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Volume 16, Issue 1

# Corrigendum to: The Dirichlet problem with mean curvature operator in Minkowski space

Cristian Bereanu
/ Petru Jebelean
/ Jean Mawhin
• Corresponding author
• Institut de Recherche en Mathématique et Physique, Université Catholique de Louvain, 1348 Louvain-la-Neuve, Belgium
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Published Online: 2016-01-23 | DOI: https://doi.org/10.1515/ans-2015-5030

We essentially use Lemma 2.1 which is proved in [1]. Accordingly, the set Ω must be “a bounded domain in ${ℝ}^{N}$ ($N\ge 2$) with boundary $\partial \mathrm{\Omega }$ of class ${C}^{2}$”, instead of “an open bounded set in ${ℝ}^{N}$ with boundary $\partial \mathrm{\Omega }$ of class ${C}^{2}$”. This change does not affect the validity of the results stated in the paper. But, it affects the proof of Lemma 3.1. In this view, the only necessary modification is the following.

#### Proof of Lemma 3.1.

It suffices to show that $u\ge 0$ in Ω. From (2.2), (3.4) and the integration by parts formula it follows

$-{\int }_{\mathrm{\Omega }}\frac{\nabla u\cdot \nabla v}{\sqrt{1-{|\nabla u|}^{2}}}={\int }_{\mathrm{\Omega }}\mu \left(x\right){|u|}^{q-1}uv-\lambda {\int }_{\mathrm{\Omega }}\overline{g}\left(x,u\right)v,$1

for all $v\in {W}^{1,\mathrm{\infty }}\left(\mathrm{\Omega }\right)$ with ${v|}_{\partial \mathrm{\Omega }}=0$. We denote

$\mathcal{𝒪}:=\left\{x\in \mathrm{\Omega }:u\left(x\right)<0\right\},{u}^{-}:=\mathrm{min}\left\{0,u\right\},{\mathcal{𝒪}}^{\prime }:=\left\{x\in \mathcal{𝒪}:|\nabla {u}^{-}\left(x\right)|>0\right\}.$

From [2, Theorem A.1] we have ${u}^{-}\in {W}^{1,\mathrm{\infty }}\left(\mathrm{\Omega }\right)$ and $\nabla {u}^{-}=\nabla u$ in $\mathcal{𝒪}$, $\nabla {u}^{-}={0}_{{ℝ}^{N}}$ in $\mathrm{\Omega }\setminus \mathcal{𝒪}$. So, taking $v={u}^{-}$ in (1) and using hypothesis (H), it follows

$-{\int }_{\mathcal{𝒪}}\frac{{|\nabla {u}^{-}|}^{2}}{\sqrt{1-{|\nabla {u}^{-}|}^{2}}}={\int }_{\mathcal{𝒪}}\mu \left(x\right){|{u}^{-}|}^{q+1}\ge 0.$2

If $meas{\mathcal{𝒪}}^{\prime }>0$, then from (2) we get the contradiction

$0>-{\int }_{{\mathcal{𝒪}}^{\prime }}\frac{{|\nabla {u}^{-}|}^{2}}{\sqrt{1-{|\nabla {u}^{-}|}^{2}}}\ge 0.$

Consequently, $meas{\mathcal{𝒪}}^{\prime }=0$ and, as $\nabla {u}^{-}={0}_{{ℝ}^{N}}$ in $\mathrm{\Omega }\setminus \mathcal{𝒪}$, we get that $\nabla {u}^{-}={0}_{{ℝ}^{N}}$ a.e. on Ω. As Ω is a domain, we infer that ${u}^{-}=\text{const.}$, hence ${u}^{-}\equiv 0$ in Ω. This means $u\ge 0$ in Ω. ∎

## References

• [1]

Corsato C., Obersnel F., Omari P. and Rivetti S., Positive solutions of the Dirichlet problem for the prescribed mean curvature equation in Minkowski space, J. Math. Anal. Appl. 405 (2013), 227–239.  Google Scholar

• [2]

Kinderlehrer D. and Stampacchia G., An Introduction to Variational Inequalities and Their Applications, Academic Press, New York, 1980.  Google Scholar

Published Online: 2016-01-23

Published in Print: 2016-02-01

Citation Information: Advanced Nonlinear Studies, Volume 16, Issue 1, Pages 173–174, ISSN (Online) 2169-0375, ISSN (Print) 1536-1365,

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