Find the bifurcation of $a$ and describe the bifurcation that take place at each value

$\displaystyle dy/dt=e^{-y^2}+a$

I let $\displaystyle e^{-y^2}+a=0$ then solve for y. I got $y^2=-\ln(a)$ What do I do next to find $a$?


1 Answer 1


You started well to let $e^{-y^2} + a =0$ and then solve for $y$. This will allow us to determine the equilibrium points. You will arrive at $$y^2 = -\ln(-a).$$ (I believe you misplaced a negative sign.) From this we see that first of all, we must have $a < 0$. In order to find bifurcation points, we need to consider what values of $a$ will yield a change in the nature of the equilibrium points. That is, what value of $a$ will cause a change in the number or behavior of the equilibrium points? Is there a value of $a$ that will cause no solution to the equation?

I leave this to you, but please ask a question if you get stuck.

  • $\begingroup$ Thank you for fixing my mistake, I did forget the negative sign infron ogf a. I try some value of $a$ and notice that If $a<-1$, no equilibrium point . There is one equilibrium point when $a=-1$. If $-1<a<0$, there are 2 equilibrias. If $a>0$, no equilibrium point. Am I correct? $\endgroup$ May 14, 2013 at 19:42
  • $\begingroup$ That is what I have found. Therefore we say that the bifurcation occurs when $a=-1$ since this is the value of $a$ that changes the nature of the equilibrium solutions. The next step is to also describe the behavior of the equilibrium solutions. That is, are they saddles, sources or sinks? $\endgroup$ May 14, 2013 at 20:08
  • $\begingroup$ I got it, Thank you very much $\endgroup$ May 14, 2013 at 20:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .