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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?

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    $\begingroup$ If $n>3$ then what is the curl? $\endgroup$ Jan 13, 2018 at 12:37
  • $\begingroup$ @LordSharktheUnknown, I am not expert in math, but I just know that Helmholtz decomposition is extended to the higher dimension as well. $\endgroup$
    – ar2015
    Jan 13, 2018 at 12:38
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    $\begingroup$ This question may be better posed to physics SE $\endgroup$
    – Myridium
    Jan 13, 2018 at 12:53
  • $\begingroup$ @Myridium, I am afraid that they give me -1000 and then kick me back as I am after higher dimensions ($n>3$) $\endgroup$
    – ar2015
    Jan 13, 2018 at 12:58

1 Answer 1

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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 $0$.

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  • $\begingroup$ I think by $V$ you mean $F$. There are also two unclear points for me. What and how does vanish at $(1,1,\dots,1)$? $\endgroup$
    – ar2015
    Jan 24, 2018 at 5:10
  • $\begingroup$ Yes, you call this variable $F$. All of the components of the vector field $V$ vanish at $(1,\cdots,1)$ since $6\cdot 1^2 - 4\cdot 1 - 2\cdot 1 = 0$. $\endgroup$
    – Bertram
    Jan 24, 2018 at 7:39
  • $\begingroup$ Tried to plot it. But, $x_{i-1}$ does not exist for when $i=1$. $\endgroup$
    – ar2015
    Jan 25, 2018 at 6:10
  • $\begingroup$ I set $x_0:= x_n$; in fact any $x_i$ with $i\neq n$ will do. The point is that any vector field $f(x_j)\partial_{x_i}$ with $i\neq j$ has zero divergence and thus is a ``curl''. You can add this to the gradient vector field $-\nabla\Phi$ and force the result to have a zero somewhere which will give a stationary flowline. $\endgroup$
    – Bertram
    Jan 25, 2018 at 15:34
  • $\begingroup$ This is true. But the point (1,1,...,1) is an unstable stationary point. Mathematica: VectorPlot[{6 y^2 - 4 y - 2 x, 6 x^2 - 4 x - 2 y}, {x, 0, 2}, {y, 0, 2}]. Is there any example where the point is stationary as well? $\endgroup$
    – ar2015
    Jan 26, 2018 at 0:59

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