Bifurcation points of differential equation (example)

Question

Assume the differential equation: $$ x'=\lambda^2-8a\lambda x+2x^2, \quad a\in \mathbb{R}. $$ The critical points are the solutions to the equation: $$ x'=0 \iff 2x^2-8a\lambda x +\lambda^2=0\tag{1} $$ which admits solutions: $$ x=\lambda \cdot \frac{8a \pm \sqrt{64a^2-8}}{2} $$ Now, having in mind that by bifurcation point, we mean a point where change of the number of equilibrium points occur, it seems to me that the number of $(1)$'s solutions depends only on $a$ and not on $\lambda$. Is this the case or did I get something wrong?

Answer 1

To study bifurcation points for $$ x'=\lambda^2-8a\lambda x+2x^2, \quad a\in \mathbb{R}, $$ we see that $x'=0$ implies that $$ x = \frac{\lambda}{2}\left(4a \pm \sqrt{2(8a^2-1)}\right). $$ $\textbf{Case 1}$. If $8a^2 < 1$, then the expression in the radical is a negative (real) number. So there are no bifurcation points, i.e., there doesn't exist an $x\in \mathbb{R}$ that will give $x'=0$.

$\textbf{Case 2}$. If $8a^2 = 1 $, then $x^*=2a\lambda$, which is a fixed point (it is a double root). If you plug in values of $x< 2a\lambda$ into $x'$, then $x'>0$. You also see that if you plug in values of $x > 2a\lambda$, then $x'>0$. So if you perturb the fixed point $x^*$ a little to the left, it appears to be an attractor. But if you perturb $x^*$ a little to the right, it appears to be a repeller. So $x^*$ is neither an attractor nor a repeller; it is a semistable point.

$\textbf{Case 3}$. If $8a^2 > 1$, then there are two subcases to consider.

$\color{green}{\textbf{Subcase 1}}$. If $\lambda\not=0$, then there are two fixed points: $$ x_1^* = \frac{\lambda}{2}\left(4a - \sqrt{2(8a^2-1)}\right), \quad x_2^* = \frac{\lambda}{2}\left(4a + \sqrt{2(8a^2-1)}\right). $$ We can use a similar analysis as in $\textbf{Case 2}$, or we can plot the differential equation on a planar graph to see that $x_1^*$ is an attracting (stable) fixed point, while $x_2^*$ is a repelling (unstable) fixed point.

$\color{green}{\textbf{Subcase 2}}$. If $\lambda=0$, then $x^*=0$ is a (double) fixed point, which is analogous to $\textbf{Case 2}$.

Answer 2

It appears to me our OP LoneBone is correct; I can find no errors in his/her logic.