order of convergence PDE

Question

If I have the generic PDE \begin{equation} u_t + u_x = f, \end{equation} approximated with a first order in time and a second order in space numerical scheme, how can I show that the solution effectively converges as $\Delta t$ and $\Delta x^2$, respectively? If I look at the plot of the error as a function of $\Delta x$, I can observe a knee $-$ it represents the fact that the order of accuracy of time takes over on the spatial one. Hence, how can I see that the scheme converges as $\Delta t$?

Answer 1

You can take $\Delta t = C \Delta x^2$. If so the decay of $O(\Delta t)$ term will be of the same magnitude as $O(\Delta x^2)$. For example, you can divide your timestep by $4$ each time you divide your spatial step by $2$. Note that the Courant's number $$ C = \frac{\Delta t}{\Delta x} \sim \Delta x $$ will not stay constant during this process.

Also note that simple explicit method of order $O(\Delta t, \Delta x^2)$ is not stable if $\Delta t \sim \Delta x$, only when $\Delta t \sim \Delta x^2$. Also, convergence will only be in $L_2$ norm, not in supremum norm.