Local truncation error vs Global truncation error

Question

I know that global truncation error is proportional to $h^p$ while local truncation error is proportional to $h^{(p+1)}$, where $h$ is the step size.

But where does this relationship come from and how can you derive it?

Answer 1

Let us assume that you use your solver $N$ times (over the global time interval $[t_0, t_{\mathrm{end}}]$), and in each time step you make an error $\tau$, which is the local truncation error. Let $\epsilon_N$ be the total error that you have in the end. As you mentioned, $\tau = \mathcal{O}(h^{p+1})$ whereas $\epsilon_N = \mathcal{O}(h^p)$.

The idea is that, if you do an error of $\tau$ in each step, and there are $N$ steps in total, your total accumulated error is going to be more or less $N\tau$. Since $h = \frac{t_{\mathrm{end}}-t_0}{N}$, this yields $$\epsilon_N \approx N\tau = \mathcal{O}(Nh^{p+1}) = \mathcal{O}((t_{\mathrm{end}}-t_0)h^p) = \mathcal{O}(h^p),$$ since $t_{\mathrm{end}}-t_0$ is a constant. So the fact that you accumulate the error 'removes one order'.

Now, the argumentation above is not, strictly speaking, correct, because the accumulated error is not additive (it is not $\tau + \tau + ...$), and is instead "multiplicative" (a bit like the accumulated interest for money you deposit at a bank, it increases exponentially over time) with a relationship of the sort $\epsilon_N \leq c\frac{\max_i \tau_i}{h}$, where $\tau_i$ is the local truncation error in the i-th step. Still, the idea is the same.