2010
DOI: 10.5186/aasfm.2010.3507
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Compact embeddings for Sobolev spaces of variable exponents and existence of solutions for nonlinear elliptic problems involving the p(x)-Laplacian and its critical exponent

Abstract: Abstract. We give a sufficient condition for the compact embedding from Wwhere Ω is a bounded open set in R N . As an application, we find a nontrivial nonnegative weak solution of the nonlinear elliptic equationWe also consider the existence of a weak solution to the problem above even if the embedding is not compact.

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Cited by 29 publications
(17 citation statements)
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“…(Ω) ֒→ L q(x) (Ω), this result was obtained in [18]. Following the same ideas we can prove a similar result for the trace immersion.…”
Section: Compact Casesupporting
confidence: 73%
See 1 more Smart Citation
“…(Ω) ֒→ L q(x) (Ω), this result was obtained in [18]. Following the same ideas we can prove a similar result for the trace immersion.…”
Section: Compact Casesupporting
confidence: 73%
“…First, employing the same ideas as in [18] we obtain some restricted conditions on the exponents p and r guarantying that the immersion W 1,p(x) (Ω) ֒→ L r(x) (∂Ω) remains compact and so the existence of an extremal for T (p(·), r(·), Ω, Γ) holds true. As in the Sobolev immersion Theorem more general conditions for the existence of extremals are needed and these are the contents of our main results.…”
Section: P(x) γmentioning
confidence: 99%
“…In fact, in [18] the authors find conditions on the exponents p and r such that A T = ∅ but the immersion remains compact. This type of conditions were first discovered in [27] where the embedding W 1,p(x) 0…”
Section: Introductionmentioning
confidence: 90%
“…In [17] the authors prove that if A is small and there exists a control on the rate of how q reaches the critical value p * , then the immersion W 1,p(·) 0 (U ) ֒→ L q(·) (U ) remains compact, and so the usual techniques can be applied. When the immersion fails to be compact they prove that if the subcriticality set U \ A contains a sufficiently large ball, then (1.1) with h = 0 has a nonnegative solution.…”
Section: Introductionmentioning
confidence: 99%