Throughout this problem we assume $f:\mathbb{R}\to \mathbb{R}$ is a smooth function, with $|f'|\leq C$ for some fixed constant $C>0$.
The constant $\lambda >0$ is defined to be a large enough number so that the map $f + \lambda I$ is nondecreasing.
I am proving the following theorem from Evans, PDE, p.544 (Section 9.3).
Theorem. Assume the existence of a weak supersolution $\overline{u}$ and a weak subsolution $\underline{u}$ to the Poisson equation
$$ \begin{cases} – \Delta u = f(u) &\text{in $U$},\\
u = 0 &\text{ on $\partial U$}\end{cases} $$
satisfying
\begin{align*}
&\underline{u}\leq 0, \, \overline{u}\geq 0 \text{ on $\partial U$ in the trace sense},\\
&\text{$\underline{u}\leq \overline{u}$ a.e. in $U$.}
\end{align*}
Then there exists a weak solution $u$ to the above Poisson equation, satisfying
\begin{align*}
\underline{u} \leq u \leq \overline{u} \quad \text{a.e. in $U$.}
\end{align*}
Here, we define a weak supersolution $\overline{u}\in H^1_0(U)$ to the Poisson equation as a function satisfying
$$\int_U \nabla \overline{u}\cdot \nabla \phi \, dx \ge \int_U f(\overline{u}) \phi \, dx$$
for all $\phi\in H^1_0(U)$ that satisfy $\phi\geq 0$ a.e. A weak subsolution $\underline{u}\in H^1_0(U)$ is defined analogously.
The key idea is as follows. We write $u_0=\underline{u}$, then inductively define $u_{k+1}\in H^1_0(U)$ from $u_k$ by setting
$$\begin{cases} -\Delta u_{k+1} + \lambda u_{k+1} = f(u_k) + \lambda u_k &\text{in $U$},\\ u_{k+1}= 0 &\text{on $\partial U$}. \end{cases} $$
By choosing appropriate functions $\phi\in H^1_0(U)$ one can show that
$$\underline{u} = u_0 \le u_1 \le \cdots \le \overline{u}.$$
Question.
Now, the only part I'm stuck on is at P. 546. From the given recursive relation for the $u_k$, and the fact that $\|f(u_k)\|_{L^2(U)} \le C(\|u_k\|_{L^2(U)} + 1)$, the author states that we can deduce
$$\sup_k \|u_k\|_{H^1_0(U)} < \infty.$$
Eh, but I have no idea how. I thought about multiplying both sides of the recursive relation by $u_{k+1}$, then integrating over $U$, which gives me
$$\int_U |\nabla u_{k+1}|^2 + \lambda |u_{k+1}|^2 dx = \int_U f(u_k) u_{k+1} + \lambda u_k u_{k+1} dx$$
so perhaps I should use the Cauchy-Schwarz inequality appropriately on the right-hand-side? I'm guessing I have to use the fact that $u_k \le u_{k+1}$… But I'm not getting anywhere. I would appreciate all help or tips.
Best Answer
Since $u_{0}\leqslant u_{k}\leqslant\bar{u}$ and $u_{0}, \bar{u}\in L^{2}(U)$, which implies $\left\{u_{k}\right\}$ is bounded in $L^{2}(U)$ ( $\lvert u_{k}\lvert\leqslant \lvert u_{0}\lvert+\lvert\bar{u}\lvert$), then$$\Vert f(u_{k})+\lambda u_{k}\Vert_{L^{2}(U)}\leqslant C.$$ And \begin{align*} \int_{U}\lvert\nabla u_{k+1}\lvert^{2}+\lambda u_{k+1}^{2}\mathrm{d}x&=\int_{U}\left(f(u_{k})+u_{k}\right)u_{k+1}\mathrm{d}x\\ &\leqslant\varepsilon\Vert u_{k}\Vert_{L^{2}(U)}^{2}+\frac{1}{4\varepsilon}\Vert f(u_{k})+ u_{k}\Vert_{L^{2}(U)}^{2}\\ &\leqslant \varepsilon\Vert u_{k}\Vert_{L^{2}(U)}^{2}+C. \end{align*} Fix $0<\varepsilon\leqslant\lambda$, then $\Vert u_{k+1}\Vert_{H_{0}^{1}(U)}\leqslant C.$