There is a system of ODEs or vector field below:
$$\dot x = x-y+x^2-x^3-xy^2$$
$$ \dot y =x+y+xy-x^2y-y^3$$
(a) let $\Sigma$ be the postive $x$ axis,$$\Sigma=\{(x,y)\in \mathbb R^2:x>0,y=0\}$$
Show for any initial value in $\Sigma$ gives a solution to the initial value problem such that it returns to $\Sigma$ after a positive finite time. For this vector field, the domain of the Poincaré map $P$ is all of $\Sigma$.$$$$
(b)The annulus $$\{(x,y):a\le\sqrt {x^2+y^2}=r \le b\}$$ is an invariant set. Show that $P(a)>a$ and $P(b)<b$. Use intermediate value theorem for continuous functions to show that this Poincaré map has at least one fixed point in $\Sigma$ and a periodic orbit in $\mathbb R^2$
My attempts:
It is obvious that $(0,0)$ is an equilibrium point and it is unstable by checking the Jacobian matrix. I have done the conversion of this system of ODEs into polar coordinates and they are
$$\dot r=r-r^3+r^2\cos \theta$$ $$\dot \theta=1$$ Also I have shown there exists a positive invariant set as an annulus
$$\{(x,y):\frac{\sqrt5 -1}2\le\sqrt {x^2+y^2}=r\le \frac{\sqrt5+1}2\}$$ There is no equilibrium point inside this annulus. How do I use these information to solve (a) and (b). Especially no initial point for the Poincare map is provided and that's where I got stuck. for (b) if we can show that $P(a)>a$ and $P(b)<b$. then it is easy to claim $(P(a)-a)(P(b)-b)<0$ so we are done. but how do I show $P(a)>a$ and $P(b)<b$?
Best Answer
First observe that $\dot \theta=1$ implies $\theta=t+C$. Thus is immediate that any nonconstant solution circles around the origin infinitely many times.
Now we have only one ODE for $r(t)$: $$\dot r=r-r^3+r^2\cos(t+C).$$ But, because we have $-1\le\cos(t+C)\le1$, it's true that $$r-r^3-r^2\le r-r^3+r^2\cos(t+C)\le r-r^3+r^2.$$ So, for $0<r< \frac{1}{2}$ we have $0<r-r^3-r^2\le r-r^3+r^2\cos(t+C)$, that is, all solutions are strictly increasing for this interval of $r$. From this follows, for example, $P(1/4)>1/4$. In the same lines we find, for $2<r$, that $r-r^3+r^2\cos(t+C)\le r-r^3+r^2<0$. From this we have that solutions are strictly decreasing for this interval of $r$. Finally we have $P(5/2)<5/2$.
I believe that your difficult lies in the fact that you're trying work with the minimal invariant annulus.