2021
DOI: 10.3934/dcdss.2021111
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Anisotropic singular double phase Dirichlet problems

Abstract: <p style='text-indent:20px;'>We consider an anisotropic double phase problem with a reaction in which we have the competing effects of a parametric singular term and a superlinear perturbation. We prove a bifurcation-type result describing the changes in the set of positive solutions as the parameter varies on <inline-formula><tex-math id="M1">\begin{document}$ \mathring{\mathbb{R}}_+ = (0, +\infty) $\end{document}</tex-math></inline-formula>. Our approach uses variational tools t… Show more

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Cited by 40 publications
(18 citation statements)
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“…Therefore by Proposition A4 of Papageorgiou-Rȃdulescu-Zhang [19], we conclude that for δ ∈ (0, 1) small m δ 0 < u(z) ∀z ∈ Ω 0 , a contradiction to the definition of m 0 . Hence λ ∈ L and we conclude that λ * λ 0 < +∞.…”
Section: We Use the Test Functionmentioning
confidence: 76%
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“…Therefore by Proposition A4 of Papageorgiou-Rȃdulescu-Zhang [19], we conclude that for δ ∈ (0, 1) small m δ 0 < u(z) ∀z ∈ Ω 0 , a contradiction to the definition of m 0 . Hence λ ∈ L and we conclude that λ * λ 0 < +∞.…”
Section: We Use the Test Functionmentioning
confidence: 76%
“…(Ω) is a positive solution of (P λ ) (λ ∈ (0, 1) small). Proposition A1 of Papageorgiou-Rȃdulescu-Zhang [19] implies that u λ ∈ L ∞ (Ω) and then the singular anisotropic regularity theory (Saoudi-Ghanmi [22]) and the singular anisotropic maximum principle (see Papageorgiou-Rȃdulescu-Zhang [19, Proposition A3]), imply that u λ ∈ intC + .…”
Section: We Use the Test Functionmentioning
confidence: 99%
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“…(Ω) is bounded. Then the anisotropic regularity theory (see [19], [38]) implies that (32) y n ∈ L ∞ (Ω), y n ∞ C 9 for some C 9 > 0, all n ∈ N.…”
Section: Analysis Of the "Frozen" Problemmentioning
confidence: 99%
“…We set xk = w k n ρ p (Dw k n ) 1/p(z) ∈ W 1,p(z) 0(Ω), k ∈ N 0 . On(34) we act with w k n ∈ W 1,p(z) 0(Ω) and obtainρ p (Dw k n ) + ρ q (Dw k n ) = Ω g vn (z, w k−1 n )w k n dz ⇒ ρ p (Dx k ) − Ĉk + vn (z, w k−1 n ) (w k−1 n ) p(z)−1 xp(z)−1 k−1 xk dz, ⇒ ρ(D xk ) Ω g vn (z, w k−1 n ) (w k−1 n ) p(z)−1 xp(z)−1 k−1 xk dz + Ĉk for all k ∈ N.(37)From the proof of Lemma 4, we know that(38) 0 < C 11 xk C 12 for all k k 0 , some 0 < C 11 C 12 .…”
mentioning
confidence: 98%