Let $U_1$ and $U_2$ be two open sets in ${\mathbb R}^n$ with $U_1\cap U_2 \neq \emptyset$. $u\in L_{loc}^1 (U_1\cup U_2)$. If the weak derivatives of $u$ of order $\alpha$ in $U_1$ and $U_2$ are $v_1$ and $v_2$, respectively, then $$v_1=v_2,\ \ \ \ a.e. \ \ x\in U_1\cap U_2.$$
I want to prove $$\int_{U_1\cap U_2}(v_1-v_2)\ dx=0.$$ However, what I know is just, $\forall \varphi\in C_c^{\infty}(U)$, $$(\int_{U_1-U_2}-\int_{U_2-U_1})\ uD^{\alpha}\varphi\ dx=(-1)^{|\alpha|}(\int_{U_1\cap U_2}(v_1-v_2)\varphi\ +\int_{U_2-U_1}v_2\varphi\ -\int_{U_1-U_2} v_1\varphi)\ dx,$$ by the definition of weak derivatives. I am wondering how to take some $\textit{nice}$ test function $\varphi$ to get the conclusion.
Thank you very much for help!
The order $\alpha$ weak derivative $D^\alpha u$ of a function $u$ in a region $U$ is characterized by $$\int_U (D^\alpha u) \phi \, dx = (-1)^{|\alpha|} \int_U u (D^\alpha \phi) \, dx$$for all $\phi \in C_0^\infty(U)$.
In your case, you can take $U = U_1 \cap U_2$ to find that if $\phi \in C_0^\infty(U)$ then $$ \int_U v_1 \phi \, dx = (-1)^{|\alpha|} \int_U u (D^\alpha \phi) \, dx = \int_U v_2 \phi \, dx.$$
Now use standard mollifiers for the $\phi$ to conclude $v_1 = v_2$ throughout $U$.