Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For a open Lipschitz domain $\Omega$, consider the space $$A =\{ u \in H^1(\Omega) \mid \Delta u \in L^2(\Omega)\}.$$

Now I heard somewhere that all the second derivatives of a function $u$ are controlled by its Laplacian (because $|u|_{L^2} + |\Delta u|_{L^2}$ is equivalent to $|u|_{H^2}$) so then is not $A=H^2(\Omega)$?

Why is this space used??

share|improve this question

2 Answers 2

up vote 1 down vote accepted

If $\Omega$ is convex or polygonal (polyhedral), one knows that the solution $v$ of \begin{align*}-\Delta v + v &= f \quad\text{in }\Omega\\\frac{\partial}{\partial n}v + v &= 0 \quad\text{on }\Gamma\end{align*} Belongs to $H^2(\Omega)$ for all $f \in L^2(\Omega)$. In this regular case, you get indeed $A = H^2(\Omega)$.

However, this is not the case, e.g., for the L-shaped domain. Then, the solution with $f = 1$ belongs to $H^1(\Omega)$ but not to $H^2(\Omega)$. However, you still have $-\Delta v = f - v \in L^2(\Omega)$.

share|improve this answer

$\Delta u\in L^2$ is equivalent to the $H^2$ seminorm in the whole space (you can use Riesz transforms), or in the periodic case (same reason).

However, if you try to obtain the same result in a bounded domain, integrating by parts you get some "extra" terms coming from the boundary. If you impose some boundary conditions (for instance $u,Du=0$) for your functions then you can handle this new terms.

In your set $A$ there is no conditions on the value of the function of its derivatives in the boundary.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.