Method of sub- and supersolutions

Question

PDE Evans, 2nd edition: Chapter 9, Exercise 6:

Assume $f : \mathbb{R} \to \mathbb{R}$ is Lipschitz continuous, bounded, with $f(0)=0$ and $f'(0)>\lambda_1$, $\lambda_1$ denoting the principal eigenvalue for $-\Delta$ on $H_0^1(U)$. Use the method of sub- and supersolutions to show there exists a weak solution $u$ of $$\begin{cases}-\Delta u = f(u) & \text{in }U \\ \quad \, \, \, \, u=0 & \text{on }\partial U \\ \quad \, \, \, \, u > 0 & \text{in }U.\end{cases}$$

My thoughts:

Given that $f(0)=0$, $f'(0)>\lambda_1>0$, and $f$ is Lipschitz continuous, should I show that $|f'|\le C$ for some constant $C$? Because $f'$ being bounded is apparently a requirement to follow, at least roughly, the proof of Theorem 1 (pages 544-546) in the textbook.

Furthermore, does $f$ being Lipschitz continuous, along with the other given estimates of $f$, imply that $f$ is smooth, or at least in $C^1$?

Answer 1

I am willing to share some idea. Since $f$ is bounded, we can let $M=:\sup |f|$. Define $$ \begin{cases} -\Delta\bar{u}=M \\ u|_{\partial_{U}}=0\\ \end{cases},\begin{cases} -\Delta\underline{u}=-M \\ u|_{\partial_{U}}=0\\ \end{cases} $$ Using maximum principle, we can directly check $\underline{u}\le\bar{u}$. Then we can obtain the existence of solution. However, we also need to prove the positivity. We only need to prove $\lVert u^{-}\rVert_{L^2(U)}=0$. we know that $$ \int_U Du Dvdx=\int_{U} f(u)vdx ,\quad\forall v\in H_0^1(U) $$ If we let $v=u^{-}$, we obtain $$ -\int_U |Du^{-}|^2dx=\int_{U} f(u)u^{-}dx $$ In addition, $$ \int_U |Du^{-}|^2dx\ge \lambda_1 \int_U |u^{-}|^2dx $$ In the end, we arrive at $$ \int_U (f(u)+\lambda_1 u^{-})u^{-}dx\le 0 $$ I don't know how to continue. However, if Lip$(f)<\lambda_1$, we can get the conclusion.