# Max/Min Principle

• May 15th 2010, 12:52 PM
lvleph
Max/Min Principle
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
$
\Delta u = -f(u) \text{ in } D \text{ and } u = 0 \text{ on } \partial D,
$

where $f(y)$ is smooth with $f(0) = f(\alpha) = 0, f(y) > 0$ for $0 < y <
\alpha$
and $f(y) < 0$ for $y < 0$ or $y > \alpha.$ Determine which one of the
following cases holds: $\max_{x\in D} u = \alpha; \max_{x\in D} u < \alpha;
\max_{x\in D} u > \alpha.$

Solution
Let's consider two cases: Case I $0 < y < \alpha$ and Case II $y < 0 \cup y
> \alpha.$

Case I: Assume $u(x_0) < \alpha$ and let $\Omega_{\alpha} = \left\{x \in
\Omega | u(x) < \alpha\right\} \ne \emptyset$
so $\Omega_{\alpha} \subset
\Omega.$
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
$
\min_{\bar{\Omega_{\alpha}}} u = \min_{\partial\Omega_{\alpha}} u,
$

since $u = \alpha$ on $\partial \Omega_{\alpha}.$ A condtradiction, since we
assumed $u < \alpha$ and therefore $\min{\partial \Omega_{\alpha}} u <
\alpha.$
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
$\Omega_{\alpha} = \left\{x \in \Omega | u(x) > \alpha\right\} \ne
\emptyset$
so $\Omega_{\alpha} \subset \Omega.$ Since $u$ is continuous,
$u=\alpha$ on $\partial \Omega_{\alpha}.$ Additionally, we see that $f(u)
\le 0$
and therefore $Lu =\Delta u \ge 0$ on $\Omega_{\alpha}.$ By the
Weak Maximum Principle
$
\max_{\bar{\Omega_{\alpha}}} u = \max_{\partial\Omega_{\alpha}} u,
$

since $u = \alpha$ on $\partial \Omega_{\alpha}.$ A condtradiction, since we
assumed $u > \alpha$ and therefore $\max{\partial \Omega_{\alpha}} u >
\alpha.$
Thus, $u(x) \le \alpha\; \forall x \in \Omega.$
Combining the results for Case I and Case II we see that $u(x) = \alpha\;
\forall x \in \Omega$
and therefore $\max_{\Omega} u = \alpha.$