Polyconvex functionals and maximum principle

Abstract

<abstract><p>Let us consider continuous minimizers $ u : \bar \Omega \subset \mathbb{R}^n \to \mathbb{R}^n $ of</p> <p><disp-formula> <label/> <tex-math id="FE1"> \begin{document}$ \mathcal{F}(v) = \int_{\Omega} [|Dv|^p \, + \, |{\rm det}\,Dv|^r] dx, $\end{document} </tex-math></disp-formula></p> <p>with $ p &gt; 1 $ and $ r &gt; 0 $; then it is known that every component $ u^\alpha $ of $ u = (u^1, ..., u^n) $ enjoys maximum principle: the set of interior points $ x $, for which the value $ u^\alpha(x) $ is greater than the supremum on the boundary, has null measure, that is, $ \mathcal{L}^n(\{ x \in \Omega: u^\alpha (x) &gt; \sup_{\partial \Omega} u^\alpha \}) = 0 $. If we change the structure of the functional, it might happen that the maximum principle fails, as in the case</p> <p><disp-formula> <label/> <tex-math id="FE2"> \begin{document}$ \mathcal{F}(v) = \int_{\Omega}[\max\{(|Dv|^p - 1); 0 \} \, + \, |{\rm det}\,Dv|^r] dx, $\end{document} </tex-math></disp-formula></p> <p>with $ p &gt; 1 $ and $ r &gt; 0 $. Indeed, for a suitable boundary value, the set of the interior points $ x $, for which the value $ u^\alpha(x) $ is greater than the supremum on the boundary, has a positive measure, that is $ \mathcal{L}^n(\{ x \in \Omega: u^\alpha (x) &gt; \sup_{\partial \Omega} u^\alpha \}) &gt; 0 $. In this paper we show that the measure of the image of these bad points is zero, that is $ \mathcal{L}^n(u(\{ x \in \Omega: u^\alpha (x) &gt; \sup_{\partial \Omega} u^\alpha \})) = 0 $, provided $ p &gt; n $. This is a particular case of a more general theorem.</p></abstract>