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I am thinking about smooth vector fields on some (open set of an) euclidean space $\mathbb{R}^n$.

I know that the integral curves of a general vector field $X$ are not defined for every time $t\in \mathbb{R}$. A simple example is given by $X=x^2 \partial _x$ on $\mathbb{R}$, whose integral curve emanating at $t=0$ from some $x_0 >0$ is given by $\gamma(t) = \frac{x_0}{1-tx_0}$. This curve is defined only on $]-\infty , \frac{1}{x_0}[$; moreover as $t$ ranges in that set we have $\gamma (t)\in ]0,+\infty[$.

I would like to see an example of a vector field $X \in \mathfrak{X}(\mathbb{R}^n)$ such that the integral curve emanating at $t=0$ from some point $p$ is defined only for bounded times $t\in]-T,T[$ $ \ $ (with $T<+\infty$) and also remains bounded "in space", i.e. there exist a compact $K\subset \mathbb{R}^n$ such that $\gamma (t) \in K, \ \ \forall t \in ]-T,T[$. Can anyone provide some example or hint to build one?

Edit from comments: it has been pointed out that, by the Escape Lemma, such a vector field cannot exist, because if the maximal domain is not the whole $\mathbb{R}$, then the curves are forced to "escape" any compact set.

So let me slightly modify my question: in the above notations, let $T=+\infty$, so that the maximal domain is $\mathbb{R}$ and the integral curves can a priori be bounded. The first example I can think of is the following:

consider $X=x\partial _y -y\partial _x \in \mathfrak{X}(\mathbb{R}^2)$ and $p=(1,0)$. Then the integral curve starting at $p$ is just $S^1$ and $X$ can be pictured as its tangent counter-clockwise unit vector field.

This would answer my question, but this is not what I was really looking for, because this curve is not simple (it's periodic indeed), so it's defined for every time and bounded, but let me say in a quite trivial way.

So what I'm actually looking for is this: a vector field $X$ that at some point has a simple integral curve (i.e. injective as a map $\gamma : \mathbb{R} \to \mathbb{R}^n$) which is bounded in a compact set $K$ (and so globally defined, by the Escape Lemma). Is this possible? (maybe there is another result I don't know which proves this cannot be the case).

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Maybe I'm forgetting (it's been years since I took manifolds), but isn't this exactly what the Escape Lemma says cannot happen? Concretely, that if the maximal domain of an integral curve is not all of $\mathbb{R}$, then it must escape to infinity (i.e. it cannot lie in any compact set). –  Matt Jan 23 '13 at 17:30
@Matt : you are right! I didn't know this lemma, thank you. I've edited my question accordingly. –  Lor Jan 23 '13 at 19:13
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up vote 2 down vote accepted

How about a vector field in the plane with a stable limit cycle?

$$ \frac{dr}{dt} = r(1-r) \\ \frac{d \theta}{dt} = r $$ If I'm not mistaken, this is a continuous vector field with an unstable singularity at the origin such that any initial condition in the punctured disk $\{1 > r > 0\}$ tends toward the periodic orbit $\{r = 1\}$.

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This is what I was looking for, so I tried to carry out the computations. It turns out (if I'm still able to solve ODEs) that the integral curve starting from $(r_0,\theta _0)$ is given by $r(t)=\frac{r_0 e^t }{1-r_0 +r_0 e^t }$ and $\theta (t)= \theta _0 + ln (1-r_0 +r_0 e^t)$. So this is a complete vector field with integral curves bounded in the unit ball and tending towards the special orbits in $0$ and $S^1$ as $t \to -\infty$ or $+\infty$. Thank you very much! –  Lor Jan 31 '13 at 11:58
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How about this flow in $\mathbb{R}^3$: $$\begin{align} \dot{x} &= -\beta y + 2zx\\ \dot{y} &= \beta x + 2zy\\ \dot{z} &= 4 - x^2 - y^2 + z^2 \end{align} $$ If one embed $\mathbb{R}^3$ into $S^3 \subset \mathbb{R}^4 \sim \mathbb{C}^2$ through the mapping: $$\begin{align} (x,y,z) &\to (X,Y,Z,W) = (\frac{x}{1+r^2/4},\frac{y}{1+r^4/4},\frac{z}{1+r^2/4},\frac{1-r^2/4}{1+r^2/4})\\ &\to (U,V) = (X+iY, Z+iW) \end{align} $$ the above flow can be rewritten as: $$\begin{align} \dot{U} &= i\beta U\\ \dot{V} &= -4 i V \end{align} $$ This flow is a rotation in $U$ direction with speed $\beta$ and $V$ direction with speed $-4$. If $\beta$ is irrational and $|U|, |V| \ne 0$, the flow line will not repeat and fill the surface of a torus.

This flow in $\mathbb{R}^3$ is possible because of the famous Hopf fibration of $S^3$ by $S^2$. If you want to have a feeling how the torus are nested, the wiki page of Hopf fibration is a possible start.

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This example is really nice, thank you! Let me see if I get the geometry of this. The integral curves starting from $(U_0,V_0)$ are given by $(U(t),V(t))=(U_0 e^{i\beta t},V_0 e^{-i4t})$, so they live on a torus in $S^3$. If I take $|U|=|V|=1$, isn't this the canonical Heegaard surface of genus 1 for $S^3$? –  Lor Jan 31 '13 at 12:53
Do you mean $|U| = |V| = \frac{1}{\sqrt{2}}$? It is clearly a torus and acted as a Heegaard surface of genius 1. I don't know whether it is the canonical one you referred to. I've only heard of their names but never studied them. –  achille hui Jan 31 '13 at 13:26
sure, I was thinking about unit vectors, but I haven't written it properly. Anyway this is what I mean with "canonical Heegaard surface of genus 1 for $S^3$", so I'm really happy with this example, thanks again! –  Lor Jan 31 '13 at 14:07
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