2021
DOI: 10.1051/cocv/2021017
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Regularity results for a class of obstacle problems with p, q−growth conditions

Abstract: In this paper we prove the the local Lipschitz continuity for solutions to a class of obstacle problems of the type[[EQUATION]] Here $\mathcal{K}_{\psi}(\Omega)$ is the set of admissible functions $z \in {u_0 + W^{1,p}(\Omega)}$ {for a given $u_0 \in W^{1,p}(\Omega)$}such that $z \ge \psi$ a.e. in $\Omega$, $\psi$ being the obstacle and $\Omega$ being an open bounded set of $\mathbb{R}^n$, $n \ge 2$.The main novelty here is that we are assuming that the integrand $ F(x, Dz)$ satisfies $(p,q)$-growth conditions… Show more

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Cited by 13 publications
(9 citation statements)
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“…Many recent works deal with regularity properties of solutions to variational problems in which the integrand depends on the x−variable trough a function that is possibly discontinuous, such as in the case of Sobolev-type dependence, under quadratic (see [34]), and super-quadratic growth conditions (see [19,20,21,22,28,33]). This kind of topics has been object of study also in the framework of obstacle problems (see [8,10,11,32]), even in the case of (p, q)-growth condition (such as in [7,14,15]).…”
Section: ˆωmentioning
confidence: 99%
“…Many recent works deal with regularity properties of solutions to variational problems in which the integrand depends on the x−variable trough a function that is possibly discontinuous, such as in the case of Sobolev-type dependence, under quadratic (see [34]), and super-quadratic growth conditions (see [19,20,21,22,28,33]). This kind of topics has been object of study also in the framework of obstacle problems (see [8,10,11,32]), even in the case of (p, q)-growth condition (such as in [7,14,15]).…”
Section: ˆωmentioning
confidence: 99%
“…In this paper we address the analogue issue in the case of constrained minimizers since, for the so called obstacle problem, the situation has not been well established yet. In the recent paper [5] the authors, dealing with the question of Lipschitz continuity for minimizers of the obstacle problem, were forced to deal with the relation between minima and extremals, in the sense of solutions to a corresponding variational inequality. In that specific situation, this problem has been solved thanks to a suitable higher differentiability result and imposing a smallness condition on the gap between the coercivity and the growth exponent of the lagrangian.…”
Section: Introductionmentioning
confidence: 99%
“…Inspired by this result, we assume that the map x −→ A(x, ξ) is weakly differentiable in a suitable sense, see assumption (A4) below, by means of a pointwise characterization of Sobolev spaces due to Hajłasz, see [26]. However our situation turns out to be even more complicated for two reasons: 1) on one hand, differently from [20], we cannot deduce the higher differentiability estimates by simply relying on a class of approximating auxiliary problems and then passing the estimates to the limit, because in the constrained case we are not allowed to use an appropriate choice of the test functions and, along the recent Lipschitz regularity results (see for istance [5]) we would need to use a linearization technique. At this point we prefer to deal with the difference quotient method, so that assumption (A4) turns to be the appropriate assumption (see for instance [2] for the corresponding hypothesis in the case of Hölder continuity results); 2) on the other hand, one of the main tools we need in order to perform our higher differentiability result is a Calderon-Zygmund regularity result, more precisely [10], see also [4,31].…”
Section: Introductionmentioning
confidence: 99%
“…5 and inserting(4.18),(4.19) and(4.26) in the previous estimate, we get|V I| ≤ C|h| 2 BR |Du(x)| log n (e + |Du| 1 ) dx + 1Inserting estimates (4.9), (4.12), (4.13), (4.22), (4.27) and (4.28) in (4.8), we infer the existence of constantsC ε ≡ C ε (ε, ν, L, ℓ, n, γ 1 , γ 2 , R) and C ≡ C(ν, L, ℓ, n, γ 1 , γ 2 , R) such that ν ˆΩ η 2 |τ h Du(x)| 2 (1 + |Du(x + h)| 2 + |Du(x)| 2 |) ˆΩ η 2 |τ h Du(x)| 2 (1 + |Du(x + h)| 2 + |Du(x)| 2 |) (H 2 + H 3 + H 4 + H 5 + H 6 ) Choosing ε = ν 6 we get ν ˆΩ η 2 |τ h Du(x)| 2 (1 + |Du(x + h)| 2 + |Du(x)| 2 |) H 2 + H 3 + H 4 + H 5 + H 6 isa finite quantity. Using Lemma 3.1 in the left hand side of previous estimate and recalling that η ≡ 1 on B R |τ h V γ1 (Du(x))| 2 dx ≤ H|h| 2 Lemma 3.6 implies that V γ1 (Du) ∈ W 1,2 (B R…”
mentioning
confidence: 99%