Proof of Dulac’s Criteria

Consider the the planar system $\dot{x}=f(x)$, where $f={(X,Y)}^t$ and $x={(x,y)}^t$. Consider the vector field $(-\rho Y,\rho X)$. Suppose that there is a periodic orbit contained in $E$ associated to the planar system. Let $\gamma$ be that periodic orbit. We have:

${\displaystyle {\int }_{\gamma }}(-\rho Y,\rho X)𝑑s={\displaystyle {\int }_0^{\tau }}(-\rho Y(x,y),\rho X(x,y))\cdot (\dot{x},\dot{y})𝑑t={\displaystyle {\int }_0^{\tau }}-\rho Y(x,y)X(x,y)+\rho X(x,y)Y(x,y)=0$

On the other hand, the region within $E$ that is limited by $\gamma$ is simply connected because $E$ is simply connected. Let $\tilde{E}$ be the region limited by $\gamma$. Then, by Green’s theorem, we have:

$${\int }_{{\gamma }^{+}}(-\rho Y,\rho X)𝑑s=\int {\int }_{\tilde{E}}\frac{\partial }{\partial x}(\rho X)-\frac{\partial }{\partial y}(-\rho Y)dxdy=\int {\int }_{\tilde{E}}\frac{\partial }{\partial x}(\rho X)+\frac{\partial }{\partial y}(\rho Y)dxdy$$

Because $\tilde{E}$ has positive area and the integrand function has constant signal, then this integral is different from zero. This is a contradiction. So there are no periodic orbits.

Title	Proof of Dulac’s Criteria
Canonical name	ProofOfDulacsCriteria
Date of creation	2013-03-11 19:17:10
Last modified on	2013-03-11 19:17:10
Owner	Filipe (28191)
Last modified by	(0)
Numerical id	9
Author	Filipe (0)
Entry type	Proof
Classification	msc 34C25
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