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).
Best Answer
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 $\{r = 1\}$.