Timeline for Which higher-order terms require 4th-order integration of quadratically-constrained dynamics?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jun 25 at 18:47 | comment | added | Not a chance | Thank you. I apologise if my question wasn't asking the thing I wanted to know exactly, but for me it was the behaviour of the integrator in the face of simple equations/constraints that surprising. I didn't see the problem with the velocity dependence on the velocity and for me this is the crux of the issue. I'm going to give the 'correct' answer to LPZ because I think he/she has explained what the cause is. | |
| Jun 23 at 16:40 | comment | added | Kyle Kanos | You have 3 questions between the title and the body; 2 of them are explicitly about the generic case I address in this answer (to wit: the title and last question). If you want a more stable integrator that involves velocity-dependent forces, you could see this answer of mine. | |
| Jun 23 at 7:52 | comment | added | Not a chance | Thanks. I added an image to show what I am talking about. My question is not about these problems in general, it is about this one. This one is a non-stiff, non-dissipative system. For me it clearly needs a symplectic integrator. My question is really directed at the second-order ODE with quadratic constraints. I would expect this to be solved by a quadratic integrator but clearly the integration is missing some important high-order terms. Eliminating z is an option, but I have essentially eliminated the constraint instead. As an exercise I find the problem interesting as stated. | |
| Jun 22 at 14:58 | comment | added | Kyle Kanos | Alternatively, since constraints reduce the degrees of freedom, you can use $g=0$ to imply that $z=\pm\left(r^2-x^2-y^2\right)^{1/2}$ to eliminate $z$ from the system of equations. | |
| Jun 22 at 14:57 | comment | added | Kyle Kanos | As an aside: since the constraint force in this case must be perpendicular to the surface, $\nabla g\cdot\dot{\mathbf{x}}=0$, then you could write $\ddot{\mathbf{x}}\cdot\dot{\mathbf{x}}=0$ which may be a more tractable way to solve the problem. | |
| Jun 22 at 14:14 | history | answered | Kyle Kanos | CC BY-SA 4.0 |