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Consider an open set $\Omega$ with smooth boundary and some smooth time-depending divergence-free vector field $u:[0,T]\times \Omega \rightarrow \mathbb{R}^d$, with $d=2$ or $3$, satisfying $u_{|\partial\Omega}=0$.

Intuitevily, because of the divergence-free condition and the vanishing at the boundary, I would guess that the vector field $u$ has to "loop" in some sense (it's even clearer in dimension $2$).

For $x\in \Omega$, I call $\mathscr{C}_x$ the trajectory issued from $x$, that is the solution of the Cauchy problem \begin{align*} Z'(t) &= u(t,Z(t)) \\ Z(0)&=x. \end{align*} My question : what can be said about the set of all $x\in \Omega$ such $\mathscr{C}_x$ is closed ? Is it always non-empty ? Can we prove that for every $a\in\Omega$, $a$ may be encircled by a trajectory of this kind ? What if I replace $u$ by $u \star \varphi$, where $\varphi$ is some smooth regularizing kernel ?

Thanks for any help !

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I don't see how you can expect periodic orbits from a time-dependent vector field which is not itself periodic. Even simple rescaling $f(t)\vec v(x)$ can destroy all periodic solutions by virtue of $f$ being non-periodic. – user31373 Aug 7 '12 at 2:58
You are completely right. I meant " closed " (i.e. $Z(t)=Z(s)$ , for two different times $s\neq t$) curve along which the field $u$ is tangent. In the case were $u$ does not depend on time, these are periodic solution of the Cauchy problem, hence my confusion. I found the term "circle cell" in a book of Showang, I will to correct it above. – xounamoun Aug 8 '12 at 15:34

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