I have the problem
$u_t+2tx^2u_x=0, x \in [0,2], t \in [0,1] \\ u(0,x)=u_0(x)=e^{-\beta \left( x-1\right)^2} , x \in [0,2]\\ u(t,0)=0, t \in [0,1]$

I have found that the cfl-condition is $2 \nu t_n x_j^2 \leq 1$. Since it holds that $x \in [0,2], t \in [0,1]$, I thought that we could use the upper bounds for x and t and pick as Courant number, $\nu=\frac{1}{8}$. Is this right? @ThomasKlimpel

For N_x=200 and N_t=800:

er = 2.2172e-004



For N_x=400 and N_t=1600:

er = 5.8732e-005



For N_x=800 and N_t=3200:

er = 2.0509e-005 @ThomasKlimpel

@ThomasKlimpel I have calculated the errors for each of the above times:

er (t_{800}=1) = 2.2172e-004


er1 (t_{100}=1/8) = 0.0154


er2 (t_{250}=5/16) = 0.0754


er3 (t_{500}=5/8) = 0.1661