# Dynamical Systems. Bendixson's and Dulac-Bendixson's theorems.

I am looking for a place to read the proofs of the Bendixson and Dulac-Bendixson theorems. Namely, let $D$ be a simply connected set and the following system be defined in $D$, and \begin{align} \dot x &=P(x,y), \\ \dot y &=Q(x,y), \end{align} then the following theorems hold.

Theorem: (Bendixson) Given that $P_x+Q_y$ doesn’t change sign in $D$, the system does not have a periodical solution.

Theorem: (Dulac-Bendixson) If a function $B(x,y)\in C^1(D)$ exists, such that $(BP)_x+(BQ)_y$ doesn't change sign in $D$. Then the system does not have a periodical solution.

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Assuming a periodic orbit $C$ with interior $U$, $$\oint_C(BQ\,dx-BP\,dy)=\oint_C B(\dot y\,dx-\dot x\,dy)=0$$ because $\dot y\,dx-\dot x\,dy=0$ along the orbit. Now use Green's theorem to conclude $$\iint_U\bigl((BP)_x+(BQ)_y\bigr)\,dx\,dy=0.$$You now have your contradiction. This proves Dulac–Bendixson; Bendixson follows by setting $B=1$.
Great timing, I was just typing essentially the same thing. While this is clear from what you wrote, I want to emphasise that $U$ has positive area because $C$ is embedded. ($C$ is embedded by uniqueness of solutions to ODE.) – Sam Lisi Apr 30 '13 at 12:11
Thank you so much, sir. You may have long forgotten this question but I still have some issues unresolved. How did you use that $P_x+Q_y$ doesn't change sing in D? Is that why $\dot y\,dx-\dot x\,dy=0$ along the orbit? Why does $\dot y\,dx-\dot x\,dy=0$ along the orbit? – Student Jul 12 '13 at 20:47
The sign question: If you integrate a positive function over $U$, you get a positive result, not zero. Similarly if the function is negative. For the other question, along an orbit you have $dx=\dot x\,dt$ and $dy=\dot y\,dt$. – Harald Hanche-Olsen Jul 13 '13 at 11:45