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

Assume that $\underline{u},\bar{u}$ are smooth, sub- and supersolutions of the boundary-value problem $(1)$ in §9.3. Use the maximum principle to verify directly $$\underline{u}=u_0\le u_1\le \cdots \le u_k \le \cdots \le \bar{u},$$ where the $\{u_k\}_{k=0}^\infty$ are defined as in §9.3.

For context, here is the essential part of §9.3 (taken from pages 543-544) of the textbook:

$\quad$We will investigate this boundary-value problem for the nonlinear Poisson equation: $$\begin{cases}-\Delta u = f(u) & \text{in }U \\ \quad \, \, \, u=0 & \text{on } \partial U,\end{cases} \tag{1}$$ where $f : \mathbb{R} \to \mathbb{R}$ is smooth, with $$|f'| \le C \quad (z \in \mathbb{R}) \tag{2}$$ for some constant $C$.

DEFINITIONS. (i) We say that $\bar{u} \in H^1(U)$ is a weak supersolution of problem $(1)$ if $$\int_U D\bar{u} \cdot Dv \, dx \ge \int_U f(\bar{u})v \, dx \tag{3}$$ for each $v \in H_0^1(U)$, $v \ge 0$ a.e.

$\quad$(ii) Similarly, $\underline{u} \in H^1(U)$ is a weak solution provided $$\int_U D\underline{u} \cdot Dv \, dx \le \int_U f(\underline{u})v \, dx \tag{4}$$ for each $v \in H_0^1(U)$, $v \ge 0$ a.e.

$\quad$(ii) We say $u \in H_0^1(U)$ is a weak supersolution of $(1)$ if $$\int_U D\underline{u} \cdot Dv \, dx \le \int_U f(\underline{u})v \, dx$$ for each $v \in H_0^1(U)$, $v \ge 0$ a.e.

Remark. If $\bar{u},\underline{u} \in C^2(U)$, then from $(3)$ and $(4)$ it follows that $$-\Delta\bar{u} \ge f(\bar{u}), -\Delta\underline{u} \le f(\underline{u}) \quad \text{in }U.$$

THEOREM 1 (Existence of a solution between sub- and supersolutions). Assume there exist a weak supersolution $\bar{u}$ and a weak subsolution $\underline{u}$ of $(1)$, satisfying $$\underline{u} \le 0, \quad \bar{u} \ge 0 \text{ on $\partial U$ in the trace sense}, \quad \underline{u} \le \bar{u} \text{ a.e. in }U. \tag{5}$$ Then there exists a weak solution $u$ of $(1)$, such that $$\underline{u} \ge u \le \bar{u} \qquad \text{a.e.} \quad \text{in }U.$$

Attempted proof (please keep in mind of my many erroneous statements to follow):

Suppose $\underline{u},\bar{u}$ are smooth, sub- and supersolutions of $(1)$ in §9.3. Then by the remark, $-\Delta\bar{u} \ge f(\bar{u}), -\Delta\underline{u} \le f(\underline{u})$ in $U$.

If $f(\bar{u}) \ge 0$, then $-\Delta u \ge 0$ in $U$. By the maximum principle, $u_k \ge \underline{u}$.

If $f(\underline{u}) \le 0$, then $-\Delta u \le 0$ in $U$. By the maximum principle, $u_k \le \bar{u}$.

And $u_k \le u_{k+1}$, or otherwise there will be a contradiction with the maximum principle.

Okay, that was not a very good proof and there's a lot for me to improve here. But FWIW at least I have some idea. How should I apply correctly the maximum principle for this?


Without loss of generality, consider $k=0$. In the Evan's proof, we have that $$ -\Delta u_{1} - \lambda_{1}u_{1} = f(u_{0}) + \lambda u_{0}$$

Since $u_{0}$ is a subsolution, then $ -\Delta u_{0} \leq f(u_{0})$.

So, $ -\Delta u_{1} - \lambda_{1}u_{1} \geq \Delta (-u_{0}) +\lambda u_{0}$.

Wich implies, $$-\Delta ( u_{1} - u_{0}) -\lambda (u_{1} - u_{0}) \geq 0$$.

By assumption, $u_{0}$ is smooth, so applying the maximun principle together with boundary conditions, we have

$$u_{1} -u_{0}\geq 0$$.

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