Showing asymptotic orbital stability

[Mattbringssoda]

Homework Statement

[/B]

Show that every trajectory, except r = 0, is asymptotically orbitally stable

Homework Equations

[/B]
I have a hint that: I should think of two initially close trajectories (say a(0) = |x(0) - y(0)| sufficiently small and see how the difference evolves over time (the rate a'(t).) I need to make |x(t)-y(t)| bounded.

The Attempt at a Solution

[/B]
I think I'm getting confused by the symbology and I'm not sure how to operate the equations.

First I tried to calculate the fixed points. The only fixed point I get is the origin, r = 0. (If I tried r = 1, then I can't get theta ' = 0, so r = 0 must be the fixed point to look at, I'm guessing.)

So, next, I can see that r' = r(1-r) will go to zero as r increases in value. I think that's something I'm looking for.

Next, I think I need theta' to be bounded. Well, I know the first part, 1 + sin^2, will always be less than 2...but there's nothing to bound (1-r)^2. As r increases, the term will just keep increasing without bound...

These are my thoughts on the problem so far and I don't know where to go with it from here...

Thanks for any help!

 

[mighty2000]

Hello,

Thank you for your post. It seems like you are on the right track with your approach. I will try to guide you through the solution step by step.

Firstly, we need to understand what the statement "every trajectory, except r = 0, is asymptotically orbitally stable" means. This means that for any initial condition (r(0), theta(0)) that is not equal to (0,0), the trajectory will eventually approach the origin (0,0) as t approaches infinity. This is what is meant by asymptotically stable.

Now, let's consider two initial conditions that are close to each other, say (r(0), theta(0)) and (r(0)+a(0), theta(0)+b(0)), where a(0) and b(0) are small perturbations. We can write the difference between these two trajectories as |x(t)-y(t)| = |(r(t)+a(t))cos(theta(t)+b(t)) - r(t)cos(theta(t))|. We can simplify this expression using the trigonometric identity cos(A+B) = cos(A)cos(B) - sin(A)sin(B), to get |x(t)-y(t)| = |a(t)cos(theta(t)) - b(t)r(t)sin(theta(t))|.

Now, let's consider the rate of change of this difference, which is given by |x'(t)-y'(t)| = |a'(t)cos(theta(t)) - b'(t)r(t)sin(theta(t))|. We can use the given equation for r' and the fact that a and b are small perturbations to write this as |x'(t)-y'(t)| ≤ |a'(t)| + |b'(t)|r(t).

Since a(0) and b(0) are small, we can assume that a'(t) and b'(t) are also small. This means that as t approaches infinity, |a'(t)| and |b'(t)| will approach 0. Also, since r(t) approaches 0 as t approaches infinity, |b'(t)|r(t) will also approach 0. This means that |x'(t)-y'(t)| will approach 0 as t approaches infinity.

Therefore, we have shown that |x(t)-y(t)| is bounded and |