# Maximum principle for weak sub- and supersolutions

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