Consider $x' = -x^4 + 5ax^2 - 4a^2$ a) Determine the equilibrium points and bifurcation value(s) for this family of DE.

First I let $y = x^2$

Then set $-y^2 + 5ay - 4a^2 = 0 $ >>> $\frac{-5a +- \sqrt(9a^2)}{2}$ So $ y = -a$ or $y = -4a$.

So $x^2 = -a $ >>> $ x=+-\sqrt(-a)$ and $x^2 = -4a $ >>> $ x=+-\sqrt(-4a)$

Is this enough for equilibrium points? If so, how to find the bifurcation values?


1 Answer 1


We want to set:

$$-x^4 + 5ax^2 - 4a^2 = 0 \implies -(-x^2+a)(-x^2+4a) = 0 \implies x = \pm~ \sqrt{a},~ \pm~ 2 \sqrt{a}$$

That is, we have four equilibrium (critical) points for this system.

Lets draw a Direction Field plot using a sample $a = 2$. Notice the four equilibrium points on the DF plot? See how they follow the analytical result?

enter image description here

Here is a DF plot for the case $a = 0$. What do you notice has changed?

enter image description here

What do you suspect happens to the DF plot for $a < 0$?

Now, how do you approach finding the bifurcation value(s)? Hint:

  • If $a > 0$, there are four equilibrium points.
  • If $a = 0$, there is one equilibrium point.
  • If $a < 0$, there are no equilibrium points.

What conclusion can be drawn about the bifurcation?

  • $\begingroup$ I'm always so impressed by your thoroughness (even in your hints!) and the care you take in your posts! Nice, Amzoti! $\endgroup$
    – amWhy
    Feb 27, 2014 at 13:29

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