# Is it possible for a node-node bifurcation to exist?

A saddle-node bifurcation for a vector field on a plane $\dot{x} = f(x, \mu)$, where $x \in \mathbb{R}^{2}$ and $\mu \in \mathbb{R}$ corresponds to a creation of two equilibria (both hyperbolic, one saddle, the other node) out of zero, as the parameter $\mu$ passes some critical value.

Is it possible to have a scenario where two hyperbolic equilibria, are born such that $\textbf{both}$ are nodes? (Say both would be (un)stable, or one stable, the other unstable). ( I call this hypothetical scenario node-node bifurcation).

This leads to a more general question: is it possible for a vector field on a plane to have $\textbf{only 2 fixed points such that both are nodes}$, and no other fixed points?

I've been unable to come up with an example of such a vector field, or come up with a reason why this is not possible.

• Yes, this is possible. All you need to do is have a 1-D saddle-node bifurcation happen simultaneously in both dimensions. – Hans Engler Dec 1 '16 at 0:32
• Thanks. Could you give an example? I was thinking of $x= \mu -x^{2}$, $y =\mu - y^{2}$ but here as $\mu$ passes through zero, we get 4 equilibria, not 2... @HansEngler – Alex Dec 1 '16 at 0:46
• Thinking about it, vector field with just two equilibria that are both nodes (of the same stability at least) does not exist since one would require another invariant set to separate the trajectories, which gives a contradiction @HansEngler – Alex Dec 10 '16 at 19:47
• That is not quite right. The vector field $\left\{\frac{-2 x^2+2 y^2+2}{\left(x^2+y^2+1\right)},-\frac{4 x y}{\left(x^2+y^2+1\right)}\right\}$ has a stable node at $(1,0)$ and an unstable node at $(-1,0)$ and no other stationary points. – Hans Engler Dec 20 '16 at 3:28

Let $H(x,y,\lambda) = \left(\lambda - x^2,-x\cdot y\right)$. Here is a phase portrait for $H(\cdot, \cdot,1)$.
As $\lambda \to 0$, the two nodes move towards each other and annihilate each other at $\lambda = 0$.