# Error comparison of one step vs two steps of a given ODE method

In general, my understanding is the global vs. one-step-error of a given method will differ by 1. Assuming this is true and that the one-step-error is order $\alpha+1$ then the error of taking a single step of size $h$ from our known initial condition, say $y(t_0)=y_0$ means that our approximation of $y(t_0+h)$, lets call it $w_1$ is $O(h^{\alpha+1})$.

Now considering cutting $h$ in half and taking 2 steps to approximate $y(t_0+h)$. Let's call this approximation $\bar{w_1}$, then this error should be $O( (h/2)^\alpha)=(1/2)^\alpha O(h^\alpha)$. For example, consider $\alpha=2$ and one step of $h=0.1$ vs two steps of $h=0.05$, then $0.1^3 < 0.05^2$ which seems to imply the accuracy went down by reducing $h$.

I'm use to just thinking that the accuracy goes up by reducing the stepwise, in particular, by a factor of the reduction of the step size to the order of the method. But this assumes that the global error is used in both cases. Considering taking a single step, vs 2, with the respective one-step and global errors, the analysis does not seem as clear.

To try and test it, I used RKmidpoint, an order 2 method, on a problem with an exact answer. Interestingly, comparing one step, vs two steps out to the same final time, generally scaled by $(1/2)^2$ (e.g. $|y(t_f)-\bar{w_1}| \approx (1/4)*|y(t_f) -w_1|)$, as would be expected by just considering global error and not the additional order from the one-step-error perspective.

Should this be expected in general? Was that just due to the problem I considered? Why shouldn't it follow the analysis that considers the one-step-error and the global error?

Consider some interval $[a,b]$ so that $L(b-a)$ is rather small and the variation in the derivatives of the exact solution is not large. Then one can assume that the local error of a single step looks like $$Ch^{α+1}+..$$ and that the errors of the steps simply add to the global error. If the integration over $[a,b]$ is preformed in $N$ steps of step size $h=\frac{b-a)}N$ the dominating term of the accumulated global error is thus $$N(Ch^{α+1}+..)=(b-a)Ch^α+..= (b-a)^{α+1}CN^{-α}+..,$$ in $N=2$ steps thus $$2(Ch^{α+1}+..)=(b-a)Ch^α+..= (b-a)^{α+1}C2^{-α}+..,$$ As this demonstrates, the number of steps combines with one factor $h$ of the local error to the interval length as additional constant factor, thus decreasing the local error order by 1 to give the global error order.
Over larger integration intervals one has to take into account that the accumulated errors of the last step are magnified, as per Grönwall, by a factor of $e^{Lh}$. If one steers the step size in such a way that the local error at step $k$ is always smaller than $\epsilon h_k$, then the accumulated errors satisfy $$e_{k+1}\le e^{Lh_k}e_k+ϵh_k\implies e^{-Lt_{k+1}}e_{k+1}\le e^{-Lt_k}e_k+e^{-Lt_{k+1}}ϵh_k$$ so that in the main terms $$e_n\leϵ\int_{t_0}^{t_n}e^{L(t_n-t)}dt=\fracϵL\left(e^{L(t_n-t_0)}-1\right)$$
• Well $2^{-\alpha}Ch^{\alpha+1}$ nicely shows considering the one-step-error throughout gives the $2^{-\alpha}$ scaling that would be expected from considering the global. But I'd still like to know why considering the one-step and the global is still wrong? What's the fallacy there? – Fractal20 Dec 11 '17 at 18:27
• The fallacy is that you assume that the constant of the big-O terms are all $1$ or at least equal. I hope I have changed the text to be better understandable. – LutzL Dec 11 '17 at 22:34