Find the values of $r$ at which bifurcation occurs

Question

Consider the system $$\dot{x}= rx - \frac{x}{1 + x^2}$$ where $r \in \mathbb R$. Find the value(s) at which bifurcations occur and where possible classify those as saddle-node, transcritical or supercritical pitchfork.

I found the fixed points of my system as

$$x = 0 $$ and $$x = \pm \frac{\sqrt{1 -r}}{\sqrt{r}}$$ Assuming $r\neq 0$ for the last equilibrium, of course.

How do I continue from here? I am not sure how to find the values of $r$ at which bifurcation occurs.

Answer 1

In order to find bifurcation, you need to study the existence and the stability of all steady states (a.k.a. equilibria, or fixed points).

Existence $$\begin{cases} x^*_1 = 0 & \text{always exists}\\ x^*_2 = \frac{\sqrt{1 -r}}{\sqrt{r}} & \text{exists for}~r \in(0, 1]\\ x^*_3 = -\frac{\sqrt{1 -r}}{\sqrt{r}} & \text{exists for}~r \in(0, 1] \end{cases}$$

Stability

Firstly, you need the derivative of $\dot{x} = f(x)$ with respect to $x$:

$$f'(x) = \frac{rx^4+(2r+1)x^2+r-1}{(1+x^2)^2}$$

The sign of $f'$ characterizes the stability of $x^*_n$. Specifically, we need to find the values of the parameter $r$ such that the derivative $f'$ evaluated in a steady state is negative:

$$\begin{cases} f'(x^*_1) = r - 1 < 0 & \forall r < 1 \\ f'(x^*_2) = -2r^2 + 2r < 0 & \forall r <0 \vee r > 1\\ f'(x^*_3) = -2r^2 + 2r < 0 & \forall r <0 \vee r > 1\\ \end{cases}$$

Critical points

• For $r < 0$, there is only one equilibrium (i.e. $x_1^*$) which is stable.
• For $0< r< 1$, there are $3$ equilibria: $x_1^*$ is stable, the others are unstable.

Then, for $r = 0$ you have a saddle-node bifurcation for $x_2^*$ and $x_3^*$, since they appear from nowhere and in $r=0$ you have that $f'(x_2^*) = f'(x_3^*) = 0$.

• For $0< r < 1$, there are $3$ equilibria: $x_1^*$ is stable, the others are unstable.
• For $r > 1$, there is only $1$ equilibrium, $x_1^*$, which is now unstable.

Then, for $r = 1$ you have a saddle-node bifurcation for $x_2^*$ and $x_3^*$, since they disappear and you have $f'(x_2^*) = f'(x_3^*) = 0$ for $r=1$.

Notice that $x_1^*$ never disappears and it changes stability only with himself, so in this case it never experiences bifurcations.