Does curl vector influence the final destination of a particle? If we have an $n$ dimensional space ($n>3$) with a continuous $n$-dimensional vector field $\boldsymbol F$ 
$$\boldsymbol F:\mathbb{R}^n\rightarrow\mathbb{R}^n$$
and for every particle in this hyper-space,
$$d\boldsymbol x=\boldsymbol F dt$$
where $t$ is time.
We assume that the origin is a stable point which means
$$
\forall t'>t\\
\boldsymbol x(t)=\boldsymbol 0 ~~ \Rightarrow ~~\boldsymbol x(t')=\boldsymbol 0
$$
According to Helmholtz decomposition (Helmholtz decomposition extended to the higher dimensions as well as curl),
$$\boldsymbol F=-\nabla\Phi+\nabla\times\boldsymbol A$$
If under rule of $\dot x=-\nabla\Phi$, any particle at any location finally terminates to origin $\boldsymbol 0$
$$\dot x=-\nabla\Phi \Rightarrow \lim_{t\to\infty} \boldsymbol x=\boldsymbol 0$$
then, can someone conclude that, under rule $\dot x=\boldsymbol F$, the particles at any place terminate to the origin too?
$$\dot x=\boldsymbol F=-\nabla\Phi+\nabla\times\boldsymbol A \Rightarrow \lim_{t\to\infty} \boldsymbol x=\boldsymbol 0$$
In another term, does the curl part influence the final destination of the particle?
 A: The following seems to be a rather trivial counterexample:
Set $\Phi = \sum_{i=1}^n x_i^2$, so that $-\nabla\Phi = -2\sum_{i=1}^n x_i\partial_{x_i}$. Its flowlines are given by $e^{-2t}(x_1,\cdots,x_n)$, in particular the $t\to\infty$ limit of any flowline is $0$. For $1\le i,j\le n$ and $i\neq j$, define the $(n-2)$-form $A_{ij} = (x_i^3 - x_i^2)\mathrm dx_1\cdots\widehat{\mathrm dx_i}\cdots\widehat{\mathrm dx_j}\cdots\mathrm dx_n$, where the hats mean that we take the product of all the $\mathrm dx_k$ except for $k = i$ and $k = j$. Then $\mathrm d A_{ij} = \pm (3x_i^2 -
 2x_i)\mathrm dx_1\cdots\widehat{\mathrm dx_j}\cdots\mathrm dx_n$, so that the corresponding vector field is $\pm*\mathrm d A_{ij} = (3x_i^2 -
 2x_i)\partial_{x_j} =: V_{ij}$. We therefore see that $V_{ij}$ is a curl (which of course follows from $\nabla\cdot V_{ij}= 0$). Then 
$$
V = -\nabla\Phi + 2\sum_{i=1}^n V_{i+1,i} = \sum_{i=1}^n\big(6x_{i-1}^2 -4x_{i-1} - 2x_i\big)\partial_{x_i}
$$
with $V_{n+1,n}:= V_{1,n}$ and $x_{0}:= x_n$ fulfills all the conditions of your theorem. Since it vanishes at $(1,1,\cdots,1)$, the constant path at this point is a flowline of $V$, and in particular this flowline does not converge to $0$.
