Timeline for Inverse problem of the calculus of variations for autonomous second-order ODEs
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 1 at 16:14 | vote | accept | A. J. Pan-Collantes | ||
| Aug 31 at 19:23 | answer | added | Robert Bryant | timeline score: 4 | |
| Aug 30 at 19:28 | comment | added | Robert Bryant | I guess the problem is that your $L$ is not even $C^1$ where $\dot q = 0$. We usually require the Lagrangian to be $\dot q$-convex, i.e., $L_{\dot q\dot q}$ exists everywhere and is nonzero. If you allow time dependence, this problem goes away: $L = \mathrm{e}^t\dot q^2$ is smooth and convex. | |
| Aug 30 at 13:23 | comment | added | A. J. Pan-Collantes | @RobertBryant Sorry, I copied it wrong from my notes. It is $L=-q+\dot{q} \ln(\dot{q})$. The Euler-Lagrange eq is $\frac{\ddot{q}}{\dot{q}}=-1$ | |
| Aug 29 at 7:27 | history | edited | A. J. Pan-Collantes |
added tags
|
|
| Aug 29 at 7:22 | comment | added | A. J. Pan-Collantes | @RobertBryant I must have misunderstood something very silly. Why is not a valid Lagrangian $L=-q+\ln(\dot{q})$? Its E-L equation is $\ddot{q}=\dot{q}$, if I am right... | |
| Aug 27 at 17:06 | comment | added | Robert Bryant | You might want to have a look at mathoverflow.net/questions/379946/…. In particular, the equation $\ddot q + \dot q = 0$ does not have a Lagrangian that is independent of time, but it is the E-L equation of the functional $\int \mathrm{e}^t {\dot q}^2\, \mathrm{d}t$. | |
| Aug 27 at 16:14 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor Math Jaxing (formula hyperlinking)
|
| Aug 27 at 16:06 | history | asked | A. J. Pan-Collantes | CC BY-SA 4.0 |