# Difference between order of convergence and order of consistency?

I understand that, as the time step is reduced:

Consistency: The local error must approach zero.
Convergence: The global error must approach zero.

But why are both the order of convergence and consistency of explicit Euler = 1, if the global error is $O(h)$ while the local error is $O(h^{2})$?

More details: The Euler method is used to approximate an initial value ODE by time stepping using the derivative.

For example, for ODE form: $y'(t) = f(t,y(t)), \qquad \qquad y(t_0)=y_0$

The solution at point $n$ is: $y_{n+1} = y_n + hf(t_n,y_n)$

This formula can be obtained from the Taylor expansion, where $h$ is your time step:

$y(t + h) = y(t) + hy'(t) + \frac{h^{2} y''(t)}{2!} + ...$

So the local error is of order $O(h^{2})$. Now the global error is one order less which is why Euler is called a first-order method. So my question is: Why is the order of consistency also 1, even though the local error is of order 2?

• @Moody Which tags would you like? This area is certainly not heavily represented on the site, to relevant tags may be missing. We can create them for you! (Regarding your edit, please make it so there is one post to read. We keep revisions, so "EDIT: ... " is redundant and, actually, harmful for the flow of reading.) – Raphael Jan 31 '16 at 14:34
• @Raphael Thanks for the advice. But I think, as the below poster has mentioned, I should have asked my question on scicomp.stackexchange.com instead of here. Cheers. – Moody Jan 31 '16 at 18:55

Let's use the definition from Wikipedia. Let $\delta^h_{n+k}$ be the local error. A method has order $\textbf{p}$ if $$\delta^h_{n+k} = O(h^{\textbf{p+1}}) \text{ as } h \to 0.$$
Notice $p$ and $p+1$ in the definition. Since the explicit Euler method has local error of order $O(h^2)$, $p+1 = 2$, hence $p = 1$.