I am examining exercise $3.1.1 $ of Strogatz in which it is asked to sketch the vector fields that occur as $r$ is varied and to show that a saddle-node bifurcation occurs at a critical value of $r$, to be determined.
$$\dot{x} = \underbrace{1+rx + x^2}_{f(x)}$$
To my understanding, a bifurcation occurs when fixed points approach, collide, and mutually annihilate each other.
I first consider the fixed points of my system: $f(x) = 0 \implies x_{1,2}^{*} = \frac{-r \pm \sqrt{r^2 – 4}}{2}$
Where my confusion comes from is that I do not understand how to sketch the vector fields as we vary the 'r' parameter.
The discriminant is zero when $r = 2$ or $r = -2$ and therefore has one real root (with multiplicity 2). The discriminant is positive for any $r \neq 2$ and $r \neq 0$ and therefore has two real roots. The discriminant is negative for $r = 0$ and therefore there are no real roots.
The "collision" of fixed points is where we have this multiplicity of $2$ root: when $r^2 – 4 = 0 \implies r = 2,-2$ and then $x_{1,2}^{*} = \pm1$.
My issue is…understanding how to draw the "arrows" to denote stability. To classify the stability of a fixed point we examine if the derivative of $f(x)$ is positive (unstable) or negative (stable).
Here, $f'(x_1^*, 0) = 2(1) > 0$ which means unstable. Likewise, $f'(x_2^*, 0) = -2 < 0$ which means stable. But now how do I draw the arrows on the graph?
Best Answer
Your analyses are spot on.
The saddle-node bifurcation requires that $f(x) = 1 + rx + x^2$ has two identical solutions, so the discriminant $∆ = r^2 − 4 = 0$. Therefore, $r = \pm2$ and $x^∗ = \mp1$.
If we draw a phase portrait, it would look like this for $r = -2$
However, I think they want the streamline, which is (compare the phase portrait for stable vs. unstable flow)
For $r = 2$, we have
The streamline is
I also think they are looking for the bifurcation diagram (blue is stable, orange is unstable)
Update I used Mathematica to draw all of the figures.