I came across this problem when I try to prove that for space $H(\Omega):=H_0^1(\Omega)\cap H^2(\Omega)$, where $\Omega$ is open bounded with nice boundary, then the norm $\|u\|_1:=\|\Delta u\|_{L^2}$ is equivalent to the norm $\|u\|_{H^2}$ in the usual sense.

This problem can be proved by using open mapping theorem and showing that $\|u\|_1$ is actually make $H$ as a Banach space.

However, I was thinking by using the fact that the norm $\|u\|_2:=\|u\|_{L^2}+\|D^2u\|_{L^2}$ is already an equivalent norm of $\|u\|_{H^2}$ and the term $\|u\|_{L^2}$ will be taken care by Poincare. Then I only need to show that $\|D^2u\|_{L^2(\Omega)}\leq C\|\Delta u\|_{L^2(\Omega)}$ for some constant $C$.

I remembered when I deal with the weak solution of Biharmonic functions, I proved that $$ \int_\Omega \partial_i\partial_j u\cdot \partial_i\partial_j u\,dx=\int_\Omega \partial_i\partial_i u\cdot \partial_j\partial_j u\,dx \tag 1$$ by using density argument.

So here I was going to use the same approach. But I can not see what function is dense in $H$ under $H^2$ norm. I don't think $C_c^\infty(\Omega)$ will work here because it will give us $H_0^2$ but not $H$. But we can not use $C^\infty(\bar \Omega)$ neither because we will lose $H_0^1$. So, what function should I use as the approximate function here?

i.e., can I somehow use smooth function to approx $u\in H$ under $H^2$ norm?


This question has been solved but I want to update that we can generalize this to $H^{2k}(\Omega)\cap H_0^k(\Omega)$. That is, the space $H^{2k}(\Omega)\cap H_0^k(\Omega)$ has equivalent norm $\| \Delta^k u\|_{L^2(\Omega)}$ and it can be proved by using regularity of Laplace equation. (The equivalent of $H^2\cap H_0^1$ can be proved by regularity in only TWO steps!)


We have that (see Evans in regularity part) for each $f\in L^2(\Omega)$, the problem

$$ \left\{ \begin{array}{cc} -\Delta u =f&\mbox{ in $\Omega$,} \\ u=0 &\mbox{ on $\partial\Omega$,} \end{array} \right. $$

has an unique solution $u\in H$ satisfying $\|u\|_{H^2}\le C\|f\|_2$.

Fix $u\in H$ and write $-\Delta u=f$. Let $f_n\in C^\infty(\overline{\Omega})$ be a sequence satisfying $f_n\to f$ in $L^2(\Omega)$. Let $u_n\in C^2(\overline{\Omega})\cap C_0(\overline{\Omega})$ be the unique solution of

$$ \left\{ \begin{array}{cc} -\Delta u_n =f_n&\mbox{ in $\Omega$,} \\ u_n=0 &\mbox{ on $\partial\Omega$,} \end{array} \right. $$

therefore $$ \left\{ \begin{array}{cc} -\Delta(u- u_n) =f-f_n&\mbox{ in $\Omega$,} \\ u- u_n=0 &\mbox{ on $\partial\Omega$.} \end{array} \right. $$

Thus, $\|u-u_n\|_{H^2(\Omega)}\le C\|f-f_n\|_2$, which implies that $u_n\to u$ in the $H^2$ norm.

Remark: Although the previous construction shows that $C^2(\overline{\Omega})\cap C_0(\overline{\Omega})$ is dense in $H$, it would be nice to do it directly, i.e. by constructing an explicitly sequence $u_n$, using a mollifier sequence.

Update: I will give an answer for the question proposed in the remark. In Burenkov's book chapter 2, there is a method to regularize a function in $W^{k,p}(\Omega)$, which preserves boundary values.

By applying this method here, we can prove that $C^2(\overline{\Omega})\cap C_0(\overline{\Omega})$ is dense in $H$.

  • $\begingroup$ Just one more question. For each $f_n\in C^\infty(\bar{\Omega})$, how could you obtain a solution in $C^2(\bar\Omega)\cap C_0^1(\bar\Omega)$? How the part of $C_0^1(\bar{\Omega})$ comes? By $C_0^1(\bar \Omega)$ you mean only $u=0$ on $\partial \Omega$ or both $u$ and $\nabla u$ is $0$ on $\partial \Omega$? Thank you! $\endgroup$ – spatially Dec 31 '14 at 0:52
  • $\begingroup$ I mean I understand there will a solution in $C^\infty(\bar\Omega)\cap C_0(\bar\Omega)$ but I don't quit see why the solution will in $C_0^1(\bar\Omega)$ $\endgroup$ – spatially Dec 31 '14 at 0:53
  • $\begingroup$ Sorry, it is as you have said in your second comment. I will change it later or if you want, you can fix it now. $\endgroup$ – Tomás Dec 31 '14 at 0:58
  • $\begingroup$ No problem. So my equation $(1)$ will not hold right? because I need both $u$ and $\nabla u$ are $0$ to derive it. $\endgroup$ – spatially Dec 31 '14 at 0:59
  • $\begingroup$ It seems that you need only $\nabla u=0$ on the boundary. $\endgroup$ – Tomás Dec 31 '14 at 10:46

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