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Let $X=P\partial_x+Q\partial_y$ be a vector field on the plane $\mathbb{R}^2$. Assume that we have :$$P_xP_y+Q_xQ_y=0$$ Does this imply that the vector field $X$ is a divergence-free vector field with respect to a Riemannian metric defined out of singularities of $X$? Obviously this would imply that any $X$ with $P_xP_y+Q_xQ_y=0$ can not have any limit cycle. But is it realy the case?

Namely, does the equality $P_xP_y+Q_xQ_y=0$ imply that $X=P\partial_x+Q\partial_y$ does not have any limit cycle?

Some approaches and motivations I have examined:

Main Motivation: To our vector field $X=P\partial_x+Q\partial_y$ we associate the complex function $X=P(z)+iQ(z)$. Then we consider the complex dilatation of $X$ with $\mu(z)=\bar\partial X(z)/\partial X(z)$.

Let $\phi$ be the flow of the vector field $X$. So with a similar argument for proof of the standard variational equation one obtain that $$\begin{cases}(\partial \phi_t)'=\partial X(\phi_t(x))\partial \phi_t+\bar \partial X(\phi_t(x)) \partial \bar \phi_t\\ (\bar \partial \phi_t)'=\partial X(\phi_t(x)\bar \partial \phi_t+\bar \partial X(\phi_t(X))\overline{\partial \phi_t} \end{cases}$$

This imply that $\mu(\phi_t)'(x)=\frac{\bar\partial X(\phi_t(x))}{(\partial \phi_t)^2}exp\int_0^t div X(\phi_t (x))dt$

I observe that:

Simple Fact: For a vector field $X=P\partial_x+Q\partial_y$ if the dilatation $\mu(X(z))=\lambda$ is a constant map in $z$ with $|\lambda|=1$ then $X$ does not have any limit cycle.

Proof:
Assume that $\mu(X(z))=\lambda$ for a fixed complex number $\lambda$ with $|\lambda|=1$ . So after a linear change of coordinate $H(z)=(\beta )z $ where $\beta$ is a complex number with $\beta^2 \lambda=1$ then we have $H^* X(z)=\beta X(z/\beta)$. Since $H$ is a linear function then it is equal to its linear part. Since both $H$ and $H^{-1}$ are holomorphic maps then we have \begin{equation}\label{eqq}\begin{split} \mu(H^*(X))(z)=\mu (H^{-1}\circ X \circ H)(z)\\&=\mu(X)(z)\times (\frac{ H'(z)}{|H'(z)|})^2=\lambda.\beta^2=1\end{split} \end{equation}.

For the dilatation of functions after right and left composition with holomorphic maps see page 9 Ahlfors lectesur on Quassi conformal mapping,Van Nostrand company,1966.

Note that $X$ and $H^*(X)$ are orbitally equivalent vector fields. So if we prove that every vector field $Y$ with $\mu(Y(z))=1$ can not have any limit cycle then the proof of the proposition would be completed. But it is an obvious fact: let $Y=P\partial_x+Q\partial_y$ be a vector field whose dilatation function is identically equal to $1$. this implies that $P_y=Q_y=0$. So $P,Q$ are functions in $x$. It is an standard fact in the theory of ordinary differential equation that a one dimensional autonomous vector field can not posses a periodic orbit. We apply this to $x'=P(x)$. So $Y$ has no periodic orbit. An alternative Proof for non existence of periodic orbit for the planar vector field $Y(x,y)=P(x)\partial_x+Q(x)\partial_y$ is the following: We have $[Y,\partial/\partial y]=0$. If $\gamma$ is a periodic orbit of $Y$, then there is a point $A=(x_A,y_A)$ on $\gamma$ whose $x$-cordinate $x_A$ is maximum on $\gamma$. Then $Y(A)$ is a vertical vector,i.e. $Y(A)$ is parallel to the $y$-axis.So $P(A)=0$. Then $P(x_A,y)=0,\;\;\forall y\in \mathbb{R}$. This shows that the vertical line $x=x_A$ is invariant under the flow of $Y$. This contradicts the fact that the periodic orbit $\gamma$ intersects this invariant line. In fact this situation violates the uniqueness of solutions of ordinary differential equation. In fact this argument can be used to give a proof for the following Proposition which I found , as an exercise, in "Differentiable Manifold, by Prof. Siavash Shahshahani(His Persian lecture).

