Let's say we have a one parameter family:

$$ \frac{dy}{dt} = y^2 + k $$

I want to find the bifurcation value. What does this mean?

It seems like I need to set dy/dt = 0 and then solve for k, but then I get a negative square root:

$$ 0 = y^2 + k $$ $$ y^2 = -k $$ $$ y=sqrt(-k) $$ or $$ k = -y^2 $$

Is this the right approach?

  • $\begingroup$ How many steady solution can you get for particular $k$? Hint: for some $k$ you get zero, for some two and for one in particular you get only one steady solutions. $\endgroup$
    – tom
    Sep 23, 2013 at 0:29
  • $\begingroup$ Well, when k=0, y^2=0, which implies y=0 (one stable solution). I'm not sure about the others. $\endgroup$ Sep 23, 2013 at 0:39
  • $\begingroup$ When does the equation $y^2 = -k$ has real solution $y\in \mathbb{R}$? $\endgroup$
    – tom
    Sep 23, 2013 at 0:40

1 Answer 1



  • Consider $k$ negative, positive and zero.
  • What happens for each of the values to the fixed points?
  • What is the bifurcation point defined as from these results?


Here are phase portraits for $k = -1, 0, 1$. What do you notice happening?

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  • $\begingroup$ Ah, so we use 0 as the base line bifurcation value in this example? And then we observe what happens with the roots when we set k > 0 and k < 0? $\endgroup$ Sep 23, 2013 at 0:40
  • $\begingroup$ @Bob: Yes, that is correct, I will post phase portraits that I just finished and you can see if you can draw 1-D phase portraits from them. $\endgroup$
    – Amzoti
    Sep 23, 2013 at 0:41
  • $\begingroup$ Thanks! When k<0, there is one stable and one unstable equilibrium solution. When k=0, there is one semi-stable equilibrium solution. When k>0, there is no equilibrium solution. $\endgroup$ Sep 23, 2013 at 0:50
  • $\begingroup$ @Bob: Spot on! Regards $\endgroup$
    – Amzoti
    Sep 23, 2013 at 0:52
  • $\begingroup$ You've been getting a lot of use out of your software for phase portraits! +1 $\endgroup$
    – amWhy
    Sep 24, 2013 at 0:14

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