# Determining the bifurcation value(s) for a one-parameter family

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?

• 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.
– tom
Sep 23, 2013 at 0:29
• Well, when k=0, y^2=0, which implies y=0 (one stable solution). I'm not sure about the others. Sep 23, 2013 at 0:39
• When does the equation $y^2 = -k$ has real solution $y\in \mathbb{R}$?
– tom
Sep 23, 2013 at 0:40

Hints:

• 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?

Update

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

• 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? Sep 23, 2013 at 0:40
• @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. Sep 23, 2013 at 0:41
• 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. Sep 23, 2013 at 0:50
• @Bob: Spot on! Regards Sep 23, 2013 at 0:52
• You've been getting a lot of use out of your software for phase portraits! +1 Sep 24, 2013 at 0:14