2017
DOI: 10.1515/fca-2017-0061
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Infinitely many sign-changing solutions for the Brézis-Nirenberg problem involving the fractional Laplacian

Abstract: In this paper, we consider the following Brézis-Nirenberg problem involving the fractional Laplacian operator:where s ∈ (0, 1), Ω is a bounded smooth domain of R N (N > 6s) and 2 * s = 2NN −2s is the critical fractional Sobolev exponent. We show that, for each λ > 0, this problem has infinitely many sign-changing solutions by using a compactness result obtained in [34] and a combination of invariant sets method and Ljusternik-Schnirelman type minimax method.MSC 2010 : Primary 35J60; Secondary 47J30, 58E05

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Cited by 15 publications
(18 citation statements)
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“…Let us compare the "solutions" in the two cases. For the "spectral formulation", the behavior of the solution (11) around the boundary |x| = 1 is quite different than (6) for the "Riesz formulation" (see Section 1.1). Also, we give this simple example to illustrate the following Caffarelli -Stinga result (see [8])…”
Section: Homogeneous Dirichlet Conditionsmentioning
confidence: 96%
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“…Let us compare the "solutions" in the two cases. For the "spectral formulation", the behavior of the solution (11) around the boundary |x| = 1 is quite different than (6) for the "Riesz formulation" (see Section 1.1). Also, we give this simple example to illustrate the following Caffarelli -Stinga result (see [8])…”
Section: Homogeneous Dirichlet Conditionsmentioning
confidence: 96%
“…then, comparing (9) with (10), it follows that (u, e k ) = 0 for k = 2m and (u, e 2m+1 ) = 2 2 π 2s+1 1 (2m + 1) 2s+1 , m = 0, 1, 2... Thus, (11) u s (x) = 2 2…”
Section: Homogeneous Dirichlet Conditionsmentioning
confidence: 97%
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“…One is the spectral fractional operator (cf. [28,58] and the references therein), the other is the integral fractional operator (cf. [17]), which is one of main topics in this paper.…”
Section: 2mentioning
confidence: 99%
“…where α 0 = λ − µ as in (28). Since φ ∈ L p (R N ) ∩ H s (R N ), then for every η > 0 there exists a R 1 = R 1 (η) ≥ 1 such that for all k ≥ R 1 ,…”
Section: 1mentioning
confidence: 99%