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Let $F : \mathbb{R}^n \to \mathbb{R}^n$ be a smooth mapping and consider the following autonomous ODE \begin{equation} y'(t)=F(y(t)) \end{equation} with the initial data $y(0)=x \in \mathbb{R}^n$.

Further assume that there exists a global unique solution for each $x$ and dependence of the solution on the initial data $x$ is smooth.

That is, we have a unique, jointly smooth mapping $y(t,x) : [0,\infty) \times \mathbb{R}^n \to \mathbb{R}^n$ which solves the above ODE.

Now, I fix $t$ and consider $y(t, \cdot ) : \mathbb{R}^n \to \mathbb{R}^n$ as a smooth vector field on $\mathbb{R}^n$.

In this case, what additional conditions on $F$ ensure that $y(t, \cdot ) : \mathbb{R}^n \to \mathbb{R}^n$ is a conservative vector field? It means that line integral of $y(t, \cdot )$ on any piecewise smooth closed curve in $\mathbb{R}^n$ is always zero.

Could anyone please provide any insight?

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  • $\begingroup$ Do you know examples where $F$ is conservative but the flow $x \mapsto y(t,x)$ is not? $\endgroup$
    – Giorgio Metafune
    Commented Jun 24, 2023 at 8:21
  • $\begingroup$ No, I don't . Could you provide more information? $\endgroup$
    – Isaac
    Commented Jun 24, 2023 at 17:56
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    $\begingroup$ Just some thoughts. If $z^i$ is the vector $\frac{\partial y}{\partial x_i}$ and $z$ is the corresponding matrix, then for any fixed $x$, $z'(t)=JF (y(t,x))z(t), z(0)=I$. Assuming that $F=\nabla \phi$ then $z$ satisfies the linear equation $z'=A(t)z, z(0)=I$ with $A(t)=D^2\phi(t,x)$. If the hessian matrices $D^2\phi$ at different points commute, then $A(t)A(s)=A(s)A(t)$ and $z(t)=\exp (\int_0^t A(s)\, ds)$ is symmetric which gives $\frac{\partial y_i}{\partial x_j}=\frac{\partial y_j}{\partial x_i}$. This should give the result for commuting Hessians but I do not know counterexamples. $\endgroup$
    – Giorgio Metafune
    Commented Jun 24, 2023 at 18:25
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    $\begingroup$ The condition that $F$ is a gradient is however necessary because for small $t$, $y(t,x) \approx x+tF(x)$. $\endgroup$
    – Giorgio Metafune
    Commented Jun 24, 2023 at 18:28

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