Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
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
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 $\{r = 1\}$.

share|cite|improve this answer
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

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.