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$$\dot{r}=r(1-r^2)+\frac{1}{2}rcos(\theta) \\ \dot{\theta}=1 $$

I want to show that this system has a periodic orbit.

I tried:

for $\dot{r}<0$: $$r^2\ge\frac{1}{2}cos(\theta)-1$$ But I don't know how to find the lower bound.

Then we find $\dot{r}>0$. Since I don't know how to do it for the first case, I won't try this case.

Using the above cases, we then find the interval that $r$ is in.

We then use Poincaré-Bendixson theorem and show that the fixed point is not in the interval, and conclude that the system has a periodic orbit.

I also don't know how to find the fixed point for this system. Does it require me to convert the system back to its non-polar form? If so, I don't know how.

John B
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sucksatmath
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1 Answers1

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For $r$ very large you have $\dot r<0$, while for $r$ very small you have $\dot r$ approximately equal to $$r(1+\cos\theta/2)>r(1-1/2)=r/2>0.$$ Since the only equilibrium point is the origin, it follows from the Poincaré-Bendixson that there is a periodic orbit.

Note that you don't need to find the exact regions where $\dot r$ is negative or positive.

John B
  • 16,854