where there is equality in Courant's nodal domain theorem for the Laplacian with a Robin boundary condition on the square. We treated the cases where the Robin parameter h > 0 is large, small in [5], [6] respectively. In this paper we investigate the case where h < 0 .MSC classification (2010): 35P99, 58J50, 58J37.We treat the particular example where Ω is the square S = (− π 2 , π 2 ) 2 ⊂ R 2 . Our key question is whether it is possible to determine the Courant-sharp eigenvalues of the R… Show more

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Cited by 5 publications
(11 citation statements)
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“…We note that if (i+j) is odd for any pair (i, j) such that λ n,h (S) = π −2 (α 2 i +α 2 j ), then we get by (2.10), u(−x, −y) = −u(x, y) and as a consequence u has an even number of nodal domains. As we shall see later, other symmetries related to the finite group generated by the identity and the symmetries (x, y) → (−x, y) and (x, y) → (x, −y) can be considered (see [20]).…”
Section: Symmetry Of Robin Eigenfunctions In 2dmentioning
confidence: 99%
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“…We note that if (i+j) is odd for any pair (i, j) such that λ n,h (S) = π −2 (α 2 i +α 2 j ), then we get by (2.10), u(−x, −y) = −u(x, y) and as a consequence u has an even number of nodal domains. As we shall see later, other symmetries related to the finite group generated by the identity and the symmetries (x, y) → (−x, y) and (x, y) → (x, −y) can be considered (see [20]).…”
Section: Symmetry Of Robin Eigenfunctions In 2dmentioning
confidence: 99%
“…In a second paper, [20], we hope to look at the situation where the Robin parameter h tends to 0 and to discuss the following conjecture. Conjecture 1.3.…”
mentioning
confidence: 99%
“…In previous work [6,7], we considered the case where h > 0. In [6], we proved the following theorem which asserts that there are finitely many Courant-sharp Robin eigenvalues when h > 0.…”
mentioning
confidence: 99%
“…For the case where h < 0, numerically we have that h * 9 ≈ −1.6293. In the case where |h| is sufficiently small, the arguments of [7] for h > 0 also apply for h < 0. In addition, the labelling of the Robin eigenvalues λ k,h (S) is the same as that for the Neumann eigenvalues λ k,0 (S) (see Remark 4.3).…”
mentioning
confidence: 99%
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