A function with weak derivatives in two dimension also has weak derivative in one dimension?

Question

Let $\Omega = (-1,1) \times (-1,1) \subset \mathbb{R}^{2}$. In $\mathbb{R}^{2}$ we use the coordinates $(r,s)$. For a given $u \in L^1_{loc}(\Omega)$, suppose it has all weak derivatives of first order, that is, there exists $v_1, v_2 \in L^1_{loc}(\Omega)$ such that $$ \int_\Omega u \partial_r \varphi = - \int_\Omega v_1 \varphi $$ and $$ \int_\Omega u \partial_s \varphi = - \int_\Omega v_2 \varphi $$ for all $\varphi \in C^\infty_0(\Omega)$. For a fixed $r \in (-1,1)$ define $w_r = u(r,\cdot)$ definied in $(-1,1)$. I am wondering if $w_r$ has weak derivative. If this is the case, I suspect its weak derivative should be $v_2(r, \cdot)$. However I could not prove or disapprove it. Any help of sugestions is welcome.