In some cases I have seen that if the local error is:

$Err = O (\Delta t^{p+1})$

where the global error is p. So if local error is 3, global will be 2.

Does somebody know where it comes? For instance in the case of multistep solver you can say that regarding wikipedia, but are there other cases. Is there a simple way to explain such thing?

If the truncation error is:

$Err = \Delta t f(t) + O (\Delta t^{3})$

might I say that my error is:

$Err = f(t) + O (\Delta t^{2})$


2 Answers 2


This is not a formal proof, but intuitively you're making a truncation error at each time step of $O(\Delta t^k)$. Your global error will be the error at the time horizon $T = N\Delta t$.

This means you're taking $N=T/\Delta t$ time steps and making an $O(\Delta t^k)$ error at each of those steps. So global error will be on the order of $(T/\Delta t) O(\Delta t^k) \Rightarrow O(\Delta t^{k-1})$.


This link comes from the Lax Equivalence Theorem and requires that the numerical algorithm is stable. Essentially the Lax Equivalence Theorem is a proof that if the numerical method is stable, then the link between local truncation error and global error that is in the OP must follow. The last step of the proof is just, because you now have a bound on the global error, this implies convergence as $\Delta t \rightarrow 0 $. So along the way it establishes the estimate you're looking for in a pretty general manner.

The way you prove this estimate is to think of the system as a discrete dynamical system, and do substitution $N$ times and collect the error terms, and do some bounds estimating.


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