Order of accuracy of a numerical method From this article, if some numerical method is second order accurate then we will obtain four times smaller error $E$ given that the step size $h$ is halved. Let say we have a numerical method for some partial differential equation. This method is second order accurate in space and time, $O(\Delta x^2+\Delta t^2)$. From the previous article I conclude: the error $E$ will be four times smaller if we halve both $\Delta x$ and $\Delta t$. Is my conclusion correct? Another question, what happens to the error $E$ if we only reduce one of the step sizes, e.g. half $\Delta x$ with same $\Delta t$ or half $\Delta t$ with same $\Delta x$?
 A: When we write that a numerical method has an error of size $\mathcal{O}(\Delta x^2) + \mathcal{O}(\Delta t^2)$ we mean that there are constants $C_x$ and $C_t$, independent of $\Delta x$ and $\Delta t$, such that asymptotically the error
$$
E(\Delta x, \Delta t) \leq C_x \Delta x^2 + C_t \Delta t^2,
$$
i.e. for $\Delta x$ and $\Delta t$ sufficiently small. This is because the error generally is a polynomial in $\Delta x$ and $\Delta t$ whose leading terms are proportional to $\Delta x^2$ and $\Delta t^2$. As such, yes it is true that halving $\Delta x$ and $\Delta t$ should reduce the error by a factor 4, assuming that $\Delta x$ and $\Delta t$ are so small that asymptotic behaviour sets in.
For your second question, what happens if either $\Delta x$ or $\Delta t$ is halved but not the other, it really depends on a few things:


*

*Sometimes $\Delta x$ and $\Delta t$ cannot be varied independently without numerical instability setting in. If this is the case, the error will increase beyond bound.

*If $C_x \Delta x^2 \sim C_t \Delta t^2$ then halving the temporal term will not significantly affect the error since the error will then become dominated by the spatial term, which we haven't changed.

*On the other hand, if say $C_x \Delta x^2 \ll C_t \Delta t^2$ and we are halving the temporal term, we might well see a significant reduction of the error. If $C_x \Delta x^2 \ll C_t \Delta (t/2)^2$ and we are in the asymptotic regime, this reduction should be by a factor 4. If not, it may be somewhat less.

