0
$\begingroup$

Let $\Omega \in \mathbb R^n$, $n>1$, be a bounded domain with smooth boundary. Let $u \in C^1 (\bar \Omega)$ be harmonic in $\Omega$.

(a) Prove $\max_{\bar \Omega} |\nabla u|^2=\max_{\partial \Omega} |\nabla u|^2$.

(b) Let $u$ satisfies $u \in C^2 (\mathbb R^n)$, $\Delta u =0$ on $\mathbb R^n$. Show that if $u \in L^2 (\mathbb R^n)$, then $u$ is identically equal to zero.

Thoughts: For part (a), I think I need to apply the maximum principle but based on the maximum principle, do I need to show that $\nabla u$ is harmonic and is in $C^2$?

Further thoughts: I can't comment because of low points. For John Ma, we know the definition of $C^1$ function, then we can just required $u$ to satisfy $\Delta u =0$.

Final thoughts: Just checked my previous homework and I think I need to show $|\nabla u |^2$ is subharmonic, then apply the maximum principle to subharmonic functions.

Any hints would be appreciated.


I think I need to show $\mid \nabla u \mid ^2$ is subharmonic, then apply the maximum principle to subharmonic functions.

$\endgroup$
  • $\begingroup$ @JohnMa I don't see how it contradicts to the condition. Like I said (in the edit), $u$ is a $C^1$ function and at the same time $\Delta u=0$. $\endgroup$ – Chris Gartland Oct 23 '15 at 8:37
2
$\begingroup$

$\nabla u$ isn't harmonic in general. $\nabla u$ is a gradient vector, $|\nabla u|^2=\sum_i (\partial_i u)^2$ thus, $$\Delta|\nabla u|^2=\sum_i \Delta (\partial_i u)^2= \sum_i \left(2 (\partial_i u)\Delta (\partial_i u)+|\nabla\partial_i u|^2 \right)= \sum_{i ,j} \partial_i\partial_j u\ne0 $$ The last equality is due to the fact that $u$ is given harmonic and $\Delta (\partial_i u) =\partial_i( \Delta u)$.

So you definitely need another idea.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.