Stability of fixed points in this example from a dynamical systems textbook

I hope people will indulge me a bit of a beginner Dynamical Systems question.

The following example was taken from Nonlinear Dynamics and Chaos (Second Edition) by Steven Strogatz (example 2.2.1).

Find all fixed points for $\dot{x}=x^2 - 1$ and classify their stability.

Solution: Here $f(x) = x^2 - 1$. To find the fixed points, we set $f(x*) = 0$ and solve for $x*$. Thus, $x* = \pm1$. To determine stability, we plot $x^2 - 1$ and then sketch the vector field (see figure below). The flow is to the right where $x^2 - 1 > 0$ and to the left where $x^2 - 1 < 0$. Thus, $x* = -1$ is stable, and $x* = 1$ is unstable.

The solid dot in the picture below (at $x = -1$) represents a stable equilibrium, whereas the open dot (at $x = +1$) represents an unstable equilibrium. The arrows represent the direction of "local flow."

So I understand that the fixed points are the roots of $f(x)=x^2 + 1$, which is obviously $\pm1$, and that the fixed points are equilibrium solutions. I understand that the equilibrium at $x = -1$ is stable because the local flow is "towards" it and the fact that the equilibrium at $x = +1$ is unstable because the equilibrium is "away from" it, but I'm confused as to how they came up with the direction of the local flow in the first place. Can someone explain why the local flow is in those directions in this case?

• Actually you can edit the typo in the last paragraph of the function $f(x)$...It is a convention i think which we use while doing vector problems in physics too there you take any direction to a be a positive one and opposite to it as a negative one ... here too you can take opposite directions as given in the book and still end up getting the same stability at the two points... – BAYMAX Mar 12 '17 at 6:12

The local flow going to the right means that $x(t)$ is increasing, which happens when $\dot x>0$.
The equation says that $$\dot x=x^2-1,$$ so $\dot x > 0$ if and only if $x^2-1 > 0$, which is true in the part of the figure where the curve $y=x^2-1$ lies above the $x$ axis.
• That's right, because $x^2 - 1$ is the derivative - now I feel like an idiot :) - thanks for helping, I understand now. – EJoshuaS Mar 12 '17 at 18:40