Timeline for WhenEvent&NDSolve: How to detect saddle point?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 7, 2019 at 18:55 | comment | added | Michael E2 |
@Ulrich, if you put a space between the @ and the Michael, I don't get notified of your response. -- BTW, the use of sgn as a discrete variable is similar to how NDSolve implements Sign and other discontinuous functions.
|
|
| Jul 7, 2019 at 12:20 | comment | added | Ulrich Neumann | @ ChrisK From the physical expectation the "Tanh version" gives more pausibel results from the engineering point of view. | |
| Jul 7, 2019 at 12:18 | comment | added | Ulrich Neumann |
@ Michael E2 Never ending story... I was looking for an analytic solution of the "Sign-ode" to verify NDSolve solution. That's why I formulated the "sgn-question".
|
|
| Jul 7, 2019 at 8:28 | comment | added | Chris K |
One more thought. There's one other atypical thing about your system that results in the continuous range of equilibria (a degenerate result). That's the fact that the x-isocline of v=0 perfectly overlaps the v-isocline given by that line segment. That explains why the Tanh approximation in your previous question leads to a unique outcome -- it slightly bends the "v-isocline" to result in a unique equilibrium point.
|
|
| Jul 6, 2019 at 21:43 | comment | added | Michael E2 |
@UlrichNeumann I believe it's the same as the problem here. I don't see how to use sgn to get around the issue with Sign. (I personally am convinced that Sign is the right way to go, and the linked problem gives the correct solution.)
|
|
| Jul 6, 2019 at 15:04 | comment | added | Chris K | I imagine NDSolve doesn’t like it when the right hand side is discontinuous like that, but the phase plane convinces me that this solution is basically correct. | |
| Jul 6, 2019 at 14:25 | comment | added | Ulrich Neumann |
No I meant something like Plot[-v'[t] + 1 - x[t] - .05 Sign[v[t]] (1 + x[t])/.sol,{t,...}]
|
|
| Jul 6, 2019 at 14:20 | comment | added | Chris K |
No worries! Plot[Evaluate[x'[t] /. sol], {t, 0, 25}] looks like I'd expect -- oscillating until it gets trapped around t=21, then flat. Is that what you meant?
|
|
| Jul 6, 2019 at 14:15 | comment | added | Ulrich Neumann | If you substitude the numerical solution found by NDSolve into the ode you will get a time dependend expression . That's what I call residuum(sorry faulty translation...) and which is called residual! Thanks for your time and effort. | |
| Jul 6, 2019 at 14:09 | comment | added | Chris K | Could you explain more about what you meant by the residuum of the solution? | |
| Jul 6, 2019 at 13:43 | comment | added | Ulrich Neumann | That's the point where I'm struckling: If you check the residuum of the solution you'll see a value near to zero for t<21, but for t>21 you see nonvanishing switching constant values. Does that mean the numerical solution doesn't approximate the ode for t>21??? | |
| Jul 6, 2019 at 13:19 | comment | added | Chris K | No prob. I think it could end up anywhere on that line segment from x=0.905 to 1.105. | |
| Jul 6, 2019 at 13:09 | comment | added | Ulrich Neumann |
Thank you very much for your helpful comprehensive answer. With my attempt I tried to verify the solution of NDSolve of the ode with Sign (expecting a asymptotic solution x[t]==1). I need some time to elaborate your answer...
|
|
| Jul 6, 2019 at 9:56 | history | edited | Chris K | CC BY-SA 4.0 |
added 18 characters in body
|
| Jul 6, 2019 at 9:26 | history | edited | Chris K | CC BY-SA 4.0 |
added black segment to last plot
|
| Jul 6, 2019 at 9:07 | history | answered | Chris K | CC BY-SA 4.0 |