# Vector fields whose divergence are proper maps

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

• @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. – Ali Taghavi Dec 17 '14 at 5:12
• yes, now I see the point – Pietro Majer Dec 17 '14 at 6:48

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.