Consider the advection pde $v_t + a v_x = 0$ and let $R = \frac{a \Delta t }{\Delta x}$. Then, the FTFS scheme is given by
$$ u_k^{n+1} = u_k^n - R (u_{k+1}^n - u_k^n ) $$
and $u_k^n = u(k \Delta x, n \Delta t)$. Suppose we take $R=-1$ and assume $a<0$, then we see that our scheme becomes
$$ u_k^{n+1} = u_{k+1}^n $$
now my books says: it is easy to see that this scheme produces exact solution.
Im having hard time trying to see why this is true.
I know that solution of this PDE is constant along characteristiscs $x-a=t$. Maybe this would help, but I still don't see how this may help.
In other, suppose $U(x,t)$ is the exact solution. We are trying to show then that $U-u=0$. If we replace this in our scheme we have
$$ U_k^{n+1} - U_{k+1}^n =0 $$
but even if we use Taylor series, i dont see how we can prove $U-u=0$. Any advice?