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 :)

  • $\begingroup$ 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

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}


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