I am studying for qualifying exams. The following problem was on my final. I am unsure if I got it right or not.

Let D be the unit ball in \mathbb{R}^n and u \not \equiv 0 satisfy
<br />
		\Delta u = -f(u) \text{ in } D \text{ and } u = 0 \text{ on } \partial D,<br />
where f(y) is smooth with f(0) = f(\alpha) = 0, f(y) > 0 for 0 < y <<br />
	\alpha and f(y) < 0 for y < 0 or y > \alpha. Determine which one of the
following cases holds: \max_{x\in D} u = \alpha; \max_{x\in D} u < \alpha;<br />
	\max_{x\in D} u > \alpha.

Solution
Let's consider two cases: Case I 0 < y < \alpha and Case II y < 0 \cup y<br />
	> \alpha.
Case I: Assume u(x_0) < \alpha and let \Omega_{\alpha} = \left\{x \in<br />
		\Omega | u(x) < \alpha\right\} \ne \emptyset so \Omega_{\alpha} \subset<br />
		\Omega. Since u is continuous, u=\alpha on \partial \Omega_{\alpha}.
Additionally, we see that f(u) \ge 0 and therefore Lu =\Delta u \le 0 on
\Omega_{\alpha}. By the Weak Minimum Principle
<br />
			\min_{\bar{\Omega_{\alpha}}} u = \min_{\partial\Omega_{\alpha}} u,<br />
since u = \alpha on \partial \Omega_{\alpha}. A condtradiction, since we
assumed u < \alpha and therefore \min{\partial \Omega_{\alpha}} u <<br />
		\alpha. Thus, u(x) \ge \alpha\; \forall x \in \Omega.

Case II: On the other hand, let's now assume u(x_0) > \alpha and let
\Omega_{\alpha} = \left\{x \in \Omega | u(x) > \alpha\right\} \ne<br />
		\emptyset so \Omega_{\alpha} \subset \Omega. Since u is continuous,
u=\alpha on \partial \Omega_{\alpha}. Additionally, we see that f(u)<br />
		\le 0 and therefore Lu =\Delta u \ge 0 on \Omega_{\alpha}. By the
Weak Maximum Principle
<br />
			\max_{\bar{\Omega_{\alpha}}} u = \max_{\partial\Omega_{\alpha}} u,<br />
since u = \alpha on \partial \Omega_{\alpha}. A condtradiction, since we
assumed u > \alpha and therefore \max{\partial \Omega_{\alpha}} u ><br />
		\alpha. Thus, u(x) \le \alpha\; \forall x \in \Omega.
Combining the results for Case I and Case II we see that u(x) = \alpha\;<br />
		\forall x \in \Omega and therefore \max_{\Omega} u = \alpha.