Explicit solution to the advection-reaction equation?

Setting $u(x,t)=v(t-x,t+x)$ transforms the equation to $$ 2\partial_2v+v=0 $$ so that $$ u(x,t)=e^{-(t+x)/2}w(t-x) $$ This implies that $$ e^{t/2+s}u(s, t+s)=w(t)=e^{t/2}u(0,t)=e^{t/2+2\pi}u(2\pi,t+2\pi) $$ which is not compatible with the boundary conditions, as they would require $1=e^{2\pi}$.

Actually, it turns out there is a solution:

$$u(x,t) = \begin{cases} e^{-x}\sin(x-t), &x<t\\ e^{-t}\sin(x-t), &x \geq t \end{cases}.$$

This is in fact a $C^1$ function, which solves the original PDE, and satisfies all the boundary conditions.