Timeline for NDSolve: Improperly assigned stepsize
Current License: CC BY-SA 3.0
11 events
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| Jul 24, 2016 at 17:23 | comment | added | Feyre | Well it's an orbit, in fact I'm evaluating one sidereal year. I'll read up on the info you've given and consider the matter settled, and accept this is just one of those things in which human intervention is required. | |
| Jul 24, 2016 at 17:10 | comment | added | Michael E2 |
I guess I should mention that it is local error that NDSolve controls. If you have many steps, those errors can add up if they're mainly in one direction.
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| Jul 24, 2016 at 16:47 | comment | added | Michael E2 |
(2) Strictly speaking the error estimate is just an estimate, so it is not guaranteed to be accurate, but usually it is. It depends on the system, and I would think that for yours, it should be pretty good. Rounding error is a separate issue (from truncation error, the inherent error of the discrete integration method used by NDSolve[]), and that sometimes is a problem. For this sort of problem, a symplectic integrator is often used; see this tutorial.
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| Jul 24, 2016 at 16:47 | comment | added | Feyre |
@MichaelE2 The solution has three components, $x,y,z$, the first two fluctuate beween magnitude 1 and value0, the z component stays in the -4 order for the duration of this calculation. However it is the x and y components that have the higher order errors.
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| Jul 24, 2016 at 16:39 | comment | added | Michael E2 |
@Feyre (1) What controls the step size depends on which is bigger, 10.^-pg * * solution or 10.^-ag: In the first case, relative error will dominate, and in the second, absolute error will dominate. When they're about the same, then the (absolute) error tolerance will be about two times 10^-ag. (This is just a consequence of the formula in the tutorial I linked above.) So the answer to your question depends on whether you're talking about absolute or relative error (if the solution has a numerical magnitude of about 1., then they're roughly the same).
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| Jul 24, 2016 at 15:48 | comment | added | Feyre |
@MichaelE2 I calculated the distance {at a time period of a body (specifically Earth) } between the expected position, and the position that the model computes at that point. It was my (maybe naive) belief that in theory the error should not exceed Precision which is half WorkingPrecision. Am I to understand that with the available information it cannot be guaranteed that the model delivers 7-digit precision as I expected?
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| Jul 24, 2016 at 15:39 | comment | added | Feyre |
@J.M. The solar system model mentioned in mathematica.stackexchange.com/questions/121035/… Though I've converted to AU and day units.
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| Jul 24, 2016 at 15:10 | comment | added | Michael E2 |
It's not clear the 2nd plot indicates the condition "that the estimated error in the solution is just within the tolerances specified by PrecisionGoal and AccuracyGoal" is violated. (1) NDSolve uses NDSolve`ScaledVectorNorm[2, {10.^-pg, 10.^-ag}], (pg, ag = PrecisionGoal, AccuracyGoal settings resp.) to measure the acceptability of the estimated error. (2) The estimated error is an estimate, and not guaranteed to bound the actual error. -- Some clarification about how σ is computed would be helpful.
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| Jul 24, 2016 at 14:43 | comment | added | J. M.'s missing motivation♦ | Can you at least mention what kind of ODEs are you solving? | |
| Jul 24, 2016 at 14:42 | history | edited | J. M.'s missing motivation♦ | CC BY-SA 3.0 |
added 1 character in body; edited tags; edited title
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| Jul 24, 2016 at 14:13 | history | asked | Feyre | CC BY-SA 3.0 |