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If $V_r$ be a vector field defined on $S^2_r$ that is always tangent to the sphere on which it is defined.Define a vector field $V$ on $R^3$ such that $V(x)$=$r^2(1-r^2)$$V_r(x)$.Prove that for each $t \in R$ the time $t$ flow associated to this vector field is defined.

Ester
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  • But there is no smooth vector field on $S^2$: http://en.wikipedia.org/wiki/Hairy_ball_theorem – Michael Hoppe Oct 20 '13 at 10:42
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    @MichaelHoppe: There are plenty of smooth vector fields on $S^2$. There are no smooth non-vanishing vector fields on $S^2$. – Jason DeVito - on hiatus Oct 20 '13 at 11:19
  • Yes, you're right, of course. – Michael Hoppe Oct 20 '13 at 11:22
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    Here's a hint for the problem. Let $\gamma$ denote the flow through some point. Prove $|\gamma|$ is constant, so the flow lies on whatever sphere bout the origin it starts on. In particular, since the sphere is compact, flows on it last for all time. – Jason DeVito - on hiatus Oct 20 '13 at 11:27
  • @Ester: The idea seems to be correct, but I'm not sure about a couple of the details: I don't understand how you go from $x_0$ to $(0,x_0)$. Further, you need to prove the the curve lies on a sphere - just because it starts on a sphere doesn't mean it lies on a sphere for all time. I also don't understand your last question - to me, the flow is nothing but the collection of all integral curves. – Jason DeVito - on hiatus Oct 24 '13 at 13:15
  • @Jason:How will I show that |gamma| is 1? – Ester Oct 24 '13 at 13:45
  • Your notation is somewhat fishy. Does $V_r$ live only one particular sphere of radius $r$? Then what does $V_r(x)$ mean for arbitrary $x\in{\mathbb R}^3$? What does the factor $r^2(1-r^2)$ ("here $r$ is varying") mean in connection with $x$? – Christian Blatter Oct 24 '13 at 14:01
  • @ChristianBlatter:Sorry for the confusion.I have edited. – Ester Oct 24 '13 at 14:13
  • @Ester: Start with $f(t) = |\gamma(t)|^2$. Then $f'(t) = 2\langle \gamma(t), \gamma'(t)\rangle = 2\langle \gamma(t), V(\gamma(t))\rangle = 2\langle \gamma(t), f(t)(1-f(t)) V_r(\gamma(t))$. Since $V_r(\gamma(t))$ is tangent to the sphere, what do you know about this? – Jason DeVito - on hiatus Oct 24 '13 at 17:26
  • @JAson:Thanks for your ideas and corrections. – Ester Oct 25 '13 at 06:38
  • @Ester: You might consider writing up your answer here. It might save you some bounty points ;-). Glad I could help! – Jason DeVito - on hiatus Oct 25 '13 at 12:49

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