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?


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

loglog error plot

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 $\frac{10^{-15}}h+0.03\cdot h^4-0.08\cdot h^5+0.0225\cdot h^6$.

  • $\begingroup$ Just to make sure I'm following this correctly....x-axis: step size, y-axis: error...."noise" on left is due to floating point. (if correct, no need to comment). By the way...thanks for the plot because this really helps to visualize my question!! $\endgroup$ – ThatsRightJack May 23 at 3:32
  • $\begingroup$ OK, so in the context of my question, would the "leading order truncation error" be that middle linear rise? Assuming the floating point error was not an issue, this would continue down to the left, showing infinitely small error, hence converge to PDE solution. This is the classic discussion and makes sense. Can you say more, in general, about the linear slope? You said RK4-type scheme...can we learn something about the order from this graph? $\endgroup$ – ThatsRightJack May 23 at 3:35
  • $\begingroup$ OK, lets go up and to the right, following the linear curve. As you show (thanks again for the plot), the curve changes and no longer is linear. What happened at this point? You gave a series expression...should I think of c's as terms in a Taylor series? Did you just factor out the leading order term to get that inequality? I'm assuming the inequality represents where the linearity is lost? Can you elaborate more on the region beyond linearity in discussion? This is what I'm curious in. It seems like there's a lot more action between linearity and the step size stability limit. $\endgroup$ – ThatsRightJack May 23 at 3:54
  • $\begingroup$ You get the shape of the first two parts as the loglog plot of $10^{-15}/h+0.03\cdot h^4$, the sum of the floating point error bound and the method truncation error. Experimentally one gets the wiggles around $h=1$ with the addition of negative and positive higher order terms, like $|10^{-15}/h+0.03\cdot h^4-0.08\cdot h^5+0.022\cdot h^6|$. $\endgroup$ – Dr. Lutz Lehmann May 23 at 6:04

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