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

Question

Consider $$ \begin{equation} \begin{cases} -\Delta{\omega}=f \ on \ \Omega \\ \ \ \ \ \ \omega=0 \ on \ \partial\Omega \end{cases} \ \ \ \ \ \ (1)\end{equation} $$ 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 $$ u(x_1,x_2)= x_1x_2(1-\sqrt{x_1^2+x_2^2}) , where \ \ (x_1,x_2)\in \overline{B_1(0)}\subset \mathbb{R^2}$$