# Max/Min Principle

#### lvleph

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

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

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

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