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I was wondering if there is a physical interpretation of ODEs of the form $$\frac d{dt}\vec x(t)=\vec y(t)$$ $$ \frac d{dt} \vec y(t) = \nabla \times \vec F(x(t))$$

(or equivalently $\frac {d^2} {dt^2}\vec x(t) = \nabla \times \vec F(x(t))$) where $x(t)\in \mathbb R^3$.

In case, is there an explicit formula for the solutions? And what are the properties of the system? Any reference is appreciated.

I came up with this equation, thinking about motion under a force field which is not conservative, since any (sufficiently regular) vector field $A = \nabla\times\xi+ \nabla f$ by Helmholtz, and $\nabla f$ gives a conservative force I thought what happens if we take the other bit?

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  • $\begingroup$ Magnetic field is a curl of vector potential, so Lorentz force could be written as something similar (though there is also velocity in there). Other place to look it is hydrodynamics and all kinds of vortices. $\endgroup$
    – Roger V.
    Commented May 11, 2022 at 14:56
  • $\begingroup$ Note that your system is not hamiltonian ($\dot x = p= \partial_p H$ is OK but the other equation does not satisfy $\dot p= -\partial_x H$), which makes me think that the Liouville theorem does not hold: phase space volume is probably not conserved. Very interesting question. $\endgroup$
    – Quillo
    Commented May 11, 2022 at 15:12
  • $\begingroup$ @Quillo thanks Indeed, it has been constructed expressly for not being conservative. $\endgroup$
    – Overflowian
    Commented May 11, 2022 at 15:14
  • $\begingroup$ Do you have a particular system in mind? I have found this paper: arxiv.org/abs/1906.04476v1 but it seems to be about $\dot x = \nabla \times F$, not $\ddot x = \nabla \times F$. $\endgroup$
    – Quillo
    Commented May 11, 2022 at 17:00
  • $\begingroup$ @Quillo that system looks very different, if $F$ is constant, then the solutions to $\dot x = \nabla \times F$ are constant while to $\ddot x = \nabla \times F$ they would be lines. More generally there should be the same difference that there is between a gradient system $\dot x = -\nabla V$ and an Hamiltonian system $\ddot x = - \nabla V$. $\endgroup$
    – Overflowian
    Commented May 11, 2022 at 17:16

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