2
$\begingroup$

Consider the dynamical system

$$ dx/dt = -\cos(r)\sin(x) $$

Clearly $x=0$ and $x=\pi$ are two fixed points of this system. The stability of these two fixed points change as r is varied. Starting from $r=0$, the zero fixed point is initially stable but becomes unstable at $r=\pi/2$ and the opposite happens for the other fixed point.

Is this a transcritical bifurcation? I have so far only seen systems where one of the fixed point approaches the other fixed point and their stability switch as they cross each other.

The usually shown bifurcation diagram for a system undergoing transcritical bifurcation

However in my system, the fixed points remain at the same values of $x$. Only stability changes. If not transcritical, what type of bifurcation is this?

$\endgroup$
1
$\begingroup$

TL;DR: I don't believe the system displays a transcritical bifurcation.

Your stability analysis is correct. For instance, for $r=0$, $\dot{x}$ is simply a negative sine function, so positive $x\approx 0$ values have negative derivatives (and decrease, i.e., move towards zero) and negative values have positive derivatives (so increase and also move towards zero), i.e., $x=x^*=0$ is stable for $r<\pi/2$. See figure below:

enter image description here

And the system does go through a bifurcation as $r$ is varied — though not one of the simple ones, due to the singular nature of the system at $r=r^*=\pi/2$: For this value of $r$ all points $x\in[0,2\pi]$ are (neutrally stable) fixed points, since the system is simply $dx/dt=0$. That's how I'd sketch its bifurcation diagram:

enter image description here

It looks like we could say there's a collision on $r^*=\pi/2$ and the fixed points "exchange stabilities — just like in a transcritical bifurcation, however, I think we have something different here, for two reasons:

I. If we check the formal conditions for the system $\dot{x}=f_r(x)$ to go through a transcritical bifurcation (e.g., Guckenheim & Holmes or Wolfram MathWorld), we'll see they are: \begin{align} f_r(x)\Bigr|_{\forall r, x=x^*}=0 \tag{equilibria}\\ \frac{\partial f_r}{\partial x}\Bigr|_{r=r^*, x=x^*} = 0 \tag{null eigenvalue}\\ \frac{\partial^2 f_r}{\partial r\partial x}\Bigr|_{r=r^*, x=x^*} \ne 0 \tag{TC1}\\ \frac{\partial^2 f_r}{\partial^2 x}\Bigr|_{r=r^*, x=x^*} \ne 0 \tag{TC2} \label{TC2} \end{align} where TC stands for transversality condition. And it's clear that our $f_r(x)$ fails to satisfy \ref{TC2}.

II. By changing variables, $r \mapsto -r + \pi/2$, the system becomes $f_r(x)=-\sin(r)\sin(x)$ and $r^*=0$. If we then try to put $f_r(x)$ in a algebraic normal form, by Taylor expanding it around $(r^*,x^*)$ we obtain, after a new change of variables $r\mapsto r-r^3/6$, $$ f_r(x) = rx -r\frac{x^3}{6} +O(5) $$ which I don't believe can be put in the normal form for the transcritical bifurcation, nor any of the other simple ones: \begin{align} \dot{x} &= r - x^2 \tag{saddle-node}\\ \dot{x} &= rx - x^2 \tag{transcritical}\\ \dot{x} &= r - x^3 \tag{pitchfork} \end{align} Which is perhaps not that surprising, given the system's bifurcation diagram.

$\endgroup$
6
  • $\begingroup$ I missed out a negative sign in the equation. Thanks for pointing it out. If not transcritical, does this type of bifurcation have some name to it that I can search for? $\endgroup$ May 17 at 18:47
  • $\begingroup$ @SnehaSrikanth Not that I know, maybe we could call it transcritical-like? :) $\endgroup$
    – stafusa
    May 17 at 18:50
  • $\begingroup$ @SnehaSrikanth I'm now convinced the system doesn't display a transcritical bifurcation, check my updated answer. $\endgroup$
    – stafusa
    May 18 at 19:24
  • $\begingroup$ Thanks for such a clear detailed answer! I still want to keep the question open for anyone who might be able to pinpoint what kind of bifurcation this is. I am new to StackExchange.. do I edit this post and change the question or make a new question? I do not want your answer to become irrelevant. $\endgroup$ May 19 at 11:44
  • 1
    $\begingroup$ Ok will make that change! I am actually unable to upvote yet (I need 15 reputations for that). I'll surely upvote when I hit 15 reputations :D $\endgroup$ May 19 at 16:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.