The following is Problem 3 from Chapter 9 of Evans's book on PDE, 2nd edition.
(Penalty method) Let $\varepsilon>0$ and define $$\beta_\varepsilon(z)=\begin{cases} 0 & z \ge 0 \\ \frac{z}{\varepsilon} & z<0 \end{cases}$$ and suppose $u_\varepsilon\in H^1_0(U)$ is the weak solution of \begin{equation} \tag{*} \begin{cases} -\Delta u_\varepsilon +\beta_\varepsilon(u_\varepsilon)=f & U \\ u_\varepsilon=0 & \partial U\end{cases} \end{equation} where $f\in L^2(U)$. Prove that as $\varepsilon \to 0$, $u_\varepsilon \rightharpoonup u$ weakly in $H^1_0(U)$, $u$ being the unique solution of the variational inequality $$\int_U \nabla u \cdot\nabla(w-u)\, dx\ge \int_U f\cdot(w-u)\, dx,\qquad w\in H^1_0(U),\ w\ge 0\ \text{a.e.}$$
My strategy to solve this is the following. From (*) we see that the family $u_\varepsilon$ is $H^1_0$-bounded, so it has weak limit points. Let $u$ be one of this limit points: our claim is that $u$ solves the variational inequality. We already know from the variational theory that the solution of the variational inequality is unique, so by a standard argument we will be able to conclude that $u_\varepsilon$ converges to $u$.
This said, we analyze $u$. We know that, up to a subsequence, $u_\varepsilon\rightharpoonup u $ in $H^1_0(U)$ and that $u_\varepsilon \to u$ strongly in $L^2(U)$. So in particular we have for the negative part $u_\varepsilon^-\to u^-$ in $L^2(U)$. From this we can infer that $u\ge 0$ because, if $v\in H^1_0$ is a test function, then \begin{equation}\tag{1} \frac{1}{\varepsilon}\int_U u_\varepsilon^- v = \int\nabla u_\varepsilon\cdot\nabla v-\int f\cdot v, \end{equation} and the right hand side converges, so $\int u_\varepsilon^-v=O(\varepsilon)$, from which we deduce that $\int u^- v=0$.
This looks like a good start, since we know from the variational theory that the unique solution of the variational inequality solves the following problem (cfr. Evans 2nd ed. pag. 494): \begin{equation}\tag{2}\begin{cases} -\Delta u = f & U\cap\{ u > 0\} \\ -\Delta u-f\ge 0\ \text{and}\ u\ge 0 & \text{a.e. on } U\end{cases}\end{equation}
Unfortunately this is where I get confused. It is natural (I think) to go back to (1) with $v=w-u$, where $w\ge 0$, obtaining $$\frac{1}{\varepsilon}\int_U u_\varepsilon^- (w-u) = \int\nabla u_\varepsilon\cdot\nabla (w-u)-\int f\cdot (w-u).$$ The problem is that the left hand side has not a sign. I would have expected it to be non-negative, but since $u\ge 0$, that is not the case. How to solve this? Perhaps we could try to show directly that $u$ solves (2)?
Thank you for reading.
