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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?

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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…

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  • $\begingroup$ thank you very much for your answer. I will read it and concentrate on your guess that $F=cx$ is the only nontrivial centralizer. $\endgroup$ Apr 14 '14 at 19:24
  • $\begingroup$ @AliTaghavi: well, I may be wrong! But the heuristics is as follows: transverse commuting vector fields give Lie symmetries and improve integrability (by quadrature on the underlying differential equation $\frac{dy}{dx}=\ldots$). But it seems to me that the system is not integrable (in that sense) when $F$ is "too complicated". Yet I'm no specialist of Liénard systems, I'd rather trust you on that subject ;) $\endgroup$ Apr 14 '14 at 19:29
  • $\begingroup$ Your second order PDE approach is very interesting. It reminds me of the following situation. some years ago I tried to approach to limit cycle of Lienard equation with this method: Assume $(x(t), y(t))$ is a periodic solution. then $x'(t), y'(t))$ is a periodic curve. Put new variable $X=x' , Y=y'$ then (X(t), Y(t)$ satisfies a new authonomous differential equation in X_Y plane. But unfortunatly this new vector field is topological equivalent to the original lienard vec. field!!!(A very bad chance, a mathematical loop). But this is a motivation to th followin question: $\endgroup$ Apr 16 '14 at 14:42
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    $\begingroup$ @AliTaghavi: no, it's not true. Take the linear center case $P(x,y)=y, Q(x,y)=-x$. Then $X\cdot P=Q$ and $X\cdot Q=-P$ so that you obtain a multiple of the radial vector field. $\endgroup$ Apr 16 '14 at 15:04
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    $\begingroup$ @AliTaghavi: the trick you mentionned with the vector field obtained by looking for an autonomous sytem satisfied by $\dot \gamma$, seems to work only in that specific case (or other «simple» cases). Otherwise you can't get rid of the variables $x, y$ in the expression of $\ddot x=\dot y\frac{\partial P}{\partial x}(x,y)+\dot y\frac{\partial P}{\partial y}(x,y)$ and the same for $\ddot y$. $\endgroup$ Apr 16 '14 at 15:22

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