Let $\sigma\in L^\infty([0,T]\times\mathbb{T}^d)$ and let $u_1,u_2\in L^\infty(0,T;W^{1,p}(\mathbb{T}^d))$ be such that $\int_{\mathbb{T}^d} u_i \; \text{d}x=0$ and $$ \text{div}u_1(t,x_1(t,y)) = \text{div}u_2(t,x_2(t,y)) = \sigma(t,y), $$ where $x_i(t,y)$ solves an ODE $$ \dot{x_i} = u_i(t,x_i), $$ $$ x_i(0,y) = y. $$
I was wondering if this implies that $u_1(t,x)=u_2(t,x)$. I know that if simply $\text{div}u_1(t,x)=\text{div}u_2(t,x)$, then $u_1=u_2$, but here I have a composition of $\text{div}u_i$ and $x_i$ and I don't know how to deal with this.
I would be very happy if you could help me find a counterexample or give some references where I could seek an answer.