In the proof of interior regularity of elliptic equation, it uses the difference quotient: $D^h_k u := \frac{u(x+he_k)-u(x)}{h}$, $e_k$ is the coordinate vector in the $k$ direction, $k=1,\ldots, n$. But for bounded domain $\Omega$, $D^h_k$ is not defined for $x$ sufficiently close to $\partial \Omega$, because $x+he_k$ will be outside of $\Omega$. And then it constructs $v=-D^{-h}_k(\eta^2 D^h_ku) \in H^1_0(\Omega)$ for a cut-off fucntion $\eta$. But $v$ is not defined on whole $\Omega$. Could anyone help me about this? Thanks!
Asked
Active
Viewed 202 times
1
-
If you're proving interior regularity, why worry about the boundary? x is bounded away from the boundary. For small enough h, $x+he_k$ is inside $\Omega$. – mathematician Dec 19 '13 at 10:01
1 Answers
0
The presence of cut-off function $\eta$ saves the day. It should be chosen so that $\eta=0$ when the distance to the boundary is less than $2h$. Although $D_k^h$ is only defined on the subset $\Omega_h=\{x: \operatorname{dist}(x,\partial \Omega)>h\}$, the product $\eta^2 D_k^h$ can be extended by $0$ to the whole domain $\Omega$; or even to all $\mathbb R^n$. Simply define it to be $=0$ where it's not defined yet. This extension is continuous because $\eta^2 D_k^h$ is zero near the boundary of $\Omega_h$.
Post No Bulls
- 9,056