0
$\begingroup$

Let $f\in\mathcal{C}^1(\mathbb{R}^n,\mathbb{R})$ $(n\geq 1)$ be an observable, and let $v^1,v^2\in\mathcal{C}^1(\mathbb{R}^n,\mathbb{R}^n)$ be two vector fields such that for any $(x^1_t)_{t\geq 0}$ and $(x^2_t)_{t\geq 0}$ solutions of

\begin{align} & \dot{x}^1_t=v^1(x^1_t) \\ & \dot{x}^2_t=v^2(x^2_t), \end{align}

$\big(f(x^1_t)\big)_{t\geq 0}$ is increasing, and $\big(f(x^2_t)\big)_{t\geq 0}$ is non-decreasing.

Question: for any $(x_t)_{t\geq 0}$ solution of $\dot{x}_t=(v^1+v^2)(x_t)$, is $\big(f(x_t)\big)_{t\geq 0}$ increasing?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$
    – leo monsaingeon
    Commented Apr 8, 2022 at 0:55
  • $\begingroup$ 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? $\endgroup$
    – G. Panel
    Commented Apr 8, 2022 at 10:17
  • 1
    $\begingroup$ 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. $\endgroup$
    – leo monsaingeon
    Commented Apr 9, 2022 at 1:53

0

Reset to default

You must log in to answer this question.