# Stability of a continuous-time dynamical system

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

-
(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

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.

-

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.

-