Truncation error with growing step size

Question

When I read about finite difference methods (or really any approximation method), truncation error is often central to the discussion, and rightfully so. But it is also most often discussed in the context of consistency/convergence where the step size decreases. My question is, what happens when you go the other way? Instead of making the step size smaller, lets make it larger! Does the idea of "leading truncation error" go out the window, in which the higher order terms dominate the error?

Answer 1

Yes, in $c_1h^p+c_2h^{p+1}+...$, the second term will dominate the first one for $h>\frac{c_1}{c_2}$.

In numerical applications, the many steps required by smaller step sizes eventually accumulate floating point noise sufficient to dominate the truncation error, so that a loglog plot of error vs. step size has a V shape with a fuzzy left leg, a middle piece on the right leg that is linearly raising and then some curved section for large $h$.

For a non-linear test problem $F[y]=F[p]$ with $F[y]=y''+1.16\sin(y)$ with exact solution $y(t)=p(t)=\cos(t)$ over the interval $[0,10]$ and using the 4th order classical Runge-Kutta method, this can look like this

The main error trends are first the accumulated floating point errors proportional to $\mu\frac{T}{h}$ where $\mu$ is the machine constant and $T$ the length of the integration interval, so that $T/h$ is the number of steps. And second the global error of the method proportional to $h^4$. A good fit was found with $h\mapsto\frac{10^{-15}}h+0.03\cdot h^4$.

Adding further higher order terms allows to reproduce the non-linear shape for larger $h$. Playing with the coefficients, a good fit was found manually with $\frac{10^{-15}}h+0.03\cdot h^4-0.08\cdot h^5+0.0225\cdot h^6$.