Proposition: let $X$ be a vector field on $\mathbb{R}^n$ such that $[X, \partial x_i]=0,\;\;i=1,2,\ldots,n-1$ then $X$ has no closed orbit.

This simple observation leads us to the following guess:

Guess: If dilatation $\mu(X)$ is a real valued map then $X$ has no any limit cycle

Guess':(Equivalent formulation): If $X=P\partial_x+Q\partial_y$ satisfies $P_xP_y+Q_xQ_y=0$ then $X$ has no any limit cycle.

1)(The approach I examined) One can define a flat surface $\delta:\mathbb{R}^2 \to \mathbb{R}^3$ with $$\delta(u,v)=(P(u,v), Q(u,v),0)$$ then the first fundamental form of this surface is a diagonal matrix(the metric is diagonal). but I do not know how to continue with this situation? how can I prove that the vector field $X$ is a Killing vector field with respect to a Riemannian metric?

The evidences showing the validity of my guess:

2)Every vector field in the form $f(x)\partial_x+g(y)\partial_y$ has no any closed orbit since a gradient vector field can not posses non constant periodic orbit. However there is example of such a gradient vector field which is not a divergence free vector field with respect to any Riemannian metric defined on the complement of the singularities

  1. Every (Hamiltonian) vector field in the form $P(y)\partial_x+Q(x)\partial_y$ obviously is divergence free and satisfies the underlying relation $P_xP_y+Q_xQ_y=0$

4)Every holomorphic function $f(z)=P(z)+iQ(z)$ determines a vector field $P\partial_x+Q\partial_y$. It satisfies $P_xP_y+Q_xQ_y=0$, thaks to CR equations. On the other hand it is obvious that any such a system does not have any limit cycle since the flow is a holomorphic map hence does not admit non isolated fixed point. but i do not know if in this case our vector field is divergence free with respect to an appropriate metric.

Remark: Can complex dilatation be helpful in a systematic study of limit cycle theory? In the literature, are there some references which use dilatation to count the number of limit cycles?

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  • $\begingroup$ You probably want some additional conditions, like that the vector field has no zeros? $\endgroup$ Jan 29, 2021 at 22:07
  • $\begingroup$ @RyanBudney No but we work on the complement of the singularities.We require that our vector field is divergence free out of singularities. $\endgroup$ Jan 29, 2021 at 22:13

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I can answer your following question. Let $X=P\partial x+Q\partial y$ be a vector field on the plane $\mathbb{R}^2$. Assume that we have $P_xP_y+Q_xQ_y=0$. Does this imply that the vector field $X$ is a divergence-free vector field with respect to a Riemannian metric defined out of singularities of X?

The answer is No.

$\operatorname{div} X = \operatorname{trace} \nabla X$ where $\nabla$ is the Levi-Civita covariant derivative of the Riemannanian metric. When $X = X^j\mathbf{d}_j$ (summation convention), then $$ \operatorname{trace}\nabla X = (\nabla_iX)^i = (\mathbf{d}_iX + X^j \Gamma_{ij}^k\mathbf{d}_k)^i = \operatorname{div}_{\text{eucl}}X + X^j\Gamma_{ij}^i. $$ The second term is in general nonzero (probably it is zero for unimodular Riemannian metrics).

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    $\begingroup$ Dear Prof. Eschenburg Thank you very much for your answer. You are well come to MO! $\endgroup$ Feb 23, 2021 at 14:09
  • $\begingroup$ Where did you used the equation $P_xP_y+Q_xQ_y=0$? $\endgroup$ Feb 24, 2021 at 5:40

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