0
$\begingroup$

If I have a PDE of the form

$\left\{\begin{array}{ll} (-\Delta+q)v_1=-(-\Delta+q)v_0, & \text{in}\quad\Omega \\ v_1=0 & \text{on}\quad\partial\Omega \end{array} \right.$

then is the following estimate correct: $||v_1||_{L^2(\Omega)}\leq C||-(-\Delta+q)v_0||_{H^{-2}(\Omega)}?$ Here the function $v_1$ solves the above PDE, $C\geq0$ and $v_0$ is an explicit function. This seems that it should work because of the way that PDE regularity usually works. I think that the form of my PDE is not important. I am also interested to find a reference for this, so I would be very thankful if someone could provide me with one :)

$\endgroup$
  • $\begingroup$ Just to clarify, $H^{-2}(\Omega)$ is the dual of $H^2_0(\Omega)$, right? $\endgroup$ – StarBug Aug 23 '19 at 8:54
  • $\begingroup$ @StarBug Yes, you are right. $\endgroup$ – MathLearner Aug 24 '19 at 6:56
  • $\begingroup$ Just to clarify, I am more interested in the reference than in an actual proof of the fact. I need this for my masters thesis. $\endgroup$ – MathLearner Aug 24 '19 at 8:51
  • $\begingroup$ I don't think the estimate is true unfortunately. I will try to give a proper answer. $\endgroup$ – StarBug Aug 24 '19 at 10:04
1
$\begingroup$

The estimate is not correct. The classical elliptic a priori estimates in Sobolov spaces do not extend to Sobolev spaces of negative order. Roughly speaking, the boundary values have to play a role in the a priori estimate, which in the setting of negative spaces they do not.

A simple counter example: Consider on $\Omega:=B_1$ some smooth function $\phi_\epsilon$ such that $\phi_\epsilon=1$ on $B_{(1-\epsilon)}$ and $\phi=0$ on $\partial B_1$. Then (trivially) \begin{align} \Delta \phi_\epsilon = \Delta \phi_\epsilon\quad \text{in } B_1,\qquad \phi_\epsilon=0 \quad \text{on } \partial B_1, \end{align} but \begin{align} &||\Delta \phi_\epsilon||_{H^{-2}} \rightarrow 0\quad\text{as}\quad \epsilon\rightarrow 0\quad\text{and}\\ &||\phi_\epsilon||_2\rightarrow |B_1|^{1/2} \quad\text{as}\quad \epsilon\rightarrow 0. \end{align}

$\endgroup$

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.