how can I conclude $\|v\|^2_{H^2(\Omega)} \leq C \int (\Delta v)^2 dx$?

Question

I need to prove that : Let $\Omega$ be a square with boundary $\Gamma$. Show that there is a constant C such that $\|v\|^2_{H^2(\Omega)} \leq C \int_\Omega (\Delta v)² dx $ $\forall x \in H^2_0(\Omega)$

prove: I konw that $\|v\|^2_{H^2(\Omega)} = \int_\Omega v^2 + |\frac{\partial v}{\partial x} |^2 + |\frac{\partial v}{\partial y} |^2 + \frac{\partial^2 v}{\partial x^2} |^2 + |\frac{\partial^2 v}{\partial y^2} |^2 + |\frac{\partial^2 v}{\partial x \partial y} |^2$

But, $v \in H^2_0(\Omega)$ so $H^2_0(\Omega) = \{v \in H^2(\Omega) : v= \frac{\partial v}{\partial n} = 0 \ \mathrm{in} \ \Gamma\}$

Then, $\|v\|^2_{H^2(\Omega)} = \int_\Omega |\frac{\partial^2 v}{\partial x^2} |^2 + |\frac{\partial^2 v}{\partial y^2} |^2 + |\frac{\partial^2 v}{\partial x \partial y} |^2$

For Green's formule:

$\|v\|^2_{H^2(\Omega)} = \int_\Omega |\frac{\partial^2 v}{\partial x^2} |^2 + |\frac{\partial^2 v}{\partial y^2} |^2 + |\frac{\partial^2 v}{\partial x^2} \frac{\partial^2 v}{\partial y^2} |^2 \\ \leq \int_\Omega |\frac{\partial^2 v}{\partial x^2} |^2 + |\frac{\partial^2 v}{\partial y^2} |^2 + 2|\frac{\partial^2 v}{\partial x^2} \frac{\partial^2 v}{\partial y^2} |^2 = \int |(\Delta v)^2|$

It's correct? How can I conclude this?

Answer 1

Your method is a little incorrect. You say $\int_{\Omega}v^2=0$ by the definition of the space. But it's not true. As this implies, $v=0$. To show the inequality you have to use Poincare Inequality twice, once for $u$ and then for $\nabla u$. Here,eqiuvalent norms in $H_0^2$ you can get more idea for the solution