# 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:

${\int}_{\gamma}}(-\rho Y,\rho X)\mathit{d}s={\displaystyle {\int}_{0}^{\tau}}(-\rho Y(x,y),\rho X(x,y))\cdot (\dot{x},\dot{y})\mathit{d}t={\displaystyle {\int}_{0}^{\tau}}-\rho Y(x,y)X(x,y)+\rho X(x,y)Y(x,y)=0$

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

$${\int}_{{\gamma}^{+}}(-\rho Y,\rho X)\mathit{d}s=\int {\int}_{\stackrel{~}{E}}\frac{\partial}{\partial x}(\rho X)-\frac{\partial}{\partial y}(-\rho Y)dxdy=\int {\int}_{\stackrel{~}{E}}\frac{\partial}{\partial x}(\rho X)+\frac{\partial}{\partial y}(\rho Y)dxdy$$ |

Because $\stackrel{~}{E}$ 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. \qed

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 |

Synonym | |

Related topic | |

Defines |