Nonlinear Variational Inequalities with Bilateral Constraints Coinciding on a Set of Positive Measure

Abstract

We consider variational inequalities with invertible operators ${{\mathcal{A}}_{s}}{\text{:}}~\,W_{0}^{{1,p}}\left( {{\Omega }} \right) \to {{W}^{{ - 1,p'}}}\left( {{\Omega }} \right),$ $s \in \mathbb{N},$ in divergence form and with constraint set $V = \{ {v} \in W_{0}^{{1,p}}\left( {{\Omega }} \right){\text{: }}\varphi \leqslant {v} \leqslant \psi ~$ a.e. in ${{\Omega }}\} ,$ where ${{\Omega }}$ is a nonempty bounded open set in ${{\mathbb{R}}^{n}}$ $\left( {n \geqslant 2} \right)$, p > 1, and $\varphi ,\psi {\text{: }\Omega } \to \bar{\mathbb{R}}$ are measurable functions. Under the assumptions that the operators ${{\mathcal{A}}_{s}}$ G-converge to an invertible operator $\mathcal{A}{\text{: }}W_{0}^{{1,p}}\left( {{\Omega }} \right) \to {{W}^{{ - 1,p'}}}\left( {{\Omega }} \right)$, ${\text{int}}\left\{ {\varphi = \psi } \right\} \ne \emptyset ,$ ${\text{meas}}\left( {\partial \left\{ {\varphi = \psi } \right\} \cap {{\Omega }}} \right)$ = 0, and there exist functions $\bar {\varphi },\bar {\psi } \in W_{0}^{{1,p}}\left( {{\Omega }} \right)$ such that $\varphi \leqslant \overline {\varphi ~} \leqslant \bar {\psi } \leqslant \psi$ a.e. in ${{\Omega }}$ and ${\text{meas}}\left( {\left\{ {\varphi \ne \psi } \right\}{{\backslash }}\left\{ {\bar {\varphi } \ne \bar {\psi }} \right\}} \right) = 0,$ we establish that the solutions u_s of the variational inequalities converge weakly in $W_{0}^{{1,p}}\left( {{\Omega }} \right)$ to the solution u of a similar variational inequality with the operator $\mathcal{A}$ and the constraint set V. The fundamental difference of the considered case from the previously studied one in which ${\text{meas}}\left\{ {\varphi = \psi } \right\} = 0$ is that, in general, the functionals ${{\mathcal{A}}_{s}}{{u}_{s}}$ do not converge to $\mathcal{A}u$ even weakly in ${{W}^{{ - 1,p'}}}\left( {{\Omega }} \right)$ and the energy integrals $\langle {{\mathcal{A}}_{s}}{{u}_{s}},{{u}_{s}}\rangle$ do not converge to $\langle \mathcal{A}u,u\rangle$.