MATLAB: How does choosing more stringent RelTol and AbsTol values in ode45 compare with choosing finer tspan integration time-steps

MATLABode45

Hi,

So I modified my ode45 error tolerances, both RelTol and AbsTol, to be at 1e-10. 1e-14 or so seems to be the most stringent value. Can I further "add more accuracy", by choosing finer tspan time-steps, say, tspan = linspace(0,10, 20,000)? Or does adding finer time-steps just bring back more solutions, but with the same accuracy, and with the same error tolerances satisfied? For example, if I play with the simple differential equation xdot = cos(t), with initial condition x(0) = sin(0) = 0, I want to see how close the solution y = sin(t) can get to 0, near pi = 3.14…

However, adjusting the RelTol and AbsTol values to be more stringent doesn't seem to help; the time-steps still aren't fine enough, and t gets to about 3.143143143143143, and so y(t) = sin(t) is not actually very close to 0, for my needs. It is only when I add finer integration time-steps, in the tspan setting, does t get much closer to approximating pi, and thus y = sin(t) gets much closer to the value of 0.

Should I simply use a combination of both, i.e. choose stringent error tolerances as well as finer tspan timesteps, in order to get the most accurate solutions? My motivation for studying RelTol, AbsTol, and tspan time-steps is to eventually return to root-finding practice using Matlab's fsolve, so this is an important digression from multivariable root-finding.

Thanks,

Best Answer

Adding finer time steps just brings back more solutions with the same accuracy.

ode45() does not typically evaluate at the exact locations given in the tspan, other than the first and last one. Instead, when it crosses a time listed in tspan and is satisfied that the step met accuracy conditions, then it synthesizes outputs for the tspan element. It would not have accepted the step unless it thought the results were within the tolerance, so it has no reason to subdivide the interval looking for a potentially more accurate solution.

Related Solutions

MATLAB: AbsTol, RelTol

The absolute tolerance of 1e-3 is meaningless, when the values are very small, e.g. 1.234e-27. Then the relative tolerance is more useful.

The precision of the result is IEEE-64bit double in every case, but the accuracy is influenced by the tolerances. The tolerances are used to limit the local discretization error: If the difference between a high-accuracy and low-accuracy integration is higher than one of the tolerances, the step size is reduced.

If the tolerance is too low, the large number of steps will increase the accumulated rounding errors, while for a to high tolerance the local discretization errors dominate the accuracy of the result.

See also: Answers: absolute and relative tolerances and doc odeset.

MATLAB: AbsTol and RelTol in vpaintegral

Internally, vpaintegral() always works with tolerances. It has default values that it uses. It is not the case that providing a tolerance as an option results in the code proceeding differently -- no

if relative_tolerance_was_specified
    do some extra work
end

So you should not worry about only providing one instead of the other to have it do less work: it is always going to be doing the work, just with default values if you did not specify the option.

Typically reltol is more important in figuring out when to stop. Typically abstol is there as a fall-back for places where the relative values get small. You know that your final result is roughly 1000, but there could still be ranges of your integral where the values are (say) near 1e-15, and over that stretch vpaintegral() might be doing a lot of fine work to try to be more precise about the small numbers, but you in your knowledge of the overall range of results might say that it is pointless to be precise down below 1e-15 about a value that is effectively going to contribute nothing to the overall result, so you might want to put in an absolute tolerance of (say) 1e-5.

Absolute tolerance of 1e-5 does not mean that it will persue the integration until it is certain that the result over the entire integration is within 1e-5 of the "correct" value: instead it means that values less than 1e-5 would more or less filtered out.

There are some cases where the absolute values might get quite small, and yet the contribution could be very important. For example the integral of 1/x, as x goes to infinity, is infinite, ln(infinity), and yet the contributions up around realmax are only about 5e-309 each, with the total contribution by realmax only being about 709 -- the contributions beyond realmax are all negligibly small from the perspective of IEEE 754 double precision, and yet in theory they contribute an infinite amount out to infinity.

Therefore, it is not always appropriate to use an absolute tolerance: you have to know enough about the function to be able to establish that the integral of the missed values will not matter to your accuracy.

In the meantime, I recommand that you use matlabFunction() to turn the vpaintegral() with symbolic upper bound into an anonymous function that calls integral()

I was about to suggest you consider using the matlabFunction() options 'File' and 'optimize', but then I remembered that I have open cases against bugs in the optimization function that can result in the optimized function being quite wrong. It might still be worth using the 'File' option, but if you do, then explicitly set 'optimize' to false.

Related Question

Error :RelTol must be a positive scalar.