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?



The centralizer of Lienard equation
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…


