Hi I am very confused about the following, I was asked to consider the system
$r^{\bullet}=r(1-r^2)$ and $\theta^{\bullet}=1-\cos(\theta)$ to find the fixed points and sketch the phase portrait and moreover explain why $(1,0)$ is attractive but not Liapunov stable.
Here is what I have tried, and what I was told.
The fixed points can easily be solved to be $(0,0)$ and $(1,0)$
now I am not confident at all at sketching phase portraits in polar coordinates as I am not sure what to expect or to look for, here are some things I noted but I do not know if they will have any relevance.
I noticed that as $\theta$ goes from $0$ to $2 \pi$, $\theta^{\bullet}$ goes continuously from $0$ to $1$ , then from $1$ to $2$ then from $2$ to $1$ and from $1$ to $0 $, ( in intervals of $\pi/2$)
$r^{\bullet}$ begins at 0 when $r$ is $0$, as r increases $r^{\bullet}$ positively increases ( untill it hits a max at $r= \sqrt{1/3}$ ) then decreases as well but once $r$ is greater than $1$, $r^{\bullet}$ is negative.
I also have formulas that state that $$r^{\bullet}r=xx^{\bullet}+yy^{\bullet}$$
and that $$\theta^{\bullet}=\frac{xy^{\bullet}-yx^{\bullet}}{r^2}$$
yet I do not know if this will really help with anything.
Finally, here is what it is supposed to look like. I was also told that (1,0) is attracting but all of the solutions except those which start on the positive x axis must traverse near the unit circle before they can reach the fixed point. (why?)
But I am still confused how that was obtained from the starting information, and why we still graph in terms of $x$ and $y$ axes.
So I am looking for some explanations and help on this. Also, this is not meant to have a completely computer generated graph. it is meant to be a rough sketch just using information we could get from it ourselves.
I was also thinking of the ideas of transferring to cartesian coordinates and then looking at the Jacobian matrix, or maybe trying to solve the ODE directly in polar and seeing if that tells us anything. I will put a bounty because I have still not understood this and have spent alot of time trying to figure it out
Thanks
You want to sketch the phase portrait of this system:
$$\begin{cases} r' = r(1 - r^2) \\ \theta' = 1 - \cos{\theta} \\ \end{cases}.$$
Before you start doing anything, you should ask yourself: what is $r$? what is $\theta$? what do they represent? Suppose $(r, \theta)$ are the polar coordinates of a point $P$; then:
Once you understand this, you can start working on your system. After finding the fixed points (which you did), you must ask yourself: where are $r'$ and $\theta'$ positive, and where are they negative? You can easily check that
$$r' > 0 \implies r < 1$$
and
$$\theta' > 0 \implies \cos{\theta} < 1.$$
It is important to know this because $r'$ and $\theta'$ are rates of change: $r$ increases in those regions where $r' > 0$, and decreases where $r' < 0$; the same can be said for $\theta'$. In this context, "increasing" and "decreasing" respectively mean that trajectories are blown away from the origin wherever $r' > 0$ ($r$ increases), whereas they get closer to it wherever $r' < 0$ ($r$ decreases); similarly, points are forced to move counterclockwise wherever $\theta' > 0$ ($\theta$ increases), while they move clockwise where $\theta' < 0$ ($\theta$ decreases).
Sketching the phase portrait
Now, recall that $r' > 0 \implies r < 1$: this means that every trajectory that starts from inside the circle of radius $1$ centered at the origin will be repelled from the origin itself, whereas trajectories starting from outside the circle will get close to it. So, no matter where you start from: your trajectories will all eventually reach the vicinities of the circle $r = 1$. What about points that lie on the circle itself? On the circle, $r' = 0$: the distance from the origin neither increases nor decreases, so any trajectory starting on the circle will forever be confined on it.
As for $\theta$, trigonometry tells you that $\cos{\theta}$ is always less than $1$ (except at points where $\theta$ is $0$, because $\cos{0} = 1$), which means that almost every trajectory in your phase portrait will be forced to move in a counterclockwise fashion. The circle $r = 1$ itself is a trajectory, so the flow moves counterlockwise on it as well. The only points for which this fails are those that lie on the positive $x$-semiaxis, i.e. those that have $\theta = 0$; for all of them, $\theta' = 0$, hence there is no motion at all in the angular direction. A trajectory starting on this semiaxis will be forced to stay there forever because it cannot move away from it (which would otherwise imply an angular shift), so the flow on this semiaxis is entirely ruled by $r'$, which you have already studied.
The last, fundamental thing you have to remember is that two different trajectories can never cross: once you pair everything I said above with the fact that trajectories must not intersect, you cannot obtain anything different from the phase portrait Jonas has already sketched and posted. Of course, the fact that $(1,0)$ is attractive but not Lyapunov stable is also easily understandable at this point.