Let's follows Evans's PDE book notation. We say $G(x,y)$ is the green function build on $\Omega\subset \mathbb R^N$ where $\Omega$ open bounded nice boundary.

Then if $u\in C^2(\bar\Omega)$, we could represent $u$ by $$ u(x)=\int_\Omega G(x,y)\Delta u(y)\,dy+\int_{\partial\Omega}\partial_\nu G(x,y)u(y)\,d\sigma(y) \tag 1$$

Ok, for $u\in C^2(\bar \Omega)$, proving $(1)$ is an easy task. However, my professor states that this formula is also true if we only have $u\in C^2(\Omega)\cap C^0(\bar\Omega)$ and we could prove this by using approximation...

I tried a while but I still got stuck on this approximation problem.

Honestly, I don't even have a good idea how to approx a function $u\in C^2(\Omega)\cap C^0(\bar\Omega)$ by $(u_n)\subset C^2(\bar\Omega)$.

I think the approximation should hold in $L^\infty$ sense so I was trying to using mollification... But I don't really see how...Any help is really welcome!

Here @Tomas suggest we go approx with respect to domain. Here is what I tried (didn't get too much however...). Define $$\Omega_\epsilon:=\{x\in\Omega,\,\,\text{dist}(x,\partial\Omega)>\epsilon\} $$ Now fix arbitrary $x\in \Omega$ and for $\epsilon>0$ small enough we have $x\in\Omega_\epsilon$. Also we have $u\in C^2(\bar \Omega_\epsilon)$. Apply green identity, we have $$ u(x)=\int_{\Omega_\epsilon}\Delta u(y)G(x,y)dy+\int_{\partial\Omega_\epsilon}\partial_\nu G(x,y)u(y)\,d\sigma - \int_{\partial\Omega_\epsilon} G(x,y)\partial_\nu u(y)\,d\sigma:=I+II+III$$

The Problem is how we take $\epsilon\to 0$. We have no problem with $II$, but $I$ and $III$ will be the problem.... In order to much to limit by using LDCT, we need to know, for $I$ $$ \int_{\Omega}\Delta u(y)G(x,y)dy $$ is well define. And hence we could use LDCT to justify $I$. Of course if $u\in C^2(\bar\Omega)$ we are good to go; if not, we have to consider the behavior of $\Delta u$ around $\partial \Omega$. As $G(x,y)=0$ over $\partial\Omega$ we know that $G(x,y)\to 0$ as $y\to \partial\Omega$. Hence $G(x,y)\to 0$ uniformly. Now the problem is $G(x,y)$ compete with $\Delta u$ around the boundary...but could $\Delta u$ just blow up at $\partial\Omega$ and hence the integration blows up?

The part $III$ actually have the same problem as part $I$...

  • $\begingroup$ Maybe the aprroximation should be done with respect to the domain $\Omega$? $\endgroup$ – Tomás Jan 4 '15 at 0:48
  • $\begingroup$ So you mean take a smaller set, such as $\Omega_\epsilon\subset \Omega$ and try to send $\epsilon\to 0$? $\endgroup$ – spatially Jan 4 '15 at 1:12
  • $\begingroup$ Yes, have you tried it? $\endgroup$ – Tomás Jan 4 '15 at 1:13
  • $\begingroup$ @Tomás I haven't. But there might be a problem is that green function will not be $0$ over $\partial \Omega_\epsilon$ $\endgroup$ – spatially Jan 4 '15 at 1:16
  • $\begingroup$ @Tomás no luck... Please check out my update... $\endgroup$ – spatially Jan 4 '15 at 3:35

I think the best you can do is the following: There exists $p_n$ such that for every $u\in W^{2,p}(\Omega)$ with $p>p_n$ Green's representation formula is valid for $u$. To see this note that for $2p>N$ $u$ is continuous in $\bar{\Omega}$ by Sobolev embedding, so that the LHS and the boundary term make sense. Moreover, since $G(x,\cdot)\in L^q(\Omega)$, uniformly in $x$, for $1<q<N/(N-2)$ we get that for $p>N/2$ the whole formula makes sense. The fact that $C^\infty(\bar{\Omega})$ is dense in $W^{2,p}(\Omega)$ follows since $\Omega$ is smooth.

To see that this is optimal consider $y_0\in \partial \Omega$ and consider $u(x)=|x-y_0|^2\sin(|x-y_0|^{2a})$. For values $a<<-2N$, we get that $u\in C^2(\Omega) \cap C(\bar{\Omega})$, but the first integral term in the representation formula doesn't make sense, since $\Delta u$ blows up too fast near $y_0$. If by $u\in C^2(\Omega)$ you mean that $D^2 u$ is bounded, then interpolation inequalities give that $u\in W^{2,p}(\Omega)$ for every finite $p$, and thus we're in the situation above.

Of course all of this handwaves the approximation, citing a more general result for Sobolev spaces. I don't know if you can prove a similar result avoiding this, and constructing a direct approximating sequence...


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.