In which sense is u a solution for (1) on $\Omega=B_1(0)$?

Consider $$$$\begin{cases} -\Delta{\omega}=f \ on \ \Omega \\ \ \ \ \ \ \omega=0 \ on \ \partial\Omega \end{cases} \ \ \ \ \ \ (1)$$$$ Definition 1 : $$Let\ \ u\in W^{1,2}_0(\Omega)\ \ such\ \ that \\ \int_{\Omega}(\nabla u*\nabla\phi-f\phi)dx=0 \ \ for \ \ all \ \ \phi\in W^{1,2}_0(\Omega)\ ,$$ $$then \ \ u \ \ is \ \ called \ \ a \ \ weak \ \ solution \ \ for \ \ (1) \ \ on \ \ \Omega \ .$$ $$\\$$ Definition 2 :$$Let\ \ u\in C_0(\overline \Omega)\ \ such\ \ that \\ \int_{\Omega}(- u*\Delta\phi-f\phi)dx=0 \ \ for \ \ all \ \ \phi\in C^{\infty}_0(\Omega) \ ,$$ $$then \ \ u \ \ is \ \ called \ \ a \ \ distributional \ \ solution \ \ for \ \ (1) \ \ on \ \ \Omega \ .$$ $$\\$$ I'm not really sure , but I think u is a distributional solution for (1) . Using a few times partial Integration and keeping in mind that the integral on the boundary disappears and that u is continuously differentiable I get the following : $$-\int_{B_1(0)}u\Delta\phi dx=-\int_{\partial B_1(0)}u\nabla\phi\nu d\sigma_x+\int_{B_1(0)}\nabla u \nabla \phi dx\\ =\int_{\partial B_1(0)}\nabla u \phi\nu d\sigma_x-\int_{B_1(0)}\Delta u\phi dx=-\int_{B_1(0)}\Delta u\phi dx=\int_{B_1(0)}f\phi dx .$$ I forgot something to mention $$u(x_1,x_2)= x_1x_2(1-\sqrt{x_1^2+x_2^2}) , where \ \ (x_1,x_2)\in \overline{B_1(0)}\subset \mathbb{R^2}$$

• I did a quick calculation, and it appears that $u\in W^{1,2}_0$ for the $u$ that you provided. This may mean it is a weak solution. Someone may need to confirm, though. – MasterYoda Sep 10 '18 at 20:46
• I don't really understand what I've to do if $$u \in W^{1,2}_0$$ . Well then it follows that $$u \in B_1(0)\{0}$$ is a distributional solution for (1) . – Matillo Sep 11 '18 at 6:17
• I don't really understand what I've to do if $$u \in W^{1,2}_0$$ . In fact I computed that u is in $$W^{3,p}(B_1(0)) \cap W^{4,1}(B_1(0))$$ . – Matillo Sep 11 '18 at 6:36
• Can someone explain me how I can show that u is a distributional solution for (1) on $$B_1(0)\diagdown (0,0)$$ ? I know that $$u\in C^{\infty}(B_1(0)\diagdown(0,0))$$ . – Matillo Sep 11 '18 at 14:11