Timeline for Sufficient condition to be increasing, following a vector field
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Apr 24, 2022 at 14:03 | history | edited | Amir Sagiv | CC BY-SA 4.0 |
ode tag + english
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| Apr 9, 2022 at 1:53 | comment | added | leo monsaingeon | You're right. So, modulo the slight issue of the regularity at the origin, your $v_1$ gives a counterexample to your conjecture and the answer to your question is NO: just take $v^2$ to be the same as $v^1$ but with opposite rotation in the $(x,y)$ plane. Starting at any initial position $(x_0,0,z_0)$ with $x_0<0$ gives a stationary point, since there $(v^1+v^2)=(0,0,2+2x/\sqrt{x^2+y^2})$ vanishes (hence $f$ is trivially not increasing). For the record: I initially thought the question was completely trivial, now I see it's not and deserved at least some care. I unvoted to close. | |
| Apr 8, 2022 at 10:17 | comment | added | G. Panel | Your assumption that $\nabla f\cdot v^1>0$ seems false to me in the general case (for $v^1:\mathbb{R}^3-\{0\}\ni (x,y,z) \longmapsto (y,-x,1+x/\sqrt(x^2+y^2)$, $(z_t)$ is increasing for any initial condition, while $\nabla z\cdot v^1(-1,0,1)=0$. Sorry, $v^1$ is not defined here on the whole $\mathbb{R}^3$). Did I understand your remark correctly? | |
| Apr 8, 2022 at 0:55 | comment | added | leo monsaingeon | Yes, because your assumptions simply mean $\nabla f\cdot v^1>0$ and $\nabla f\cdot v^2\geq 0$ everywhere. Hence for any solution $x(t)$ of your ODE you have $\frac{d}{dt}f(x_t)=\nabla f(x_t)\cdot \dot x_t=\nabla f(x_t)\cdot [v^1(x_t)+v^2(x_t)]>0$. This does not really belong to MO, vote to close and crosspost to SE. | |
| Apr 7, 2022 at 22:35 | review | Close votes | |||
| Apr 24, 2022 at 15:01 | |||||
| Apr 7, 2022 at 19:48 | history | asked | G. Panel | CC BY-SA 4.0 |