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Let $X$ be a polynomial vector field of degree $2$ on $\mathbb{R}^{2}$. Does there exist a nonvanishing smooth function $g$ such that $Div(gX)$ is a proper map?Or at least the zero locus of $Div(gX)=0$ is a compact set? Can we find this $g$ an algebraic map or at least in the form $e^{P(x,y)}$ where $P$ is a polynomial?

Is there a uniform upper bound $PDH(n)$ for the number of limit cycles of a those polynomial vector field of degree $n$ for which the divergence is proper or at least $Div=0$ is a compact curve. For $n=3$, two what extent thses vector field are classified in term of their coefficients

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  • $\begingroup$ @PietroMajer Thanks for the comments. Could you please more explain. for example the div of $(x^{2}+cy^{2})\partial_{x}+y^{2}\partial_{y}$ is not proper. $\endgroup$ – Ali Taghavi Dec 17 '14 at 5:12
  • $\begingroup$ yes, now I see the point $\endgroup$ – Pietro Majer Dec 17 '14 at 6:48
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Since $\mathbb R^2$ has one end and $\mathbb R^1$ has two ends, a proper map $\mathbb R^2 \to \mathbb R^1$ must send the end of $\mathbb R^2$ to one of the two ends of $\mathbb R^1$ - that is, $f(x,y)$ is either a large positive number of a large negative number when $(x,y)$ is large.

But we can force $\nabla \cdot gX$ to oscillate arbitrarily many times. Choose a vector field like $ \sin(x^2+y^2)( xdx + ydy)$. Then the integral of the divergence over a disc is the integral of the vector field dot the normal vector over a circle, which switches sign periodically in the radius of the function, so the divergence must switch sign arbitrarily often, so it can't be proper.

Moreover because it switches sign with arbitrarily large radius, the set where it is 0 cannot be compact.

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  • $\begingroup$ Thank you for your answer. But I can not see why this is the answer to the first part of my question.Could you please more explain. $\endgroup$ – Ali Taghavi Dec 17 '14 at 5:17

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