In the Poisson equation $\Delta u = f $ in $ \Omega = (0,1)^2, u=0$ in $\partial \Omega$
With $u \in H^2(\Omega)\cap H_0^1(\Omega)$ Prove
$ |u|_{H^2(\Omega)} =||\Delta u||_{L^2(\Omega)} $.
S far I have that $ |u|_{H^2(\Omega)}^2= ||u_{xx} ||_{L^2(\Omega)}^2 +||u_{yy} ||_{L^2(\Omega)}^2 + 2||u_{xy} ||_{L^2(\Omega)}^2$ U and $||\Delta u||_{L^2(\Omega)}^2=||u_{xx} ||_{L^2(\Omega)}^2 +||u_{yy} ||_{L^2(\Omega)}^2 + 2<u_{xx}, u_{yy} >_{L^2(\Omega)}$
Looks like I have to prove that the last term of the first expression equals the last term of the second one... I am stuck here... Any ideas?
You are correct, you must show that $$\int_\Omega u_{xy}u_{xy} = \int_\Omega u_{xx}u_{yy}.$$ Starting with the left hand side, assuming the function is sufficiently differentiable, by integration by parts $$\int_\Omega u_{xy}u_{xy} = \int_\Omega \partial_x(u_{y}u_{xy}) - u_yu_{xyx}$$ $$= \int_\Omega \partial_x(u_{y}u_{xy}) - \partial_y(u_yu_{xx}) + u_{xx}u_{yy}$$ $$=\int_{\partial\Omega} (u_{y}u_{xy})n_x - (u_yu_{xx})n_y + \int_\Omega u_{xx}u_{yy}$$ $$=\int_\Omega u_{xx}u_{yy}.$$ The first boundary term vanishes due to the following observations: $n_x=0$ on when $y\in \{0,1\}$ and $u_y=0$ when $x\in \{0,1\}$ since it is a tangential derivative and $u$ is zero on $\partial\Omega$. The second boundary term vanishes since $u_{xx}=0$ when $y\in \{0,1\}$ (second order tangential derivative) and $u_y=0$ when $x\in \{0,1\}$.