Max/Min Principle

Mar 2009
378
68
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 \(\displaystyle \max_{\Omega} u = \alpha.\)