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Suppose I am given a vector field $X$ over a two-dimensional Euclidean Space. Suppose that the equation $\{f=0\}$ (which defines a curve in the plane) is the equation of a orbit of the vector fields $X$.

What does this means in terms of $X(f)$ I am guessing that the set $\{f=0\}$ must be $X$-invariant, therefore $f$ satisfies $X(f)=p\cdot f$, $p$ being a constant. is this right?

EDIT: I should also add that $f$ is an irreductible polynomial! We can also consider that $X$ is a polynomial field of the form $X=px\partial_x+qy\partial_y$ where $p$ and $q$ are positive integer which are relatively primes.

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It should be zero. Take the 2 dimensional plane $X= \partial / \partial x$. Then the orbits are horizontal lines. Take $f(y) : y^{2}= 0 $ then $X(f) = 0$. Unless I am misunderstanding what you mean by an orbit. – DBS Jul 26 '13 at 6:35
Humm... ok. I need to add some context. – OhMyGod Jul 26 '13 at 6:37
up vote 2 down vote accepted

If orbit belongs to some curve $f(x, y) = \rm{const}$, then at least you could obtain by differentiantion that $(\nabla f, X) = 0$ along this curve. It's easy to show that by chain rule: suppose that $(x(t), y(t))$ is a trajectory that belongs to curve $f(x, y)= \rm const$, then $f(x(t), y(t)) = \rm{const}$ and $\frac{\partial f}{\partial x} \cdot x'_t + \frac{\partial f}{\partial y} \cdot y'_t = 0$ which can be rewritten as $(\nabla f, X)$.

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I assume you are writting the vector field as $X=(x',y')$. Is that right? – OhMyGod Jul 26 '13 at 6:46
Also that $f'_x=\dfrac{\partial f}{\partial x}$; in this case you should exclude the ' from $f'$ – OhMyGod Jul 26 '13 at 6:48
Thanks, I've corrected notation. I meant that derivative of solution w.r.t. to $t$ coincides with vector field $X$ along the solution. – Evgeny Jul 26 '13 at 6:53
Yes, I got it now. Thank you very much. – OhMyGod Jul 26 '13 at 6:54

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