Consider function $u(x)=\text{sgn}\ x_1+\text{sgn}\ x_2$ in the unit disc $B(0,1)\subset \mathbb{R}^2.$ Show that the weak derivative $u_{x_1x_2}$ exists, but the weak derivative $u_{x_1}$ does not exist.
My attempt: For all $\phi\in C_0^2(\Omega)$, we have $\int_\Omega u\phi_{x_1 x_2}dx=2\int_{\Omega_1}\phi_{x_1x_2}dx-2\int_{\Omega_3}\phi_{x_1 x_2}dx=0$ I don't understand why this has to equal zero. Am I missing something?
For $u_{x_1}=v$, there exists $\int_{\Omega}u\phi_{x_1}=-\int_{\Omega}v\phi,$ for all $\phi\in C_0^\infty(\Omega).$ How do we show that the weak derivative $u_{x_1}$ exists, and why does it equal 0?
$2\int_{\Omega_1}\phi_{x_1}dx=0, \phi\in C_0^\infty(\Omega_1)$ $\int_{\Omega_i}v\phi dx=0,$ for all $\phi\in C_0^\infty(\Omega_i)$ which implies $v=0$ in $\Omega_i.$
Not really sure how to finish up this part...
I got that $\int_0^1 (\int_0^{\sqrt{1-x_2^2}}\phi_{x_1}dx_1)dx_2=\int_0^1(\phi(\sqrt{1-x_2^2},x_2)-\phi(0,x_2))dx_2)=-\int_0^1\phi(0,x_2)dx_2$ $\int_{-1}^0(\int_{-\sqrt{1-x_2^2}^0}\phi_{x_1}dx_1)dx_2=\int_{-1}^0(\phi(0,x_2)+\phi(-\sqrt{1-x_2^2},x_2))dx_2=\int_{-1}^0\phi(0,x_2)dx_2$
These imply that $\int_{-1}^1\phi(0,x_2)dx_2=0,$ but this cannot happen. But why is this zero for all $\phi\in C_0^\infty(\Omega)?$