Timeline for Existence and uniqueness of solutions to a distributional ordinary differential equation
Current License: CC BY-SA 4.0
20 events
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| Apr 23 at 10:07 | answer | added | crow | timeline score: 0 | |
| Mar 22 at 19:58 | comment | added | cheshircat | That makes sense to me. And it also makes sense that moving to cadlag gives the math a similar ability to "bias the race condition" in the same way that a numerical method would. | |
| Mar 22 at 4:35 | comment | added | Willie Wong | Yes, I am sure that in numerical simulations you don't actually get a race condition. My point is more that in principle there MAY be a race condition, and hence your simulation code will have to define the algorithm in such a way that the outcome is repeatable. What you outlined is exactly one such solution. However, note that from the "pure mathematics" those two things (root finding and checking the call back) should "happen at the same time"; the numerics are designed to resolve the ambiguity that appears in the "pure math" (at least its naive interpretation). | |
| Mar 22 at 1:00 | vote | accept | cheshircat | ||
| Mar 22 at 0:39 | comment | added | cheshircat | With regard to the race condition, the way that this typically works is that you run a root finding algorithm on the differential equation without the callback to figure out when exactly you will hit the callback, and then you run the callback at the time that the root finding algorithm returns. So in practice there isn't a race condition... I don't know if this addresses what you were concerned about though. | |
| Mar 22 at 0:36 | comment | added | cheshircat | Ah, the word cadlag brings up memories of an analysis class long past... this is exactly the next clue I was looking for! | |
| Mar 21 at 8:16 | history | edited | YCor | CC BY-SA 4.0 |
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| Mar 21 at 7:13 | history | became hot network question | |||
| Mar 21 at 4:53 | comment | added | Willie Wong | That said, I think if you insist all of your functions are one-sided differentiable and cadlag, there is probably a suitable theory that you can develop. (In fact, it probably has been, by folks studying stochastic processes.) | |
| Mar 21 at 4:46 | comment | added | Willie Wong | I guess the programming term for this is: you've set up a race condition and the output is ill defined. | |
| Mar 21 at 4:45 | comment | added | Willie Wong | My second answer can be interpreted as the following, then: you have a continuous call back function $v$ that does something when its argument hits the value 1. You have a process that approaches the value 1 as time increases toward 0, but jumps to 2 right after time 0. Does your call back fire? To implement the process numerically, you have it increase toward 1. And at time $t = 0$, it starts at $1$ and jumps to $2$ all the while $t$ hasn't changed. Whether your callback fires is now dependent on, when evaluating it at $t = 0$, whether the callback is used before or after the jump. | |
| Mar 21 at 4:35 | answer | added | Willie Wong | timeline score: 4 | |
| Mar 21 at 4:08 | comment | added | cheshircat | Really what I'm looking for is a suitable definition, or really what I was looking for is if experts in the theory of distributions had studied this kind of problem before. People do scientific simulations all the time with jumps in them, where a "discrete event" happens when you hit a certain threshold, see: docs.sciml.ai/DiffEqDocs/stable/features/callback_functions, or elastic collisions in physics software. So I was trying to figure out if this could be "geometrized" using distribution theory, but I guess not. | |
| Mar 21 at 4:03 | comment | added | Willie Wong | With your update, you run into a different problem: $\delta_1(x)$ is not well-defined as a distribution. In standard theory $\delta_1(x(t))$ is a well-defined distribution for $x$ with the property "if $x(t) =1$, then $x'(t)$ exists and is not equal to zero." This doesn't hold in your case. Please provide a definition of what $v(x)$ means in this case (and in general). | |
| Mar 21 at 3:46 | history | edited | cheshircat | CC BY-SA 4.0 |
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| Mar 21 at 3:16 | answer | added | Willie Wong | timeline score: 6 | |
| Mar 21 at 2:25 | comment | added | Willie Wong | By the way, the integral expression you write doesn't make sense in general. Given a distribution $v$, the composition $v(x(t))$ is only guaranteed to make sense if $x$ is sufficiently smooth at all $t$ such that $x(t)$ is in the singular support of $v$. I doubt that is the correct way to interpret what you want. | |
| Mar 21 at 2:19 | comment | added | Willie Wong | What you wrote cannot be a solution. You seem to be under the impression that $\int_{-\infty}^{\infty} \delta_1(x(t)) \omega(t) ~dt = \omega(1)$ for the particular choice of function $x$ that you wrote. But that is not the case. This integration should only factor in those values of $t$ for which $x(t) = 1$. Luckily for you, your function $x(t) = 1$ when $t = 0$, and $x$ is smooth near there (with derivative $1$). So in fact you have, for your given $x$, that $$ \int \delta_1(x(t)) \omega(t) ~dt = \omega(0)$$. And hence what you wrote is not a solution. | |
| Mar 20 at 23:15 | history | edited | cheshircat | CC BY-SA 4.0 |
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| Mar 20 at 23:08 | history | asked | cheshircat | CC BY-SA 4.0 |