# Penalty method for finding a weak solution

From PDE Evans, 2nd edition: Chapter 9, Problem 3

1. (Penalty method) Let $\epsilon > 0$. Define (for $z \in \mathbb{R}$)$$\beta_\epsilon (z) := \begin{cases}0 & \text{if }z \ge 0 \\ \frac z{\epsilon} & \text{if }z \le 0, \end{cases}$$ and suppose $u_\epsilon \subset H_0^1(U)$ is the weak solution of $$\begin{cases}-\Delta u_\epsilon + \beta_\epsilon (u_\epsilon)=f & \text{in }U \\ \qquad \qquad \quad \, \, \, u_\epsilon = 0 & \text{on }\partial U, \end{cases} \tag{*}$$ where $f \in L^2(U)$. Prove that as $\epsilon \to 0$, $u^\epsilon \rightharpoonup u$ weakly in $H_0^1(U)$, $u$ being the unique nonnegative solution of the variational inequality $$\int_U Du \cdot D(w-u) \, dx \ge \int_U f(w-u) \, dx$$ for all $w \in H_0^1(U)$ with $w \ge 0$ a.e.

$\quad \, \, \, \,$Approximating the variational inequality by $(*)$ is the penalty method.

From the PDE, I moved $\beta_\epsilon(u_\epsilon)$ over to the other side to obtain $$\begin{cases}-\Delta u_\epsilon = \hat{f}(u_\epsilon) & \text{in }U \\ \qquad u=0 & \text{on }\partial U, \end{cases} \tag{**}$$ where $\hat{f}(u_\epsilon) := f-\beta_\epsilon(u_\epsilon)$. Then, since $u$ is the weak solution to $(*)$, and so to $(**)$, we can express $B[u,v]=(f,v)$ as $$\int_U Du_\epsilon \cdot Dv \, dx = \int_U \hat{f}(u_\epsilon) v \, dx.$$ To show that $u^\epsilon \rightharpoonup u$ weakly in $H_0^1(U)$, I must show that $$\langle u^*, u_\epsilon \rangle \to \langle u^*, u \rangle$$ for each $u^* \in H_0^{1*}(U)=H_0^1(U)$ (since Hilbert spaces are self-dual).

Am I on the right track with this?