Timeline for Euler method (and others) for unbounded intervals
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Jan 20, 2022 at 15:45 | comment | added | alhal | @DieterKadelka I thought about your suggestion some more - but I'm not sure how it is supposed to work in detail. If the solution of my ODE is defined on all of $\mathbb{R}$ (or even $[0,\infty)$, it seems to me there is no way I can reparametrize time to obtain a solution, having the same image, that is defined on a closed bounded interval (which is what I would need from a numerical perspective). | |
| Jun 2, 2021 at 22:48 | answer | added | Richard Diagram | timeline score: 2 | |
| Jun 2, 2021 at 12:27 | vote | accept | alhal | ||
| Jun 2, 2021 at 9:50 | comment | added | alhal | @RichardDiagram I would be interestedd in knowing about it, if you can phrase it in a not too technical way and its revelant to question 1., which seems to me to still be partially open. | |
| Jun 2, 2021 at 9:39 | comment | added | alhal | @DavidKetcheson Yes, but my conundrum was that the numerical method may not be valid beyond a certain finite interval, since errors might accumulate, as you mentioned in your answer too. Your latest comment to your answer, where you pointed me to an arXiv article, where a numerical method for unbounded intervals was given, is what I was after in this regard. | |
| Jun 2, 2021 at 1:07 | comment | added | Richard Diagram | If you are willing to forgo the matrix approach to Euler's method; which is typically how Runge and everybody does it; and switch to an analytic scenario, I have much to say about "Euler's Method" with holomorphic functions on $\phi(s,z) : \mathcal{S} \times \mathbb{C} \to \mathbb{C}$. Though, I phrased it as taking contour/line integrals in non-abelian settings. | |
| Jun 1, 2021 at 14:34 | comment | added | David Ketcheson | "an explicit formula for the recurrence equation(s) that determine the approximation" is what you already have in the numerical method itself. In some case (e.g. linear ODEs) you can also explicitly solve this recurrence. | |
| Jun 1, 2021 at 13:09 | comment | added | alhal | @someuser I saw you removed your comment; I would have been interested to understand what you meant by your example. | |
| Jun 1, 2021 at 12:29 | comment | added | alhal | @DavidKetcheson I made an edit to my question to address your question. (And thanks for pointing me towards scicomp.SE, I will keep that in mind for future questions.) | |
| Jun 1, 2021 at 12:27 | history | edited | alhal | CC BY-SA 4.0 |
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| Jun 1, 2021 at 5:31 | answer | added | David Ketcheson | timeline score: 5 | |
| Jun 1, 2021 at 5:12 | comment | added | David Ketcheson | Can you explain what it is that you want to know about the solution behavior over the unbounded interval? Often solutions approach a steady state or periodic orbit, and there are techniques for determining that kind of thing. Also, you will probably get more answers on scicomp.SE (though there are people like me who read both that and this). | |
| May 31, 2021 at 15:24 | comment | added | alhal | @DieterKadelka I guess that would be possible, though it would then not be a priori clear to me what other properties of the original solution such a transform would destroy/preserve. On the other hand, if there is a result that state that "nothing goes wrong" (I 'm myself unsure what a good formalization of this would be), I'd be happy with such an approach. | |
| May 31, 2021 at 15:21 | history | edited | alhal | CC BY-SA 4.0 |
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| May 31, 2021 at 15:14 | comment | added | Dieter Kadelka | One possible solution (of course not the best) is simply to transform the unbounded interval into a bounded one. | |
| May 31, 2021 at 15:05 | history | asked | alhal | CC BY-SA 4.0 |