# Does $P_xP_y+Q_xQ_y=0 \implies$ “non-existence of limit cycle” for $P\partial_x+Q\partial_y$"? (Complex dilatation and limit cycle theory)

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 $$$$\label{eqq}\begin{split} \mu(H^*(X))(z)=\mu (H^{-1}\circ X \circ H)(z)\\&=\mu(X)(z)\times (\frac{ H'(z)}{|H'(z)|})^2=\lambda.\beta^2=1\end{split}$$$$.

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?

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

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?
$$\operatorname{div} X = \operatorname{trace} \nabla X$$ 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).
• Where did you used the equation $P_xP_y+Q_xQ_y=0$? – Ali Taghavi Feb 24 at 5:40