Solution of PDE in the sense of multiplication by $\eta \in W^{1,2}_0$ and a uniform boundedness observation In this book (on page 20), it is asserted that $u_{\epsilon} \in W^{1,2}$ will solve $\Delta u_{\epsilon} = f \chi_{\epsilon}(u_{\epsilon})$ in a bounded domain $D \subset \mathbb{R}^n$; $u_{\epsilon} = g$ on $\partial D$ in the sense that it will minimize the following functional for all $\eta \in W^{1,2}_0$:
$$\int_D (\nabla u_{\epsilon}\nabla\eta + f \chi_{\epsilon}(u_{\epsilon})\eta) \, \mathrm dx = 0$$
Loosely speaking, $\chi_{\epsilon}$ is a characteristic function on some function, $u_{\epsilon}$ -- but the sharp corner is "smoothed out" (see book for precise detail, picture below).
My Questions:

*

*I'm familiar with solving PDEs in the sense of $\eta \in C^\infty_0$ -- but this seems different. Can someone comment why using $\eta \in W^{1,2}_0$ still makes sense as a solution? And why not using compact smooth functions?


*They go on to say that they take $\eta = u_{\epsilon} - g$ and Poincare inequality to get: $$\int_D|\nabla (u_{\epsilon}-g)|^2 \leq C(f,g) \,\, \forall \epsilon \in (0,1)$$
Can someone provide more detail on how this inequality was obtained?

 A: (1) It does not matter: Since $C^\infty_0 (\Omega)$ is a dense subset of $W^{1, 2}_0 (\Omega)$ (with the $W^{1, 2}$ norm), for all $\eta \in W_0^{1, 2}$, let $\eta_n \in C^\infty_0 (\Omega)$ be a sequence which converges to $\eta$. Then
$$ \nabla u_{\epsilon}\nabla\eta_n + f \chi_{\epsilon}(u_{\epsilon})\eta_n \to \nabla u_{\epsilon}\nabla\eta + f \chi_{\epsilon}(u_{\epsilon})\eta  $$
in $L^2$. Thus the fact
$$\int_D (\nabla u_{\epsilon}\nabla\eta + f \chi_{\epsilon}(u_{\epsilon})\eta) \, \mathrm dx = 0,  \ \ \ \forall \eta \in C^\infty_0 (\Omega)$$
is sufficient to conclude
$$\int_D (\nabla u_{\epsilon}\nabla\eta + f \chi_{\epsilon}(u_{\epsilon})\eta) \, \mathrm dx = 0,  \ \ \ \forall \eta \in W^{1,2}_0 (\Omega)$$
(2) Setting $\eta = u_\epsilon - g \in W^{1,2}_0(\Omega)$, we obtain
$$\int_D \big(\nabla u_{\epsilon}\cdot\nabla (u_\epsilon - g)+ f \chi_{\epsilon}(u_{\epsilon})(u_\epsilon -g)\big) \, \mathrm dx = 0$$
which implies
\begin{align*}
\int_D |\nabla (u_\epsilon -g)|^2 &= - \int _D f \chi_{\epsilon}(u_{\epsilon})(u_\epsilon -g) - \int_D \nabla g \cdot \nabla (u_\epsilon -g)\\
& \le \int_D |f| |u_\epsilon -g| + \int_D |\nabla g| | \nabla (u_\epsilon - g)| \\
& \le \| f\|_{L^2} \| u_\epsilon-g\|_{L^2} + \int _D \left(\frac 14| \nabla (u_\epsilon - g)|^2 + |\nabla g|^2  \right)
\end{align*}
in the last line we used the Holder's inequality and
$$\tag{1} ab \le \frac{1}{2\delta} a^2+ \frac{\delta}{2} b^2$$
with $\delta = 2$. Using Poincare inequality and (1) again with suitable $\delta$,
$$\| f\|_{L^2} \| u_\epsilon-g\|_{L^2} \le C\|f\|_{L^2} \| \nabla (u_\epsilon -g)\|_{L^2} \le \frac 14 \int_D |\nabla (u_\epsilon -g)|^2 + C' \int_D |f|^2.$$
Finally,
\begin{align*}\int_D |\nabla (u_\epsilon -g)|^2 &\le \frac 12\int_D |\nabla (u_\epsilon -g)|^2 + \int_D |\nabla g|^2 + C' |f|^2 \\
\Rightarrow \int_D |\nabla (u_\epsilon -g)|^2 &\le 2 \left(\int_D |\nabla g|^2 + C' |f|^2\right).
\end{align*}
