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Considering a vector field with a source and a sink in a finite comact space, are there any bounds on the length of the integral curves?

Specifically, I am interested in the average length of all the integral curves.

To make it easier, we could restrict ourselfs to everything being smooth and two dimensions.


Here a more physical motivation, so one can picture the problem better:

Say you have a finite volume with one inlet and one outlet. I have some physical gas in mind and the flow $\phi(t)$ through any fixed point is determined by the velocity field $X$. The average length would be associated with an average residence time of a particle in the volume. The geometry of that volume can be as you wish. I'd be interested in cylinders, but any sharp corners can be smoothed out and there are no loops.

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My routh idea is (and all the following equations are merely intuitive verbalizations of my thoughts) that the integral lines $\phi_s(t)$ parameterized in $t$ are a familay in $s$ (or more parameters in a higher dimensional space) and once the governing equation, say $\dot \phi_s(t)=X(x_{t,s},y_{t,s})$ is given the deviation from on integral curve $\phi_{s_0}(t)$ to the next $\phi_{s_0+\text ds}(t)$ must be determined by the field $X$. If $L_s=\int\phi_s(t)\text d t$ is the length of the flow line/integral curve with parameter $s$, then I assume the average length is $\langle L_s\rangle=\frac{\int\int\phi_s(t)\text d t\text d s}{\int\text d s}$.

What I can see, and what agrees with this are curves, which effectly don't depend on $s$. For example, if we consider a sphere or radius $R$ and a source at one point, a sink at its antipode and the flow lines along the longitude, then $\langle L_s\rangle=\frac{\int\int\phi_s(t)\text d t\text d s}{\int\text d s}=\frac{\int\text d s\int\phi_s(t)\text d t}{\int\text d s}=\int\phi_s(t)\text d t=\pi R$. The step $\frac{\int\int\phi_s(t)\text d t\text d s}{\int\text d s}=\frac{\int\text d s\int\phi_s(t)\text d t}{\int\text d s}$ should generalize to all curve for which the Lie-derivative along $s$ vanishes.

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It's quite off-topic, but I thought this could interest you anyway: there are flows on compact manifolds whose all orbits are periodic but whose periods are unbounded. (Denis Sullivan, A counterexample to the periodic orbit conjecture, Publ. IHÉS 46 (1976), an exquisite little paper). –  PseudoNeo Aug 16 '12 at 14:25
    
Your physical motivation is a little tricky, because the average length only gives you a bound on the average time if the magnitude of velocity is bounded from below. –  Rahul Aug 19 '12 at 23:12
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1 Answer 1

Given any number $M$, there exists a smooth vector field $X$ on $\mathbb S^2$ with one source at North pole and one sink at South pole, such that all trajectories have length at least $M$. Moreover, all derivatives of $X$ are bounded independently of $M$.

The idea is simple: in a neighborhood of equator $X$ is nearly parallel to the equator. This forces all trajectories to travel around the world many times before leaving the tropics. To be more specific, take your favorite smooth field that points toward South pole everywhere, and rotate each vector to west by $g-\epsilon$, where $g:\mathbb S^2\to [0,\pi/2]$ is a smooth function which is equal to $\pi/2$ near the equator and vanishes near the poles. As $\epsilon\to 0$, the length of trajectories tends to infinity.

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