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In this problem we consider one-dimensional continuous-time dynamical system $x'=f(x)$ with fixed point $u$ at which $f(u)=0$. For each of the following systems, discuss the stability of the fixed point $u=0$.

  1. $f(x)=x^2$.

  2. $f(x)=-x^2$.

  3. $f(x)=x^3$.

  4. $f(x)=-x^3$.

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closed as too broad by Did, Normal Human, Strants, J. W. Perry, Batominovski Aug 21 at 2:23

There are either too many possible answers, or good answers would be too long for this format. Please add details to narrow the answer set or to isolate an issue that can be answered in a few paragraphs. If this question can be reworded to fit the rules in the help center, please edit the question.

(1.) The problem looks like it's from a homework. Is it? If so, please add the [homework] tag. (2.) Also, do not post the message as an order (e.g., "show that..." or "discuss..."); ask a question instead. (3.) Have you done anything at all for the problem? Can you share your work with us? –  Srivatsan Oct 29 '11 at 15:41
Thank for help me editing this post.Yes, it's in fact a homework exercise. I try to do, but I struggle with $\epsilon,\delta$ language and I found nothing. Sorry because I am not good at mathematics and my question bother you. –  Arsenaler Oct 29 '11 at 15:44
Apologies, I did not notice the [homework] tag before somehow. Please ignore the first item in my comment. –  Srivatsan Oct 29 '11 at 15:47
Ok. But still no one want to help me ? Please help me, because I have to prepare for the midterm test. Thanks –  Arsenaler Oct 30 '11 at 1:20
It's not a "time dynamical system" that is continuous; it's a dynamical system in continuous time. Hence I added a hyphen. –  Michael Hardy Oct 30 '11 at 1:39

2 Answers 2

You mention $\epsilon$ language, so I'm imagining that the class requires you to apply whatever method they teach. I think such things can often over-complicate these problems. Before you apply specific mathematical tools, I think you should use extremely simple logic to set expectations.

Particularly, why don't you just make a table posing the question of "what does the particle do when moving in X direction?" Let's look at this one for example:

$$x' = f(x) = -x^2$$

please excuse my substitution of left and right for negative and positive respectively

  • Displaced right of origin $\rightarrow$ Is moving left
  • Displaced left of origin $\rightarrow$ Is moving left

Is this stable? The system will return to $x=0$ in the case that it starts with $x>0$. It is right of origin and moving left. In the other case it is left of origin and moving left, so it is headed for $-\infty$. No, this is not stable.

Rinse and repeat.

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To get some intuition in equilibria of 1-dimensional dynamical systems try this. You already know that $x=0$ is the (only) equilibrium. So set an $x$ and see if $x>0$, $x$ increases or decreases? If $x<0$?

To find the stability, you can draw $f$ and view this as the velocity of a particle, so given the starting position, you can find the trajectory.

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