The first question is whether or not conservative systems have attractors
And the answer is no, conservative systems do not have attractors
However, even though conservative systems do not have attractors, they can be chaotic.
For example, the motion of Pluto is chaotic, but our solar system, on human time scales, is a conservative system
So conservative systems can be chaotic
Question 3 asks us which one of these would let us know that were at a fixed point of a dynamical system
The last option, If the system is at that state and you perturb it, that perturbation will shrink, is a characterization of stability
While stability conditions are often associated with fixed points, that is not the definition of a fixed point, so this condition is not sufficient to let us know that were at a fixed point of the dynamical system
The trajectory starting there is also not good enough; the trajectory must stay there
The definition of a fixed point is that the dynamics of f take the fixed point x* back to x*
That is, that once a trajectory has reached that state, a trajectory will never leave that state
So, the answer to question 3 is that trajectories never leave that state is a sufficient condition to tell us that were at a fixed point of a dynamical system
The next question asks which of these characterizes a stable fixed point
We know from the previous question that the second answer characterizes a fixed point, but it doesnt talk about stability
We also know that the first answer wasnt even enough to characterize a fixed point, let alone a stable fixed point
The first part of the last option characterizes stability; that is, if a system is in the state and you slightly perturb it, that perturbation does not grow
That characterizes the stability portion
And trajectories never leave that state without external influence characterizes the fixed point portion
For these reasons, this last option is correct
In the last problem, we need to characterize each of these points of the damped pendulum as stable, unstable, or not a fixed point
Part a corresponds to this pendulum, where theta is equal to zero and its not moving
As we saw in the lecture, this is a stable fixed point
Part b is a little bit trickier
So theta equals zero, so it does correspond to same pendulum; however, omega, which is the velocity, is pi
So while the pendulum is in this bottom position, it is moving at a velocity of pi
Because of its angular velocity, even though the pendulum is in this bottom position, at the next time step the pendulum will continue to move
And so f(x*) will not be x*
So, this is not a fixed point
In this next problem, the pendulum is now in this state
As you saw in the lecture, this was a fixed point the pendulum will balance here but its very unstable; even blowing on the pendulum will cause it to fall
For parts d and e, all we need to know is all even multiples of pi with no angular velocity are stable fixed points
And all odd multiples of pi with no angular velocity are unstable fixed points