Why is $f_\epsilon(u) \in W_0^{1,2}(\Omega)$?

Question

For $\epsilon>0$ let $f_\epsilon(u)=\sqrt{\epsilon^2+u^2}-\epsilon$

One calculates that $\nabla f_\epsilon(u)=\frac{u}{\sqrt{\epsilon^2+u^2}}\nabla u $ , for $\epsilon$ to 0 this term goes to $\nabla |u|$

and one finds$ f_\epsilon(u) \in W_0^{1,2}(\Omega)$.

We need to know more about $\Omega$. Why is $f_{\epsilon}$ compactly supported? — fGDu94, May 06 '19 at 20:52
$f_\epsilon(u)$ should give an approximation of |u| with $u \in W_0^{1,2}(\Omega)$ — AnabolicHorse, May 06 '19 at 20:59
oh ok, so we have some $f_{\epsilon}(u)[x]$, that's to say we still evaluate on our space with variable $x$. — fGDu94, May 06 '19 at 21:01
What is wrong with the answer to this given in your other question here? It would help answers if we knew what it is you dont understand about that — Rhys Steele, May 06 '19 at 21:15
@RhysSteele the answer I gave there seems to be exactly the same (in content) as the accepted answer here, with a similar discussion in the comments... — Calvin Khor, May 08 '19 at 17:48

score 2 · Accepted Answer · answered May 06 '19 at 21:16

2

Ok so it is clear that $f_{\epsilon, u}$ is compactly supported from the fact that $u$ is.

Now we do $\int_{\Omega}|f_{\epsilon, u}|^2(x)dx+\int_{\Omega}|\nabla f_{\epsilon, u}|^2(x)dx$. It is our goal to show these integrals converge.

$\int_{\Omega}|f_{\epsilon, u}|^2(x)dx = \int_{\Omega}u^2 - 2\epsilon \sqrt{\epsilon^2+u^2} \leq \int_{\Omega}u^2 < \infty$ by defn of $u \in W_0^{1,2}$

$\int_{\Omega}|\nabla f_{\epsilon, u}|^2(x)dx = \int_{\Omega}\frac{u^2}{{u^2+\epsilon^2}}||\nabla u||_2^2 dx \leq \int_{\Omega}||\nabla u||_2^2dx <\infty$ again by $u \in W_0^{1,2}$

answered May 06 '19 at 21:16

fGDu94

3,916

The calculation shows that $f\epsilon(u) \in W^{1,2}(\Omega)$ – AnabolicHorse May 06 '19 at 21:32
Sure but $u \in W_0^{1,2}(\Omega)$, and wherever $u = 0$ so is $f_{\epsilon}u$ – fGDu94 May 06 '19 at 21:33
The fact that u is compactly supported plus the calculation show that $f\epsilon(u) \in W_0^{1,2}(\Omega)$? – AnabolicHorse May 06 '19 at 21:34
yep that's right, the above space is just the compactly supported functions in $W^{1,2}(\Omega)$ – fGDu94 May 06 '19 at 21:35
thank you very much for your help . – AnabolicHorse May 06 '19 at 21:36
I just noticed that $|f_{\epsilon,u}|^2(x)=2 \epsilon^2+u^2-2\epsilon \sqrt{\epsilon^2+u^2}$ – AnabolicHorse May 07 '19 at 06:25
it's fine as $2\epsilon^2 - 2\epsilon\sqrt{\epsilon^2+u^2} \leq 0$ – fGDu94 May 07 '19 at 12:31

Why is $f_\epsilon(u) \in W_0^{1,2}(\Omega)$?

1 Answers1