So the rest remains unchanged and we just say that the problem is written equivalently as follows, right?
$\left\{\begin{matrix}
u_t+2u_x=0, & x \in [0,1], & t \in [0,1] \\
u(x,0)=\overline{u_0}(x)=e^{-\beta \left(x-\lfloor x \rfloor-\frac{1}{2} \right )^2}, & x \in [0,1] & \\
u(0,t)=u(1,t), & t \in [0,1] &
\end{matrix}\right.$