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I'm reading the book "Nonlinear dynamics and Chaos" by S Strogatz.

In section 2.2, titled "Fixed points and stability", he defines equilibrium points as solutions where

...all sufficiently small disturbances away from it dampen away in time.

As an example, he demonstrates this for a system described by $\dot{x}=x^2-1$.

Here's the vector field which can be plotted:

Strogatz 2.2.2

(the arrows indicate the direction of motion for the relevant domains)

Clearly the definition of equilibrium points justifies that $x=-1$ is an equilibrium point while $x=1$ is not: for a 'small' displacement of about $0.1$ (or actually any displacement in this case), the arrows clearly show that a body initially at $x=1$ will not be able to move back to $x=1$.

But when discussing the equilibrium at $x=-1$, he says that after a displacement of more than $+2$, an object initially at $x=-1$ will not return to the initial position. Again, the arrows make this obvious.


What qualifies as a 'small' displacement?

I believe that it would be relative (i.e. no universal constant value can be defined): with $\dot{x}=x^2-1$, a change of $+3$ is not small, but if you're considering $\dot{x}=x^2-1000$, the same change of $+3$ is small. I considered using limits/calculus, but I wasn't able to write a nice formal condition in the form of

The point $x=a$ represents equilibrium if $$\lim_{\delta x\rightarrow 0}f(a\pm \delta x)\geq M$$ (where $f(x)$ and $M$ are functions of $x$ and $\dot{x}$)

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Small means arbitrarily small.

Thus, for an introduction, calculus is indeed the way to go:

  • every point $x^*$ where the derivative vanishes ($\dot x = f(x^*)=0$) is a fixed point;
  • if the derivative of $f$ is positive ($\dot f(x^*) > 0$) then the fixed point is unstable (any perturbation, no matter how small leads the system away from this point;
  • if $\dot f(x^*) < 0$, then the fixed point $x^*$ is stable, or an equilibrium.

The explanation why it's so has already been given:

remember that the derivative $f'$ is the slope of $f$. So at a fixed point where $f(x) = 0$, if $f'(x) > 0$ we have that $f$ is increasing at $x$, or that $f(x+\epsilon) > 0 > f(x-\epsilon)$ for all sufficiently small, positive $\epsilon$. This shows that if you start with initial value $x_0 > x$, but close to $x$, since $f(x_0) > 0$ we will have that the ODE forces the particle to increase its value of $x$, and move away from the fixed point. If you start with $x_0 < x$, but close to $x$, the ODE will now force the particle to decrease its value of $x$, and move away from the fixed point. Hence if $f'(x) > 0$ we say that the fixed point is unstable.

Similarly you can work it out to show that if $f'(x) < 0$, the fixed point is stable.

Rigorously we're only talking about linear stability, which is often the dominant term, but not always. For instance, if $\dot f(x^*)=0$, then the criterion above is insufficient (one might talk about neutral stability): you need to consider higher orders of derivatives. Also, a linearly stable fixed point might have a basin of attraction so small that, for any practical purposes, the point is effectively unstable.

Other types of stability exist, even more in systems of higher dimensionality. You can find more about it searching for stability theory.

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