0
$\begingroup$

When I read PDE books as a novice, $\phi \in C^\infty_0(\Omega)$ often appears.

Accidentally, given that condition I saw that function values in $\partial\Omega$ are zero.

For example, a note from my class is as follow.

My question is,

(1) Is it true that given compact support condition, the function values on boundary are zero?

(2) It is very hard for me to understand compact support. Could you explain that concept in terms of the following proof or in general?

Let $u\in L^1_{loc}(\Omega)$. u is ``weakly'' harmonic if $$\int_\Omega u\Delta \phi \, dx=0, \quad\forall \phi \in > C^\infty_0(\Omega).$$

$u\in C^2(\Omega)$ a weakly harmonic $\implies$ u is harmonic.

(proof) $\nabla \cdot (u\nabla\phi) = \nabla u \cdot \nabla \phi + u \Delta \phi$ $$U = \int_\Omega u\Delta \phi \,dx = \int_\Omega \nabla \cdot (u\nabla \phi )dx - \int_\Omega \nabla u\cdot \nabla \phi\,dx $$ By divergence theorem, $$ = \int_{\partial \Omega} \nu \cdot (u\nabla\phi)ds - \int_{\partial\Omega}\nu\cdot(\phi\nabla u)ds + \int_\Omega \phi \Delta u dx $$

$$ \therefore \int_\Omega \phi \Delta u dx=0, \quad\forall \phi \in > C^\infty_0(\Omega)\\ \text{note: I think that the fact in question is used.}$$

Therefore, $\Delta u=0$ on $\Omega$.

$\because$ if $\exists x_0 \in \Omega \mid \Delta u(x_0) \neq 0$, we may assume $\Delta u(x_0) > 0$. By continuity of $\Delta u$, $$\exists B_\delta(x_0)\subset\Omega \mid \Delta u(x)>0 \quad\forall x\in B_\delta(x_0)$$ We choose $\phi\in C^\infty_0$ so that $$\phi(x) = \begin{cases}1 & |x-x_0| < \frac \delta 2 \\ 0 & otherwise\end{cases}$$ Then $$\underbrace{0=\int_\Omega\phi\Delta u dx}_{assumption} = \int_{B_\delta (x_0)} \phi\Delta u dx \geq \int_{B_{\delta/2}(x_0)}\phi\Delta u\,dx$$ $$=\int_{B_{\delta/2}(x_0)}\Delta u\,dx > 0$$ contradiction.

$\endgroup$
  • $\begingroup$ Yes, the values are zero on the boundary (this follows from continuity!). $\endgroup$ – user98602 Mar 24 '14 at 16:53
  • $\begingroup$ If $u$ has compact support then this means that by definition the set where $u\neq 0$ is a compact subset $K$ of $\Omega$ (which may depend on $u$). Since for such a set the distance of $K$ to $\partial \Omega$ is positive, $u$ has to be $=0$ even in a whole boundary of $\partial \Omega$ $\endgroup$ – Thomas Mar 24 '14 at 16:56
  • $\begingroup$ I would add that in the general framework, $\Omega$ is open. $K\subset \Omega$ compact implies that the function is 0 not only on $\partial \Omega$ but on one neighborhood of it: in particular, the derivatives are also $=0$. $\endgroup$ – mookid Mar 24 '14 at 17:01

Your Answer

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

Browse other questions tagged or ask your own question.