The centralizer of Lienard equation Consider the lienard vector field $\cases{
x'=y -F(x)  \\
y'=-x }
$ in  $\mathbb{R}^{2}$,  where $F$ is  a polynomial fuction with $F(0)=0$. Assume that $Y$ is a smooth vector field globally defined on $R^{2}$ such that $[X,Y]=0$. 
can we conclude that $Y$ is tangent to solutions of lienard vector field?
Namely we ask: 
Does the lienard vector field has trivial centralizer?
The first motivation: Two vector fields $X$ and $Y$ with $[X,Y]=0$, have the same number of limit cycles. So this question help us to study the Lienard system, from the view of limit cycle theory.
The second motivation: When $F$ is an even polynomial, the system has a non isochrounous center. On the other hand a non isochronouse center has locally trivial centralizer/  
Can we use some methods of PDE for a possible solution to this question?
 A: This is not a complete answer, but it should give some elements towards it (especially regarding the link with PDEs). First, the Liénard system has a non-trivial centralizer in the case $F''=0$, i.e. $F(x)=cx$ for a real constant $c$, as the radial vector field belongs to it. This trival case put aside, write $Y=A\partial_x+B\partial_y$ so that $[X,Y]=0$ is equivalent to the system $$ X\cdot B=-A \\ X\cdot A = B-AF'$$ where $X\cdot f$ denotes the Lie (directional) derivative of $f$ along $X$. From this system you deduce that $B$ must satisfy the linear PDE $$X\cdot X\cdot B + (X\cdot B)F'+B=0.$$ Any solution $B$ to this PDE determines uniquely $A$, according to the first line of the above system. 
Let's write this neat PDE into its explicit (and ugly) form, namely (if my computations are correct) $$(y-F)(F\partial_{xx}^2B+x\partial^2_{xy}B+\partial_y B)+x(-x\partial_{xx}^2B+ F'\partial_y B+\partial_x B) = B.$$ Letting $x:=0$ gives $$y\partial_y B(0,y)=B(0,y),$$ that is (since $B$ is $C^\infty$) $$B(x,y)=cy+xb(x,y)$$ for some real constant $c$ and $C^\infty$ function $b$. Now one can play around with the equation satisfied by $b$, but that's besides the point. The point is that $B$ satisfies a second order linear PDE and an explicit initial condition $B(0,y)=cy$, so one might want to conclude that a solution $B$ exists for any $c$ and reach the conclusion that the centralizer is not reduced to tangent-to-$X$ vector fields (consider $c\neq0$). Unfortunately there are two catches:


*

*the line $\{x=0\}$ is a characteristic of the PDE (save for the case $F'(0)\neq0$), meaning that it is not a good choice for an initial value problem,

*also the PDE is of mixed-type, since its discriminant is $$\Delta=x^2 (y-F(x))(y+3F(x)),$$ in particular it is parabolic along the line $\{x=0\}$, which is an annoying fact regarding existence. This fact actually disallows the given argument to work when $F''=0$, so that the centralizer can be trivial.


I'm no specialist of PDEs, so I don't have anything smarter to suggest, but it seems that the approach via general PDEs might prove difficult.
Yet, the fact that the PDE is not any PDE, but comes from a Lie derivative applied twice, might help. Notice in particular that the integral of $A\tau$ and $(B-AF')\tau$ on any tangent cycle must vanish, where $\tau$ is any $1$-form such that $\tau(X)=1$ (a time form of $X$), thanks to the Stokes-like formula (valid for any tangent curve $\gamma$): $$\int_\gamma (X\cdot F) \tau = F(\gamma(1))-F(\gamma(0)).$$ This should give mild constraints on $A$ and $B$.
Taking all these elements into account, my bet is that the case $F''=0$ is the only one for which the centralizer is not trivial. If this were indeed the case, it wouldn't help you count limit cycles of the general Liénard system, I'm afraid…
